+1234 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 $ABC$.
+9673 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:
\(\sqrt{(3 \sqrt{26} + 9)(3\sqrt{26} - 9)(9^2)} = 27\sqrt{17}\).
