MathDB
Find the angle alpha [Iran Second Round 1994]

Source:

November 26, 2010
geometry proposedgeometry

Problem Statement

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]