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2021USATSTSTp1
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Let $ABCD$ be a quadrilateral inscribed in a circle with center $O$. Points $X$ and $Y$ lie on sides $AB$ and $CD$, respectively. Suppose the circumcircles of $ADX$ and $BCY$ meet line $XY$ again at $P$ and $Q$, respectively. Show that $OP=OQ$.
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a b c = triangle; d = on_circum a b c; o = circumcenter a b c; x = on_line a b; y = on_line c d; o1 = circumcenter a d x; p = on_line x y, on_circle o1 a; o2 = circumcenter b c y; q = on_line x y, on_circle o2 b ? cong o p o q
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A B C = triangle; (ABC) = circle A B C; D = on_circle (ABC); O = circumcenter A B C; AB = line A B; X = on_line AB; CD = line C D; Y = on_line CD; XY = line X Y; O1 = circumcenter A D X; (O1) = circle_center_point O1 A; P = intersection XY (O1); O2 = circumcenter B C Y; (O2) = circle_center_point O2 B; Q = intersection XY (O2); Prove: cong O P O Q
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2021USEMOp3
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Let $A_1C_2B_1A_2C_1B_2$ be an equilateral hexagon. Let $O_1$ and $H_1$ denote the circumcenter and orthocenter of $\triangle A_1B_1C_1$, and let $O_2$ and $H_2$ denote the circumcenter and orthocenter of $\triangle A_2B_2C_2$. Suppose that $O_1 \ne O_2$ and $H_1 \ne H_2$. Prove that the lines $O_1O_2$ and $H_1H_2$ are either parallel or coincide.
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a1 b1 c1 = triangle; a2 = on_bline b1 c1; b2 = on_circle c1 a2, on_bline a1 c1; c2 = on_circle b1 a2, on_bline a1 b1; o1 = circumcenter a1 b1 c1; h1 = orthocenter a1 b1 c1; o2 = circumcenter a2 b2 c2; h2 = orthocenter a2 b2 c2 ? para o1 o2 h1 h2
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A1 B1 C1 = triangle; l1 = perpendicular_bisector B1 C1; A2 = on_line l1; l2 = perpendicular_bisector A1 C1; (C1) = circle_center_point C1 A2; B2 = intersection l2 (C1); l3 = perpendicular_bisector A1 B1; (B1) = circle_center_point B1 A2; C2 = intersection l3 (B1); O1 = circumcenter A1 B1 C1; H1 = orthocenter A1 B1 C1; O2 = circumcenter A2 B2 C2; H2 = orthocenter A2 B2 C2; O1O2 = line O1 O2; H1H2 = line H1 H2; Prove: parallel O1O2 H1H2
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2022ARMOg10p2
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On side $BC$ of an acute triangle $ABC$ are marked points $D$ and $E$ so that $BD = CE$. On the arc $DE$ of the circumscribed circle of triangle $ADE$ that does not contain the point $A$, there are points $P$ and $Q$ such that $AB = PC$ and $AC = BQ$. Prove that $AP=AQ$.
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a b c = triangle; d = on_line b c; m = midpoint b c; e = mirror d m; o1 = circumcenter a d e; p = on_circle o1 a, eqdistance p c a b; q = on_circle o1 a, eqdistance q b a c ? cong a p a q
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A B C = acute_triangle; BC = line B C; D = on_line BC; M = midpoint B C; E = reflect_point_wrt_point D M; O1 = circumcenter A D E; (O1) = circle_center_point O1 A; (C,AB) = circle_center_radius C A B; P = intersection (O1) (C,AB); (B,AC) = circle_center_radius B A C; Q = intersection (O1) (B,AC); Prove: cong A P A Q
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2022BalkanMOp1
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Let $ABC$ be an acute triangle such that $CA \neq CB$ with circumcircle $\omega$ and circumcentre $O$. Let $t_A$ and $t_B$ be the tangents to $\omega$ at $A$ and $B$ respectively, which meet at $X$. Let $Y$ be the foot of the perpendicular from $O$ onto the line segment $CX$. The line through $C$ parallel to line $AB$ meets $t_A$ at $Z$. Prove that the line $YZ$ passes through the midpoint of the line segment $AC$.
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a b c = triangle; o = circumcenter a b c; x = on_tline a o a, on_tline b o b; y = foot o c x; z = on_line a x, on_pline c a b; m = midpoint a c ? coll y z m
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A B C = acute_triangle; O = circumcenter A B C; (O) = circle_center_point O A; OA = line O A; OB = line O B; ta = perpendicular_line A OA; tb = perpendicular_line B OB; X = intersection ta tb; CX = line C X; Y = foot O CX; AB = line A B; l1 = parallel_line C AB; Z = intersection l1 ta; M = midpoint A C; Prove: collinear Y Z M
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2022CGMOp3
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In triangle $ABC,AB>AC,I$ is the incenter, $AM$ is the midline. The line crosses $I$ and is perpendicular to $BC $ intersect $AM$ at point $L$, and the symmetry of $I$ with respect to point $A$ is $J$ Prove that: $\angle ABJ= \angle LBI$.
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a b c = triangle; i = incenter a b c; m = midpoint b c; l = on_tline i b c, on_line a m; j = mirror i a ? eqangle a b b j i b b l
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A B C = triangle; I = incenter A B C; M = midpoint B C; BC = line B C; l1 = perpendicular_line I BC; AM = line A M; L = intersection AM l1; J = reflect_point_wrt_point I A; Prove: equal_angle A B J I B L
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2022CHNSouthEastMOg11p6
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Let $O$ be the circumcenter of $\triangle ABC$. A circle with center $P$ pass through $A$ and $O$ and $OP$//$BC$. $D$ is a point such that $\angle DBA = \angle DCA = \angle BAC$. Prove that: Circle $(P)$, circle $(BCD)$ and the circle with diameter $(AD)$ share a common point.
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a b c = triangle; o = circumcenter a b c; p = on_bline a o, on_pline o b c; d = on_aline d b a b a c, on_aline d c a c a b; q = midpoint a d; k = on_circle p a, on_circle q a ? cyclic k b c d
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A B C = triangle; O = circumcenter A B C; l1 = perpendicular_bisector A O; BC = line B C; l2 = parallel_line O BC; P = intersection l1 l2; l3 = angle_equal1 B A B A C; l4 = angle_equal1 C A C A B; D = intersection l3 l4; Q = midpoint A D; (P) = circle_center_point P A; (Q) = circle_center_point Q A; K = intersection (P) (Q); Prove: concyclic K B C D
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2022CTSTp1
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In a cyclic convex hexagon $ABCDEF$, $AB$ and $DC$ intersect at $G$, $AF$ and $DE$ intersect at $H$. Let $M, N$ be the circumcenters of $BCG$ and $EFH$, respectively. Prove that the $BE$, $CF$ and $MN$ are concurrent.
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a b c = triangle; o = circumcenter a b c; d = on_circle o a; e = on_circle o a; f = on_circle o a; g = on_line a b, on_line d c; h = on_line a f, on_line d e; m = circumcenter b c g; n = circumcenter e f h; x = on_line b e, on_line c f ? coll m n x
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A B C = triangle; O = circumcenter A B C; (O) = circle_center_point O A; D = on_circle (O); E = on_circle (O); F = on_circle (O); AB = line A B; CD = line C D; G = intersection AB CD; AF = line A F; DE = line D E; H = intersection AF DE; M = circumcenter B C G; N = circumcenter E F H; BE = line B E; CF = line C F; X = intersection BE CF; Prove: collinear M N X
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2022CzechAPSlovakp5
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Let $ABC$ be a triangle with $AB < AC$ and circumcenter $O$. The angle bisector of $\angle BAC$ meets the side $BC$ at $D$. The line through $D$ perpendicular to $BC$ meets the segment $AO$ at $X$. Furthermore, let $Y$ be the midpoint of segment $AD$. Prove that points $B, C, X, Y$ are concyclic.
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a b c = triangle; o = circumcenter a b c; d = on_line b c, angle_bisector b a c; x = on_line a o, on_tline d b c; y = midpoint a d ? cyclic b c x y
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A B C = triangle; O = circumcenter A B C; l1 = angle_bisector B A C; BC = line B C; D = intersection l1 BC; l2 = perpendicular_line D BC; AO = line A O; X = intersection l2 AO; Y = midpoint A D; Prove: concyclic B C X Y
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2022EGMOp1
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Let $ABC$ be an acute-angled triangle in which $BC<AB$ and $BC<CA$. Let point $P$ lie on segment $AB$ and point $Q$ lie on segment $AC$ such that $P \neq B$, $Q \neq C$ and $BQ = BC = CP$. Let $T$ be the circumcenter of triangle $APQ$, $H$ the orthocenter of triangle $ABC$, and $S$ the point of intersection of the lines $BQ$ and $CP$. Prove that $T$, $H$, and $S$ are collinear.
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a b c = triangle; q = on_line a c, on_circle b c; p = on_line a b, on_circle c b; t = circumcenter a p q; h = orthocenter a b c; s = on_line b q, on_line c p ? coll t h s
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A B C = acute_triangle; AB = line A B; (C) = circle_center_point C B; P = intersection AB (C); AC = line A C; (B) = circle_center_point B C; Q = intersection AC (B); T = circumcenter A P Q; H = orthocenter A B C; BQ = line B Q; CP = line C P; S = intersection BQ CP; Prove: collinear T H S
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2022FakeCMOp3
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In $\triangle ABC$ with excenters $I_A,I_B,I_C$, let $P$ be any point. Lines $AP,BP,CP$ meet the circumcircles of $\triangle BPC,\triangle CPA,\triangle APB$ again at $P_A,P_B,P_C$. The circumcircles of $\triangle I_BI_CP_A$ and $\triangle I_AI_BP_C$ meet again at $Q\ne I_B$. Lines $AQ,BQ,CQ$ meet the circumcircles of $\triangle BQC,\triangle CQA,\triangle AQB$ again at $Q_A,Q_B,Q_C$. Prove that $P_AQ_A$, $P_BQ_B$, $P_CQ_C$ are concurrent.
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a b c = triangle; i_a = excenter a b c; i_b = excenter b c a; i_c = excenter c a b; p = free; p_a = on_line a p, on_circum b p c; p_b = on_line b p, on_circum c p a; p_c = on_line c p, on_circum a p b; q = on_circum i_b i_c p_a, on_circum i_a i_b p_c; q_a = on_line a q, on_circum b q c; q_b = on_line b q, on_circum a q c; q_c = on_line c q, on_circum a q b; k = on_line p_a q_a, on_line p_b q_b ? coll k p_c q_c
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A B C = triangle; Ia = excenter A B C; Ib = excenter B C A; Ic = excenter C A B; P = point; AP = line A P; (BPC) = circumcircle B P C; Pa = intersection AP (BPC); BP = line B P; (CPA) = circumcircle C P A; Pb = intersection BP (CPA); CP = line C P; (APB) = circumcircle A P B; Pc = intersection CP (APB); (IbIcPa) = circumcircle Ib Ic Pa; (IaIbPc) = circumcircle Ia Ib Pc; Q = intersection (IbIcPa) (IaIbPc); AQ = line A Q; (BQC) = circumcircle B Q C; Qa = intersection AQ (BQC); BQ = line B Q; (AQC) = circumcircle A Q C; Qb = intersection BQ (AQC); CQ = line C Q; (AQB) = circumcircle A Q B; Qc = intersection CQ (AQB); PaQa = line Pa Qa; PbQb = line Pb Qb; K = intersection PaQa PbQb; Prove: collinear K Pc Qc
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2022G2
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In the acute-angled triangle $ABC$, the point $F$ is the foot of the altitude from $A$, and $P$ is a point on the segment $AF$. The lines through $P$ parallel to $AC$ and $AB$ meet $BC$ at $D$ and $E$, respectively. Points $X \ne A$ and $Y \ne A$ lie on the circles $ABD$ and $ACE$, respectively, such that $DA = DX$ and $EA = EY$. Prove that $B, C, X,$ and $Y$ are concyclic.
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a b c = triangle; p = on_tline a b c; d = on_line b c, on_pline p a c; e = on_line b c, on_pline p a b; o1 = circumcenter a b d; o2 = circumcenter a c e; x = on_circle d a, on_circle o1 a; y = on_circle e a, on_circle o2 a ? cyclic b c x y
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A B C = acute_triangle; BC = line B C; l = perpendicular_line A BC; P = on_line l; AC = line A C; l1 = parallel_line P AC; D = intersection l1 BC; AB = line A B; l2 = parallel_line P AB; E = intersection l2 BC; O1 = circumcenter A B D; (O1) = circle_center_point O1 A; (D) = circle_center_point D A; X = intersection (O1) (D); O2 = circumcenter A C E; (O2) = circle_center_point O2 A; (E) = circle_center_point E A; Y = intersection (O2) (E); Prove: concyclic B C X Y
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2022G5
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Let $ABC$ be a triangle and $\ell_1,\ell_2$ be two parallel lines. Let $\ell_i$ intersects line $BC,CA,AB$ at $X_i,Y_i,Z_i$, respectively. Let $\Delta_i$ be the triangle formed by the line passed through $X_i$ and perpendicular to $BC$, the line passed through $Y_i$ and perpendicular to $CA$, and the line passed through $Z_i$ and perpendicular to $AB$. Prove that the circumcircles of $\Delta_1$ and $\Delta_2$ are tangent.
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a b c = triangle; x1 = on_line b c; y1 = on_line a c; z1 = on_line a b, on_line x1 y1; x2 = on_line b c; y2 = on_line a c, on_pline x2 x1 y1; z2 = on_line a b, on_pline x2 x1 y1; u1 = on_tline y1 a c, on_tline z1 a b; v1 = on_tline x1 b c, on_tline z1 a b; w1 = on_tline x1 b c, on_tline y1 a c; u2 = on_tline y2 a c, on_tline z2 a b; v2 = on_tline x2 b c, on_tline z2 a b; w2 = on_tline x2 b c, on_tline y2 a c; o1 = circumcenter u1 v1 w1; o2 = circumcenter u2 v2 w2; t = on_circle o1 u1, on_circle o2 u2 ? coll t o1 o2
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A B C = triangle; BC = line B C; X1 = on_line BC; AC = line A C; Y1 = on_line AC; l1 = line X1 Y1; AB = line A B; Z1 = intersection l1 AB; X2 = on_line BC; l2 = parallel_line X2 l1; Y2 = intersection l2 AC; Z2 = intersection l2 AB; l3 = perpendicular_line X1 BC; l4 = perpendicular_line Y1 AC; l5 = perpendicular_line Z1 AB; U1 = intersection l4 l5; V1 = intersection l3 l5; W1 = intersection l3 l4; l6 = perpendicular_line X2 BC; l7 = perpendicular_line Y2 AC; l8 = perpendicular_line Z2 AB; U2 = intersection l7 l8; V2 = intersection l6 l8; W2 = intersection l6 l7; O1 = circumcenter U1 V1 W1; O2 = circumcenter U2 V2 W2; (O1) = circle_center_point O1 U1; (O2) = circle_center_point O2 U2; T = intersection (O1) (O2); Prove: collinear T O1 O2
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2022IZhautykovOp3
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In parallelogram $ABCD$ with acute angle $A$ a point $N$ is chosen on the segment $AD$, and a point $M$ on the segment $CN$ so that $AB = BM = CM$. Point $K$ is the reflection of $N$ in line $MD$. The line $MK$ meets the segment $AD$ at point $L$. Let $P$ be the common point of the circumcircles of $AMD$ and $CNK$ such that $A$ and $P$ share the same side of the line $MK$. Prove that $\angle CPM = \angle DPL$.
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a b c = triangle; d = parallelogram a b c d; m = on_bline b c, eqdistance m b a b; n = on_line c m, on_line a d; k = reflect n m d; l = on_line m k, on_line a d; o1 = circumcenter a m d; o2 = circumcenter c n k; p = on_circle o1 a, on_circle o2 c ? eqangle c p p m d p p l
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A B C = triangle; D = parallelogram A B C; l1 = perpendicular_bisector B C; (B) = circle_center_point B A; M = intersection l1 (B); CM = line C M; AD = line A D; N = intersection CM AD; MD = line M D; K = reflect_point_wrt_line N MD; MK = line M K; L = intersection MK AD; O1 = circumcenter A M D; O2 = circumcenter C N K; (O1) = circle_center_point O1 A; (O2) = circle_center_point O2 C; P = intersection (O1) (O2); Prove: equal_angle C P M D P L
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2022IranGOAp2
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We are given an acute triangle $ABC$ with $AB\neq AC$. Let $D$ be a point of $BC$ such that $DA$ is tangent to the circumcircle of $ABC$. Let $E$ and $F$ be the circumcenters of triangles $ABD$ and $ACD$, respectively, and let $M$ be the midpoints $EF$. Prove that the line tangent to the circumcircle of $AMD$ through $D$ is also tangent to the circumcircle of $ABC$.
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a b c = triangle; o = circumcenter a b c; d = on_line b c, on_tline a o a; e = circumcenter a b d; f = circumcenter a c d; m = midpoint e f; o1 = circumcenter a m d; t = on_tline d o1 d; x = foot o d t ? cong o x o a
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A B C = acute_triangle; O = circumcenter A B C; OA = line O A; l1 = perpendicular_line A OA; BC = line B C; D = intersection l1 BC; E = circumcenter A B D; F = circumcenter A C D; M = midpoint E F; O1 = circumcenter A D M; O1D = line O1 D; l = perpendicular_line D O1D; X = foot O l; Prove: cong O X O A
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2022IranGOAp3
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In triangle $ABC$ $(\angle A\neq 90^\circ)$, let $O$, $H$ be the circumcenter and the foot of the altitude from $A$ respectively. Suppose $M$, $N$ are the midpoints of $BC$, $AH$ respectively. Let $D$ be the intersection of $AO$ and $BC$ and let $H'$ be the reflection of $H$ about $M$. Suppose that the circumcircle of $OH'D$ intersects the circumcircle of $BOC$ at $E$. Prove that $NO$ and $AE$ are concurrent on the circumcircle of $BOC$.
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a b c = triangle; o = circumcenter a b c; h = foot a b c; m = midpoint b c; n = midpoint a h; d = on_line a o, on_line b c; h1 = mirror h m; o1 = circumcenter o h1 d; o2 = circumcenter b o c; e = on_circle o1 o, on_circle o2 o; p = on_line n o, on_line a e ? cyclic p b o c
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A B C = triangle; O = circumcenter A B C; BC = line B C; H = foot A BC; M = midpoint B C; N = midpoint A H; AO = line A O; D = intersection AO BC; H1 = reflect H M; O1 = circumcenter O H1 D; O2 = circumcenter B O C; (DH1O) = circle_center_point O1 O; (BCO) = circle_center_point O2 O; E = intersection (DH1O) (BCO); NO = line N O; AE = line A E; P = intersection NO AE; Prove: concyclic P B O C
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2022IranGOMp3
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Let $O$ be the circumcenter of triangle $ABC$. Arbitrary points $M$ and $N$ lie on the sides $AC$ and $BC$, respectively. Points $P$ and $Q$ lie in the same half-plane as point $C$ with respect to the line $MN$, and satisfy $\triangle CMN \sim \triangle PAN \sim \triangle QMB$ (in this exact order). Prove that $OP=OQ$.
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a b c = triangle; o = circumcenter a b c; m = on_line a c; n = on_line b c; p = on_aline p a n c m n, on_aline p n a c n m; q = on_aline q m b c m n, on_aline q b m c n m ? cong o p o q
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A B C = triangle; O = circumcenter A B C; AC = line A C; M = on_line AC; BC = line B C; N = on_line BC; l1 = angle_equal1 A N C M N; l2 = angle_equal1 N A C N M; P = intersection l1 l2; l3 = angle_equal1 M B C M N; l4 = angle_equal1 B M C N M; Q = intersection l3 l4; Prove: cong O P O Q
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2022IranGOMp5
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Let $ABCD$ be a quadrilateral inscribed in a circle $\omega$ with center $O$. Let $P$ be the intersection of two diagonals $AC$ and $BD$. Let $Q$ be a point lying on the segment $OP$. Let $E$ and $F$ be the orthogonal projections of $Q$ on the lines $AD$ and $BC$, respectively. The points $M$ and $N$ lie on the circumcircle of triangle $QEF$ such that $QM \parallel AC$ and $QN \parallel BD$. Prove that the two lines $ME$ and $NF$ meet on the perpendicular bisector of segment $CD$.
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a b c = triangle; d = on_circum a b c; o = circumcenter a b c; p = on_line a c, on_line b d; q = on_line o p; e = foot q a d; f = foot q b c; o1 = circumcenter q e f; m = on_circle o1 q, on_pline q a c; n = on_circle o1 q, on_pline q b d; x = on_line m e, on_line n f ? cong x c x d
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A B C = triangle; (ABC) = circle A B C; D = on_circle (ABC); O = circumcenter A B C; AC = line A C; BD = line B D; P = intersection AC BD; OP = line O P; Q = on_line OP; AD = line A D; BC = line B C; E = foot Q AD; F = foot Q BC; O1 = circumcenter Q E F; (O1) = circle_center_point O1 Q; l1 = parallel_line Q AC; M = intersection (O1) l1; l2 = parallel_line Q BD; N = intersection (O1) l2; ME = line M E; NF = line N F; X = intersection ME NF; Prove: cong X C X D
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2022KoMaLA805
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In acute triangle $ABC,$ the feet of the altitudes are $A_1,B_1,$ and $C_1$ (with the usual notations on sides $BC,CA,$ and $AB$ respectively). The circumcircles of triangles $AB_1C_1$ and $BC_1A_1$ intersect at the circumcircle of triangle $ABC$ ar points $P\neq A$ and $Q\neq B,$ respectively. Prove that lines $AQ, BP$ and the Euler line of triangle $ABC$ are either concurrent or parallel to each other.
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a b c = triangle; a1 = foot a b c; b1 = foot b a c; c1 = foot c a b; o = circumcenter a b c; o1 = circumcenter a b1 c1; p = on_circle o a, on_circle o1 a; o2 = circumcenter b c1 a1; q = on_circle o b, on_circle o2 b; h = orthocenter a b c; r = on_line a q, on_line b p ? coll r o h
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A B C = acute_triangle; BC = line B C; A1 = foot A BC; AC = line A C; B1 = foot B AC; AB = line A B; C1 = foot C AB; O = circumcenter A B C; O1 = circumcenter A B1 C1; (O) = circle_center_point O A; (O1) = circle_center_point O1 A; P = intersection (O1) (O); O2 = circumcenter B C1 A1; (O2) = circle_center_point O2 B; Q = intersection (O2) (O); H = orthocenter A B C; AQ = line A Q; BP = line B P; R = intersection AQ BP; Prove: collinear R O H
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2022SilkRoadp1
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Convex quadrilateral $ABCD$ is inscribed in circle $w.$Rays $AB$ and $DC$ intersect at $K.\ L$ is chosen on the diagonal $BD$ so that $\angle BAC= \angle DAL.\ M$ is chosen on the segment $KL$ so that $CM \mid\mid BD.$ Prove that line $BM$ touches $w.$
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a b c = triangle; d = on_circum a b c; k = on_line a b, on_line c d; l = on_line b d, on_aline l a d b a c; m = on_line k l, on_pline c b d; o = circumcenter a b c ? perp o b b m
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A B C = triangle; O = circumcenter A B C; (ABC) = circle A B C; D = on_circle (ABC); AB = line A B; CD = line C D; K = intersection AB CD; BD = line B D; l1 = angle_equal1 A D B A C; L = intersection BD l1; KL = line K L; l2 = parallel_line C BD; M = intersection KL l2; OB = line O B; BM = line B M; Prove: perpendicular BM OB
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2022TurkeyTSTp3
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In a triangle $ABC$, the incircle centered at $I$ is tangent to the sides $BC, AC$ and $AB$ at $D, E$ and $F$, respectively. Let $X, Y$ and $Z$ be the feet of the perpendiculars drawn from $A, B$ and $C$ to a line $\ell$ passing through $I$. Prove that $DX, EY$ and $FZ$ are concurrent.
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a b c = triangle; i = incenter a b c; d = foot i b c; e = foot i a c; f = foot i a b; p = free; x = foot a i p; y = foot b i p; z = foot c i p; t = on_line d x, on_line e y ? coll t f z
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A B C = triangle; I = incenter A B C; BC = line B C; AC = line A C; AB = line A B; D = foot I BC; E = foot I AC; F = foot I AB; P = point; IP = line I P; X = foot A IP; Y = foot B IP; Z = foot C IP; DX = line D X; EY = line E Y; T = intersection DX EY; Prove: collinear T F Z
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2022USATSTSTp6
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Let $O$ and $H$ be the circumcenter and orthocenter, respectively, of an acute scalene triangle $ABC$. The perpendicular bisector of $\overline{AH}$ intersects $\overline{AB}$ and $\overline{AC}$ at $X_A$ and $Y_A$ respectively. Let $K_A$ denote the intersection of the circumcircles of triangles $OX_AY_A$ and $BOC$ other than $O$. Define $K_B$ and $K_C$ analogously by repeating this construction two more times. Prove that $K_A$, $K_B$, $K_C$, and $O$ are concyclic.
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a b c = triangle; o = circumcenter a b c; h = orthocenter a b c; xa = on_bline a h, on_line a b; ya = on_bline a h, on_line a c; o_a = circumcenter o xa ya; o1 = circumcenter b o c; ka = on_circle o_a o, on_circle o1 o; xb = on_bline b h, on_line b c; yb = on_bline b h, on_line b a; o_b = circumcenter o xb yb; o2 = circumcenter a o c; kb = on_circle o_b o, on_circle o2 o; xc = on_bline c h, on_line c a; yc = on_bline c h, on_line c b; o_c = circumcenter o xc yc; o3 = circumcenter a o b; kc = on_circle o_c o, on_circle o3 o ? cyclic ka kb kc o
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A B C = acute_triangle; O = circumcenter A B C; H = orthocenter A B C; l1 = perpendicular_bisector A H; AB = line A B; Xa = intersection l1 AB; AC = line A C; Ya = intersection l1 AC; Oa = circumcenter O Xa Ya; (OXaYa) = circle_center_point Oa O; O1 = circumcenter B O C; (BCO) = circle_center_point O1 O; Ka = intersection (OXaYa) (BCO); l2 = perpendicular_bisector B H; BC = line B C; Xb = intersection l2 BC; Yb = intersection l2 AB; Ob = circumcenter O Xb Yb; (OXbYb) = circle_center_point Ob O; O2 = circumcenter A O C; (ACO) = circle_center_point O2 O; Kb = intersection (OXbYb) (ACO); l3 = perpendicular_bisector C H; Xc = intersection l3 AC; Yc = intersection l3 BC; Oc = circumcenter O Xc Yc; (OXcYc) = circle_center_point Oc O; O3 = circumcenter A O B; (ABO) = circle_center_point O3 O; Kc = intersection (OXcYc) (ABO); Prove: concyclic Ka Kb Kc O
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2023ARMOg11p4
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Let $\omega$ be the circumcircle of triangle $ABC$ with $AB<AC$. Let $I$ be its incenter and let $M$ be the midpoint of $BC$. The foot of the perpendicular from $M$ to $AI$ is $H$. The lines $MH, BI, AB$ form a triangle $T_b$ and the lines $MH, CI, AC$ form a triangle $T_c$. The circumcircle of $T_b$ meets $\omega$ at $B'$ and the circumcircle of $T_c$ meets $\omega$ at $C'$. Prove that $B', H, C'$ are collinear.
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a b c = triangle; o = circumcenter a b c; i = incenter a b c; m = midpoint b c; h = foot m a i; v1 = on_line m h, on_line b i; v2 = on_line m h, on_line a b; o2 = circumcenter v1 v2 b; b1 = on_circle o2 b, on_circle o a; w1 = on_line m h, on_line c i; w2 = on_line m h, on_line a c; o3 = circumcenter w1 w2 c; c1 = on_circle o3 c, on_circle o a ? coll b1 h c1
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A B C = triangle; O = circumcenter A B C; (O) = circle_center_point O A; I = incenter A B C; M = midpoint B C; AI = line A I; H = foot M AI; HM = line H M; BI = line B I; AB = line A B; V1 = intersection HM BI; V2 = intersection HM AB; O2 = circumcenter V1 V2 B; (O2) = circle_center_point O2 B; B1 = intersection (O2) (O); CI = line C I; AC = line A C; W1 = intersection HM CI; W2 = intersection HM AC; O3 = circumcenter W1 W2 C; (O3) = circle_center_point O3 C; C1 = intersection (O) (O3); Prove: collinear B1 H C1
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2023CHNSouthEastMOg10p3
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In acute triangle $ABC$ ($\triangle ABC$ is not an isosceles triangle), $I$ is its incentre, and circle $ \omega$ is its inscribed circle. $\odot\omega$ touches $BC, CA, AB$ at $D, E, F$ respectively. $AD$ intersects with $\odot\omega$ at $J$ ($J\neq D$), and the circumcircle of $\triangle BCJ$ intersects $\odot\omega$ at $K$ ($K\neq J$). The circumcircle of $\triangle BFK$ and $\triangle CEK$ meet at $L$ ($L\neq K$). Let $M$ be the midpoint of the major arc $BAC$.Prove that $M, I, L$ are collinear.
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a b c = triangle; i = incenter a b c; d = foot i b c; e = foot i a c; f = foot i a b; j = on_line a d, on_circle i d; o1 = circumcenter b c j; k = on_circle o1 b, on_circle i d; o2 = circumcenter b f k; o3 = circumcenter c e k; l = on_circle o2 b, on_circle o3 c; o = circumcenter a b c; m = on_bline b c, on_circle o a ? coll m i l
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A B C = acute_triangle; I = incenter A B C; BC = line B C; D = foot I BC; (I) = circle_center_point I D; AC = line A C; E = foot I AC; AB = line A B; F = foot I AB; AD = line A D; J = intersection AD (I); O1 = circumcenter B C J; (BCJ) = circle_center_point O1 B; K = intersection (BCJ) (I); O2 = circumcenter B F K; (BFK) = circle_center_point O2 B; O3 = circumcenter C E K; (CEK) = circle_center_point O3 C; L = intersection (BFK) (CEK); M = midarc B A C; Prove: collinear M I L
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2023CHNSouthEastMOg11p5
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As shown in the figure, in $\vartriangle ABC$, $AB>AC$, the inscribed circle $I$ is tangent to the sides $BC$, $CA$, $AB$ at points $D$, $E$, $F$ respectively, and the straight lines $BC$ and $EF$ intersect at point $K$, $DG \perp EF$ at point $G$, ray $IG$ intersects the circumscribed circle of $\vartriangle ABC$ at point $H$. Prove that points $H$, $G$, $D$, $K$ lie on a circle.
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a b c = triangle; i = incenter a b c; d = foot i b c; e = foot i c a; f = foot i a b; k = on_line b c, on_line e f; g = foot d e f; h = on_line i g, on_circum a b c ? cyclic h g d k
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A B C = triangle; I = incenter A B C; BC = line B C; D = foot I BC; AC = line A C; E = foot I AC; AB = line A B; F = foot I AB; EF = line E F; K = intersection BC EF; G = foot D EF; GI = line G I; (ABC) = circle A B C; H = intersection GI (ABC); Prove: concyclic H G D K
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2023CHNWesternMOp6
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As shown in the figure, let point $E$ be the intersection of the diagonals $AC$ and $BD$ of the cyclic quadrilateral $ABCD$. The circumcenter of triangle $ABE$ is denoted as $K$. Point $X$ is the reflection of point $B$ with respect to the line $CD$, and point $Y$ is the point on the plane such that quadrilateral $DKEY$ is a parallelogram. Prove that the points $D,E,X,Y$ are concyclic.
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a b c = triangle; d = on_circum a b c; e = on_line a c, on_line b d; k = circumcenter a b e; x = reflect b c d; y = parallelogram d k e y ? cyclic d e x y
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A B C = triangle; (ABC) = circle A B C; D = on_circle (ABC); AC = line A C; BD = line B D; E = intersection AC BD; K = circumcenter A B E; CD = line C D; X = reflect_point_wrt_line B CD; Y = parallelogram D K E; Prove: concyclic D E X Y
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2023ELMOSLp2
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Let \(ABC\) be an acute scalene triangle with orthocenter \(H\). Line \(BH\) intersects \(\overline{AC}\) at \(E\) and line \(CH\) intersects \(\overline{AB}\) at \(F\). Let \(X\) be the foot of the perpendicular from \(H\) to the line through \(A\) parallel to \(\overline{EF}\). Point \(B_1\) lies on line \(XF\) such that \(\overline{BB_1}\) is parallel to \(\overline{AC}\), and point \(C_1\) lies on line \(XE\) such that \(\overline{CC_1}\) is parallel to \(\overline{AB}\). Prove that points \(B\), \(C\), \(B_1\), \(C_1\) are concyclic.
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a b c = triangle; h = orthocenter a b c; e = on_line a c, on_line b h; f = on_line a b, on_line c h; x = on_pline a e f, on_tline h e f; b1 = on_line x f, on_pline b a c; c1 = on_line x e, on_pline c a b ? cyclic b c b1 c1
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A B C = acute_triangle; H = orthocenter A B C; BH = line B H; AC = line A C; E = intersection BH AC; CH = line C H; AB = line A B; F = intersection CH AB; EF = line E F; la = parallel_line A EF; X = foot H la; l1 = parallel_line B AC; FX = line F X; B1 = intersection FX l1; l2 = parallel_line C AB; EX = line E X; C1 = intersection EX l2; Prove: concyclic B C B1 C1
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2023EuropeanMCupSp2
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Let $ABC$ be a triangle such that $\angle BAC = 90^{\circ}$. The incircle of triangle $ABC$ is tangent to the sides $\overline{BC}$, $\overline{CA}$, $\overline{AB}$ at $D,E,F$ respectively. Let $M$ be the midpoint of $\overline{EF}$. Let $P$ be the projection of $A$ onto $BC$ and let $K$ be the intersection of $MP$ and $AD$. Prove that the circumcircles of triangles $AFE$ and $PDK$ have equal radius.
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a b c = r_triangle; i = incenter a b c; d = foot i b c; e = foot i a c; f = foot i a b; m = midpoint e f; p = foot a b c; k = on_line m p, on_line a d; o2 = circumcenter p d k ? cong m a o2 p
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A B C = right_triangle; I = incenter A B C; BC = line B C; D = foot I BC; AC = line A C; E = foot I AC; AB = line A B; F = foot I AB; M = midpoint E F; P = foot A BC; MP = line M P; AD = line A D; K = intersection MP AD; O2 = circumcenter P D K; Prove: cong A M O2 P
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2023G1
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Let $ABCDE$ be a convex pentagon such that $\angle ABC = \angle AED = 90^\circ$. Suppose that the midpoint of $CD$ is the circumcenter of triangle $ABE$. Let $O$ be the circumcenter of triangle $ACD$. Prove that line $AO$ passes through the midpoint of segment $BE$.
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a c d = triangle; m = midpoint c d; b = on_circle m a, on_dia a c; e = on_circle m a, on_dia a d; o = circumcenter a c d; n = midpoint b e ? coll a o n
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A C D = triangle; M = midpoint C D; (AC) = circle_diameter A C; (M,A) = circle_center_point M A; B = intersection (AC) (M,A); (AD) = circle_diameter A D; E = intersection (AD) (M,A); O = circumcenter A C D; N = midpoint B E; Prove: collinear A O N
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2023G2
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Let $ABC$ be a triangle with $AC > BC,$ let $\omega$ be the circumcircle of $\triangle ABC,$ and let $r$ be its radius. Point $P$ is chosen on $\overline{AC}$ such taht $BC=CP,$ and point $S$ is the foot of the perpendicular from $P$ to $\overline{AB}$. Ray $BP$ mets $\omega$ again at $D$. Point $Q$ is chosen on line $SP$ such that $PQ = r$ and $S,P,Q$ lie on a line in that order. Finally, let $E$ be a point satisfying $\overline{AE} \perp \overline{CQ}$ and $\overline{BE} \perp \overline{DQ}$. Prove that $E$ lies on $\omega$.
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a b c = triangle; o = circumcenter a b c; p = on_line a c, on_circle c b; s = foot p a b; d = on_line b p, on_circle o a; q = on_line s p, eqdistance p o a; e = on_tline a c q, on_tline b d q ? cyclic e a b c
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A B C = triangle; O = circumcenter A B C; AC = line A C; (C) = circle_center_point C B; P = intersection AC (C); AB = line A B; S = foot P AB; BP = line B P; (O) = circle_center_point O A; D = intersection BP (O); PS = line P S; (P,AO) = circle_center_radius P A O; Q = intersection PS (P,AO); CQ = line C Q; l1 = perpendicular_line A CQ; DQ = line D Q; l2 = perpendicular_line B DQ; E = intersection l1 l2; Prove: concyclic E A B C
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2023G5
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Let $ABC$ be an acute-angled triangle with circumcircle $\omega$ and circumcentre $O$. Points $D\neq B$ and $E\neq C$ lie on $\omega$ such that $BD\perp AC$ and $CE\perp AB$. Let $CO$ meet $AB$ at $X$, and $BO$ meet $AC$ at $Y$. Prove that the circumcircles of triangles $BXD$ and $CYE$ have an intersection lie on line $AO$.
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a b c = triangle; o = circumcenter a b c; d = on_circle o a, on_tline b a c; e = on_circle o a, on_tline c a b; x = on_line c o, on_line a b; y = on_line b o, on_line a c; o1 = circumcenter b x d; o2 = circumcenter c y e; p = on_circle o1 b, on_circle o2 c ? coll p a o
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A B C = acute_triangle; O = circumcenter A B C; (O) = circle_center_point O A; AC = line A C; l1 = perpendicular_line B AC; D = intersection (O) l1; AB = line A B; l2 = perpendicular_line C AB; E = intersection (O) l2; CO = line C O; X = intersection CO AB; BO = line B O; Y = intersection BO AC; O1 = circumcenter B D X; O2 = circumcenter C E Y; (O1) = circle_center_point O1 B; (O2) = circle_center_point O2 C; Z = intersection (O1) (O2); Prove: collinear A O Z
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2023G7
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Let $ABC$ be an acute, scalene triangle with orthocentre $H$. Let $\ell_a$ be the line through the reflection of $B$ with respect to $CH$ and the reflection of $C$ with respect to $BH$. Lines $\ell_b$ and $\ell_c$ are defined similarly. Suppose lines $\ell_a$, $\ell_b$, and $\ell_c$ determine a triangle $\mathcal T$. Prove that the orthocentre of $\mathcal T$, the circumcentre of $\mathcal T$, and $H$ are collinear.
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a b c = triangle; h = orthocenter a b c; b1 = reflect b c h; c1 = reflect c b h; c2 = reflect c a h; a2 = reflect a c h; a3 = reflect a b h; b3 = reflect b a h; p = on_line b1 c1, on_line c2 a2; q = on_line c2 a2, on_line a3 b3; r = on_line a3 b3, on_line b1 c1; h1 = orthocenter p q r; o1 = circumcenter p q r ? coll h1 o1 h
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A B C = acute_triangle; H = orthocenter A B C; CH = line C H; B1 = reflect_point_wrt_line B CH; BH = line B H; C1 = reflect_point_wrt_line C BH; la = line B1 C1; A1 = reflect A BH; AH = line A H; B2 = reflect B AH; lc = line A1 B2; A2 = reflect A CH; C2 = reflect C AH; lb = line A2 C2; X = intersection lb lc; Y = intersection la lc; Z = intersection la lb; H1 = orthocenter X Y Z; O1 = circumcenter X Y Z; Prove: collinear H H1 O1
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2023IMOp6
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Let $ABC$ be an equilateral triangle. Let $A_1,B_1,C_1$ be interior points of $ABC$ such that $BA_1=A_1C$, $CB_1=B_1A$, $AC_1=C_1B$, and $$\angle BA_1C+\angle CB_1A+\angle AC_1B=480^\circ$$Let $BC_1$ and $CB_1$ meet at $A_2,$ let $CA_1$ and $AC_1$ meet at $B_2,$ and let $AB_1$ and $BA_1$ meet at $C_2.$ Prove that if triangle $A_1B_1C_1$ is scalene, then the three circumcircles of triangles $AA_1A_2, BB_1B_2$ and $CC_1C_2$ all pass through two common points.
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a b c = ieq_triangle; o = circumcenter a b c; a1 = on_bline b c; b1 = on_bline c a; c0 = on_aline c0 a b a1 c b1; c1 = on_bline a b, on_aline c1 a b c0 a o; a2 = on_line b c1, on_line c b1; b2 = on_line c a1, on_line a c1; c2 = on_line a b1, on_line b a1; o1 = circumcenter a a1 a2; o2 = circumcenter b b1 b2; o3 = circumcenter c c1 c2; x = on_circle o1 a, on_circle o2 b ? cyclic x c c1 c2
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A B C = equilateral_triangle; O = circumcenter A B C; l1 = perpendicular_bisector B C; A1 = on_line l1; l2 = perpendicular_bisector A C; B1 = on_line l2; l3 = perpendicular_bisector A B; l4 = angle_equal1 A B A1 C B1; C0 = on_line l4; l5 = angle_equal1 A B C0 A O; C1 = intersection l3 l5; BC1 = line B C1; CB1 = line C B1; A2 = intersection BC1 CB1; CA1 = line C A1; AC1 = line A C1; B2 = intersection CA1 AC1; AB1 = line A B1; BA1 = line B A1; C2 = intersection AB1 BA1; O1 = circumcenter A A1 A2; O2 = circumcenter B B1 B2; O3 = circumcenter C C1 C2; (O1) = circle_center_point O1 A; (O2) = circle_center_point O2 B; X = intersection (O1) (O2); Prove: concyclic X C C1 C2
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2023IZhautykovOp2
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The tangent at $C$ to $\Omega$, the circumcircle of scalene triangle $ABC$ intersects $AB$ at $D$. Through point $D$, a line is drawn that intersects segments $AC$ and $BC$ at $K$ and $L$ respectively. On the segment $AB$ points $M$ and $N$ are marked such that $AC \parallel NL$ and $BC \parallel KM$. Lines $NL$ and $KM$ intersect at point $P$ lying inside the triangle $ABC$. Let $\omega$ be the circumcircle of $MNP$. Suppose $CP$ intersects $\omega$ again at $Q$. Show that $DQ$ is tangent to $\omega$.
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a b c = triangle; o = circumcenter a b c; d = on_tline c o c, on_line a b; k = on_line a c; l = on_line d k, on_line b c; n = on_line a b, on_pline l a c; m = on_line a b, on_pline k b c; p = on_line n l, on_line k m; o1 = circumcenter m n p; q = on_line c p, on_circle o1 m ? perp d q o1 q
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A B C = triangle; O = circumcenter A B C; OC = line O C; l1 = perpendicular_line C OC; AB = line A B; D = intersection l1 AB; AC = line A C; K = on_line AC; BC = line B C; DK = line D K; L = intersection DK BC; l2 = parallel_line K BC; M = intersection AB l2; l3 = parallel_line L AC; N = intersection AB l3; LN = line L N; KM = line K M; P = intersection LN KM; O1 = circumcenter M N P; (O1) = circle_center_point O1 M; CP = line C P; Q = intersection CP (O1); DQ = line D Q; O1Q = line O1 Q; Prove: perpendicular DQ O1Q
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2023IranGOAp1
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We are given an acute triangle $ABC$. The angle bisector of $\angle BAC$ cuts $BC$ at $P$. Points $D$ and $E$ lie on segments $AB$ and $AC$, respectively, so that $BC \parallel DE$. Points $K$ and $L$ lie on segments $PD$ and $PE$, respectively, so that points $A$, $D$, $E$, $K$, $L$ are concyclic. Prove that points $B$, $C$, $K$, $L$ are also concyclic.
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a b c = triangle; p = on_line b c, angle_bisector b a c; d = on_line a b; e = on_line a c, on_pline d b c; o1 = circumcenter a d e; k = on_line p d, on_circle o1 a; l = on_line p e, on_circle o1 a ? cyclic b c k l
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A B C = acute_triangle; l1 = angle_bisector B A C; BC = line B C; P = intersection l1 BC; AB = line A B; D = on_line AB; l2 = parallel_line D BC; AC = line A C; E = intersection AC l2; DP = line D P; (ADE) = circle A D E; K = intersection DP (ADE); EP = line E P; L = intersection EP (ADE); Prove: concyclic B C K L
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2023IranGOAp4
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Let $ABC$ be a triangle with bisectors $BE$ and $CF$ meet at $I$. Let $D$ be the projection of $I$ on the $BC$. Let M and $N$ be the orthocenters of triangles $AIF$ and $AIE$, respectively. Lines $EM$ and $FN$ meet at $P.$ Let $X$ be the midpoint of $BC$. Let $Y$ be the point lying on the line $AD$ such that $XY \perp IP$. Prove that line $AI$ bisects the segment $XY$.
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a b c = triangle; e = on_line a c, angle_bisector a b c; f = on_line a b, angle_bisector a c b; i = incenter a b c; d = foot i b c; m = orthocenter a i f; n = orthocenter a i e; p = on_line e m, on_line f n; x = midpoint b c; y = on_line a d, on_tline x i p; m_xy = midpoint x y ? coll m_xy a i
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A B C = triangle; l1 = angle_bisector A B C; AC = line A C; E = intersection l1 AC; l2 = angle_bisector B C A; AB = line A B; F = intersection l2 AB; I = incenter A B C; BC = line B C; D = foot I BC; M = orthocenter A I F; N = orthocenter A I E; EM = line E M; FN = line F N; P = intersection EM FN; X = midpoint B C; IP = line I P; l3 = perpendicular_line X IP; AD = line A D; Y = intersection AD l3; L = midpoint X Y; Prove: collinear L A I
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2023IranGOAp5
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In triangle $ABC$ points $M$ and $N$ are the midpoints of sides $AC$ and $AB$, respectively and $D$ is the projection of $A$ into $BC$. Point $O$ is the circumcenter of $ABC$ and circumcircles of $BOC$, $DMN$ intersect at points $R, T$. Lines $DT$, $DR$ intersect line $MN$ at $E$ and $F$, respectively. Lines $CT$, $BR$ intersect at $K$. A point $P$ lies on $KD$ such that $PK$ is the angle bisector of $\angle BPC$. Prove that the circumcircles of $ART$ and $PEF$ are tangent.
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a b c = triangle; m = midpoint a c; n = midpoint a b; d = foot a b c; o = circumcenter a b c; r = on_circum b o c, on_circum d m n; t = on_circum b o c, on_circum d m n; e = on_line d t, on_line m n; f = on_line d r, on_line m n; k = on_line c t, on_line b r; x = on_bline b c, on_line k d; p = on_line k d, on_circum b c x; o3 = circumcenter a r t; o4 = circumcenter p e f; a1 = on_line o3 o4, on_circle o4 p ? midp o4 a a1
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A B C = triangle; M = midpoint A C; N = midpoint A B; BC = line B C; D = foot A BC; O = circumcenter A B C; (BOC) = circumcircle B O C; (DMN) = circumcircle D M N; R = intersection (BOC) (DMN); T = intersection (BOC) (DMN); DT = line D T; MN = line M N; E = intersection DT MN; DR = line D R; F = intersection DR MN; CT = line C T; BR = line B R; K = intersection CT BR; l1 = perpendicular_bisector B C; KD = line K D; X = intersection l1 KD; (BCX) = circumcircle B C X; P = intersection KD (BCX); O3 = circumcenter A R T; O4 = circumcenter P E F; O3O4 = line O3 O4; (O4) = circle_center_point O4 P; A1 = intersection O3O4 (O4); Prove: midpoint O4 A A1
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2023IranGOMp4
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Let $ABC$ be a triangle and $P$ be the midpoint of arc $BAC$ of circumcircle of triangle $ABC$ with orthocenter $H$. Let $Q, S$ be points such that $HAPQ$ and $SACQ$ are parallelograms. Let $T$ be the midpoint of $AQ$, and $R$ be the intersection point of the lines $SQ$ and $PB$. Prove that $AB$, $SH$ and $TR$ are concurrent.
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a b c = triangle; h = orthocenter a b c; o = circumcenter a b c; p = on_bline b c, on_circle o a; q = parallelogram h a p q; s = parallelogram a c q s; t = midpoint a q; r = on_line s q, on_line p b; x = on_line a b, on_line s h ? coll x t r
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A B C = triangle; H = orthocenter A B C; O = circumcenter A B C; P = midarc_no B C A; Q = parallelogram H A P; S = parallelogram A C Q; T = midpoint A Q; QS = line Q S; BP = line B P; R = intersection QS BP; AB = line A B; HS = line H S; X = intersection AB HS; Prove: collinear X T R
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2023IranTSTp6
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$ABC$ is an acute triangle with orthocenter $H$. Point $P$ is in triangle $BHC$ that $\angle HPC = 3 \angle HBC $ and $\angle HPB =3 \angle HCB $. Reflection of point $P$ through $BH,CH$ is $X,Y$. if $S$ is the center of circumcircle of $AXY$ , Prove that: $$\angle BAS = \angle CAP$$
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a b c = triangle; h = orthocenter a b c; h1 = reflect h b c; c1 = reflect c h b; b1 = reflect b h c; p = eqangle3 p h b c b1 h1, eqangle3 p h c b c1 h1; x = reflect p b h; y = reflect p c h; s = circumcenter a x y ? eqangle b a a s p a a c
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A B C = triangle; H = orthocenter A B C; BC = line B C; H1 = reflect_point_wrt_line H BC; BH = line B H; C1 = reflect_point_wrt_line C BH; CH = line C H; B1 = reflect_point_wrt_line B CH; Ο1 = angle_equal2 H B B1 C H1; Ο2 = angle_equal2 H C C1 B H1; P = intersection Ο1 Ο2; X = reflect_point_wrt_line P BH; Y = reflect_point_wrt_line P CH; S = circumcenter A X Y; Prove: equal_angle B A S P A C
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2023KoMaLA840
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The incircle of triangle $ABC$ touches the sides in $X$, $Y$ and $Z$. In triangle $XYZ$ the feet of the altitude from $X$ and $Y$ are $X'$ and $Y'$, respectively. Let line $X'Y'$ intersect the circumcircle of triangle $ABC$ at $P$ and $Q$. Prove that points $X$, $Y$, $P$ and $Q$ are concyclic.
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a b c = triangle; i = incenter a b c; x = foot i b c; y = foot i a c; z = foot i a b; x1 = foot x y z; y1 = foot y x z; p = on_line x1 y1, on_circum a b c; q = on_line x1 y1, on_circum a b c ? cyclic x y p q
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A B C = triangle; I = incenter A B C; BC = line B C; X = foot I BC; AC = line A C; Y = foot I AC; AB = line A B; Z = foot I AB; YZ = line Y Z; X1 = foot X YZ; XZ = line X Z; Y1 = foot Y XZ; X1Y1 = line X1 Y1; (ABC) = circle A B C; P = intersection X1Y1 (ABC); Q = intersection X1Y1 (ABC); Prove: concyclic X Y P Q
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2023KoMaLA844
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The inscribed circle of triangle $ABC$ is tangent to sides $BC$, $AC$ and $AB$ at points $D$, $E$ and $F$, respectively. Let $E'$ be the reflection of point $E$ across line $DF$, and $F'$ be the reflection of point $F$ across line $DE$. Let line $EF$ intersect the circumcircle of triangle $AE'F'$ at points $X$ and $Y$. Prove that $DX=DY$.
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a b c = triangle; i = incenter a b c; d = foot i b c; e = foot i a c; f = foot i a b; e1 = reflect e d f; f1 = reflect f d e; o1 = circumcenter a e1 f1; x = on_line e f, on_circle o1 a; y = on_line e f, on_circle o1 a ? cong d x d y
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A B C = triangle; I = incenter A B C; BC = line B C; D = foot I BC; AC = line A C; E = foot I AC; AB = line A B; F = foot I AB; DF = line D F; E1 = reflect_point_wrt_line E DF; DE = line D E; F1 = reflect_point_wrt_line F DE; EF = line E F; O1 = circumcenter A E1 F1; (O1) = circle_center_point O1 A; X Y = intersection EF (O1); Prove: cong D X D Y
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2023KoMaLA855
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In scalene triangle $ABC$ the shortest side is $BC$. Let points $M$ and $N$ be chosen on sides $AB$ and $AC$, respectively, such that $BM=CN=BC$. Let $I$ and $O$ denote the incenter and circumcentre of triangle $ABC$, and let $D$ and $E$ denote the incenter and circumcenter of triangle $AMN$. Prove that lines $IO$ and $DE$ intersect each other on the circumcircle of triangle $ABC$.
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a b c = triangle; m = on_line a b, eqdistance m b c b; n = on_line a c, eqdistance n c c b; i = incenter a b c; o = circumcenter a b c; d = incenter a m n; e = circumcenter a m n; x = on_line i o, on_line d e ? cyclic x a b c
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A B C = triangle; AB = line A B; (B,BC) = circle_center_radius B B C; M = intersection AB (B,BC); AC = line A C; (C,BC) = circle_center_radius C B C; N = intersection AC (C,BC); I = incenter A B C; O = circumcenter A B C; D = incenter A M N; E = circumcenter A M N; IO = line I O; DE = line D E; X = intersection IO DE; Prove: concyclic X A B C
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2023MOSTMockp2
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Let $\triangle A_1A_3A_5 \sim \triangle A_4A_6A_2$ with sides intersecting at $X_1,\dots,X_6$. For $1\leq i\leq6$, let $O_i$ be the circumcenter of $\triangle A_iX_iA_{i+1}$ ($A_7=A_1$). Prove $O_1O_4$, $O_2O_5$, $O_3O_6$ are concurrent.
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a1 a3 a5 = triangle; a4 = free; a6 = free; a2 = on_aline a2 a4 a6 a5 a1 a3, on_aline a2 a6 a4 a5 a3 a1; x1 = on_line a1 a3, on_line a2 a6; x2 = on_line a1 a3, on_line a2 a4; x3 = on_line a2 a4, on_line a3 a5; x4 = on_line a3 a5, on_line a4 a6; x5 = on_line a1 a5, on_line a6 a4; x6 = on_line a1 a5, on_line a2 a6; o1 = circumcenter a1 x1 a2; o2 = circumcenter a2 x2 a3; o3 = circumcenter a3 x3 a4; o4 = circumcenter a4 x4 a5; o5 = circumcenter a5 x5 a6; o6 = circumcenter a6 x6 a1; k = on_line o1 o4, on_line o2 o5 ? coll k o3 o6
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A1 A3 A5 = triangle; A4 = point; A6 = point; l1 = angle_equal1 A4 A6 A5 A1 A3; l2 = angle_equal1 A6 A4 A5 A3 A1; A2 = intersection l1 l2; A1A3 = line A1 A3; A2A6 = line A2 A6; X1 = intersection A1A3 A2A6; A2A4 = line A2 A4; X2 = intersection A1A3 A2A4; A3A5 = line A3 A5; X3 = intersection A2A4 A3A5; A4A6 = line A4 A6; X4 = intersection A3A5 A4A6; A1A5 = line A1 A5; X5 = intersection A1A5 A4A6; X6 = intersection A1A5 A2A6; O1 = circumcenter A1 X1 A2; O2 = circumcenter A2 X2 A3; O3 = circumcenter A3 X3 A4; O4 = circumcenter A4 X4 A5; O5 = circumcenter A5 X5 A6; O6 = circumcenter A6 X6 A1; O1O4 = line O1 O4; O2O5 = line O2 O5; K = intersection O1O4 O2O5; Prove: collinear K O3 O6
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2023PKUSummerp3
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In $\triangle ABC$, $BC$ is the longest side. The perpendicular bisector of $AC$ intersects $BC, AB$ at $D,E$ respectively, and the symmetric point of $B$ w.r.t. this perpendicular bisector is $F$. The perpendicular bisector of $AB$ intersects $BC,AC$ at $J,K$ respectively, and the symmetric point of $C$ w.r.t. this perpendicular bisector is $L$. $BL \cap CF = N$, $\odot(BJL) \cap JN = J,R$, $\odot(CDF) \cap DN = D,Q$. The parallel line of $BC$ passes through $N$ and intersects $EK$ at $P$. $FL \cap BC = M$. $l$ is the diameter of $\odot(ABC)$ parallel to BC. Prove that $QR, MP$ and $l$ are cocurrent.
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a b c = triangle; d = on_bline a c, on_line b c; e = on_bline a c, on_line a b; f = reflect b d e; j = on_bline a b, on_line b c; k = on_bline a b, on_line a c; l = reflect c j k; n = on_line b l, on_line c f; o1 = circumcenter b j l; r = on_circle o1 b, on_line j n; o2 = circumcenter c d f; q = on_circle o2 c, on_line d n; p = on_pline n b c, on_line e k; m = on_line f l, on_line b c; o = circumcenter a b c; x = on_line q r, on_line m p ? para o x b c
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A B C = triangle; l1 = perpendicular_bisector A C; BC = line B C; D = intersection l1 BC; AB = line A B; E = intersection l1 AB; DE = line D E; F = reflect_point_wrt_line B DE; l2 = perpendicular_bisector A B; J = intersection l2 BC; AC = line A C; K = intersection l2 AC; JK = line J K; L = reflect_point_wrt_line C JK; BL = line B L; CF = line C F; N = intersection BL CF; O1 = circumcenter B J L; (O1) = circle_center_point O1 B; JN = line J N; R = intersection (O1) JN; O2 = circumcenter C D F; (O2) = circle_center_point O2 C; DN = line D N; Q = intersection (O2) DN; l3 = parallel_line N BC; EK = line E K; P = intersection l3 EK; FL = line F L; M = intersection FL BC; O = circumcenter A B C; QR = line Q R; MP = line M P; X = intersection QR MP; OX = line O X; Prove: parallel OX BC
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2023PlanetCupp10
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In obtuse $\triangle ABC$ ($\angle BAC>90^\circ$) with centroid $G$, let the nine-point circle and circumcircle intersect at $P,Q$. Lines $PG$ and $QG$ meet $BC$ at $X,Y$ respectively. Prove that the intersection point of $PY$, $QX$, and $AG$ lies on the circumcircle of $\triangle ABC$.
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a b c = triangle; m1 m2 m3 g = centroid a b c; o = circumcenter a b c; ni = circumcenter m1 m2 m3; p = on_circle o a, on_circle ni m1; q = on_circle o a, on_circle ni m1; x = on_line p g, on_line b c; y = on_line q g, on_line b c; k = on_line p y, on_line a g ? cyclic k a b c
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A B C = obtuse_triangle; G M1 M2 M3 = centroid A B C; O = circumcenter A B C; Ni = circumcenter M1 M2 M3; (O) = circle_center_point O A; (Ni) = circle_center_point Ni M1; P Q = intersection (O) (Ni); PG = line P G; BC = line B C; X = intersection PG BC; QG = line Q G; Y = intersection QG BC; PY = line P Y; AG = line A G; K = intersection PY AG; Prove: concyclic K A B C
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2023PlanetCupp9
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For $\triangle ABC$ with circumcircle $\Omega$, orthocenter $H$, and circumcenter $O$, line $OH$ meets $AC,AB$ at $E,F$. The circumcircle of $\triangle AHO$ meets $\Omega$ at $K$, and $KH$ extended meets $\Omega$ at $L$. With $M$ as $BC$'s midpoint, let $P\in HM$ satisfy $PE=PF$. If $PL$ meets $BC$ at $Q$, prove $QH=QO$.
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a b c = triangle; o = circumcenter a b c; h = orthocenter a b c; e = on_line o h, on_line a c; f = on_line o h, on_line a b; o1 = circumcenter a h o; k = on_circle o1 a, on_circle o a; l = on_line k h, on_circle o a; m = midpoint b c; p = on_line h m, on_bline e f; q = on_line p l, on_line b c ? cong q h q o
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A B C = triangle; O = circumcenter A B C; H = orthocenter A B C; OH = line O H; AC = line A C; E = intersection OH AC; AB = line A B; F = intersection OH AB; O1 = circumcenter A H O; (O1) = circle_center_point O1 A; (O) = circle_center_point O A; K = intersection (O1) (O); KH = line K H; L = intersection KH (O); M = midpoint B C; HM = line H M; l1 = perpendicular_bisector E F; P = intersection HM l1; PL = line P L; BC = line B C; Q = intersection PL BC; Prove: cong Q H Q O
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2023RMMSLG1
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Let $ABC$ be a triangle with incentre $I$ and circumcircle $\omega$. The incircle of the triangle $ABC$ touches the sides $BC$, $CA$ and $AB$ at $D$, $E$ and $F$, respectively. The circumcircle of triangle $ADI$ crosses $\omega$ again at $P$, and the lines $PE$ and $PF$ cross $\omega$ again at $X$and $Y$, respectively. Prove that the lines $AI$, $BX$ and $CY$ are concurrent.
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a b c = triangle; i = incenter a b c; o = circumcenter a b c; d = foot i b c; e = foot i c a; f = foot i a b; o1 = circumcenter a d i; p = on_circle o a, on_circle o1 a; x = on_line p e, on_circle o a; y = on_line p f, on_circle o a; z = on_line a i, on_line b x ? coll z c y
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A B C = triangle; I = incenter A B C; O = circumcenter A B C; BC = line B C; D = foot I BC; AC = line A C; E = foot I AC; AB = line A B; F = foot I AB; O1 = circumcenter A D I; (O1) = circle_center_point O1 A; (O) = circle_center_point O A; P = intersection (O1) (O); EP = line E P; X = intersection EP (O); FP = line F P; Y = intersection FP (O); AI = line A I; BX = line B X; Z = intersection AI BX; Prove: collinear Z C Y
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2023RMMSLG3
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A point $P$ is chosen inside a triangle $ABC$ with circumcircle $\Omega$. Let $\Gamma$ be the circle passing through the circumcenters of the triangles $APB$, $BPC$, and $CPA$. Let $\Omega$ and $\Gamma$ intersect at points $X$ and $Y$. Let $Q$ be the reflection of $P$ in the line $XY$ . Prove that $\angle BAP = \angle CAQ$.
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a b c = triangle; p = free; o = circumcenter a b c; o1 = circumcenter a p b; o2 = circumcenter b p c; o3 = circumcenter c p a; o_g = circumcenter o1 o2 o3; x = on_circle o a, on_circle o_g o1; y = on_circle o a, on_circle o_g o1; q = reflect p x y ? eqangle b a a p q a a c
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A B C = triangle; O = circumcenter A B C; (O) = circumcircle A B C; P = point; O1 = circumcenter A P B; O2 = circumcenter B P C; O3 = circumcenter C P A; O4 = circumcenter O1 O2 O3; (O1O2O3) = circle_center_point O4 O1; X Y = intersection (O) (O1O2O3); XY = line X Y; Q = reflect_point_wrt_line P XY; Prove: equal_angle B A P Q A C
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2023SAGFp2
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In an acute-angled triangle, $ABC$ $w$ is a circumscribed circle, and $O$ is inside the triangle and $OB = OC$. The point $D$ is selected on the side of $AB$,so that $OD \parallel BC$. The straight line $AO$ repeatedly intersects the circles $w$ and $(COD)$ at the points $A',P$, respectively. $M$ is the middle of the side of $BC$. Circle $(MAA')$ intersects the line $BC$ at the point $L$. The straight line $AL$ repeatedly intersects the circle $w$ at the point $K$. Prove that $\angle APK = \angle AA'M$.
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a b c = triangle; u = circumcenter a b c; o = on_bline b c; d = on_line a b, on_pline o b c; a1 = on_line a o, on_circle u a; o1 = circumcenter c o d; p = on_line a o, on_circle o1 o; m = midpoint b c; o2 = circumcenter m a a1; l = on_line b c, on_circle o2 m; k = on_line a l, on_circle u a ? eqangle a p p k m a1 a1 a
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A B C = acute_triangle; U = circumcenter A B C; (ABC) = circle_center_point U A; l = perpendicular_bisector B C; O = on_line l; BC = line B C; l1 = parallel_line O BC; AB = line A B; D = intersection l1 AB; AO = line A O; A1 = intersection AO (ABC); O1 = circumcenter C O D; (CDO) = circle_center_point O1 O; P = intersection AO (CDO); M = midpoint B C; O2 = circumcenter M A A1; (AA1M) = circle_center_point O2 M; L = intersection (AA1M) BC; AL = line A L; K = intersection AL (ABC); Prove: equal_angle A P K M A1 A
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2023SAGFp3
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Let $P$ be a point on the bisector of $\angle BAC$. Line $BP,CP$ intersect $AC,AB$ at $E,D$, respectively. Tangents at $A$ WRT $\odot(ABC),\odot(ADE)$ intersect $BC,DE$ at $S,T$, respectively. $M$ is the Miquel point of $ADBPCE$ $(M = (CPE) \cap (BDP))$. Prove that $A,M,S,T$ are concyclic
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a b c = triangle; p = angle_bisector b a c; e = on_line b p, on_line a c; d = on_line c p, on_line a b; o1 = circumcenter a b c; s = on_tline a o1 a, on_line b c; o2 = circumcenter a d e; t = on_tline a o2 a, on_line d e; o3 = circumcenter c p e; o4 = circumcenter b d p; m = on_circle o3 p, on_circle o4 p ? cyclic a m s t
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A B C = triangle; l = angle_bisector B A C; P = on_line l; BP = line B P; AC = line A C; E = intersection BP AC; CP = line C P; AB = line A B; D = intersection CP AB; O1 = circumcenter A B C; O1A = line O1 A; l1 = perpendicular_line A O1A; BC = line B C; S = intersection l1 BC; O2 = circumcenter A D E; O2A = line O2 A; l2 = perpendicular_line A O2A; DE = line D E; T = intersection l2 DE; O3 = circumcenter C P E; O4 = circumcenter B D P; (CEP) = circle_center_point O3 P; (BDP) = circle_center_point O4 P; M = intersection (CEP) (BDP); Prove: concyclic A M S T
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2023SAGFp8
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On Circumcircle of $\Delta ABC$ , $D,E,F$ are midpoint of arcs $BC,CA,AB$ not containing the vertices. $R,S,T$ antipodes of $D,E,F$. $D',E',F'$ are reflections of $D,E,F$ across $BC,CA,AB$ and $R',S',T'$ are reflections of $R,S,T$ across $BC,CA,AB$. Let $H_1$ be orthocenter $\Delta D'E'F'$ and $O_1$ be circumcenter of $\Delta R'S'T'$. Prove that is the Euler line of $\Delta ABC\parallel H_1O_1$. $(b)$ $H_2 - $ be orthocenter $\Delta R'S'T'$,$O_2$ - be circumcenter of $\Delta D'E'F'$. Prove it: $H_1O_1 \parallel O_2H_2$.
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a b c = triangle; o = circumcenter a b c; d = on_bline b c, on_circle o a; e = on_bline c a, on_circle o a; f = on_bline a b, on_circle o a; r = mirror d o; s = mirror e o; t = mirror f o; d1 = reflect d b c; e1 = reflect e c a; f1 = reflect f a b; r1 = reflect r b c; s1 = reflect s c a; t1 = reflect t a b; h1 = orthocenter d1 e1 f1; o1 = circumcenter r1 s1 t1; h = orthocenter a b c ? para h1 o1 o h
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A B C = triangle; O = circumcenter A B C; D = midarc_no B C A; E = midarc_no C A B; F = midarc_no A B C; R = reflect D O; S = reflect E O; T = reflect F O; BC = line B C; CA = line C A; AB = line A B; D1 = reflect_point_wrt_line D BC; E1 = reflect_point_wrt_line E CA; F1 = reflect_point_wrt_line F AB; R1 = reflect_point_wrt_line R BC; S1 = reflect_point_wrt_line S CA; T1 = reflect_point_wrt_line T AB; H1 = orthocenter D1 E1 F1; O1 = circumcenter R1 S1 T1; H = orthocenter A B C; HO = line H O; H1O1 = line H1 O1; Prove: parallel HO H1O1
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2023SerbiaMOp6
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Given is a triangle $ABC$ with incenter $I$ and circumcircle $\omega$. The incircle is tangent to $BC$ at $D$. The perpendicular at $I$ to $AI$ meets $AB, AC$ at $E, F$ and the circle $(AEF)$ meets $\omega$ and $AI$ at $G, H$. The tangent at $G$ to $\omega$ meets $BC$ at $J$ and $AJ$ meets $\omega$ at $K$. Prove that $(DJK)$ and $(GIH)$ are tangent to each other.
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a b c = triangle; i = incenter a b c; o = circumcenter a b c; d = foot i b c; e = on_line a b, on_tline i a i; f = on_line a c, on_tline i a i; o1 = circumcenter a e f; g = on_circle o1 a, on_circle o a; h = on_circle o1 a, on_line a i; j = on_tline g o g, on_line b c; k = on_line a j, on_circle o a; o3 = circumcenter d j k; o4 = circumcenter g i h; t = on_circle o4 i, on_circle o3 d ? coll t o4 o3
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A B C = triangle; I = incenter A B C; O = circumcenter A B C; BC = line B C; D = foot I BC; AB = line A B; AI = line A I; l1 = perpendicular_line I AI; E = intersection AB l1; AC = line A C; F = intersection AC l1; O1 = circumcenter A E F; (O1) = circle_center_point O1 A; (O) = circle_center_point O A; G = intersection (O1) (O); AI = line A I; H = intersection (O1) AI; OG = line O G; l2 = perpendicular_line G OG; J = intersection l2 BC; AJ = line A J; K = intersection AJ (O); O3 = circumcenter D J K; O4 = circumcenter G I H; (GIH) = circle_center_point O4 I; (DJK) = circle_center_point O3 D; T = intersection (GIH) (DJK); Prove: collinear T O4 O3
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2023SpFMOp5
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In $\triangle ABC$, let $N,S$ be the north/south poles, $M$ the midpoint of $BC$, and $D\in NS$ with $ND=MS$. A circle centered at $D$ with radius $AD$ meets the midline parallel to $BC$ at $X,Y$. Prove $X$, the Feuerbach point $F_e$, $Y$, and the $A$-excircle's touchpoint $P$ with $BC$ are concyclic.
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a b c = triangle; o = circumcenter a b c; s = on_bline b c, on_circle o a; n = on_bline b c, on_circle o a; m1 = midpoint b c; d = mirror m1 o; m3 = midpoint a b; m2 = midpoint a c; x = on_circle d a, on_line m2 m3; y = on_circle d a, on_line m2 m3; ni = circumcenter m1 m2 m3; i = incenter a b c; d1 = foot i b c; f_e = on_circle i d1, on_circle ni m1; i_a = excenter a b c; p = foot i_a b c ? cyclic x f_e y p
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A B C = triangle; O = circumcenter A B C; l1 = perpendicular_bisector B C; (O) = circle_center_point O A; S = intersection l1 (O); N = intersection l1 (O); M1 = midpoint B C; D = reflect_point_wrt_point M1 O; M3 = midpoint A B; M2 = midpoint A C; (D,A) = circle_center_point D A; M2M3 = line M2 M3; X = intersection (D,A) M2M3; Y = intersection (D,A) M2M3; Ni = circumcenter M1 M2 M3; I = incenter A B C; BC = line B C; D1 = foot I BC; (I) = circle_center_point I D1; (Ni) = circle_center_point Ni M1; Fe = intersection (I) (Ni); Ia = excenter A B C; P = foot Ia BC; Prove: concyclic X Fe Y P
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2023TurkeyTSTp5
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Let $ABC$ be a scalene triangle with circumcentre $O$, incentre $I$ and orthocentre $H$. Let the second intersection point of circle which passes through $O$ and tangent to $IH$ at point $I$, and the circle which passes through $H$ and tangent to $IO$ at point $I$ be $M$. Prove that $M$ lies on circumcircle of $ABC$.
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a b c = triangle; o = circumcenter a b c; i = incenter a b c; h = orthocenter a b c; o1 = on_bline o i, on_tline i i h; o2 = on_bline h i, on_tline i i o; m = on_circle o1 o, on_circle o2 h ? cyclic m a b c
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A B C = triangle; O = circumcenter A B C; I = incenter A B C; H = orthocenter A B C; l1 = perpendicular_bisector O I; IH = line I H; l2 = perpendicular_line I IH; O1 = intersection l1 l2; l3 = perpendicular_bisector H I; IO = line I O; l4 = perpendicular_line I IO; O2 = intersection l3 l4; (O1) = circle_center_point O1 O; (O2) = circle_center_point O2 H; M = intersection (O1) (O2); Prove: concyclic M A B C
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2023USAMOp1
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In an acute triangle $ABC$, let $M$ be the midpoint of $\overline{BC}$. Let $P$ be the foot of the perpendicular from $C$ to $AM$. Suppose that the circumcircle of triangle $ABP$ intersects line $BC$ at two distinct points $B$ and $Q$. Let $N$ be the midpoint of $\overline{AQ}$. Prove that $NB=NC$.
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a b c = triangle; m = midpoint b c; p = foot c a m; q = on_line b c, on_circum a b p; n = midpoint a q ? cong n b n c
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A B C = acute_triangle; M = midpoint B C; AM = line A M; P = foot C AM; (ABP) = circle A B P; BC = line B C; Q = intersection (ABP) BC; N = midpoint A Q; Prove: cong B N C N
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2023USAMOp6
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Let $ABC$ be a triangle with incenter $I$ and excenters $I_a$, $I_b$, and $I_c$ opposite $A$, $B$, and $C$, respectively. Let $D$ be an arbitrary point on the circumcircle of $\triangle{ABC}$ that does not lie on any of the lines $II_a$, $I_bI_c$, or $BC$. Suppose the circumcircles of $\triangle{DII_a}$ and $\triangle{DI_bI_c}$ intersect at two distinct points $D$ and $F$. If $E$ is the intersection of lines $DF$ and $BC$, prove that $\angle{BAD} = \angle{EAC}$.
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a b c = triangle; i = incenter a b c; ia = excenter a b c; ib = excenter b c a; ic = excenter c a b; o = circumcenter a b c; d = on_circle o a; o1 = circumcenter d i ia; o2 = circumcenter d ib ic; f = on_circle o1 d, on_circle o2 d; e = on_line d f, on_line b c ? eqangle b a a d e a a c
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A B C = triangle; I = incenter A B C; Ia = excenter A B C; Ib = excenter B C A; Ic = excenter C A B; O = circumcenter A B C; (ABC) = circle_center_point O A; D = on_circle (ABC); O1 = circumcenter D I Ia; O2 = circumcenter D Ib Ic; (DIIa) = circle_center_point O1 D; (DIbIc) = circle_center_point O2 D; F = intersection (DIIa) (DIbIc); DF = line D F; BC = line B C; E = intersection DF BC; Prove: equal_angle B A D E A C
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2023VietnamTSTp3
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Let $ABC$ be an acute, non-isosceles triangle with circumcircle $(O)$. $BE, CF$ are the heights of $\triangle ABC$, and $BE, CF$ intersect at $H$. Let $M$ be the midpoint of $AH$, and $K$ be the point on $EF$ such that $HK \perp EF$. A line not going through $A$ and parallel to $BC$ intersects the minor arc $AB$ and $AC$ of $(O)$ at $P$, $Q$, respectively. Show that the tangent line of $(CQE)$ at $E$, the tangent line of $(BPF)$ at $F$, and $MK$ concur.
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a b c = triangle; o = circumcenter a b c; e = foot b a c; f = foot c a b; h = on_line b e, on_line c f; m = midpoint a h; k = foot h e f; p = on_circle o a; q = on_circle o a, on_pline p b c; o1 = circumcenter c q e; o2 = circumcenter b p f; x = on_tline e o1 e, on_tline f o2 f ? coll x m k
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A B C = triangle; O = circumcenter A B C; AC = line A C; E = foot B AC; AB = line A B; F = foot C AB; BE = line B E; CF = line C F; H = intersection BE CF; M = midpoint A H; EF = line E F; K = foot H EF; (O) = circle_center_point O A; P = on_circle (O); BC = line B C; l1 = parallel_line P BC; Q = intersection (O) l1; O1 = circumcenter C Q E; O2 = circumcenter B P F; O1E = line O1 E; l2 = perpendicular_line E O1E; O2F = line O2 F; l3 = perpendicular_line F O2F; X = intersection l2 l3; Prove: collinear X M K
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2024ARMOg10p6
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Let $ABCD$ be a parallelogram. Let $M$ be the midpoint of the arc $AC$ containing $B$ of the circumcircle of $ABC$ . Let $E$ be a point on segment $AD$ and $F$ a point on segment $CD$ such that $ME=MD=MF$. Show that $BMEF$ is cyclic.
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a b c = triangle; d = parallelogram a b c d; o = circumcenter a b c; m = on_circle o a, on_bline a c; e = on_line a d, on_circle m d; f = on_line c d, on_circle m d ? cyclic b m e f
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A B C D = parallelogram; O = circumcenter A B C; M = midarc A B C; AD = line A D; (M,D) = circle_center_point M D; E = intersection AD (M,D); CD = line C D; F = intersection CD (M,D); Prove: concyclic B M E F
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2024ARMOg9p4
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In cyclic quadrilateral $ABCD$, $\angle A+ \angle D=\frac{\pi}{2}$. $AC$ intersects $BD$ at ${E}$. A line ${l}$ cuts segment $AB, CD, AE, DE$ at $X, Y, Z, T$ respectively. If $AZ=CE$ and $BE=DT$, prove that the diameter of the circumcircle of $\triangle EZT$ equals $XY$.
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a b c = triangle; d = on_circum a b c, on_tline c a b; e = on_line a c, on_line b d; z = on_line a c, eqdistance z a c e; t = on_line d b, eqdistance t d b e; x = on_line a b, on_line z t; y = on_line c d, on_line z t; o1 = circumcenter e z t; e1 = mirror e o1 ? cong e e1 x y
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A B C = triangle; (ABC) = circumcircle A B C; AB = line A B; l1 = perpendicular_line C AB; D = intersection (ABC) l1; AC = line A C; BD = line B D; E = intersection AC BD; (A) = circle_center_radius A C E; Z = intersection AC (A); (D) = circle_center_radius D B E; DB = line D B; T = intersection DB (D); AB = line A B; ZT = line Z T; X = intersection AB ZT; CD = line C D; Y = intersection CD ZT; O1 = circumcenter E Z T; E1 = reflect_point_wrt_point E O1; Prove: cong E E1 X Y
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2024ARMOg9p6
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The altitudes of an acute triangle $ABC$ with $AB<AC$ intersect at a point $H$, and $O$ is the center of the circumcircle $\Omega$. The segment $OH$ intersects the circumcircle of $BHC$ at a point $X$, different from $O$ and $H$. The circumcircle of $AOX$ intersects the smaller arc $AB$ of $\Omega$ at point $Y$. Prove that the line $XY$ bisects the segment $BC$.
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a b c = triangle; h = orthocenter a b c; o = circumcenter a b c; o1 = circumcenter b h c; x = on_line o h, on_circle o1 b; o2 = circumcenter a o x; y = on_circle o a, on_circle o2 a; m = midpoint b c ? coll x y m
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A B C = acute_triangle; H = orthocenter A B C; O = circumcenter A B C; (O) = circle_center_point O A; HO = line H O; O1 = circumcenter B C H; (O1) = circle_center_point O1 B; X = intersection HO (O1); O2 = circumcenter A O X; (O2) = circle_center_point O2 A; Y = intersection (O2) (O); XY = line X Y; BC = line B C; M = midpoint B C; Prove: collinear M X Y
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2024AsiaPacificMOp1
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Let $ABC$ be an acute triangle. Let $D$ be a point on side $AB$ and $E$ be a point on side $AC$ such that lines $BC$ and $DE$ are parallel. Let $X$ be an interior point of $BCED$. Suppose rays $DX$ and $EX$ meet side $BC$ at points $P$ and $Q$, respectively, such that both $P$ and $Q$ lie between $B$ and $C$. Suppose that the circumcircles of triangles $BQX$ and $CPX$ intersect at a point $Y \neq X$. Prove that the points $A, X$, and $Y$ are collinear.
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a b c = triangle; d = on_line a b; e = on_line a c, on_pline d b c; x = free; p = on_line b c, on_line d x; q = on_line b c, on_line e x; y = on_circum b q x, on_circum c p x ? coll a x y
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A B C = acute_triangle; AB = line A B; D = on_line AB; BC = line B C; l1 = parallel_line D BC; AC = line A C; E = intersection AC l1; X = point; DX = line D X; P = intersection DX BC; EX = line E X; Q = intersection EX BC; (BQX) = circle B Q X; (CPX) = circle C P X; Y = intersection (BQX) (CPX); Prove: collinear A X Y
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2024AsiaPacificMOp5
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Line $\ell$ intersects sides $BC$ and $AD$ of cyclic quadrilateral $ABCD$ in its interior points $R$ and $S$, respectively, and intersects ray $DC$ beyond point $C$ at $Q$, and ray $BA$ beyond point $A$ at $P$. Circumcircles of the triangles $QCR$ and $QDS$ intersect at $N \neq Q$, while circumcircles of the triangles $PAS$ and $PBR$ intersect at $M\neq P$. Let lines $MP$ and $NQ$ meet at point $X$, lines $AB$ and $CD$ meet at point $K$ and lines $BC$ and $AD$ meet at point $L$. Prove that point $X$ lies on line $KL$.
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a b c = triangle; d = on_circum a b c; p1 q1 = segment; p = on_line a b, on_line p1 q1; q = on_line c d, on_line p1 q1; r = on_line b c, on_line p1 q1; s = on_line a d, on_line p1 q1; o1 = circumcenter q c r; o2 = circumcenter q d s; n = on_circle o1 q, on_circle o2 q; o3 = circumcenter p a s; o4 = circumcenter p b r; m = on_circle o3 p, on_circle o4 p; x = on_line m p, on_line n q; k = on_line a b, on_line c d; l = on_line b c, on_line a d ? coll x k l
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A B C = triangle; (ABC) = circle A B C; D = on_circle (ABC); l = line; BC = line B C; R = intersection l BC; AD = line A D; S = intersection l AD; CD = line C D; Q = intersection l CD; AB = line A B; P = intersection l AB; O1 = circumcenter C Q R; (CQR) = circle_center_point O1 Q; O2 = circumcenter D Q S; (DQS) = circle_center_point O2 Q; N = intersection (CQR) (DQS); O3 = circumcenter A P S; (APS) = circle_center_point O3 P; O4 = circumcenter B P R; (BPR) = circle_center_point O4 P; M = intersection (APS) (BPR); MP = line M P; NQ = line N Q; X = intersection MP NQ; K = intersection AB CD; L = intersection BC AD; KL = line K L; Prove: collinear X K L
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2024BalkanMOp1
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Let $ABC$ be an acute-angled triangle with $AC > AB$ and let $D$ be the foot of the $A$-angle bisector on $BC$. The reflections of lines $AB$ and $AC$ in line $BC$ meet $AC$ and $AB$ at points $E$ and $F$ respectively. A line through $D$ meets $AC$ and $AB$ at $G$ and $H$ respectively such that $G$ lies strictly between $A$ and $C$ while $H$ lies strictly between $B$ and $F$. Prove that the circumcircles of $\triangle EDG$ and $\triangle FDH$ are tangent to each other.
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a b c = triangle; d = on_line b c, angle_bisector b a c; a1 = reflect a b c; e = on_line a c, on_line a1 b; f = on_line a b, on_line a1 c; g = on_line a c; h = on_line a b, on_line d g; o1 = circumcenter e d g; o2 = circumcenter f d h ? coll o1 o2 d
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A B C = acute_triangle; l1 = angle_bisector C A B; BC = line B C; D = intersection l1 BC; AB = line A B; A1 = reflect_point_wrt_line A BC; A1B = line A1 B; AC = line A C; E = intersection A1B AC; A1C = line A1 C; F = intersection AB A1C; l = line_through D; G = intersection l AC; H = intersection l AB; O1 = circumcenter D E G; O2 = circumcenter D F H; Prove: collinear O1 O2 D
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2024EGMOp2
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Let $ABC$ be a triangle with $AC>AB$ , and denote its circumcircle by $\Omega$ and incentre by $I$. Let its incircle meet sides $BC,CA,AB$ at $D,E,F$ respectively. Let $X$ and $Y$ be two points on minor arcs $\widehat{DF}$ and $\widehat{DE}$ of the incircle, respectively, such that $\angle BXD = \angle DYC$. Let line $XY$ meet line $BC$ at $K$. Let $T$ be the point on $\Omega$ such that $KT$ is tangent to $\Omega$ and $T$ is on the same side of line $BC$ as $A$. Prove that lines $TD$ and $AI$ meet on $\Omega$.
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a b c = triangle; i = incenter a b c; d = foot i b c; e = foot i a c; f = foot i a b; o = circumcenter a b c; x = on_circle i d; y = on_circle i d, eqangle3 y d c x b d; k = on_line x y, on_line b c; t t1 = tangent k o a; u = on_line t d, on_line a i ? cyclic a b c u
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A B C = triangle; I = incenter A B C; BC = line B C; CA = line C A; AB = line A B; D = foot I BC; E = foot I CA; F = foot I AB; (I) = circle_center_point I D; X = on_circle (I); Ο1 = angle_equal2 D C B X D; Y = intersection Ο1 (I); XY = line X Y; K = intersection XY BC; O = circumcenter A B C; (O) = circle_center_point O A; T T1 = tangent_point K (O); TD = line T D; AI = line A I; U = intersection TD AI; Prove: concyclic A B C U
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2024ELMOSLp1
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In convex quadrilateral $ABCD$, let diagonals $\overline{AC}$ and $\overline{BD}$ intersect at $E$. Let the circumcircles of $ADE$ and $BCE$ intersect $\overline{AB}$ again at $P \neq A$ and $Q \neq B$, respectively. Let the circumcircle of $ACP$ intersect $\overline{AD}$ again at $R \neq A$, and let the circumcircle of $BDQ$ intersect $\overline{BC}$ again at $S \neq B$. Prove that $A$, $B$, $R$, and $S$ are concyclic.
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a b c d = quadrangle; e = on_line a c, on_line b d; p = on_line a b, on_circum a d e; q = on_line a b, on_circum b c e; r = on_line a d, on_circum a c p; s = on_line b c, on_circum b d q ? cyclic a b r s
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A B C D = quadrilateral; AC = line A C; BD = line B D; E = intersection AC BD; AB = line A B; (ADE) = circle A D E; P = intersection AB (ADE); (BCE) = circle B C E; Q = intersection AB (BCE); AD = line A D; (ACP) = circle A C P; R = intersection AD (ACP); BC = line B C; (BDQ) = circle B D Q; S = intersection BC (BDQ); Prove: concyclic A B R S
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2024ELMOSLp4
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In quadrilateral $ABCD$ with incenter $I$, points $W,X,Y,Z$ lie on sides $AB, BC,CD,DA$ with $AZ=AW$, $BW=BX$, $CX=CY$, $DY=DZ$. Define $T=\overline{AC}\cap\overline{BD}$ and $L=\overline{WY}\cap\overline{XZ}$. Let points $O_a,O_b,O_c,O_d$ be such that $\angle O_aZA=\angle O_aWA=90^\circ$ (and cyclic variants), and $G=\overline{O_aO_c}\cap\overline{O_bO_d}$. Prove that $\overline{IL}\parallel\overline{TG}$.
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t1 t2 t3 = triangle; i = circumcenter t1 t2 t3; t4 = on_circle i t1; a = on_tline t1 i t1, on_tline t2 i t2; b = on_tline t2 i t2, on_tline t3 i t3; c = on_tline t3 i t3, on_tline t4 i t4; d = on_tline t1 i t1, on_tline t4 i t4; w = on_line a b; x = on_line b c, on_circle b w; y = on_line c d, on_circle c x; z = on_line d a, on_circle d y; t = on_line a c, on_line b d; l = on_line w y, on_line x z; oa = on_tline z a z, on_tline w a w; ob = on_tline x b x, on_tline w b w; oc = on_tline y c y, on_tline x c x; od = on_tline z d z, on_tline y d y; g = on_line oa oc, on_line ob od ? para i l t g
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T1 T2 T3 = triangle; I = circumcenter T1 T2 T3; (I) = circle_center_point I T1; T4 = on_circle (I); IT1 = line I T1; IT2 = line I T2; IT3 = line I T3; IT4 = line I T4; l1 = perpendicular_line T1 IT1; l2 = perpendicular_line T2 IT2; A = intersection l1 l2; l3 = perpendicular_line T3 IT3; B = intersection l2 l3; l4 = perpendicular_line T4 IT4; C = intersection l3 l4; D = intersection l1 l4; AB = line A B; W = on_line AB; BC = line B C; (B,W) = circle_center_point B W; X = intersection BC (B,W); CD = line C D; (C,X) = circle_center_point C X; Y = intersection CD (C,X); AD = line A D; (D,Y) = circle_center_point D Y; Z = intersection AD (D,Y); AC = line A C; BD = line B D; T = intersection AC BD; WY = line W Y; XZ = line X Z; L = intersection WY XZ; l5 = perpendicular_line Z AD; l6 = perpendicular_line W AB; l7 = perpendicular_line X BC; l8 = perpendicular_line Y CD; Oa = intersection l5 l6; Ob = intersection l6 l7; Oc = intersection l7 l8; Od = intersection l5 l8; OaOc = line Oa Oc; ObOd = line Ob Od; G = intersection OaOc ObOd; IL = line I L; TG = line T G; Prove: parallel IL TG
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2024IranTSTp2
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For a right angled triangle $\triangle ABC$ with $\angle A=90$ we have $AC=2AB$. Point $M$ is the midpoint of side $BC$ and $I$ is incenter of triangle $\triangle ABC$. The line passing trough $M$ and perpendicular to $BI$ intersect with lines $BI$ and $AC$ at points $H$ and $K$ respectively. If the semi-line $IK$ cuts circumcircle of triangle $\triangle ABC$ at $F$ and $S$ be the second intersection point of line $FH$ with circumcircle of triangle $\triangle ABC$ , then prove that $SM$ is tangent to the incircle of triangle $\triangle ABC$.
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a b x = risos; c = mirror a x; m = midpoint b c; i = incenter a b c; h = foot m b i; k = on_tline m b i, on_line a c; f = on_line i k, on_circle m a; s = on_line f h, on_circle m a; d = foot i a b; t = foot i s m ? cong i t i d
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A B X = right_isos_triangle A B X; C = reflect_point_wrt_point A X; M = midpoint B C; I = incenter A B C; BI = line B I; H = foot M BI; l1 = perpendicular_line M BI; AC = line A C; K = intersection l1 AC; IK = line I K; (ABC) = circle_center_point M A; F = intersection IK (ABC); FH = line F H; S = intersection FH (ABC); AB = line A B; D = foot I AB; SM = line S M; T = foot I SM; Prove: cong I T I D
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2024KoMaLA857
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Let $ABC$ be a given acute triangle, in which $BC$ is the longest side. Let $H$ be the orthocenter of the triangle, and let $D$ and $E$ be the feet of the altitudes from $B$ and $C$, respectively. Let $F$ and $G$ be the midpoints of sides $AB$ and $AC$, respectively. $X$ is the point of intersection of lines $DF$ and $EG$. Let $O_1$ and $O_2$ be the circumcenters of triangles $EFX$ and $DGX$, respectively. Finally, $M$ is the midpoint of line segment $O_1O_2$. Prove that points $X, H$ and $M$ are collinear.
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a b c = triangle; h = orthocenter a b c; d = foot b a c; e = foot c a b; f = midpoint a b; g = midpoint a c; x = on_line d f, on_line e g; o1 = circumcenter e f x; o2 = circumcenter d g x; m = midpoint o1 o2 ? coll x h m
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A B C = acute_triangle; H = orthocenter A B C; AC = line A C; D = foot B AC; AB = line A B; E = foot C AB; F = midpoint A B; G = midpoint A C; DF = line D F; EG = line E G; X = intersection DF EG; O1 = circumcenter E F X; O2 = circumcenter D G X; M = midpoint O1 O2; Prove: collinear X H M
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2024KoMaLA864
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Let $ABC$ be a triangle and $O$ be its circumcenter. Let $D$, $E$ and $F$ be the respective tangent points of the incircle of $\triangle ABC$, and sides $BC$, $CA$ and $AB$. Let $M$ and $N$ be the respective midpoints of sides $AB$ and $AC$. Let $M'$ and $N'$ be the respective reflections of points $M$ and $N$ across lines $DE$ and $DF$. Let lines $CM'$ and $BN'$ intersect lines $DE$ and $DF$ at points $H$ and $J$, respectively. Prove that the points $H$, $J$ and $O$ are collinear.
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a b c = triangle; o = circumcenter a b c; i = incenter a b c; d = foot i b c; e = foot i a c; f = foot i a b; m = midpoint a b; n = midpoint a c; m1 = reflect m d e; n1 = reflect n d f; h = on_line c m1, on_line d e; j = on_line b n1, on_line d f ? coll h j o
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A B C = triangle; O = circumcenter A B C; I = incenter A B C; BC = line B C; D = foot I BC; AC = line A C; E = foot I AC; AB = line A B; F = foot I AB; M = midpoint A B; N = midpoint A C; DE = line D E; M1 = reflect_point_wrt_line M DE; DF = line D F; N1 = reflect_point_wrt_line N DF; CM1 = line C M1; H = intersection CM1 DE; BN1 = line B N1; J = intersection BN1 DF; Prove: collinear H J O
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2024KoMaLA877
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A convex quadrilateral $ABCD$ is circumscribed about circle $\omega$. A tangent to $\omega$ parallel to $AC$ intersects $BD$ at a point $P$ outside of $\omega$. The second tangent from $P$ to $\omega$ touches $\omega$ at a point $T$. Prove that $\omega$ and circumcircle of $ATC$ are tangent.
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t1 t2 t3 = triangle; i = circumcenter t1 t2 t3; t4 = on_circle i t1; a = on_tline t1 i t1, on_tline t2 i t2; b = on_tline t2 i t2, on_tline t3 i t3; c = on_tline t3 i t3, on_tline t4 i t4; d = on_tline t1 i t1, on_tline t4 i t4; t5 = on_circle i t1, on_tline i a c; p = on_tline t5 i t5, on_line b d; t = reflect t5 i p; o = circumcenter a t c ? coll i o t
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T1 T2 T3 = triangle; I = circumcenter T1 T2 T3; (I) = circle_center_point I T1; T4 = on_circle (I); IT1 = line I T1; IT2 = line I T2; IT3 = line I T3; IT4 = line I T4; l1 = perpendicular_line T1 IT1; l2 = perpendicular_line T2 IT2; A = intersection l1 l2; l3 = perpendicular_line T3 IT3; B = intersection l2 l3; l4 = perpendicular_line T4 IT4; C = intersection l3 l4; D = intersection l1 l4; AC = line A C; l5 = perpendicular_line I AC; T5 = intersection (I) l5; IT5 = line I T5; l6 = perpendicular_line T5 IT5; BD = line B D; P = intersection l6 BD; IP = line I P; T = reflect_point_wrt_line T5 IP; O = circumcenter A T C; Prove: collinear I O T
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2024KoMaLA878
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Let point $A$ be one of the intersections of circles $c$ and $k$. Let $X_1$ and $X_2$ be arbitrary points on circle $c$. Let $Y_i$ denote the intersection of line $AX_i$ and circle $k$ for $i=1,2$. Let $P_1$, $P_2$ and $P_3$ be arbitrary points on circle $k$, and let $O$ denote the center of circle $k$. Let $K_{ij}$ denote the center of circle $(X_iY_iP_j)$ for $i=1,2$ and $j=1,2,3$. Let $L_j$ denote the center of circle $(OK_{1j}K_{2j})$ for $j=1,2,3$. Prove that points $L_1$, $L_2$ and $L_3$ are collinear.
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a = free; oc = free; o = free; x1 = on_circle oc a; x2 = on_circle oc a; y1 = on_line a x1, on_circle o a; y2 = on_line a x2, on_circle o a; p1 = on_circle o a; p2 = on_circle o a; p3 = on_circle o a; k11 = circumcenter x1 y1 p1; k21 = circumcenter x2 y2 p1; k12 = circumcenter x1 y1 p2; k22 = circumcenter x2 y2 p2; k13 = circumcenter x1 y1 p3; k23 = circumcenter x2 y2 p3; l1 = circumcenter o k11 k21; l2 = circumcenter o k12 k22; l3 = circumcenter o k13 k23 ? coll l1 l2 l3
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A = point; Oc = point; O = point; (Oc) = circle_center_point Oc A; (O) = circle_center_point O A; A = intersection (Oc) (O); X1 = on_circle (Oc); X2 = on_circle (Oc); AX1 = line A X1; Y1 = intersection AX1 (O); AX2 = line A X2; Y2 = intersection AX2 (O); P1 = on_circle (O); P2 = on_circle (O); P3 = on_circle (O); K11 = circumcenter X1 Y1 P1; K12 = circumcenter X1 Y1 P2; K13 = circumcenter X1 Y1 P3; K21 = circumcenter X2 Y2 P1; K22 = circumcenter X2 Y2 P2; K23 = circumcenter X2 Y2 P3; L1 = circumcenter O K11 K21; L2 = circumcenter O K12 K22; L3 = circumcenter O K13 K23; Prove: collinear L1 L2 L3
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2024PlanetCupp10
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Let $\triangle DEF$ and $\triangle D'E'F'$ be the orthic triangles of $\triangle ABC$ and $\triangle DEF$ respectively. With $N$ as the nine-point center and $M$ as $BC$'s midpoint, suppose $E'F'$ and $D'N$ meet $BC$ at $J,K$. Let $O$ be $\triangle AJK$'s circumcenter. Line $AM$ meets $\odot O$ and $EF$ at $T,P$. If $TK\cap AJ=Q$, prove $PQ\perp OM$.
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a b c = triangle; d = foot a b c; e = foot b a c; f = foot c a b; d1 = foot d e f; e1 = foot e d f; f1 = foot f d e; n1 = circumcenter d e f; m = midpoint b c; j = on_line e1 f1, on_line b c; k = on_line d1 n1, on_line b c; o = circumcenter a j k; t = on_line a m, on_circle o a; p = on_line a m, on_line e f; q = on_line t k, on_line a j ? perp p q o m
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A B C = triangle; BC = line B C; D = foot A BC; AC = line A C; E = foot B AC; AB = line A B; F = foot C AB; EF = line E F; DF = line D F; DE = line D E; D1 = foot D EF; E1 = foot E DF; F1 = foot F DE; N1 = circumcenter D E F; M = midpoint B C; E1F1 = line E1 F1; BC = line B C; J = intersection E1F1 BC; D1N1 = line D1 N1; K = intersection D1N1 BC; O = circumcenter A J K; AM = line A M; (O) = circle_center_point O A; T = intersection AM (O); EF = line E F; P = intersection AM EF; TK = line T K; AJ = line A J; Q = intersection TK AJ; PQ = line P Q; OM = line O M; Prove: perpendicular PQ OM
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2024PolandMOp6
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Let $ABCD$ be a parallelogram. Let $X \notin AC $ lie inside $ABCD$ so that $\angle AXB = \angle CXD = 90^ {\circ}$ and $\Omega$ denote the circumcircle of $AXC$. Consider a diameter $EF$ of $\Omega$ and assume neither $E, \ X, \ B$ nor $F, \ X, \ D$ are collinear. Let $K \neq X$ be an intersection point of circumcircles of $BXE$ and $DXF$ and $L \neq X$ be such point on $\Omega$ so that $\angle KXL = 90^{\circ}$. Prove that $AB = KL$.
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a b c = triangle; d = parallelogram a b c d; m_ab = midpoint a b; m_cd = midpoint c d; x = on_circle m_ab a, on_circle m_cd c; o = circumcenter a x c; e = on_circle o a; f = mirror e o; o1 = circumcenter b x e; o2 = circumcenter d x f; k = on_circle o1 b, on_circle o2 d; l = on_circle o a, on_tline x x k ? cong a b k l
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A B C = triangle; D = parallelogram A B C; M_ab = midpoint A B; M_cd = midpoint C D; (AB) = circle_center_point M_ab A; (CD) = circle_center_point M_cd C; X = intersection (AB) (CD); O = circumcenter A X C; (AXC) = circle_center_point O A; E = on_circle (AXC); F = reflect_point_wrt_point E O; O1 = circumcenter B X E; O2 = circumcenter D X F; (BXE) = circle_center_point O1 B; (DXF) = circle_center_point O2 D; K = intersection (BXE) (DXF); XK = line X K; l1 = perpendicular_line X XK; L = intersection (AXC) l1; Prove: cong A B K L
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2024SAGFp6
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$BD$ is an altitude of $\triangle ABC$. $H$ is the orthocenter. $P$ is a point on $\odot(BC)$. Tangent of $\odot(BC)$ at $P$ intersects $BD$ at $Q$. $X$ is on line $AB$ and $PX\perp BC$. $PH, QX$ intersect at point $R$. Line $AR$ intersects $\odot(AH)$ again at $U$. Prove that $U,P,D,Q$ are concyclic.
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a b c = triangle; d = foot b a c; h = orthocenter a b c; m1 = midpoint b c; p = on_circle m1 b; q = on_tline p m1 p, on_line b d; x = on_line a b, on_tline p b c; r = on_line p h, on_line q x; m2 = midpoint a h; u = on_line a r, on_circle m2 a ? cyclic u p d q
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A B C = triangle; AC = line A C; D = foot B AC; H = orthocenter A B C; M1 = midpoint B C; (BC) = circle_center_point M1 B; P = on_circle (BC); M1P = line M1 P; l1 = perpendicular_line P M1P; BD = line B D; Q = intersection l1 BD; BC = line B C; l2 = perpendicular_line P BC; AB = line A B; X = intersection AB l2; HP = line H P; QX = line Q X; R = intersection HP QX; AR = line A R; M2 = midpoint A H; (AH) = circle_center_point M2 A; U = intersection AR (AH); Prove: concyclic U P D Q
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2024ShuZhiMiMOM0p3
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Given rhombus $ABCD$ and point $E$, let $EB,EC$ meet $DA$ at $F,G$. The common external tangents (other than $DA$) of incircles of $\triangle ABF$ and $\triangle CDG$ meet $AB,CD$ at $H,K$ and $EB,EC$ at $I,J$. Prove the Newton lines of quadrilaterals $AHIF$, $DKJG$, $BCJI$ are concurrent.
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b a c = iso_triangle; d = parallelogram a b c d; f = on_line d a; g = on_line d a; i1 = incenter a b f; i2 = incenter c d g; t1 = foot i1 a d; t2 = foot i2 a d; t3 = reflect t1 i1 i2; t4 = reflect t2 i1 i2; h = on_line t3 t4, on_line a b; i = on_line t3 t4, on_line b f; j = on_line t3 t4, on_line c g; k = on_line t3 t4, on_line c d; m1 = midpoint a i; m2 = midpoint h f; m3 = midpoint d j; m4 = midpoint k g; m5 = midpoint b j; m6 = midpoint c i; p = on_line m1 m2, on_line m3 m4 ? coll p m5 m6
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B A C = isos_triangle; D = parallelogram A B C; AD = line A D; F = on_line AD; G = on_line AD; I1 = incenter A B F; I2 = incenter C D G; T1 = foot I1 AD; T2 = foot I2 AD; I1I2 = line I1 I2; T3 = reflect_point_wrt_line T1 I1I2; T4 = reflect_point_wrt_line T2 I1I2; T3T4 = line T3 T4; AB = line A B; H = intersection T3T4 AB; BF = line B F; I = intersection T3T4 BF; CG = line C G; J = intersection T3T4 CG; CD = line C D; K = intersection T3T4 CD; M1 = midpoint A I; M2 = midpoint H F; M3 = midpoint D J; M4 = midpoint K G; M5 = midpoint B J; M6 = midpoint C I; M1M2 = line M1 M2; M3M4 = line M3 M4; P = intersection M1M2 M3M4; Prove: collinear P M5 M6
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2024ShuZhiMiMOM12p2
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In $\triangle ABC$, let $X$ be a point distinct from the orthocenter. Lines $AX$, $BX$, $CX$ meet the circumcircle at $X_A$, $X_B$, $X_C$. Let $Y$ be the isogonal conjugate of $X$ in $\triangle X_AX_BX_C$. Lines $AX$ and $YX$ meet the circumcircle of $\triangle BXC$ at $P$ and $Q$ (other than $X$). Line $PQ$ meets $BC$ at $R$. Prove $PR = AR$.
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a b c = triangle; x = free; o = circumcenter a b c; x1 = on_line a x, on_circle o a; x2 = on_line b x, on_circle o a; x3 = on_line c x, on_circle o a; y = on_aline y x1 x2 x3 x1 x, on_aline y x2 x1 x3 x2 x; o1 = circumcenter b x c; p = on_line a x, on_circle o1 b; q = on_line y x, on_circle o1 b; r = on_line p q, on_line b c ? cong p r a r
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A B C = triangle; X = point; O = circumcenter A B C; AX = line A X; (O) = circle_center_point O A; X1 = intersection AX (O); BX = line B X; X2 = intersection BX (O); CX = line C X; X3 = intersection CX (O); l1 = angle_equal1 X1 X2 X3 X1 X; l2 = angle_equal1 X2 X1 X3 X2 X; Y = intersection l1 l2; O1 = circumcenter B X C; (O1) = circle_center_point O1 B; P = intersection AX (O1); YX = line Y X; Q = intersection YX (O1); PQ = line P Q; BC = line B C; R = intersection PQ BC; Prove: cong P R A R
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2024ShuZhiMiMOM3p5
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In $\triangle ABC$, let $M$ be the midpoint of $BC$. The line $AM$ meets the circumcircle of $\triangle ABC$ at $E$. Denote by $N$ the midpoint of the arc $BAC$ (not containing $BC$). The line $NE$ intersects $BC$ at $L$. The line $AB$ meets $CN$ at $S$ and $MN$ at $T$. Let $\Gamma_1$ be the circle with diameter $BN$, and $\Gamma_2$ be the circumcircle of $\triangle BLT$; they intersect at point $U$. The lines $MS$ and $LT$ meet at $V$. Prove that the circumcircles of $\triangle CLV$ and $\triangle NTU$ are tangent to each other.
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a b c = triangle; o = circumcenter a b c; m = midpoint b c; e = on_line a m, on_circle o a; n = on_bline b c, on_circle o a; l = on_line n e, on_line b c; s = on_line a b, on_line c n; t = on_line a b, on_line m n; o2 = circumcenter b l t; u = on_dia b n, on_circle o2 b; v = on_line m s, on_line l t; o3 = circumcenter c l v; o4 = circumcenter n t u; k = on_circle o3 c, on_circle o4 n ? coll k o3 o4
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A B C = triangle; O = circumcenter A B C; M = midpoint B C; AM = line A M; (O) = circle_center_point O A; E = intersection AM (O); l1 = perpendicular_bisector B C; N = intersection l1 (O); NE = line N E; BC = line B C; L = intersection NE BC; AB = line A B; CN = line C N; S = intersection AB CN; MN = line M N; T = intersection AB MN; O2 = circumcenter B L T; (BN) = circle_diameter B N; (O2) = circle_center_point O2 B; U = intersection (BN) (O2); MS = line M S; LT = line L T; V = intersection MS LT; O3 = circumcenter C L V; O4 = circumcenter N T U; (O3) = circle_center_point O3 C; (O4) = circle_center_point O4 N; K = intersection (O3) (O4); Prove: collinear K O3 O4
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2024ShuZhiMiMOM4p6
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In $\triangle ABC$ with incircle $(I)$ tangent to $AB,AC$ at $F,E$, line $EF$ meets $(ABC)$ at $P,Q$. Let $BI,CI$ meet $(IPQ)$ again at $L,K$, and the external bisector of $\angle BAC$ meet $(ABC)$ at $N$ and the Euler line at $M$. Prove $K,L,M,N$ are concyclic.
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a b c = triangle; i = incenter a b c; f = foot i a b; e = foot i a c; o = circumcenter a b c; p = on_line e f, on_circle o a; q = on_line e f, on_circle o a; o1 = circumcenter i p q; l = on_line b i, on_circle o1 i; k = on_line c i, on_circle o1 i; n = on_tline a a i, on_circle o a; h = orthocenter a b c; m = on_tline a a i, on_line o h ? cyclic k l m n
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A B C = triangle; I = incenter A B C; AC = line A C; AB = line A B; F = foot I AB; E = foot I AC; O = circumcenter A B C; EF = line E F; (O) = circle_center_point O A; P = intersection EF (O); Q = intersection EF (O); O1 = circumcenter I P Q; BI = line B I; (O1) = circle_center_point O1 I; L = intersection BI (O1); CI = line C I; K = intersection CI (O1); AI = line A I; l1 = perpendicular_line A AI; N = intersection l1 (O); H = orthocenter A B C; OH = line O H; M = intersection l1 OH; Prove: concyclic K L M N
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2024TurkeyMO2ndRp2
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Let $\triangle ABC$ be an acute triangle, where $H$ is the orthocenter and $D,E,F$ are the feet of the altitudes from $A,B,C$ respectively. A circle tangent to $(DEF)$ at $D$ intersects the line $EF$ at $P$ and $Q$. Let $R$ and $S$ be the second intersection points of the circumcircle of triangle $\triangle BHC$ with $PH$ and $QH$, respectively. Let $T$ be the point on the line $BC$ such that $AT\perp EF$. Prove that the points $R,S,D,T$ are concyclic.
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a b c = triangle; h = orthocenter a b c; d = foot a b c; e = foot b a c; f = foot c a b; o1 = circumcenter d e f; o2 = on_line o1 d; p = on_circle o2 d, on_line e f; q = on_circle o2 d, on_line e f; o3 = circumcenter b h c; r = on_line p h, on_circle o3 b; s = on_line q h, on_circle o3 b; t = on_line b c, on_tline a e f ? cyclic r s d t
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A B C = triangle; H = orthocenter A B C; BC = line B C; AC = line A C; AB = line A B; D = foot A BC; E = foot B AC; F = foot C AB; O1 = circumcenter D E F; O1D = line O1 D; O2 = on_line O1D; (O2) = circle_center_point O2 D; EF = line E F; P = intersection (O2) EF; Q = intersection (O2) EF; O3 = circumcenter B H C; PH = line P H; (O3) = circle_center_point O3 B; R = intersection PH (O3); QH = line Q H; S = intersection QH (O3); l1 = perpendicular_line A EF; T = intersection BC l1; Prove: concyclic R S D T
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2024TurkeyTSTp5
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In a scalene triangle $ABC$, $H$ is the orthocenter, and $G$ is the centroid. Let $A_b$ and $A_c$ be points on $AB$ and $AC$, respectively, such that $B$, $C$, $A_b$, $A_c$ are cyclic, and the points $A_b$, $A_c$, $H$ are collinear. $O_a$ is the circumcenter of the triangle $AA_bA_c$. $O_b$ and $O_c$ are defined similarly. Prove that the centroid of the triangle $O_aO_bO_c$ lies on the line $HG$.
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a b c = triangle; o = circumcenter a b c; h = orthocenter a b c; m1 m2 m3 g = centroid a b c; ab = on_line a b, on_tline h o a; ac = on_line a c, on_tline h o a; oa = circumcenter a ab ac; bc = on_line b c, on_tline h o b; ba = on_line b a, on_tline h o b; ob = circumcenter b bc ba; ca = on_line c a, on_tline h o c; cb = on_line c b, on_tline h o c; oc = circumcenter c ca cb; n1 n2 n3 g1 = centroid oa ob oc ? coll g1 h g
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A B C = triangle; O = circumcenter A B C; H = orthocenter A B C; G M1 M2 M3 = centroid A B C; AB = line A B; BC = line B C; AC = line A C; OA = line O A; l1 = perpendicular_line H OA; Ab = intersection AB l1; Ac = intersection AC l1; Oa = circumcenter A Ab Ac; OB = line O B; l2 = perpendicular_line H OB; Bc = intersection BC l2; Ba = intersection AB l2; Ob = circumcenter B Bc Ba; OC = line O C; l3 = perpendicular_line H OC; Ca = intersection AC l3; CB = line C B; Cb = intersection BC l3; Oc = circumcenter C Ca Cb; G1 N1 N2 N3 = centroid Oa Ob Oc; Prove: collinear G1 H G
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2024TurkeyTSTp9
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In a scalene triangle $ABC,$ $I$ is the incenter and $O$ is the circumcenter. The line $IO$ intersects the lines $BC,CA,AB$ at points $D,E,F$ respectively. Let $A_1$ be the intersection of $BE$ and $CF$. The points $B_1$ and $C_1$ are defined similarly. The incircle of $ABC$ is tangent to sides $BC,CA,AB$ at points $X,Y,Z$ respectively. Let the lines $XA_1, YB_1$ and $ZC_1$ intersect $IO$ at points $A_2,B_2,C_2$ respectively. Prove that the circles with diameters $AA_2,BB_2$ and $CC_2$ have a common point.
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a b c = triangle; i = incenter a b c; o = circumcenter a b c; d = on_line i o, on_line b c; e = on_line i o, on_line a c; f = on_line i o, on_line a b; a1 = on_line b e, on_line c f; b1 = on_line c f, on_line a d; c1 = on_line a d, on_line b e; x = foot i b c; y = foot i a c; z = foot i a b; a2 = on_line x a1, on_line i o; b2 = on_line y b1, on_line i o; c2 = on_line z c1, on_line i o; k = on_dia a a2, on_dia b b2 ? perp c k k c2
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A B C = triangle; I = incenter A B C; O = circumcenter A B C; IO = line I O; BC = line B C; D = intersection IO BC; AC = line A C; E = intersection IO AC; AB = line A B; F = intersection IO AB; BE = line B E; CF = line C F; A1 = intersection BE CF; AD = line A D; B1 = intersection CF AD; C1 = intersection AD BE; X = foot I BC; Y = foot I AC; Z = foot I AB; XA1 = line X A1; A2 = intersection XA1 IO; YB1 = line Y B1; B2 = intersection YB1 IO; ZC1 = line Z C1; C2 = intersection ZC1 IO; (AA2) = circle_diameter A A2; (BB2) = circle_diameter B B2; K = intersection (AA2) (BB2); CK = line C K; C2K = line C2 K; Prove: perpendicular CK C2K
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2024USATSTSTp4
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Let $ABCD$ be a quadrilateral inscribed in a circle with center $O$ and $E$ be the intersection of segments $AC$ and $BD$. Let $\omega_1$ be the circumcircle of $ADE$ and $\omega_2$ be the circumcircle of $BCE$. The tangent to $\omega_1$ at $A$ and the tangent to $\omega_2$ at $C$ meet at $P$. The tangent to $\omega_1$ at $D$ and the tangent to $\omega_2$ at $B$ meet at $Q$. Show that $OP=OQ$.
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a b c = triangle; d = on_circum a b c; o = circumcenter a b c; e = on_line a c, on_line b d; o1 = circumcenter a d e; o2 = circumcenter b c e; p = on_tline a o1 a, on_tline c o2 c; q = on_tline d o1 d, on_tline b o2 b ? cong o p o q
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A B C = triangle; O = circumcenter A B C; (ABC) = circle A B C; D = on_circle (ABC); AC = line A C; BD = line B D; E = intersection AC BD; O1 = circumcenter A D E; O1A = line O1 A; l1 = perpendicular_line A O1A; O2 = circumcenter B C E; O2C = line O2 C; l2 = perpendicular_line C O2C; P = intersection l1 l2; O1D = line O1 D; l3 = perpendicular_line D O1D; O2B = line O2 B; l4 = perpendicular_line B O2B; Q = intersection l3 l4; Prove: cong O P O Q
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2024VietnamTSTp3
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Let $ABC$ be an acute scalene triangle. Incircle of $ABC$ touches $BC,CA,AB$ at $D,E,F$ respectively. Let $X,Y,Z$ be feet the altitudes of from $A,B,C$ to the sides $BC,CA,AB$ respectively. Let $A',B',C'$ be the reflections of $X,Y,Z$ in $EF,FD,DE$ respectively. Prove that triangles $ABC$ and $A'B'C'$ are similar.
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a b c = triangle; i = incenter a b c; d = foot i b c; e = foot i a c; f = foot i a b; x = foot a b c; y = foot b a c; z = foot c a b; a1 = reflect x e f; b1 = reflect y f d; c1 = reflect z d e ? simtri a b c a1 b1 c1
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A B C = triangle; I = incenter A B C; BC = line B C; AC = line A C; AB = line A B; D = foot I BC; E = foot I AC; F = foot I AB; X = foot A BC; Y = foot B AC; Z = foot C AB; EF = line E F; FD = line F D; DE = line D E; A1 = reflect_point_wrt_line X EF; B1 = reflect_point_wrt_line Y FD; C1 = reflect_point_wrt_line Z DE; Prove: similar A B C A1 B1 C1
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2024VietnamTSTp5
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Let incircle $(I)$ of triangle $ABC$ touch the sides $BC,CA,AB$ at $D,E,F$ respectively. Let $(O)$ be the circumcircle of $ABC$. Ray $EF$ meets $(O)$ at $M$. Tangents at $M$ and $A$ of $(O)$ meet at $S$. Tangents at $B$ and $C$ of $(O)$ meet at $T$. Line $TI$ meets $OA$ at $J$. Prove that $\angle ASJ=\angle IST$.
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a b c = triangle; i = incenter a b c; d = foot i b c; e = foot i a c; f = foot i a b; o = circumcenter a b c; m = on_line e f, on_circle o a; s = on_tline m o m, on_tline a o a; t = on_tline b o b, on_tline c o c; j = on_line t i, on_line o a ? eqangle a s s j i s s t
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A B C = triangle; I = incenter A B C; BC = line B C; AC = line A C; AB = line A B; D = foot I BC; E = foot I AC; F = foot I AB; O = circumcenter A B C; EF = line E F; (O) = circle_center_point O A; M = intersection EF (O); OM = line O M; OA = line O A; l1 = perpendicular_line M OM; l2 = perpendicular_line A OA; S = intersection l1 l2; OB = line O B; OC = line O C; l3 = perpendicular_line B OB; l4 = perpendicular_line C OC; T = intersection l3 l4; TI = line T I; J = intersection TI OA; Prove: equal_angle A S J I S T
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2025CTSTp8
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Let quadrilateral $A_1A_2A_3A_4$ be not cyclic and haves edges not parallel to each other. Denote $B_i$ as the intersection of the tangent line at $A_i$ with respect to circle $A_{i-1}A_iA_{i+1}$ and the $A_{i+2}$-symmedian with respect to triangle $A_{i+1}A_{i+2}A_{i+3}$ and $C_i$ as the intersection of lines $A_iA_{i+1}$ and $B_iB_{i+1}$, where all indexes taken cyclically. Prove that $C_1$, $C_2$, $C_3$, and $C_4$ are collinear.
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a1 a2 a3 a4 = quadrangle; o1 = circumcenter a4 a1 a2; o2 = circumcenter a1 a2 a3; o3 = circumcenter a2 a3 a4; o4 = circumcenter a3 a4 a1; m1 = midpoint a1 a3; m2 = midpoint a2 a4; b1 = on_tline a1 o1 a1, on_aline b1 a3 a2 a4 a3 m2; b2 = on_tline a2 o2 a2, on_aline b2 a4 a3 a1 a4 m1; b3 = on_tline a3 o3 a3, on_aline b3 a1 a4 a2 a1 m2; b4 = on_tline a4 o4 a4, on_aline b4 a2 a1 a3 a2 m1; c1 = on_line a1 a2, on_line b1 b2; c2 = on_line a2 a3, on_line b2 b3; c3 = on_line a3 a4, on_line b3 b4 ? coll c1 c2 c3
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A1 A2 A3 A4 = quadrilateral; O1 = circumcenter A4 A1 A2; O2 = circumcenter A1 A2 A3; O3 = circumcenter A2 A3 A4; O4 = circumcenter A3 A4 A1; M1 = midpoint A1 A3; M2 = midpoint A2 A4; O1A1 = line O1 A1; O2A2 = line O2 A2; O3A3 = line O3 A3; O4A4 = line O4 A4; l1 = perpendicular_line A1 O1A1; l2 = perpendicular_line A2 O2A2; l3 = perpendicular_line A3 O3A3; l4 = perpendicular_line A4 O4A4; m1 = angle_equal1 A3 A2 A4 A3 M2; m2 = angle_equal1 A4 A3 A1 A4 M1; m3 = angle_equal1 A1 A4 A2 A1 M2; m4 = angle_equal1 A2 A1 A3 A2 M1; B1 = intersection l1 m1; B2 = intersection l2 m2; B3 = intersection l3 m3; B4 = intersection l4 m4; A1A2 = line A1 A2; B1B2 = line B1 B2; C1 = intersection A1A2 B1B2; A2A3 = line A2 A3; B2B3 = line B2 B3; C2 = intersection A2A3 B2B3; A3A4 = line A3 A4; B3B4 = line B3 B4; C3 = intersection A3A4 B3B4; Prove: collinear C1 C2 C3
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2025IranTSTp12
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In a scalene triangle $ABC$, points $Y$ and $X$ lie on $AC$ and $BC$ respectively such that $BC \perp XY$. Points $Z$ and $T$ are the reflections of $X$ and $Y$ with respect to the midpoints of sides $BC$ and $AC$, respectively. Point $P$ lies on segment $ZT$ such that the circumcenter of triangle $XZP$ coincides with the circumcenter of triangle $ABC$. Prove that the nine-point circle of triangle $ABC$ passes through the midpoint of segment $XP$.
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a b c = triangle; y = on_line a c; x = on_line b c, on_tline y b c; m1 m2 m3 n = ninepoints a b c; z = mirror x m1; t = mirror y m2; o = circumcenter a b c; p = on_line z t, on_circle o x; m4 = midpoint x p ? cyclic m4 m1 m2 m3
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A B C = triangle; AC = line A C; Y = on_line AC; BC = line B C; l1 = perpendicular_line Y BC; X = intersection BC l1; N M1 M2 M3 = ninepointcenter A B C; Z = reflect_point_wrt_point X M1; T = reflect_point_wrt_point Y M2; O = circumcenter A B C; ZT = line Z T; (O,X) = circle_center_point O X; P = intersection ZT (O,X); M4 = midpoint X P; Prove: concyclic M4 M1 M2 M3
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2025KoeraFinalRoundp3
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An acute triangle $\bigtriangleup ABC$ is given which $BC>CA>AB$. $I$ is the interior and the incircle of $\bigtriangleup ABC$ meets $BC, CA, AB$ at $D,E,F$. $AD$ and $BE$ meet at $P$. Let $l_{1}$ be a tangent from D to the circumcircle of $\bigtriangleup DIP$, and define $l_{2}$ and $l_{3}$ on $E$ and $F$, respectively. Prove $l_{1},l_{2},l_{3}$ meet at one point.
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a b c = triangle; i = incenter a b c; d = foot i b c; e = foot i a c; f = foot i a b; p = on_line a d, on_line b e; o1 = circumcenter d i p; o2 = circumcenter e i p; o3 = circumcenter f i p; q = on_tline d o1 d, on_tline e o2 e ? perp q f o3 f
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A B C = triangle; I = incenter A B C; BC = line B C; AC = line A C; AB = line A B; D = foot I BC; E = foot I AC; F = foot I AB; AD = line A D; BE = line B E; P = intersection AD BE; O1 = circumcenter D I P; O2 = circumcenter E I P; O3 = circumcenter F I P; O1D = line O1 D; l1 = perpendicular_line D O1D; O2E = line O2 E; l2 = perpendicular_line E O2E; Q = intersection l1 l2; QF = line Q F; O3F = line O3 F; Prove: perpendicular QF O3F
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2025ShuZhiMiMOM3p5
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In non-isosceles acute-angled $\triangle ABC$, $\Omega$ is the circumcircle and $\omega$ is the nine-point circle. The tangents from $B$ and $C$ to $\Omega$ intersect at point $P$. Let $Q$ be a point on $\omega$ such that $BQ = QC$ and $Q$ is not on $BC$. Construct a circle $\Gamma$ symmetrical to $\Omega$ about $BC$, and let $\Gamma$ intersect $\omega$ at points $D$ and $E$. Let $F$ be the isogonal conjugate of $D$ with respect to $\triangle ABC$. Prove that $E$, $F$, $P$, $Q$ are concyclic.
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a b c = triangle; o = circumcenter a b c; m1 = midpoint b c; m2 = midpoint a c; m3 = midpoint a b; n = circumcenter m1 m2 m3; p = on_tline b o b, on_tline c o c; q = on_circle n m1, on_bline b c; o1 = reflect o b c; d = on_circle o1 b, on_circle n m1; e = on_circle o1 b, on_circle n m1; f = on_aline f a c b a d, on_aline f b c a b d ? cyclic e f p q
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A B C = triangle; O = circumcenter A B C; M1 = midpoint B C; M2 = midpoint A C; M3 = midpoint A B; N = circumcenter M1 M2 M3; OB = line O B; OC = line O C; l1 = perpendicular_line B OB; l2 = perpendicular_line C OC; P = intersection l1 l2; (N) = circle_center_point N M1; l3 = perpendicular_bisector B C; Q = intersection (N) l3; BC = line B C; O1 = reflect_point_wrt_line O BC; (O1) = circle_center_point O1 B; D E = intersection (O1) (N); l4 = angle_equal1 A C B A D; l5 = angle_equal1 B C A B D; F = intersection l4 l5; Prove: concyclic E F P Q
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2025VietnamTSTp2
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Let $ABC$ be an acute, non-isosceles triangle with orthocenter $H$. Let $D, E, F$ be the reflections of $H$ over $BC, CA, AB$, respectively, and let $A', B', C'$ be the reflections of $A, B, C$ over $BC, CA, AB$, respectively. Let $S$ be the circumcenter of triangle $A'B'C'$, and let $H'$ be the orthocenter of triangle $DEF$. Define $J$ as the center of the circle passing through the three projections of $H$ onto the lines $B'C', C'A', A'B'$. Prove that $HJ$ is parallel to $H'S$.
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a b c = triangle; h = orthocenter a b c; d = reflect h b c; e = reflect h c a; f = reflect h a b; a1 = reflect a b c; b1 = reflect b c a; c1 = reflect c a b; s = circumcenter a1 b1 c1; h1 = orthocenter d e f; p1 = foot h b1 c1; p2 = foot h c1 a1; p3 = foot h a1 b1; j = circumcenter p1 p2 p3 ? para h j h1 s
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A B C = triangle; H = orthocenter A B C; BC = line B C; AC = line A C; AB = line A B; D = reflect_point_wrt_line H BC; E = reflect_point_wrt_line H AC; F = reflect_point_wrt_line H AB; A1 = reflect_point_wrt_line A BC; B1 = reflect_point_wrt_line B AC; C1 = reflect_point_wrt_line C AB; S = circumcenter A1 B1 C1; H1 = orthocenter D E F; B1C1 = line B1 C1; C1A1 = line C1 A1; A1B1 = line A1 B1; P1 = foot H B1C1; P2 = foot H C1A1; P3 = foot H A1B1; J = circumcenter P1 P2 P3; HJ = line H J; H1S = line H1 S; Prove: parallel HJ H1S
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2025XinXingSpringMOp5
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In non-isosceles acute $\triangle ABC$ with circumcenter $O$ and orthocenter $H$, let $EF\parallel AO$ through $H$ meet $AC,AB$ at $E,F$. The second intersection of $(ABC)$ and $(AEF)$ is $G$, with $AG\cap BC=J$. Let $H'$ be the orthocenter of $\triangle AEF$, $K$ lie on $(AEF)$ with $HA=HK$, $L$ be $A$'s antipodal point on $(AEF)$, and $X=AG\cap HH'$. Prove the intersection of $LX$ and $JK$ lies on $(AEF)$.
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a b c = triangle; o = circumcenter a b c; h = orthocenter a b c; e = on_pline h a o, on_line a c; f = on_pline h a o, on_line a b; o1 = circumcenter a e f; g = on_circle o a, on_circle o1 a; j = on_line a g, on_line b c; h1 = orthocenter a e f; k = on_circle o1 a, on_circle h a; l = mirror a o1; x = on_line a g, on_line h h1; y = on_line l x, on_line j k ? cyclic y a e f
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A B C = triangle; O = circumcenter A B C; H = orthocenter A B C; AO = line A O; l1 = parallel_line H AO; AC = line A C; E = intersection l1 AC; AB = line A B; F = intersection l1 AB; O1 = circumcenter A E F; (O) = circle_center_point O A; (O1) = circle_center_point O1 A; G = intersection (O) (O1); AG = line A G; BC = line B C; J = intersection AG BC; H1 = orthocenter A E F; (H) = circle_center_point H A; K = intersection (O1) (H); L = reflect_point_wrt_point A O1; HH1 = line H H1; X = intersection AG HH1; LX = line L X; JK = line J K; Y = intersection LX JK; Prove: concyclic Y A E F
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2ndFNGOp9
|
For fixed $a>0$, let circles $c_1,c_2,c_3$ be centered at $A,B,C$ with radius $a$. Points $D,E$ lie on the circumcircle of $\triangle ABC$, with $F$ as the midpoint of arc $DE$. Rays $EA,DA$ meet $c_1,c_2$ at $A_1,A_2$ respectively with $FA_1=FA_2$ (similarly define $B_1,B_2,C_1,C_2$). Prove that for any $a$ and $D,E$, the radical axis of the circumcircles of $\triangle A_1B_1C_1$ and $\triangle A_2B_2C_2$ always passes through either the incenter or one excenter of $\triangle ABC$.
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a b c = triangle; e = on_circum a b c; f = on_circum a b c; a1 = on_line a e; a2 = reflect a1 a f; b1 = on_line b e, eqdistance b1 b a a1; b2 = reflect b1 b f; c1 = on_line c e, eqdistance c1 c a a1; c2 = reflect c1 c f; x = on_circum a1 b1 c1, on_circum a2 b2 c2; y = on_circum a1 b1 c1, on_circum a2 b2 c2; i = incenter a b c ? coll x y i
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A B C = triangle; (ABC) = circle A B C; E = on_circle (ABC); F = on_circle (ABC); AE = line A E; A1 = on_line AE; AF = line A F; A2 = reflect_point_wrt_line A1 AF; BE = line B E; (B) = circle_center_radius B A A1; B1 = intersection BE (B); BF = line B F; B2 = reflect_point_wrt_line B1 BF; CE = line C E; (C) = circle_center_radius C A A1; C1 = intersection CE (C); CF = line C F; C2 = reflect_point_wrt_line C1 CF; (A1B1C1) = circumcircle A1 B1 C1; (A2B2C2) = circumcircle A2 B2 C2; X Y = intersection (A1B1C1) (A2B2C2); I = incenter A B C; Prove: collinear X Y I
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JCGV2p2(2023ShuZhiMiMOM9p2)
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Given $\triangle ABC$ and $\triangle XYZ$ sharing a circumcircle and isogonal conjugates $P,Q$. Let $YZ\cap AB=W$, $YZ\cap AC=V$, and define $U,T,S,R$ similarly. The circumcircles of $\triangle ZRW$, $\triangle XST$, $\triangle YUV$ meet $\odot ABC$ again at $M,N,O$. Prove $AN$, $BO$, $CM$ concur at a point depending only on $P,Q$.
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a b c = triangle; o = circumcenter a b c; p = free; x = on_circle o a, on_line a p; y = on_circle o a, on_line b p; z = on_circle o a, on_line c p; w = on_line y z, on_line a b; v = on_line y z, on_line a c; s = on_line x z, on_line b c; r = on_line x z, on_line b a; u = on_line x y, on_line c a; t = on_line x y, on_line c b; o1 = circumcenter z r w; m = on_circle o1 z, on_circle o a; o2 = circumcenter x s t; n = on_circle o2 x, on_circle o a; o3 = circumcenter y u v; q = on_circle o3 y, on_circle o a; k = on_line a n, on_line b q ? coll k c m
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A B C = triangle; O = circumcenter A B C; P = point; (O) = circle_center_point O A; AP = line A P; X = intersection (O) AP; BP = line B P; Y = intersection (O) BP; CP = line C P; Z = intersection (O) CP; YZ = line Y Z; AB = line A B; W = intersection YZ AB; AC = line A C; V = intersection YZ AC; XZ = line X Z; BC = line B C; S = intersection XZ BC; R = intersection XZ AB; XY = line X Y; U = intersection XY AC; T = intersection XY BC; O1 = circumcenter Z R W; (O1) = circle_center_point O1 Z; M = intersection (O1) (O); O2 = circumcenter X S T; (O2) = circle_center_point O2 X; N = intersection (O2) (O); O3 = circumcenter Y U V; (O3) = circle_center_point O3 Y; Q = intersection (O3) (O); AN = line A N; BQ = line B Q; K = intersection AN BQ; Prove: collinear K C M
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MouseMOp3
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In acute $\triangle ABC$, let $X_1,X_2$ be the 1st and 2nd Fermat points. For $S$ on the circumcircle of $\triangle AX_1X_2$, line $AS$ meets the circumcircle of $\triangle BSC$ at $T$. Prove that the Euler lines of $\triangle ABC$ and $\triangle TBC$ are parallel.
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a b c = triangle; d1 = eq_triangle b c; e1 = eq_triangle c a; f1 = eq_triangle a b; x1 = on_line a d1, on_line b e1; d2 = eq_triangle c b; e2 = eq_triangle a c; f2 = eq_triangle b a; x2 = on_line a d2, on_line b e2; o1 = circumcenter a x1 x2; s = on_circle o1 a; o2 = circumcenter b s c; t = on_line a s, on_circle o2 b; o = circumcenter a b c; h = orthocenter a b c; h1 = orthocenter t b c ? para o h o2 h1
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A B C = triangle; D1 = equilateral_triangle B C; E1 = equilateral_triangle C A; F1 = equilateral_triangle A B; AD1 = line A D1; BE1 = line B E1; X1 = intersection AD1 BE1; D2 = equilateral_triangle C B; E2 = equilateral_triangle A C; F2 = equilateral_triangle B A; AD2 = line A D2; BE2 = line B E2; X2 = intersection AD2 BE2; O1 = circumcenter A X1 X2; (O1) = circle_center_point O1 A; S = on_circle (O1); O2 = circumcenter B S C; AS = line A S; (O2) = circle_center_point O2 B; T = intersection AS (O2); O = circumcenter A B C; H = orthocenter A B C; H1 = orthocenter T B C; OH = line O H; O2H1 = line O2 H1; Prove: parallel OH O2H1
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ShuZhiMi2023SpFp8
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For $\triangle ABC$ with circumcenter $O$, incenter $I$, orthocenter $H$, and $A$-excenter $J$, let $A_1$ be $A$'s reflection over $BC$, $A_2=AO\cap(BOC)\setminus A$, $R=OI\cap JA_1$, $Q=IH\cap JA_2$, and $P$ be the second intersection of $(AOH)$'s tangent at $A$ with $(ABC)$. Prove $Q,P,R$ are collinear.
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a b c = triangle; o = circumcenter a b c; i = incenter a b c; h = orthocenter a b c; j = excenter a b c; a1 = reflect a b c; o1 = circumcenter b o c; a2 = on_line a o, on_circle o1 b; r = on_line o i, on_line j a1; q = on_line i h, on_line j a2; o2 = circumcenter a o h; p = on_tline a o2 a, on_circle o a ? coll q p r
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A B C = triangle; O = circumcenter A B C; I = incenter A B C; H = orthocenter A B C; J = excenter A B C; BC = line B C; A1 = reflect_point_wrt_line A BC; O1 = circumcenter B O C; AO = line A O; (O1) = circle_center_point O1 B; A2 = intersection AO (O1); OI = line O I; JA1 = line J A1; R = intersection OI JA1; IH = line I H; JA2 = line J A2; Q = intersection IH JA2; O2 = circumcenter A O H; O2A = line O2 A; l1 = perpendicular_line A O2A; (O) = circle_center_point O A; P = intersection l1 (O); Prove: collinear Q P R
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ShuZhiMi2024SpFp19
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In $\triangle ABC$ with altitudes $AD,BE,CF$, let $I_1,I_2,I_3$ be incenters of $\triangle AEF,\triangle BFD,\triangle CDE$. For $l_a$ as the external common tangent of $(I_2),(I_3)$ distinct from $BC$ (similarly $l_b,l_c$), prove $l_a,l_b,l_c$ and $OI$ concur.
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a b c = triangle; o = circumcenter a b c; i = incenter a b c; d = foot a b c; e = foot b a c; f = foot c a b; i1 = incenter a e f; i2 = incenter b f d; i3 = incenter c d e; d2 = foot i2 b c; d3 = foot i3 b c; e1 = foot i1 a c; e3 = foot i3 a c; p2 = reflect d2 i2 i3; p3 = reflect d3 i2 i3; q1 = reflect e1 i1 i3; q3 = reflect e3 i1 i3; x = on_line p3 p2, on_line q1 q3 ? coll x i o
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A B C = triangle; O = circumcenter A B C; BC = line B C; AC = line A C; AB = line A B; I = incenter A B C; D = foot A BC; E = foot B AC; F = foot C AB; I1 = incenter A E F; I2 = incenter B F D; I3 = incenter C D E; D2 = foot I2 BC; D3 = foot I3 BC; E1 = foot I1 AC; E3 = foot I3 AC; I2I3 = line I2 I3; P2 = reflect_point_wrt_line D2 I2I3; P3 = reflect_point_wrt_line D3 I2I3; I1I3 = line I1 I3; Q1 = reflect_point_wrt_line E1 I1I3; Q3 = reflect_point_wrt_line E3 I1I3; P3P2 = line P3 P2; Q1Q3 = line Q1 Q3; X = intersection P3P2 Q1Q3; Prove: collinear X I O
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ShuZhiMiGeo109
|
For $\triangle ABC$ with incenter $I$, excenters $I_a,I_b,I_c$, and $\triangle I_aI_bI_c$'s circumcenter $B_e$, construct circle with diameter $IB_e$. Lines through $I\parallel BC,CA,AB$ meet this circle at $D,E,F$. Prove $\triangle DEF$'s incenter is $\triangle ABC$'s Nagel point.
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a b c = triangle; i = incenter a b c; i_a = excenter a b c; i_b = excenter b c a; i_c = excenter c a b; b_e = circumcenter i_a i_b i_c; d = on_pline i b c, on_dia i b_e; e = on_pline i c a, on_dia i b_e; f = on_pline i a b, on_dia i b_e; d1 = foot i_a b c; e1 = foot i_b a c; n1 = on_line a d1, on_line b e1 ? eqangle d e e n1 n1 e e f
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A B C = triangle; I = incenter A B C; Ia = excenter A B C; Ib = excenter B C A; Ic = excenter C A B; Be = circumcenter Ia Ib Ic; BC = line B C; AC = line A C; AB = line A B; l1 = parallel_line I BC; (IBe) = circle_diameter I Be; D = intersection l1 (IBe); l2 = parallel_line I AC; E = intersection l2 (IBe); l3 = parallel_line I AB; F = intersection l3 (IBe); D1 = foot Ia BC; E1 = foot Ib AC; AD1 = line A D1; BE1 = line B E1; N1 = intersection AD1 BE1; Prove: equal_angle D E N1 N1 E F
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ShuZhiMiGeo117
|
In $\triangle ABC$ with circumcenter $O$, incenter $I$, and A-excenter $I_A$, let $BI\cap AC=E$, $CI\cap AB=F$, and $K$ be the isogonal conjugate of $I$ wrt $\triangle I_AEF$. If $X$ is $K$'s projection on $EF$, prove $OI$ bisects $KX$.
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a b c = triangle; o = circumcenter a b c; i = incenter a b c; ia = excenter a b c; e = on_line b i, on_line a c; f = on_line c i, on_line a b; k = on_aline k f e ia f i, on_aline k e f ia e i; x = foot k e f; m = midpoint k x ? coll m o i
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A B C = triangle; O = circumcenter A B C; I = incenter A B C; Ia = excenter A B C; BI = line B I; AC = line A C; E = intersection BI AC; CI = line C I; AB = line A B; F = intersection CI AB; l1 = angle_equal1 F E Ia F I; l2 = angle_equal1 E F Ia E I; K = intersection l1 l2; EF = line E F; X = foot K EF; M = midpoint K X; Prove: collinear M O I
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ShuZhiMiGeo128
|
For point $P$ in $\triangle ABC$ with Cevian triangle $DEF$ and Miquel point $Q$ of $DEF$, take $R\in PQ$ with $AR\cap\odot(AEF)=J$, $BR\cap\odot(BFD)=K$, $CR\cap\odot(CDE)=L$. Prove $\odot(AJD)$, $\odot(BKE)$, $\odot(CLF)$ are coaxial.
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a b c = triangle; p = free; d = on_line a p, on_line b c; e = on_line b p, on_line a c; f = on_line c p, on_line a b; o1 = circumcenter a e f; o2 = circumcenter b d f; q = on_circle o1 a, on_circle o2 b; r = on_line p q; j = on_line a r, on_circle o1 a; k = on_line b r, on_circle o2 b; o3 = circumcenter c d e; l = on_line c r, on_circle o3 c; oa = circumcenter a j d; ob = circumcenter b k e; t = on_circle oa a, on_circle ob b ? cyclic t c l f
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A B C = triangle; P = point; AP = line A P; BC = line B C; D = intersection AP BC; BP = line B P; AC = line A C; E = intersection BP AC; CP = line C P; AB = line A B; F = intersection CP AB; O1 = circumcenter A E F; O2 = circumcenter B D F; (O1) = circle_center_point O1 A; (O2) = circle_center_point O2 B; Q = intersection (O1) (O2); PQ = line P Q; R = on_line PQ; AR = line A R; J = intersection AR (O1); BR = line B R; K = intersection BR (O2); O3 = circumcenter C D E; CR = line C R; (O3) = circle_center_point O3 C; L = intersection CR (O3); Oa = circumcenter A J D; Ob = circumcenter B K E; (Oa) = circle_center_point Oa A; (Ob) = circle_center_point Ob B; T = intersection (Oa) (Ob); Prove: concyclic T C L F
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ShuZhiMiGeo178
|
In $\triangle ABC$ with orthocenter $H$ and nine-point center $V$, let $\odot(AVH)\cap\odot(BVC)=X$ (similarly define $Y,Z$). Prove $AX,BY,CZ$ are concurrent.
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a b c = triangle; h = orthocenter a b c; m1 = midpoint b c; m2 = midpoint c a; m3 = midpoint a b; v = circumcenter m1 m2 m3; o1 = circumcenter a v h; o2 = circumcenter b v c; x = on_circle o1 a, on_circle o2 b; o3 = circumcenter b v h; o4 = circumcenter c v a; y = on_circle o3 b, on_circle o4 c; o5 = circumcenter c v h; o6 = circumcenter a v b; z = on_circle o5 c, on_circle o6 a; k = on_line a x, on_line b y ? coll k c z
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A B C = triangle; H = orthocenter A B C; M1 = midpoint B C; M2 = midpoint C A; M3 = midpoint A B; V = circumcenter M1 M2 M3; O1 = circumcenter A V H; O2 = circumcenter B V C; (O1) = circle_center_point O1 A; (O2) = circle_center_point O2 B; X = intersection (O1) (O2); O3 = circumcenter B V H; O4 = circumcenter C V A; (O3) = circle_center_point O3 B; (O4) = circle_center_point O4 C; Y = intersection (O3) (O4); O5 = circumcenter C V H; O6 = circumcenter A V B; (O5) = circle_center_point O5 C; (O6) = circle_center_point O6 A; Z = intersection (O5) (O6); AX = line A X; BY = line B Y; K = intersection AX BY; Prove: collinear K C Z
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ShuZhiMiGeo180
|
Let $D,E,F$ be contact points of $\triangle ABC$'s excircles. Points $X,Y,Z,U,V,W$ satisfy $DY=DW=EZ=EU=FX=FV$. If $O_1,\dots,O_6$ are circumcenters of $\triangle AUV,\triangle BVW,\triangle CWU,\triangle AZX,\triangle BXY,\triangle CYZ$ respectively, then line $IO$ (where $I,O$ are incenter and circumcenter) is the radical axis of $\triangle O_1O_2O_3$ and $\triangle O_4O_5O_6$'s circumcircles
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a b c = triangle; i = incenter a b c; o = circumcenter a b c; i_a = excenter a b c; d = foot i_a b c; i_b = excenter b c a; e = foot i_b a c; i_c = excenter c a b; f = foot i_c a b; u = on_line a c; z = mirror u e; v = on_line a b, eqdistance v f e u; x = mirror v f; w = on_line b c, eqdistance w d u e; y = mirror w d; o1 = circumcenter a u v; o2 = circumcenter b v w; o3 = circumcenter c w u; o4 = circumcenter a z x; o5 = circumcenter b x y; o6 = circumcenter c y z; oa = circumcenter o1 o2 o3; ob = circumcenter o4 o5 o6; t = on_circle oa o1, on_circle ob o4 ? coll i o t
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A B C = triangle; I = incenter A B C; BC = line B C; AC = line A C; AB = line A B; O = circumcenter A B C; Ia = excenter A B C; D = foot Ia BC; Ib = excenter B C A; E = foot Ib AC; I_c = excenter C A B; F = foot I_c AB; U = on_line AC; Z = reflect_point_wrt_point U E; AB = line A B; (F) = circle_center_radius F E U; V = intersection AB (F); X = reflect_point_wrt_point V F; BC = line B C; (D) = circle_center_radius D E U; W = intersection BC (D); Y = reflect_point_wrt_point W D; O1 = circumcenter A U V; O2 = circumcenter B V W; O3 = circumcenter C W U; O4 = circumcenter A Z X; O5 = circumcenter B X Y; O6 = circumcenter C Y Z; Oa = circumcenter O1 O2 O3; Ob = circumcenter O4 O5 O6; (Oa) = circle_center_point Oa O1; (Ob) = circle_center_point Ob O4; T = intersection (Oa) (Ob); Prove: collinear I O T
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ShuZhiMiGeo209
|
In $\triangle ABC$ with angle bisectors $AD,BE,CF$ and altitudes $AX,BY,CZ$, let $P,Q,R$ be orthocenters of $\triangle AEF,\triangle BDF,\triangle CDE$, and $K,L,M$ be incenters of $\triangle AYZ,\triangle BZX,\triangle CXY$. Prove $PK,QL,MR$ concur on a common tangent to the incircle and nine-point circle.
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a b c = triangle; i = incenter a b c; d = on_line b c, on_line a i; e = on_line c a, on_line b i; f = on_line a b, on_line c i; x = foot a b c; y = foot b c a; z = foot c a b; p = orthocenter a e f; q = orthocenter b f d; r = orthocenter c d e; k = incenter a y z; l = incenter b z x; m = incenter c x y; t = on_line p k, on_line q l; ni = circumcenter x y z; d1 = foot i b c; fe = on_circle i d1, on_circle ni x ? perp t fe i fe
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A B C = triangle; I = incenter A B C; BC = line B C; AI = line A I; D = intersection BC AI; CA = line C A; BI = line B I; E = intersection CA BI; AB = line A B; CI = line C I; F = intersection AB CI; X = foot A BC; Y = foot B CA; Z = foot C AB; P = orthocenter A E F; Q = orthocenter B F D; R = orthocenter C D E; K = incenter A Y Z; L = incenter B Z X; M = incenter C X Y; PK = line P K; QL = line Q L; T = intersection PK QL; Ni = circumcenter X Y Z; D1 = foot I BC; (I) = circle_center_point I D1; (Ni) = circle_center_point Ni X; Fe = intersection (I) (Ni); TFe = line T Fe; IFe = line I Fe; Prove: perpendicular TFe IFe
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