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

Solution 2

$[asy] import three; currentprojection = orthographic(1,1,1); triple O = (0,0,0); triple A = (0,2,0); triple B = (0,0,1); triple C = (3,0,0); triple D = (3,2,1); triple E = (3,2,0); triple F = (0,2,1); triple G = (3,0,1); draw(A--B--C--cycle, red); draw(A--B--D--cycle, red); draw(A--C--D--cycle, red); draw(B--C--D--cycle, red); draw(E--A--O--C--cycle); draw(D--F--B--G--cycle); draw(O--B); draw(A--F); draw(E--D); draw(C--G); label("$O$", O, SW); label("$A$", A, NW); label("$B$", B, W); label("$C$", C, S); label("$D$", D, NE); label("$E$", E, SE); label("$F$", F, NW); label("$G$", G, NE); [/asy]$

Inscribe tetrahedron $ABCD$ in an rectangular prism as shown above.

By the Pythagorean theorem, we note

$OA^2 + OB^2 = AB^2 = 41,$ $OA^2 + OC^2 = AC^2 = 80, \text{and}$ $OB^2 + OC^2 = BC^2 = 89.$

Solving yields $OA = 4, OB = 5,$ and $OC = 8.$

Since each face of the tetrahedron is congruent, we know the point we seek is the center of the circumsphere of $ABCD.$ We know all rectangular prisms can be inscribed in a circumsphere, therefore the circumsphere of the rectangular prism is also the circumsphere of $ABCD.$

We know that the distance from all $4$ faces must be the same, so we only need to find the distance from the center to plane $ABC$ .

Let $O = (0,0,0), A = (4,0,0), B = (0,5,0),$ and $C = (0,0,8).$ We obtain that the plane of $ABC$ can be marked as $\frac{x}{4} + \frac{y}{5} + \frac{z}{8} = 1,$ or $10x + 8y + 5z - 40 = 0,$ and the center of the prism is $(2,\frac{5}{2},4).$

Using the Point-to-Plane distance formula, our distance is

$d = \frac{|10\cdot 2 + 8\cdot \frac{5}{2} + 5\cdot 4 - 40|}{\sqrt{10^2 + 8^2 + 5^2}} = \frac{20}{\sqrt{189}} = \frac{20\sqrt{21}}{63}.$

Our answer is $20 + 21 + 63 = \boxed{104}.$

- spectraldragon8

Solution 3(Formula Abuse)

We use the formula for the volume of iscoceles tetrahedron. $V = \sqrt{(a^2 + b^2 - c^2)(b^2 + c^2 - a^2)(a^2 + c^2 - b^2)/72}$

Note that all faces have equal area due to equal side lengths. By Law of Cosines, we find $\cos{\angle ACB} = \frac{80 + 89 - 41}{2\sqrt{80\cdot 89}}= \frac{16}{9\sqrt{5}}.$ .

From this, we find $\sin{\angle ACB} = \sqrt{1-\cos^2{\angle ACB}} = \sqrt{1 - \frac{256}{405}} = \sqrt{\frac{149}{405}}$ and can find the area of $\triangle ABC$ as $A = \frac{1}{2} \sqrt{89\cdot 80}\cdot \sin{\angle ACB} = 6\sqrt{21}.$

Let $R$ be the distance we want to find. By taking the sum of (equal) volumes $[ABCI] + [ABDI] + [ACDI] + [BCDI] = V,$ We have $V = \frac{4AR}{3}.$ Plugging in and simplifying, we get $R = \frac{20\sqrt{21}}{63}$ for an answer of $20 + 21 + 63 = \boxed{104}$

~AtharvNaphade

Video Solution 1 by OmegaLearn.org

https://youtu.be/qtu0HTFCsqc

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