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Difference between revisions of "2024 AIME I Problems/Problem 14"

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~Aaryabhatta1
 
~Aaryabhatta1
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==Video Solution==
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https://youtu.be/tq6lraC5prQ?si=5q1NX80POeR949qs
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~MathProblemSolvingSkills.com
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==Video Solution 1 by OmegaLearn.org==
 
==Video Solution 1 by OmegaLearn.org==
 
https://youtu.be/qtu0HTFCsqc
 
https://youtu.be/qtu0HTFCsqc

Revision as of 14:49, 12 March 2024

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)

See also

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