MathDB

1994 Iran MO (2nd round)

Part of Iran MO (2nd Round)

Subcontests

(3)
2
2

Find the angle alpha [Iran Second Round 1994]

In the following diagram, OO is the center of the circle. If three angles α,β\alpha, \beta and γ\gamma be equal, find α.\alpha. [asy] unitsize(40); import graph; size(300); real lsf = 0.5; pen dp = linewidth(0.7) + fontsize(10); defaultpen(dp); pen ds = black; pen ttttff = rgb(0.2,0.2,1); pen ffttww = rgb(1,0.2,0.4); pen qqwuqq = rgb(0,0.39,0); draw(circle((0,0),2.33),ttttff+linewidth(2.8pt)); draw((-1.95,-1.27)--(0.64,2.24),ffttww+linewidth(2pt)); draw((0.64,2.24)--(1.67,-1.63),ffttww+linewidth(2pt)); draw((-1.95,-1.27)--(1.06,0.67),ffttww+linewidth(2pt)); draw((1.67,-1.63)--(-0.6,0.56),ffttww+linewidth(2pt)); draw((-0.6,0.56)--(1.06,0.67),ffttww+linewidth(2pt)); pair parametricplot0_cus(real t){ return (0.6*cos(t)+0.64,0.6*sin(t)+2.24); } draw(graph(parametricplot0_cus,-2.2073069497794027,-1.3111498158746024)--(0.64,2.24)--cycle,qqwuqq); pair parametricplot1_cus(real t){ return (0.6*cos(t)+-0.6,0.6*sin(t)+0.56); } draw(graph(parametricplot1_cus,0.06654165390165974,0.9342857038103908)--(-0.6,0.56)--cycle,qqwuqq); pair parametricplot2_cus(real t){ return (0.6*cos(t)+-0.6,0.6*sin(t)+0.56); } draw(graph(parametricplot2_cus,-0.766242589858673,0.06654165390165967)--(-0.6,0.56)--cycle,qqwuqq); dot((0,0),ds); label("OO", (-0.2,-0.38), NE*lsf); dot((0.64,2.24),ds); label("AA", (0.72,2.36), NE*lsf); dot((-1.95,-1.27),ds); label("BB", (-2.2,-1.58), NE*lsf); dot((1.67,-1.63),ds); label("CC", (1.78,-1.96), NE*lsf); dot((1.06,0.67),ds); label("EE", (1.14,0.78), NE*lsf); dot((-0.6,0.56),ds); label("DD", (-0.92,0.7), NE*lsf); label("α\alpha", (0.48,1.38),NE*lsf); label("β\beta", (-0.02,0.94),NE*lsf); label("γ\gamma", (0.04,0.22),NE*lsf); clip((-8.84,-9.24)--(-8.84,8)--(11.64,8)--(11.64,-9.24)--cycle); [/asy]

Prove that the points are collinear [Iran Second Round 1994]

The incircle of triangle ABCABC meet the sides AB,ACAB, AC and BCBC in M,NM,N and PP, respectively. Prove that the orthocenter of triangle MNP,MNP, the incenter and the circumcenter of triangle ABCABC are collinear.
[asy] import graph; size(300); real lsf = 0.5; pen dp = linewidth(0.7) + fontsize(10); defaultpen(dp); pen ds = black; pen ttttff = rgb(0.2,0.2,1); pen ffwwww = rgb(1,0.4,0.4); pen xdxdff = rgb(0.49,0.49,1); draw((8,17.58)--(2.84,9.26)--(20.44,9.21)--cycle); draw((8,17.58)--(2.84,9.26),ttttff+linewidth(2pt)); draw((2.84,9.26)--(20.44,9.21),ttttff+linewidth(2pt)); draw((20.44,9.21)--(8,17.58),ttttff+linewidth(2pt)); draw(circle((9.04,12.66),3.43),blue+linewidth(1.2pt)+linetype("8pt 8pt")); draw((6.04,14.42)--(8.94,9.24),ffwwww+linewidth(1.2pt)); draw((8.94,9.24)--(11.12,15.48),ffwwww+linewidth(1.2pt)); draw((11.12,15.48)--(6.04,14.42),ffwwww+linewidth(1.2pt)); draw((8.94,9.24)--(7.81,14.79)); draw((11.12,15.48)--(6.95,12.79)); draw((6.04,14.42)--(10.12,12.6)); dot((8,17.58),ds); label("AA", (8.11,18.05),NE*lsf); dot((2.84,9.26),ds); label("BB", (2.11,8.85), NE*lsf); dot((20.44,9.21),ds); label("CC", (20.56,8.52), NE*lsf); dot((9.04,12.66),ds); label("OO", (8.94,12.13), NE*lsf); dot((6.04,14.42),ds); label("MM", (5.32,14.52), NE*lsf); dot((11.12,15.48),ds); label("NN", (11.4,15.9), NE*lsf); dot((8.94,9.24),ds); label("PP", (8.91,8.58), NE*lsf); dot((7.81,14.79),ds); label("DD", (7.81,15.14),NE*lsf); dot((6.95,12.79),ds); label("FF", (6.64,12.07),NE*lsf); dot((10.12,12.6),ds); label("GG", (10.41,12.35),NE*lsf); dot((8.07,13.52),ds); label("HH", (8.11,13.88),NE*lsf); clip((-0.68,-0.96)--(-0.68,25.47)--(30.71,25.47)--(30.71,-0.96)--cycle); [/asy]