In triangle ABC, point X is on side BC such that AX=13,BX=10,CX=10, and the circumcircles of triangles ABX and ACX have the same radius. Find the area of triangle ABC.

Guest May 9, 2023

#2**+1 **

This question was posted a few days ago, but in that version CX = 4.

I'll sketch out a solution for that one. (This new version is marginally easier.)

For a triangle ABC, the sine rule says that

\(\displaystyle \frac{a}{\sin A}=\frac{b}{\sin B} =\frac{c}{\sin C}= 2R,\)

where R is the radius of the circumscribing circle (of the triangle).

Applying that to the triangles ACX and ABX, we have,

\(\displaystyle \frac{AX}{\sin C}=2R_{1} \quad \text{ and } \quad\frac{AX}{\sin B}=2R_{2}\)

and since the two radii are equal, it follows that the angles B and C are equal meaning that the triangle ABC is isosceles.

Drop a perpendicular from A to BC, meeting BC at Y, say, then XY = 3, and calculate AY using Pythagoras.

The area of ABC will then be equal to (1/2)*BC*AY.

Tiggsy May 9, 2023