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Problem 1153

AIME late

Geometry Difficulty 5.1 Prove it Iranian Mathematical Olympiad · Iran

Points $X$ and $Y$ respectively lie on the tangent lines to the circumcircle of triangle $ABC$ passing through $B$ , $C$ such that $AB = BX$ and $AC = CY$ . (Points $X$ , $Y$ , $A$ are on the same side of line $BC$ .) Let $I$ be the incenter of triangle $ABC$ , prove that
$\widehat{BAC} + \widehat{XIY} = 180^\circ$

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

We know that $I$ is the incenter of triangle $ABC$ , thus we have $\widehat{AIC} = 90^\circ + \frac{\widehat{ABC}}{2}$ . Moreover we have
$\left. \begin{array}{l} \widehat{ABC} = \widehat{ACY} \\ AC = AY \\ \widehat{AYC} = 90^\circ - \frac{\widehat{ACY}}{2} \end{array} \right\} \implies \widehat{AIC} + \widehat{AYC} = 180^\circ \\ \implies AICY \text{ is cyclic}$

In a similar way $AIBX$ is also cyclic. Hence we have

$\begin{aligned} \widehat{XIY} &= \widehat{AIX} + \widehat{AIY} = \widehat{ABX} + \widehat{ACY} = \widehat{ABC} + \widehat{ACB} = 180^\circ - \widehat{BAC} \\ \implies \widehat{BAC} + \widehat{XIY} &= 180^\circ \end{aligned}$

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