The incircle of the triangle $\triangle{ABC}$ has center at $O$ and it is tangent to the sides $BC$, $AC$ and $AB$ at the points $X$, $Y$ and $Z$, respectively. The lines $BO$ and $CO$ intersect the line $YZ$ at the points $P$ and $Q$, respectively. Show that if the segments $XP$ and $XQ$ has the same length, then the triangle $\triangle ABC$ is isosceles.
Problem
Source: Spanish Communities
Tags: geometry, incenter, angle bisector, geometry unsolved
grobber
15.04.2006 04:28
A simple angle computation shows that $\angle BPZ=\frac{\angle ACB}2$. This means that $\angle OPY+\angle OCY=\pi$, i.e. $COPY$ is cyclic. Since $OY\perp CY$, we get $BP=OP\perp CP$. In the same way we get $CQ\perp BQ$, and this means that if $D=BQ\cap CP$, then $O$ is the orthocenter of $BCD$, whereas $XPQ$ is the orthic triangle of $BCD$, i.e. $O$ is the incenter of $XPQ$. Assume now that $XP=XQ$. Then $XO$, which is the angle bisector of $\angle PXQ$, is also the $X$-altitude of the isosceles triangle $XPQ$, i.e. $XO\perp PQ=YZ$. But $YZ\perp AO\Rightarrow AO\|XO\Rightarrow AO\perp BC$, i.e. the $A$-bisector of $ABC$ is also its $A$-altitude. We thus get $AB=AC$.
Vo Duc Dien
25.11.2010 19:58
http://www.cut-the-knot.org/wiki-math/index.php?n=MathematicalOlympiads.IbAMO2001Problem2 Vo Duc Dien
Packito
25.07.2016 03:15
If $M$ si the midpoint of $BC$ then $MP=MQ$ then $X=M$ and we are done.
baenanabread
18.07.2021 00:34
We claim that $\triangle QOX \cong \triangle QOY$. To see this, we observe that $OX \cong OY$ and $\angle QOX = 180° - \angle XOC = 180° - \angle COY = \angle YOZ$. From this it follows that $\angle OXQ = \angle QYO =\frac{\angle A}{2}$. Analogously we have $\angle OXP = \frac{\angle A}{2}$. Next we observe that $\angle QXB = 90°-\angle OXQ = 90°-\angle OXP = \angle CXP$ so $2\angle QXB=180°-\angle PXQ$ which is equal to $\angle XQP$ if we assume $XQ \cong XP$. This result implies that $ZY \| BC$, so $\triangle ABC \sim \triangle AZY$, which is known to be an isosceles triangle. $\square$