[Incircle touch point to midpoint of opposite side]

With $a = BC = 13$, $b = CA = 15$, $c = AB = 14$, semi-perimeter $s = 21$. Tangent length from $A$: $AP = s - a = 8$.

Median length. By Apollonius for median $AM$ from $A$ to mid-$BC$:

$

AM^{2} = \frac{2b^{2} + 2c^{2} - a^{2}}{4} = \frac{2(225) + 2(196) - 169}{4} = \frac{673}{4}.

$

Angle $\angle BAM$. Apply the law of cosines in $\triangle ABM$, with $BM = 13/2$:

$

BM^{2} = AB^{2} + AM^{2} - 2 AB\cdot AM\cos\angle BAM \Rightarrow \tfrac{169}{4} = 196 + \tfrac{673}{4} - 14\sqrt{673}\cos\angle BAM,

$

$

\Rightarrow 14\sqrt{673}\cos\angle BAM = 322 \Rightarrow \cos\angle BAM = \tfrac{23}{\sqrt{673}}.

$

Apply law of cosines in $\triangle AMP$: $AP = 8$, $AM = \tfrac{\sqrt{673}}{2}$:

$

MP^{2} = 64 + \tfrac{673}{4} - 2 \cdot 8 \cdot \tfrac{\sqrt{673}}{2} \cdot \tfrac{23}{\sqrt{673}} = 64 + \tfrac{673}{4} - 184 = \tfrac{193}{4}.

$

So $4 MP^{2} = 193$. Answer: 193.

Why CAT-level: Coordinate geometry works but produces fractions with denominator $169$. The slick path requires combining median formula + law of cosines, an approach not directly signalled by the problem.