Difference between revisions of "2024 AIME I Problems/Problem 14"
m (→problem) |
|||
Line 1: | Line 1: | ||
==Problem== | ==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>. | 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== | ||
+ | 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 | ||
+ | <cmath> | ||
+ | \mathbf n:=\frac14\overrightarrow{BC}\times\overrightarrow{CD}=\frac14 | ||
+ | </cmath> | ||
+ | Hence, the distance from <math>A</math> to plane <math>BCD</math>, or the height of the tetrahedron, is | ||
+ | <cmath> | ||
+ | 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}. | ||
+ | </cmath> | ||
+ | Each side of the tetrahedron has the same area due to congruency by "S-S-S", and we call it <math>S</math>. 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, <math>r=\tfrac h4=\tfrac{20\sqrt{21}}{63}</math>, and so the answer is <math>20+21+63=\boxed{104}</math>. | ||
+ | |||
+ | <font size=2>Solution by Quantum-Phantom</font> | ||
==See also== | ==See also== |
Revision as of 19:50, 2 February 2024
Problem
Let be a tetrahedron such that , , and . There exists a point inside the tetrahedron such that the distances from to each of the faces of the tetrahedron are all equal. This distance can be written in the form , where , , and are positive integers, and are relatively prime, and is not divisible by the square of any prime. Find .
Solution 1
Notice that , , and , let , , , and . Then the plane has a normal
Hence, the distance from to plane , or the height of the tetrahedron, is
Each side of the tetrahedron has the same area due to congruency by "S-S-S", and we call it . Then by the volume formula for cones,
Solution by Quantum-Phantom
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.