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Difference between revisions of "2024 AIME I Problems/Problem 14"

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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 \(41=4^2+5^2\), \(89=5^2+8^2\), and \(80=8^2+4^2\), let \(A~(0,0,0)\), \(B~(4,5,0)\), \(C~(0,5,8)\), and \(D~(4,0,8)\). Then the plane \(BCD\) has a normal
−
<cmath>
+
\begin{equation*}
 
\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}.
 
\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}.
−
</cmath>
+
\end{equation*}
−
Hence, the distance from <math>A</math> to plane <math>BCD</math>, or the height of the tetrahedron, is
+
Hence, the distance from \(A\) to plane \(BCD\), or the height of the tetrahedron, is
−
<cmath>
+
\begin{equation*}
 
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}.
 
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}.
−
</cmath>
+
\end{equation*}
−
Each side of the tetrahedron has the same area due to congruency by "S-S-S", and we call it <math>S</math>. Then by the volume formula for cones,
+
Each side of the tetrahedron has the same area due to congruency by "S-S-S", and we call it \(S\). Then by the volume formula for pyramids,
 
\begin{align*}
 
\begin{align*}
 
\frac13Sh&=V_{D\text-ABC}=V_{I\text-ABC}+V_{I\text-BCD}+V_{I\text-CDA}+V_{I\text-DAB}\\
 
\frac13Sh&=V_{D\text-ABC}=V_{I\text-ABC}+V_{I\text-BCD}+V_{I\text-CDA}+V_{I\text-DAB}\\
 
&=\frac13Sr\cdot4.
 
&=\frac13Sr\cdot4.
 
\end{align*}
 
\end{align*}
−
Hence, <math>r=\tfrac h4=\tfrac{20\sqrt{21}}{63}</math>, and so the answer is <math>20+21+63=\boxed{104}</math>.
+
Hence, \(r=\tfrac h4=\tfrac{20\sqrt{21}}{63}\), and so the answer is \(20+21+63=\boxed{104}\).
  
 
<font size=2>Solution by Quantum-Phantom</font>
 
<font size=2>Solution by Quantum-Phantom</font>
Line 81: Line 81:
 
Our answer is <math>20 + 21 + 63 = \boxed{104}.</math>
 
Our answer is <math>20 + 21 + 63 = \boxed{104}.</math>
  
−
- spectraldragon8
+
- [https://artofproblemsolving.com/wiki/index.php/User:Spectraldragon8 spectraldragon8]
 +
 
 +
==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
 +
 
 +
==Solution 4==
 +
 
 +
Let <math>AH</math> be perpendicular to <math>BCD</math> that meets this plane at point <math>H</math>.
 +
Let <math>AP</math>, <math>AQ</math>, and <math>AR</math> be heights to lines <math>BC</math>, <math>CD</math>, and <math>BD</math> with feet <math>P</math>, <math>Q</math>, and <math>R</math>, respectively.
 +
 
 +
We notice that all faces are congruent. Following from Heron's formula, the area of each face, denoted as <math>A</math>, is <math>A = 6 \sqrt{21}</math>.
 +
 
 +
Hence, by using this area, we can compute <math>AP</math>, <math>AQ</math> and <math>AR</math>.
 +
We have <math>AP = \frac{2 A}{BC} = \frac{2A}{\sqrt{89}}</math>, <math>AQ = \frac{2 A}{CD} = \frac{2A}{\sqrt{41}}</math>, and <math>AR = \frac{2 A}{BC} = \frac{2A}{\sqrt{80}}</math>.
 +
 
 +
Because <math>AH \perp BCD</math>, we have <math>AH \perp BC</math>. Recall that <math>AP \perp BC</math>.
 +
Hence, <math>BC \perp APH</math>. Hence, <math>BC \perp HP</math>.
 +
 
 +
Analogously, <math>CD \perp HQ</math> and <math>BD \perp HR</math>.
 +
 
 +
We introduce a function <math>\epsilon \left( l \right)</math> for <math>\triangle BCD</math> that is equal to 1 (resp. -1) if point <math>H</math> and the opposite vertex of side <math>l</math> are on the same side (resp. opposite sides) of side <math>l</math>.
 +
 
 +
The area of <math>\triangle BCD</math> is
 +
\begin{align*}
 +
A & = \epsilon_{BC} {\rm Area} \ \triangle HBC
 +
+ \epsilon_{CD} {\rm Area} \ \triangle HCD
 +
+ \epsilon_{BD} {\rm Area} \ \triangle HBD \\
 +
& = \frac{1}{2} \epsilon_{BC} BC \cdot HP
 +
+ \frac{1}{2} \epsilon_{CD} CD \cdot HQ +
 +
\frac{1}{2} \epsilon_{BD} BD \cdot HR \\
 +
& = \frac{1}{2} \epsilon_{BC} BC \cdot \sqrt{AP^2 - AH^2}
 +
+ \frac{1}{2} \epsilon_{CD} CD \cdot \sqrt{AQ^2 - AH^2} \\
 +
& \quad + \frac{1}{2} \epsilon_{BD} CD \cdot \sqrt{AR^2 - AH^2} . \hspace{1cm} (1)
 +
\end{align*}
 +
 
 +
Denote <math>B = 2A</math>.
 +
The above equation can be organized as
 +
\begin{align*}
 +
B & = \epsilon_{BC} \sqrt{B^2 - 89 AH^2}
 +
+ \epsilon_{CD} \sqrt{B^2 - 41 AH^2} \\
 +
& \quad + \epsilon_{BD} \sqrt{B^2 - 80 AH^2} .
 +
\end{align*}
 +
 
 +
This can be further reorganized as
 +
\begin{align*}
 +
B - \epsilon_{BC} \sqrt{B^2 - 89 AH^2}
 +
& = \epsilon_{CD} \sqrt{B^2 - 41 AH^2}
 +
+ \epsilon_{BD} \sqrt{B^2 - 80 AH^2} .
 +
\end{align*}
 +
 
 +
Taking squares on both sides and reorganizing terms, we get
 +
\begin{align*}
 +
& 16 AH^2 - \epsilon_{BC} B \sqrt{B^2 - 89 AH^2} \\
 +
& = \epsilon_{CD} \epsilon_{BD}
 +
\sqrt{\left( B^2 - 41 AH^2 \right) \left( B^2 - 80 AH^2 \right)} .
 +
\end{align*}
 +
 
 +
Taking squares on both sides and reorganizing terms, we get
 +
<cmath>
 +
\[
 +
- \epsilon_{BC} 2 B \sqrt{B^2 - 89 AH^2} = - 2 B^2 + 189 AH^2 .
 +
\]
 +
</cmath>
 +
 
 +
Taking squares on both sides, we finally get
 +
\begin{align*}
 +
AH & = \frac{20B}{189} \\
 +
& = \frac{40A}{189}.
 +
\end{align*}
 +
 
 +
Now, we plug this solution to Equation (1). We can see that <math>\epsilon_{BC} = -1</math>, <math>\epsilon_{CD} = \epsilon_{BD} = 1</math>.
 +
This indicates that <math>H</math> is out of <math>\triangle BCD</math>. To be specific, <math>H</math> and <math>D</math> are on opposite sides of <math>BC</math>, <math>H</math> and <math>C</math> are on the same side of <math>BD</math>, and <math>H</math> and <math>B</math> are on the same side of <math>CD</math>.
 +
 
 +
Now, we compute the volume of the tetrahedron <math>ABCD</math>, denoted as <math>V</math>. We have <math>V = \frac{1}{3} A \cdot AH = \frac{40 A^2}{3 \cdot 189}</math>.
 +
 
 +
Denote by <math>r</math> the inradius of the inscribed sphere in <math>ABCD</math>.
 +
Denote by <math>I</math> the incenter.
 +
Thus, the volume of <math>ABCD</math> can be alternatively calculated as
 +
\begin{align*}
 +
V & = {\rm Vol} \ IABC + {\rm Vol} \ IACD + {\rm Vol} \ IABD + {\rm Vol} \ IBCD \\
 +
& = \frac{1}{3} r \cdot 4A .
 +
\end{align*}
 +
 
 +
From our two methods to compute the volume of <math>ABCD</math> and equating them, we get
 +
\begin{align*}
 +
r & = \frac{10A}{189} \\
 +
& = \frac{20 \sqrt{21}}{63} .
 +
\end{align*}
 +
 
 +
Therefore, the answer is <math>20 + 21 + 63 = \boxed{\textbf{(104) }}</math>.
 +
 
 +
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 +
 
 +
==Solution 5(A quicker method to compute the height from <math>A</math> to plane <math>BCD</math>)==
 +
 
 +
We put the solid to a 3-d coordinate system. Let <math>B = \left( 0, 0, 0 \right)</math>, <math>D = \left( \sqrt{80}, 0, 0 \right)</math>.
 +
We put <math>C</math> on the <math>x-o-y</math> plane.
 +
Now ,we compute the coordinates of <math>C</math>.
 +
 
 +
Applying the law of cosines on <math>\triangle BCD</math>, we get <math>\cos \angle CBD = \frac{4}{\sqrt{41 \cdot 5}}</math>.
 +
Thus, <math>\sin \angle CBD = \frac{3 \sqrt{21}}{\sqrt{41 \cdot 5}}</math>.
 +
Thus, <math>C = \left( \frac{4}{\sqrt{5}} , \frac{3 \sqrt{21}}{\sqrt{5}} , 0 \right)</math>.
 +
 
 +
Denote <math>A = \left( x, y , z \right)</math> with <math>z > 0</math>.
 +
 
 +
Because <math>AB = \sqrt{89}</math>, we have
 +
<math>x^2 + y^2 + z^2 = 89 </math>
 +
 
 +
Because <math>AD = \sqrt{41}</math>, we have
 +
<cmath>
 +
\left( x - \sqrt{80} \right)^2 + y^2 + z^2 = 41 \hspace{1cm} (2)
 +
</cmath>
 +
 
 +
Because <math>AC = \sqrt{80}</math>, we have
 +
<cmath>
 +
\left( x - \frac{4}{\sqrt{5}} \right)^2 + \left( y - \frac{3 \sqrt{21}}{\sqrt{5}} \right)^2
 +
+ z^2 = 80 \hspace{1cm} (3)
 +
</cmath>
 +
 
 +
Now, we compute <math>x</math>, <math>y</math> and <math>z</math>.
 +
 
 +
Taking <math>(1)-(2)</math>, we get
 +
<cmath>
 +
2 \sqrt{80} x = 128 .
 +
</cmath>
 +
 
 +
Thus, <math>x = \frac{16}{\sqrt{5}}</math>.
 +
 
 +
Taking <math>(1) - (3)</math>, we get
 +
<cmath>
 +
2 \cdot \frac{4}{\sqrt{5}} x + 2 \cdot \frac{3 \sqrt{21}}{\sqrt{5}} y
 +
= 50 .
 +
</cmath>
 +
 
 +
Thus, <math>y = \frac{61}{3 \sqrt{5 \cdot 21}}</math>.
 +
 
 +
Plugging <math>x</math> and <math>y</math> into Equation (1), we get <math>z = \frac{80 \sqrt{21}}{63}</math>.
 +
 
 +
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 +
~numerophile (formatting edits)
 +
 
 +
==Solution 6 (Different Perspective)==
 +
Consider the following construction of the tetrahedron. Place <math>AB</math> on the floor. Construct an isosceles vertical triangle with <math>AB</math> as its base and <math>M</math> as the top vertex. Place <math>CD</math> on the top vertex parallel to the ground with midpoint <math>M.</math> Observe that <math>CD</math> can rotate about its midpoint. At a certain angle, we observe that the lengths satisfy those given in the problem. If we project <math>AB</math> onto the plane of <math>CD</math>, let the minor angle <math>\theta</math> be this discrepancy.
 +
 
 +
By Median formula or Stewart's theorem, <math>AM = \frac{1}{2}\sqrt{2AC^2 + 2AD^2 - CD^2} = \frac{3\sqrt{33}}{2}.</math> Consequently the area of <math>\triangle AMB</math> is <math>\frac{\sqrt{41}}{2} \left (\sqrt{(\frac{3\sqrt{33}}{2})^2 - (\frac{\sqrt{41}}{2})^2} \right ) = 4\sqrt{41}.</math> Note the altitude <math>8</math> is also the distance between the parallel planes containing <math>AB</math> and <math>CD.</math>
 +
 
 +
By Distance Formula,
 +
<cmath>\begin{align*}
 +
(\frac{\sqrt{41}}{2} - \frac{1}{2}CD \cos{\theta})^2 + (\frac{1}{2}CD \sin{\theta}^2) + (8)^2 &= AC^2 = 80 \\
 +
(\frac{\sqrt{41}}{2} + \frac{1}{2}CD \cos{\theta})^2 + (\frac{1}{2}CD \sin{\theta}^2) + (8)^2 &= AD^2 = 89 \\
 +
\implies CD \cos{\theta} \sqrt{41} &= 9 \\
 +
\sin{\theta} &= \sqrt{1 - (\frac{9}{41})^2} = \frac{40}{41}.
 +
\end{align*}</cmath>
 +
Then the volume of the tetrahedron is given by <math>\frac{1}{3} [AMB] \cdot CD \sin{\theta} = \frac{160}{3}.</math>
 +
 
 +
The volume of the tetrahedron can also be segmented into four smaller tetrahedrons from <math>I</math> w.r.t each of the faces. If <math>r</math> is the inradius, i.e the distance to the faces, then <math>\frac{1}{3} r([ABC] + [ABD] + [ACD] + [BCD])</math> must the volume. Each face has the same area by SSS congruence, and by Heron's it is <math>\frac{1}{4}\sqrt{(a + b + c)(a + b - c)(c + (a-b))(c -(a - b))} = 6\sqrt{21}.</math>
 +
 
 +
Therefore the answer is, <math>\dfrac{3 \frac{160}{3}}{24 \sqrt{21}} = \frac{20\sqrt{21}}{63} \implies \boxed{104}.</math>
 +
 
 +
~Aaryabhatta1
 +
 
 +
==Solution 7 (simplest way to calculate [ABCD])==
 +
After finding that <math>ABCD</math> can be inscribed in a <math>4\times5\times8</math> box, note that <math>[ABCD]=\mathbf{box}-\frac{4}{3}(4)(5)(8)=\frac{160}{3}</math>, where the third term describes the area of 4 congruent right tetrahedrons whose right angle is at the apex and whose apex lengths are 4, 5, and 8. Then proceed as in Solution 3.
 +
 
 +
~clarkculus
 +
 
 +
==Solution 8 (fast)==
 +
Observe that <math>ABCD</math> can be inscribed in a <math>4\times5\times8</math> box, which gives the coordinates of the points: we have <math>A=(8,0,0)</math>, <math>B=(8,5,4)</math>, <math>C=(0,0,4)</math> and <math>D=(0,5,0)</math>. Recall the formula <math>V=\frac{1}{3}RS</math>, where <math>V</math> is the volume of the tetrahedron, <math>S</math> is the surface area, and <math>R</math> is the insphere radius. Using the fact that the faces of <math>ABCD</math> have equal area and that <math>V=\frac{1}{3}[ACD] \times h</math>, where <math>h</math> is the distance from plane <math>ACD</math> to <math>B</math>, we can rearrange to get <math>R=\frac{h}{4}</math>. We compute that the the equation of plane <math>ACD</math> is <math>5x+8y+10z-40=0</math>, and applying point-to-plane gives <math>h=\frac{\vert 5(8)+8(5)+10(4)-40 \vert}{\sqrt{5^2+8^2+10^2}}=\frac{80}{\sqrt{189}} \implies R=\frac{20\sqrt{21}}{63} \implies \boxed{104}</math>.
 +
 
 +
~bomberdoodles
 +
 
 +
==Solution 9 (Analytical Geometry)==
 +
Like in Solution 1, we have <math>A = (0,0,0)</math>, <math>B = (4,5,0)</math>, <math>C = (0,5,8)</math>, <math>D = (4,0,8)</math>. If we let the inradius be <math>r</math>, then the volume of the tetrahedron is equal to <math>\frac{1}{3}[ABC]r + \frac{1}{3}[BCD]r + \frac{1}{3}[ABD]r + \frac{1}{3}[ACD]r = V</math> by the formula <math>V = \frac{1}{3}Ah</math>. Also note that <math>[ABC] = [BCD] = [ABD] = [ACD]</math> because they all have the same side lengths. So, if we let <math>[ABC] = A</math>, then <math>\frac{4}{3}Ar = V</math>, so <math>r = \frac{3V}{4A}</math>. Now, all we need to do is to find V and A. We note that the area of the triangle <math>[ABC] = |\frac{\overrightarrow{AB}\times\overrightarrow{AC}}{2}|</math>. Also, if we let the normal vector from <math>\Delta ABC</math> to be <math>\bold{n}</math>, the height <math>h</math> from <math>D</math> to <math>\Delta ABC</math> is <math>\overrightarrow{AD} \cdot \hat{\bold{n}} = \frac{\overrightarrow{AD} \cdot \bold{n}}{\bold{|n|}}</math>. Therefore, the volume of the tetrahedron is equal to <math>\frac{1}{3}[ABC]h = \frac{1}{3}|\frac{\overrightarrow{AB}\times\overrightarrow{AC}}{2}|\frac{\overrightarrow{AD} \cdot \bold{n}}{\bold{|n|}} = \frac{1}{6}|\overrightarrow{AB}\times\overrightarrow{AC}|\frac{\overrightarrow{AD} \cdot \bold{n}}{\bold{|n|}}</math>. Now, we notice that <math>\overrightarrow{AB}\times\overrightarrow{AC}</math> is a normal vector of the <math>\Delta ABC</math> because the cross product of two vectors is always perpendicular to both of the vectors, and both of the vectors lie in the plane of <math>\Delta ABC</math>. This means that we can let <math>\bold{n} = \overrightarrow{AB}\times\overrightarrow{AC}</math>. So, the volume of the tetrahedron is <math>V = \frac{1}{6}|\overrightarrow{AB}\times\overrightarrow{AC}|\frac{\overrightarrow{AD} \cdot (\bold{\overrightarrow {AB}\times\overrightarrow{AC}})}{\bold{|\overrightarrow{AB}\times\overrightarrow{AC}|}} = \frac{1}{6}\overrightarrow{AD} \cdot (\bold{\overrightarrow{AB}\times\overrightarrow{AC}})</math>. The expression <math>\overrightarrow{AD} \cdot (\bold{\overrightarrow{AB}\times\overrightarrow{AC}})</math> is more commonly known as a box product and can be used to find the volume of any parallelepiped. We know that <math>\overrightarrow{AB} = (4,5,0)</math>,
 +
<math>\overrightarrow{AC} = (0,5,8)</math>, and <math>\overrightarrow{AD} = (4,0,8)</math>. Now, <math>\overrightarrow{AB} \times \overrightarrow{AC} = (8*5-5*0, 0*0-8*4, 4*5-5*0) = (40, -32, 20)</math>. So, <math>\overrightarrow{AD} \cdot (\bold{\overrightarrow{AB}\times\overrightarrow{AC}}) = (4,0,8) \cdot (40,-32,20) = 4*40 + 0*-32 + 8*20 = 320</math>. So, <math>V = \frac{1}{6}\overrightarrow{AD} \cdot (\bold{\overrightarrow{AB}\times\overrightarrow{AC}}) = \frac{320}{6} = \frac{160}{3}</math>. Now <math>[ABC]</math> is just equal to <math>\frac{1}{2}|\overrightarrow{AB} \times \overrightarrow{AC}|</math> as it is half the parallelogram. So, we have <math>A = \frac{1}{2}|\overrightarrow{AB} \times \overrightarrow{AC}| = \frac{1}{2}|(40,-32,20)| = \frac{1}{2}\sqrt{3024}</math>. So, <math>r = \frac{3V}{4A} = \frac{160}{2\sqrt{3024}}</math>. This expression can be simplified to <math>\frac{20\sqrt{21}}{63}</math>. So, our answer is <math>20 + 63 + 21 = \boxed{104}</math>.
 +
 
 +
- Rutvik Arora (plz sub to https://www.youtube.com/@themathguy6302)
 +
 
 +
==Video Solution==
 +
https://youtu.be/tq6lraC5prQ?si=5q1NX80POeR949qs
 +
 
 +
~MathProblemSolvingSkills.com
 +
 
 +
 
 +
 
 +
==Video Solution 1 by OmegaLearn.org==
 +
https://youtu.be/qtu0HTFCsqc
 +
 
 +
==Video Solution 2==
 +
 
 +
https://youtu.be/1ZtLfJ77Ycg
 +
 
 +
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
  
 
==See also==
 
==See also==
 
{{AIME box|year=2024|n=I|num-b=13|num-a=15}}
 
{{AIME box|year=2024|n=I|num-b=13|num-a=15}}
 +
 +
[[Category:Intermediate Geometry Problems]]
 +
[[Category:3D Geometry Problems]]
  
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 01:35, 21 November 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 $m+n+p$.

Solution 1

Notice that \(41=4^2+5^2\), \(89=5^2+8^2\), and \(80=8^2+4^2\), let \(A~(0,0,0)\), \(B~(4,5,0)\), \(C~(0,5,8)\), and \(D~(4,0,8)\). Then the plane \(BCD\) has a normal \begin{equation*} \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}. \end{equation*} Hence, the distance from \(A\) to plane \(BCD\), or the height of the tetrahedron, is \begin{equation*} 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}. \end{equation*} Each side of the tetrahedron has the same area due to congruency by "S-S-S", and we call it \(S\). Then by the volume formula for pyramids, \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, \(r=\tfrac h4=\tfrac{20\sqrt{21}}{63}\), and so the answer is \(20+21+63=\boxed{104}\).

Solution by Quantum-Phantom

Solution 2

[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]


Inscribe tetrahedron $ABCD$ in an rectangular prism as shown above.

By the Pythagorean theorem, we note

\[OA^2 + OB^2 = AB^2 = 41,\] \[OA^2 + OC^2 = AC^2 = 80, \text{and}\] \[OB^2 + OC^2 = BC^2 = 89.\]

Solving yields $OA = 4, OB = 5,$ and $OC = 8.$

Since each face of the tetrahedron is congruent, we know the point we seek is the center of the circumsphere of $ABCD.$ We know all rectangular prisms can be inscribed in a circumsphere, therefore the circumsphere of the rectangular prism is also the circumsphere of $ABCD.$

We know that the distance from all $4$ faces must be the same, so we only need to find the distance from the center to plane $ABC$.

Let $O = (0,0,0), A = (4,0,0), B = (0,5,0),$ and $C = (0,0,8).$ We obtain that the plane of $ABC$ can be marked as $\frac{x}{4} + \frac{y}{5} + \frac{z}{8} = 1,$ or $10x + 8y + 5z - 40 = 0,$ and the center of the prism is $(2,\frac{5}{2},4).$

Using the Point-to-Plane distance formula, our distance is

\[d = \frac{|10\cdot 2 + 8\cdot \frac{5}{2} + 5\cdot 4 - 40|}{\sqrt{10^2 + 8^2 + 5^2}} = \frac{20}{\sqrt{189}} = \frac{20\sqrt{21}}{63}.\]

Our answer is $20 + 21 + 63 = \boxed{104}.$

- spectraldragon8

Solution 3(Formula Abuse)

We use the formula for the volume of iscoceles tetrahedron. $V = \sqrt{(a^2 + b^2 - c^2)(b^2 + c^2 - a^2)(a^2 + c^2 - b^2)/72}$

Note that all faces have equal area due to equal side lengths. By Law of Cosines, we find \[\cos{\angle ACB} = \frac{80 + 89 - 41}{2\sqrt{80\cdot 89}}= \frac{16}{9\sqrt{5}}.\].

From this, we find \[\sin{\angle ACB} = \sqrt{1-\cos^2{\angle ACB}} = \sqrt{1 - \frac{256}{405}} = \sqrt{\frac{149}{405}}\] and can find the area of $\triangle ABC$ as \[A = \frac{1}{2} \sqrt{89\cdot 80}\cdot \sin{\angle ACB} = 6\sqrt{21}.\]

Let $R$ be the distance we want to find. By taking the sum of (equal) volumes \[[ABCI] + [ABDI] + [ACDI] + [BCDI] = V,\] We have \[V = \frac{4AR}{3}.\] Plugging in and simplifying, we get $R = \frac{20\sqrt{21}}{63}$ for an answer of $20 + 21 + 63 = \boxed{104}$

~AtharvNaphade

Solution 4

Let $AH$ be perpendicular to $BCD$ that meets this plane at point $H$. Let $AP$, $AQ$, and $AR$ be heights to lines $BC$, $CD$, and $BD$ with feet $P$, $Q$, and $R$, respectively.

We notice that all faces are congruent. Following from Heron's formula, the area of each face, denoted as $A$, is $A = 6 \sqrt{21}$.

Hence, by using this area, we can compute $AP$, $AQ$ and $AR$. We have $AP = \frac{2 A}{BC} = \frac{2A}{\sqrt{89}}$, $AQ = \frac{2 A}{CD} = \frac{2A}{\sqrt{41}}$, and $AR = \frac{2 A}{BC} = \frac{2A}{\sqrt{80}}$.

Because $AH \perp BCD$, we have $AH \perp BC$. Recall that $AP \perp BC$. Hence, $BC \perp APH$. Hence, $BC \perp HP$.

Analogously, $CD \perp HQ$ and $BD \perp HR$.

We introduce a function $\epsilon \left( l \right)$ for $\triangle BCD$ that is equal to 1 (resp. -1) if point $H$ and the opposite vertex of side $l$ are on the same side (resp. opposite sides) of side $l$.

The area of $\triangle BCD$ is \begin{align*} A & = \epsilon_{BC} {\rm Area} \ \triangle HBC + \epsilon_{CD} {\rm Area} \ \triangle HCD + \epsilon_{BD} {\rm Area} \ \triangle HBD \\ & = \frac{1}{2} \epsilon_{BC} BC \cdot HP + \frac{1}{2} \epsilon_{CD} CD \cdot HQ + \frac{1}{2} \epsilon_{BD} BD \cdot HR \\ & = \frac{1}{2} \epsilon_{BC} BC \cdot \sqrt{AP^2 - AH^2} + \frac{1}{2} \epsilon_{CD} CD \cdot \sqrt{AQ^2 - AH^2} \\ & \quad + \frac{1}{2} \epsilon_{BD} CD \cdot \sqrt{AR^2 - AH^2} . \hspace{1cm} (1) \end{align*}

Denote $B = 2A$. The above equation can be organized as \begin{align*} B & = \epsilon_{BC} \sqrt{B^2 - 89 AH^2} + \epsilon_{CD} \sqrt{B^2 - 41 AH^2} \\ & \quad + \epsilon_{BD} \sqrt{B^2 - 80 AH^2} . \end{align*}

This can be further reorganized as \begin{align*} B - \epsilon_{BC} \sqrt{B^2 - 89 AH^2} & = \epsilon_{CD} \sqrt{B^2 - 41 AH^2} + \epsilon_{BD} \sqrt{B^2 - 80 AH^2} . \end{align*}

Taking squares on both sides and reorganizing terms, we get \begin{align*} & 16 AH^2 - \epsilon_{BC} B \sqrt{B^2 - 89 AH^2} \\ & = \epsilon_{CD} \epsilon_{BD} \sqrt{\left( B^2 - 41 AH^2 \right) \left( B^2 - 80 AH^2 \right)} . \end{align*}

Taking squares on both sides and reorganizing terms, we get \[ - \epsilon_{BC} 2 B \sqrt{B^2 - 89 AH^2} = - 2 B^2 + 189 AH^2 . \]

Taking squares on both sides, we finally get \begin{align*} AH & = \frac{20B}{189} \\ & = \frac{40A}{189}. \end{align*}

Now, we plug this solution to Equation (1). We can see that $\epsilon_{BC} = -1$, $\epsilon_{CD} = \epsilon_{BD} = 1$. This indicates that $H$ is out of $\triangle BCD$. To be specific, $H$ and $D$ are on opposite sides of $BC$, $H$ and $C$ are on the same side of $BD$, and $H$ and $B$ are on the same side of $CD$.

Now, we compute the volume of the tetrahedron $ABCD$, denoted as $V$. We have $V = \frac{1}{3} A \cdot AH = \frac{40 A^2}{3 \cdot 189}$.

Denote by $r$ the inradius of the inscribed sphere in $ABCD$. Denote by $I$ the incenter. Thus, the volume of $ABCD$ can be alternatively calculated as \begin{align*} V & = {\rm Vol} \ IABC + {\rm Vol} \ IACD + {\rm Vol} \ IABD + {\rm Vol} \ IBCD \\ & = \frac{1}{3} r \cdot 4A . \end{align*}

From our two methods to compute the volume of $ABCD$ and equating them, we get \begin{align*} r & = \frac{10A}{189} \\ & = \frac{20 \sqrt{21}}{63} . \end{align*}

Therefore, the answer is $20 + 21 + 63 = \boxed{\textbf{(104) }}$.

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Solution 5(A quicker method to compute the height from $A$ to plane $BCD$)

We put the solid to a 3-d coordinate system. Let $B = \left( 0, 0, 0 \right)$, $D = \left( \sqrt{80}, 0, 0 \right)$. We put $C$ on the $x-o-y$ plane. Now ,we compute the coordinates of $C$.

Applying the law of cosines on $\triangle BCD$, we get $\cos \angle CBD = \frac{4}{\sqrt{41 \cdot 5}}$. Thus, $\sin \angle CBD = \frac{3 \sqrt{21}}{\sqrt{41 \cdot 5}}$. Thus, $C = \left( \frac{4}{\sqrt{5}} , \frac{3 \sqrt{21}}{\sqrt{5}} , 0 \right)$.

Denote $A = \left( x, y , z \right)$ with $z > 0$.

Because $AB = \sqrt{89}$, we have $x^2 + y^2 + z^2 = 89$

Because $AD = \sqrt{41}$, we have \[\left( x - \sqrt{80} \right)^2 + y^2 + z^2 = 41 \hspace{1cm} (2)\]

Because $AC = \sqrt{80}$, we have \[\left( x - \frac{4}{\sqrt{5}} \right)^2 + \left( y - \frac{3 \sqrt{21}}{\sqrt{5}} \right)^2 + z^2 = 80 \hspace{1cm} (3)\]

Now, we compute $x$, $y$ and $z$.

Taking $(1)-(2)$, we get \[2 \sqrt{80} x = 128 .\]

Thus, $x = \frac{16}{\sqrt{5}}$.

Taking $(1) - (3)$, we get \[2 \cdot \frac{4}{\sqrt{5}} x + 2 \cdot \frac{3 \sqrt{21}}{\sqrt{5}} y = 50 .\]

Thus, $y = \frac{61}{3 \sqrt{5 \cdot 21}}$.

Plugging $x$ and $y$ into Equation (1), we get $z = \frac{80 \sqrt{21}}{63}$.

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) ~numerophile (formatting edits)

Solution 6 (Different Perspective)

Consider the following construction of the tetrahedron. Place $AB$ on the floor. Construct an isosceles vertical triangle with $AB$ as its base and $M$ as the top vertex. Place $CD$ on the top vertex parallel to the ground with midpoint $M.$ Observe that $CD$ can rotate about its midpoint. At a certain angle, we observe that the lengths satisfy those given in the problem. If we project $AB$ onto the plane of $CD$, let the minor angle $\theta$ be this discrepancy.

By Median formula or Stewart's theorem, $AM = \frac{1}{2}\sqrt{2AC^2 + 2AD^2 - CD^2} = \frac{3\sqrt{33}}{2}.$ Consequently the area of $\triangle AMB$ is $\frac{\sqrt{41}}{2} \left (\sqrt{(\frac{3\sqrt{33}}{2})^2 - (\frac{\sqrt{41}}{2})^2} \right ) = 4\sqrt{41}.$ Note the altitude $8$ is also the distance between the parallel planes containing $AB$ and $CD.$

By Distance Formula, \begin{align*} (\frac{\sqrt{41}}{2} - \frac{1}{2}CD \cos{\theta})^2 + (\frac{1}{2}CD \sin{\theta}^2) + (8)^2 &= AC^2 = 80 \\ (\frac{\sqrt{41}}{2} + \frac{1}{2}CD \cos{\theta})^2 + (\frac{1}{2}CD \sin{\theta}^2) + (8)^2 &= AD^2 = 89 \\ \implies CD \cos{\theta} \sqrt{41} &= 9 \\ \sin{\theta} &= \sqrt{1 - (\frac{9}{41})^2} = \frac{40}{41}. \end{align*} Then the volume of the tetrahedron is given by $\frac{1}{3} [AMB] \cdot CD \sin{\theta} = \frac{160}{3}.$

The volume of the tetrahedron can also be segmented into four smaller tetrahedrons from $I$ w.r.t each of the faces. If $r$ is the inradius, i.e the distance to the faces, then $\frac{1}{3} r([ABC] + [ABD] + [ACD] + [BCD])$ must the volume. Each face has the same area by SSS congruence, and by Heron's it is $\frac{1}{4}\sqrt{(a + b + c)(a + b - c)(c + (a-b))(c -(a - b))} = 6\sqrt{21}.$

Therefore the answer is, $\dfrac{3 \frac{160}{3}}{24 \sqrt{21}} = \frac{20\sqrt{21}}{63} \implies \boxed{104}.$

~Aaryabhatta1

Solution 7 (simplest way to calculate [ABCD])

After finding that $ABCD$ can be inscribed in a $4\times5\times8$ box, note that $[ABCD]=\mathbf{box}-\frac{4}{3}(4)(5)(8)=\frac{160}{3}$, where the third term describes the area of 4 congruent right tetrahedrons whose right angle is at the apex and whose apex lengths are 4, 5, and 8. Then proceed as in Solution 3.

~clarkculus

Solution 8 (fast)

Observe that $ABCD$ can be inscribed in a $4\times5\times8$ box, which gives the coordinates of the points: we have $A=(8,0,0)$, $B=(8,5,4)$, $C=(0,0,4)$ and $D=(0,5,0)$. Recall the formula $V=\frac{1}{3}RS$, where $V$ is the volume of the tetrahedron, $S$ is the surface area, and $R$ is the insphere radius. Using the fact that the faces of $ABCD$ have equal area and that $V=\frac{1}{3}[ACD] \times h$, where $h$ is the distance from plane $ACD$ to $B$, we can rearrange to get $R=\frac{h}{4}$. We compute that the the equation of plane $ACD$ is $5x+8y+10z-40=0$, and applying point-to-plane gives $h=\frac{\vert 5(8)+8(5)+10(4)-40 \vert}{\sqrt{5^2+8^2+10^2}}=\frac{80}{\sqrt{189}} \implies R=\frac{20\sqrt{21}}{63} \implies \boxed{104}$.

~bomberdoodles

Solution 9 (Analytical Geometry)

Like in Solution 1, we have $A = (0,0,0)$, $B = (4,5,0)$, $C = (0,5,8)$, $D = (4,0,8)$. If we let the inradius be $r$, then the volume of the tetrahedron is equal to $\frac{1}{3}[ABC]r + \frac{1}{3}[BCD]r + \frac{1}{3}[ABD]r + \frac{1}{3}[ACD]r = V$ by the formula $V = \frac{1}{3}Ah$. Also note that $[ABC] = [BCD] = [ABD] = [ACD]$ because they all have the same side lengths. So, if we let $[ABC] = A$, then $\frac{4}{3}Ar = V$, so $r = \frac{3V}{4A}$. Now, all we need to do is to find V and A. We note that the area of the triangle $[ABC] = |\frac{\overrightarrow{AB}\times\overrightarrow{AC}}{2}|$. Also, if we let the normal vector from $\Delta ABC$ to be $\bold{n}$, the height $h$ from $D$ to $\Delta ABC$ is $\overrightarrow{AD} \cdot \hat{\bold{n}} = \frac{\overrightarrow{AD} \cdot \bold{n}}{\bold{|n|}}$. Therefore, the volume of the tetrahedron is equal to $\frac{1}{3}[ABC]h = \frac{1}{3}|\frac{\overrightarrow{AB}\times\overrightarrow{AC}}{2}|\frac{\overrightarrow{AD} \cdot \bold{n}}{\bold{|n|}} = \frac{1}{6}|\overrightarrow{AB}\times\overrightarrow{AC}|\frac{\overrightarrow{AD} \cdot \bold{n}}{\bold{|n|}}$. Now, we notice that $\overrightarrow{AB}\times\overrightarrow{AC}$ is a normal vector of the $\Delta ABC$ because the cross product of two vectors is always perpendicular to both of the vectors, and both of the vectors lie in the plane of $\Delta ABC$. This means that we can let $\bold{n} = \overrightarrow{AB}\times\overrightarrow{AC}$. So, the volume of the tetrahedron is $V = \frac{1}{6}|\overrightarrow{AB}\times\overrightarrow{AC}|\frac{\overrightarrow{AD} \cdot (\bold{\overrightarrow {AB}\times\overrightarrow{AC}})}{\bold{|\overrightarrow{AB}\times\overrightarrow{AC}|}} = \frac{1}{6}\overrightarrow{AD} \cdot (\bold{\overrightarrow{AB}\times\overrightarrow{AC}})$. The expression $\overrightarrow{AD} \cdot (\bold{\overrightarrow{AB}\times\overrightarrow{AC}})$ is more commonly known as a box product and can be used to find the volume of any parallelepiped. We know that $\overrightarrow{AB} = (4,5,0)$, $\overrightarrow{AC} = (0,5,8)$, and $\overrightarrow{AD} = (4,0,8)$. Now, $\overrightarrow{AB} \times \overrightarrow{AC} = (8*5-5*0, 0*0-8*4, 4*5-5*0) = (40, -32, 20)$. So, $\overrightarrow{AD} \cdot (\bold{\overrightarrow{AB}\times\overrightarrow{AC}}) = (4,0,8) \cdot (40,-32,20) = 4*40 + 0*-32 + 8*20 = 320$. So, $V = \frac{1}{6}\overrightarrow{AD} \cdot (\bold{\overrightarrow{AB}\times\overrightarrow{AC}}) = \frac{320}{6} = \frac{160}{3}$. Now $[ABC]$ is just equal to $\frac{1}{2}|\overrightarrow{AB} \times \overrightarrow{AC}|$ as it is half the parallelogram. So, we have $A = \frac{1}{2}|\overrightarrow{AB} \times \overrightarrow{AC}| = \frac{1}{2}|(40,-32,20)| = \frac{1}{2}\sqrt{3024}$. So, $r = \frac{3V}{4A} = \frac{160}{2\sqrt{3024}}$. This expression can be simplified to $\frac{20\sqrt{21}}{63}$. So, our answer is $20 + 63 + 21 = \boxed{104}$.

- Rutvik Arora (plz sub to https://www.youtube.com/@themathguy6302)

Video Solution

https://youtu.be/tq6lraC5prQ?si=5q1NX80POeR949qs

~MathProblemSolvingSkills.com


Video Solution 1 by OmegaLearn.org

https://youtu.be/qtu0HTFCsqc

Video Solution 2

https://youtu.be/1ZtLfJ77Ycg

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

See also

2024 AIME I (Problems • Answer Key • Resources)
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Problem 13 		Followed by
Problem 15
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