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Problem 1494

AIME late

Geometry Difficulty 5.5 Find the answer Junior Balkan MO

Let $ABC$ be an isosceles triangle with $AC=BC$ , let $M$ be the midpoint of its side $AC$ , and let $Z$ be the line through $C$ perpendicular to $AB$ . The circle through the points $B$ , $C$ , and $M$ intersects the line $Z$ at the points $C$ and $Q$ . Find the radius of the circumcircle of the triangle $ABC$ in terms of $m = CQ$ .

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Official solution

Let length of side $CB = x$ and length of $QM = a$ . We shall first prove that $QM = QB$ .
Let $O$ be the circumcenter of $\triangle ACB$ which must lie on line $Z$ as $Z$ is a perpendicular bisector of isosceles $\triangle ACB$ .
So, we have $\angle ACO = \angle BCO = \angle C/2$ .
Now $MQBC$ is a cyclic quadrilateral by definition, so we have: $\angle QMB = \angle QCB = \angle C/2$ and, $\angle QBM = \angle QCM = \angle C/2$ , thus $\angle QMB = \angle QBM$ , so $QM = QB = a$ .
Therefore in isosceles $\triangle QMB$ we have that $MB = 2 QB \cos C/2 = 2 a \cos C/2$ .
Let $R$ be the circumradius of $\triangle ACB$ .
So we have $CM = x/2 = R \cos C/2$ or $x = 2R \cos C/2$
Now applying Ptolemy's theorem in cyclic quadrilateral $MQBC$ , we get:
$m . MB = x . QM + (x/2) . QB$ or,
$m . (2 a \cos C/2) = (3/2 x) . a = (3/2).(2Ra) \cos C/2 = 3Ra \cos C/2$ or,
$R = (2/3)m$
$Kris17$

Source: Omni-MATH, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.