Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 1)
(→‎Solution 1)
Line 1: Line 1:
 +
==Problem==
 +
Let <math>ABCD</math> be a tetrahedron such that <math>AB=CD= \sqrt{41}</math>, <math>AC=BD= \sqrt{80}</math>, and <math>BC=AD= \sqrt{89}</math>. There exists a point <math>I</math> inside the tetrahedron such that the distances from <math>I</math> to each of the faces of the tetrahedron are all equal. This distance can be written in the form <math>\frac{m \sqrt n}{p}</math>, where <math>m</math>, <math>n</math>, and <math>p</math> are positive integers, <math>m</math> and <math>p</math> are relatively prime, and <math>n</math> is not divisible by the square of any prime. Find <math>m+n+p</math>.
 +
 
==Solution 1==
 
==Solution 1==
 
Notice that <math>41=4^2+5^2</math>, <math>89=5^2+8^2</math>, and <math>80=8^2+4^2</math>, let <math>A~(0,0,0)</math>, <math>B~(4,5,0)</math>, <math>C~(0,5,8)</math>, and <math>D~(4,0,8)</math>. Then the plane <math>BCD</math> has a normal
 
Notice that <math>41=4^2+5^2</math>, <math>89=5^2+8^2</math>, and <math>80=8^2+4^2</math>, let <math>A~(0,0,0)</math>, <math>B~(4,5,0)</math>, <math>C~(0,5,8)</math>, and <math>D~(4,0,8)</math>. Then the plane <math>BCD</math> has a normal
Line 16: Line 19:
  
 
<font size=2>Solution by Quantum-Phantom</font>
 
<font size=2>Solution by Quantum-Phantom</font>
−
 
  
 
==Solution 2==
 
==Solution 2==

Revision as of 20:05, 2 February 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 \[\mathbf n:=\frac14\overrightarrow{BC}\times\overrightarrow{CD}=\frac14\begin{pmatrix}-4\\0\\8\end{pmatrix}\times\begin{pmatrix}4\\-5\\0\end{pmatrix}=\begin{pmatrix}10\\8\\5\end{pmatrix}.\] 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

See also

2024 AIME I (Problems • Answer Key • Resources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2024_AIME_I_Problems/Problem_14&oldid=214329"