Competition problem — Geometry — Maths Olympiad Prep

Geometry Difficulty 6.2 National Olympiad Prove it Spain

Let $\triangle ABC$ be an acute triangle and let $X$ be the foot of the height drawn from $A$ and $Y$ be the intersection of the perpendicular to $AC$ drawn from $X$ . If the circumcircle of triangle $ABX$ meets $BY$ at point $Z$ (distinct of $B$ ) and the extension of $AZ$ meets $XY$ at point $P$ , then prove that $BX \cdot XP = PY \cdot XC$ .

Solution

Since $\triangle AXY$ and $\triangle AXC$ are both right triangles in $Y$ and $X$ respectively, then $\angle AXP = 90^\circ - \angle XAC = \angle YCB$ . On the other hand, since $AZXB$ is a cyclic quadrilateral, then $\angle XAP = \angle YBC$ . Therefore,

$\triangle AXP \sim \triangle BCY$ and we have $\frac{BC}{AX} = \frac{YC}{XP}$ . Likewise, $\triangle AXC \sim \triangle XYC$ because both are right angled triangles with a common acute angle. So, we have
$\frac{AX}{XC} = \frac{XY}{YC}$
From the preceding ratios we get
$BC \cdot XP = AX \cdot YC = XC \cdot XY$
from which follows
$\frac{BC}{XC} = \frac{XY}{XP}$
Subtracting one to both sides of the above equality, yields
$\frac{BC}{XC} - 1 = \frac{XY}{XP} - 1 \Leftrightarrow \frac{BC - XC}{XC} = \frac{XY - XP}{XP} \Leftrightarrow \frac{BX}{XC} = \frac{PY}{XP}$
on account that $BC = BX + XC$ and $XY = XP + PY$ respectively.

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