A pyramid has a triangular base with side lengths $20$, $20$, and $24$. The three edges of the pyramid from the three corners of the base to the fourth vertex of the pyramid all have length $25$. The volume of the pyramid is $m\sqrt{n}$, where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime. Find $m+n$.
Let the triangular base be $\triangle ABC$, with $\overline {AB} = 24$. We find that the altitude to side $\overline {AB}$ is $16$, so the area of $\triangle ABC$ is $(24*16)/2 = 192$.
Let the fourth vertex of the tetrahedron be $P$, and let the midpoint of $\overline {AB}$ be $M$. Since $P$ is equidistant from $A$, $B$, and $C$, the line through $P$ perpendicular to the plane of $\triangle ABC$ will pass through the circumcenter of $\triangle ABC$, which we will call $O$. Note that $O$ is equidistant from each of $A$, $B$, and $C$. Then,
\[\overline {OM} + \overline {OC} = \overline {CM} = 16\]
Let $\overline {OM} = d$. Then $OC=OA=\sqrt{d^2+12^2}.$
Equation $(1)$:
\[d + \sqrt {d^2 + 144} = 16\]
Squaring both sides, we have
\[d^2 + 144 + 2d\sqrt {d^2+144} + d^2 = 256\]
\[2d^2 + 2d\sqrt {d^2+144} = 112\]
\[2d(d + \sqrt {d^2+144}) = 112\]
Substituting with equation $(1)$:
\[2d(16) = 112\]
\[d = 7/2\]
We now find that $\sqrt{d^2 + 144} = 25/2$.
Let the distance $\overline {OP} = h$. Using the Pythagorean Theorem on triangle $AOP$, $BOP$, or $COP$ (all three are congruent by SSS):
\[25^2 = h^2 + (25/2)^2\]
\[625 = h^2 + 625/4\]
\[1875/4 = h^2\]
\[25\sqrt {3} / 2 = h\]

Finally, by the formula for volume of a pyramid,
\[V = Bh/3\]
\[V = (192)(25\sqrt{3}/2)/3\]
This simplifies to $V = 800\sqrt {3}$, so $m+n = \boxed {803}$.

NOTE : If you don’t know or remember the formula for the volume of a triangular pyramid, you can derive it using calculus as follows :
Take a small triangular element in the pyramid. We know that it’s area is proportional to the height from the vertex to the base. Hence, we know that $\frac{A_{small element}}{A} = \frac{h^2}{H^2} \implies A_{small element} = \frac{Ah^2}{H^2}$. Now integrate it taking the limits $0$ to $H$

Shortcut
Here is a shortcut for finding the radius $R$ of the circumcenter of $\triangle ABC$.
As before, we find that the foot of the altitude from $P$ lands on the circumcenter of $\triangle ABC$. Let $BC=a$, $AC=b$, and $AB=c$.
Then we write the area of $\triangle ABC$ in two ways:
\[[ABC]= \frac{1}{2} \cdot 24 \cdot 16 = \frac{abc}{4R}\]
Plugging in $20$, $20$, and $24$ for $a$, $b$, and $c$ respectively, and solving for $R$, we obtain $R= \frac{25}{2}=OA=OB=OC$.
Then continue as before to use the Pythagorean Theorem on $\triangle AOP$, find $h$, and find the volume of the pyramid.

Another Shortcut (Extended Law of Sines)
Take the base $\triangle ABC$, where $AB = BC = 20$ and $AC = 24$. Draw an altitude from $B$ to $AC$ that bisects $AC$ at point $D$. Then the altitude has length $\sqrt{20^2 - 12^2} = \sqrt{16^2} = 16$. Next, let $\angle BCA = \theta$. Then from the right triangle $\triangle BDC$, $\sin \theta = 4/5$. From the extended law of sines, the circumradius is $20 \cdot \dfrac{5}{4} \cdot \dfrac{1}{2} = \dfrac{25}{2}$.