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| ==Solution 1==
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| <asy>
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| import three;
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| currentprojection = orthographic(1,1,1);
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| triple O = (0,0,0);
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| triple A = (0,2,0);
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| triple B = (0,0,1);
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| triple C = (3,0,0);
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| triple D = (3,2,1);
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| triple E = (3,2,0);
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| triple F = (0,2,1);
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| triple G = (3,0,1);
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| draw(A--B--C--cycle, red);
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| draw(A--B--D--cycle, red);
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| draw(A--C--D--cycle, red);
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| draw(B--C--D--cycle, red);
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| draw(E--A--O--C--cycle);
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| draw(D--F--B--G--cycle);
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| draw(O--B);
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| draw(A--F);
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| draw(E--D);
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| draw(C--G);
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| label("$O$", O, SW);
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| label("$A$", A, NW);
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| label("$B$", B, W);
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| label("$C$", C, S);
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| label("$D$", D, NE);
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| label("$E$", E, SE);
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| label("$F$", F, NW);
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| label("$G$", G, NE);
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| </asy>
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| Inscribe tetrahedron <math>ABCD</math> in an rectangular prism as shown above.
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| By the Pythagorean theorem, we note
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| <cmath>OA^2 + OB^2 = AB^2 = 41,</cmath>
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| <cmath>OA^2 + OC^2 = AC^2 = 80, \text{and}</cmath>
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| <cmath>OB^2 + OC^2 = BC^2 = 89.</cmath>
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| Solving yields <math>OA = 4, OB = 5,</math> and <math>OC = 8.</math>
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| Since each face of the tetrahedron is congruent, we know the point we seek is the center of the circumsphere of <math>ABCD.</math> We know all rectangular prisms can be inscribed in a circumsphere, therefore the circumsphere of the rectangular prism is also the circumsphere of <math>ABCD.</math>
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| We know that the distance from all <math>4</math> faces must be the same, so we only need to find the distance from the center to plane <math>ABC</math>.
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| Let <math>O = (0,0,0), A = (4,0,0), B = (0,5,0),</math> and <math>C = (0,0,8).</math> We obtain that the plane of <math>ABC</math> can be marked as <math>\frac{x}{4} + \frac{y}{5} + \frac{z}{8} = 1,</math> or <math>10x + 8y + 5z - 40 = 0,</math> and the center of the prism is <math>(2,\frac{5}{2},4).</math>
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| Using the Point-to-Plane distance formula, our distance is
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| <cmath>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}.</cmath>
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| Our answer is <math>20 + 21 + 63 = \boxed{104}.</math>
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| - spectraldragon8
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| ==Solution 1== | | ==Solution 1== |
| Notice that <math>41=4^2+5^2</math>, <math>89=5^2+8^2</math>, and <math>80=8^2+4^2</math>, let <math>A~(0,0,0)</math>, <math>B~(4,5,0)</math>, <math>C~(0,5,8)</math>, and <math>D~(4,0,8)</math>. Then the plane <math>BCD</math> has a normal | | Notice that <math>41=4^2+5^2</math>, <math>89=5^2+8^2</math>, and <math>80=8^2+4^2</math>, let <math>A~(0,0,0)</math>, <math>B~(4,5,0)</math>, <math>C~(0,5,8)</math>, and <math>D~(4,0,8)</math>. Then the plane <math>BCD</math> has a normal |
Solution 1
Notice that 
, 
, and 
, let 
, 
, 
, and 
. Then the plane 
has a normal
![\[\mathbf n:=\frac14\overrightarrow{BC}\times\overrightarrow{CD}=\frac14\begin{pmatrix}-4\\0\\8\end{pmatrix}\times\begin{pmatrix}4\\-5\\0\end{pmatrix}=\begin{pmatrix}10\\8\\5\end{pmatrix}.\]](//latex.artofproblemsolving.com/3/1/6/316fcbf7c566933ff5acb19e217494b57429b1c9.png)
Hence, the distance from 
to plane , or the height of the tetrahedron, is
![\[h:=\frac{\mathbf n\cdot\overrightarrow{AB}}{|\mathbf n|}=\frac{10\times4+8\times5+5\times0}{\sqrt{10^2+8^2+5^2}}=\frac{80\sqrt{21}}{63}.\]](//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
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing
, 
, 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 ![\[h:=\frac{\mathbf n\cdot\overrightarrow{AB}}{|\mathbf n|}=\frac{10\times4+8\times5+5\times0}{\sqrt{10^2+8^2+5^2}}=\frac{80\sqrt{21}}{63}.\]](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 
.
