2024 AIME I Problems/Problem 14
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[hide]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
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
Solution 3(Formula Abuse)
We use the formula for the volume of iscoceles tetrahedron.
Note that all faces have equal area due to equal side lengths. By Law of Cosines, we find .
From this, we find and can find the area of as
Let be the distance we want to find. By taking the sum of (equal) volumes We have Plugging in and simplifying, we get for an answer of
~AtharvNaphade
Video Solution 1 by OmegaLearn.org
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 | ||
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