In triangle ABC, point X is on side 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.
I have had a look at similar questions to this, and I was unable to figure it out.
Since the circumcircles of triangles ABX and ACX have the same radius, then triangle AXB is similar to triangle ACX.
Let r be the radius of the circumcircles of triangles AXB and ACX. Then the circumradius of triangle ABC is also r.
By Power of a Point,
\begin{align*} (AX)^2 &= r^2 + 10^2 \ (CX)^2 &= r^2 + 4^2 \ (BX)^2 &= r^2 + (AX - CX)^2 \ &= r^2 + 13^2 - 8^2 \ &= r^2 + 144 - 64 \ &= r^2 + 80. \end{align*}Adding these equations, we get
[279 = 3r^2.]Then r=93.
The area of triangle ABC is
[\frac{1}{2} \cdot BC \cdot r = \frac{1}{2} \cdot 10 \cdot \sqrt{93} = \boxed{5 \sqrt{93}}.]
