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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 $ABC$. 

 Feb 15, 2024
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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.

 Feb 16, 2024

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