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Problem 1550

National Olympiad, first round

Geometry Difficulty 6.0 Prove it Iranian Mathematical Olympiad · Iran

Let $\omega$ be the circumcircle of isosceles triangle $ABC$ where $AB = AC$ . Points $P$ and $Q$ lie on $\omega$ and $BC$ respectively such that $AP = AQ$ . Lines $AP$ and $BC$ intersect at $R$ . Prove that the tangents from $B$ and $C$ to the incircle of triangle $AQR$ (different from $BC$ ) are concurrent on $\omega$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Let $I, I_a$ be the incenter and $A$ -excenter of triangle $AQR$ . To prove the claim of the problem, it suffices to show that $\widehat{BIC} = 90^\circ + \frac{1}{2}\widehat{BAC}$ .

We have
$\begin{aligned} \widehat{ARC} &= \widehat{PCB} - \widehat{RPC} = \widehat{PCA} + \widehat{ACB} - \widehat{ABC} = \widehat{PCA} \\ \implies AC^2 &= AP \cdot AR. \end{aligned}$
On the other hand, note that points $I, R, I_a, Q$ lie on circle with diameter $II_a$ , and
$\widehat{QIR} = 90^\circ + \frac{1}{2} \widehat{QAR} = \widehat{QPR}.$
Therefore $PIQR$ is a cyclic quadrilateral and $P$ also lies on circle with diameter $II_a$ . Now we have
$AI \cdot AI_a = AP \cdot AR = AC^2.$
Let $D$ be the intersection point of $AI, QR$ , and let $B'$ be the second intersection point of the incircle of triangle $ICI_a$ with $BC$ . We know that this circle is tangent to $AC$ , therefore
$-1 = C(AD, II_a) = C(CB', II_a).$
So $AB'$ is also tangent to the incircle of triangle $ICI_a$ which implies $AB' = AC$ , therefore $B'$ and $B$ are coincident. Finally we conclude that
$\widehat{BIC} = 180^\circ - \widehat{ABC} = 90^\circ + \frac{1}{2} \widehat{BAC}.$
Hence the claim of the problem.

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