Let $ABC$ be an acute triangle, which is not equilateral. Denote by $O$ and $H$ its circumcenter and orthocenter, respectively. The circle $k$ passes through $B$ and touches the line $AC$ at $A$. The circle $l$ with center on the ray $BH$ touhes the line $AB$ at $A$. The circles $k$ and $l$ meet in $X$ ($X\ne A$). Show that $\angle HXO=180^\circ-\angle BAC$. Proposed by Josef Tkadlec
Problem
Source: Czech-Polish-Slovak Match 2015, Problem 5
Tags: geometry
Higgsboson79
20.06.2015 01:20
We have that $\angle{XAH}=\angle{CAB}+\angle{CBA}-90^{\circ}-\angle{XAC}$ and $\angle{XBO}=\angle{CAB}+\angle{CBA}-90^{\circ}-\angle{XBA}$, but $\angle{XAC}=\angle{XBA}$, hence $\angle{XAH}=\angle{XBO}$. Let $l\cap AC$=$\lbrace A,D \rbrace$. We have that, since $\triangle{XBA} \sim \triangle{XAD}$, then $\frac{XA}{XB}=\frac{AD}{AB}$. Let $O'$ be the reflection of $O$ wrt segment $BC$. It is a known fact the one that $OO'=AH$. Moreover, it can be checked by an easy angle chase that $\triangle{OO'B}$ and $\triangle{ADB}$ are similar, hence $\frac{OO'}{OB}=\frac{AH}{OB}=\frac{AD}{AB}=\frac{XA}{XB}$. So, $\triangle{XAH} \sim \triangle{XBO}$ by $SAS$ criterion. This implies that $\triangle{XHO}$ and $\triangle{XAB}$ are similar too, getting the desired result.
tranquanghuy7198
20.06.2015 10:59
My solution: Let $K, L$ be the centers of $k, l$ We have: $(AO, AK) = -(AH, AL), (OA, OK) = -(HA, HL)$ (mod $\pi$) $\Rightarrow \triangle{AOK} \sim \triangle{AHL}$ (1) On the other hand: $OK\parallel{AL}(\perp{AB}), HL\parallel{AK}$ $\Rightarrow (LH, KO) = (AK, AL) = (XL, XK)$ (mod $\pi$) (2) And: $\frac{LH}{KO} = \frac{AL}{AK} (\because (1)) = \frac{XL}{XK}$ (3) (2), (3) $\Rightarrow \triangle{XLH} \sim \triangle{XKO}$ (same direction) $\Rightarrow \angle{HXO} = \angle{LXK} = \angle{LAK} = 180^\circ-\angle{BAC}$ Q.E.D
Attachments:
