2024 AIME I Problems/Problem 14
Solution 1
![[asy] import three; currentprojection = orthographic(1,1,1); triple O = (0,0,0); triple A = (0,2,0); triple B = (0,0,1); triple C = (3,0,0); triple D = (3,2,1); triple E = (3,2,0); triple F = (0,2,1); triple G = (3,0,1); draw(A--B--C--cycle, red); draw(A--B--D--cycle, red); draw(A--C--D--cycle, red); draw(B--C--D--cycle, red); draw(E--A--O--C--cycle); draw(D--F--B--G--cycle); draw(O--B); draw(A--F); draw(E--D); draw(C--G); label("$O$", O, SW); label("$A$", A, NW); label("$B$", B, W); label("$C$", C, S); label("$D$", D, NE); label("$E$", E, SE); label("$F$", F, NW); label("$G$", G, NE); [/asy]](http://latex.artofproblemsolving.com/3/8/b/38b15fd4095742c938655456e20a818653bb85f9.png)
Inscribe tetrahedron 
in an rectangular prism as shown above.
By the Pythagorean theorem, we note
![\[OA^2 + OB^2 = AB^2 = 41,\]](http://latex.artofproblemsolving.com/3/e/f/3efbb2c3db616a05e81df28b7d67c38dc7b14a4e.png)
![\[OA^2 + OC^2 = AC^2 = 80, \text{and}\]](http://latex.artofproblemsolving.com/f/c/b/fcb3b9ca8c219cd6e531e555ff5e54f72b2bfd15.png)
![\[OB^2 + OC^2 = BC^2 = 89.\]](http://latex.artofproblemsolving.com/5/0/3/5039ee201ca034874b49ba63c178463efb436de7.png)
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
![\[d = \frac{|10\cdot 2 + 8\cdot \frac{5}{2} + 5\cdot 4 - 40|}{\sqrt{10^2 + 8^2 + 5^2}} = \frac{20}{\sqrt{189}} = \frac{20\sqrt{21}}{63}.\]](http://latex.artofproblemsolving.com/4/c/a/4ca419e936ae97cc08313d4ff4e5679b65efb331.png)
Our answer is 
- spectraldragon8
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 | ||
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