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China Western Mathematical Olympiad

Geometry Difficulty 7.1 National Olympiad, round 2 Prove it China

Let $D$ be a point on the side $BC$ of an acute triangle $ABC$ . The circle with diameter $BD$ meets the lines $AB$ and $AD$ respectively at the points $X$ and $P$ , which are different from the points $B$ and $D$ . The circle with diameter $CD$ meets the lines $AC$ and $AD$ respectively at the points $Y$ and $Q$ , which are different from the points $C$ and $D$ . Through the point $A$ draw two lines which are perpendicular to $PX$ and $QY$ with the feet of perpendicular $M$ and $N$ respectively.
Prove that $\triangle AMN \sim \triangle ABC$ if and only if the line $AD$ passes through the circumcenter of $\triangle ABC$ . (Posed by Li Qiusheng)

Solution

Proof Join the segments $XY$ and $DX$ . It follows from the given conditions that $B$ , $P$ , $D$ , $X$ are concyclic, and $C$ , $Y$ , $Q$ , $D$ are concyclic. Then
$\angle AXM = \angle BXP = \angle BDP \\ = \angle QDC = \angle AYN.$

And it follows from $\angle AMX = \angle ANY = 90^\circ$ that $\triangle AMC \sim \triangle ANY$ , so $\angle MAX = \angle NAY$ and $\frac{AM}{AX} = \frac{AN}{AY}$ , and hence $\angle MAN = \angle XAY$ .

Combining the two results above, we have $\triangle AMN \sim \triangle AXY$ . So we get
$\begin{aligned} \triangle AMN \sim \triangle ABC &\Leftrightarrow \triangle AXY \sim \triangle ABC \\ &\Leftrightarrow XY \parallel BC \Leftrightarrow \angle DXY = \angle XDB. \end{aligned}$
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As $A$ , $X$ , $D$ , $Y$ are concyclic, we have $\angle DXY = \angle DAY$ .
As $\angle XDB = 90^\circ - \angle ABC$ ,
$\angle DXY = \angle XDB \iff \angle DAC = 90^\circ - \angle ABC,$
which is equivalent to the fact that the line $AD$ passes through the circumcenter of $\triangle ABC$ .
Hence, $\triangle AMN \sim \triangle ABC$ if and only if $AD$ passes through the circumcenter of $\triangle ABC$ .

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