Difference between revisions of "2024 AIME I Problems/Problem 14"
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+ | ==Solution 3(Formula Abuse)== | ||
+ | We use the formula for the volume of iscoceles tetrahedron. | ||
+ | <math>V = \sqrt{(a^2 + b^2 - c^2)(b^2 + c^2 - a^2)(a^2 + c^2 - b^2)/72}</math> | ||
+ | |||
+ | Note that all faces have equal area due to equal side lengths. By Law of Cosines, we find <cmath>\cos{\angle ACB} = \frac{80 + 89 - 41}{2\sqrt{80\cdot 89}}= \frac{16}{9\sqrt{5}}.</cmath>. | ||
+ | |||
+ | From this, we find <cmath>\sin{\angle ACB} = \sqrt{1-\cos^2{\angle ACB}} = \sqrt{1 - \frac{256}{405}} = \sqrt{\frac{149}{405}}</cmath> and can find the area of <math>\triangle ABC</math> as <cmath>A = \frac{1}{2} \sqrt{89\cdot 80}\cdot \sin{\angle ACB} = 6\sqrt{21}.</cmath> | ||
+ | |||
+ | Let <math>R</math> be the distance we want to find. By taking the sum of (equal) volumes <cmath>[ABCI] + [ABDI] + [ACDI] + [BCDI] = V,</cmath> We have <cmath>V = \frac{4AR}{3}.</cmath> Plugging in and simplifying, we get <math>R = \frac{20\sqrt{21}}{63}</math> for an answer of <math>20 + 21 + 63 = \boxed{104}</math> | ||
+ | |||
+ | ~AtharvNaphade | ||
==See also== | ==See also== |
Revision as of 14:23, 3 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,
\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
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
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.