Find the area of a triangle $\triangle MNT$ I've found this geometry problem in a math competition:
Let $\ \triangle ABC$ be a triangle, with $AB=52$, $BC=56$, $AC=60$, $\ \Gamma$ be the circumscribed circle of $\ \triangle ABC$, $\ M$ and $N$ be the midpoints of $AB$ and $AC$, respectively. The point $T$ lies on the minor arc $\overset{\frown}{BC}$, so that the circumscribed circle of the triangle $\ \triangle MNT$ is internally tangent to $\Gamma$. What's the area of $MNT$?
The answer to this problem is $525$.
I have drawn this figure of the situation described in the problem.

Applying the intercept theorem, I know that $MN=\dfrac{BC}{2}=28$.
Applying the law of cosines, I've also found the cosine of the three angles of the triangle $ABC$, $\cos\hat{A}=\dfrac{33}{65}$, $\cos \hat{B}=\dfrac{5}{13}$, $\cos \hat{C}=\dfrac{3}{5}$.
Now, how can I use the condition of tangency in order to resolve the problem?


As $a = 56, b =  60, c = 52$, $s = 84$
Using Heron's formula, $\Delta = 1344$
Circumradius $ \displaystyle R = \frac{abc}{4 \Delta} = \frac{65}{2}$
If altitude from $A$ to $BC$ is $h$, $1344 = \frac 12 \cdot a \cdot h$
$~~~~\implies h = 48$
If the perp from $A$ to $BC$ meets $BC$ at $H$ and midpoint of $BC$ is $G$,
$BH^2 = c^2 - h^2 \implies BH = 20$, $GH = 8$
That shows that the horizontal distance between midpoints of $MN$ and $BC$ or in other words, $OI = 4$
$ \displaystyle OG^2 + (a/2)^2 = R^2 \implies OG = \frac{39}{2}$. So, $IJ = \displaystyle 24 - \frac{39}{2} = \frac 92 $
If $r$ is circumradius of $\triangle MNT$ and $OO' = x$,
$ \displaystyle O'M = r = \frac{65}{2} - x$
$ \displaystyle \left(\frac{65}{2} - x\right)^2 = 14^2 + \left(\frac 92 + \sqrt{x^2 - 16}\right)^2$
Solve for $x$,
Then altitude from $T$ to $MN$ is,
$\displaystyle h_1 = \frac{9}{2} + \frac{\sqrt{x^2-16}}{x} \cdot \frac {65}{2}$
Finally, $\Delta_{MNT} = \frac 12 \cdot 28 \cdot h_1$
