2024 AIME I Problems/Problem 14

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, \begin{align*} \frac13Sh&=V_{D\text-ABC}=V_{I\text-ABC}+V_{I\text-BCD}+V_{I\text-CDA}+V_{I\text-DAB}\\ &=\frac13Sr\cdot4. \end{align*} Hence, $r=\tfrac h4=\tfrac{20\sqrt{21}}{63}$ , and so the answer is $20+21+63=\boxed{104}$ .

Solution by Quantum-Phantom

2024 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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2024 AIME I Problems/Problem 14

Problem

Solution 1

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