+101 In triangle $ABC$, point $X$ is on side $\overline{BC}$ such that $AX = 13,$ $BX = 10,$ $CX = 4,$ and the circumcircles of triangles $ABX$ and $ACX$ have the same radius. Find the area of triangle $ABC$.
+743 In triangle ABC, let O1 and O2 be the centers of the circumcircles of triangles ABX and ACX, respectively.
By Power of a Point, [AC \cdot AO_1 = AB \cdot AO_2.]
Since the radii of the circumcircles of ABX and ACX are equal, we have AO1=AO2. Therefore, AC=AB. This implies that triangle ABC is isosceles.
Let D be the foot of the altitude from A to BC. Then
[\triangle ACD \sim \triangle ABX \implies \frac{CD}{AX} = \frac{AC}{AB} = 1.]
Therefore, CD=13. By the Pythagorean Theorem on right triangle ACD, AD = sqrt(AC2−CD2) = sqrt(2⋅132−132) = sqrt(13^2) = 13.
Thus, the area of triangle ABC is \frac{1}{2} \cdot AC \cdot AD = \frac{1}{2} \cdot 13 \cdot 13 = 169/2.
