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1995USAMOp3
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Given a nonisosceles, nonright triangle ABC, let O denote the center of its circumscribed circle, and let $A_1$, $B_1$, and $C_1$ be the midpoints of sides BC, CA, and AB, respectively. Point $A_2$ is located on the ray $OA_1$ so that $OAA_1$ is similar to $OA_2A$. Points $B_2$ and $C_2$ on rays $OB_1$ and $OC_1$, respectively, are defined similarly. Prove that lines $AA_2$, $BB_2$, and $CC_2$ are concurrent
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a b c = triangle; o = circumcenter a b c; a1 = midpoint b c; b1 = midpoint a c; c1 = midpoint a b; a2 = on_line o a1, on_aline a2 a o o a1 a; b2 = on_line o b1, on_aline b2 b o o b1 b; c2 = on_line o c1, on_aline c2 c o o c1 c; p = on_line a a2, on_line b b2 ? coll p c c2
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A B C = acute_triangle; O = circumcenter A B C; A1 = midpoint B C; B1 = midpoint C A; C1 = midpoint A B; l1 = angle_equal1 A O O A1 A; OA1 = line O A1; A2 = intersection OA1 l1; l2 = angle_equal1 B O O B1 B; OB1 = line O B1; B2 = intersection OB1 l2; l3 = angle_equal1 C O O C1 C; OC1 = line O C1; C2 = intersection OC1 l3; AA2 = line A A2; BB2 = line B B2; P = intersection AA2 BB2; Prove: collinear P C C2
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1999USAMOp6
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Let $ABCD$ be an isosceles trapezoid with $AB \parallel CD$. The inscribed circle $\omega$ of triangle $BCD$ meets $CD$ at $E$. Let $F$ be a point on the (internal) angle bisector of $\angle DAC$ such that $EF \perp CD$. Let the circumscribed circle of triangle $ACF$ meet line $CD$ at $C$ and $G$. Prove that the triangle $AFG$ is isosceles.
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a b c d = eq_trapezoid; i = incenter b c d; e = foot i c d; f = angle_bisector d a c, on_tline e c d; o = circumcenter a c f; g = on_line c d, on_circle o a ? cong a f f g
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D A B C = isos_trapezoid; I = incenter B C D; CD = line C D; E = foot I CD; l1 = angle_bisector D A C; l2 = perpendicular_line E CD; F = intersection l1 l2; O1 = circumcenter A C F; (O1) = circle_center_point O1 A; G = intersection (O1) CD; Prove: cong F A F G
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20**_11thXueErSiMO
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Let $\triangle ABC$ have circumcircle $\Gamma$. The angle bisector of $\angle BAC$ intersects $BC$ at $D$. Point $P$ lies on segment $AD$. Straight line $BP$ intersects $AC$ at $E$, and straight line $CP$ intersects $AB$ at $F$. Let the tangent of the circle $\Gamma$ at point $A$ intersect the line $EF$ at the point $Q$. Prove that $PQ\parallel BC$.
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a b c = triangle; o = circumcenter a b c; d = on_line b c, angle_bisector b a c; p = on_line a d; e = on_line b p, on_line a c; f = on_line c p, on_line a b; q = on_line e f, on_tline a o a ? para p q b c
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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; AD = line A D; P = on_line AD; BP = line B P; AC = line A C; E = intersection BP AC; CP = line C P; AB = line A B; F = intersection CP AB; OA = line O A; l = perpendicular_line A OA; EF = line E F; Q = intersection l EF; PQ = line P Q; Prove: parallel PQ BC
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2000USATSTp2
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Let $ ABCD$ be a cyclic quadrilateral and let $ E$ and $ F$ be the feet of perpendiculars from the intersection of diagonals $ AC$ and $ BD$ to $ AB$ and $ CD$, respectively. Prove that $ EF$ is perpendicular to the line through the midpoints of $ AD$ and $ BC$.
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a b c = triangle; d = on_circum a b c; p = on_line a c, on_line b d; e = foot p a b; f = foot p c d; m = midpoint a d; n = midpoint b c ? perp e f m n
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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; P = intersection AC BD; AB = line A B; E = foot P AB; CD = line C D; F = foot P CD; M = midpoint A D; N = midpoint B C; EF = line E F; MN = line M N; Prove: perpendicular EF MN
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2002CTSTp25
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Let $E$ and $F$ be the intersections of opposite sides of a convex quadrilateral $ABCD$. The two diagonals meet at $P$. Let $O$ be the foot of the perpendicular from $P$ to $EF$. Show that $\angle BOC=\angle AOD$.
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a b c d = quadrangle; e = on_line a b, on_line c d; f = on_line a d, on_line b c; p = on_line a c, on_line b d; o = foot p e f ? eqangle b o o c a o o d
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A B C D = quadrilateral; AB = line A B; CD = line C D; E = intersection AB CD; AD = line A D; BC = line B C; F = intersection AD BC; AC = line A C; BD = line B D; P = intersection AC BD; EF = line E F; O = foot P EF; Prove: equal_angle B O C A O D
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2003CHNGaoLian
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From point $P$ outside a circle draw two tangents to the circle touching at $A$ and $B$. Draw a secant line intersecting the circle at points $C$ and $D$, with $C$ between $P$ and $D$. Choose point $Q$ on the chord $CD$ such that $\angle DAQ=\angle PBC$. Prove that $\angle DBQ=\angle PAC$.
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o = free; a0 = free; p = free; a b = tangent a b p o a0; c = on_circle o a0; d = on_line p c, on_circle o a0; q = on_line c d, on_aline q a d c b p ? eqangle d b b q p a a c
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(O) = circle; P = outside (O); l1 l2 = tangent P (O); A = intersection l1 (O); B = intersection l2 (O); C = on_circle (O); CP = line C P; D = intersection CP (O); CD = line C D; l3 = angle_equal1 A D C B P; Q = intersection CD l3; Prove: equal_angle D B Q P A C
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2003USATSTp6
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Let $\overline{AH_1}, \overline{BH_2}$, and $\overline{CH_3}$ be the altitudes of an acute scalene triangle $ABC$. The incircle of triangle $ABC$ is tangent to $\overline{BC}, \overline{CA},$ and $\overline{AB}$ at $T_1, T_2,$ and $T_3$, respectively. For $k = 1, 2, 3$, let $P_i$ be the point on line $H_iH_{i+1}$ (where $H_4 = H_1$) such that $H_iT_iP_i$ is an acute isosceles triangle with $H_iT_i = H_iP_i$. Prove that the circumcircles of triangles $T_1P_1T_2$, $T_2P_2T_3$, $T_3P_3T_1$ pass through a common point.
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a b c = triangle; h1 = foot a b c; h2 = foot b a c; h3 = foot c a b; i = incenter a b c; t1 = foot i b c; t2 = foot i a c; t3 = foot i a b; p1 = on_line h1 h2, on_circle h1 t1; p2 = on_line h2 h3, on_circle h2 t2; p3 = on_line h3 h1, on_circle h3 t3; o1 = circumcenter t1 p1 t2; o2 = circumcenter t2 p2 t3; o3 = circumcenter t3 p3 t1; x = on_circle o1 t1, on_circle o2 t2 ? cong x o3 t3 o3
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A B C = acute_triangle; BC = line B C; H1 = foot A BC; AC = line A C; H2 = foot B AC; AB = line A B; H3 = foot C AB; I = incenter A B C; T1 = foot I BC; T2 = foot I AC; T3 = foot I AB; H1H2 = line H1 H2; (H1) = circle_center_point H1 T1; P1 = intersection H1H2 (H1); H2H3 = line H2 H3; (H2) = circle_center_point H2 T2; P2 = intersection H2H3 (H2); H1H3 = line H1 H3; (H3) = circle_center_point H3 T3; P3 = intersection H1H3 (H3); O1 = circumcenter T1 P1 T2; O2 = circumcenter T2 P2 T3; O3 = circumcenter T3 P3 T1; (O1) = circle_center_point O1 T1; (O2) = circle_center_point O2 T2; X = intersection (O1) (O2); Prove: cong X O3 T3 O3
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2004CTSTp1
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Using $ AB$ and $ AC$ as diameters, two semi-circles are constructed respectively outside the acute triangle $ ABC$. $ AH \perp BC$ at $ H$, $ D$ is any point on side $ BC$ ($ D$ is not coinside with $ B$ or $ C$ ), through $ D$, construct $ DE \parallel AC$ and $ DF \parallel AB$ with $ E$ and $ F$ on the two semi-circles respectively. Show that $ D$, $ E$, $ F$ and $ H$ are concyclic.
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a b c = triangle; o1 = midpoint a b; o2 = midpoint a c; h = foot a b c; d = on_line b c; e = on_circle o1 a, on_pline d a c; f = on_circle o2 a, on_pline d a b ? cyclic d e f h
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A B C = acute_triangle; BC = line B C; H = foot A BC; D = on_line BC; AC = line A C; l1 = parallel_line D AC; O1 = midpoint A B; (O1) = circle_center_point O1 A; E = intersection l1 (O1); AB = line A B; l2 = parallel_line D AB; O2 = midpoint A C; (O2) = circle_center_point O2 A; F = intersection l2 (O2); Prove: concyclic D E F H
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2004CTSTp22
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Points $D,E,F$ are on the sides $BC, CA$ and $AB$, respectively which satisfy $EF || BC$, $D_1$ is a point on $BC,$ Make $D_1E_1 || D_E, D_1F_1 || DF$ which intersect $AC$ and $AB$ at $E_1$ and $F_1$, respectively. Make $\bigtriangleup PBC \sim \bigtriangleup DEF$ such that $P$ and $A$ are on the same side of $BC.$ Prove that $E, E_1F_1, PD_1$ are concurrent.
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a b c = triangle; e = on_line c a; f = on_line a b, on_pline e b c; d = on_line b c; d1 = on_line b c; e1 = on_line a c, on_pline d1 d e; f1 = on_line a b, on_pline d1 d f; p = on_aline p b c d e f, on_aline p c b d f e; x = on_line e1 f1, on_line e f ? coll p d1 x
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A B C = triangle; BC = line B C; D = foot A BC; AC = line A C; E = on_line AC; l1 = parallel_line E BC; AB = line A B; F = intersection AB l1; D1 = on_line BC; DE = line D E; l2 = parallel_line D1 DE; E1 = intersection AC l2; DF = line D F; l3 = parallel_line D1 DF; F1 = intersection AB l3; l4 = angle_equal1 B C D E F; l5 = angle_equal1 C B D F E; P = intersection l4 l5; EF = line E F; E1F1 = line E1 F1; X = intersection E1F1 EF; Prove: collinear P D1 X
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2005CHNSouthEastMOp2
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Circle $C$ (with center $O$) does not have common point with line $l$. Draw $OP$ perpendicular to $l$, $P \in l$. Let $Q$ be a point on $l$ ($Q$ is different from $P$), $QA$ and $QB$ are tangent to circle $C$, and intersect the circle at $A$ and $B$ respectively. $AB$ intersects $OP$ at $K$. $PM$, $PN$ are perpendicular to $QB$, $QA$, respectively, $M \in QB$, $N \in QA$. Prove that segment $KP$ is bisected by line $MN$.
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o = free; r1 = free; p = free; q = on_tline p o p; a b = tangent a b q o r1; k = on_line a b, on_line o p; m = foot p q b; n = foot p q a; x = midpoint k p ? coll x m n
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(O) = circle; O = circumcenter (O); l = line; P = foot O l; Q = on_line l; A B = tangent_point Q (O); OP = line O P; AB = line A B; K = intersection OP AB; BQ = line B Q; M = foot P BQ; AQ = line A Q; N = foot P AQ; X = midpoint K P; Prove: collinear X M N
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2005CHNWesternMOp2
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Given three points $P$, $A$, $B$ and a circle such that the lines $PA$ and $PB$ are tangent to the circle at the points $A$ and $B$, respectively. A line through the point $P$ intersects that circle at two points $C$ and $D$. Through the point $B$, draw a line parallel to $PA$; let this line intersect the lines $AC$ and $AD$ at the points $E$ and $F$, respectively. Prove that $BE = BF$.
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o a b = iso_triangle; p = on_tline a o a, on_tline b o b; t = free; c = on_circle o a; d = on_line p c, on_circle o a; e = on_line a c, on_pline b p a; f = on_line a d, on_pline b p a ? cong b e b f
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O A B = isos_triangle; OA = line O A; OB = line O B; l1 = perpendicular_line A OA; l2 = perpendicular_line B OB; P = intersection l1 l2; (O) = circle_center_point O A; C = on_circle (O); CP = line C P; D = intersection CP (O); AP = line A P; l3 = parallel_line B AP; AC = line A C; E = intersection l3 AC; AD = line A D; F = intersection l3 AD; Prove: cong B E B F
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2005CMOp2
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A circle meets the three sides $BC,CA,AB$ of a triangle $ABC$ at points $D_1,D_2;E_1,E_2; F_1,F_2$ respectively. Furthermore, line segments $D_1E_1$ and $D_2F_2$ intersect at point $L$, line segments $E_1F_1$ and $E_2D_2$ intersect at point $M$, line segments $F_1D_1$ and $F_2E_2$ intersect at point $N$. Prove that the lines $AL,BM,CN$ are concurrent.
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a b c = triangle; o = free; d1 = on_line b c; d2 = on_line b c, on_circle o d1; e1 = on_line a c, on_circle o d1; e2 = on_line a c, on_circle o d1; f1 = on_line a b, on_circle o d1; f2 = on_line a b, on_circle o d1; l = on_line d1 e1, on_line d2 f2; m = on_line e1 f1, on_line e2 d2; n = on_line f1 d1, on_line f2 e2; p = on_line a l, on_line b m ? coll c n p
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A B C = triangle; O = point; BC = line B C; D1 = on_line BC; (O) = circle_center_point O D1; D2 = intersection (O) BC; AC = line A C; E1 E2 = intersection (O) AC; AB = line A B; F1 F2 = intersection (O) AB; D1E1 = line D1 E1; D2F2 = line D2 F2; L = intersection D1E1 D2F2; E1F1 = line E1 F1; D2E2 = line D2 E2; M = intersection E1F1 D2E2; D1F1 = line D1 F1; E2F2 = line E2 F2; N = intersection D1F1 E2F2; AL = line A L; BM = line B M; P = intersection AL BM; Prove: collinear C N P
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2005CTSTp1
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Point $P$ lies inside triangle $ABC$. Let the projections of $P$ onto sides $BC$,$CA$,$AB$ be $D$, $E$, $F$ respectively. Let the projections from $A$ to the lines $BP$ and $CP$ be $M$ and $N$ respectively. Prove that $ME$, $NF$ and $BC$ are concurrent.
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a b c = triangle; p = free; d = foot p b c; e = foot p a c; f = foot p a b; m = foot a b p; n = foot a c p; t = on_line m e, on_line b c ? coll t n f
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A B C = triangle; P = point; BC = line B C; D = foot P BC; AC = line A C; E = foot P AC; AB = line A B; F = foot P AB; BP = line B P; M = foot A BP; CP = line C P; N = foot A CP; EM = line E M; FN = line F N; T = intersection EM BC; Prove: collinear F N T
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2005CTSTp11b
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Let $\omega$ be the circumcircle of acute triangle $ABC$. Two tangents of $\omega$ from $B$ and $C$ intersect at $P$, $AP$ and $BC$ intersect at $D$. Point $E$, $F$ are on $AC$ and $AB$ such that $DE \parallel BA$ and $DF \parallel CA$. (1) Prove that $F,B,C,E$ are concyclic. (2) Denote $A_{1}$ the centre of the circle passing through $F,B,C,E$. $B_{1}$, $C_{1}$ are difined similarly. Prove that $AA_{1}$, $BB_{1}$, $CC_{1}$ are concurrent.
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a b c = triangle; o = circumcenter a b c; p = on_tline b o b, on_tline c o c; q = on_tline a o a, on_tline c o c; r = on_tline b o b, on_tline a o a; d1 = on_line a p, on_line b c; e1 = on_line a c, on_pline d1 a b; f1 = on_line a b, on_pline d1 a c; d2 = on_line b q, on_line a c; e2 = on_line b a, on_pline d2 b c; f2 = on_line b c, on_pline d2 b a; d3 = on_line c r, on_line b a; e3 = on_line c b, on_pline d3 c a; f3 = on_line c a, on_pline d3 c b; a1 = circumcenter f1 b c; b1 = circumcenter f2 a c; c1 = circumcenter f3 a b; x = on_line a a1, on_line b b1 ? coll x c c1
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A B C = acute_triangle; O = circumcenter A B C; OA = line O A; OB = line O B; OC = line O C; l1 = perpendicular_line A OA; l2 = perpendicular_line B OB; l3 = perpendicular_line C OC; P = intersection l2 l3; Q = intersection l1 l3; R = intersection l1 l2; AP = line A P; BC = line B C; D1 = intersection AP BC; BQ = line B Q; AC = line A C; D2 = intersection BQ AC; CR = line C R; AB = line A B; D3 = intersection CR AB; l4 = parallel_line D1 AB; E1 = intersection l4 AC; l5 = parallel_line D1 AC; F1 = intersection l5 AB; l6 = parallel_line D2 BC; E2 = intersection l6 AB; l7 = parallel_line D2 AB; F2 = intersection l7 BC; l8 = parallel_line D3 AC; E3 = intersection l8 BC; l9 = parallel_line D3 BC; F3 = intersection l9 AC; A1 = circumcenter F1 B C; B1 = circumcenter F2 C A; C1 = circumcenter F3 A B; AA1 = line A A1; BB1 = line B B1; X = intersection AA1 BB1; Prove: collinear X C C1
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2005CTSTp19
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Convex quadrilateral $ABCD$ is cyclic in circle $(O)$, $P$ is the intersection of the diagonals $AC$ and $BD$. Circle $(O_{1})$ passes through $P$ and $B$, circle $(O_{2})$ passes through $P$ and $A$, Circles $(O_{1})$ and $(O_{2})$ intersect at $P$ and $Q$. $(O_{1})$, $(O_{2})$ intersect $(O)$ at another points $E$, $F$ (besides $B$, $A$), respectively. Prove that $PQ$, $CE$, $DF$ are concurrent or parallel.
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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; o1 = on_bline p b; o2 = on_bline p a; q = on_circle o1 p, on_circle o2 p; e = on_circle o1 p, on_circle o a; f = on_circle o2 p, on_circle o a; x = on_line p q, on_line c e ? coll x d f
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A B C = triangle; O = circumcenter A B C; (O) = circle_center_point O A; D = on_circle (O); AC = line A C; BD = line B D; P = intersection AC BD; l1 = perpendicular_bisector P B; O1 = on_line l1; (O1) = circle_center_point O1 P; l2 = perpendicular_bisector P A; O2 = on_line l2; (O2) = circle_center_point O2 P; Q = intersection (O1) (O2); E = intersection (O1) (O); F = intersection (O2) (O); PQ = line P Q; CE = line C E; DF = line D F; X = intersection PQ CE; Prove: collinear X D F
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2005G6
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Let $ABC$ be a triangle, and $M$ the midpoint of its side $BC$. Let $\gamma$ be the incircle of triangle $ABC$. The median $AM$ of triangle $ABC$ intersects the incircle $\gamma$ at two points $K$ and $L$. Let the lines passing through $K$ and $L$, parallel to $BC$, intersect the incircle $\gamma$ again in two points $X$ and $Y$. Let the lines $AX$ and $AY$ intersect $BC$ again at the points $P$ and $Q$. Prove that $BP = CQ$.
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a b c = triangle; m = midpoint b c; i = incenter a b c; d = foot i b c; k = on_line a m, on_circle i d; l = on_line a m, on_circle i d; x = on_pline k b c, on_circle i d; y = on_pline l b c, on_circle i d; p = on_line a x, on_line b c; q = on_line a y, on_line b c ? cong b p c q
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A B C = triangle; M = midpoint B C; I = incenter A B C; BC = line B C; D = foot I BC; (I) = circle_center_point I D; AM = line A M; K = intersection AM (I); L = intersection AM (I); BC = line B C; l1 = parallel_line K BC; X = intersection l1 (I); l2 = parallel_line L BC; Y = intersection l2 (I); AX = line A X; P = intersection AX BC; AY = line A Y; Q = intersection AY BC; Prove: cong B P C Q
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2005USAMOp3
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Let $ABC$ be an acute-angled triangle, and let $P$ and $Q$ be two points on its side $BC$. Construct a point $C_{1}$ in such a way that the convex quadrilateral $APBC_{1}$ is cyclic, $QC_{1}\parallel CA$, and $C_{1}$ and $Q$ lie on opposite sides of line $AB$. Construct a point $B_{1}$ in such a way that the convex quadrilateral $APCB_{1}$ is cyclic, $QB_{1}\parallel BA$, and $B_{1}$ and $Q$ lie on opposite sides of line $AC$. Prove that the points $B_{1}$, $C_{1}$, $P$, and $Q$ lie on a circle.
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a b c = triangle; p = on_line b c; q = on_line b c; c1 = on_circum a p b, on_pline q c a; b1 = on_circum a p c, on_pline q b a ? cyclic b1 c1 p q
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A B C = acute_triangle; BC = line B C; P = on_line BC; Q = on_line BC; AC = line A C; l1 = parallel_line Q AC; (ABP) = circle A B P; C1 = intersection (ABP) l1; AB = line A B; l2 = parallel_line Q AB; (ACP) = circle A C P; B1 = intersection (ACP) l2; Prove: concyclic B1 C1 P Q
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2006CGMOp2
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Let $O$ be the intersection of the diagonals of convex quadrilateral $ABCD$. The circumcircles of $\triangle{OAD}$ and $\triangle{OBC}$ meet at $O$ and $M$. Line $OM$ meets the circumcircles of $\triangle{OAB}$ and $\triangle{OCD}$ at $T$ and $S$ respectively. Prove that $M$ is the midpoint of $ST$.
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a b c d = quadrangle; o = on_line a c, on_line b d; o1 = circumcenter o a d; o2 = circumcenter o b c; m = on_circle o1 o, on_circle o2 o; o3 = circumcenter o a b; t = on_line o m, on_circle o3 o; o4 = circumcenter o c d; s = on_line o m, on_circle o4 o ? midp m s t
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A B C D = quadrilateral; AC = line A C; BD = line B D; O = intersection AC BD; (ADO) = circle A D O; (BCO) = circle B C O; M = intersection (ADO) (BCO); MO = line M O; (ABO) = circle A B O; T = intersection MO (ABO); (CDO) = circle C D O; S = intersection MO (CDO); Prove: midpoint M S T
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2006CTSTp19
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$H$ is the orthocentre of $\triangle{ABC}$. $D$, $E$, $F$ are on the circumcircle of $\triangle{ABC}$ such that $AD \parallel BE \parallel CF$. $S$, $T$, $U$ are the semetrical points of $D$, $E$, $F$ with respect to $BC$, $CA$, $AB$. Show that $S, T, U, H$ lie on the same circle.
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a b c = triangle; o = circumcenter a b c; h = orthocenter a b c; d = on_circum a b c; e = on_circum a b c, on_pline b a d; f = on_circum a b c, on_pline c a d; s = reflect d b c; t = reflect e c a; u = reflect f a b ? cyclic s t u h
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A B C = triangle; H = orthocenter A B C; O = circumcenter A B C; (ABC) = circle A B C; D = on_circle (ABC); AD = line A D; l1 = parallel_line B AD; E = intersection (ABC) l1; l2 = parallel_line C AD; F = intersection (ABC) l2; BC = line B C; S = reflect_point_wrt_line D BC; AC = line A C; T = reflect_point_wrt_line E AC; AB = line A B; U = reflect_point_wrt_line F AB; Prove: concyclic S T U H
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2006CTSTp5
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Let $\omega$ be the circumcircle of $\triangle{ABC}$. $P$ is an interior point of $\triangle{ABC}$. $A_{1}, B_{1}, C_{1}$ are the intersections of $AP, BP, CP$ respectively and $A_{2}, B_{2}, C_{2}$ are the symmetrical points of $A_{1}, B_{1}, C_{1}$ with respect to the midpoints of side $BC, CA, AB$. Show that the circumcircle of $\triangle{A_{2}B_{2}C_{2}}$ passes through the orthocentre of $\triangle{ABC}$.
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a b c = triangle; o = circumcenter a b c; p = free; a1 = on_line a p, on_circle o a; b1 = on_line b p, on_circle o a; c1 = on_line c p, on_circle o a; ma = midpoint b c; mb = midpoint c a; mc = midpoint a b; a2 = mirror a1 ma; b2 = mirror b1 mb; c2 = mirror c1 mc; h = orthocenter a b c ? cyclic h a2 b2 c2
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A B C = triangle; O = circumcenter A B C; (ABC) = circle_center_point O A; P = point; AP = line A P; A1 = intersection AP (ABC); BP = line B P; B1 = intersection BP (ABC); CP = line C P; C1 = intersection CP (ABC); L = midpoint B C; M = midpoint C A; N = midpoint A B; A2 = reflect A1 L; B2 = reflect B1 M; C2 = reflect C1 N; H = orthocenter A B C; Prove: concyclic H A2 B2 C2
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2006CTSTp7
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The centre of the circumcircle of quadrilateral $ABCD$ is $O$ and $O$ is not on any of the sides of $ABCD$. $P=AC \cap BD$. The circumecentres of $\triangle{OAB}$, $\triangle{OBC}$, $\triangle{OCD}$ and $\triangle{ODA}$ are $O_1$, $O_2$, $O_3$ and $O_4$ respectively. Prove that $O_1O_3$, $O_2O_4$ and $OP$ are concurrent.
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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; o1 = circumcenter o a b; o2 = circumcenter o b c; o3 = circumcenter o c d; o4 = circumcenter o d a; x = on_line o1 o3, on_line o2 o4 ? coll o p x
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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; O1 = circumcenter O A B; O2 = circumcenter O B C; O3 = circumcenter O C D; O4 = circumcenter O D A; O1O3 = line O1 O3; O2O4 = line O2 O4; X = intersection O1O3 O2O4; Prove: collinear O P X
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2006G4
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A point $D$ is chosen on the side $AC$ of a triangle $ABC$ with $\angle C < \angle A < 90^\circ$ in such a way that $BD=BA$. The incircle of $ABC$ is tangent to $AB$ and $AC$ at points $K$ and $L$, respectively. Let $J$ be the incenter of triangle $BCD$. Prove that the line $KL$ intersects the line segment $AJ$ at its midpoint.
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a b c = triangle; d = on_line a c, eqdistance d b b a; i = incenter a b c; k = foot i a b; l = foot i a c; j = incenter b c d; m = on_line k l, on_line a j ? midp m a j
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A B C = acute_triangle; AC = line A C; (B) = circle_center_point B A; D = intersection AC (B); I = incenter A B C; AB = line A B; K = foot I AB; L = foot I AC; J = incenter B C D; KL = line K L; AJ = line A J; M = intersection KL AJ; Prove: midpoint M A J
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2006G9
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Points $ A_{1}$, $ B_{1}$, $ C_{1}$ are chosen on the sides $ BC$, $ CA$, $ AB$ of a triangle $ ABC$ respectively. The circumcircles of triangles $ AB_{1}C_{1}$, $ BC_{1}A_{1}$, $ CA_{1}B_{1}$ intersect the circumcircle of triangle $ ABC$ again at points $ A_{2}$, $ B_{2}$, $ C_{2}$ respectively ($ A_{2}\neq A, B_{2}\neq B, C_{2}\neq C$). Points $ A_{3}$, $ B_{3}$, $ C_{3}$ are symmetric to $ A_{1}$, $ B_{1}$, $ C_{1}$ with respect to the midpoints of the sides $ BC$, $ CA$, $ AB$ respectively. Prove that the triangles $ A_{2}B_{2}C_{2}$ and $ A_{3}B_{3}C_{3}$ are similar.
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a b c = triangle; a1 = on_line b c; b1 = on_line a c; c1 = on_line a b; o = circumcenter a b c; o1 = circumcenter a b1 c1; a2 = on_circle o1 a, on_circle o a; o2 = circumcenter b c1 a1; b2 = on_circle o2 b, on_circle o b; o3 = circumcenter c a1 b1; c2 = on_circle o3 c, on_circle o c; m1 = midpoint b c; a3 = mirror a1 m1; m2 = midpoint a c; b3 = mirror b1 m2; m3 = midpoint a b; c3 = mirror c1 m3 ? simtri a2 b2 c2 a3 b3 c3
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A B C = triangle; BC = line B C; A1 = on_line BC; AC = line A C; B1 = on_line AC; AB = line A B; C1 = on_line AB; O = circumcenter A B C; (ABC) = circle_center_point O A; O1 = circumcenter A B1 C1; (AB1C1) = circle_center_point O1 A; A2 = intersection (ABC) (AB1C1); O2 = circumcenter B C1 A1; (A1BC1) = circle_center_point O2 B; B2 = intersection (ABC) (A1BC1); O3 = circumcenter C A1 B1; (A1B1C) = circle_center_point O3 C; C2 = intersection (ABC) (A1B1C); L = midpoint B C; A3 = reflect A1 L; M = midpoint C A; B3 = reflect B1 M; N = midpoint A B; C3 = reflect C1 N; Prove: similar A2 B2 C2 A3 B3 C3
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2006USATSTp6
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Let $ABC$ be a triangle. Triangles $PAB$ and $QAC$ are constructed outside of triangle $ABC$ such that $AP = AB$ and $AQ = AC$ and $\angle{BAP}= \angle{CAQ}$. Segments $BQ$ and $CP$ meet at $R$. Let $O$ be the circumcenter of triangle $BCR$. Prove that $AO \perp PQ.$
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a b c = triangle; p = on_circle a b; q = on_circle a c, on_aline q a c b a p; r = on_line b q, on_line c p; o = circumcenter b c r ? perp a o p q
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A B C = triangle; (A,B) = circle_center_point A B; P = on_circle (A,B); (A,C) = circle_center_point A C; l1 = angle_equal1 A C B A P; Q = intersection (A,C) l1; BQ = line B Q; CP = line C P; R = intersection BQ CP; O = circumcenter B C R; AO = line A O; PQ = line P Q; Prove: perpendicular AO PQ
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2007CHNSouthEastMOp2
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$AB$ is the diameter of semicircle $O$. $C$,$D$ are two arbitrary points on semicircle $O$. Point $P$ lies on line $CD$ such that line $PB$ is tangent to semicircle $O$ at $B$. Line $PO$ intersects line $CA$, $AD$ at point $E$, $F$ respectively. Prove that $OE$=$OF$.
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a b = segment; o = midpoint a b; c = on_circle o a; d = on_circle o a; p = on_line c d, on_tline b o b; e = on_line c a, on_line p o; f = on_line a d, on_line p o ? cong o e o f
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A = point; B = point; O = midpoint A B; (O) = circle_center_point O A; C = on_circle (O); D = on_circle (O); OB = line O B; l1 = perpendicular_line B OB; CD = line C D; P = intersection CD l1; OP = line O P; AC = line A C; E = intersection OP AC; AD = line A D; F = intersection OP AD; Prove: cong E O F O
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2007CMOp4
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Let $O, I$ be the circumcenter and incenter of triangle $ABC$. The incircle of $\triangle ABC$ touches $BC, CA, AB$ at points $D, E, F$ repsectively. $FD$ meets $CA$ at $P$, $ED$ meets $AB$ at $Q$. $M$ and $N$ are midpoints of $PE$ and $QF$ respectively. Show that $OI \perp MN$.
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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; p = on_line f d, on_line a c; q = on_line e d, on_line a b; m = midpoint p e; n = midpoint q f ? perp o i m n
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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; DF = line D F; P = intersection DF AC; DE = line D E; Q = intersection DE AB; M = midpoint P E; N = midpoint Q F; IO = line I O; MN = line M N; Prove: perpendicular IO MN
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2007CTSTp2
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Let $ I$ be the incenter of triangle $ ABC.$ Let $ M,N$ be the midpoints of $ AB,AC,$ respectively. Points $ D,E$ lie on $ AB,AC$ respectively such that $ BD=CE=BC.$ The line perpendicular to $ IM$ through $ D$ intersects the line perpendicular to $ IN$ through $ E$ at $ P.$ Prove that
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a b c = triangle; i = incenter a b c; m = midpoint a b; n = midpoint a c; d = on_line a b, on_circle b c; e = on_line a c, on_circle c b; p = on_tline d i m, on_tline e i n ? perp a p b c
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A B C = triangle; I = incenter A B C; M = midpoint A B; N = midpoint A C; AB = line A B; (B,C) = circle_center_point B C; D = intersection AB (B,C); AC = line A C; (C,B) = circle_center_point C B; E = intersection AC (C,B); IM = line I M; l1 = perpendicular_line D IM; IN = line I N; l2 = perpendicular_line E IN; P = intersection l1 l2; AP = line A P; BC = line B C; Prove: perpendicular AP BC
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2007USATSTp1
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Circles $ \omega_1$ and $ \omega_2$ meet at $ P$ and $ Q$. Segments $ AC$ and $ BD$ are chords of $ \omega_1$ and $ \omega_2$ respectively, such that segment $ AB$ and ray $ CD$ meet at $ P$. Ray $ BD$ and segment $ AC$ meet at $ X$. Point $ Y$ lies on $ \omega_1$ such that $ PY \parallel BD$. Point $ Z$ lies on $ \omega_2$ such that $ PZ \parallel AC$. Prove that points $ Q,X,Y,Z$ are collinear.
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p q = segment; o1 = on_bline p q; o2 = on_bline p q; a = on_circle o1 p; b = on_line a p, on_circle o2 q; d = on_circle o2 q; c = on_line p d, on_circle o1 p; x = on_line b d, on_line a c; y = on_circle o1 p, on_pline p b d ? coll q x y
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P = point; Q = point; l = perpendicular_bisector P Q; O1 = on_line l; O2 = on_line l; (O1) = circle_center_point O1 P; (O2) = circle_center_point O2 P; A = on_circle (O1); C = on_circle (O1); AP = line A P; B = intersection (O2) AP; CP = line C P; D = intersection (O2) CP; AC = line A C; BD = line B D; X = intersection AC BD; l1 = parallel_line P BD; Y = intersection (O1) l1; Prove: collinear Q X Y
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2007USATSTp5
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Triangle $ ABC$ is inscribed in circle $ \omega$. The tangent lines to $ \omega$ at $ B$ and $ C$ meet at $ T$. Point $ S$ lies on ray $ BC$ such that $ AS \perp AT$. Points $ B_1$ and $ C_1$ lie on ray $ ST$ (with $ C_1$ in between $ B_1$ and $ S$) such that $ B_1T = BT = C_1T$. Prove that triangles $ ABC$ and $ AB_1C_1$ are similar to each other.
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a b c = triangle; o = circumcenter a b c; t = on_tline b o b, on_tline c o c; s = on_line b c, on_tline a a t; b1 = on_line s t, on_circle t b; c1 = on_line s t, on_circle t b ? simtri a b c a b1 c1
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A B C = triangle; O = circumcenter A B C; OB = line O B; OC = line O C; l1 = perpendicular_line B OB; l2 = perpendicular_line C OC; T = intersection l1 l2; AT = line A T; l3 = perpendicular_line A AT; BC = line B C; S = intersection BC l3; ST = line S T; (T) = circle_center_point T B; B1 C1 = intersection ST (T); Prove: similar A B C A B1 C1
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2008ARMOg10p6
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In a scalene triangle $ ABC$ the altitudes $ AA_{1}$ and $ CC_{1}$ intersect at $ H, O$ is the circumcenter, and $ B_{0}$ the midpoint of side $ AC$. The line $ BO$ intersects side $ AC$ at $ P$, while the lines $ BH$ and $ A_{1}C_{1}$ meet at $ Q$. Prove that the lines $ HB_{0}$ and $ PQ$ are parallel.
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a b c = triangle; a1 = foot a b c; c1 = foot c a b; h = orthocenter a b c; o = circumcenter a b c; b0 = midpoint a c; p = on_line b o, on_line a c; q = on_line b h, on_line a1 c1 ? para h b0 p q
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A B C = triangle; BC = line B C; A1 = foot A BC; AB = line A B; C1 = foot C AB; H = orthocenter A B C; O = circumcenter A B C; B0 = midpoint A C; BO = line B O; AC = line A C; P = intersection BO AC; BH = line B H; A1C1 = line A1 C1; Q = intersection BH A1C1; B0H = line B0 H; PQ = line P Q; Prove: parallel B0H PQ
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2008ARMOg9p3
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In a scalene triangle $ ABC, H$ and $ M$ are the orthocenter an centroid respectively. Consider the triangle formed by the lines through $ A,B$ and $ C$ perpendicular to $ AM,BM$ and $ CM$ respectively. Prove that the centroid of this triangle lies on the line $ MH$.
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a b c = triangle; h = orthocenter a b c; x y z m = centroid a b c; p = on_tline b b m, on_tline c c m; q = on_tline c c m, on_tline a a m; r = on_tline a a m, on_tline b b m; x1 y1 z1 m1 = centroid p q r ? coll m1 m h
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A B C = triangle; H = orthocenter A B C; M X Y Z = centroid A B C; AM = line A M; l1 = perpendicular_line A AM; BM = line B M; l2 = perpendicular_line B BM; CM = line C M; l3 = perpendicular_line C CM; P = intersection l2 l3; Q = intersection l1 l3; R = intersection l1 l2; M1 X1 Y1 Z1 = centroid P Q R; Prove: collinear M H M1
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2008AsiaPacificMOp3
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Let $ \Gamma$ be the circumcircle of a triangle $ ABC$. A circle passing through points $ A$ and $ C$ meets the sides $ BC$ and $ BA$ at $ D$ and $ E$, respectively. The lines $ AD$ and $ CE$ meet $ \Gamma$ again at $ G$ and $ H$, respectively. The tangent lines of $ \Gamma$ at $ A$ and $ C$ meet the line $ DE$ at $ L$ and $ M$, respectively. Prove that the lines $ LH$ and $ MG$ meet at $ \Gamma$.
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a b c = triangle; o = circumcenter a b c; o1 = on_bline a c; d = on_line b c, on_circle o1 a; e = on_line b a, on_circle o1 a; g = on_line a d, on_circum a b c; h = on_line c e, on_circum a b c; l = on_line d e, on_tline a o a; m = on_line d e, on_tline c o c; x = on_line l h, on_line m g ? cyclic a b c x
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A B C = triangle; O = circumcenter A B C; l1 = perpendicular_bisector A C; O1 = on_line l1; (O1) = circle_center_point O1 A; BC = line B C; D = intersection (O1) BC; AB = line A B; E = intersection (O1) AB; AD = line A D; (ABC) = circle A B C; G = intersection AD (ABC); CE = line C E; H = intersection CE (ABC); OA = line O A; t1 = perpendicular_line A OA; DE = line D E; L = intersection t1 DE; OC = line O C; t2 = perpendicular_line C OC; M = intersection t2 DE; HL = line H L; GM = line G M; X = intersection HL GM; Prove: concyclic A B C X
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2008CHNSouthEastMOp3
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In $\triangle ABC$, side $BC>AB$. Point $D$ lies on side $AC$ such that $\angle ABD=\angle CBD$. Points $Q,P$ lie on line $BD$ such that $AQ\bot BD$ and $CP\bot BD$. $M,E$ are the midpoints of side $AC$ and $BC$ respectively. Circle $O$ is the circumcircle of $\triangle PQM$ intersecting side $AC$ at $H$. Prove that $O,H,E,M$ lie on a circle
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a b c = triangle; d = on_line a c, angle_bisector a b c; q = foot a b d; p = foot c b d; m = midpoint a c; e = midpoint b c; o = circumcenter p q m; h = on_line a c, on_circle o p ? cyclic o h e m
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A B C = triangle; l1 = angle_bisector A B C; AC = line A C; D = intersection l1 AC; BD = line B D; Q = foot A BD; P = foot C BD; M = midpoint A C; E = midpoint B C; O = circumcenter P Q M; (O) = circle_center_point O P; H = intersection AC (O); Prove: concyclic O H E M
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2008CHNSouthEastMOp6
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Circle $I$ is the incircle of $\triangle ABC$. Circle $I$ is tangent to sides $BC$ and $AC$ at $M,N$ respectively. $E,F$ are midpoints of sides $AB$ and $AC$ respectively. Lines $EF, BI$ intersect at $D$. Show that $M,N,D$ are collinear.
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a b c = triangle; i = incenter a b c; m = foot i b c; n = foot i a c; e = midpoint a b; f = midpoint a c; d = on_line e f, on_line b i ? coll m n d
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A B C = triangle; I = incenter A B C; BC = line B C; M = foot I BC; AC = line A C; N = foot I AC; E = midpoint A B; F = midpoint A C; EF = line E F; BI = line B I; D = intersection EF BI; Prove: collinear M N D
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2008CTSTp13
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Let $ ABC$ be an acute triangle, let $ M,N$ be the midpoints of minor arcs $ \widehat{CA},\widehat{AB}$ of the circumcircle of triangle $ ABC,$ point $ D$ is the midpoint of segment $ MN,$ point $ G$ lies on minor arc $ \widehat{BC}.$ Denote by $ I,I_{1},I_{2}$ the incenters of triangle $ ABC,ABG,ACG$ respectively.Let $ P$ be the second intersection of the circumcircle of triangle $ GI_{1}I_{2}$ with the circumcircle of triangle $ ABC.$ Prove that three points $ D,I,P$ are collinear.
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a b c = triangle; o = circumcenter a b c; m = on_circum a b c, on_bline a c; n = on_circum a b c, on_bline a b; d = midpoint m n; g = on_circum a b c; i = incenter a b c; i1 = incenter a b g; i2 = incenter a c g; o1 = circumcenter g i1 i2; p = on_circum g i1 i2, on_circum a b c ? coll d i p
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A B C = acute_triangle; O = circumcenter A B C; M = midarc_no C A B; N = midarc_no A B C; D = midpoint M N; (ABC) = circle A B C; G = on_circle (ABC); I = incenter A B C; I1 = incenter A B G; I2 = incenter A C G; O1 = circumcenter G I1 I2; (GI1I2) = circle G I1 I2; P = intersection (GI1I2) (ABC); Prove: collinear D I P
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2008CTSTp19
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Let $ ABC$ be a triangle, let $ AB > AC$. Its incircle touches side $ BC$ at point $ E$. Point $ D$ is the second intersection of the incircle with segment $ AE$ (different from $ E$). Point $ F$ (different from $ E$) is taken on segment $ AE$ such that $ CE = CF$. The ray $ CF$ meets $ BD$ at point $ G$. Show that $ CF = FG$.
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a b c = triangle; i = incenter a b c; e = foot i b c; d = on_line a e, on_circle i e; f = on_line a e, on_circle c e; g = on_line c f, on_line b d ? cong c f f g
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A B C = triangle; I = incenter A B C; BC = line B C; E = foot I BC; (I) = circle_center_point I E; AE = line A E; D = intersection AE (I); (C) = circle_center_point C E; F = intersection AE (C); BD = line B D; CF = line C F; G = intersection BD CF; Prove: cong C F F G
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2008CTSTp4
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Let $ ABC$ be a triangle, line $ l$ cuts its sides $ BC,CA,AB$ at $ D,E,F$, respectively. Denote by $ O_{1},O_{2},O_{3}$ the circumcenters of triangle $ AEF,BFD,CDE$, respectively. Prove that the orthocenter of triangle $ O_{1}O_{2}O_{3}$ lies on line $ l$.
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a b c = triangle; d = on_line b c; e = on_line a c; f = on_line a b, on_line d e; o1 = circumcenter a e f; o2 = circumcenter b f d; o3 = circumcenter c d e; h = orthocenter o1 o2 o3 ? coll d e h
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A B C = triangle; l = line; BC = line B C; D = intersection l BC; AC = line A C; E = intersection l AC; AB = line A B; F = intersection l AB; O1 = circumcenter A E F; O2 = circumcenter B F D; O3 = circumcenter C D E; H = orthocenter O1 O2 O3; Prove: collinear D E H
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2008CTSTp7
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Let $ P$ be the the isogonal conjugate of $ Q$ with respect to triangle $ ABC$, and $ P,Q$ are in the interior of triangle $ ABC$. Denote by $ O_{1},O_{2},O_{3}$ the circumcenters of triangle $ PBC,PCA,PAB$, $ O'_{1},O'_{2},O'_{3}$ the circumcenters of triangle $ QBC,QCA,QAB$, $ O$ the circumcenter of triangle $ O_{1}O_{2}O_{3}$, $ O'$ the circumcenter of triangle $ O'_{1}O'_{2}O'_{3}$. Prove that $ OO'$ is parallel to $ PQ$.
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a b c = triangle; q = free; p = on_aline p a c b a q, on_aline p b a c b q; o1 = circumcenter p b c; o2 = circumcenter p c a; o3 = circumcenter p a b; o4 = circumcenter q b c; o5 = circumcenter q c a; o6 = circumcenter q a b; o = circumcenter o1 o2 o3; u = circumcenter o4 o5 o6 ? para o u p q
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A B C = triangle; P = point; l1 = angle_equal1 A B C A P; l2 = angle_equal1 B C A B P; Q = intersection l1 l2; O1 = circumcenter P B C; O2 = circumcenter P C A; O3 = circumcenter P A B; O4 = circumcenter Q B C; O5 = circumcenter Q C A; O6 = circumcenter Q A B; O = circumcenter O1 O2 O3; U = circumcenter O4 O5 O6; OU = line O U; PQ = line P Q; Prove: parallel OU PQ
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2008USAMOp2
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Let $ ABC$ be an acute, scalene triangle, and let $ M$, $ N$, and $ P$ be the midpoints of $ \overline{BC}$, $ \overline{CA}$, and $ \overline{AB}$, respectively. Let the perpendicular bisectors of $ \overline{AB}$ and $ \overline{AC}$ intersect ray $ AM$ in points $ D$ and $ E$ respectively, and let lines $ BD$ and $ CE$ intersect in point $ F$, inside of triangle $ ABC$. Prove that points $ A$, $ N$, $ F$, and $ P$ all lie on one circle.
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a b c = triangle; m = midpoint b c; n = midpoint c a; p = midpoint a b; d = on_bline a b, on_line a m; e = on_bline a c, on_line a m; f = on_line b d, on_line c e ? cyclic a n f p
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A B C = acute_triangle; M = midpoint B C; N = midpoint C A; P = midpoint A B; AB = line A B; l1 = perpendicular_bisector A B; AM = line A M; D = intersection l1 AM; AC = line A C; l2 = perpendicular_bisector A C; E = intersection l2 AM; BD = line B D; CE = line C E; F = intersection BD CE; Prove: concyclic A N F P
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2009ARMOg11p7
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Let be given a parallelogram $ ABCD$ and two points $ A_1$, $ C_1$ on its sides $ AB$, $ BC$, respectively. Lines $ AC_1$ and $ CA_1$ meet at $ P$. Assume that the circumcircles of triangles $ AA_1P$ and $ CC_1P$ intersect at the second point $ Q$ inside triangle $ ACD$. Prove that $ \angle PDA = \angle QBA$.
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a b c = triangle; d = parallelogram a b c d; a1 = on_line a b; c1 = on_line b c; p = on_line a c1, on_line c a1; q = on_circum a a1 p, on_circum c c1 p ? eqangle p d d a a b b q
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A B C D = parallelogram; AB = line A B; A1 = on_line AB; BC = line B C; C1 = on_line BC; AC1 = line A C1; A1C = line A1 C; P = intersection AC1 A1C; (AA1P) = circle A A1 P; (CC1P) = circle C C1 P; Q = intersection (AA1P) (CC1P); Prove: equal_angle P D A A B Q
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2009ARMOg9p2
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Let be given a triangle $ ABC$ and its internal angle bisector $ BD$ $ (D\in BC)$. The line $ BD$ intersects the circumcircle $ \Omega$ of triangle $ ABC$ at $ B$ and $ E$. Circle $ \omega$ with diameter $ DE$ cuts $ \Omega$ again at $ F$. Prove that $ BF$ is the symmedian line of triangle $ ABC$.
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a b c = triangle; d = on_line a c, angle_bisector a b c; o = circumcenter a b c; e = on_line b d, on_circle o a; f = on_circle o a, on_dia d e; m = midpoint a c; f1 = reflect f b d ? coll b f1 m
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A B C = triangle; l1 = angle_bisector A B C; AC = line A C; D = intersection l1 AC; O = circumcenter A B C; (O) = circle_center_point O A; BD = line B D; E = intersection BD (O); (DE) = circle_diameter D E; F = intersection (DE) (O); F1 = reflect_point_wrt_line F BD; M = midpoint A C; Prove: collinear M B F1
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2009CGMOp2
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Right triangle $ ABC,$ with $ \angle A=90^{\circ},$ is inscribed in circle $ \Gamma.$ Point $ E$ lies on the interior of arc $ {BC}$ (not containing $ A$) with $ EA>EC.$ Point $ F$ lies on ray $ EC$ with $ \angle EAC = \angle CAF.$ Segment $ BF$ meets $ \Gamma$ again at $ D$ (other than $ B$). Let $ O$ denote the circumcenter of triangle $ DEF.$ Prove that $ A,C,O$ are collinear.
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a b c = r_triangle; o = midpoint b c; e = on_circle o a; f = on_line c e, on_aline f a c c a e; d = on_line b f, on_circle o a; o1 = circumcenter d e f ? coll a c o1
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A B C = right_triangle; O = midpoint B C; (O) = circle_center_point O A; E = on_circle (O); CE = line C E; l = angle_equal1 A C C A E; F = intersection CE l; BF = line B F; D = intersection BF (O); O1 = circumcenter D E F; Prove: collinear A C O1
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2009CTSTp19
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Let $ ABC$ be a triangle. Point $ D$ lies on its sideline $ BC$ such that $ \angle CAD = \angle CBA.$ Circle $ (O)$ passing through $ B,D$ intersects $ AB,AD$ at $ E,F$, respectively. $ BF$ meets $ DE$ at $ G$.Denote by$ M$ the midpoint of $ AG.$ Show that $ CM\perp AO.$
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a b c = triangle; d = on_line b c, on_aline d a c c b a; e = on_line a b; o = circumcenter b d e; f = on_line a d, on_circle o b; g = on_line b f, on_line d e; m = midpoint a g ? perp c m a o
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A B C = triangle; BC = line B C; l1 = angle_equal1 A C C B A; D = intersection BC l1; AB = line A B; E = on_line AB; O = circumcenter B D E; (O) = circle_center_point O B; AD = line A D; F = intersection AD (O); BF = line B F; DE = line D E; G = intersection BF DE; M = midpoint A G; CM = line C M; AO = line A O; Prove: perpendicular CM AO
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2009G3
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Let $ABC$ be a triangle. The incircle of $ABC$ touches the sides $AB$ and $AC$ at the points $Z$ and $Y$, respectively. Let $G$ be the point where the lines $BY$ and $CZ$ meet, and let $R$ and $S$ be points such that the two quadrilaterals $BCYR$ and $BCSZ$ are parallelogram. Prove that $GR=GS$.
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a b c = triangle; i = incenter a b c; z = foot i a b; y = foot i a c; g = on_line b y, on_line c z; r = parallelogram y c b r; s = parallelogram z b c s ? cong g r g s
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A B C = triangle; I = incenter A B C; AB = line A B; Z = foot I AB; AC = line A C; Y = foot I AC; BY = line B Y; CZ = line C Z; G = intersection BY CZ; R = parallelogram B C Y; S = parallelogram Z B C; Prove: cong G R G S
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2009G4
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Given a cyclic quadrilateral $ABCD$, let the diagonals $AC$ and $BD$ meet at $E$ and the lines $AD$ and $BC$ meet at $F$. The midpoints of $AB$ and $CD$ are $G$ and $H$, respectively. Show that $EF$ is tangent at $E$ to the circle through the points $E$, $G$ and $H$.
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a b c = triangle; o = circumcenter a b c; d = on_circle o a; e = on_line a c, on_line b d; f = on_line a d, on_line b c; g = midpoint a b; h = midpoint c d; o1 = circumcenter e g h ? perp o1 e e f
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A B C = triangle; O = circumcenter A B C; (O) = circle_center_point O A; D = on_circle (O); AC = line A C; BD = line B D; E = intersection AC BD; AD = line A D; BC = line B C; F = intersection AD BC; G = midpoint A B; H = midpoint C D; EF = line E F; O1 = circumcenter E G H; O1E = line O1 E; Prove: perpendicular O1E EF
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2009G6
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Let the sides $AD$ and $BC$ of the quadrilateral $ABCD$ (such that $AB$ is not parallel to $CD$) intersect at point $P$. Points $O_1$ and $O_2$ are circumcenters and points $H_1$ and $H_2$ are orthocenters of triangles $ABP$ and $CDP$, respectively. Denote the midpoints of segments $O_1H_1$ and $O_2H_2$ by $E_1$ and $E_2$, respectively. Prove that the perpendicular from $E_1$ on $CD$, the perpendicular from $E_2$ on $AB$ and the lines $H_1H_2$ are concurrent.
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a b c d = quadrangle; p = on_line a d, on_line b c; o1 = circumcenter a b p; h1 = orthocenter a b p; o2 = circumcenter c d p; h2 = orthocenter c d p; e1 = midpoint o1 h1; e2 = midpoint o2 h2; x = on_tline e1 c d, on_tline e2 a b ? coll x h1 h2
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A B C D = quadrilateral; AD = line A D; BC = line B C; P = intersection AD BC; O1 = circumcenter A B P; H1 = orthocenter A B P; O2 = circumcenter C D P; H2 = orthocenter C D P; E1 = midpoint O1 H1; E2 = midpoint O2 H2; CD = line C D; l1 = perpendicular_line E1 CD; AB = line A B; l2 = perpendicular_line E2 AB; X = intersection l1 l2; Prove: collinear X H1 H2
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2009G8
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Let $ABCD$ be a circumscribed quadrilateral. Let $g$ be a line through $A$ which meets the segment $BC$ in $M$ and the line $CD$ in $N$. Denote by $I_1$, $I_2$ and $I_3$ the incenters of $\triangle ABM$, $\triangle MNC$ and $\triangle NDA$, respectively. Prove that the orthocenter of $\triangle I_1I_2I_3$ lies on $g$.
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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; m = on_line b c; n = on_line a m, on_line c d; i1 = incenter a b m; i2 = incenter m n c; i3 = incenter n d a; h = orthocenter i1 i2 i3 ? coll h a m
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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; BC = line B C; M = on_line BC; AM = line A M; CD = line C D; N = intersection AM CD; I1 = incenter A B M; I2 = incenter M N C; I3 = incenter N D A; H = orthocenter I1 I2 I3; Prove: collinear H A M
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2010ARMOg10p3
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Let $O$ be the circumcentre of the acute non-isosceles triangle $ABC$. Let $P$ and $Q$ be points on the altitude $AD$ such that $OP$ and $OQ$ are perpendicular to $AB$ and $AC$ respectively. Let $M$ be the midpoint of $BC$ and $S$ be the circumcentre of triangle $OPQ$. Prove that $\angle BAS =\angle CAM$.
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a b c = triangle; o = circumcenter a b c; d = foot a b c; p = on_line a d, on_tline o a b; q = on_line a d, on_tline o a c; m = midpoint b c; s = circumcenter o p q ? eqangle b a a s m a a c
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A B C = acute_triangle; O = circumcenter A B C; BC = line B C; D = foot A BC; AB = line A B; l1 = perpendicular_line O AB; AD = line A D; P = intersection l1 AD; AC = line A C; l2 = perpendicular_line O AC; Q = intersection l2 AD; M = midpoint B C; S = circumcenter O P Q; Prove: equal_angle B A S M A C
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2010ARMOg10p6
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Into triangle $ABC$ gives point $K$ lies on bisector of $ \angle BAC$. Line $CK$ intersect circumcircle $ \omega$ of triangle $ABC$ at $M \neq C$. Circle $ \Omega$ passes through $A$, touch $CM$ at $K$ and intersect segment $AB$ at $P \neq A$ and $\omega $ at $Q \neq A$. Prove, that $P$, $Q$, $M$ lies at one line.
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a b c = triangle; k = angle_bisector b a c; m = on_line c k, on_circum a b c; o = on_tline k c m, on_bline a k; p = on_line a b, on_circle o a; q = on_circle o a, on_circum a b c ? coll p q m
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A B C = triangle; la = angle_bisector B A C; K = on_line la; CK = line C K; (ABC) = circle A B C; M = intersection CK (ABC); CM = line C M; l1 = perpendicular_line K CM; l2 = perpendicular_bisector A K; O = intersection l1 l2; (O) = circle_center_point O A; AB = line A B; P = intersection AB (O); Q = intersection (ABC) (O); Prove: collinear P Q M
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2010ARMOg11p3
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Quadrilateral $ABCD$ is inscribed into circle $\omega$, $AC$ intersect $BD$ in point $K$. Points $M_1$, $M_2$, $M_3$, $M_4$-midpoints of arcs $AB$, $BC$, $CD$, and $DA$ respectively. Points $I_1$, $I_2$, $I_3$, $I_4$-incenters of triangles $ABK$, $BCK$, $CDK$, and $DAK$ respectively. Prove that lines $M_1I_1$, $M_2I_2$, $M_3I_3$, and $M_4I_4$ all intersect in one point.
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a b c = triangle; d = on_circum a b c; o = circumcenter a b c; k = on_line a c, on_line b d; m1 = on_bline a b, on_circle o a; m2 = on_bline b c, on_circle o a; m3 = on_bline c d, on_circle o a; m4 = on_bline d a, on_circle o a; i1 = incenter a b k; i2 = incenter b c k; i3 = incenter c d k; i4 = incenter d a k; x = on_line m1 i1, on_line m2 i2 ? coll x m3 i3
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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; K = intersection AC BD; M1 = midarc_no A B C; M2 = midarc_no B C D; M3 = midarc_no C D A; M4 = midarc_no D A B; I1 = incenter A B K; I2 = incenter B C K; I3 = incenter C D K; I4 = incenter D A K; I1M1 = line I1 M1; I2M2 = line I2 M2; X = intersection I1M1 I2M2; Prove: collinear X M3 I3
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2010AsiaPacificMOp1
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Let $ABC$ be a triangle with $\angle BAC \neq 90^{\circ}.$ Let $O$ be the circumcenter of the triangle $ABC$ and $\Gamma$ be the circumcircle of the triangle $BOC.$ Suppose that $\Gamma$ intersects the line segment $AB$ at $P$ different from $B$, and the line segment $AC$ at $Q$ different from $C.$ Let $ON$ be the diameter of the circle $\Gamma.$ Prove that the quadrilateral $APNQ$ is a parallelogram.
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a b c = triangle; o = circumcenter a b c; o1 = circumcenter o b c; p = on_line a b, on_circle o1 o; q = on_line a c, on_circle o1 o; n = mirror o o1 ? para a p n q
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A B C = triangle; O = circumcenter A B C; O1 = circumcenter O B C; (O1) = circle_center_point O1 O; AB = line A B; P = intersection AB (O1); AC = line A C; Q = intersection AC (O1); N = reflect_point_wrt_point O O1; AP = line A P; NQ = line N Q; Prove: parallel AP NQ
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2010AsiaPacificMOp4
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Let $ABC$ be an acute angled triangle satisfying the conditions $AB>BC$ and $AC>BC$. Denote by $O$ and $H$ the circumcentre and orthocentre, respectively, of the triangle $ABC.$ Suppose that the circumcircle of the triangle $AHC$ intersects the line $AB$ at $M$ different from $A$, and the circumcircle of the triangle $AHB$ intersects the line $AC$ at $N$ different from $A.$ Prove that the circumcentre of the triangle $MNH$ lies on the line $OH$.
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a b c = triangle; o = circumcenter a b c; h = orthocenter a b c; o1 = circumcenter a h c; m = on_line a b, on_circle o1 a; o2 = circumcenter a h b; n = on_line a c, on_circle o2 a; o3 = circumcenter m n h ? coll o3 o h
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A B C = acute_triangle; O = circumcenter A B C; H = orthocenter A B C; AB = line A B; O1 = circumcenter A C H; (O1) = circle_center_point O1 A; M = intersection AB (O1); AC = line A C; O2 = circumcenter A B H; (O2) = circle_center_point O2 A; N = intersection AC (O2); P = circumcenter M N H; Prove: collinear P H O
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2010CGMOp2
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In triangle $ABC$, $AB = AC$. Point $D$ is the midpoint of side $BC$. Point $E$ lies outside the triangle $ABC$ such that $CE \perp AB$ and $BE = BD$. Let $M$ be the midpoint of segment $BE$. Point $F$ lies on the minor arc $\widehat{AD}$ of the circumcircle of triangle $ABD$ such that $MF \perp BE$. Prove that $ED \perp FD.$
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a b c = iso_triangle; d = midpoint b c; e = on_tline c a b, on_circle b d; m = midpoint b e; o = circumcenter a b d; f = on_circle o a, on_tline m b e ? perp e d f d
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A B C = isos_triangle; D = midpoint B C; AB = line A B; l1 = perpendicular_line C AB; (B,D) = circle_center_point B D; E = intersection l1 (B,D); M = midpoint B E; BE = line B E; l2 = perpendicular_line M BE; O1 = circumcenter A B D; (ABD) = circle_center_point O1 A; F = intersection (ABD) l2; DE = line D E; DF = line D F; Prove: perpendicular DE DF
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2010CHNWesternMOp2
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$AB$ is a diameter of a circle with center $O$. Let $C$ and $D$ be two different points on the circle on the same side of $AB$, and the lines tangent to the circle at points $C$ and $D$ meet at $E$. Segments $AD$ and $BC$ meet at $F$. Lines $EF$ and $AB$ meet at $M$. Prove that $E,C,M$ and $D$ are concyclic.
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a b = segment; o = midpoint a b; c = on_circle o a; d = on_circle o a; e = on_tline c o c, on_tline d o d; f = on_line a d, on_line b c; m = on_line e f, on_line a b ? cyclic e c m d
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A = point; B = point; O = midpoint A B; (O) = circle_center_point O A; C = on_circle (O); D = on_circle (O); l1 = tangent_line C (O); l2 = tangent_line D (O); E = intersection l1 l2; AD = line A D; BC = line B C; F = intersection AD BC; EF = line E F; AB = line A B; M = intersection EF AB; Prove: concyclic E C M D
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2010CHNWesternMOp6
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$\Delta ABC$ is a right-angled triangle, $\angle C = 90^{\circ}$. Draw a circle centered at $B$ with radius $BC$. Let $D$ be a point on the side $AC$, and $DE$ is tangent to the circle at $E$. The line through $C$ perpendicular to $AB$ meets line $BE$ at $F$. Line $AF$ meets $DE$ at point $G$. The line through $A$ parallel to $BG$ meets $DE$ at $H$. Prove that $GE = GH$.
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c a b = r_triangle; e = on_circle b c; d = on_line a c, on_tline e b e; f = on_line b e, on_tline c a b; g = on_line a f, on_line d e; h = on_line d e, on_pline a b g ? midp g e h
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C A B = right_triangle; (B) = circle_center_point B C; E = on_circle (B); AC = line A C; BE = line B E; l1 = perpendicular_line E BE; D = intersection AC l1; AB = line A B; l2 = perpendicular_line C AB; BE = line B E; F = intersection l2 BE; AF = line A F; DE = line D E; G = intersection AF DE; BG = line B G; l3 = parallel_line A BG; H = intersection l3 DE; Prove: midpoint G E H
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2010CTSTp19
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Given acute triangle $ABC$ with $AB>AC$, let $M$ be the midpoint of $BC$. $P$ is a point in triangle $AMC$ such that $\angle MAB=\angle PAC$. Let $O,O_1,O_2$ be the circumcenters of $\triangle ABC,\triangle ABP,\triangle ACP$ respectively. Prove that line $AO$ passes through the midpoint of $O_1 O_2$.
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a b c = triangle; m = midpoint b c; p = on_aline p a c b a m; o = circumcenter a b c; o1 = circumcenter a b p; o2 = circumcenter a c p; n = midpoint o1 o2 ? coll a o n
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A B C = acute_triangle; M = midpoint B C; l1 = angle_equal1 A C B A M; P = on_line l1; O = circumcenter A B C; O1 = circumcenter A B P; O2 = circumcenter A C P; N = midpoint O1 O2; Prove: collinear A O N
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2010CTSTp4
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Let $\triangle ABC$ be an acute triangle with $AB>AC$, let $I$ be the center of the incircle. Let $M,N$ be the midpoint of $AC$ and $AB$ respectively. $D,E$ are on $AC$ and $AB$ respectively such that $BD\parallel IM$ and $CE\parallel IN$. A line through $I$ parallel to $DE$ intersects $BC$ in $P$. Let $Q$ be the projection of $P$ on line $AI$. Prove that $Q$ is on the circumcircle of $\triangle ABC$.
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a b c = triangle; i = incenter a b c; m = midpoint a c; n = midpoint a b; d = on_line a c, on_pline b i m; e = on_line a b, on_pline c i n; p = on_line b c, on_pline i d e; q = foot p a i ? cyclic q a b c
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A B C = acute_triangle; I = incenter A B C; M = midpoint A C; N = midpoint A B; IM = line I M; l1 = parallel_line B IM; AC = line A C; D = intersection AC l1; IN = line I N; l2 = parallel_line C IN; AB = line A B; E = intersection AB l2; DE = line D E; DE = parallel_line I DE; l3 = parallel_line I DE; BC = line B C; P = intersection l3 BC; AI = line A I; Q = foot P AI; Prove: concyclic Q A B C
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2010G1
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Let $ABC$ be an acute triangle with $D, E, F$ the feet of the altitudes lying on $BC, CA, AB$ respectively. One of the intersection points of the line $EF$ and the circumcircle is $P.$ The lines $BP$ and $DF$ meet at point $Q.$ Prove that $AP = AQ.$
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a b c = triangle; d = foot a b c; e = foot b a c; f = foot c a b; o = circumcenter a b c; p = on_line e f, on_circle o a; q = on_line b p, on_line d f ? cong a p a q
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A B C = acute_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; O = circumcenter A B C; (O) = circle_center_point O A; P = intersection EF (O); BP = line B P; DF = line D F; Q = intersection BP DF; Prove: cong A P A Q
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2011ARMOg10p6
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Given is an acute triangle $ABC$. Its heights $BB_1$ and $CC_1$ are extended past points $B_1$ and $C_1$. On these extensions, points $P$ and $Q$ are chosen, such that angle $PAQ$ is right. Let $AF$ be a height of triangle $APQ$. Prove that angle $BFC$ is a right angle.
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a b c = triangle; b1 = foot b a c; c1 = foot c a b; p = on_line b b1; q = on_line c c1, on_tline a a p; f = foot a p q ? perp b f f c
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A B C = acute_triangle; AB = line A B; AC = line A C; B1 = foot B AC; C1 = foot C AB; BB1 = line B B1; P = on_line BB1; AP = line A P; l = perpendicular_line A AP; CC1 = line C C1; Q = intersection CC1 l; PQ = line P Q; F = foot A PQ; BF = line B F; FC = line F C; Prove: perpendicular BF FC
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2011ARMOg11p2
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On side $BC$ of parallelogram $ABCD$ ($A$ is acute) lies point $T$ so that triangle $ATD$ is an acute triangle. Let $O_1$, $O_2$, and $O_3$ be the circumcenters of triangles $ABT$, $DAT$, and $CDT$ respectively. Prove that the orthocenter of triangle $O_1O_2O_3$ lies on line $AD$.
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a b c = triangle; d = parallelogram a b c d; t = on_line b c; o1 = circumcenter a b t; o2 = circumcenter d a t; o3 = circumcenter c d t; h = orthocenter o1 o2 o3 ? coll a d h
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A B C D = parallelogram; BC = line B C; T = on_line BC; O1 = circumcenter A B T; O2 = circumcenter D A T; O3 = circumcenter C D T; H = orthocenter O1 O2 O3; Prove: collinear A D H
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2011ARMOg11p8
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Let $N$ be the midpoint of arc $ABC$ of the circumcircle of triangle $ABC$, let $M$ be the midpoint of $AC$ and let $I_1, I_2$ be the incentres of triangles $ABM$ and $CBM$. Prove that points $I_1, I_2, B, N$ lie on a circle.
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a b c = triangle; o = circumcenter a b c; n = on_circle o a, on_bline a c; m = midpoint a c; i1 = incenter a b m; i2 = incenter c b m ? cyclic i1 i2 b n
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A B C = triangle; O = circumcenter A B C; N = midarc A B C; M = midpoint A C; I1 = incenter A B M; I2 = incenter C B M; Prove: concyclic I1 I2 B N
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2011ARMOg9p2
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Given is an acute angled triangle $ABC$. A circle going through $B$ and the triangle's circumcenter, $O$, intersects $BC$ and $BA$ at points $P$ and $Q$ respectively. Prove that the intersection of the heights of the triangle $POQ$ lies on line $AC$.
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a b c = triangle; o = circumcenter a b c; k = on_bline b o; p = on_line b c, on_circle k b; q = on_line b a, on_circle k b; t = orthocenter p o q ? coll a c t
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A B C = acute_triangle; O = circumcenter A B C; l = perpendicular_bisector O B; K = on_line l; (K) = circle_center_point K O; BC = line B C; P = intersection (K) BC; AB = line A B; Q = intersection (K) AB; H = orthocenter P O Q; Prove: collinear H A C
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2011AsiaPacificMOp3
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Let $ABC$ be an acute triangle with $\angle BAC=30^{\circ}$. The internal and external angle bisectors of $\angle ABC$ meet the line $AC$ at $B_1$ and $B_2$, respectively, and the internal and external angle bisectors of $\angle ACB$ meet the line $AB$ at $C_1$ and $C_2$, respectively. Suppose that the circles with diameters $B_1B_2$ and $C_1C_2$ meet inside the triangle $ABC$ at point $P$. Prove that $\angle BPC=90^{\circ}$ .
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o b c = ieq_triangle; a = on_circle o b; b1 = on_line a c, angle_bisector a b c; b2 = on_line a c, on_tline b b b1; c1 = on_line a b, angle_bisector a c b; c2 = on_line a b, on_tline c c c1; m1 = midpoint b1 b2; m2 = midpoint c1 c2; p = on_circle m1 b1, on_circle m2 c1 ? perp b p c p
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O B C = equilateral_triangle; (O) = circle_center_point O B; A = on_circle (O); l1 = angle_bisector A B C; AC = line A C; B1 = intersection l1 AC; l2 = angle_exbisector A B C; B2 = intersection l2 AC; l3 = angle_bisector A C B; AB = line A B; C1 = intersection l3 AB; l4 = angle_exbisector A C B; C2 = intersection l4 AB; M1 = midpoint B1 B2; (M1) = circle_center_point M1 B1; M2 = midpoint C1 C2; (M2) = circle_center_point M2 C1; P = intersection (M1) (M2); BP = line B P; CP = line C P; Prove: perpendicular BP CP
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2011BalkanMOp1
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Let $ABCD$ be a cyclic quadrilateral which is not a trapezoid and whose diagonals meet at $E$. The midpoints of $AB$ and $CD$ are $F$ and $G$ respectively, and $\ell$ is the line through $G$ parallel to $AB$. The feet of the perpendiculars from E onto the lines $\ell$ and $CD$ are $H$ and $K$, respectively. Prove that the lines $EF$ and $HK$ are perpendicular.
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a b c = triangle; d = on_circum a b c; e = on_line a c, on_line b d; f = midpoint a b; g = midpoint c d; s = on_pline g a b; h = foot e g s; k = foot e c d ? perp e f h k
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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; F = midpoint A B; G = midpoint C D; AB = line A B; l = parallel_line G AB; H = foot E l; CD = line C D; K = foot E CD; EF = line E F; HK = line H K; Prove: perpendicular EF HK
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2011CGMOp8
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The $A$-excircle $(O)$ of $\triangle ABC$ touches $BC$ at $M$. The points $D,E$ lie on the sides $AB,AC$ respectively such that $DE\parallel BC$. The incircle $(O_1)$ of $\triangle ADE$ touches $DE$ at $N$. If $BO_1\cap DO=F$ and $CO_1\cap EO=G$, prove that the midpoint of $FG$ lies on $MN$.
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a b c = triangle; o = excenter a b c; m = foot o b c; d = on_line a b; e = on_line a c, on_pline d b c; o1 = incenter a d e; n = foot o1 d e; f = on_line b o1, on_line d o; g = on_line c o1, on_line e o; m1 = midpoint f g ? coll m1 m n
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A B C = triangle; O = excenter A B C; BC = line B C; M = foot O BC; AB = line A B; D = on_line AB; l1 = parallel_line D BC; AC = line A C; E = intersection AC l1; O1 = incenter A D E; DE = line D E; N = foot O1 DE; BO1 = line B O1; DO = line D O; F = intersection BO1 DO; CO1 = line C O1; EO = line E O; G = intersection CO1 EO; L = midpoint F G; Prove: collinear L M N
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2011CHNSouthEastMOp4
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Let $O$ be the circumcenter of triangle $ABC$ , a line passes through $O$ intersects sides $AB,AC$ at points $M,N$ , $E$ is the midpoint of $MC$ , $F$ is the midpoint of $NB$ , prove that : $\angle FOE= \angle BAC$
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a b c = triangle; o = circumcenter a b c; m = on_line a b; n = on_line a c, on_line o m; e = midpoint n b; f = midpoint m c ? eqangle f o o e c a a b
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A B C = triangle; O = circumcenter A B C; AB = line A B; M = on_line AB; AC = line A C; OM = line O M; N = intersection OM AC; E = midpoint M C; F = midpoint N B; Prove: equal_angle F O E B A C
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2011CHNWesternMOp7
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In triangle $ABC$ with $AB>AC$ and incenter $I$, the incircle touches $BC,CA,AB$ at $D,E,F$ respectively. $M$ is the midpoint of $BC$, and the altitude at $A$ meets $BC$ at $H$. Ray $AI$ meets lines $DE$ and $DF$ at $K$ and $L$, respectively. Prove that the points $M,L,H,K$ are concyclic.
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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; m = midpoint b c; h = foot a b c; k = on_line a i, on_line d e; l = on_line a i, on_line d f ? cyclic m l h 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; M = midpoint B C; H = foot A BC; AI = line A I; DE = line D E; K = intersection AI DE; DF = line D F; L = intersection AI DF; Prove: concyclic M L H K
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2011CTSTp10
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Let $AA',BB',CC'$ be three diameters of the circumcircle of an acute triangle $ABC$. Let $P$ be an arbitrary point in the interior of $\triangle ABC$, and let $D,E,F$ be the orthogonal projection of $P$ on $BC,CA,AB$, respectively. Let $X$ be the point such that $D$ is the midpoint of $A'X$, let $Y$ be the point such that $E$ is the midpoint of $B'Y$, and similarly let $Z$ be the point such that $F$ is the midpoint of $C'Z$. Prove that triangle $XYZ$ is similar to triangle $ABC$.
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a b c = triangle; o = circumcenter a b c; a1 = mirror a o; b1 = mirror b o; c1 = mirror c o; p = free; d = foot p b c; e = foot p a c; f = foot p a b; x = mirror a1 d; y = mirror b1 e; z = mirror c1 f ? simtri x y z a b c
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A B C = acute_triangle; O = circumcenter A B C; A1 = reflect_point_wrt_point A O; B1 = reflect_point_wrt_point B O; C1 = reflect_point_wrt_point C O; P = point; BC = line B C; D = foot P BC; AC = line A C; E = foot P AC; AB = line A B; F = foot P AB; X = reflect A1 D; Y = reflect B1 E; Z = reflect C1 F; Prove: similar X Y Z A B C
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2011CTSTp16
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Let $H$ be the orthocenter of an acute trangle $ABC$ with circumcircle $\Gamma$. Let $P$ be a point on the arc $BC$ (not containing $A$) of $\Gamma$, and let $M$ be a point on the arc $CA$ (not containing $B$) of $\Gamma$ such that $H$ lies on the segment $PM$. Let $K$ be another point on $\Gamma$ such that $KM$ is parallel to the Simson line of $P$ with respect to triangle $ABC$. Let $Q$ be another point on $\Gamma$ such that $PQ \parallel BC$. Segments $BC$ and $KQ$ intersect at a point $J$. Prove that $\triangle KJM$ is an isosceles triangle.
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a b c = triangle; h = orthocenter a b c; o = circumcenter a b c; p = on_circle o a; m = on_line p h, on_circle o a; p1 = foot p b c; p2 = foot p c a; k = on_circle o a, on_pline m p1 p2; q = on_circle o a, on_pline p b c; j = on_line b c, on_line k q ? cong k j m j
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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; P = on_circle (O); HP = line H P; M = intersection (O) HP; BC = line B C; P1 = foot P BC; AC = line A C; P2 = foot P AC; P1P2 = line P1 P2; l1 = parallel_line M P1P2; K = intersection (O) l1; l2 = parallel_line P BC; Q = intersection (O) l2; KQ = line K Q; J = intersection BC KQ; Prove: cong J K J M
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2011G3
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Let $ABCD$ be a convex quadrilateral whose sides $AD$ and $BC$ are not parallel. Suppose that the circles with diameters $AB$ and $CD$ meet at points $E$ and $F$ inside the quadrilateral. Let $\omega_E$ be the circle through the feet of the perpendiculars from $E$ to the lines $AB,BC$ and $CD$. Let $\omega_F$ be the circle through the feet of the perpendiculars from $F$ to the lines $CD,DA$ and $AB$. Prove that the midpoint of the segment $EF$ lies on the line through the two intersections of $\omega_E$ and $\omega_F$.
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a b c d = quadrangle; m1 = midpoint a b; m2 = midpoint c d; e = on_circle m1 a, on_circle m2 c; f = on_circle m1 a, on_circle m2 c; e1 = foot e a b; e2 = foot e b c; e3 = foot e c d; o_e = circumcenter e1 e2 e3; f1 = foot f c d; f2 = foot f d a; f3 = foot f a b; o_f = circumcenter f1 f2 f3; k1 = on_circle o_e e1, on_circle o_f f1; k2 = on_circle o_e e1, on_circle o_f f1; m = midpoint e f ? coll m k1 k2
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A B C D = quadrilateral; M1 = midpoint A B; M2 = midpoint C D; (AB) = circle_center_point M1 A; (CD) = circle_center_point M2 C; E F = intersection (AB) (CD); AB = line A B; P = foot E AB; BC = line B C; Q = foot E BC; CD = line C D; R = foot E CD; S = foot F CD; AD = line A D; T = foot F AD; U = foot F AB; O1 = circumcenter P Q R; (PQR) = circle_center_point O1 P; O2 = circumcenter S T U; (STU) = circle_center_point O2 S; K1 K2 = intersection (PQR) (STU); M = midpoint E F; Prove: collinear M K1 K2
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2011G5
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Let $ABC$ be a triangle with incentre $I$ and circumcircle $\omega$. Let $D$ and $E$ be the second intersection points of $\omega$ with $AI$ and $BI$, respectively. The chord $DE$ meets $AC$ at a point $F$, and $BC$ at a point $G$. Let $P$ be the intersection point of the line through $F$ parallel to $AD$ and the line through $G$ parallel to $BE$. Suppose that the tangents to $\omega$ at $A$ and $B$ meet at a point $K$. Prove that the three lines $AE,BD$ and $KP$ are either parallel or concurrent.
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a b c = triangle; i = incenter a b c; o = circumcenter a b c; d = on_line a i, on_circle o a; e = on_line b i, on_circle o b; f = on_line d e, on_line a c; g = on_line d e, on_line b c; p = on_pline f a d, on_pline g b e; k = on_tline a o a, on_tline b o b; x = on_line a e, on_line b d ? coll x k p
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A B C = triangle; I = incenter A B C; O = circumcenter A B C; (O) = circle_center_point O A; AI = line A I; D = intersection (O) AI; BI = line B I; E = intersection (O) BI; DE = line D E; AC = line A C; F = intersection DE AC; BC = line B C; G = intersection DE BC; AD = line A D; l1 = parallel_line F AD; BE = line B E; l2 = parallel_line G BE; P = intersection l1 l2; OA = line O A; OB = line O B; l3 = perpendicular_line A OA; l4 = perpendicular_line B OB; K = intersection l3 l4; AE = line A E; BD = line B D; X = intersection AE BD; Prove: collinear X K P
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2011G6
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Let $ABC$ be a triangle with $AB=AC$ and let $D$ be the midpoint of $AC$. The angle bisector of $\angle BAC$ intersects the circle through $D,B$ and $C$ at the point $E$ inside the triangle $ABC$. The line $BD$ intersects the circle through $A,E$ and $B$ in two points $B$ and $F$. The lines $AF$ and $BE$ meet at a point $I$, and the lines $CI$ and $BD$ meet at a point $K$. Show that $I$ is the incentre of triangle $KAB$.
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a b c = iso_triangle; d = midpoint a c; o1 = circumcenter d b c; e = on_circle o1 d, angle_bisector b a c; o2 = circumcenter a e b; f = on_line b d, on_circle o2 a; i = on_line a f, on_line b e; k = on_line c i, on_line b d ? eqangle b a a i a i a k
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A B C = isos_triangle; D = midpoint A C; l1 = angle_bisector B A C; O1 = circumcenter D B C; (BCD) = circle_center_point O1 D; E = intersection l1 (BCD); BD = line B D; O2 = circumcenter A E B; (ABE) = circle_center_point O2 A; F = intersection BD (ABE); AF = line A F; BE = line B E; I = intersection AF BE; CI = line C I; K = intersection CI BD; Prove: equal_angle B A I I A K
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2012ARMOg10p2
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The inscribed circle $\omega$ of the non-isosceles acute-angled triangle $ABC$ touches the side $BC$ at the point $D$. Suppose that $I$ and $O$ are the centres of inscribed circle and circumcircle of triangle $ABC$ respectively. The circumcircle of triangle $ADI$ intersects $AO$ at the points $A$ and $E$. Prove that $AE$ is equal to the radius $r$ of $\omega$.
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a b c = triangle; i = incenter a b c; d = foot i b c; o = circumcenter a b c; e = on_line a o, on_circum a d i ? cong a e i d
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A B C = acute_triangle; I = incenter A B C; BC = line B C; D = foot I BC; O = circumcenter A B C; AO = line A O; (ADI) = circle A D I; E = intersection AO (ADI); AE = line A E; DI = line D I; Prove: cong A E D I
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2012ARMOg10p8
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The point $E$ is the midpoint of the segment connecting the orthocentre of the scalene triangle $ABC$ and the point $A$. The incircle of triangle $ABC$ incircle is tangent to $AB$ and $AC$ at points $C'$ and $B'$ respectively. Prove that point $F$, the point symmetric to point $E$ with respect to line $B'C'$, lies on the line that passes through both the circumcentre and the incentre of triangle $ABC$.
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a b c = triangle; h = orthocenter a b c; e = midpoint a h; i = incenter a b c; b1 = foot i a c; c1 = foot i a b; f = reflect e b1 c1; o = circumcenter a b c ? coll f o i
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A B C = triangle; H = orthocenter A B C; E = midpoint H A; I = incenter A B C; AB = line A B; B1 = foot I AB; AC = line A C; C1 = foot I AC; B1C1 = line B1 C1; F = reflect_point_wrt_line E B1C1; O = circumcenter A B C; Prove: collinear O I F
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2012ARMOg9p3
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Consider the parallelogram $ABCD$ with obtuse angle $A$. Let $H$ be the feet of perpendicular from $A$ to the side $BC$. The median from $C$ in triangle $ABC$ meets the circumcircle of triangle $ABC$ at the point $K$. Prove that points $K,H,C,D$ lie on the same circle.
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a b c = triangle; d = parallelogram a b c d; h = foot a b c; m = midpoint a b; o = circumcenter a b c; k = on_line c m, on_circle o a ? cyclic k h c d
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A B C D = parallelogram; BC = line B C; H = foot A BC; M = midpoint A B; CM = line C M; O = circumcenter A B C; (ABC) = circle_center_point O A; K = intersection CM (ABC); Prove: concyclic K H C D
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2012AsiaPacificMOp4
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Let $ ABC $ be an acute triangle. Denote by $ D $ the foot of the perpendicular line drawn from the point $ A $ to the side $ BC $, by $M$ the midpoint of $ BC $, and by $ H $ the orthocenter of $ ABC $. Let $ E $ be the point of intersection of the circumcircle $ \Gamma $ of the triangle $ ABC $ and the half line $ MH $, and $ F $ be the point of intersection (other than $E$) of the line $ ED $ and the circle $ \Gamma $. Prove that $ \tfrac{BF}{CF} = \tfrac{AB}{AC} $ must hold.
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a b c = triangle; d = foot a b c; m = midpoint b c; h = orthocenter a b c; e = on_line m h, on_circum a b c; f = on_line e d, on_circum a b c ? eqratio b f f c a b a c
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A B C = acute_triangle; BC = line B C; D = foot A BC; M = midpoint B C; H = orthocenter A B C; HM = line H M; (ABC) = circle A B C; E = intersection HM (ABC); DE = line D E; F = intersection DE (ABC); Prove: equal_ratio B F C F A B A C
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2012CGMOp5
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As shown in the figure below, the in-circle of $ABC$ is tangent to sides $AB$ and $AC$ at $D$ and $E$ respectively, and $O$ is the circumcenter of $BCI$. Prove that $\angle ODB = \angle OEC$.
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a b c = triangle; i = incenter a b c; d = foot i a b; e = foot i a c; o = circumcenter b c i ? eqangle o d d b c e e o
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A B C = triangle; I = incenter A B C; AB = line A B; D = foot I AB; AC = line A C; E = foot I AC; O = circumcenter B C I; Prove: equal_angle O D B C E O
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2012CHNGaoLian
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In an acute-angled triangle $ABC$, $AB>AC$. $M,N$ are distinct points on side $BC$ such that $\angle BAM=\angle CAN$. Let $O_1,O_2$ be the circumcentres of $\triangle ABC, \triangle AMN$, respectively. Prove that $O_1,O_2,A$ are collinear.
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a b c = triangle; m = on_line b c; n = on_line b c, on_aline n a c b a m; o1 = circumcenter a b c; o2 = circumcenter a m n ? coll o1 o2 a
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A B C = acute_triangle; BC = line B C; M = on_line BC; l = angle_equal1 A C B A M; N = intersection BC l; O1 = circumcenter A B C; O2 = circumcenter A M N; Prove: collinear O1 O2 A
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2012CHNSouthEastMOp2
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The incircle $I$ of $\triangle ABC$ is tangent to sides $AB,BC,CA$ at $D,E,F$ respectively. Line $EF$ intersects lines $AI,BI,DI$ at $M,N,K$ respectively. Prove that $DM\cdot KE=DN\cdot KF$.
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a b c = triangle; i = incenter a b c; d = foot i a b; e = foot i b c; f = foot i a c; m = on_line e f, on_line a i; n = on_line e f, on_line b i; k = on_line e f, on_line d i ? eqratio d m d n k f k e
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A B C = triangle; I = incenter A B C; AB = line A B; D = foot I AB; BC = line B C; E = foot I BC; AC = line A C; F = foot I AC; EF = line E F; AI = line A I; M = intersection EF AI; BI = line B I; N = intersection EF BI; DI = line D I; K = intersection EF DI; Prove: equal_ratio D M D N F K E K
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2012CHNWesternMOp5
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$O$ is the circumcenter of acute $\Delta ABC$, $H$ is the Orthocenter. $AD \bot BC$, $EF$ is the perpendicular bisector of $AO$,$D,E$ on the $BC$. Prove that the circumcircle of $\Delta ADE$ through the midpoint of $OH$.
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a b c = triangle; o = circumcenter a b c; h = orthocenter a b c; d = foot a b c; e = on_line b c, on_bline a o; n = midpoint o h ? cyclic a d e n
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A B C = acute_triangle; O = circumcenter A B C; H = orthocenter A B C; BC = line B C; D = foot A BC; l1 = perpendicular_bisector A O; E = intersection l1 BC; N = midpoint O H; Prove: concyclic A D E N
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2012CTSTp2
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Given a scalene triangle $ABC$. Its incircle touches $BC,AC,AB$ at $D,E,F$ respectvely. Let $L,M,N$ be the symmetric points of $D$ with $EF$,of $E$ with $FD$,of $F$ with $DE$,respectively. Line $AL$ intersects $BC$ at $P$,line $BM$ intersects $CA$ at $Q$,line $CN$ intersects $AB$ at $R$. Prove that $P,Q,R$ 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; l = reflect d e f; m = reflect e f d; n = reflect f d e; p = on_line a l, on_line b c; q = on_line b m, on_line a c; r = on_line c n, on_line a b ? coll p q r
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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; L = reflect_point_wrt_line D EF; DF = line D F; M = reflect_point_wrt_line E DF; DE = line D E; N = reflect F DE; AL = line A L; P = intersection AL BC; BM = line B M; Q = intersection BM AC; CN = line C N; R = intersection CN AB; Prove: collinear P Q R
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2012EGMOp1
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Let $ABC$ be a triangle with circumcentre $O$. The points $D,E,F$ lie in the interiors of the sides $BC,CA,AB$ respectively, such that $DE$ is perpendicular to $CO$ and $DF$ is perpendicular to $BO$. (By interior we mean, for example, that the point $D$ lies on the line $BC$ and $D$ is between $B$ and $C$ on that line.) Let $K$ be the circumcentre of triangle $AFE$. Prove that the lines $DK$ and $BC$ are perpendicular.
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a b c = triangle; o = circumcenter a b c; d = on_line b c; e = on_line a c, on_tline d c o; f = on_line a b, on_tline d b o; k = circumcenter a e f ? perp d k b c
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A B C = triangle; O = circumcenter A B C; BC = line B C; D = foot O BC; CO = line C O; E = foot D CO; BO = line B O; F = foot D BO; K = circumcenter A F E; DK = line D K; Prove: perpendicular DK BC
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2012EGMOp7
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Let $ABC$ be an acute-angled triangle with circumcircle $\Gamma$ and orthocentre $H$. Let $K$ be a point of $\Gamma$ on the other side of $BC$ from $A$. Let $L$ be the reflection of $K$ in the line $AB$, and let $M$ be the reflection of $K$ in the line $BC$. Let $E$ be the second point of intersection of $\Gamma $ with the circumcircle of triangle $BLM$. Show that the lines $KH$, $EM$ and $BC$ are concurrent. (The orthocentre of a triangle is the point on all three of its altitudes.)
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a b c = triangle; o = circumcenter a b c; h = orthocenter a b c; k = on_circle o a; l = reflect k a b; m = reflect k b c; o1 = circumcenter b l m; e = on_circle o b, on_circle o1 b; p = on_line k h, on_line e m ? coll p b c
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A B C = acute_triangle; O = circumcenter A B C; (O) = circle_center_point O A; H = orthocenter A B C; K = on_circle (O); AB = line A B; L = reflect_point_wrt_line K AB; BC = line B C; M = reflect_point_wrt_line K BC; O1 = circumcenter B L M; (O1) = circle_center_point O1 B; E = intersection (O) (O1); HK = line H K; EM = line E M; P = intersection HK EM; Prove: collinear P B C
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2012G2
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Let $ABCD$ be a cyclic quadrilateral whose diagonals $AC$ and $BD$ meet at $E$. The extensions of the sides $AD$ and $BC$ beyond $A$ and $B$ meet at $F$. Let $G$ be the point such that $ECGD$ is a parallelogram, and let $H$ be the image of $E$ under reflection in $AD$. Prove that $D,H,F,G$ 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; f = on_line a d, on_line b c; g = parallelogram d e c g; h = reflect e a d ? cyclic d h f g
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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; AD = line A D; BC = line B C; F = intersection AD BC; G = parallelogram D E C; H = reflect_point_wrt_line E AD; Prove: concyclic D H F G
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2012G3
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In an acute triangle $ABC$ the points $D,E$ and $F$ are the feet of the altitudes through $A,B$ and $C$ respectively. The incenters of the triangles $AEF$ and $BDF$ are $I_1$ and $I_2$ respectively; the circumcenters of the triangles $ACI_1$ and $BCI_2$ are $O_1$ and $O_2$ respectively. Prove that $I_1I_2$ and $O_1O_2$ are parallel.
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a b c = triangle; d = foot a b c; e = foot b a c; f = foot c a b; i1 = incenter a e f; i2 = incenter b d f; o1 = circumcenter a c i1; o2 = circumcenter b c i2 ? para i1 i2 o1 o2
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A B C = acute_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; I1 = incenter A E F; I2 = incenter B D F; O1 = circumcenter A C I1; O2 = circumcenter B C I2; I1I2 = line I1 I2; O1O2 = line O1 O2; Prove: parallel I1I2 O1O2
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2012G4
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Let $ABC$ be a triangle with $AB \neq AC$ and circumcenter $O$. The bisector of $\angle BAC$ intersects $BC$ at $D$. Let $E$ be the reflection of $D$ with respect to the midpoint of $BC$. The lines through $D$ and $E$ perpendicular to $BC$ intersect the lines $AO$ and $AD$ at $X$ and $Y$ respectively. Prove that the quadrilateral $BXCY$ is cyclic.
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a b c = triangle; o = circumcenter a b c; d = on_line b c, angle_bisector b a c; m = midpoint b c; e = mirror d m; x = on_line a o, on_tline d b c; y = on_line a d, on_tline e b c ? cyclic b x c 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; M = midpoint B C; E = reflect D M; l2 = perpendicular_line D BC; AO = line A O; X = intersection l2 AO; l3 = perpendicular_line E BC; AD = line A D; Y = intersection l3 AD; Prove: concyclic B X C Y
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2012G8
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Let $ABC$ be a triangle with circumcircle $\omega$ and $\ell$ a line without common points with $\omega$. Denote by $P$ the foot of the perpendicular from the center of $\omega$ to $\ell$. The side-lines $BC,CA,AB$ intersect $\ell$ at the points $X,Y,Z$ different from $P$. Prove that the circumcircles of the triangles $AXP$, $BYP$ and $CZP$ have a common point different from $P$ or are mutually tangent at $P$.
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a b c = triangle; o = circumcenter a b c; p = free; p1 = on_tline p o p; x = on_line b c, on_line p p1; y = on_line c a, on_line p p1; z = on_line a b, on_line p p1; o1 = circumcenter a x p; o2 = circumcenter b y p; o3 = circumcenter c z p ? coll o1 o2 o3
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A B C = triangle; O = circumcenter A B C; l = line; P = foot O l; BC = line B C; X = intersection BC l; AC = line A C; Y = intersection AC l; AB = line A B; Z = intersection AB l; O1 = circumcenter A P X; O2 = circumcenter B P Y; O3 = circumcenter C P Z; Prove: collinear O1 O2 O3
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2012RMMp2
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Given a non-isosceles triangle $ABC$, let $D,E$, and $F$ denote the midpoints of the sides $BC,CA$, and $AB$ respectively. The circle $BCF$ and the line $BE$ meet again at $P$, and the circle $ABE$ and the line $AD$ meet again at $Q$. Finally, the lines $DP$ and $FQ$ meet at $R$. Prove that the centroid $G$ of the triangle $ABC$ lies on the circle $PQR$.
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a b c = triangle; d e f g = centroid a b c; p = on_line b e, on_circum b c f; q = on_line a d, on_circum a b e; r = on_line d p, on_line f q ? cyclic g p q r
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A B C = triangle; G D E F = centroid A B C; (BCF) = circle B C F; BE = line B E; P = intersection (BCF) BE; (ABE) = circle A B E; AD = line A D; Q = intersection (ABE) AD; DP = line D P; FQ = line F Q; R = intersection DP FQ; Prove: concyclic G P Q R
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2012USAMOp5
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Let $P$ be a point in the plane of $\triangle ABC$, and $\gamma$ a line passing through $P$. Let $A', B', C'$ be the points where the reflections of lines $PA, PB, PC$ with respect to $\gamma$ intersect lines $BC, AC, AB$ respectively. Prove that $A', B', C'$ are collinear.
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a b c = triangle; p = free; t = free; a1 = reflect a p t; a2 = on_line b c, on_line p a1; b1 = reflect b p t; b2 = on_line a c, on_line p b1; c1 = reflect c p t; c2 = on_line a b, on_line p c1 ? coll a2 b2 c2
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A B C = triangle; P = point; T = point; PT = line P T; A1 = reflect_point_wrt_line A PT; B1 = reflect_point_wrt_line B PT; C1 = reflect_point_wrt_line C PT; A1P = line A1 P; BC = line B C; A2 = intersection A1P BC; B1P = line B1 P; AC = line A C; B2 = intersection B1P AC; C1P = line C1 P; AB = line A B; C2 = intersection C1P AB; Prove: collinear A2 B2 C2
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2012USATSTp6
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In cyclic quadrilateral $ABCD$, diagonals $AC$ and $BD$ intersect at $P$. Let $E$ and $F$ be the respective feet of the perpendiculars from $P$ to lines $AB$ and $CD$. Segments $BF$ and $CE$ meet at $Q$. Prove that lines $PQ$ and $EF$ are perpendicular to each other.
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a b c = triangle; d = on_circum a b c; p = on_line a c, on_line b d; e = foot p a b; f = foot p c d; q = on_line b f, on_line c e ? perp p q e f
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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; P = intersection AC BD; AB = line A B; E = foot P AB; CD = line C D; F = foot P CD; BF = line B F; CE = line C E; Q = intersection BF CE; PQ = line P Q; EF = line E F; Prove: perpendicular PQ EF
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2013ARMOg10p7
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The incircle of triangle $ ABC $ has centre $I$ and touches the sides $ BC $, $ CA $, $ AB $ at points $ A_1 $, $ B_1 $, $ C_1 $, respectively. Let $ I_a $, $ I_b $, $ I_c $ be excentres of triangle $ ABC $, touching the sides $ BC $, $ CA $, $ AB $ respectively. The segments $ I_aB_1 $ and $ I_bA_1 $ intersect at $ C_2 $. Similarly, segments $ I_bC_1 $ and $ I_cB_1 $ intersect at $ A_2 $, and the segments $ I_cA_1 $ and $ I_aC_1 $ at $ B_2 $. Prove that $ I $ is the center of the circumcircle of the triangle $ A_2B_2C_2 $.
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a b c = triangle; i = incenter a b c; a1 = foot i b c; b1 = foot i a c; c1 = foot i a b; ia = excenter a b c; ib = excenter b c a; ic = excenter c a b; c2 = on_line ia b1, on_line ib a1; a2 = on_line ib c1, on_line ic b1; b2 = on_line ic a1, on_line ia c1 ? cong i a2 i b2
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A B C = triangle; I = incenter A B C; BC = line B C; A1 = foot I BC; AC = line A C; B1 = foot I AC; AB = line A B; C1 = foot I AB; Ia = excenter A B C; Ib = excenter B C A; Ic = excenter C A B; B1Ia = line B1 Ia; A1Ib = line A1 Ib; C2 = intersection B1Ia A1Ib; C1Ib = line C1 Ib; B1Ic = line B1 Ic; A2 = intersection C1Ib B1Ic; A1Ic = line A1 Ic; C1Ia = line C1 Ia; B2 = intersection A1Ic C1Ia; Prove: cong I A2 I B2
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2013ARMOg9p2
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Acute-angled triangle $ABC$ is inscribed into circle $\Omega$. Lines tangent to $\Omega$ at $B$ and $C$ intersect at $P$. Points $D$ and $E$ are on $AB$ and $AC$ such that $PD$ and $PE$ are perpendicular to $AB$ and $AC$ respectively. Prove that the orthocentre of triangle $ADE$ is the midpoint of $BC$.
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a b c = triangle; o = circumcenter a b c; p = on_tline b o b, on_tline c o c; d = foot p a b; e = foot p a c; h = orthocenter a d e ? midp h b c
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A B C = acute_triangle; O = circumcenter A B C; OB = line O B; OC = line O C; l1 = perpendicular_line B OB; l2 = perpendicular_line C OC; P = intersection l1 l2; AB = line A B; D = foot P AB; AC = line A C; E = foot P AC; H = orthocenter A D E; Prove: midpoint H B C
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2013ARMOg9p7
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Squares $CAKL$ and $CBMN$ are constructed on the sides of acute-angled triangle $ABC$, outside of the triangle. Line $CN$ intersects line segment $AK$ at $X$, while line $CL$ intersects line segment $BM$ at $Y$. Point $P$, lying inside triangle $ABC$, is an intersection of the circumcircles of triangles $KXN$ and $LYM$. Point $S$ is the midpoint of $AB$. Prove that angle $\angle ACS=\angle BCP$.
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a b c = triangle; l k = square a c l k; m n = square c b m n; x = on_line c n, on_line a k; y = on_line c l, on_line b m; o1 = circumcenter k x n; o2 = circumcenter l y m; p = on_circle o1 k, on_circle o2 l; s = midpoint a b ? eqangle a c c s p c b c
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A B C = acute_triangle; L K = square A C; M N = square C B; CN = line C N; AK = line A K; X = intersection CN AK; CL = line C L; BM = line B M; Y = intersection CL BM; O1 = circumcenter K N X; (O1) = circle_center_point O1 K; O2 = circumcenter L M Y; (O2) = circle_center_point O2 L; P = intersection (O1) (O2); S = midpoint A B; Prove: equal_angle A C S P C B
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2013BalkanMOp1
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In a triangle $ABC$, the excircle $\omega_a$ opposite $A$ touches $AB$ at $P$ and $AC$ at $Q$, while the excircle $\omega_b$ opposite $B$ touches $BA$ at $M$ and $BC$ at $N$. Let $K$ be the projection of $C$ onto $MN$ and let $L$ be the projection of $C$ onto $PQ$. Show that the quadrilateral $MKLP$ is cyclic.
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a b c = triangle; i1 = excenter a b c; p = foot i1 a b; q = foot i1 a c; i2 = excenter b c a; m = foot i2 a b; n = foot i2 b c; k = foot c m n; l = foot c p q ? cyclic m k l p
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A B C = triangle; I1 = excenter A B C; I2 = excenter B A C; AB = line A B; P = foot I1 AB; AC = line A C; Q = foot I1 AC; M = foot I2 AB; BC = line B C; N = foot I2 BC; MN = line M N; K = foot C MN; PQ = line P Q; L = foot C PQ; Prove: concyclic M K L P
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2013CGMOp2
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As shown in the figure below, $ABCD$ is a trapezoid, $AB \parallel CD$. The sides $DA$, $AB$, $BC$ are tangent to $\odot O_1$ and $AB$ touches $\odot O_1$ at $P$. The sides $BC$, $CD$, $DA$ are tangent to $\odot O_2$, and $CD$ touches $\odot O_2$ at $Q$. Prove that the lines $AC$, $BD$, $PQ$ meet at the same point.
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a b c d = trapezoid; e = on_line a d, on_line b c; o1 = excenter e a b; p = foot o1 a b; o2 = incenter e c d; q = foot o2 c d; x = on_line a c, on_line b d ? coll x p q
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A D C B = trapezoid; AD = line A D; BC = line B C; E = intersection AD BC; O1 = excenter E A B; AB = line A B; P = foot O1 AB; O2 = incenter E C D; CD = line C D; Q = foot O2 CD; AC = line A C; BD = line B D; PQ = line P Q; X = intersection AC BD; Prove: collinear X P Q
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2013CHNGaoLianA
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$AB$ is a chord of circle $\omega$, $P$ is a point on minor arc $AB$, $E,F$ are on segment $AB$ such that $AE=EF=FB$. $PE,PF$ meets $\omega$ at $C,D$ respectively. Prove that $EF\cdot CD=AC\cdot BD$.
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a b = segment; o = on_bline a b; p = on_circle o a; e f = trisegment a b; c = on_line p e, on_circle o a; d = on_line p f, on_circle o a ? eqratio e f a c b d c d
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A = point; B = point; l = perpendicular_bisector A B; O = on_line l; (O) = circle_center_point O A; P = on_circle (O); E F = trisegment A B; EP = line E P; C = intersection EP (O); FP = line F P; D = intersection FP (O); Prove: equal_ratio E F A C B D C D
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2013CHNWesternMOp3
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Let $ABC$ be a triangle, and $B_1,C_1$ be its excenters opposite $B,C$. $B_2,C_2$ are reflections of $B_1,C_1$ across midpoints of $AC,AB$. Let $D$ be the extouch at $BC$. Show that $AD$ is perpendicular to $B_2C_2$
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a b c = triangle; b1 = excenter b c a; c1 = excenter c a b; m1 = midpoint a c; b2 = mirror b1 m1; m2 = midpoint a b; c2 = mirror c1 m2; a1 = excenter a b c; d = foot a1 b c ? perp a d b2 c2
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A B C = triangle; B1 = excenter B C A; C1 = excenter C A B; M = midpoint A C; N = midpoint A B; B2 = reflect B1 M; C2 = reflect C1 N; A1 = excenter A B C; BC = line B C; D = foot A1 BC; AD = line A D; B2C2 = line B2 C2; Prove: perpendicular AD B2C2
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2013CTSTp1
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The quadrilateral $ABCD$ is inscribed in circle $\omega$. $F$ is the intersection point of $AC$ and $BD$. $BA$ and $CD$ meet at $E$. Let the projection of $F$ on $AB$ and $CD$ be $G$ and $H$, respectively. Let $M$ and $N$ be the midpoints of $BC$ and $EF$, respectively. If the circumcircle of $\triangle MNG$ only meets segment $BF$ at $P$, and the circumcircle of $\triangle MNH$ only meets segment $CF$ at $Q$, prove that $PQ$ is parallel to $BC$.
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a b c = triangle; d = on_circum a b c; f = on_line a c, on_line b d; e = on_line b a, on_line c d; g = foot f a b; h = foot f c d; m = midpoint b c; n = midpoint e f; o1 = circumcenter m n g; p = on_line b f, on_circle o1 m; o2 = circumcenter m n h; q = on_line c f, on_circle o2 m ? para p q b c
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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; F = intersection AC BD; AB = line A B; CD = line C D; E = intersection AB CD; G = foot F AB; H = foot F CD; M = midpoint B C; N = midpoint E F; BF = line B F; O1 = circumcenter G M N; (O1) = circle_center_point O1 M; P = intersection BF (O1); CF = line C F; O2 = circumcenter H M N; (O2) = circle_center_point O2 M; Q = intersection CF (O2); PQ = line P Q; BC = line B C; Prove: parallel PQ BC
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2013G2
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Let $\omega$ be the circumcircle of a triangle $ABC$. Denote by $M$ and $N$ the midpoints of the sides $AB$ and $AC$, respectively, and denote by $T$ the midpoint of the arc $BC$ of $\omega$ not containing $A$. The circumcircles of the triangles $AMT$ and $ANT$ intersect the perpendicular bisectors of $AC$ and $AB$ at points $X$ and $Y$, respectively; assume that $X$ and $Y$ lie inside the triangle $ABC$. The lines $MN$ and $XY$ intersect at $K$. Prove that $KA=KT$.
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a b c = triangle; o = circumcenter a b c; m = midpoint a b; n = midpoint a c; t = on_circle o a, angle_bisector b a c; o1 = circumcenter a m t; o2 = circumcenter a n t; x = on_circle o1 a, on_bline a c; y = on_circle o2 a, on_bline a b; k = on_line m n, on_line x y ? cong k a k t
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A B C = triangle; O = circumcenter A B C; M = midpoint A B; N = midpoint A C; l1 = angle_bisector B A C; (ABC) = circle_center_point O A; T = intersection l1 (ABC); l2 = perpendicular_bisector A C; O1 = circumcenter A M T; (AMT) = circle_center_point O1 A; X = intersection (AMT) l2; l3 = perpendicular_bisector A B; O2 = circumcenter A N T; (ANT) = circle_center_point O2 A; Y = intersection (ANT) l3; MN = line M N; XY = line X Y; K = intersection MN XY; Prove: cong K A K T
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2013G4
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Let $ABC$ be a triangle with $\angle B > \angle C$. Let $P$ and $Q$ be two different points on line $AC$ such that $\angle PBA = \angle QBA = \angle ACB $ and $A$ is located between $P$ and $C$. Suppose that there exists an interior point $D$ of segment $BQ$ for which $PD=PB$. Let the ray $AD$ intersect the circle $ABC$ at $R \neq A$. Prove that $QB = QR$.
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a b c = triangle; o = circumcenter a b c; p = on_line a c, on_aline p b a b c a; q = on_line a c, on_aline q b a a c b; d = on_line b q, on_circle p b; r = on_line a d, on_circle o a ? cong q b q r
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A B C = triangle; O = circumcenter A B C; (O) = circle_center_point O A; AC = line A C; l1 = angle_equal1 B A B C A; P = intersection AC l1; l2 = angle_equal1 B A A C B; Q = intersection AC l2; BQ = line B Q; (P) = circle_center_point P B; D = intersection BQ (P); AD = line A D; R = intersection AD (O); Prove: cong Q B Q R
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Dataset Card for the HAGeo-409 Benchmark
[🤗 Benchmark] • [📜 Paper] • [🐱 GitHub] • [🐦 Twitter] • [📕 Rednote]
The HAGeo-409 benchmark includes 409 IMO-level geometry theorem-proving problems and typically presents greater difficulty than the widely used IMO-30 benchmark.
Figure 1: Problem difficulity distribution of the IMO-30 benchmark and our new HAGeo-409 benchmark.
Data Loading
from datasets import load_dataset
dataset = load_dataset("HAGeo-IMO/HAGeo-409")
Data Instance
Each instance contains the following fields: 'Problem_ID', 'Natural_Language', 'AlphaGeometry', and 'HAGeo'.
- Problem_ID: The unique identifier of each problem, indicating the source competition and the corresponding problem index.
- Natural_Language: The problem statement written in natural language.
- AlphaGeometry: The problem expressed in AlphaGeometry’s geometry-specific representation language.
- HAGeo: The problem expressed in HAGeo’s geometry-specific representation language (Cases in Appendix C, D).
Cite Us
If you find this dataset helpful, please consider citing our paper:
@misc{duan2025goldmedallevelolympiadgeometrysolving,
title={Gold-Medal-Level Olympiad Geometry Solving with Efficient Heuristic Auxiliary Constructions},
author={Boyan Duan and Xiao Liang and Shuai Lu and Yaoxiang Wang and Yelong Shen and Kai-Wei Chang and Ying Nian Wu and Mao Yang and Weizhu Chen and Yeyun Gong},
year={2025},
eprint={2512.00097},
archivePrefix={arXiv},
primaryClass={cs.AI},
url={https://arxiv.org/abs/2512.00097},
}
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