Problem: Let $ABCD$ be a tetrahedron such that $AB=CD= \sqrt{41}$ $AC=BD= \sqrt{80}$ , and $BC=AD= \sqrt{89}$ . There exists a point $I$ inside the tetrahedron such that the distances from $I$ to each of the faces of the tetrahedron are all equal. This distance can be written in the form $\frac{m \sqrt n}{p}$ , where $m$ $n$ , and $p$ are positive integers, $m$ and $p$ are relatively prime, and $n$ is not divisible by the square of any prime. Find $m+n+p$

Solution: Let $AH$ be perpendicular to $BCD$ that meets this plane at point $H$ .
Let $AP$ $AQ$ , and $AR$ be heights to lines $BC$ $CD$ , and $BD$ with feet $P$ $Q$ , and $R$ , respectively.
We notice that all faces are congruent. Following from Heron's formula, the area of each face, denoted as $A$ , is $A = 6 \sqrt{21}$
Hence, by using this area, we can compute $AP$ $AQ$ and $AR$ .
We have $AP = \frac{2 A}{BC} = \frac{2A}{\sqrt{89}}$ $AQ = \frac{2 A}{CD} = \frac{2A}{\sqrt{41}}$ , and $AR = \frac{2 A}{BC} = \frac{2A}{\sqrt{80}}$
Because $AH \perp BCD$ , we have $AH \perp BC$ . Recall that $AP \perp BC$ .
Hence, $BC \perp APH$ . Hence, $BC \perp HP$
Analogously, $CD \perp HQ$ and $BD \perp HR$
We introduce a function $\epsilon \left( l \right)$ for $\triangle BCD$ that is equal to 1 (resp. -1) if point $H$ and the opposite vertex of side $l$ are on the same side (resp. opposite sides) of side $l$
The area of $\triangle BCD$ is
\begin{align*}
A & = \epsilon_{BC} {\rm Area} \ \triangle HBC
+ \epsilon_{CD} {\rm Area} \ \triangle HCD
+ \epsilon_{BD} {\rm Area} \ \triangle HBD \\
& = \frac{1}{2} \epsilon_{BC} BC \cdot HP
+ \frac{1}{2} \epsilon_{CD} CD \cdot HQ +
\frac{1}{2} \epsilon_{BD} BD \cdot HR \\
& = \frac{1}{2} \epsilon_{BC} BC \cdot \sqrt{AP^2 - AH^2}
+ \frac{1}{2} \epsilon_{CD} CD \cdot \sqrt{AQ^2 - AH^2} \\
& \quad + \frac{1}{2} \epsilon_{BD} CD \cdot \sqrt{AR^2 - AH^2} . \hspace{1cm} (1)
\end{align*}
Denote $B = 2A$ .
The above equation can be organized as
\begin{align*}
B & = \epsilon_{BC} \sqrt{B^2 - 89 AH^2}
+ \epsilon_{CD} \sqrt{B^2 - 41 AH^2} \\
& \quad + \epsilon_{BD} \sqrt{B^2 - 80 AH^2} .
\end{align*}
This can be further reorganized as
\begin{align*}
B - \epsilon_{BC} \sqrt{B^2 - 89 AH^2}
& = \epsilon_{CD} \sqrt{B^2 - 41 AH^2}
+ \epsilon_{BD} \sqrt{B^2 - 80 AH^2} .
\end{align*}
Taking squares on both sides and reorganizing terms, we get
\begin{align*}
& 16 AH^2 - \epsilon_{BC} B \sqrt{B^2 - 89 AH^2} \\
& = \epsilon_{CD} \epsilon_{BD}
\sqrt{\left( B^2 - 41 AH^2 \right) \left( B^2 - 80 AH^2 \right)} .
\end{align*}
Taking squares on both sides and reorganizing terms, we get \[ - \epsilon_{BC} 2 B \sqrt{B^2 - 89 AH^2} = - 2 B^2 + 189 AH^2 . \]
Taking squares on both sides, we finally get
\begin{align*}
AH & = \frac{20B}{189} \\
& = \frac{40A}{189}.
\end{align*}
Now, we plug this solution to Equation (1). We can see that $\epsilon_{BC} = -1$ $\epsilon_{CD} = \epsilon_{BD} = 1$ .
This indicates that $H$ is out of $\triangle BCD$ . To be specific, $H$ and $D$ are on opposite sides of $BC$ $H$ and $C$ are on the same side of $BD$ , and $H$ and $B$ are on the same side of $CD$
Now, we compute the volume of the tetrahedron $ABCD$ , denoted as $V$ . We have $V = \frac{1}{3} A \cdot AH = \frac{40 A^2}{3 \cdot 189}$
Denote by $r$ the inradius of the inscribed sphere in $ABCD$ .
Denote by $I$ the incenter.
Thus, the volume of $ABCD$ can be alternatively calculated as
\begin{align*}
V & = {\rm Vol} \ IABC + {\rm Vol} \ IACD + {\rm Vol} \ IABD + {\rm Vol} \ IBCD \\
& = \frac{1}{3} r \cdot 4A .
\end{align*}
From our two methods to compute the volume of $ABCD$ and equating them, we get
\begin{align*}
r & = \frac{10A}{189} \\
& = \frac{20 \sqrt{21}}{63} .
\end{align*}
Therefore, the answer is $20 + 21 + 63 = \boxed{104}$

Answer:  104