, 2024

Geometry Difficulty 5.2 AIME, harder Prove it Czech-Polish-Slovak Mathematical Match

Let $ABCD$ be a quadrilateral, such that $AB = BC = CD$ . There are points $X, Y$ on rays $CA, BD$ , respectively, such that $BX = CY$ . Let $P, Q, R, S$ be the midpoints of segments $BX, CY, XD, YA$ , respectively. Prove that points $P, Q, R, S$ lie on a circle.

Solution

Let $M$ be the midpoint of $XY$ . Note that $PR$ is midline in triangles $XBD$ and $XBY$ , hence $M$ lies on $PR$ . Analogously $M$ lies on $QS$ .

Let $\omega_1$ be a circle with center $B$ and radius $AB = BC$ and $\omega_2$ be a circle with center $C$ and radius $BC = CD$ .
Distance of $X$ from center of $\omega_1$ is the same as distance of $Y$ from center of $\omega_2$ and also $\omega_1$ and $\omega_2$ have radius of same size, hence power of $X$ with respect to $\omega_1$ is the same as power of $Y$ with respect to $\omega_2$ , so
$XA \cdot XC = YD \cdot YB.$
Using homotheties centered at $X, Y$ we get that $MS \cdot MQ = MR \cdot MP$ and thus points $P, Q, R, S$ lie on a circle.

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Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.