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 $ABC$.

Vxritate Feb 15, 2024

#1**-1 **

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.

bingboy Feb 16, 2024