NZMO Round Two

Geometry Difficulty 7.2 National Olympiad, round 2 Prove it New Zealand

Problem:

Let $ABCD$ be a square (vertices labelled in clockwise order). Let $Z$ be any point on diagonal $AC$ between $A$ and $C$ such that $AZ > ZC$ . Points $X$ and $Y$ exist such that $AXYZ$ is a square (vertices labelled in clockwise order) and point $B$ lies inside $AXYZ$ . Let $M$ be the point of intersection of lines $BX$ and $DZ$ (extended if necessary). Prove that $C$ , $M$ and $Y$ are collinear.

Solution

Solution:

Since $Z$ lies on diagonal $AC$ , we have $\angle DAZ = 45^{\circ}$ and $\angle ZAB = 45^{\circ}$ . Therefore $B$ lies on diagonal $AY$ of square $AXYZ$ and $\angle BAX = 45^{\circ}$ .

Since $AB = AD$ and $AX = AZ$ and $\angle BAX = 45^{\circ} = \angle DAZ$ , we have congruent triangles
$\triangle DAZ \equiv \triangle BAX \qquad (SAS).$
Therefore let $x = \angle ZDA = \angle XBA$ and $y = \angle AZD = \angle AXB$ . By the angle sum in triangle $DAZ$ we have $\angle DAZ + \angle AZD + \angle ZDA = 180^{\circ}$ . Therefore $x + y = 135^{\circ}$ . Now by the angle sum in quadrilateral $DAXM$ we get $\angle DAX + \angle AXM + \angle XMD + \angle MDA = 360^{\circ}$ . Therefore
$\angle BMD = 90^{\circ}.$
Hence $ABMCD$ is cyclic (the circle with diameter $BD$ ). Therefore
$\angle DMC = \angle DAC = 45^{\circ}.$
Also $AXYMZ$ is cyclic (the circle with diameter $XZ$ ). Therefore
$\angle YMX = \angle YAX = 45^{\circ}.$
Hence $\angle YMC = \angle YMX + \angle YMD + \angle DMC = 45^{\circ} + 90^{\circ} + 45^{\circ} = 180^{\circ}$ as required.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.