2024 AIME I Problems/Problem 14
Solution 1
Notice that 
, 
, and 
, let 
, 
, 
, and 
. Then the plane 
has a normal
![\[\mathbf n:=\frac14\overrightarrow{BC}\times\overrightarrow{CD}=\frac14\begin{pmatrix}-4\\0\\8\end{pmatrix}\times\begin{pmatrix}4\\-5\\0\end{pmatrix}=\begin{pmatrix}10\\8\\5\end{pmatrix}.\]](http://latex.artofproblemsolving.com/3/1/6/316fcbf7c566933ff5acb19e217494b57429b1c9.png)
Hence, the distance from 
to plane , or the height of the tetrahedron, is
![\[h:=\frac{\mathbf n\cdot\overrightarrow{AB}}{|\mathbf n|}=\frac{10\times4+8\times5+5\times0}{\sqrt{10^2+8^2+5^2}}=\frac{80\sqrt{21}}{63}.\]](http://latex.artofproblemsolving.com/7/f/a/7fa653bb222d5513eb6827f5386cca34ed77abc2.png)
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
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. 
