In triangle $ABC$, where $AB<AC$, let $X$, $Y$, $Z$ denote the points where the incircle is tangent to $BC$, $CA$, $AB$, respectively. On the circumcircle of $ABC$, let $U$ denote the midpoint of the arc $BC$ that contains the point $A$. The line $UX$ meets the circumcircle again at the point $K$. Let $T$ denote the point of intersection of $AK$ and $YZ$. Prove that $XT$ is perpendicular to $YZ$.
Problem
Source: XVII Olimpíada Matemática Rioplatense (2008)
Tags: geometry, circumcircle, trigonometry, cyclic quadrilateral, angle bisector, geometry unsolved
Luis González
25.07.2011 08:32
$T$ is the orthogonal projection of $X$ onto $YZ$ and ray $AT$ cuts the circumcircle $(O,R)$ of $\triangle ABC$ at $K.$ Then, we shall show that $K,X$ and the midpoint $U$ of the arc $BAC$ of $(O)$ are collinear. Since the pencil $T(B,C,X,Y)$ is clearly harmonic and $XT \perp YZ,$ then $TX,YZ$ bisect $\angle BTC.$ Thus, $\angle BZT=\angle CYT$ implies that $\triangle TZB$ and $\triangle TYC$ are similar. Hence, we deduce that $\frac{KB}{KC}=\frac{2R \cdot \sin \widehat{ZAT}}{ 2R \cdot \sin \widehat{YAT}}=\frac{AY}{AZ} \cdot \frac{TZ}{TY}=\frac{BZ}{CY}=\frac{XB}{XC}$ $\Longrightarrow KX$ bisects $\angle BKC$ internally, i.e. $K,X,U$ are collinear, as desired.
littletush
04.12.2011 07:21
again Kevin Yu's theorem: in cyclic quadrilateral ABCD,AC intersects BD at E,then $\frac{AE}{EC}=\frac{AB*AD}{BC*DC}$ is useful.
ThirdTimeLucky
28.03.2013 22:20
Sorry for reviving, This sure can be proved directly too.. Let $ YZ \cap BC =L $. Then we have $ T \{BC,XL\} $ is harmonic. So $ K\{BC,XL\} $ is harmonic too. By angle chasing, we have $\angle TKX= \angle AKU= \angle ACU = \angle UCB - \angle ACB = \frac{B-C}{2} $ and also $ \angle TLX = \angle ZLB = 180^{\circ} - \angle LZB - \angle LBZ = 180^{\circ} - (90^{\circ}-\frac{A}{2}) - (180^{\circ}-B) = \frac{B-C}{2} $. So $ ZLKX $ is cyclic. Since $ KU $ bisects $ \angle BKC $, it follows that $ KL$ is the external angle bisector of $ \angle BKC $ and hence $ \angle XKL = \angle XTL = 90^{\circ} $ That is $ XT \perp YZ $.
Math_CYCR
01.07.2016 19:23
Sorry for reviving but I don't think we need the harmonic bundles. We easily get by law of sines and the general angle bisector theorem: $\frac{ZT}{TY} = \frac{ \sin \angle ZAT}{ \sin \angle YAT} = \frac{BK}{KC}$ On the other hand, by the angle bisector theorem in $\triangle BCK$ $\frac{BK}{KC} = \frac{BX}{XC} = \frac{BZ}{CY}$ $\Longrightarrow \frac{ZT}{TY} = \frac{BZ}{CY}$ By SAS we get $\triangle BZT \sim \triangle CYT$. Furthermore $\angle ZTB= \angle YTB (\star)$ $\Longrightarrow \frac{BX}{XC} = \frac{BZ}{YC} = \frac{BT}{CT}$ $\Longrightarrow \angle BTX = \angle CTX (\star \star)$ By $(\star)$ and $(\star \star)$ we get $\angle ZTX= \angle YTX= \tfrac{ \angle ZTY}{2} = 90$ Done!
Lcz
24.10.2022 19:37
Let $T$ be the foot from $X$ to $YZ$ and $K=AT \cap (ABC)$, $K'=UX \cap (ABC)$. By some ratio lemma and cool ratio lemma, $$\frac{BK}{KC}=\frac{\sin(\angle ZAT)}{\sin(\angle TAY)} = \frac{ZT}{TY} = \frac{\tan\frac{C}{2}}{\tan\frac{B}{2}} = \frac{BX}{XC} = \frac{BK'}{K'C},$$win. mira74 :star_struck: