Prove that the points are collinear [Iran Second Round 1994]
Source:
November 26, 2010
geometryincentercircumcircleinradiusgeometry proposed
Problem Statement
The incircle of triangle meet the sides and in and , respectively. Prove that the orthocenter of triangle the incenter and the circumcenter of triangle 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("", (8.11,18.05),NE*lsf); dot((2.84,9.26),ds); label("", (2.11,8.85), NE*lsf); dot((20.44,9.21),ds); label("", (20.56,8.52), NE*lsf); dot((9.04,12.66),ds); label("", (8.94,12.13), NE*lsf); dot((6.04,14.42),ds); label("", (5.32,14.52), NE*lsf); dot((11.12,15.48),ds); label("", (11.4,15.9), NE*lsf); dot((8.94,9.24),ds); label("", (8.91,8.58), NE*lsf); dot((7.81,14.79),ds); label("", (7.81,15.14),NE*lsf); dot((6.95,12.79),ds); label("", (6.64,12.07),NE*lsf); dot((10.12,12.6),ds); label("", (10.41,12.35),NE*lsf); dot((8.07,13.52),ds); label("", (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]