2024 AIME I Problems/Problem 14

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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 1

Notice that $41=4^2+5^2$, $89=5^2+8^2$, and $80=8^2+4^2$, let $A~(0,0,0)$, $B~(4,5,0)$, $C~(0,5,8)$, and $D~(4,0,8)$. Then the plane $BCD$ has a normal \[\mathbf n:=\frac14\overrightarrow{BC}\times\overrightarrow{CD}=\frac14\begin{pmatrix}-4\\0\\8\end{pmatrix}\times\begin{pmatrix}4\\-5\\0\end{pmatrix}=\begin{pmatrix}10\\8\\5\end{pmatrix}.\] Hence, the distance from $A$ to plane $BCD$, or the height of the tetrahedron, is \[h:=\frac{\mathbf n\cdot\overrightarrow{AB}}{|\mathbf n|}=\frac{10\times4+8\times5+5\times0}{\sqrt{10^2+8^2+5^2}}=\frac{80\sqrt{21}}{63}.\] Each side of the tetrahedron has the same area due to congruency by "S-S-S", and we call it $S$. Then by the volume formula for cones, 13Sh=VD-ABC=VI-ABC+VI-BCD+VI-CDA+VI-DAB=13Sr4. Hence, $r=\tfrac h4=\tfrac{20\sqrt{21}}{63}$, and so the answer is $20+21+63=\boxed{104}$.

Solution by Quantum-Phantom

See also

2024 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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