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. (Someone else asked the same question a bit ago, but nobody answered it, so I'm asking on a new strand)
Let $O_1$ and $O_2$ be the circumcenters of triangles $ABX$ and $ACX$, respectively. Since the circumcircles of triangles $ABX$ and $ACX$ have the same radius, $O_1O_2$ is the perpendicular bisector of segment $AX$. Let $O_1O_2$ intersect $AX$ at $M$. Then, $AM=13/2$.
[asy]
pair o1, o2, a, b, c, x, m;
o1 = (0,0);
a = (-11.25,9.55);
b = (0,0);
c = (14.63,0);
x = (10,0);
o2 = (10,-5.22);
m = (5,-1.35);
draw(circumcircle(a,b,x), dotted);
draw(circumcircle(a,c,x), dotted);
draw(o1--o2, dashed);
draw(a--b--c--cycle);
draw(a--x);
draw(o1--b);
draw(o2--c);
label("$O_1$", o1, W);
label("$O_2$", o2, E);
label("$A$", a, NW);
label("$B$", b, SW);
label("$C$", c, SE);
label("$X$", x, S);
label("$M$", m, S);
[/asy]
Let $O$ be the circumcenter of triangle $ABC$, and let $R$ be the circumradius of triangle $ABC$. Then, $OM=R\cos\angle A$, and $O_1O=R$. Note that $\angle O_1BO=\angle O_2CO=90^\circ$, so $OBO_1X$ and $OCO_2X$ are cyclic. It follows that $\angle BAX=\angle XBO_1=\angle XCO_2=\angle CAX$, so triangle $ABC$ is isosceles with $AB=AC$.
Also, $MX$ is the perpendicular bisector of segment $BC$, so $BM=CM=10$.
[asy]
pair o, a, b, c, x, m;
o = (0,0);
a = (-11.25,9.55);
b = (-10,0);
c = (10,0);
x = (0,0);
m = (0,-5);
draw(o--a);
draw(o--b);
draw(o--c);
draw(a--b--c--cycle);
draw(x--m);
label("$O$", o, N);
label("$A$", a, NW);
label("$B$", b, SW);
label("$C$", c, SE);
label("$X$", x, S);
label("$M$", m, S);
[/asy]
By the Law of Cosines on triangles $ABX$ and $ACX$, we get
\begin{align*}
\cos \angle BAC &= \frac{AB^2+AC^2-BC^2}{2\cdot AB\cdot AC} \\
&= \frac{(AX^2+BX^2-2\cdot AX\cdot BX\cdot \cos \angle BAX)+(AX^2+CX^2-2\cdot AX\cdot CX\cdot \cos \angle CAX)-BC^2}{2\cdot AX\cdot BX} \\
&= \frac{2\cdot AX^2+2\cdot BX\cdot CX-BC^2}{2\cdot AX\cdot BX} \\
&= \frac{2\cdot AX^2+2\cdot\cdot 100 - (13+20)^2}{2\cdot AX \cdot 10} \\
&= -\frac{189}{260}.
\end{align*}
It follows that $\sin \angle BAC=\sqrt{1-\cos^2 \angle BAC}=\frac{527}{260}$. Finally, the area of triangle $ABC$ is
\[\frac{1}{2} \cdot AB\cdot AC\cdot\sin \angle BAC = \boxed{\frac{527}{8}}.\]