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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>.
 
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>.
  

Revision as of 17:13, 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 $m+n+p$.

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

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