Difference between revisions of "2024 AIME I Problems/Problem 14"
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+ | ==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 | ||
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<font size=2>Solution by Quantum-Phantom</font> | <font size=2>Solution by Quantum-Phantom</font> | ||
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==Solution 2== | ==Solution 2== |
Revision as of 21:05, 2 February 2024
Contents
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,
\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,
, and so the answer is
.
Solution by Quantum-Phantom
Solution 2
Inscribe tetrahedron in an rectangular prism as shown above.
By the Pythagorean theorem, we note
Solving yields and
Since each face of the tetrahedron is congruent, we know the point we seek is the center of the circumsphere of We know all rectangular prisms can be inscribed in a circumsphere, therefore the circumsphere of the rectangular prism is also the circumsphere of
We know that the distance from all faces must be the same, so we only need to find the distance from the center to plane
.
Let and
We obtain that the plane of
can be marked as
or
and the center of the prism is
Using the Point-to-Plane distance formula, our distance is
Our answer is
- 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.