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 \begin{equation*} \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}. \end{equation*} Hence, the distance from $A$ to plane $BCD$, or the height of the tetrahedron, is \begin{equation*} 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}. \end{equation*} 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

Solution 4

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 $20 + 21 + 63 = \boxed{\textbf{(104) }}$ .

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Solution 5(A quicker method to compute the height from $A$ to plane $BCD$ )

We put the solid to a 3-d coordinate system. Let $B = \left( 0, 0, 0 \right)$ , $D = \left( \sqrt{80}, 0, 0 \right)$ . We put $C$ on the $x-o-y$ plane. Now ,we compute the coordinates of $C$ .

Applying the law of cosines on $\triangle BCD$ , we get $\cos \angle CBD = \frac{4}{\sqrt{41 \cdot 5}}$ . Thus, $\sin \angle CBD = \frac{3 \sqrt{21}}{\sqrt{41 \cdot 5}}$ . Thus, $C = \left( \frac{4}{\sqrt{5}} , \frac{3 \sqrt{21}}{\sqrt{5}} , 0 \right)$ .

Denote $A = \left( x, y , z \right)$ with $z > 0$ .

Because $AB = \sqrt{89}$ , we have \[ x^2 + y^2 + z^2 = 89 \hspace{1cm} (1) \]

Because $AD = \sqrt{41}$ , we have \[ \left( x - \sqrt{80} \right)^2 + y^2 + z^2 = 41 \hspace{1cm} (2) \]

Because $AC = \sqrt{80}$ , we have \[ \left( x - \frac{4}{\sqrt{5}} \right)^2 + \left( y - \frac{3 \sqrt{21}}{\sqrt{5}} \right)^2 + z^2 = 80 \hspace{1cm} (3) \]

Now, we compute $x$ , $y$ and $z$ .

Taking $(1)-(2)$ , we get \[ 2 \sqrt{80} x = 128 . \]

Thus, $x = \frac{16}{\sqrt{5}}$ .

Taking $(1) - (3)$ , we get \[ 2 \cdot \frac{4}{\sqrt{5}} x + 2 \cdot \frac{3 \sqrt{21}}{\sqrt{5}} y = 50 . \]

Thus, $y = \frac{61}{3 \sqrt{5 \cdot 21}}$ .

Plugging $x$ and $y$ into Equation (1), we get $z = \frac{80 \sqrt{21}}{63}$ .

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Solution 6 (Different Perspective)

Consider the following construction of the tetrahedron. Place $AB$ on the floor. Construct an isosceles vertical triangle with $AB$ as its base and $M$ as the top vertex. Place $CD$ on the top vertex parallel to the ground with midpoint $M.$ Observe that $CD$ can rotate about its midpoint. At a certain angle, we observe that the lengths satisfy those given in the problem. If we project $AB$ onto the plane of $CD$ , let the minor angle $\theta$ be this discrepancy.

By Median formula or Stewart's theorem, $AM = \frac{1}{2}\sqrt{2AC^2 + 2AD^2 - CD^2} = \frac{3\sqrt{33}}{2}.$ Consequently the area of $\triangle AMB$ is $\frac{\sqrt{41}}{2} \left (\sqrt{(\frac{3\sqrt{33}}{2})^2 - (\frac{\sqrt{41}}{2})^2} \right ) = 4\sqrt{41}.$ Note the altitude $8$ is also the distance between the parallel planes containing $AB$ and $CD.$

By Distance Formula, $\begin{align*} (\frac{\sqrt{41}}{2} - \frac{1}{2}CD \cos{\theta})^2 + (\frac{1}{2}CD \sin{\theta}^2) + (8)^2 &= AC^2 = 80 \\ (\frac{\sqrt{41}}{2} + \frac{1}{2}CD \cos{\theta})^2 + (\frac{1}{2}CD \sin{\theta}^2) + (8)^2 &= AD^2 = 89 \\ \implies CD \cos{\theta} \sqrt{41} &= 9 \\ \sin{\theta} &= \sqrt{1 - (\frac{9}{41})^2} = \frac{40}{41}. \end{align*}$ Then the volume of the tetrahedron is given by $\frac{1}{3} [AMB] \cdot CD \sin{\theta} = \frac{160}{3}.$

The volume of the tetrahedron can also be segmented into four smaller tetrahedrons from $I$ w.r.t each of the faces. If $r$ is the inradius, i.e the distance to the faces, then $\frac{1}{3} r([ABC] + [ABD] + [ACD] + [BCD])$ must the volume. Each face has the same area by SSS congruence, and by Heron's it is $\frac{1}{4}\sqrt{(a + b + c)(a + b - c)(c + (a-b))(c -(a - b))} = 6\sqrt{21}.$

Therefore the answer is, $\dfrac{3 \frac{160}{3}}{24 \sqrt{21}} = \frac{20\sqrt{21}}{63} \implies \boxed{104}.$

~Aaryabhatta1

Video Solution

https://youtu.be/tq6lraC5prQ?si=5q1NX80POeR949qs

~MathProblemSolvingSkills.com

Video Solution 1 by OmegaLearn.org

https://youtu.be/qtu0HTFCsqc

Video Solution 2

https://youtu.be/1ZtLfJ77Ycg

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

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

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2024 AIME I Problems/Problem 14

Contents

Problem

Solution 1

Solution 2

Solution 3(Formula Abuse)

Solution 4

Solution 5(A quicker method to compute the height from $A$ to plane $BCD$ )

Solution 6 (Different Perspective)

Video Solution

Video Solution 1 by OmegaLearn.org

Video Solution 2

See also

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Search

2024 AIME I Problems/Problem 14

Contents

Problem

Solution 1

Solution 2

Solution 3(Formula Abuse)

Solution 4

Solution 5(A quicker method to compute the height from to plane )

Solution 6 (Different Perspective)

Video Solution

Video Solution 1 by OmegaLearn.org

Video Solution 2

See also

Solution 5(A quicker method to compute the height from $A$ to plane $BCD$ )