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Iranian Mathematical Olympiad

Geometry Difficulty 5.4 AIME, harder Prove it Iran

In the isosceles triangle $ABC$ , we have $AB = AC$ and $BC > AB$ . $D$ and $M$ are the midpoints of $BC$ and $AB$ respectively. $X$ is a point that $BX \perp AC$ and $XD \parallel AB$ . $H$ is the intersection of $BX$ and $AD$ . If $P$ be the intersection of $DX$ with the circumcircle of $AHX$ (not $X$ ), prove that the tangent line in $A$ to the circumcircle of $AMP$ is parallel to $BC$ .

Solution

Obviously $X, P, D, M$ are collinear. Since $HAPX$ is an inscribed quadrilateral, then $\angle APM = \angle AHX$ , and since $HCDT$ is inscribed ( $T$ is the foot of the altitude from $B$ on $AC$ ), then $\angle AHX = \angle AHT = \angle ACD$ . Hence, $\angle APM = \angle C = \angle B$ . Now if we draw ray $Ax$ parallel to $BC$ such that $\angle CAx = \angle C$ , then in the circumcircle of $AMP$ : (let $s$ be the measure of the arc $AM$ )
$\angle MAx = \angle C = \angle APM = \frac{s}{2}$
Hence we deduce that $Ax$ is tangent to circumcircle of $AMP$ .

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