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2013RMMp3
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Let $ABCD$ be a quadrilateral inscribed in a circle $\omega$. The lines $AB$ and $CD$ meet at $P$, the lines $AD$ and $BC$ meet at $Q$, and the diagonals $AC$ and $BD$ meet at $R$. Let $M$ be the midpoint of the segment $PQ$, and let $K$ be the common point of the segment $MR$ and the circle $\omega$. Prove that the circumcircle of the triangle $KPQ$ and $\omega$ are tangent to one another.
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a b c = triangle; d = on_circum a b c; o = circumcenter a b c; p = on_line a b, on_line c d; q = on_line a d, on_line b c; r = on_line a c, on_line b d; m = midpoint p q; k = on_line m r, on_circle o a; o1 = circumcenter k p q ? coll o o1 k
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A B C = triangle; O = circumcenter A B C; (ABC) = circle_center_point O A; D = on_circle (ABC); AB = line A B; CD = line C D; P = intersection AB CD; AD = line A D; BC = line B C; Q = intersection AD BC; AC = line A C; BD = line B D; R = intersection AC BD; M = midpoint P Q; MR = line M R; K = intersection MR (ABC); O1 = circumcenter K P Q; Prove: collinear O O1 K
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2013USAMOp1
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In triangle $ABC$, points $P$, $Q$, $R$ lie on sides $BC$, $CA$, $AB$ respectively. Let $\omega_A$, $\omega_B$, $\omega_C$ denote the circumcircles of triangles $AQR$, $BRP$, $CPQ$, respectively. Given the fact that segment $AP$ intersects $\omega_A$, $\omega_B$, $\omega_C$ again at $X$, $Y$, $Z$, respectively, prove that $YX/XZ=BP/PC$.
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a b c = triangle; p = on_line b c; q = on_line a c; r = on_line a b; x = on_line a p, on_circum a q r; y = on_line a p, on_circum b r p; z = on_line a p, on_circum c p q ? eqratio y x x z b p p c
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A B C = triangle; BC = line B C; P = on_line BC; AC = line A C; Q = on_line AC; AB = line A B; R = on_line AB; ωa = circumcircle A Q R; ωb = circumcircle B R P; ωc = circumcircle C P Q; AP = line A P; X = intersection AP ωa; Y = intersection AP ωb; Z = intersection AP ωc; Prove: equal_ratio X Y X Z B P C P
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2013USATSTp2
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Let $ABC$ be a scalene triangle with $\angle BCA = 90^{\circ}$, and let $D$ be the foot of the altitude from $C$. Let $X$ be a point in the interior of the segment $CD$. Let $K$ be the point on the segment $AX$ such that $BK = BC$. Similarly, let $L$ be the point on the segment $BX$ such that $AL = AC$. The circumcircle of triangle $DKL$ intersects segment $AB$ at a second point $T$ (other than $D$). Prove that $\angle ACT = \angle BCT$.
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c b a = r_triangle; d = foot c a b; x = on_line c d; k = on_line a x, eqdistance k b b c; l = on_line b x, eqdistance l a a c; o1 = circumcenter d k l; t = on_line a b, on_circle o1 d ? eqangle a c c t t c c b
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C A B = right_triangle; AB = line A B; D = foot C AB; CD = line C D; X = on_line CD; AX = line A X; (B,C) = circle_center_point B C; K = intersection AX (B,C); BX = line B X; (A,C) = circle_center_point A C; L = intersection BX (A,C); O1 = circumcenter D K L; (DKL) = circle_center_point O1 D; T = intersection AB (DKL); Prove: equal_angle A C T T C B
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2014ARMOg10p4
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Given a triangle $ABC$ with $AB>BC$, let $ \Omega $ be the circumcircle. Let $M$, $N$ lie on the sides $AB$, $BC$ respectively, such that $AM=CN$. Let $K$ be the intersection of $MN$ and $AC$. Let $P$ be the incentre of the triangle $AMK$ and $Q$ be the $K$-excentre of the triangle $CNK$. If $R$ is midpoint of the arc $ABC$ of $ \Omega $ then prove that $RP=RQ$.
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a b c = triangle; o = circumcenter a b c; m = on_line a b; n = on_line b c, eqdistance c a m; k = on_line m n, on_line a c; p = incenter a m k; q = excenter k c n; x = angle_bisector a b c; r = on_circum a b c, on_tline b b x ? cong r p r q
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A B C = triangle; O = circumcenter A B C; AB = line A B; M = on_line AB; (C,AM) = circle_center_radius C A M; BC = line B C; N = intersection (C,AM) BC; MN = line M N; AC = line A C; K = intersection MN AC; P = incenter A M K; Q = excenter K C N; R = midarc A B C; Prove: cong R P R Q
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2014ARMOg9p4
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Let $M$ be the midpoint of the side $AC$ of acute-angled triangle $ABC$ with $AB>BC$. Let $\Omega $ be the circumcircle of $ ABC$. The tangents to $ \Omega $ at the points $A$ and $C$ meet at $P$, and $BP$ and $AC$ intersect at $S$. Let $AD$ be the altitude of the triangle $ABP$ and $\omega$ the circumcircle of the triangle $CSD$. Suppose $ \omega$ and $ \Omega $ intersect at $K\not= C$. Prove that $ \angle CKM=90^\circ $.
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a b c = triangle; m = midpoint a c; o = circumcenter a b c; p = on_tline a o a, on_tline c o c; s = on_line b p, on_line a c; d = foot a b p; o1 = circumcenter c s d; k = on_circle o1 c, on_circle o a ? perp c k k m
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A B C = acute_triangle; M = midpoint A C; (ABC) = circumcircle A B C; l1 = tangent A (ABC); l2 = tangent C (ABC); P = intersection l1 l2; BP = line B P; AC = line A C; S = intersection BP AC; D = foot A BP; (CSD) = circumcircle C S D; K = intersection (CSD) (ABC); CK = line C K; KM = line K M; Prove: perpendicular CK KM
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2014ARMOg9p6
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Let $ABCD$ be a trapezoid with $AB\parallel CD$ and $ \Omega $ is a circle passing through $A,B,C,D$. Let $ \omega $ be the circle passing through $C,D$ and intersecting with $CA,CB$ at $A_1$, $B_1$ respectively. $A_2$ and $B_2$ are the points symmetric to $A_1$ and $B_1$ respectively, with respect to the midpoints of $CA$ and $CB$. Prove that the points $A,B,A_2,B_2$ are concyclic.
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a b c d = eq_trapezoid; o = on_bline c d; a1 = on_line a c, on_circle o c; b1 = on_line b c, on_circle o c; m1 = midpoint c a; m2 = midpoint c b; a2 = mirror a1 m1; b2 = mirror b1 m2 ? cyclic a b a2 b2
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A D C B = isos_trapezoid; l = perpendicular_bisector C D; O = on_line l; (O) = circle_center_point O C; AC = line A C; A1 = intersection AC (O); BC = line B C; B1 = intersection BC (O); M1 = midpoint C A; M2 = midpoint C B; A2 = reflect_point_wrt_point A1 M1; B2 = reflect_point_wrt_point B1 M2; Prove: concyclic A B A2 B2
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2014AsiaPacificMOp5
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Circles $\omega$ and $\Omega$ meet at points $A$ and $B$. Let $M$ be the midpoint of the arc $AB$ of circle $\omega$ ($M$ lies inside $\Omega$). A chord $MP$ of circle $\omega$ intersects $\Omega$ at $Q$ ($Q$ lies inside $\omega$). Let $\ell_P$ be the tangent line to $\omega$ at $P$, and let $\ell_Q$ be the tangent line to $\Omega$ at $Q$. Prove that the circumcircle of the triangle formed by the lines $\ell_P$, $\ell_Q$ and $AB$ is tangent to $\Omega$.
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a b = segment; o1 = on_bline a b; o2 = on_bline a b; m = on_circle o1 a, on_tline o1 a b; p = on_circle o1 a; q = on_line m p, on_circle o2 a; x = on_line a b, on_tline p o1 p; y = on_line a b, on_tline q o2 q; z = on_tline p o1 p, on_tline q o2 q; o3 = circumcenter x y z; t = on_circle o3 x, on_circle o2 a ? coll o2 o3 t
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A = point; B = point; l = perpendicular_bisector A B; O1 = on_line l; O2 = on_line l; (O1) = circle_center_point O1 A; (O2) = circle_center_point O2 A; M = intersection l (O1); P = on_circle (O1); MP = line M P; Q = intersection MP (O2); O1P = line O1 P; lp = perpendicular_line P O1P; O2Q = line O2 Q; lq = perpendicular_line Q O2Q; AB = line A B; X = intersection AB lp; Y = intersection AB lq; Z = intersection lp lq; O3 = circumcenter X Y Z; (XYZ) = circle_center_point O3 X; T = intersection (XYZ) (O2) Prove: collinear O2 O3 T
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2014BulgariaMOp6
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Let $ABCD$ be a quadrilateral inscribed in a circle $k$. $AC$ and $BD$ meet at $E$. The rays $\overrightarrow{CB}, \overrightarrow{DA}$ meet at $F$. Prove that the line through the incenters of $\triangle ABE\,,\, \triangle ABF$ and the line through the incenters of $\triangle CDE\,,\, \triangle CDF$ meet at a point lying on the circle $k$.
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a b c = triangle; d = on_circum a b c; o = circumcenter a b c; e = on_line a c, on_line b d; f = on_line c b, on_line d a; i1 = incenter a b e; i2 = incenter a b f; i3 = incenter c d e; i4 = incenter c d f; p = on_line i1 i2, on_line i3 i4 ? cyclic p a b c
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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; E = intersection AC BD; CB = line C B; DA = line D A; F = intersection CB DA; I1 = incenter A B E; I2 = incenter A B F; I3 = incenter C D E; I4 = incenter C D F; I1I2 = line I1 I2; I3I4 = line I3 I4; P = intersection I1I2 I3I4; Prove: concyclic P A B C
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2014CGMOp6
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In acute triangle $ABC$, $AB > AC$. $D$ and $E$ are the midpoints of $AB$, $AC$ respectively. The circumcircle of $ADE$ intersects the circumcircle of $BCE$ again at $P$. The circumcircle of $ADE$ intersects the circumcircle $BCD$ again at $Q$. Prove that $AP = AQ$.
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a b c = triangle; d = midpoint a b; e = midpoint a c; o1 = circumcenter a d e; o2 = circumcenter b c e; p = on_circle o1 a, on_circle o2 b; o3 = circumcenter b c d; q = on_circle o1 a, on_circle o3 b ? cong a p a q
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A B C = acute_triangle; D = midpoint A B; E = midpoint A C; O1 = circumcenter A D E; (O1) = circle_center_point O1 A; O2 = circumcenter B C E; (O2) = circle_center_point O2 B; P = intersection (O1) (O2); O3 = circumcenter B C D; (O3) = circle_center_point O3 B; Q = intersection (O1) (O3); Prove: cong A P A Q
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2014CHNGaoLian
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Let $ABC$ be an acute triangle such that $\angle BAC \neq 60^\circ$. Let $D,E$ be points such that $BD,CE$ are tangent to the circumcircle of $ABC$ and $BD=CE=BC$ ($A$ is on one side of line $BC$ and $D,E$ are on the other side). Let $F,G$ be intersections of line $DE$ and lines $AB,AC$. Let $M$ be intersection of $CF$ and $BD$, and $N$ be intersection of $CE$ and $BG$. Prove that $AM=AN$.
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a b c = triangle; o = circumcenter a b c; d = on_tline b o b, on_circle b c; e = on_tline c o c, on_circle c b; f = on_line a b, on_line d e; g = on_line a c, on_line d e; m = on_line c f, on_line b d; n = on_line c e, on_line b g ? cong a m a n
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A B C = acute_triangle; O = circumcenter A B C; OB = line O B; l1 = perpendicular_line B OB; (B) = circle_center_point B C; D = intersection l1 (B); OC = line O C; l2 = perpendicular_line C OC; (C) = circle_center_point C B; E = intersection l2 (C); DE = line D E; AB = line A B; F = intersection DE AB; AC = line A C; G = intersection DE AC; CF = line C F; BD = line B D; M = intersection CF BD; CE = line C E; BG = line B G; N = intersection CE BG; AM = line A M; AN = line A N; Prove: cong A M A N
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2014CHNSouthEastMOg10p3
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In an obtuse triangle $ABC$ $(AB>AC)$,$O$ is the circumcentre and $D,E,F$ are the midpoints of $BC,CA,AB$ respectively.Median $AD$ intersects $OF$ and $OE$ at $M$ and $N$ respectively.$BM$ meets $CN$ at point $P$.Prove that
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a b c = triangle; o = circumcenter a b c; d = midpoint b c; e = midpoint c a; f = midpoint a b; m = on_line a d, on_line o f; n = on_line a d, on_line o e; p = on_line b m, on_line c n ? perp a p o p
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A B C = obtuse_triangle; O = circumcenter A B C; D = midpoint B C; E = midpoint C A; F = midpoint A B; AD = line A D; FO = line F O; M = intersection AD FO; EO = line E O; N = intersection AD EO; BM = line B M; CN = line C N; P = intersection BM CN; OP = line O P; AP = line A P; Prove: perpendicular OP AP
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2014CHNSouthEastMOg11p1
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Let $ABC$ be a triangle with $AB<AC$ and let $M $ be the midpoint of $BC$. $MI$ ($I$ incenter) intersects $AB$ at $D$ and $CI$ intersects the circumcircle of $ABC$ at $E$. Prove that $\frac{ED }{ EI} = \frac{IB }{IC}$
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a b c = triangle; m = midpoint b c; i = incenter a b c; d = on_line a b, on_line m i; o = circumcenter a b c; e = on_line c i, on_circle o a ? eqratio e d e i i b i c
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A B C = triangle; M = midpoint B C; I = incenter A B C; IM = line I M; AB = line A B; D = intersection IM AB; CI = line C I; O = circumcenter A B C; (O) = circle_center_point O A; E = intersection CI (O); Prove: equal_ratio D E E I B I C I
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2014CHNWesternMOp2
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Let $ AB$ be the diameter of semicircle $O$ , $C, D $ be points on the arc $AB$, $P, Q$ be respectively the circumcenter of $\triangle OAC $ and $\triangle OBD $ . Prove that:$CP\cdot CQ=DP \cdot DQ$.
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a b = segment; o = midpoint a b; c = on_circle o a; d = on_circle o a; p = circumcenter o a c; q = circumcenter o b d ? eqratio c p d p d q c q
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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); P = circumcenter O A C; Q = circumcenter O B D; Prove: equal_ratio C P D P D Q C Q
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2014CTSTp13
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Let the circumcenter of triangle $ABC$ be $O$. $H_A$ is the projection of $A$ onto $BC$. The extension of $AO$ intersects the circumcircle of $BOC$ at $A'$. The projections of $A'$ onto $AB, AC$ are $D,E$, and $O_A$ is the circumcentre of triangle $DH_AE$. Define $H_B, O_B, H_C, O_C$ similarly. Prove: $H_AO_A, H_BO_B, H_CO_C$ are concurrent
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a b c = triangle; o = circumcenter a b c; h1 = foot a b c; h2 = foot b a c; h3 = foot c a b; o1 = circumcenter o b c; o2 = circumcenter o a c; o3 = circumcenter o b a; a1 = on_line a o, on_circle o1 o; b1 = on_line b o, on_circle o2 o; c1 = on_line c o, on_circle o3 o; d1 = foot a1 a b; d2 = foot a1 a c; u1 = circumcenter d1 h1 d2; e1 = foot b1 b c; e2 = foot b1 b a; u2 = circumcenter e1 h2 e2; f1 = foot c1 c a; f2 = foot c1 c b; u3 = circumcenter f1 h3 f2; t = on_line u1 h1, on_line u2 h2 ? coll t u3 h3
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A B C = triangle; O = circumcenter A B C; BC = line B C; AC = line A C; AB = line A B; Ha = foot A BC; Hb = foot B AC; Hc = foot C AB; O1 = circumcenter O B C; O2 = circumcenter O A C; O3 = circumcenter O B A; AO = line A O; (BCO) = circle_center_point O1 O; BO = line B O; (ACO) = circle_center_point O2 O; CO = line C O; (ABO) = circle_center_point O3 O; A1 = intersection AO (BCO); B1 = intersection BO (ACO); C1 = intersection CO (ABO); D1 = foot A1 AB; D2 = foot A1 AC; Oa = circumcenter D1 Ha D2; E1 = foot B1 BC; E2 = foot B1 AB; Ob = circumcenter E1 Hb E2; F1 = foot C1 AC; F2 = foot C1 BC; Oc = circumcenter F1 Hc F2; HaOa = line Ha Oa; HbOb = line Hb Ob; T = intersection HaOa HbOb; Prove: collinear T Hc Oc
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2014EGMOp2
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Let $D$ and $E$ be points in the interiors of sides $AB$ and $AC$, respectively, of a triangle $ABC$, such that $DB = BC = CE$. Let the lines $CD$ and $BE$ meet at $F$. Prove that the incentre $I$ of triangle $ABC$, the orthocentre $H$ of triangle $DEF$ and the midpoint $M$ of the arc $BAC$ of the circumcircle of triangle $ABC$ are collinear.
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a b c = triangle; d = on_line a b, on_circle b c; e = on_line a c, on_circle c b; f = on_line c d, on_line b e; i = incenter a b c; h = orthocenter d e f; m = on_bline b c, on_circum a b c ? coll i h m
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A B C = triangle; 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); CD = line C D; BE = line B E; F = intersection CD BE; I = incenter A B C; H = orthocenter D E F; M = midarc B A C; Prove: collinear I H M
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2014G3
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Let $\Omega$ and $O$ be the circumcircle and the circumcentre of an acute-angled triangle $ABC$ with $AB > BC$. The angle bisector of $\angle ABC$ intersects $\Omega$ at $M \ne B$. Let $\Gamma$ be the circle with diameter $BM$. The angle bisectors of $\angle AOB$ and $\angle BOC$ intersect $\Gamma$ at points $P$ and $Q,$ respectively. The point $R$ is chosen on the line $P Q$ so that $BR = MR$. Prove that $BR\parallel AC$.
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a b c = triangle; o = circumcenter a b c; m = on_circle o a, angle_bisector a b c; p = on_dia b m, on_bline a b; q = on_dia b m, on_bline b c; r = on_line p q, on_bline b m ? para b r a c
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A B C = acute_triangle; O = circumcenter A B C; M = midarc_no A C B; (BM) = circle_diameter B M; AB = line A B; l1 = perpendicular_bisector A B; P = intersection l1 (BM); BC = line B C; l2 = perpendicular_bisector B C; Q = intersection l2 (BM); l3 = perpendicular_bisector B M; PQ = line P Q; R = intersection PQ l3; BR = line B R; AC = line A C; Prove: parallel BR AC
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2014IranGOAp3
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Let $ABC$ be an acute triangle.A circle with diameter $BC$ meets $AB$ and $AC$ at $E$ and $F$,respectively. $M$ is midpoint of $BC$ and $P$ is point of intersection $AM$ with $EF$. $X$ is an arbitary point on arc $EF$ and $Y$ is the second intersection of $XP$ with a circle with diameter $BC$.Prove that $ \measuredangle XAY=\measuredangle XYM $.
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a b c = triangle; m = midpoint b c; e = on_line a b, on_circle m b; f = on_line a c, on_circle m b; p = on_line a m, on_line e f; x = on_circle m b; y = on_line x p, on_circle m b ? eqangle x a a y x y y m
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A B C = acute_triangle; AB = line A B; M = midpoint B C; (M) = circle_center_point M B; E = intersection AB (M); AC = line A C; F = intersection AC (M); AM = line A M; EF = line E F; P = intersection AM EF; X = on_circle (M); PX = line P X; Y = intersection PX (M); Prove: equal_angle X A Y X Y M
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2014IranGOAp4
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A tangent line to circumcircle of acute triangle $ABC$ ($AC>AB$) at $A$ intersects with the extension of $BC$ at $P$. $O$ is the circumcenter of triangle $ABC$.Point $X$ lying on $OP$ such that $\measuredangle AXP=90^\circ$.Points $E$ and $F$ lying on $AB$ and $AC$,respectively,and they are in one side of line $OP$ such that $ \measuredangle EXP=\measuredangle ACX $ and $\measuredangle FXO=\measuredangle ABX $. $K$,$L$ are points of intersection $EF$ with circumcircle of triangle $ABC$.prove that $OP$ is tangent to circumcircle of triangle $KLX$.
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a b c = triangle; o = circumcenter a b c; p = on_tline a o a, on_line b c; x = foot a o p; e = on_line a b, on_aline e x p a c x; f = on_line a c, on_aline f x o a b x; k = on_line e f, on_circle o a; l = on_line e f, on_circle o a; o1 = circumcenter k l x ? perp o x o1 x
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A B C = acute_triangle; O = circumcenter A B C; OA = line O A; ta = perpendicular_line A OA; BC = line B C; P = intersection ta BC; OP = line O P; X = foot A OP; AB = line A B; AC = line A C; l1 = angle_equal1 X P A C X; E = intersection l1 AB; l2 = angle_equal1 X O A B X; F = intersection l2 AC; EF = line E F; (ABC) = circle_center_point O A; K L = intersection EF (ABC); O1 = circumcenter K L X; O1X = line O1 X; Prove: perpendicular O1X OP
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2014IranTSTp13
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The incircle of a non-isosceles triangle $ABC$ with the center $I$ touches the sides $BC,AC,AB$ at $A_{1},B_{1},C_{1}$ . let $AI,BI,CI$ meets $BC,AC,AB$ at $A_{2},B_{2},C_{2}$. let $A'$ is a point on $AI$ such that $A_{1}A'\perp B_{2}C_{2}$ .$B',C'$ respectively. prove that two triangle $A'B'C',A_{1}B_{1}C_{1}$ are equal.
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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; a2 = on_line b c, on_line a i; b2 = on_line a c, on_line b i; c2 = on_line a b, on_line c i; a3 = on_line a i, on_tline a1 b2 c2; b3 = on_line b i, on_tline b1 c2 a2; c3 = on_line c i, on_tline c1 a2 b2 ? contri a3 b3 c3 a1 b1 c1
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A B C = triangle; I = incenter A B C; BC = line B C; AC = line A C; AB = line A B; A1 = foot I BC; B1 = foot I AC; C1 = foot I AB; BC = line B C; AI = line A I; A2 = intersection BC AI; AC = line A C; BI = line B I; B2 = intersection AC BI; AB = line A B; CI = line C I; C2 = intersection AB CI; B2C2 = line B2 C2; l1 = perpendicular_line A1 B2C2; A3 = intersection AI l1; C2A2 = line C2 A2; l2 = perpendicular_line B1 C2A2; B3 = intersection BI l2; A2B2 = line A2 B2; l3 = perpendicular_line C1 A2B2; C3 = intersection CI l3; Prove: congurrent A3 B3 C3 A1 B1 C1
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2015ARMOg10p7
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In an acute-angled and not isosceles triangle $ABC,$ we draw the median $AM$ and the height $AH.$ Points $Q$ and $P$ are marked on the lines $AB$ and $AC$, respectively, so that the $QM \perp AC$ and $PM \perp AB$. The circumcircle of $PMQ$ intersects the line $BC$ for second time at point $X.$ Prove that $BH = CX.$
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a b c = triangle; m = midpoint b c; h = foot a b c; q = on_line a b, on_tline m a c; p = on_line a c, on_tline m a b; o1 = circumcenter p m q; x = on_line b c, on_circle o1 p ? cong b h c x
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A B C = acute_triangle; M = midpoint B C; BC = line B C; H = foot A BC; AC = line A C; l1 = perpendicular_line M AC; AB = line A B; Q = intersection AB l1; l2 = perpendicular_line M AB; P = intersection AC l2; O1 = circumcenter M P Q; (O1) = circle_center_point O1 P; X = intersection BC (O1); Prove: cong B H C X
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2015ARMOg11p7
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A scalene triangle $ABC$ is inscribed within circle $\omega$. The tangent to the circle at point $C$ intersects line $AB$ at point $D$. Let $I$ be the center of the circle inscribed within $\triangle ABC$. Lines $AI$ and $BI$ intersect the bisector of $\angle CDB$ in points $Q$ and $P$, respectively. Let $M$ be the midpoint of $QP$. Prove that $MI$ passes through the middle of arc $ACB$ of circle $\omega$.
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a b c = triangle; o = circumcenter a b c; d = on_line a b, on_tline c o c; i = incenter a b c; q = on_line a i, angle_bisector c d b; p = on_line b i, angle_bisector c d b; m = midpoint q p; n = on_circle o a, on_tline c c i ? coll m i n
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A B C = triangle; O = circumcenter A B C; OC = line O C; l1 = perpendicular_line C OC; AB = line A B; D = intersection l1 AB; I = incenter A B C; l2 = angle_bisector C D B; AI = line A I; Q = intersection l2 AI; BI = line B I; P = intersection l2 BI; M = midpoint Q P; N = midarc A C B; Prove: collinear M I N
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2015ARMOg9p7
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An acute-angled $ABC \ (AB<AC)$ is inscribed into a circle $\omega$. Let $M$ be the centroid of $ABC$, and let $AH$ be an altitude of this triangle. A ray $MH$ meets $\omega$ at $A'$. Prove that the circumcircle of the triangle $A'HB$ is tangent to $AB$.
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a b c = triangle; o = circumcenter a b c; x y z m = centroid a b c; h = foot a b c; a1 = on_line m h, on_circle o a; o1 = circumcenter a1 h b ? perp o1 b a b
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A B C = acute_triangle; O = circumcenter A B C; (O) = circle_center_point O A; M X Y Z = centroid A B C; BC = line B C; H = foot A BC; HM = line H M; A1 = intersection HM (O); O1 = circumcenter A1 B H; O1B = line O1 B; AB = line A B; Prove: perpendicular O1B AB
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2015AsiaPacificMOp1
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Let $ABC$ be a triangle, and let $D$ be a point on side $BC$. A line through $D$ intersects side $AB$ at $X$ and ray $AC$ at $Y$ . The circumcircle of triangle $BXD$ intersects the circumcircle $\omega$ of triangle $ABC$ again at point $Z$ distinct from point $B$. The lines $ZD$ and $ZY$ intersect $\omega$ again at $V$ and $W$ respectively. Prove that $AB = V W$
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a b c = triangle; d = on_line b c; x = on_line a b; y = on_line a c, on_line d x; z = on_circum b x d, on_circum a b c; v = on_line z d, on_circum a b c; w = on_line z y, on_circum a b c ? cong a b v w
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A B C = triangle; BC = line B C; D = on_line BC; l = line_through D; AB = line A B; X = intersection l AB; AC = line A C; Y = intersection l AC; (ABC) = circumcircle A B C; (BDX) = circle B D X; Z = intersection (BDX) (ABC); DZ = line D Z; V = intersection DZ (ABC); YZ = line Y Z; W = intersection YZ (ABC); VW = line V W; Prove: cong A B V W
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2015BalkanMOp2
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Let $\triangle{ABC}$ be a scalene triangle with incentre $I$ and circumcircle $\omega$. Lines $AI, BI, CI$ intersect $\omega$ for the second time at points $D, E, F$, respectively. The parallel lines from $I$ to the sides $BC, AC, AB$ intersect $EF, DF, DE$ at points $K, L, M$, respectively. Prove that the points $K, L, M$ are collinear.
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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 c i, on_circle o c; k = on_line e f, on_pline i b c; l = on_line d f, on_pline i a c; m = on_line d e, on_pline i a b ? coll k l m
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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 AI (O); BI = line B I; E = intersection BI (O); CI = line C I; F = intersection CI (O); BC = line B C; l1 = parallel_line I BC; EF = line E F; K = intersection l1 EF; AC = line A C; l2 = parallel_line I AC; DF = line D F; L = intersection l2 DF; AB = line A B; l3 = parallel_line I AB; DE = line D E; M = intersection l3 DE; Prove: collinear K L M
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2015CGMOp1
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Let $\triangle ABC$ be an acute-angled triangle with $AB>AC$, $O$ be its circumcenter and $D$ the midpoint of side $BC$. The circle with diameter $AD$ meets sides $AB,AC$ again at points $E,F$ respectively. The line passing through $D$ parallel to $AO$ meets $EF$ at $M$. Show that $EM=MF$.
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a b c = triangle; o = circumcenter a b c; d = midpoint b c; o1 = midpoint a d; e = on_line a b, on_circle o1 a; f = on_line a c, on_circle o1 a; m = on_line e f, on_pline d a o ? cong e m m f
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A B C = acute_triangle; O = circumcenter A B C; D = midpoint B C; (AD) = circle_diameter A D; AB = line A B; E = intersection (AD) AB; AC = line A C; F = intersection (AD) AC; AO = line A O; l1 = parallel_line D AO; EF = line E F; M = intersection l1 EF; Prove: cong E M M F
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2015CTSTp9
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Let $ \triangle ABC $ be an acute triangle with circumcenter $ O $ and centroid $ G .$ Let $ D $ be the midpoint of $ BC $ and $ E\in \odot (BC) $ be a point inside $ \triangle ABC $ such that $ AE \perp BC . $ Let $ F=EG \cap OD $ and $ K, L $ be the point lie on $ BC $ such that $ FK \parallel OB, FL \parallel OC . $ Let $ M \in AB $ be a point such that $ MK \perp BC $ and $ N \in AC $ be a point such that $ NL \perp BC . $ Let $ \omega $ be a circle tangent to $ OB, OC $ at $ B, C, $ respectively $ . $ Prove that $ \odot (AMN) $ is tangent to $ \omega $
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a b c = triangle; o = circumcenter a b c; d d1 d2 g = centroid a b c; e = on_circle d b, on_tline a b c; f = on_line e g, on_line o d; k = on_line b c, on_pline f o b; l = on_line b c, on_pline f o c; m = on_line a b, on_tline k b c; n = on_line a c, on_tline l b c; o1 = on_bline b c, on_tline b o b; o2 = circumcenter a m n; t = on_circle o1 b, on_circle o2 a ? coll o2 o1 t
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A B C = acute_triangle; O = circumcenter A B C; G D D1 D2 = centroid A B C; BC = line B C; l1 = perpendicular_line A BC; (BC) = circle_center_point D B; E = intersection (BC) l1; EG = line E G; DO = line D O; F = intersection EG DO; BO = line B O; l2 = parallel_line F BO; K = intersection BC l2; CO = line C O; l3 = parallel_line F CO; L = intersection BC l3; l4 = perpendicular_line K BC; AB = line A B; M = intersection AB l4; l5 = perpendicular_line L BC; AC = line A C; N = intersection AC l5; l6 = perpendicular_line B BO; l7 = perpendicular_line C CO; X = intersection l6 l7; (X) = circle_center_point X B; O2 = circumcenter A M N; (AMN) = circle_center_point O2 A; T = intersection (X) (AMN); Prove: collinear O2 X T
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2015ELMOp3
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Let $\omega$ be a circle and $C$ a point outside it; distinct points $A$ and $B$ are selected on $\omega$ so that $\overline{CA}$ and $\overline{CB}$ are tangent to $\omega$. Let $X$ be the reflection of $A$ across the point $B$, and denote by $\gamma$ the circumcircle of triangle $BXC$. Suppose $\gamma$ and $\omega$ meet at $D \neq B$ and line $CD$ intersects $\omega$ at $E \neq D$. Prove that line $EX$ is tangent to the circle $\gamma$.
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a b = segment; o = on_bline a b; c = on_tline a o a, on_tline b o b; x = mirror a b; o1 = circumcenter b x c; d = on_circle o a, on_circle o1 b; e = on_line c d, on_circle o a ? perp o1 x e x
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A = point; B = point; l = perpendicular_bisector A B; O = on_line l; (O) = circle_center_point O A; OA = line O A; OB = line O B; l1 = perpendicular_line A OA; l2 = perpendicular_line B OB; C = intersection l1 l2; X = reflect A B; O1 = circumcenter B X C; (O1) = circle_center_point O1 B; D = intersection (O1) (O); CD = line C D; E = intersection CD (O); EX = line E X; O1X = line O1 X; Prove: perpendicular EX O1X
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2015EuropeanMCupSp3
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Circles $k_1$ and $k_2$ intersect in points $A$ and $B$, such that $k_1$ passes through the center $O$ of the circle $k_2$. The line $p$ intersects $k_1$ in points $K$ and $O$ and $k_2$ in points $L$ and $M$, such that the point $L$ is between $K$ and $O$. The point $P$ is orthogonal projection of the point $L$ to the line $AB$. Prove that the line $KP$ is parallel to the $M-$median of the triangle $ABM$.
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a b = segment; o = on_bline a b; t = free; k = on_line o t, on_circum a b o; l = on_line o t, on_circle o a; m = mirror l o; p = foot l a b; n = midpoint a b ? para k p m n
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A = point; B = point; l = perpendicular_bisector A B; O = on_line l; (ABO) = circumcircle A B O; K = on_circle (ABO); OK = line O K; (O) = circle_center_point O A; L M = intersection OK (O); AB = line A B; P = foot L AB; N = midpoint A B; KP = line K P; MN = line M N; Prove: parallel KP MN
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2015IranGOAp1
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let $ w_1 $ and $ w_2 $ two circles such that $ w_1 \cap w_2 = \{ A , B \} $ let $ X $ a point on $ w_2 $ and $ Y $ on $ w_1 $ such that $ BY \bot BX $ suppose that $ O $ is the center of $ w_1 $ and $ X' = w_2 \cap OX $ now if $ K = w_2 \cap X'Y $ prove $ X $ is the midpoint of arc $ AK $
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a b = segment; o1 = on_bline a b; o2 = on_bline a b; x = on_circle o2 a; y = on_circle o1 a, on_tline b x b; x1 = on_line o1 x, on_circle o2 a; k = on_line x1 y, on_circle o2 a ? cong a x k x
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A = point; B = point; l = perpendicular_bisector A B; O = on_line l; P = on_line l; (O) = circle_center_point O A; (P) = circle_center_point P A; X = on_circle (P); BX = line B X; l1 = perpendicular_line B BX; Y = intersection (O) l1; OX = line O X; X1 = intersection OX (P); X1Y = line X1 Y; K = intersection (P) X1Y; Prove: cong A X X K
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2015IranGOAp3
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let $ H $ the orthocenter of the triangle $ ABC $ pass two lines $ l_1 $ and $ l_2 $ through $ H $ such that $ l_1 \bot l_2 $ we have $ l_1 \cap BC = D $ and $ l_1 \cap AB = Z $ also $ l_2 \cap BC = E $ and $ l_2 \cap AC = X $ like this picture pass a line $ d_1$ through $ D $ parallel to $ AC $ and another line $ d_2 $ through $ E $ parallel to $ AB $ let $ d_1 \cap d_2 = Y $ prove $ X $ $ , $ $ Y $ and $ Z $ are on a same line
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a b c = triangle; h = orthocenter a b c; p = free; d = on_line b c, on_line h p; z = on_line a b, on_line h p; q = on_tline h h p; e = on_line b c, on_line h q; x = on_line a c, on_line h q; y = on_pline d a c, on_pline e a b ? coll x y z
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A B C = triangle; H = orthocenter A B C; l1 = line_through H; l2 = perpendicular_line H l1; BC = line B C; D = intersection l1 BC; AB = line A B; Z = intersection l1 AB; E = intersection l2 BC; AC = line A C; X = intersection l2 AC; m1 = parallel_line D AC; m2 = parallel_line E AB; Y = intersection m1 m2; Prove: collinear X Y Z
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2015IranGOAp5
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we have a triangle $ ABC $ and make rectangles $ ABA_1B_2 $ , $ BCB_1C_2 $ and $ CAC_1A_2 $ out of it. then pass a line through $ A_2 $ perpendicular to $ C_1A_2 $ and pass another line through $ A_1 $ perpendicular to $ A_1B_2 $. let $ A' $ the common point of this two lines. like this we make $ B' $ and $ C' $. prove $ AA' $ , $ BB' $ and $ CC' $ intersect each other in a same point.
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a b c = triangle; a1 = on_tline b a b; b2 = parallelogram a b a1 b2; c2 = on_tline b c b; b1 = parallelogram c b c2 b1; a2 = on_tline c a c; c1 = parallelogram a c a2 c1; a3 = on_tline c2 a1 c2, on_tline b1 a2 b1; b3 = on_tline a2 b1 a2, on_tline c1 b2 c1; c3 = on_tline b2 c1 b2, on_tline a1 a1 c2; x = on_line a a3, on_line b b3 ? coll x c c3
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A B C = triangle; AB = line A B; t1 = perpendicular_line B AB; A1 = on_line t1; B2 = parallelogram A B A1; BC = line B C; t2 = perpendicular_line B BC; C2 = on_line t2; B1 = parallelogram C B C2; AC = line A C; t3 = perpendicular_line C AC; A2 = on_line t3; C1 = parallelogram A C A2; A1C2 = line A1 C2; l1 = perpendicular_line C2 A1C2; A2B1 = line A2 B1; l2 = perpendicular_line B1 A2B1; A3 = intersection l1 l2; l3 = perpendicular_line A2 A2B1; B2C1 = line B2 C1; l4 = perpendicular_line C1 B2C1; B3 = intersection l3 l4; l5 = perpendicular_line B2 B2C1; l6 = perpendicular_line A1 A1C2; C3 = intersection l5 l6; AA3 = line A A3; BB3 = line B B3; X = intersection AA3 BB3; Prove: collinear X C C3
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2015IranTSTp18
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$AH$ is the altitude of triangle $ABC$ and $H^\prime$ is the reflection of $H$ trough the midpoint of $BC$. If the tangent lines to the circumcircle of $ABC$ at $B$ and $C$, intersect each other at $X$ and the perpendicular line to $XH^\prime$ at $H^\prime$, intersects $AB$ and $AC$ at $Y$ and $Z$ respectively, prove that $\angle ZXC=\angle YXB$.
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a b c = triangle; h = foot a b c; m_bc = midpoint b c; h1 = mirror h m_bc; o = circumcenter a b c; x = on_tline b o b, on_tline c o c; y = on_line a b, on_tline h1 x h1; z = on_line a c, on_tline h1 x h1 ? eqangle z x x c y x x b
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A B C = triangle; BC = line B C; H = foot A BC; M_bc = midpoint B C; H1 = reflect_point_wrt_point H M_bc; O = circumcenter A B C; OB = line O B; l1 = perpendicular_line B OB; OC = line O C; l2 = perpendicular_line C OC; X = intersection l1 l2; AB = line A B; XH1 = line X H1; l3 = perpendicular_line H1 XH1; Y = intersection AB l3; AC = line A C; Z = intersection AC l3; Prove: equal_angle Z X C Y X B
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2015RMMp4
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Let $ABC$ be a triangle, and let $D$ be the point where the incircle meets side $BC$. Let $J_b$ and $J_c$ be the incentres of the triangles $ABD$ and $ACD$, respectively. Prove that the circumcentre of the triangle $AJ_bJ_c$ lies on the angle bisector of $\angle BAC$.
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a b c = triangle; i = incenter a b c; d = foot i b c; j1 = incenter a b d; j2 = incenter a c d; o = circumcenter a j1 j2 ? eqangle b a a o a o a c
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A B C = triangle; I = incenter A B C; BC = line B C; D = foot I BC; J1 = incenter A B D; J2 = incenter A C D; O = circumcenter A J1 J2; Prove: equal_angle B A O O A C
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2015USATSTp1
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Let $ABC$ be a non-isosceles triangle with incenter $I$ whose incircle is tangent to $\overline{BC}$, $\overline{CA}$, $\overline{AB}$ at $D$, $E$, $F$, respectively. Denote by $M$ the midpoint of $\overline{BC}$. Let $Q$ be a point on the incircle such that $\angle AQD = 90^{\circ}$. Let $P$ be the point inside the triangle on line $AI$ for which $MD = MP$. Prove that either $\angle PQE = 90^{\circ}$ or $\angle PQF = 90^{\circ}$.
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a b c = triangle; i = incenter a b c; d = foot i b c; e = foot i c a; f = foot i a b; m = midpoint b c; o1 = midpoint a d; q = on_circle i d, on_circle o1 a; p = on_line a i, on_circle m d ? perp p q q e
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A B C = triangle; I = incenter A B C; BC = line B C; CA = line C A; AB = line A B; D = foot I BC; E = foot I CA; F = foot I AB; (I) = circle_center_point I D; M = midpoint B C; O1 = midpoint A D; (O1) = circle_center_point O1 A; Q = intersection (I) (O1); (M,D) = circle_center_point M D; AI = line A I; P = intersection AI (M,D); PQ = line P Q; QF = line Q F; Prove: perpendicular PQ QF
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2015USATSTp6
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Let $ABC$ be a non-equilateral triangle and let $M_a$, $M_b$, $M_c$ be the midpoints of the sides $BC$, $CA$, $AB$, respectively. Let $S$ be a point lying on the Euler line. Denote by $X$, $Y$, $Z$ the second intersections of $M_aS$, $M_bS$, $M_cS$ with the nine-point circle. Prove that $AX$, $BY$, $CZ$ are concurrent.
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a b c = triangle; ma = midpoint b c; mb = midpoint c a; mc = midpoint a b; o = circumcenter a b c; h = orthocenter a b c; s = on_line o h; n = circumcenter ma mb mc; x = on_line ma s, on_circle n ma; y = on_line mb s, on_circle n mb; z = on_line mc s, on_circle n mc; k = on_line a x, on_line b y ? coll k c z
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A B C = triangle; Ma = midpoint B C; Mb = midpoint C A; Mc = midpoint A B; Ni = circumcenter Ma Mb Mc; (Ni) = circle_center_point Ni Ma; O = circumcenter A B C; H = orthocenter A B C; OH = line O H; S = on_line O H; MaS = line Ma S; X = intersection MaS (Ni); MbS = line Mb S; Y = intersection MbS (Ni); McS = line Mc S; Z = intersection McS (Ni); AX = line A X; BY = line B Y; W = intersection AX BY; Prove: collinear W C Z
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2016ARMOg10p2
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Diagonals $AC,BD$ of cyclic quadrilateral $ABCD$ intersect at $P$.Point $Q$ is on$BC$ (between$B$ and $C$) such that $PQ \perp AC$.Prove that the line passes through the circumcenters of triangles $APD$ and $BQD$ is parallel to $AD$.
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a b c = triangle; d = on_circum a b c; p = on_line a c, on_line b d; q = on_line b c, on_tline p a c; o1 = circumcenter a p d; o2 = circumcenter b q d ? para o1 o2 a d
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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; l1 = perpendicular_line P AC; BC = line B C; Q = intersection BC l1; Oa = circumcenter A P D; Ob = circumcenter B Q D; OaOb = line Oa Ob; AD = line A D; Prove: parallel OaOb AD
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2016ARMOg10p8
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In acute triangle $ABC$,$AC<BC$,$M$ is midpoint of $AB$ and $\Omega$ is it's circumcircle.Let $C^\prime$ be antipode of $C$ in $\Omega$. $AC^\prime$ and $BC^\prime$ intersect with $CM$ at $K,L$,respectively.The perpendicular drawn from $K$ to $AC^\prime$ and perpendicular drawn from $L$ to $BC^\prime$ intersect with $AB$ and each other and form a triangle $\Delta$.Prove that circumcircles of $\Delta$ and $\Omega$ are tangent.
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a b c = triangle; m = midpoint a b; o = circumcenter a b c; c1 = mirror c o; k = on_line a c1, on_line c m; l = on_line b c1, on_line c m; p = on_line a b, on_tline k a c1; q = on_line a b, on_tline l b c1; r = on_tline k a c1, on_tline l b c1; o1 = circumcenter p q r; t = on_circle o a, on_circle o1 p ? coll o o1 t
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A B C = acute_triangle; M = midpoint A B; O = circumcenter A B C; (ABC) = circle_center_point O A; C1 = reflect_point_wrt_point C O; AC1 = line A C1; CM = line C M; K = intersection AC1 CM; BC1 = line B C1; L = intersection BC1 CM; p1 = perpendicular_line K AC1; p2 = perpendicular_line L BC1; AB = line A B; X = intersection p1 AB; Y = intersection p2 AB; Z = intersection p1 p2; O1 = circumcenter X Y Z; (XYZ) = circle_center_point O1 X; T = intersection (ABC) (XYZ); Prove: collinear O O1 T
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2016ARMOg11p8
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Medians $AM_A,BM_B,CM_C$ of triangle $ABC$ intersect at $M$.Let $\Omega_A$ be circumcircle of triangle passes through midpoint of $AM$ and tangent to $BC$ at $M_A$.Define $\Omega_B$ and $\Omega_C$ analogusly.Prove that $\Omega_A,\Omega_B$ and $\Omega_C$ intersect at one point.
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a b c = triangle; ma mb mc m = centroid a b c; pa = midpoint a m; pb = midpoint b m; pc = midpoint c m; oa = on_tline ma b c, on_bline ma pa; ob = on_tline mb a c, on_bline mb pb; oc = on_tline mc a b, on_bline mc pc; x = on_circle oa ma, on_circle ob mb ? cong x oc oc mc
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A B C = triangle; M Ma Mb Mc = centroid A B C; Pa = midpoint A M; Pb = midpoint B M; Pc = midpoint C M; BC = line B C; l1 = perpendicular_line Ma BC; l2 = perpendicular_bisector Ma Pa; Oa = intersection l1 l2; (Oa) = circle_center_point Oa Ma; AC = line A C; l3 = perpendicular_line Mb AC; l4 = perpendicular_bisector Mb Pb; Ob = intersection l3 l4; (Ob) = circle_center_point Ob Mb; AB = line A B; l5 = perpendicular_line Mc AB; l6 = perpendicular_bisector Mc Pc; Oc = intersection l5 l6; X = intersection (Oa) (Ob); Prove: cong X Oc Oc Mc
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2016BalkanMOp2
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Let $ABCD$ be a cyclic quadrilateral with $AB<CD$. The diagonals intersect at the point $F$ and lines $AD$ and $BC$ intersect at the point $E$. Let $K$ and $L$ be the orthogonal projections of $F$ onto lines $AD$ and $BC$ respectively, and let $M$, $S$ and $T$ be the midpoints of $EF$, $CF$ and $DF$ respectively. Prove that the second intersection point of the circumcircles of triangles $MKT$ and $MLS$ lies on the segment $CD$.
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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 a d, on_line b c; k = foot f a d; l = foot f b c; m = midpoint e f; s = midpoint c f; t = midpoint d f; o1 = circumcenter m k t; o2 = circumcenter m l s; p = on_circle o1 m, on_circle o2 m ? coll c d p
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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; AD = line A D; BC = line B C; E = intersection AD BC; K = foot F AD; L = foot F BC; M = midpoint E F; S = midpoint C F; T = midpoint D F; O1 = circumcenter K M T; O2 = circumcenter L M S; (O1) = circle_center_point O1 M; (O2) = circle_center_point O2 M; X = intersection (O1) (O2); CD = line C D; Prove: collinear X C D
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2016CGMOp7
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In acute triangle $ABC, AB<AC$, $I$ is its incenter, $D$ is the foot of perpendicular from $I$ to $BC$, altitude $AH$ meets $BI,CI$ at $P,Q$ respectively. Let $O$ be the circumcenter of $\triangle IPQ$, extend $AO$ to meet $BC$ at $L$. Circumcircle of $\triangle AIL$ meets $BC$ again at $N$. Prove that $\frac{BD}{CD}=\frac{BN}{CN}$.
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a b c = triangle; i = incenter a b c; d = foot i b c; h = foot a b c; p = on_line a h, on_line b i; q = on_line a h, on_line c i; o = circumcenter i p q; l = on_line a o, on_line b c; n = on_line b c, on_circum a i l ? eqratio b d c d b n c n
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A B C = acute_triangle; I = incenter A B C; BC = line B C; D = foot I BC; H = foot A BC; AH = line A H; BI = line B I; P = intersection AH BI; CI = line C I; Q = intersection AH CI; O = circumcenter I P Q; AO = line A O; L = intersection AO BC; (AIL) = circle A I L; N = intersection (AIL) BC; Prove: equal_ratio B D C D B N C N
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2016CTSTp5
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Refer to the diagram below. Let $ABCD$ be a cyclic quadrilateral with center $O$. Let the internal angle bisectors of $\angle A$ and $\angle C$ intersect at $I$ and let those of $\angle B$ and $\angle D$ intersect at $J$. Now extend $AB$ and $CD$ to intersect $IJ$ and $P$ and $R$ respectively and let $IJ$ intersect $BC$ and $DA$ at $Q$ and $S$ respectively. Let the midpoints of $PR$ and $QS$ be $M$ and $N$ respectively. Given that $O$ does not lie on the line $IJ$, show that $OM$ and $ON$ are perpendicular.
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a b c = triangle; d = on_circum a b c; o = circumcenter a b c; i = angle_bisector d a b, angle_bisector b c d; j = angle_bisector a b c, angle_bisector c d a; p = on_line a b, on_line i j; r = on_line c d, on_line i j; q = on_line b c, on_line i j; s = on_line d a, on_line i j; m = midpoint p r; n = midpoint q s ? perp o m o n
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A B C = triangle; (ABC) = circle A B C; D = on_circle (ABC); O = circumcenter A B C; l1 = angle_bisector B A D; l2 = angle_bisector B C D; I = intersection l1 l2; l3 = angle_bisector A B C; l4 = angle_bisector C D A; J = intersection l3 l4; AB = line A B; IJ = line I J; P = intersection AB IJ; CD = line C D; R = intersection CD IJ; BC = line B C; Q = intersection IJ BC; AD = line A D; S = intersection IJ AD; M = midpoint P R; N = midpoint Q S; MO = line M O; NO = line N O; Prove: perpendicular MO NO
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2016CTSTp7
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$P$ is a point in the interior of acute triangle $ABC$. $D,E,F$ are the reflections of $P$ across $BC,CA,AB$ respectively. Rays $AP,BP,CP$ meet the circumcircle of $\triangle ABC$ at $L,M,N$ respectively. Prove that the circumcircles of $\triangle PDL,\triangle PEM,\triangle PFN$ meet at a point $T$ different from $P$.
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a b c = triangle; p = free; d = reflect p b c; e = reflect p c a; f = reflect p a b; o = circumcenter a b c; l = on_line a p, on_circle o a; m = on_line b p, on_circle o a; n = on_line c p, on_circle o a; o1 = circumcenter p d l; o2 = circumcenter p e m; t = on_circle o1 p, on_circle o2 p ? cyclic t p f n
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A B C = acute_triangle; P = point; BC = line B C; D = reflect_point_wrt_line P BC; AC = line A C; E = reflect_point_wrt_line P AC; AB = line A B; F = reflect_point_wrt_line P AB; AP = line A P; O = circumcenter A B C; (O) = circle_center_point O A; L = intersection AP (O); BP = line B P; M = intersection BP (O); CP = line C P; N = intersection CP (O); O1 = circumcenter D L P; (O1) = circle_center_point O1 P; O2 = circumcenter E M P; (O2) = circle_center_point O2 P; X = intersection (O1) (O2); Prove: concyclic X F N P
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2016CanadaMOp5
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Let $\triangle ABC$ be an acute-angled triangle with altitudes $AD$ and $BE$ meeting at $H$. Let $M$ be the midpoint of segment $AB$, and suppose that the circumcircles of $\triangle DEM$ and $\triangle ABH$ meet at points $P$ and $Q$ with $P$ on the same side of $CH$ as $A$. Prove that the lines $ED, PH,$ and $MQ$ all pass through a single point on the circumcircle of $\triangle ABC$.
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a b c = triangle; d = foot a b c; e = foot b a c; h = orthocenter a b c; m = midpoint a b; o1 = circumcenter d e m; o2 = circumcenter a b h; p = on_circle o1 d, on_circle o2 a; q = on_circle o1 d, on_circle o2 a; x = on_line e d, on_line p h ? coll x m q
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A B C = triangle; BC = line B C; AC = line A C; D = foot A BC; E = foot B AC; H = orthocenter A B C; M = midpoint A B; O1 = circumcenter D E M; O2 = circumcenter A B H; (O1) = circle_center_point O1 D; (O2) = circle_center_point O2 A; P Q = intersection (O1) (O2); ED = line E D; PH = line P H; X = intersection ED PH; Prove: collinear X M Q
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2016CzechAPSlovakp6
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Let $ABC$ be an acute-angled triangle with $AB < AC$. Tangent to its circumcircle $\Omega$ at $A$ intersects the line $BC$ at $D$. Let $G$ be the centroid of $\triangle ABC$ and let $AG$ meet $\Omega$ again at $H \neq A$. Suppose the line $DG$ intersects the lines $AB$ and $AC$ at $E$ and $F$, respectively. Prove that $\angle EHG = \angle GHF$
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a b c = triangle; o = circumcenter a b c; d = on_line b c, on_tline a o a; x y z g = centroid a b c; h = on_line a g, on_circle o a; e = on_line a b, on_line d g; f = on_line a c, on_line d g ? eqangle e h h g g h h f
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A B C = acute_triangle; O = circumcenter A B C; BC = line B C; OA = line O A; l = perpendicular_line A OA; D = intersection l BC; G X Y Z = centroid A B C; AG = line A G; (O) = circle_center_point O A; H = intersection AG (O); DG = line D G; AB = line A B; E = intersection DG AB; AC = line A C; F = intersection DG AC; Prove: equal_angle E H G G H F
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2016EGMOp2
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Let $ABCD$ be a cyclic quadrilateral, and let diagonals $AC$ and $BD$ intersect at $X$.Let $C_1,D_1$ and $M$ be the midpoints of segments $CX,DX$ and $CD$, respecctively. Lines $AD_1$ and $BC_1$ intersect at $Y$, and line $MY$ intersects diagonals $AC$ and $BD$ at different points $E$ and $F$, respectively. Prove that line $XY$ is tangent to the circle through $E,F$ and $X$.
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a b c = triangle; d = on_circum a b c; x = on_line a c, on_line b d; c1 = midpoint c x; d1 = midpoint d x; m = midpoint c d; y = on_line a d1, on_line b c1; e = on_line a c, on_line m y; f = on_line b d, on_line m y; o1 = circumcenter e f x ? perp o1 x x y
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A B C = triangle; (ABC) = circle A B C; D = on_circle (ABC); AC = line A C; BD = line B D; X = intersection AC BD; C1 = midpoint C X; D1 = midpoint D X; M = midpoint C D; AD1 = line A D1; BC1 = line B C1; Y = intersection AD1 BC1; MY = line M Y; E = intersection MY AC; F = intersection MY BD; XY = line X Y; O1 = circumcenter E F X; O1X = line O1 X; Prove: perpendicular O1X XY
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2016ELMOp2
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Oscar is drawing diagrams with trash can lids and sticks. He draws a triangle $ABC$ and a point $D$ such that $DB$ and $DC$ are tangent to the circumcircle of $ABC$. Let $B'$ be the reflection of $B$ over $AC$ and $C'$ be the reflection of $C$ over $AB$. If $O$ is the circumcenter of $DB'C'$, help Oscar prove that $AO$ is perpendicular to $BC$.
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a b c = triangle; o = circumcenter a b c; d = on_tline b o b, on_tline c o c; b1 = reflect b a c; c1 = reflect c a b; o1 = circumcenter d b1 c1 ? perp a o1 b c
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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; D = intersection l1 l2; AC = line A C; B1 = reflect_point_wrt_line B AC; AB = line A B; C1 = reflect_point_wrt_line C AB; O1 = circumcenter D B1 C1; AO1 = line A O1; BC = line B C; Prove: perpendicular AO1 BC
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2016G2
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Let $ABC$ be a triangle with circumcircle $\Gamma$ and incenter $I$ and let $M$ be the midpoint of $\overline{BC}$. The points $D$, $E$, $F$ are selected on sides $\overline{BC}$, $\overline{CA}$, $\overline{AB}$ such that $\overline{ID} \perp \overline{BC}$, $\overline{IE}\perp \overline{AI}$, and $\overline{IF}\perp \overline{AI}$. Suppose that the circumcircle of $\triangle AEF$ intersects $\Gamma$ at a point $X$ other than $A$. Prove that lines $XD$ and $AM$ meet on $\Gamma$.
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a b c = triangle; o = circumcenter a b c; i = incenter a b c; m = midpoint b c; d = foot i b c; e = on_line a c, on_tline i a i; f = on_line a b, on_tline i a i; o1 = circumcenter a e f; x = on_circle o1 a, on_circle o a; y = on_line x d, on_line a m ? cyclic y a b c
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A B C = triangle; I = incenter A B C; M = midpoint B C; BC = line B C; D = foot I BC; AI = line A I; l = perpendicular_line I AI; AB = line A B; F = intersection l AB; AC = line A C; E = intersection l AC; O1 = circumcenter A E F; (O1) = circle_center_point O1 A; O = circumcenter A B C; (O) = circle_center_point O A; X = intersection (O1) (O); DX = line D X; AM = line A M; T = intersection DX AM; Prove: concyclic A B C T
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2016G4
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Let $ABC$ be a triangle with $AB = AC \neq BC$ and let $I$ be its incentre. The line $BI$ meets $AC$ at $D$, and the line through $D$ perpendicular to $AC$ meets $AI$ at $E$. Prove that the reflection of $I$ in $AC$ lies on the circumcircle of triangle $BDE$.
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a b c = iso_triangle; i = incenter a b c; d = on_line a c, on_line b i; e = on_line a i, on_tline d a c; i1 = reflect i a c ? cyclic b d e i1
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A B C = isos_triangle; I = incenter A B C; BI = line B I; AC = line A C; D = intersection BI AC; l1 = perpendicular_line D AC; AI = line A I; E = intersection l1 AI; I1 = reflect_point_wrt_line I AC; Prove: concyclic B D E I1
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2016G5
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Let $D$ be the foot of perpendicular from $A$ to the Euler line (the line passing through the circumcentre and the orthocentre) of an acute scalene triangle $ABC$. A circle $\omega$ with centre $S$ passes through $A$ and $D$, and it intersects sides $AB$ and $AC$ at $X$ and $Y$ respectively. Let $P$ be the foot of altitude from $A$ to $BC$, and let $M$ be the midpoint of $BC$. Prove that the circumcentre of triangle $XSY$ is equidistant from $P$ and $M$.
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a b c = triangle; o = circumcenter a b c; h = orthocenter a b c; d = foot a o h; s = on_bline a d; x = on_circle s a, on_line a b; y = on_circle s a, on_line a c; p = foot a b c; m = midpoint b c; o1 = circumcenter x s y ? cong o1 p o1 m
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A B C = triangle; O = circumcenter A B C; H = orthocenter A B C; OH = line O H; D = foot A OH; l = perpendicular_bisector A D; S = on_line l; (S,A) = circle_center_point S A; AB = line A B; X = intersection (S,A) AB; AC = line A C; Y = intersection (S,A) AC; BC = line B C; P = foot A BC; M = midpoint B C; O1 = circumcenter X S Y; Prove: cong O1 P O1 M
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2016G6
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Let $ABCD$ be a convex quadrilateral with $\angle ABC = \angle ADC < 90^{\circ}$. The internal angle bisectors of $\angle ABC$ and $\angle ADC$ meet $AC$ at $E$ and $F$ respectively, and meet each other at point $P$. Let $M$ be the midpoint of $AC$ and let $\omega$ be the circumcircle of triangle $BPD$. Segments $BM$ and $DM$ intersect $\omega$ again at $X$ and $Y$ respectively. Denote by $Q$ the intersection point of lines $XE$ and $YF$. Prove that $PQ \perp AC$.
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a b c = triangle; d = eqangle3 d a c b c a; e = on_line a c, angle_bisector a b c; f = on_line a c, angle_bisector a d c; p = on_line b e, on_line d f; m = midpoint a c; o = circumcenter b p d; x = on_line b m, on_circle o b; y = on_line d m, on_circle o b; q = on_line x e, on_line y f ? perp p q a c
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A B C = triangle; ω1 = angle_equal2 A C C B A; D = on_circle ω1; AC = line A C; l1 = angle_bisector A B C; E = intersection AC l1; l2 = angle_bisector A D C; F = intersection AC l2; BE = line B E; DF = line D F; P = intersection BE DF; M = midpoint A C; O = circumcenter B P D; BM = line B M; (BPD) = circle_center_point O B; X = intersection BM (BPD); DM = line D M; Y = intersection DM (BPD); XE = line X E; YF = line Y F; Q = intersection XE YF; PQ = line P Q; Prove: perpendicular PQ AC
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2016IranGOAp1
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Let the circles $\omega$ and $\omega^ \prime$ intersect in $A$ and $B$. Tangent to circle$\omega$ at $A$ intersects$\omega^ \prime$ in $C$ and tangent to circle $\omega^ \prime$ at $A$ intersects $\omega$ in $D$. Suppose that $CD$ intersects$\omega$ and $\omega^ \prime$ in $E$ and $F$, respectively (assume that $E$ is between $F$ and $C$). The perpendicular to $AC$ from $E$ intersects $\omega^ \prime$ in point $P$ and perpendicular to $AD$ from $F$ intersects$\omega$ in point $Q$ (The points $A, P$ and $Q$ lie on the same side of the line $CD$). Prove that the points $A, P$ and $Q$ are collinear.
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a b = segment; o1 = on_bline a b; o2 = on_bline a b; c = on_circle o2 a, on_tline a o1 a; d = on_circle o1 a, on_tline a o2 a; e = on_line c d, on_circle o1 a; f = on_line c d, on_circle o2 a; p = on_circle o2 a, on_tline e a c; q = on_circle o1 a, on_tline f a d ? coll a p q
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A = point; B = point; l = perpendicular_bisector A B; O1 = on_line l; O2 = on_line l; (O1) = circle_center_point O1 A; (O2) = circle_center_point O2 A; O1A = line O1 A; l1 = perpendicular_line A O1A; C = intersection l1 (O2); O2A = line O2 A; l2 = perpendicular_line A O2A; D = intersection l2 (O1); CD = line C D; E = intersection CD (O1); F = intersection CD (O2); AC = line A C; l3 = perpendicular_line E AC; P = intersection l3 (O2); AD = line A D; l4 = perpendicular_line F AD; Q = intersection l4 (O1); Prove: collinear A P Q
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2016IranGOMp2
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Let two circles $C_1$ and $C_2$ intersect in points $A$ and $B$. The tangent to $C_1$ at $A$ intersects $C_2$ in $P$ and the line $PB$ intersects $C_1$ for the second time in $Q$ (suppose that $Q$ is outside $C_2$). The tangent to $C_2$ from $Q$ intersects $C_1$ and $C_2$ in $C$ and $D$, respectively. (The points $A$ and $D$ lie on different sides of the line $PQ$.) Show that $AD$ is the bisector of $\angle CAP$.
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a b = segment; o1 = on_bline a b; o2 = on_bline a b; p = on_tline a o1 a, on_circle o2 a; q = on_line p b, on_circle o1 a; d1 d2 = tangent q o2 a; c = on_line q d1, on_circle o1 a ? eqangle c a a d1 d1 a a p
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A = point; B = point; l = perpendicular_bisector A B; O1 = on_line l; O2 = on_line l; (O1) = circle_center_point O1 A; (O2) = circle_center_point O2 A; l1 = tangent_line A (O1); P = intersection l1 (O2); BP = line B P; Q = intersection BP (O1); l2 = tangent_line Q (O2); C = intersection l2 (O1); D = intersection l2 (O2); Prove: equal_angle C A D D A P
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2016IranGOMp4
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Let $\omega$ be the circumcircle of right-angled triangle $ABC$ ($\angle A = 90^{\circ}$). The tangent to $\omega$ at point $A$ intersects the line $BC$ at point $P$. Suppose that $M$ is the midpoint of the minor arc $AB$, and $PM$ intersects $\omega$ for the second time in $Q$. The tangent to $\omega$ at point $Q$ intersects $AC$ at $K$. Prove that $\angle PKC = 90^{\circ}$.
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a b c = r_triangle; o = circumcenter a b c; p = on_line b c, on_tline a o a; m = on_circle o a, on_bline a b; q = on_line p m, on_circle o a; k = on_line a c, on_tline q o q ? perp p k a c
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A B C = right_triangle; O = circumcenter A B C; (O) = circle_center_point O A; l1 = tangent_line A (O); BC = line B C; P = intersection l1 BC; M = midarc_no A B C; MP = line M P; Q = intersection MP (O); l2 = tangent_line Q (O); AC = line A C; K = intersection l2 AC; KP = line K P; CK = line C K; Prove: perpendicular KP CK
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2016IranGOMp5
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Let the circles $\omega$ and $\omega'$ intersect in points $A$ and $B$. The tangent to circle $\omega$ at $A$ intersects $\omega'$ at $C$ and the tangent to circle $\omega'$ at $A$ intersects $\omega$ at $D$. Suppose that the internal bisector of $\angle CAD$ intersects $\omega$ and $\omega'$ at $E$ and $F$, respectively, and the external bisector of $\angle CAD$ intersects $\omega$ and $\omega'$ at $X$ and $Y$, respectively. Prove that the perpendicular bisector of $XY$ is tangent to the circumcircle of triangle $BEF$.
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a b = segment; o1 = on_bline a b; o2 = on_bline a b; c = on_circle o2 a, on_tline a o1 a; d = on_circle o1 a, on_tline a o2 a; r = angle_bisector c a d; s = on_tline a a r; e = on_line a r, on_circle o1 a; f = on_line a r, on_circle o2 a; x = on_line a s, on_circle o1 a; y = on_line a s, on_circle o2 a; o3 = circumcenter b e f; m = midpoint x y; t = on_bline x y, on_pline o3 x y ? cyclic b e f t
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A = point; B = point; l = perpendicular_bisector A B; O1 = on_line l; O2 = on_line l; (O1) = circle_center_point O1 A; (O2) = circle_center_point O2 A; O1A = line O1 A; l1 = perpendicular_line A O1A; C = intersection l1 (O2); O2A = line O2 A; l2 = perpendicular_line A O2A; D = intersection l2 (O1); l3 = angle_bisector C A D; E = intersection l3 (O1); F = intersection l3 (O2); l4 = angle_exbisector C A D; X = intersection l4 (O1); Y = intersection l4 (O2); l5 = perpendicular_bisector X Y; O3 = circumcenter B E F; XY = line X Y; l6 = parallel_line O3 XY; T = intersection l5 l6; Prove: concyclic B E F T
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2016RMMSLG1
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Two circles, $\omega_1$ and $\omega_2$, centred at $O_1$ and $O_2$, respectively, meet at points $A$ and $B$. A line through $B$ meets $\omega_1$ again at $C$, and $\omega_2$ again at $D$. The tangents to $\omega_1$ and $\omega_2$ at $C$ and $D$, respectively, meet at $E$, and the line $AE$ meets the circle $\omega$ through $A, O_1, O_2$ again at $F$. Prove that the length of the segment $EF$ is equal to the diameter of $\omega$.
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a b = segment; o1 = on_bline a b; o2 = on_bline a b; c = on_circle o1 a; d = on_line b c, on_circle o2 a; e = on_tline c o1 c, on_tline d o2 d; o_w = circumcenter a o1 o2; f = on_line a e, on_circle o_w a; f1 = mirror f o_w ? cong e f f f1
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A = point; B = point; l = perpendicular_bisector A B; O1 = on_line l; O2 = on_line l; (O1) = circle_center_point O1 A; (O2) = circle_center_point O2 A; C = on_circle (O1); BC = line B C; D = intersection BC (O2); O1C = line O1 C; O2D = line O2 D; l1 = perpendicular_line C O1C; l2 = perpendicular_line D O2D; E = intersection l1 l2; O_w = circumcenter A O1 O2; AE = line A E; (O_w) = circle_center_point O_w A; F = intersection AE (O_w); F1 = reflect_point_wrt_point F O_w; Prove: cong E F F F1
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2016RMMp1
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Let $ABC$ be a triangle and let $D$ be a point on the segment $BC, D\neq B$ and $D\neq C$. The circle $ABD$ meets the segment $AC$ again at an interior point $E$. The circle $ACD$ meets the segment $AB$ again at an interior point $F$. Let $A'$ be the reflection of $A$ in the line $BC$. The lines $A'C$ and $DE$ meet at $P$, and the lines $A'B$ and $DF$ meet at $Q$. Prove that the lines $AD, BP$ and $CQ$ are concurrent (or all parallel).
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a b c = triangle; d = on_line b c; o1 = circumcenter a b d; e = on_line a c, on_circle o1 a; o2 = circumcenter a c d; f = on_line a b, on_circle o2 a; a1 = reflect a b c; p = on_line a1 c, on_line d e; q = on_line a1 b, on_line d f; x = on_line a d, on_line b p ? coll x c q
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A B C = triangle; BC = line B C; D = on_line BC; O1 = circumcenter A B D; (O1) = circle_center_point O1 A; AC = line A C; E = intersection (O1) AC; O2 = circumcenter A C D; (O2) = circle_center_point O2 A; AB = line A B; F = intersection (O2) AB; A1 = reflect_point_wrt_line A BC; A1C = line A1 C; DE = line D E; P = intersection A1C DE; A1B = line A1 B; DF = line D F; Q = intersection A1B DF; AD = line A D; BP = line B P; X = intersection AD BP; Prove: collinear X C Q
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2016SilkRoadp2
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Around the acute-angled triangle $ABC$ ($AC>CB$) a circle is circumscribed, and the point $N$ is midpoint of the arc $ACB$ of this circle. Let the points $A_1$ and $B_1$ be the feet of perpendiculars on the straight line $NC$, drawn from points $A$ and $B$ respectively (segment $NC$ lies inside the segment $A_1B_1$). Altitude $A_1A_2$ of triangle $A_1AC$ and altitude $B_1B_2$ of triangle $B_1BC$ intersect at a point $K$ . Prove that $\angle A_1KN=\angle B_1KM$, where $M$ is midpoint of the segment $A_2B_2$ .
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a b c = triangle; o = circumcenter a b c; n = on_bline a b, on_circum a b c; a1 = foot a n c; b1 = foot b n c; a2 = foot a1 a c; b2 = foot b1 b c; k = on_line a1 a2, on_line b1 b2; m = midpoint a2 b2 ? eqangle a1 k k n k m b1 k
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A B C = acute_triangle; N = midarc A C B; CN = line C N; A1 = foot A CN; B1 = foot B CN; AC = line A C; A2 = foot A1 AC; BC = line B C; B2 = foot B1 BC; A1A2 = line A1 A2; B1B2 = line B1 B2; K = intersection A1A2 B1B2; M = midpoint A2 B2; Prove: equal_angle A1 K N M K B1
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2016USAMOp3
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Let $\triangle ABC$ be an acute triangle, and let $I_B, I_C,$ and $O$ denote its $B$-excenter, $C$-excenter, and circumcenter, respectively. Points $E$ and $Y$ are selected on $\overline{AC}$ such that $\angle ABY=\angle CBY$ and $\overline{BE}\perp\overline{AC}$. Similarly, points $F$ and $Z$ are selected on $\overline{AB}$ such that $\angle ACZ=\angle BCZ$ and $\overline{CF}\perp\overline{AB}$. Lines $\overleftrightarrow{I_BF}$ and $\overleftrightarrow{I_CE}$ meet at $P$. Prove that $\overline{PO}$ and $\overline{YZ}$ are perpendicular.
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a b c = triangle; ib = excenter b c a; ic = excenter c a b; o = circumcenter a b c; e = foot b a c; y = on_line a c, angle_bisector a b c; f = foot c a b; z = on_line a b, angle_bisector a c b; p = on_line ib f, on_line ic e ? perp p o y z
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A B C = acute_triangle; Ib = excenter B C A; Ic = excenter C A B; O = circumcenter A B C; l1 = angle_bisector A B C; AC = line A C; Y = intersection AC l1; l2 = angle_bisector A C B; AB = line A B; Z = intersection AB l2; E = foot B AC; F = foot C AB; FIb = line F Ib; EIc = line E Ic; P = intersection FIb EIc; OP = line O P; YZ = line Y Z; Prove: perpendicular OP YZ
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2016USATSTSTp2
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Let $ABC$ be a scalene triangle with orthocenter $H$ and circumcenter $O$. Denote by $M$, $N$ the midpoints of $\overline{AH}$, $\overline{BC}$. Suppose the circle $\gamma$ with diameter $\overline{AH}$ meets the circumcircle of $ABC$ at $G \neq A$, and meets line $AN$ at a point $Q \neq A$. The tangent to $\gamma$ at $G$ meets line $OM$ at $P$. Show that the circumcircles of $\triangle GNQ$ and $\triangle MBC$ intersect at a point $T$ on $\overline{PN}$.
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a b c = triangle; h = orthocenter a b c; o = circumcenter a b c; m = midpoint a h; n = midpoint b c; g = on_circle m a, on_circle o a; q = on_line a n, on_circle m a; p = on_tline g m g, on_line o m; o1 = circumcenter g n q; o2 = circumcenter m b c; t = on_circle o1 g, on_circle o2 m ? coll t p n
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A B C = triangle; H = orthocenter A B C; O = circumcenter A B C; M = midpoint A H; N = midpoint B C; (M) = circle_center_point M A; (O) = circle_center_point O A; G = intersection (M) (O); AN = line A N; Q = intersection (M) AN; MG = line M G; l1 = perpendicular_line G MG; MO = line M O; P = intersection l1 MO; O1 = circumcenter G N Q; (O1) = circle_center_point O1 G; O2 = circumcenter B C M; (O2) = circle_center_point O2 M; T = intersection (O1) (O2); Prove: collinear P N T
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2016USATSTSTp6
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Let $ABC$ be a triangle with incenter $I$, and whose incircle is tangent to $\overline{BC}$, $\overline{CA}$, $\overline{AB}$ at $D$, $E$, $F$, respectively. Let $K$ be the foot of the altitude from $D$ to $\overline{EF}$. Suppose that the circumcircle of $\triangle AIB$ meets the incircle at two distinct points $C_1$ and $C_2$, while the circumcircle of $\triangle AIC$ meets the incircle at two distinct points $B_1$ and $B_2$. Prove that the radical axis of the circumcircles of $\triangle BB_1B_2$ and $\triangle CC_1C_2$ passes through the midpoint $M$ of $\overline{DK}$.
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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; k = foot d e f; o1 = circumcenter a i b; c1 = on_circle o1 a, on_circle i d; c2 = on_circle o1 a, on_circle i d; o2 = circumcenter a i c; b1 = on_circle o2 a, on_circle i d; b2 = on_circle o2 a, on_circle i d; o3 = circumcenter b b1 b2; o4 = circumcenter c c1 c2; p1 = on_circle o3 b, on_circle o4 c; p2 = on_circle o3 b, on_circle o4 c; m = midpoint d k ? coll m p1 p2
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A B C = triangle; I = incenter A B C; BC = line B C; D = foot I BC; (I) = circle_center_point I D; AC = line A C; E = foot I AC; AB = line A B; F = foot I AB; EF = line E F; K = foot D EF; O1 = circumcenter A I B; (ABI) = circle_center_point O1 A; C1 C2 = intersection (ABI) (I); O2 = circumcenter A I C; (ACI) = circle_center_point O2 A; B1 B2 = intersection (ACI) (I); M = midpoint D K; O3 = circumcenter B B1 B2; (BB1B2) = circle_center_point O3 B; O4 = circumcenter C C1 C2; (CC1C2) = circle_center_point O4 C; P1 P2 = intersection (BB1B2) (CC1C2); Prove: collinear M P1 P2
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2017ARMOg10p2
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Let $ABC$ be an acute angled isosceles triangle with $AB=AC$ and circumcentre $O$. Lines $BO$ and $CO$ intersect $AC, AB$ respectively at $B', C'$. A straight line $l$ is drawn through $C'$ parallel to $AC$. Prove that the line $l$ is tangent to the circumcircle of $\triangle B'OC$.
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a b c = iso_triangle; o = circumcenter a b c; b1 = on_line a c, on_line b o; c1 = on_line a b, on_line c o; o2 = circumcenter b1 o c; s = parallelogram a c c1 s; f = foot o2 c1 s ? cong o2 f o2 b1
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A B C = isos_triangle; O = circumcenter A B C; BO = line B O; AC = line A C; B1 = intersection BO AC; CO = line C O; AB = line A B; C1 = intersection CO AB; l = parallel_line C1 AC; O2 = circumcenter B1 O C; F = foot O2 l; Prove: cong O2 F O2 B1
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2017ARMOg10p8
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In a non-isosceles triangle $ABC$,$O$ and $I$ are circumcenter and incenter,respectively.$B^\prime$ is reflection of $B$ with respect to $OI$ and lies inside the angle $ABI$.Prove that the tangents to circumcirle of $\triangle BB^\prime I$ at $B^\prime$,$I$ intersect on $AC$.
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a b c = triangle; o = circumcenter a b c; i = incenter a b c; b1 = reflect b o i; o1 = circumcenter b b1 i; t = on_tline b1 o1 b1, on_tline i o1 i ? coll a c t
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A B C = triangle; O = circumcenter A B C; I = incenter A B C; IO = line I O; B1 = reflect_point_wrt_line B IO; O1 = circumcenter B B1 I; O1B1 = line O1 B1; l1 = perpendicular_line B1 O1B1; O1I = line O1 I; l2 = perpendicular_line I O1I; T = intersection l1 l2; Prove: collinear A C T
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2017ARMOg9p2
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$ABCD$ is an isosceles trapezoid with $BC || AD$. A circle $\omega$ passing through $B$ and $C$ intersects the side $AB$ and the diagonal $BD$ at points $X$ and $Y$ respectively. Tangent to $\omega$ at $C$ intersects the line $AD$ at $Z$. Prove that the points $X$, $Y$, and $Z$ are collinear.
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d a b c = eq_trapezoid; o = on_bline b c; x = on_line a b, on_circle o b; y = on_line b d, on_circle o b; z = on_line a d, on_tline c o c ? coll x y z
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A B C D = isos_trapezoid; l = perpendicular_bisector B C; O = on_line l; (O) = circle_center_point O B; AB = line A B; X = intersection AB (O); BD = line B D; Y = intersection BD (O); OC = line O C; l1 = perpendicular_line C OC; AD = line A D; Z = intersection l1 AD; Prove: collinear X Y Z
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2017AsiaPacificMOp2
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Let $ABC$ be a triangle with $AB < AC$. Let $D$ be the intersection point of the internal bisector of angle $BAC$ and the circumcircle of $ABC$. Let $Z$ be the intersection point of the perpendicular bisector of $AC$ with the external bisector of angle $\angle{BAC}$. Prove that the midpoint of the segment $AB$ lies on the circumcircle of triangle $ADZ$.
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a b c = triangle; d = on_circum a b c, angle_bisector b a c; z = on_tline a a d, on_bline a c; n = midpoint a b ? cyclic a d z n
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A B C = triangle; l1 = angle_bisector B A C; (ABC) = circle A B C; D = intersection (ABC) l1; l2 = perpendicular_bisector A C; l3 = angle_exbisector B A C; Z = intersection l2 l3; N = midpoint A B; Prove: concyclic A D Z N
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2017BalkanMOp2
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Consider an acute-angled triangle $ABC$ with $AB<AC$ and let $\omega$ be its circumscribed circle. Let $t_B$ and $t_C$ be the tangents to the circle $\omega$ at points $B$ and $C$, respectively, and let $L$ be their intersection. The straight line passing through the point $B$ and parallel to $AC$ intersects $t_C$ in point $D$. The straight line passing through the point $C$ and parallel to $AB$ intersects $t_B$ in point $E$. The circumcircle of the triangle $BDC$ intersects $AC$ in $T$, where $T$ is located between $A$ and $C$. The circumcircle of the triangle $BEC$ intersects the line $AB$ (or its extension) in $S$, where $B$ is located between $S$ and $A$. Prove that $ST$, $AL$, and $BC$ are concurrent.
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a b c = triangle; o = circumcenter a b c; l = on_tline b o b, on_tline c o c; d = on_pline b a c, on_tline c o c; e = on_pline c a b, on_tline b o b; o1 = circumcenter b d c; t = on_line a c, on_circle o1 b; o2 = circumcenter b e c; s = on_line a b, on_circle o2 b; x = on_line s t, on_line a l ? coll x b c
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A B C = acute_triangle; O = circumcenter A B C; (O) = circle_center_point O A; OB = line O B; OC = line O C; t1 = perpendicular_line B OB; t2 = perpendicular_line C OC; L = intersection t1 t2; AC = line A C; l1 = parallel_line B AC; D = intersection l1 t2; AB = line A B; l2 = parallel_line C AB; E = intersection l2 t1; O1 = circumcenter B C D; (O1) = circle_center_point O1 B; T = intersection AC (O1); O2 = circumcenter B C E; (O2) = circle_center_point O2 B; S = intersection AB (O2); ST = line S T; AL = line A L; X = intersection ST AL; Prove: collinear X B C
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2017CHNGaoLian
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Given an isocleos triangle $ABC$ with equal sides $AB=AC$ and incenter $I$.Let $\Gamma_1$be the circle centered at $A$ with radius $AB$,$\Gamma_2$ be the circle centered at $I$ with radius $BI$.A circle $\Gamma_3$ passing through $B,I$ intersects $\Gamma_1$,$\Gamma_2$ again at $P,Q$ (different from $B$) respectively.Let $R$ be the intersection of $PI$ and $BQ$.Show that $BR \perp CR$.
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a b c = iso_triangle; i = incenter a b c; o3 = on_bline b i; p = on_circle a b, on_circle o3 b; q = on_circle i b, on_circle o3 b; r = on_line p i, on_line b q ? perp b r c r
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A B C = isos_triangle; I = incenter A B C; l = perpendicular_bisector B I; O3 = on_line l; (A) = circle_center_point A B; (A) = circle_center_point A B; (I) = circle_center_point I B; (O3) = circle_center_point O3 B; P = intersection (O3) (A); Q = intersection (O3) (I); IP = line I P; BQ = line B Q; R = intersection IP BQ; BR = line B R; CR = line C R; Prove: perpendicular BR CR
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2017CHNSouthEastMOg10p2
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Let $ABC$ be an acute-angled triangle. In $ABC$, $AB \neq AB$, $K$ is the midpoint of the the median $AD$, $DE \perp AB$ at $E$, $DF \perp AC$ at $F$. The lines $KE$, $KF$ intersect the line $BC$ at $M$, $N$, respectively. The circumcenters of $\triangle DEM$, $\triangle DFN$ are $O_1, O_2$, respectively. Prove that $O_1 O_2 \parallel BC$.
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a b c = triangle; d = midpoint b c; k = midpoint a d; e = foot d a b; f = foot d a c; m = on_line b c, on_line k e; n = on_line b c, on_line k f; o1 = circumcenter d e m; o2 = circumcenter d f n ? para o1 o2 b c
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A B C = acute_triangle; D = midpoint B C; K = midpoint A D; AB = line A B; E = foot D AB; AC = line A C; F = foot D AC; EK = line E K; BC = line B C; M = intersection EK BC; FK = line F K; N = intersection FK BC; O1 = circumcenter D E M; O2 = circumcenter D F N; O1O2 = line O1 O2; Prove: parallel O1O2 BC
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2017CHNWesternMOp3
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In triangle $ABC$, let $D$ be a point on $BC$. Let $I_1$ and $I_2$ be the incenters of triangles $ABD$ and $ACD$ respectively. Let $O_1$ and $O_2$ be the circumcenters of triangles $AI_1D$ and $AI_2D$ respectively. Let lines $I_1O_2$ and $I_2O_1$ meet at $P$. Show that $PD\perp BC$.
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a b c = triangle; d = on_line b c; i1 = incenter a b d; i2 = incenter a c d; o1 = circumcenter a i1 d; o2 = circumcenter a i2 d; p = on_line i1 o2, on_line i2 o1 ? perp p d b c
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A B C = triangle; BC = line B C; D = on_line BC; I1 = incenter A B D; I2 = incenter A C D; O1 = circumcenter A I1 D; O2 = circumcenter A I2 D; I1O2 = line I1 O2; I2O1 = line I2 O1; P = intersection I1O2 I2O1; DP = line D P; Prove: perpendicular DP BC
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2017CTSTp14
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Let $ABCD$ be a non-cyclic convex quadrilateral. The feet of perpendiculars from $A$ to $BC,BD,CD$ are $P,Q,R$ respectively, where $P,Q$ lie on segments $BC,BD$ and $R$ lies on $CD$ extended. The feet of perpendiculars from $D$ to $AC,BC,AB$ are $X,Y,Z$ respectively, where $X,Y$ lie on segments $AC,BC$ and $Z$ lies on $BA$ extended. Let the orthocenter of $\triangle ABD$ be $H$. Prove that the common chord of circumcircles of $\triangle PQR$ and $\triangle XYZ$ bisects $BH$.
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a b c d = quadrangle; p = foot a b c; q = foot a b d; r = foot a c d; x = foot d a c; y = foot d b c; z = foot d a b; h = orthocenter a b d; o1 = circumcenter p q r; o2 = circumcenter x y z; k1 = on_circle o1 p, on_circle o2 x; k2 = on_circle o1 p, on_circle o2 x; m = midpoint b h ? coll m k1 k2
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A B C D = quadrilateral; BC = line B C; P = foot A BC; BD = line B D; Q = foot A BD; CD = line C D; R = foot A CD; AC = line A C; X = foot D AC; Y = foot D BC; AB = line A B; Z = foot D AB; H = orthocenter A B D; O1 = circumcenter P Q R; O2 = circumcenter X Y Z; (PQR) = circle_center_point O1 P; (XYZ) = circle_center_point O2 X; K1 K2 = intersection (PQR) (XYZ); M = midpoint B H; Prove: collinear M K1 K2
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2017CTSTp9
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Let $ABCD$ be a quadrilateral and let $l$ be a line. Let $l$ intersect the lines $AB,CD,BC,DA,AC,BD$ at points $X,X',Y,Y',Z,Z'$ respectively. Given that these six points on $l$ are in the order $X,Y,Z,X',Y',Z'$, show that the circles with diameter $XX',YY',ZZ'$ are coaxal.
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a b c d = quadrangle; p1 p2 = segment; x = on_line p1 p2, on_line a b; x1 = on_line p1 p2, on_line c d; y = on_line p1 p2, on_line b c; y1 = on_line p1 p2, on_line d a; z = on_line p1 p2, on_line a c; z1 = on_line p1 p2, on_line b d; m1 = midpoint x x1; m2 = midpoint y y1; m3 = midpoint z z1; t = on_circle m1 x, on_circle m2 y ? perp t z t z1
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A B C D = quadrilateral; l = line; AB = line A B; X = intersection l AB; CD = line C D; X1 = intersection l CD; BC = line B C; Y = intersection l BC; AD = line A D; Y1 = intersection l AD; AC = line A C; Z = intersection l AC; BD = line B D; Z1 = intersection l BD; M1 = midpoint X X1; M2 = midpoint Y Y1; M3 = midpoint Z Z1; (M1) = circle_center_point M1 X; (M2) = circle_center_point M2 Y; T = intersection (M1) (M2); TZ = line T Z; TZ1 = line T Z1; Prove: perpendicular TZ TZ1
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2017CzechAPSlovakp2
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Let ${\omega}$ be the circumcircle of an acute-angled triangle ${ABC}$. Point ${D}$ lies on the arc ${BC}$ of ${\omega}$ not containing point ${A}$. Point ${E}$ lies in the interior of the triangle ${ABC}$, does not lie on the line ${AD}$, and satisfies ${\angle DBE =\angle ACB}$ and ${\angle DCE = \angle ABC}$. Let ${F}$ be a point on the line ${AD}$ such that lines ${EF}$ and ${BC}$ are parallel, and let ${G}$ be a point on ${\omega}$ different from ${A}$ such that ${AF = FG}$. Prove that points ${D,E, F,G}$ lie on one circle.
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a b c = triangle; d = on_circum a b c; e = on_aline e b d b c a, on_aline e c d c b a; f = on_line a d, on_pline e b c; g = on_circle f a, on_circum a b c ? cyclic d e f g
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A B C = acute_triangle; (ABC) = circumcircle A B C; D = on_circle (ABC); l1 = angle_equal1 B D B C A; l2 = angle_equal1 C D C B A; E = intersection l1 l2; BC = line B C; l3 = parallel_line E BC; AD = line A D; F = intersection AD l3; (F) = circle_center_point F A; G = intersection (ABC) (F); Prove: concyclic D E F G
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2017CzechAPSlovakp4
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Let ${ABC}$ be a triangle. Line l is parallel to ${BC}$ and it respectively intersects side ${AB}$ at point ${D}$, side ${AC}$ at point ${E}$, and the circumcircle of the triangle ${ABC}$ at points ${F}$ and ${G}$, where points ${F,D,E,G}$ lie in this order on l. The circumcircles of triangles ${FEB}$ and ${DGC}$ intersect at points ${P}$ and ${Q}$. Prove that points ${A, P,Q}$ are collinear.
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a b c = triangle; d = on_line a b; e = on_line a c, on_pline d b c; o = circumcenter a b c; f = on_line d e, on_circle o a; g = on_line d e, on_circle o a; o1 = circumcenter f e b; o2 = circumcenter d g c; p = on_circle o1 f, on_circle o2 d; q = on_circle o1 f, on_circle o2 d ? coll a p q
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A B C = triangle; AB = line A B; D = on_line AB; AC = line A C; BC = line B C; l = parallel_line D BC; E = intersection l AC; O = circumcenter A B C; (O) = circle_center_point O A; F G = intersection l (O); O1 = circumcenter B E F; (O1) = circle_center_point O1 F; O2 = circumcenter C D G; (O2) = circle_center_point O2 D; P = intersection (O1) (O2); Q = intersection (O1) (O2); Prove: collinear A P Q
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2017EGMOp6
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Let $ABC$ be an acute-angled triangle in which no two sides have the same length. The reflections of the centroid $G$ and the circumcentre $O$ of $ABC$ in its sides $BC,CA,AB$ are denoted by $G_1,G_2,G_3$ and $O_1,O_2,O_3$, respectively. Show that the circumcircles of triangles $G_1G_2C$, $G_1G_3B$, $G_2G_3A$, $O_1O_2C$, $O_1O_3B$, $O_2O_3A$ and $ABC$ have a common point.
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a b c = triangle; x y z g = centroid a b c; o = circumcenter a b c; g1 = reflect g b c; g2 = reflect g a c; g3 = reflect g a b; o1 = reflect o b c; o2 = reflect o a c; o3 = reflect o a b; p = on_circum a b c, on_circum g1 g2 c ? cyclic p o2 o3 a
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A B C = acute_triangle; G X Y Z = centroid A B C; O = circumcenter A B C; BC = line B C; G1 = reflect_point_wrt_line G BC; AC = line A C; G2 = reflect_point_wrt_line G AC; AB = line A B; G3 = reflect_point_wrt_line G AB; O1 = reflect_point_wrt_line O BC; O2 = reflect_point_wrt_line O AC; O3 = reflect_point_wrt_line O AB; (CG1G2) = circle C G1 G2; (ABC) = circle A B C; P = intersection (ABC) (CG1G2); Prove: concyclic P O2 O3 A
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2017EuropeanMCupSp3
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Let $ABC$ be a scalene triangle and let its incircle touch sides $BC$, $CA$ and $AB$ at points $D$, $E$ and $F$ respectively. Let line $AD$ intersect this incircle at point $X$. Point $M$ is chosen on the line $FX$ so that the quadrilateral $AFEM$ is cyclic. Let lines $AM$ and $DE$ intersect at point $L$ and let $Q$ be the midpoint of segment $AE$. Point $T$ is given on the line $LQ$ such that the quadrilateral $ALDT$ is cyclic. Let $S$ be a point such that the quadrilateral $TFSA$ is a parallelogram, and let $N$ be the second point of intersection of the circumcircle of triangle $ASX$ and the line $TS$. Prove that the circumcircles of triangles $TAN$ and $LSA$ are tangent to each other.
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a b c = triangle; i = incenter a b c; d = foot i b c; e = foot i a c; f = foot i a b; x = on_line a d, on_circle i d; o1 = circumcenter a f e; m = on_line f x, on_circle o1 a; l = on_line a m, on_line d e; q = midpoint a e; o2 = circumcenter a l d; t = on_line l q, on_circle o2 a; s = parallelogram a t f s; o3 = circumcenter a s x; n = on_line t l, on_circle o3 a; u = circumcenter t a n; v = circumcenter l s a ? coll a u v
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A B C = triangle; I = incenter A B C; BC = line B C; D = foot I BC; (I) = circle_center_point I D; AC = line A C; E = foot I AC; AB = line A B; F = foot I AB; AD = line A D; X = intersection AD (I); FX = line F X; O1 = circumcenter A F E; (AEF) = circle_center_point O1 A; M = intersection FX (AEF); AM = line A M; DE = line D E; L = intersection AM DE; Q = midpoint A E; LQ = line L Q; O2 = circumcenter A L D; (ADL) = circle_center_point O2 A; T = intersection LQ (ADL); S = parallelogram A T F; O3 = circumcenter A S X; (ASX) = circle_center_point O3 A; ST = line S T; N = intersection (ASX) ST; U = circumcenter A N T; V = circumcenter A L S; Prove: collinear A U V
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2017G3
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Let $O$ be the circumcenter of an acute triangle $ABC$. Line $OA$ intersects the altitudes of $ABC$ through $B$ and $C$ at $P$ and $Q$, respectively. The altitudes meet at $H$. Prove that the circumcenter of triangle $PQH$ lies on a median of triangle $ABC$.
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a b c = triangle; o = circumcenter a b c; h = orthocenter a b c; p = on_line o a, on_line b h; q = on_line o a, on_line c h; k = circumcenter p q h; m = midpoint b c ? coll k a m
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A B C = acute_triangle; O = circumcenter A B C; H = orthocenter A B C; M = midpoint B C; AO = line A O; BH = line B H; P = intersection AO BH; CH = line C H; Q = intersection AO CH; O1 = circumcenter P Q H; Prove: collinear O1 A M
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2017G4
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In triangle $ABC$, let $\omega$ be the excircle opposite to $A$. Let $D, E$ and $F$ be the points where $\omega$ is tangent to $BC, CA$, and $AB$, respectively. The circle $AEF$ intersects line $BC$ at $P$ and $Q$. Let $M$ be the midpoint of $AD$. Prove that the circle $MPQ$ is tangent to $\omega$.
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a b c = triangle; i1 = excenter a b c; d = foot i1 b c; e = foot i1 a c; f = foot i1 a b; o1 = circumcenter a e f; p = on_line b c, on_circle o1 a; q = on_line b c, on_circle o1 a; m = midpoint a d; o2 = circumcenter m p q; u = on_circle o2 m, on_circle i1 d ? coll i1 u o2
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A B C = triangle; I1 = excenter A B C; BC = line B C; D = foot I1 BC; AC = line A C; E = foot I1 AC; AB = line A B; F = foot I1 AB; (I1) = circle_center_point I1 D; O1 = circumcenter A E F; (AEF) = circle_center_point O1 A; P = intersection BC (AEF); Q = intersection BC (AEF); M = midpoint A D; O2 = circumcenter M P Q; (MPQ) = circle_center_point O2 M; U = intersection (MPQ) (I1); Prove: collinear I1 U O2
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2017G5
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Let $ABCC_1B_1A_1$ be a convex hexagon such that $AB=BC$, and suppose that the line segments $AA_1, BB_1$, and $CC_1$ have the same perpendicular bisector. Let the diagonals $AC_1$ and $A_1C$ meet at $D$, and denote by $\omega$ the circle $ABC$. Let $\omega$ intersect the circle $A_1BC_1$ again at $E \neq B$. Prove that the lines $BB_1$ and $DE$ intersect on $\omega$.
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b a c = iso_triangle; p1 p2 = segment; a1 = reflect a p1 p2; b1 = reflect b p1 p2; c1 = reflect c p1 p2; d = on_line a c1, on_line a1 c; o1 = circumcenter a b c; o2 = circumcenter a1 b c1; e = on_circle o1 a, on_circle o2 b; x = on_line b b1, on_line d e ? cyclic x a b c
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l = line; B C A = isos_triangle; A1 = reflect A l; B1 = reflect B l; C1 = reflect C l; AC1 = line A C1; A1C = line A1 C; D = intersection AC1 A1C; O1 = circumcenter A B C; O2 = circumcenter A1 B C1; (A1BC1) = circle_center_point O2 B; (ABC) = circle_center_point O1 A; E = intersection (A1BC1) (ABC); BB1 = line B B1; DE = line D E; X = intersection BB1 DE; Prove: concyclic X A B C
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2017IOMetrop1
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Let $ABCD$ be a parallelogram in which angle at $B$ is obtuse and $AD>AB$. Points $K$ and $L$ on $AC$ such that $\angle ADL=\angle KBA$(the points $A, K, C, L$ are all different, with $K$ between $A$ and $L$). The line $BK$ intersects the circumcircle $\omega$ of $ABC$ at points $B$ and $E$, and the line $EL$ intersects $\omega$ at points $E$ and $F$. Prove that $BF||AC$.
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a b c = triangle; d = parallelogram a b c d; k = on_line a c; l = on_line a c, on_aline l d a a b k; o = circumcenter a b c; e = on_line b k, on_circle o a; f = on_line e l, on_circle o a ? para b f a c
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A B C D = parallelogram; AC = line A C; K = on_line AC; l = angle_equal1 D A A B K; L = intersection AC l; BK = line B K; O = circumcenter A B C; (O) = circle_center_point O A; E = intersection BK (O); EL = line E L; F = intersection EL (O); BF = line B F; Prove: parallel BF AC
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2017IranGOAp3
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Let $O$ be the circumcenter of triangle $ABC$. Line $CO$ intersects the altitude from $A$ at point $K$. Let $P,M$ be the midpoints of $AK$, $AC$ respectively. If $PO$ intersects $BC$ at $Y$, and the circumcircle of triangle $BCM$ meets $AB$ at $X$, prove that $BXOY$ is cyclic.
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a b c = triangle; o = circumcenter a b c; k = on_line c o, on_tline a b c; p = midpoint a k; m = midpoint a c; y = on_line b c, on_line p o; o1 = circumcenter b c m; x = on_line a b, on_circle o1 b ? cyclic b x o y
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A B C = triangle; O = circumcenter A B C; BC = line B C; l1 = perpendicular_line A BC; CO = line C O; K = intersection CO l1; P = midpoint A K; M = midpoint A C; OP = line O P; Y = intersection OP BC; AB = line A B; O1 = circumcenter B C M; (BCM) = circle_center_point O1 B; X = intersection AB (BCM); Prove: concyclic B X O Y
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2017IranGOMp2
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Two circles $\omega_1,\omega_2$ intersect at $A,B$. An arbitrary line through $B$ meets $\omega_1,\omega_2$ at $C,D$ respectively. The points $E,F$ are chosen on $\omega_1,\omega_2$ respectively so that $CE=CB,\ BD=DF$. Suppose that $BF$ meets $\omega_1$ at $P$, and $BE$ meets $\omega_2$ at $Q$. Prove that $A,P,Q$ are collinear.
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a b = segment; o1 = on_bline a b; o2 = on_bline a b; t = free; c = on_line b t, on_circle o1 a; d = on_line b t, on_circle o2 a; e = on_circle o1 a, on_circle c b; f = on_circle o2 a, on_circle d b; p = on_line b f, on_circle o1 a; q = on_line b e, on_circle o2 a ? coll a p q
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A = point; B = point; l = perpendicular_bisector A B; O1 = on_line l; O2 = on_line l; (O1) = circle_center_point O1 A; (O2) = circle_center_point O2 A; T = point; BT = line B T; C = intersection BT (O1); D = intersection BT (O2); (C) = circle_center_point C B; E = intersection (O1) (C); (D) = circle_center_point D B; F = intersection (O2) (D); BF = line B F; P = intersection BF (O1); BE = line B E; Q = intersection BE (O2); Prove: collinear A P Q
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2017IranTSTp18
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In triangle $ABC$ let $O$ and $H$ be the circumcenter and the orthocenter. The point $P$ is the reflection of $A$ with respect to $OH$. Assume that $P$ is not on the same side of $BC$ as $A$. Points $E,F$ lie on $AB,AC$ respectively such that $BE=PC \ , CF=PB$. Let $K$ be the intersection point of $AP,OH$. Prove that $\angle EKF = 90 ^{\circ}$
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a b c = triangle; o = circumcenter a b c; h = orthocenter a b c; p = reflect a o h; e = on_line a b, eqdistance e b p c; f = on_line a c, eqdistance f c p b; k = on_line a p, on_line o h ? perp e k f k
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A B C = triangle; O = circumcenter A B C; H = orthocenter A B C; OH = line O H; P = reflect_point_wrt_line A OH; AB = line A B; (B) = circle_center_radius B P C; E = intersection AB (B); AC = line A C; (C) = circle_center_radius C P B; F = intersection AC (C); AP = line A P; OH = line O H; K = intersection AP OH; EK = line E K; FK = line F K; Prove: perpendicular EK FK
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2017RMMSLG1
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Let $ABCD$ be a trapezium, $AD\parallel BC$, and let $E,F$ be points on the sides$AB$ and $CD$, respectively. The circumcircle of $AEF$ meets $AD$ again at $A_1$, and the circumcircle of $CEF$ meets $BC$ again at $C_1$. Prove that $A_1C_1,BD,EF$ are concurrent.
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d a b c = trapezoid; e = on_line a b; f = on_line c d; a1 = on_line a d, on_circum a e f; c1 = on_line b c, on_circum c e f; x = on_line a1 c1, on_line b d ? coll x e f
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A B C D = trapezoid; AB = line A B; E = on_line AB; CD = line C D; F = on_line CD; AD = line A D; (AEF) = circle A E F; A1 = intersection AD (AEF); BC = line B C; (CEF) = circle C E F; C1 = intersection BC (CEF); A1C1 = line A1 C1; BD = line B D; X = intersection A1C1 BD; Prove: collinear X E F
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2017SilkRoadp2
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The quadrilateral $ABCD$ is inscribed in the circle $\omega$. The diagonals $AC$ and $BD$ intersect at the point $O$. On the segments $AO$ and $DO$, the points $E$ and $F$ are chosen, respectively. The straight line $EF$ intersects $\omega$ at the points $E_1$ and $F_1$. The circumscribed circles of the triangles $ADE$ and $BCF$ intersect the segment $EF$ at the points $E_2$ and $F_2$ respectively (assume that all the points $E, F, E_1, F_1, E_2$ and $F_2$ are different). Prove that $E_1E_2 = F_1F_2$.
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a b c = triangle; d = on_circum a b c; o = on_line a c, on_line b d; e = on_line a o; f = on_line d o; e1 = on_line e f, on_circum a b c; f1 = on_line e f, on_circum a b c; o2 = circumcenter a d e; o3 = circumcenter b c f; e2 = on_line e f, on_circle o2 a; f2 = on_line e f, on_circle o3 b ? cong e1 e2 f1 f2
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A B C = triangle; (ABC) = circumcircle A B C; D = on_circle (ABC); AC = line A C; BD = line B D; O = intersection AC BD; AO = line A O; E = on_line AO; DO = line D O; F = on_line DO; EF = line E F; E1 F1 = intersection EF (ABC); (ADE) = circle A D E; E2 = intersection (ADE) EF; (BCF) = circle B C F; F2 = intersection (BCF) EF; Prove: cong E1 E2 F1 F2
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2017USAMOp3
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Let $ABC$ be a scalene triangle with circumcircle $\Omega$ and incenter $I$. Ray $AI$ meets $\overline{BC}$ at $D$ and meets $\Omega$ again at $M$; the circle with diameter $\overline{DM}$ cuts $\Omega$ again at $K$. Lines $MK$ and $BC$ meet at $S$, and $N$ is the midpoint of $\overline{IS}$. The circumcircles of $\triangle KID$ and $\triangle MAN$ intersect at points $L_1$ and $L_2$. Prove that $\Omega$ passes through the midpoint of either $\overline{IL_1}$ or $\overline{IL_2}$.
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a b c = triangle; o = circumcenter a b c; i = incenter a b c; d = on_line a i, on_line b c; m = on_line a i, on_circle o a; k = on_dia m d, on_circle o a; s = on_line m k, on_line b c; n = midpoint i s; o1 = circumcenter k i d; o2 = circumcenter m a n; l = on_circle o1 k, on_circle o2 m; p = midpoint i l ? cyclic a b c p
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A B C = triangle; I = incenter A B C; AI = line A I; BC = line B C; D = intersection AI BC; O = circumcenter A B C; (O) = circle_center_point O A; M = intersection AI (O); (DM) = circle_diameter D M; K = intersection (DM) (O); KM = line K M; S = intersection KM BC; N = midpoint I S; O1 = circumcenter D I K; (O1) = circle_center_point O1 K; O2 = circumcenter A M N; (O2) = circle_center_point O2 M; L = intersection (O1) (O2); P = midpoint I L; Prove: concyclic A B C P
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2017USATSTSTp5
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Let $ABC$ be a triangle with incenter $I$. Let $D$ be a point on side $BC$ and let $\omega_B$ and $\omega_C$ be the incircles of $\triangle ABD$ and $\triangle ACD$, respectively. Suppose that $\omega_B$ and $\omega_C$ are tangent to segment $BC$ at points $E$ and $F$, respectively. Let $P$ be the intersection of segment $AD$ with the line joining the centers of $\omega_B$ and $\omega_C$. Let $X$ be the intersection point of lines $BI$ and $CP$ and let $Y$ be the intersection point of lines $CI$ and $BP$. Prove that lines $EX$ and $FY$ meet on the incircle of $\triangle ABC$.
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a b c = triangle; i = incenter a b c; d = on_line b c; i2 = incenter a b d; i3 = incenter a c d; e = foot i2 b d; f = foot i3 c d; p = on_line a d, on_line i2 i3; x = on_line b i, on_line c p; y = on_line c i, on_line b p; z = on_line e x, on_line f y; d1 = foot i b c ? cong z i d1 i
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A B C = triangle; I = incenter A B C; BC = line B C; D = on_line BC; Ib = incenter A B D; Ic = incenter A C D; E = foot Ib BC; F = foot Ic BC; AD = line A D; IbIc = line Ib Ic; P = intersection AD IbIc; BI = line B I; CP = line C P; X = intersection BI CP; CI = line C I; BP = line B P; Y = intersection CI BP; EX = line E X; FY = line F Y; Q = intersection EX FY; D1 = foot I BC; Prove: cong Q I D1 I
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2018ARMOg10p2
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Let $\triangle ABC$ be an acute-angled triangle with $AB<AC$. Let $M$ and $N$ be the midpoints of $AB$ and $AC$, respectively; let $AD$ be an altitude in this triangle. A point $K$ is chosen on the segment $MN$ so that $BK=CK$. The ray $KD$ meets the circumcircle $\Omega$ of $ABC$ at $Q$. Prove that $C, N, K, Q$ are concyclic.
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a b c = triangle; m = midpoint a b; n = midpoint a c; d = foot a b c; k = on_line m n, on_bline b c; q = on_line k d, on_circum a b c ? cyclic c n k q
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A B C = acute_triangle; M = midpoint A B; N = midpoint A C; BC = line B C; D = foot A BC; l1 = perpendicular_bisector B C; MN = line M N; K = intersection MN l1; DK = line D K; (ABC) = circle A B C; Q = intersection DK (ABC); Prove: concyclic C N K Q
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2018ARMOg11p4
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On the sides $AB$ and $AC$ of the triangle $ABC$, the points $P$ and $Q$ are chosen, respectively, so that $PQ\parallel BC$. Segments $BQ$ and $CP$ intersect at point $O$. Point $A'$ is symmetric to point $A$ relative to line $BC$. The segment $A'O$ intersects the circumcircle $w$ of the triangle $APQ$ at the point $S$. Prove that circumcircle of $BSC$ is tangent to the circle $w$.
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a b c = triangle; p = on_line a b; q = on_line a c, on_pline p b c; o = on_line b q, on_line c p; a1 = reflect a b c; o1 = circumcenter a p q; s = on_line a1 o, on_circle o1 a; o2 = circumcenter b c s ? coll s o1 o2
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A B C = triangle; AB = line A B; P = on_line AB; BC = line B C; l1 = parallel_line P BC; AC = line A C; Q = intersection AC l1; BQ = line B Q; CP = line C P; O = intersection BQ CP; A1 = reflect_point_wrt_line A BC; A1O = line A1 O; O1 = circumcenter A P Q; (O1) = circle_center_point O1 A; S = intersection A1O (O1); O2 = circumcenter B C S; Prove: collinear S O1 O2
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2018CGMOp2
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Points $D,E$ lie on segments $AB,AC$ of $\triangle ABC$ such that $DE\parallel BC$. Let $O_1,O_2$ be the circumcenters of $\triangle ABE, \triangle ACD$ respectively. Line $O_1O _2$ meets $AC$ at $P$, and $AB$ at $Q$. Let $O$ be the circumcenter of $\triangle APQ$, and $M$ be the intersection of $AO$ extended and $BC$. Prove that $M$ is the midpoint of $BC$.
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a b c = triangle; d = on_line a b; e = on_line a c, on_pline d b c; o1 = circumcenter a b e; o2 = circumcenter a c d; p = on_line o1 o2, on_line a c; q = on_line o1 o2, on_line a b; o3 = circumcenter a p q; m = on_line a o3, on_line b c ? midp m b c
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A B C = triangle; AB = line A B; D = on_line AB; BC = line B C; l1 = parallel_line D BC; AC = line A C; E = intersection AC l1; O1 = circumcenter A B E; O2 = circumcenter A C D; O1O2 = line O1 O2; P = intersection O1O2 AC; Q = intersection O1O2 AB; O3 = circumcenter A P Q; AO3 = line A O3; M = intersection AO3 BC; Prove: midpoint M B C
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2018CHNWesternMOp5
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In acute triangle $ABC,$ $AB<AC,$ $O$ is the circumcenter of the triangle. $M$ is the midpoint of segment $BC,$ $(AOM)$ intersects the line $AB$ again at $D$ and intersects the segment $AC$ at $E.$ Prove that $DM=EC.$
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a b c = triangle; o = circumcenter a b c; m = midpoint b c; o1 = circumcenter a o m; d = on_line a b, on_circle o1 a; e = on_line a c, on_circle o1 a ? cong d m e c
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A B C = acute_triangle; O = circumcenter A B C; M = midpoint B C; O1 = circumcenter A O M; (O1) = circle A M O; AB = line A B; D = intersection (O1) AB; AC = line A C; E = intersection (O1) AC; Prove: cong D M E C
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2018CMOp4
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$ABCD$ is a cyclic quadrilateral whose diagonals intersect at $P$. The circumcircle of $\triangle APD$ meets segment $AB$ at points $A$ and $E$. The circumcircle of $\triangle BPC$ meets segment $AB$ at points $B$ and $F$. Let $I$ and $J$ be the incenters of $\triangle ADE$ and $\triangle BCF$, respectively. Segments $IJ$ and $AC$ meet at $K$. Prove that the points $A,I,K,E$ are cyclic.
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a b c = triangle; d = on_circum a b c; p = on_line a c, on_line b d; o1 = circumcenter a p d; e = on_line a b, on_circle o1 a; o2 = circumcenter b p c; f = on_line a b, on_circle o2 b; i = incenter a d e; j = incenter b c f; k = on_line i j, on_line a c ? cyclic a i k e
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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; O1 = circumcenter A P D; (ADP) = circle_center_point O1 A; AB = line A B; E = intersection (ADP) AB; O2 = circumcenter B P C; (BCP) = circle_center_point O2 B; F = intersection (BCP) AB; I = incenter A D E; J = incenter B C F; IJ = line I J; K = intersection IJ AC; Prove: concyclic A I K E
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2018CTSTp21
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In isosceles $\triangle ABC$, $AB=AC$, points $D,E,F$ lie on segments $BC,AC,AB$ such that $DE\parallel AB$, $DF\parallel AC$. The circumcircle of $\triangle ABC$ $\omega_1$ and the circumcircle of $\triangle AEF$ $\omega_2$ intersect at $A,G$. Let $DE$ meet $\omega_2$ at $K\neq E$. Points $L,M$ lie on $\omega_1,\omega_2$ respectively such that $LG\perp KG, MG\perp CG$. Let $P,Q$ be the circumcenters of $\triangle DGL$ and $\triangle DGM$ respectively. Prove that $A,G,P,Q$ are concyclic.
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a b c = iso_triangle; d = on_line b c; e = on_line a c, on_pline d a b; f = on_line a b, on_pline d a c; o1 = circumcenter a b c; o2 = circumcenter a e f; g = on_circle o1 a, on_circle o2 a; k = on_line d e, on_circle o2 a; l = on_circle o1 a, on_tline g k g; m = on_circle o2 a, on_tline g c g; p = circumcenter d g l; q = circumcenter d g m ? cyclic a g p q
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A B C = isos_triangle; BC = line B C; D = on_line BC; AC = line A C; AB = line A B; l1 = parallel_line D AB; E = intersection AC l1; l2 = parallel_line D AC; F = intersection AB l2; O1 = circumcenter A B C; O2 = circumcenter A E F; (O1) = circle_center_point O1 A; (O2) = circle_center_point O2 A; G = intersection (O1) (O2); DE = line D E; K = intersection DE (O2); KG = line K G; l3 = perpendicular_line G KG; L = intersection (O1) l3; CG = line C G; l4 = perpendicular_line G CG; M = intersection (O2) l4; P = circumcenter D G L; Q = circumcenter D G M; Prove: concyclic A G P Q
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2018CzechAPSlovakp2-
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Let $ABC$ be an acute scalene triangle. Let $D$ and $E$ be points on the sides $AB$ and $AC$, respectively, such that $BD=CE$. Denote by $O_1$ and $O_2$ the circumcentres of the triangles $ABE$ and $ACD$, respectively. Prove that the circumcircles of the triangles $ABC, ADE$, and $AO_1O_2$ have a common point different from $A$.
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a b c = triangle; d = on_line a b; e = on_line a c, eqdistance c b d; o1 = circumcenter a b e; o2 = circumcenter a c d; o = circumcenter a b c; o3 = circumcenter a d e; x = on_circle o a, on_circle o3 a; o4 = circumcenter a o1 o2 ? cong x o4 a o4
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A B C = acute_triangle; AB = line A B; D = on_line AB; AC = line A C; (C) = circle_center_radius C B D; E = intersection AC (C); O1 = circumcenter A B E; O2 = circumcenter A C D; O = circumcenter A B C; (O) = circle_center_point O A; O3 = circumcenter A D E; (O3) = circle_center_point O3 A; X = intersection (O) (O3); O4 = circumcenter A O1 O2; Prove: cong X O4 A O4
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2018EuropeanMCupSp2
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Let ABC be a triangle with$|AB|< |AC|. $ Let $k$ be the circumcircle of $\triangle ABC$ and let $O$ be the center of $k$. Point $M$ is the midpoint of the arc $BC $ of $k$ not containing $A$. Let $D $ be the second intersection of the perpendicular line from $M$ to $AB$ with $ k$ and $E$ be the second intersection of the perpendicular line from $M$ to $AC $ with $k$. Points $X $and $Y $ are the intersections of $CD$ and $BE$ with $OM$ respectively. Denote by $k_b$ and $k_c$ circumcircles of triangles $BDX$ and $CEY$ respectively. Let $G$ and $H$ be the second intersections of $k_b$ and $k_c $ with $AB$ and $AC$ respectively. Denote by ka the circumcircle of triangle $AGH.$ Prove that $O$ is the circumcenter of $\triangle O_aO_bO_c, $where $O_a, O_b, O_c $ are the centers of $k_a, k_b, k_c$ respectively.
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a b c = triangle; o = circumcenter a b c; m = on_circle o a, angle_bisector b a c; d = on_circle o a, on_tline m a b; e = on_circle o a, on_tline m a c; x = on_line c d, on_line o m; y = on_line b e, on_line o m; o2 = circumcenter b d x; o3 = circumcenter c e y; g = on_line a b, on_circle o2 b; h = on_line a c, on_circle o3 c; o1 = circumcenter a g h ? cong o o1 o o2
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A B C = triangle; O = circumcenter A B C; (O) = circle_center_point O A; M = midarc_no B C A; AB = line A B; l1 = perpendicular_line M AB; D = intersection l1 (O); AC = line A C; l2 = perpendicular_line M AC; E = intersection l2 (O); CD = line C D; MO = line M O; X = intersection CD MO; BE = line B E; Y = intersection BE MO; Ob = circumcenter B D X; (Ob) = circle_center_point Ob B; (CEY) = circumcircle C E Y; G = intersection (Ob) AB; H = intersection (CEY) AC; Oa = circumcenter A G H; Prove: cong O Oa O Ob
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2018G2
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Let $ABC$ be a triangle with $AB=AC$, and let $M$ be the midpoint of $BC$. Let $P$ be a point such that $PB<PC$ and $PA$ is parallel to $BC$. Let $X$ and $Y$ be points on the lines $PB$ and $PC$, respectively, so that $B$ lies on the segment $PX$, $C$ lies on the segment $PY$, and $\angle PXM=\angle PYM$. Prove that the quadrilateral $APXY$ is cyclic.
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a b c = iso_triangle; m = midpoint b c; p = on_pline a b c; x = on_line p b; y = on_line p c, eqangle3 y p m x m p ? cyclic a p x y
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A B C = isos_triangle; M = midpoint B C; BC = line B C; l1 = parallel_line A BC; P = on_line l1; PB = line P B; X = on_line PB; ω1 = angle_equal2 M P P X M; PC = line P C; Y = intersection PC ω1; Prove: concyclic A P X Y
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2018G4
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A point $T$ is chosen inside a triangle $ABC$. Let $A_1$, $B_1$, and $C_1$ be the reflections of $T$ in $BC$, $CA$, and $AB$, respectively. Let $\Omega$ be the circumcircle of the triangle $A_1B_1C_1$. The lines $A_1T$, $B_1T$, and $C_1T$ meet $\Omega$ again at $A_2$, $B_2$, and $C_2$, respectively. Prove that the lines $AA_2$, $BB_2$, and $CC_2$ are concurrent on $\Omega$.
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a b c = triangle; t = free; a1 = reflect t b c; b1 = reflect t a c; c1 = reflect t a b; o = circumcenter a1 b1 c1; a2 = on_line a1 t, on_circle o a1; b2 = on_line b1 t, on_circle o b1; c2 = on_line c1 t, on_circle o c1; x = on_line a a2, on_circle o a1 ? coll x b b2
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A B C = triangle; T = point; BC = line B C; A1 = reflect_point_wrt_line T BC; AC = line A C; B1 = reflect_point_wrt_line T AC; AB = line A B; C1 = reflect_point_wrt_line T AB; O = circumcenter A1 B1 C1; (O) = circle_center_point O A1; A1T = line A1 T; A2 = intersection A1T (O); B1T = line B1 T; B2 = intersection B1T (O); C1T = line C1 T; C2 = intersection C1T (O); AA2 = line A A2; X = intersection AA2 (O); Prove: collinear X B B2
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2018G5
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Let $ABC$ be a triangle with circumcircle $\Omega$ and incentre $I$. A line $\ell$ intersects the lines $AI$, $BI$, and $CI$ at points $D$, $E$, and $F$, respectively, distinct from the points $A$, $B$, $C$, and $I$. The perpendicular bisectors $x$, $y$, and $z$ of the segments $AD$, $BE$, and $CF$, respectively determine a triangle $\Theta$. Show that the circumcircle of the triangle $\Theta$ is tangent to $\Omega$.
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a b c = triangle; o = circumcenter a b c; i = incenter a b c; p1 p2 = segment; d = on_line p1 p2, on_line a i; e = on_line p1 p2, on_line b i; f = on_line p1 p2, on_line c i; p = on_bline a d, on_bline b e; q = on_bline b e, on_bline c f; r = on_bline c f, on_bline a d; o1 = circumcenter p q r; t = on_circle o1 p, on_circle o a ? coll t o1 o
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A B C = triangle; O = circumcenter A B C; (ABC) = circle_center_point O A; I = incenter A B C; l = line; AI = line A I; D = intersection l AI; BI = line B I; E = intersection l BI; CI = line C I; F = intersection l CI; l1 = perpendicular_bisector A D; l2 = perpendicular_bisector B E; l3 = perpendicular_bisector C F; P = intersection l2 l3; Q = intersection l1 l3; R = intersection l1 l2; O1 = circumcenter P Q R; (PQR) = circle_center_point O1 P; T = intersection (ABC) (PQR); Prove: collinear T O1 O
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2018IranGOAp1
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Two circles $\omega_1,\omega_2$ intersect each other at points $A,B$. Let $PQ$ be a common tangent line of these two circles with $P \in \omega_1$ and $Q \in \omega_2$. An arbitrary point $X$ lies on $\omega_1$. Line $AX$ intersects $ \omega_2$ for the second time at $Y$ . Point $Y'\ne Y$ lies on $\omega_2$ such that $QY = QY'$. Line $Y'B$ intersects $ \omega_1$ for the second time at $X'$. Prove that $PX = PX'$.
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a b = segment; o1 = on_bline a b; o2 = on_bline a b; p q p1 q1 = cc_tangent p q p1 q1 o1 a o2 a; x = on_circle o1 a; y = on_line a x, on_circle o2 a; y1 = on_circle o2 a, on_circle q y; x1 = on_line y1 b, on_circle o1 a ? cong p x p x1
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A = point; B = point; l = perpendicular_bisector A B; O1 = on_line l; O2 = on_line l; (O1) = circle_center_point O1 A; (O2) = circle_center_point O2 A; l1 = ex_common_tangent (O1) (O2); P = intersection l1 (O1); Q = intersection l1 (O2); X = on_circle (O1); AX = line A X; Y = intersection AX (O2); (Q) = circle_center_point Q Y; Y1 = intersection (O2) (Q); BY1 = line B Y1; X1 = intersection BY1 (O1); Prove: cong P X P X1
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2018IranGOMp3
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Let $\omega_1,\omega_2$ be two circles with centers $O_1$ and $O_2$, respectively. These two circles intersect each other at points $A$ and $B$. Line $O_1B$ intersects $\omega_2$ for the second time at point $C$, and line $O_2A$ intersects $\omega_1$ for the second time at point $D$ . Let $X$ be the second intersection of $AC$ and $\omega_1$. Also $Y$ is the second intersection point of $BD$ and $\omega_2$. Prove that $CX = DY$ .
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a b = segment; o1 = on_bline a b; o2 = on_bline a b; c = on_line o1 b, on_circle o2 a; d = on_line o2 a, on_circle o1 a; x = on_line a c, on_circle o1 a; y = on_line b d, on_circle o2 a ? cong c x d y
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A = point; B = point; l = perpendicular_bisector A B; O1 = on_line l; O2 = on_line l; (O1) = circle_center_point O1 A; (O2) = circle_center_point O2 A; BO1 = line B O1; C = intersection BO1 (O2); AO2 = line A O2; D = intersection AO2 (O1); AC = line A C; X = intersection AC (O1); BD = line B D; Y = intersection BD (O2); Prove: cong C X D Y
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2018KoMaLA719
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Let $ABC$ be a scalene triangle with circumcenter $O$ and incenter $I$. The $A$-excircle, $B$-excircle, and $C$-excircle of triangle $ABC$ touch $BC$, $CA$, and $AB$ at points $A_1$, $B_1$, and $C_1$, respectively. Let $P$ be the orthocenter of $AB_1C_1$ and $H$ be the orthocenter of $ABC$. Show that if $M$ is the midpoint of $PA_1$, then lines $HM$ and $OI$ are parallel.
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a b c = triangle; o = circumcenter a b c; i = incenter a b c; ia = excenter a b c; a1 = foot ia b c; ib = excenter b c a; b1 = foot ib c a; ic = excenter c a b; c1 = foot ic a b; p = orthocenter a b1 c1; h = orthocenter a b c; m = midpoint p a1 ? para h m o i
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A B C = triangle; O = circumcenter A B C; I = incenter A B C; Ia = excenter A B C; Ib = excenter B C A; Ic = excenter C A B; BC = line B C; A1 = foot Ia BC; AC = line A C; B1 = foot Ib AC; AB = line A B; C1 = foot Ic AB; P = orthocenter A B1 C1; H = orthocenter A B C; M = midpoint P A1; HM = line H M; IO = line I O; Prove: parallel HM IO
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2018KoMaLA726
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In triangle $ABC$ with incenter $I$, line $AI$ intersects the circumcircle of $ABC$ at $S\ne A$. Let the reflection of $I$ with respect to $BC$ be $J$, and suppose that line $SJ$ intersects the circumcircle of $ABC$ for the second time at point $P\ne S$. Show that $AI=PI.$
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a b c = triangle; i = incenter a b c; o = circumcenter a b c; s = on_line a i, on_circle o a; j = reflect i b c; p = on_line s j, on_circle o a ? cong a i p i
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A B C = triangle; I = incenter A B C; AI = line A I; O = circumcenter A B C; (O) = circle_center_point O A; S = intersection AI (O); BC = line B C; J = reflect_point_wrt_line I BC; JS = line J S; P = intersection JS (O); Prove: cong A I I P
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