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IMO Team Selection Contest

Geometry Difficulty 8.4 Shortlist Prove it Estonia

Let $D$ be the point different from $B$ on the hypotenuse $AB$ of a right triangle $ABC$ such that $|CB| = |CD|$ . Let $O$ be the circumcenter of triangle $ACD$ . Rays $OD$ and $CB$ intersect at point $P$ , and the line through point $O$ perpendicular to side $AB$ and ray $CD$ intersect at point $Q$ . Points $A$ , $C$ , $P$ , $Q$ are concyclic. Does this imply that $ACPQ$ is a square?

Solution

Figure 32

As $OQ$ is the perpendicular bisector of $AD$ , one has $\angle QAD = \angle ADQ = \angle BDC = \angle CBD$ (Fig. 32). Therefore $AQ \parallel BC$ , whence $\angle QAC = 180^\circ - \angle ACB = 90^\circ$ . From the cyclic quadrilateral $APCQ$ one also gets $\angle CPQ = \angle PQA = 90^\circ$ , i.e., $ACPQ$ is a rectangle.

As $\angle DOC = 2\angle DAC$ , one obtains
$\begin{aligned} \angle DOC &= 2\angle BAC = 2(90^\circ - \angle CBA) = 180^\circ - 2\angle CBA = \\ &= 180^\circ - \angle CBD - \angle BDC = \angle DCB, \end{aligned}$
which implies that isosceles triangles $BDC$ and $DCO$ are similar. Thus $\angle BDC = \angle DCO$ , i.e., $OC \parallel AB$ , whence $\angle QOC = 90^\circ = \angle QAC$ . So $O$ lies on the circle determined by $A$ , $C$ , $P$ , $Q$ . Therefore
$\angle ACQ = \angle AOQ = \frac{1}{2}\angle AOD = \frac{1}{2}\angle AOP = \frac{1}{2}\angle ACP.$
Consequently, the diagonal of the rectangle $ACPQ$ bisects the angle of the rectangle, whence $ACPQ$ is a square.

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