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In triangle $ABC$, point $X$ is on side $\overline{BC}$ such that $AX = 13,$ $BX = 13,$ $CX = 5,$ and the circumcircles of triangles $ABX$ and $ACX$ have the same radius. Find the area of triangle $ABC$.

May 10, 2024

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Let AB = b, and AC = c. Moreover, let $$[ABX] = A_1$$ and $$[ACX] = A_2$$, where $$[\cdot]$$ denote "area of".

By Stewart's theorem, we have

$$(13 + 5)(13^2 + 13(5)) = 5b^2 + 13c^2\\ 5b^2 + 13c^2 = 4212$$

By basic geometry, we have $$\dfrac{A_1}{A_2} = \dfrac{13}5$$

Recall that the circumradius of a triangle with sides a, b, c and area A is $$\dfrac{abc}{4A}$$. Then, since the circumradius of triangles ABX and ACX are the same, we have

$$\dfrac{13^2 b}{4A_1} = \dfrac{5(13)c}{4A_2}\\ \dfrac{A_1}{A_2} = \dfrac{13b}{5c}\\ \dfrac{13}5 = \dfrac{13b}{5c}\\ b = c$$.

Hence,

$$18b^2 = 18c^2 = 4212\\ b = c = 3 \sqrt{26}$$

We calculate the half perimeter of triangle ABC to be $$\dfrac{18 + b + c}2 = 3 \sqrt{26} + 9$$. Then, by Heron's formula, the area required is:

$$\sqrt{(3 \sqrt{26} + 9)(3\sqrt{26} - 9)(9^2)} = 27\sqrt{17}$$

May 10, 2024
edited by MaxWong  May 10, 2024