2004 JBMO Problems/Problem 2

Problem

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 $m = CQ$ .

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 $QM = QB = a$ .