Solution:
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 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, \(r=\tfrac h4=\tfrac{20\sqrt{21}}{63}\), and so the answer is \(20+21+63=\boxed{104}\).
Solution by Quantum-Phantom
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](https://artofproblemsolving.comhttps://artofproblemsolving.com/wiki/index.php/User:Spectraldragon8)
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
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)
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
\[
x^2 + y^2 + z^2 = 89 \hspace{1cm} (1)
\]
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)
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

Answer: 104

<｜begin▁of▁sentence｜><｜User｜>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 $m+n+p$.<｜Assistant｜><think>

Okay, so I have this problem about a tetrahedron ABCD with some specific edge lengths, and I need to find the distance from the incenter (the point equidistant to all faces) to each face. The answer needs to be in the form m√n/p and then find m + n + p. Hmm, tetrahedrons can be tricky, but let me see if I can work through this step by step.

First, let me recall that in a tetrahedron, the incenter is the point equidistant to all four faces, and that distance is called the inradius. The formula for the inradius r is given by the volume V divided by the surface area (sum of the areas of all four faces) divided by 3. Wait, actually, let me confirm that. For a tetrahedron, the formula is similar to the 2D case for a triangle's inradius. In 3D, the inradius r = 3V / A, where A is the total surface area. Yeah, that sounds right because Volume = (1/3)*surface area * r. So, rearranged, r = 3V / A. So I need to compute the volume and the total surface area of this tetrahedron.

The problem gives me the edge lengths in a symmetric way: AB = CD = √41, AC = BD = √80, and BC = AD = √89. This symmetry might mean that the tetrahedron is "isohedral" or at least has some symmetry, which could help in calculations. Maybe it's a kind of disphenoid? Let me recall that a disphenoid is a tetrahedron where each pair of opposite edges are equal. That seems to be the case here since AB=CD, AC=BD, BC=AD. So yes, this is a disphenoid. Disphenoids have some nice properties, like their opposite edges are equal, and they can be inscribed in a rectangular box in a certain way. Maybe that can help?

Alternatively, perhaps I can place the tetrahedron in a coordinate system to compute the volume and face areas. Let me try that approach. Let me see if I can assign coordinates to the vertices such that the edge lengths are satisfied. Since it's a disphenoid, maybe there's a symmetric coordinate system.

Let me think: in a disphenoid, sometimes you can place the vertices at (a, b, c), (-a, -b, c), (-a, b, -c), (a, -b, -c) or something like that. Let me see if that works here. Let me try to set coordinates such that the midpoints of the edges are at the origin or something. Alternatively, maybe set coordinates so that the centroid is at the origin. Hmm, perhaps another way.

Alternatively, since all opposite edges are equal, maybe the tetrahedron can be embedded in a box such that each edge corresponds to a face diagonal. Let me recall that in a rectangular box with edges of length p, q, r, the face diagonals would be sqrt(p² + q²), sqrt(p² + r²), sqrt(q² + r²). If we can set the edges of the tetrahedron as these face diagonals, then the edges of the tetrahedron would correspond to the face diagonals of the box. Since in a disphenoid, all opposite edges are equal, so maybe each pair of opposite edges corresponds to the same face diagonal. Let me see.

Suppose the edges AB and CD are both sqrt(41), AC and BD are sqrt(80), and AD and BC are sqrt(89). Let me see if these can be face diagonals of a box. Let me denote the edges of the box as x, y, z. Then, the face diagonals would be sqrt(x² + y²), sqrt(x² + z²), sqrt(y² + z²). Since each pair of opposite edges of the tetrahedron corresponds to the same face diagonal, so perhaps:

Let me suppose that AB and CD are both sqrt(x² + y²) = sqrt(41),

AC and BD are sqrt(x² + z²) = sqrt(80),

AD and BC are sqrt(y² + z²) = sqrt(89).

So then, we can set up equations:

x² + y² = 41,

x² + z² = 80,

y² + z² = 89.

Hmm, let me check if these equations can be solved. Let me subtract the first equation from the second: (x² + z²) - (x² + y²) = 80 - 41 ⇒ z² - y² = 39.

But from the third equation, y² + z² = 89. So we have:

z² - y² = 39,

z² + y² = 89.

Adding these two equations: 2 z² = 128 ⇒ z² = 64 ⇒ z = 8.

Then, substituting back, y² = 89 - z² = 89 - 64 = 25 ⇒ y = 5.

Then from the first equation, x² + 25 = 41 ⇒ x² = 16 ⇒ x = 4.

So, the edges of the box would be x = 4, y = 5, z = 8. That works! So this tetrahedron can be inscribed in a box with dimensions 4, 5, 8, such that each edge of the tetrahedron is a face diagonal of the box. That seems promising. So now, perhaps I can assign coordinates to the vertices based on this box.

Let me recall how the disphenoid is placed in the box. In such a case, the vertices can be placed at the vertices of the box such that each edge of the tetrahedron is a face diagonal. Let me think of the box with opposite corners at (0,0,0) and (4,5,8). Then, the four vertices of the tetrahedron can be chosen as four vertices of the box such that no two are adjacent in the box. Wait, actually, in a disphenoid inscribed in a box, the vertices are typically pairs of opposite edges. Let me recall the exact coordinates.

Alternatively, perhaps the coordinates can be set as follows: Let me consider the box with edges along the coordinate axes, with one vertex at (0,0,0). Then, the four vertices of the tetrahedron can be (x, y, 0), (-x, -y, 0), (0,0,z), and (something else). Hmm, maybe I need a better approach.

Alternatively, since the edges correspond to face diagonals, perhaps the coordinates can be set such that each vertex is a combination of the box's edges. Let me think of the four vertices as follows:

Let me suppose the four vertices of the tetrahedron are:

A = (x, y, 0),

B = (-x, -y, 0),

C = (0, 0, z),

D = (0, 0, -z).

Wait, but then edges AB would be the distance between (x,y,0) and (-x,-y,0), which is sqrt{(2x)^2 + (2y)^2}, which would be 2*sqrt(x² + y²). But in our case, AB is sqrt(41). Hmm, so that would require 2*sqrt(x² + y²) = sqrt(41), so sqrt(x² + y²) = sqrt(41)/2, but in our earlier equations, x² + y² = 41. So that approach might not be directly applicable.

Alternatively, maybe the coordinates are set such that each coordinate is a combination of the box edges. Let me think again. Since the edges of the tetrahedron are face diagonals of the box, perhaps each edge corresponds to a different face. Let me see.

Suppose the box has dimensions a, b, c. Then, the face diagonals are sqrt(a² + b²), sqrt(a² + c²), sqrt(b² + c²). In our case, the edges of the tetrahedron are sqrt(41), sqrt(80), sqrt(89). So, as before, setting:

sqrt(a² + b²) = sqrt(41),

sqrt(a² + c²) = sqrt(80),

sqrt(b² + c²) = sqrt(89).

Which gives the same equations as before, leading to a=4, b=5, c=8. So the box has dimensions 4,5,8.

Now, to place the tetrahedron in the box, the vertices can be chosen such that each edge of the tetrahedron is a face diagonal of the box. Let me recall that in such a disphenoid, the vertices can be placed at four vertices of the box such that each pair connected by an edge of the tetrahedron is a face diagonal of the box. Let me try to assign coordinates accordingly.

Let me denote the box with one corner at (0,0,0) and the opposite corner at (a,b,c) = (4,5,8). Then, the vertices of the tetrahedron can be:

A = (a, 0, 0),

B = (0, b, 0),

C = (0, 0, c),

D = (something else). Wait, maybe not. Alternatively, perhaps the four vertices are (a,0,0), (0,b,0), (0,0,c), and (-a,-b,-c)? Hmm, not sure. Alternatively, maybe the four vertices are (a,b,0), (0,0,c), (a,0,b), (0,a,c)? Hmm, this might get complicated. Maybe there's a standard way.

Alternatively, perhaps the vertices are (a, b, c), (-a, -b, c), (-a, b, -c), (a, -b, -c). Let me check the distances between these points. Let me compute the distance between (a,b,c) and (-a,-b,c). That would be sqrt{(2a)^2 + (2b)^2 + 0} = 2*sqrt(a² + b²). Similarly, the distance between (a,b,c) and (-a,b,-c) would be sqrt{(2a)^2 + 0 + (2c)^2} = 2*sqrt(a² + c²). And the distance between (a,b,c) and (a,-b,-c) would be sqrt{0 + (2b)^2 + (2c)^2} = 2*sqrt(b² + c²).

Ah, so in this case, the edges from the first vertex to the other three would be 2*sqrt(a² + b²), 2*sqrt(a² + c²), 2*sqrt(b² + c²). But in our problem, the edges are sqrt(41), sqrt(80), sqrt(89). So if we set 2*sqrt(a² + b²) = sqrt(41), then sqrt(a² + b²) = sqrt(41)/2, but in our earlier equations, we had a² + b² = 41. So that would not align. Therefore, perhaps this coordinate system is not the right one.

Alternatively, maybe the edges of the tetrahedron are exactly the face diagonals, so each edge is sqrt(a² + b²), sqrt(a² + c²), etc., but then each edge is a face diagonal, so each edge is between two vertices that are on the same face of the box but not adjacent. Wait, perhaps the edges of the tetrahedron are the face diagonals of the box, so each edge is a diagonal of a face, but then the tetrahedron is formed by connecting these diagonals. Hmm, maybe I need to think differently.

Alternatively, perhaps the tetrahedron can be considered as two triangles in perpendicular planes. Let me think. Since the edges are face diagonals of a box, maybe the tetrahedron is formed by connecting opposite edges of the box. Let me try to think of the coordinates again.

Suppose the box has dimensions 4,5,8. Let me pick four vertices of the box such that each pair connected by an edge of the tetrahedron is a face diagonal. Let me try:

Let me choose vertex A as (4,5,0). Then, vertex B could be (0,0,8), since the distance between A and B would be sqrt{(4)^2 + (5)^2 + (8)^2}? Wait, no, that would be a space diagonal. Hmm, not helpful.

Alternatively, let me pick A as (4,0,0), B as (0,5,0), C as (0,0,8), and D as (-4,-5,-8). Hmm, not sure. Let me compute AB distance: between (4,0,0) and (0,5,0) is sqrt(16 + 25) = sqrt(41). That's good, since AB is supposed to be sqrt(41). Then AC would be between (4,0,0) and (0,0,8): sqrt(16 + 64) = sqrt(80). Perfect, that's AC = sqrt(80). Then AD would be between (4,0,0) and (-4,-5,-8): sqrt{(8)^2 + (5)^2 + (8)^2} = sqrt(64 +25 +64) = sqrt(153). But AD is supposed to be sqrt(89). Hmm, that's not right. So that coordinate choice doesn't work.

Hmm, maybe another approach. Let me think of the coordinates such that the centroid of the tetrahedron is at the origin. Since it's a disphenoid, it might have its centroid at the intersection of its edges' midpoints. Let me see.

Alternatively, since we know the edge lengths, maybe I can compute the volume using the formula for a tetrahedron with given edge lengths. There's a formula called the Cayley-Menger determinant which can compute the volume given all the edge lengths. Maybe that's the way to go here, since setting coordinates might be error-prone.

The Cayley-Menger determinant for a tetrahedron with edge lengths AB = a, AC = b, AD = c, BC = d, BD = e, CD = f is given by:

Volume² = (1/288) * determinant of the matrix:

0 1 1 1 1

1 0 a² b² c²

1 a² 0 d² e²

1 b² d² 0 f²

1 c² e² f² 0

So if I can plug in the given edge lengths into this determinant, I can compute the volume. Let me note down the edge lengths again:

AB = sqrt(41) ⇒ AB² = 41,

AC = sqrt(80) ⇒ AC² = 80,

AD = sqrt(89) ⇒ AD² = 89,

BC = sqrt(89) ⇒ BC² = 89,

BD = sqrt(80) ⇒ BD² = 80,

CD = sqrt(41) ⇒ CD² = 41.

So plugging into the Cayley-Menger determinant matrix:

The matrix is 5x5:

Row 0: 0, 1, 1, 1, 1

Row 1: 1, 0, 41, 80, 89

Row 2: 1, 41, 0, 89, 80

Row 3: 1, 80, 89, 0, 41

Row 4: 1, 89, 80, 41, 0

Wait, let me confirm the positions. The first row and column are the "0" row and column. The entries are as follows:

The (i,j) entry for i,j ≥1 corresponds to the squared distance between points i-1 and j-1. So the first point (A) is between rows/columns 1 and 2, etc. Wait, maybe I need to be careful here.

Actually, the Cayley-Menger determinant is constructed as follows:

The first row and first column are all 1s except the (0,0) entry which is 0.

The entries in row i and column j (for i,j from 1 to 4) correspond to the squared distance between points i and j. Wait, actually, the exact indexing might depend on the source, but let me check the standard formula.

Alternatively, according to the formula, the determinant is:

| 0   1    1    1    1  |

| 1   0  AB² AC² AD² |

| 1  AB²  0  BC² BD² |

| 1  AC² BC²  0  CD² |

| 1  AD² BD² CD²  0  |

Yes, that seems right. So with the given edge lengths:

AB² = 41,

AC² = 80,

AD² = 89,

BC² = 89,

BD² = 80,

CD² = 41.

So plugging in:

Row 0: 0, 1, 1, 1, 1

Row 1: 1, 0, 41, 80, 89

Row 2: 1, 41, 0, 89, 80

Row 3: 1, 80, 89, 0, 41

Row 4: 1, 89, 80, 41, 0

Now, I need to compute the determinant of this 5x5 matrix. That might be a bit tedious, but let's try.

First, let me denote the matrix as M, so determinant(M). Let me write it out:

M = [
[0, 1, 1, 1, 1],
[1, 0, 41, 80, 89],
[1, 41, 0, 89, 80],
[1, 80, 89, 0, 41],
[1, 89, 80, 41, 0]
]

Calculating a 5x5 determinant is going to be time-consuming, but maybe there's some symmetry here we can exploit. Let me see if the matrix has some symmetric properties.

Looking at rows 1 to 4 (excluding the first row and column), the matrix seems to have a symmetric structure. For instance, row 2 and row 4 are similar but swapped? Let me check:

Row 2 (after the first element): 41, 0, 89, 80

Row 4 (after the first element): 89, 80, 41, 0

Hmm, not exactly symmetric, but there's a pattern. Similarly, row 3 is 80, 89, 0, 41 and row 2 is 41, 0, 89, 80. Maybe there's some permutation symmetry here. Alternatively, perhaps we can perform row or column operations to simplify the determinant.

Alternatively, since the problem is symmetric, maybe the volume can be calculated more easily using the box dimensions. Since we already found that the tetrahedron can be inscribed in a box with sides 4,5,8, perhaps the volume is 4/3 of the box's volume? Wait, no, that's not right. Wait, actually, in a disphenoid formed by a box, the volume can be calculated as (4/3) times the volume of the box? Wait, no, let me think again.

Wait, actually, when a tetrahedron is inscribed in a rectangular box such that its vertices are at four of the box's vertices, with each edge of the tetrahedron being a face diagonal, then the volume of the tetrahedron is (box volume)/3. Wait, let me confirm.

Suppose the box has dimensions a, b, c. Then the volume is abc. If the tetrahedron is formed by four vertices such that each edge is a face diagonal, then perhaps the volume is (a*b*c)/3. Let me see with an example. Let me take a cube of side length 1. Then a tetrahedron formed by four alternate vertices (a regular tetrahedron) would have volume 1/3? Wait, actually in a cube of edge length 1, the regular tetrahedron formed by four vertices (like (0,0,0), (1,1,0), (1,0,1), (0,1,1)) has volume 1/3. Yes, that's correct. So in general, for a box with sides a, b, c, the volume of such a tetrahedron would be (a*b*c)/3.

Wait, so in our case, since the box has dimensions 4,5,8, then the volume of the tetrahedron would be (4*5*8)/3 = 160/3. But let me confirm if this applies here.

Wait, but in the cube example, the tetrahedron is formed by four vertices where each edge of the tetrahedron is a face diagonal of the cube. So in our case, since the tetrahedron is a disphenoid with edges equal to the face diagonals of the box, then yes, the same formula should apply. Therefore, the volume V = (4*5*8)/3 = 160/3. That would be much easier than computing the Cayley-Menger determinant.

But let me verify this with the Cayley-Menger determinant to make sure. Let me see if the determinant gives the same result. Let's try to compute the determinant.

First, the formula says Volume squared is (1/288)*det(M). Let me compute det(M). Since this is a 5x5 determinant, perhaps expanding along the first row.

The determinant of M is:

0 * det(minor) - 1 * det(minor) + 1 * det(minor) - 1 * det(minor) + 1 * det(minor)

Wait, actually, the cofactor expansion along the first row would be:

0*M_{00} - 1*M_{01} + 1*M_{02} - 1*M_{03} + 1*M_{04}

But since the first element is 0, that term drops out. So:

det(M) = -1*M_{01} + 1*M_{02} -1*M_{03} +1*M_{04}

Where M_{0j} are the minors corresponding to the first row and column j.

Each minor is the determinant of the 4x4 matrix obtained by removing the first row and the j-th column.

Let me compute each minor:

First, M_{01}: remove first row and first column. The minor matrix is:

Row1: 1, 0, 41, 80, 89 → but removing first column, so:

Rows 1-4 (original rows 2-5) with columns 2-5 (original columns 1-4 except first column). Wait, maybe this is getting confusing. Let me reindex.

Wait, the original matrix is 5x5. Let me label rows 0 to 4 and columns 0 to 4.

The minor M_{01} is the determinant of the submatrix obtained by removing row 0 and column 1. So the remaining rows are 1-4 and columns 0,2,3,4.

So the minor matrix for M_{01} is:

Row1 (original row1): [1, 41, 80, 89]

Row2 (original row2): [1, 0, 89, 80]

Row3 (original row3): [1, 89, 0, 41]

Row4 (original row4): [1, 80, 41, 0]

Wait, let me confirm:

Original columns after removing column1 (index1):

Columns 0,2,3,4 (original columns 0,2,3,4). So for each row from 1 to 4:

Row1 (originally row1): entries [1 (col0), 41 (col2), 80 (col3), 89 (col4)]

Wait, original row1 was [1,0,41,80,89]. Removing column1 (the second element, which is 0), so the remaining entries are [1,41,80,89].

Similarly, original row2 was [1,41,0,89,80]. Removing column1 (41), so [1,0,89,80].

Original row3: [1,80,89,0,41]. Removing column1 (80), so [1,89,0,41].

Original row4: [1,89,80,41,0]. Removing column1 (89), so [1,80,41,0].

Therefore, the minor M_{01} matrix is:

[
[1, 41, 80, 89],
[1, 0, 89, 80],
[1, 89, 0, 41],
[1, 80, 41, 0]
]

Similarly, the other minors would be complicated. This is going to take a while. Maybe there's a pattern here. Let me see if this determinant can be simplified.

Alternatively, perhaps I can compute this determinant using some symmetry or row operations.

Let me denote the minor matrix as N:

N = [
[1, 41, 80, 89],
[1, 0, 89, 80],
[1, 89, 0, 41],
[1, 80, 41, 0]
]

Let me subtract the first row from the other rows to create zeros in the first column.

Row2 = Row2 - Row1: [1-1, 0-41, 89-80, 80-89] = [0, -41, 9, -9]

Row3 = Row3 - Row1: [1-1, 89-41, 0-80, 41-89] = [0, 48, -80, -48]

Row4 = Row4 - Row1: [1-1, 80-41, 41-80, 0-89] = [0, 39, -39, -89]

So the transformed matrix is:

Row1: [1, 41, 80, 89]

Row2: [0, -41, 9, -9]

Row3: [0, 48, -80, -48]

Row4: [0, 39, -39, -89]

Now, the determinant of N is the same as this transformed matrix (since we did row operations). To compute the determinant, we can expand along the first column now, since there are three zeros.

The determinant is 1 * det(minor11) - 0 + 0 - 0 = det of the 3x3 matrix:

[
[-41, 9, -9],
[48, -80, -48],
[39, -39, -89]
]

Let me compute this 3x3 determinant.

Let me denote the matrix as:

| -41   9    -9 |

| 48  -80   -48 |

| 39  -39   -89 |

The determinant is:

-41 * [(-80)(-89) - (-48)(-39)] - 9 * [48*(-89) - (-48)*39] + (-9) * [48*(-39) - (-80)*39]

Let me compute each term step by step.

First term: -41 * [ (80*89) - (48*39) ]

Compute 80*89: 80*90 = 7200, minus 80 = 7200 - 80 = 7120

Wait, 80*89 = 80*(90 -1) = 7200 -80 = 7120

48*39: 48*(40 -1) = 1920 -48 = 1872

So first bracket: 7120 - 1872 = 5248

First term: -41 * 5248 = let's compute 41*5000=205000, 41*248= 41*(200+48)= 8200 + 1968= 10168, so total 205000 + 10168 = 215,168. But since it's -41 * 5248, it's -215,168.

Second term: -9 * [48*(-89) - (-48)*39 ]

Compute inside the brackets:

48*(-89) = -4272,

-(-48)*39 = +48*39 = 1872,

So total: -4272 + 1872 = -2400

Multiply by -9: -9*(-2400) = +21,600

Third term: -9 * [48*(-39) - (-80)*39 ]

Compute inside the brackets:

48*(-39) = -1872,

-(-80)*39 = +80*39 = 3120,

Total: -1872 + 3120 = 1248

Multiply by -9: -9*(1248) = -11,232

Now, adding all three terms:

First term: -215,168

Second term: +21,600 ⇒ total so far: -215,168 +21,600 = -193,568

Third term: -11,232 ⇒ total: -193,568 -11,232 = -204,800

Therefore, determinant of the 3x3 matrix is -204,800.

Therefore, determinant of N (the minor M_{01}) is -204,800.

Wait, but let me check my calculations again because this seems like a big number and I might have made an arithmetic error.

First term:

First bracket: (80*89) - (48*39) = 7120 - 1872 = 7120 - 1800 = 5320, then minus 72 more: 5320 -72 = 5248. That's correct.

Then -41 * 5248: Let me compute 5248 * 40 = 209,920 and 5248 *1 = 5,248, so total 209,920 +5,248 = 215,168. So with the negative, it's -215,168. Correct.

Second term:

Inside the brackets: 48*(-89) = -4272, and (-48)*39 is -1872, but the expression is 48*(-89) - (-48)*39 = -4272 + 1872 = -2400. Then multiplied by -9 gives +21,600. Correct.

Third term:

Inside the brackets: 48*(-39) = -1872, and (-80)*39 is -3120, but the expression is 48*(-39) - (-80)*39 = -1872 + 3120 = 1248. Then multiplied by -9 gives -11,232. Correct.

Total: -215,168 +21,600 = -193,568; then -193,568 -11,232 = -204,800. Yes.

So determinant of N is -204,800. Therefore, M_{01} = determinant(N) = -204,800.

Now, moving on to the next minor, M_{02}, which is the minor for the first row and second column. So removing column2 and row0.

The minor M_{02} is the determinant of the matrix obtained by removing column2 and row0. So columns 0,1,3,4 and rows1-4.

Original columns after removing column2 (index2):

Columns 0,1,3,4.

So for each row1-4:

Row1 (original row1): entries [1,0,80,89] (since column2 was the third element (index2), which was 41, so removed)

Wait, let me recheck:

Original row1 was [1,0,41,80,89]. Removing column2 (the third element, 41), so the remaining entries are [1,0,80,89].

Similarly, original row2: [1,41,0,89,80]. Removing column2 (0), so [1,41,89,80].

Original row3: [1,80,89,0,41]. Removing column2 (89), so [1,80,0,41].

Original row4: [1,89,80,41,0]. Removing column2 (80), so [1,89,41,0].

Therefore, the minor matrix M_{02} is:

[
[1,0,80,89],
[1,41,89,80],
[1,80,0,41],
[1,89,41,0]
]

Hmm, this looks similar to the previous minor but shifted. Let me see if I can compute its determinant similarly.

Let me call this matrix P.

P = [
[1,0,80,89],
[1,41,89,80],
[1,80,0,41],
[1,89,41,0]
]

Again, let's perform row operations to simplify. Subtract row1 from the other rows.

Row2 = Row2 - Row1: [1-1,41-0,89-80,80-89] = [0,41,9,-9]

Row3 = Row3 - Row1: [1-1,80-0,0-80,41-89] = [0,80,-80,-48]

Row4 = Row4 - Row1: [1-1,89-0,41-80,0-89] = [0,89,-39,-89]

So the transformed matrix is:

Row1: [1,0,80,89]

Row2: [0,41,9,-9]

Row3: [0,80,-80,-48]

Row4: [0,89,-39,-89]

Now, compute determinant by expanding along the first column. The determinant is 1 * det of the 3x3 matrix:

[
[41, 9, -9],
[80, -80, -48],
[89, -39, -89]
]

Let me compute this determinant.

Let me denote:

| 41   9    -9 |

|80  -80   -48 |

|89  -39   -89 |

The determinant is:

41*[(-80)(-89) - (-48)(-39)] - 9*[80*(-89) - (-48)*89] + (-9)*[80*(-39) - (-80)*89]

Compute term by term:

First term: 41*[ (80*89) - (48*39) ]

Wait, same as before. 80*89 = 7120, 48*39 = 1872, so 7120 - 1872 = 5248. So first term: 41 * 5248 = 215,168.

Second term: -9*[80*(-89) - (-48)*89]

Compute inside the brackets:

80*(-89) = -7120,

-(-48)*89 = +48*89 = 4272,

Total: -7120 + 4272 = -2848

Multiply by -9: -9*(-2848) = +25,632

Third term: -9*[80*(-39) - (-80)*89]

Compute inside:

80*(-39) = -3120,

-(-80)*89 = +80*89 = 7120,

Total: -3120 +7120 = 4000

Multiply by -9: -9*(4000) = -36,000

Now, adding all terms:

First term: 215,168

Second term: +25,632 ⇒ total 240,800

Third term: -36,000 ⇒ total 240,800 -36,000 = 204,800

Therefore, determinant of the 3x3 matrix is 204,800. Hence determinant of P is 204,800. So M_{02} = 204,800.

Proceeding to the next minor, M_{03}, which is the minor for column3. So removing column3 and row0.

The minor M_{03} corresponds to removing column3 (index3), so columns 0,1,2,4 and rows1-4.

Original rows:

Row1 (original row1): [1,0,41,80,89] → removing column3 (80), so [1,0,41,89]

Row2 (original row2): [1,41,0,89,80] → removing column3 (89), so [1,41,0,80]

Row3 (original row3): [1,80,89,0,41] → removing column3 (0), so [1,80,89,41]

Row4 (original row4): [1,89,80,41,0] → removing column3 (41), so [1,89,80,0]

Thus, the minor matrix Q is:

[
[1,0,41,89],
[1,41,0,80],
[1,80,89,41],
[1,89,80,0]
]

Hmm, this is getting complex again. Let me see if there's a pattern here. Alternatively, proceed with row operations.

Let me denote Q as:

Row1: [1,0,41,89]

Row2: [1,41,0,80]

Row3: [1,80,89,41]

Row4: [1,89,80,0]

Again, subtract row1 from the other rows:

Row2 = Row2 - Row1: [0,41,-41,-9]

Row3 = Row3 - Row1: [0,80,48,-48]

Row4 = Row4 - Row1: [0,89,39,-89]

So the transformed matrix:

Row1: [1,0,41,89]

Row2: [0,41,-41,-9]

Row3: [0,80,48,-48]

Row4: [0,89,39,-89]

Compute determinant by expanding along first column:

1 * det of the 3x3 matrix:

[
[41, -41, -9],
[80, 48, -48],
[89, 39, -89]
]

Compute this determinant:

|41  -41   -9|

|80   48  -48|

|89   39  -89|

Determinant:

41*(48*(-89) - (-48)*39) - (-41)*(80*(-89) - (-48)*89) + (-9)*(80*39 -48*89)

Let me compute term by term.

First term: 41*[48*(-89) - (-48)*39]

Compute inside:

48*(-89) = -4272,

-(-48)*39 = +1872,

Total: -4272 + 1872 = -2400,

So first term: 41*(-2400) = -98,400

Second term: -(-41)*[80*(-89) - (-48)*89]

Compute inside:

80*(-89) = -7120,

-(-48)*89 = +4272,

Total: -7120 +4272 = -2848,

Multiply by -(-41): 41*(-2848) = -116,768? Wait, wait:

Wait, the second term is -(-41) multiplied by the bracket. So:

Second term: +41*(-2848) = -116,768

Wait, let me re-express:

The second term is:

- (-41) * [result] = +41 * [result]

But [result] was -2848, so 41*(-2848) = -116,768. So the second term is -116,768.

Third term: -9*[80*39 -48*89]

Compute inside:

80*39 = 3120,

48*89 = 4272,

So 3120 -4272 = -1152,

Multiply by -9: -9*(-1152) = +10,368

Now, total determinant:

First term: -98,400

Second term: -116,768 ⇒ total so far: -98,400 -116,768 = -215,168

Third term: +10,368 ⇒ total: -215,168 +10,368 = -204,800

Therefore, determinant of the 3x3 matrix is -204,800, so determinant of Q is -204,800. Thus, M_{03} = -204,800.

Now, moving on to the last minor, M_{04}, which is the minor for column4. Removing column4 (index4) and row0.

The minor matrix R is formed by columns 0,1,2,3 and rows1-4.

Original rows:

Row1 (original row1): [1,0,41,80,89] → remove column4 (89) → [1,0,41,80]

Row2 (original row2): [1,41,0,89,80] → remove column4 (80) → [1,41,0,89]

Row3 (original row3): [1,80,89,0,41] → remove column4 (41) → [1,80,89,0]

Row4 (original row4): [1,89,80,41,0] → remove column4 (0) → [1,89,80,41]

Thus, the minor matrix R is:

[
[1,0,41,80],
[1,41,0,89],
[1,80,89,0],
[1,89,80,41]
]

Again, let's perform row operations. Subtract row1 from the others:

Row2 = Row2 - Row1: [0,41,-41,9]

Row3 = Row3 - Row1: [0,80,48,-80]

Row4 = Row4 - Row1: [0,89,39, -39]

So the transformed matrix:

Row1: [1,0,41,80]

Row2: [0,41,-41,9]

Row3: [0,80,48,-80]

Row4: [0,89,39,-39]

Compute determinant by expanding along first column:

1 * det of the 3x3 matrix:

[
[41, -41, 9],
[80, 48, -80],
[89, 39, -39]
]

Compute determinant:

|41  -41    9|

|80   48  -80|

|89   39  -39|

Determinant:

41*(48*(-39) - (-80)*39) - (-41)*(80*(-39) - (-80)*89) +9*(80*39 -48*89)

Let me compute term by term.

First term: 41*[48*(-39) - (-80)*39]

Compute inside:

48*(-39) = -1872,

-(-80)*39 = +3120,

Total: -1872 +3120 = 1248,

First term: 41*1248 = let's compute 40*1248=49,920 and 1*1248=1,248 ⇒ total 51,168.

Second term: -(-41)*[80*(-39) - (-80)*89]

Compute inside:

80*(-39) = -3120,

-(-80)*89 = +7120,

Total: -3120 +7120 = 4000,

Multiply by -(-41): 41*4000 = 164,000.

Third term: 9*[80*39 -48*89]

Compute inside:

80*39 = 3120,

48*89 = 4272,

3120 -4272 = -1152,

Multiply by 9: 9*(-1152) = -10,368.

Now, adding all terms:

First term: 51,168

Second term: +164,000 ⇒ total 215,168

Third term: -10,368 ⇒ total 215,168 -10,368 = 204,800.

Therefore, determinant of the 3x3 matrix is 204,800. Hence determinant of R is 204,800. So M_{04} = 204,800.

Now, putting it all back into the determinant formula:

det(M) = -1*M_{01} +1*M_{02} -1*M_{03} +1*M_{04}

Plugging in the values:

= -1*(-204,800) + 1*(204,800) -1*(-204,800) +1*(204,800)

Compute term by term:

First term: +204,800

Second term: +204,800

Third term: +204,800

Fourth term: +204,800

Total: 204,800 *4 = 819,200.

Therefore, determinant(M) = 819,200.

Then, Volume squared is (1/288)*det(M) = (1/288)*819,200.

Compute that:

819,200 ÷ 288.

Let me divide numerator and denominator by 16: 819,200 ÷16 = 51,200; 288 ÷16 = 18.

So 51,200 /18 = 28,444.444... Wait, let me do it step by step.

Alternatively, 288 = 16 * 18.

819,200 ÷ 288 = (819,200 ÷ 16) ÷ 18 = 51,200 ÷ 18.

51,200 ÷ 18 = (51,200 ÷ 2) ÷9 = 25,600 ÷9 ≈ 2844.444...

But let's keep it as a fraction: 51,200 /18 = 25,600/9.

Therefore, Volume squared = 25,600 /9 ⇒ Volume = sqrt(25,600 /9) = (sqrt(25,600))/3 = 160/3.

Which matches the earlier result from the box method! So Volume is indeed 160/3. Great, so that checks out.

Now, moving on to compute the total surface area. Since it's a tetrahedron, it has four triangular faces. Each pair of opposite edges are equal, so maybe the faces are congruent in pairs?

Let me see. The edges are AB=CD=√41, AC=BD=√80, AD=BC=√89.

Therefore, each face is a triangle with sides:

Face ABC: edges AB=√41, BC=√89, AC=√80.

Face ABD: edges AB=√41, BD=√80, AD=√89.

Face ACD: edges AC=√80, CD=√41, AD=√89.

Face BCD: edges BC=√89, CD=√41, BD=√80.

So each face is a triangle with sides √41, √80, √89. Wait, all four faces are congruent triangles? Because each face has the same edge lengths: 41,80,89. Let me confirm:

Yes, each face is a triangle with sides sqrt41, sqrt80, sqrt89. Therefore, all four faces are congruent triangles. Therefore, the surface area is 4 times the area of one such triangle.

Therefore, if I can compute the area of one face, multiply by 4, and that's the total surface area.

Let me compute the area of triangle with sides a=√41, b=√80, c=√89.

Using Heron's formula. First compute the semi-perimeter:

s = (a + b + c)/2 = (√41 + √80 + √89)/2. Hmm, but this might be messy. Alternatively, perhaps use the coordinates to compute the area.

Alternatively, since we know the edges correspond to the face diagonals of the box with sides 4,5,8, maybe the area can be related to the box's faces.

Wait, each face of the tetrahedron is a triangle whose edges are three face diagonals of the box. Wait, but actually, each face of the tetrahedron is a triangle lying on a plane, but not necessarily on a face of the box.

Alternatively, since all the faces are congruent triangles with sides sqrt41, sqrt80, sqrt89, perhaps we can compute the area using vectors.

Alternatively, let me note that in the box with sides 4,5,8, the edges of the tetrahedron are face diagonals. Let me consider one face of the tetrahedron, say triangle ABC. The edges AB, BC, and AC are face diagonals of the box.

Wait, let me think of the coordinates again. Let me try to assign coordinates based on the box.

Suppose the box has dimensions 4,5,8. Let me assign coordinates to the tetrahedron's vertices such that each edge is a face diagonal. Let me try this approach again, more carefully.

Let me consider the box with opposite corners at (0,0,0) and (4,5,8). Let me choose the four vertices of the tetrahedron as follows:

Let me take two opposite edges of the box and connect their endpoints. Wait, perhaps the vertices are:

A = (4,5,0),

B = (0,0,8),

C = (4,0,5),

D = (0,5,8).

Wait, not sure. Alternatively, perhaps the standard coordinates for a disphenoid in a box are:

Let me recall that in a disphenoid, the four vertices can be placed at (a,b,0), (-a,-b,0), (0,0,c), and (0,0,-c). Wait, but then edges between (a,b,0) and (0,0,c) would have length sqrt(a² + b² + c²), which is a space diagonal, not a face diagonal. Hmm, not helpful.

Alternatively, perhaps the coordinates are (a,0,0), (0,b,0), (0,0,c), and (-a,-b,-c). Let me check the distances:

Distance between (a,0,0) and (0,b,0) is sqrt(a² + b²),

Between (a,0,0) and (0,0,c) is sqrt(a² + c²),

Between (a,0,0) and (-a,-b,-c) is sqrt{(2a)^2 + b² + c²},

Hmm, that might not give the desired edge lengths.

Alternatively, perhaps the coordinates are (a,b,c), (-a,-b,c), (-a,b,-c), (a,-b,-c). Let me check the distance between (a,b,c) and (-a,-b,c):

sqrt{(2a)^2 + (2b)^2 +0} = 2*sqrt(a² + b²),

Similarly between (a,b,c) and (-a,b,-c): sqrt{(2a)^2 +0 + (2c)^2} = 2*sqrt(a² + c²),

And between (a,b,c) and (a,-b,-c): sqrt{0 + (2b)^2 + (2c)^2} = 2*sqrt(b² + c²).

So in this case, the edges from the first vertex to the other three are 2*sqrt(a² + b²), 2*sqrt(a² + c²), 2*sqrt(b² + c²).

But in our problem, the edges are sqrt41, sqrt80, sqrt89. So if we set:

2*sqrt(a² + b²) = sqrt41 ⇒ sqrt(a² + b²) = sqrt(41)/2 ⇒ a² + b² = 41/4,

Similarly,

2*sqrt(a² + c²) = sqrt80 ⇒ sqrt(a² + c²) = sqrt(80)/2 ⇒ a² + c² = 20,

2*sqrt(b² + c²) = sqrt89 ⇒ sqrt(b² + c²) = sqrt(89)/2 ⇒ b² + c² = 89/4.

But earlier, we found that in the box approach, a=4, b=5, c=8, but those led to edges of 2*sqrt(a² + b²) etc., which would be larger than the given edges. So perhaps this coordinate system is different.

Alternatively, maybe the edges of the tetrahedron are the face diagonals, so each edge is sqrt(a² + b²), etc., without the factor of 2. So perhaps the coordinates are different.

Wait, perhaps the coordinates are (a, b, 0), ( -a, -b, 0), (0,0,c), and (0,0,-c). Let me check the distance between (a,b,0) and (-a,-b,0): sqrt{(2a)^2 + (2b)^2} = 2*sqrt(a² + b²),

Between (a,b,0) and (0,0,c): sqrt(a² + b² + c²),

Hmm, not face diagonals. Maybe this isn't the right approach.

Alternatively, perhaps I should just compute the area using Heron's formula for one face.

Let me try that. Let's compute the area of a triangle with sides sqrt41, sqrt80, sqrt89.

Let me denote the sides as a = sqrt41, b = sqrt80, c = sqrt89.

Compute semi-perimeter s = (a + b + c)/2.

But this might involve square roots. Alternatively, compute the area squared using Heron's formula:

Area² = s(s - a)(s - b)(s - c).

Alternatively, perhaps use the formula for area in terms of coordinates. Wait, but since I don't have coordinates, maybe use the formula for area based on edge lengths:

Alternatively, use the formula:

Area = (1/4) * sqrt( (a + b + c)(-a + b + c)(a - b + c)(a + b - c) )

But that's the same as Heron's formula.

Alternatively, since the triangle has sides sqrt41, sqrt80, sqrt89, let me compute the squares:

a² =41, b²=80, c²=89.

Let me see if this triangle is right-angled. Let me check if 41 +80 = 121, which is 11², but 89 is not 121. 41 +80=121? 41+80=121? 40+80=120, plus 1 is 121. Wait, yes! 41 +80 = 121 = 11², but c² is 89, which is not 121. Wait, so that's not a right triangle. Wait, but 41 + 80 = 121, which is greater than 89, so the triangle is acute?

Wait, let me see:

Wait, in a triangle with sides a,b,c, if a² + b² > c², it's acute, etc.

Here, the largest side is sqrt89 ≈9.43, while sqrt41≈6.4, sqrt80≈8.94. Wait, actually sqrt89 is the largest. Let me check:

a² + b² =41 +80=121, which is greater than c²=89, so angle opposite c is acute.

Hmm, okay. So not a right triangle. So Heron's formula it is.

Let me compute s = (sqrt41 + sqrt80 + sqrt89)/2. This is going to be messy, but maybe there's a better way.

Alternatively, perhaps use vectors. Since the edges correspond to face diagonals of the box with sides 4,5,8, maybe the area can be calculated as half the magnitude of the cross product of two edges.

Wait, let me try this approach. Let me suppose that the tetrahedron is embedded in the box with dimensions 4,5,8. Let me assign coordinates to the vertices such that each edge is a face diagonal.

Let me consider the following coordinates for the tetrahedron vertices:

Let me set vertex A at (4,5,0). Then, since AB is a face diagonal of the box's front face (assuming the box is along the x,y,z axes), then moving along the x and y directions. Wait, but the face diagonals would be on different faces.

Alternatively, let me think of the edges as face diagonals on different faces. Let me try to define the coordinates such that:

Let me suppose the four vertices are:

A = (4, 0, 0),

B = (0, 5, 0),

C = (0, 0, 8),

D = (-4, -5, -8).

Wait, but then the distance AB would be sqrt( (4)^2 + (5)^2 ) = sqrt(41), which is good.

AC would be sqrt(4² + 8²) = sqrt(16 +64)=sqrt(80), which is correct.

AD would be sqrt( (8)^2 + (5)^2 + (8)^2 )? Wait, coordinates of D are (-4,-5,-8). So distance from A (4,0,0) to D (-4,-5,-8):

sqrt( (8)^2 + (5)^2 + (8)^2 ) = sqrt(64 +25 +64)=sqrt(153). But AD is supposed to be sqrt(89). Not good.

Hmm, perhaps another coordinate system. Let me try:

Let me set the box with dimensions 4,5,8. Let me place the tetrahedron's vertices at four vertices of the box such that each edge of the tetrahedron is a face diagonal.

Let me choose the vertices as follows:

A = (4,5,0),

B = (0,0,8),

C = (4,0,5),

D = (0,5,8).

Let me compute the distances:

AB: between (4,5,0) and (0,0,8):

sqrt( (4)^2 +5^2 +8^2 ) = sqrt(16+25+64)=sqrt(105). Not sqrt41. Not good.

Hmm, perhaps this is not the right approach. Let me think differently.

Since the edges are face diagonals of the box with sides 4,5,8, then each edge of the tetrahedron is a face diagonal. So for example, AB is a face diagonal of the face with sides 4 and 5, so length sqrt(41). AC is a face diagonal of the face with sides 4 and 8, so sqrt(80). AD is a face diagonal of the face with sides 5 and 8, so sqrt(89). So that works.

Therefore, the edges AB, AC, AD are face diagonals of three different faces of the box. Therefore, the vertices A, B, C, D can be placed at the corners of the box such that each edge from A is a face diagonal.

Let me set vertex A at (0,0,0). Then:

- Point B is at the opposite corner of the face with sides 4 and 5, so (4,5,0).

- Point C is at the opposite corner of the face with sides 4 and 8, so (4,0,8).

- Point D is at the opposite corner of the face with sides 5 and 8, so (0,5,8).

Wait, let me check the distances:

AB: from (0,0,0) to (4,5,0): sqrt(16+25) = sqrt(41). Good.

AC: from (0,0,0) to (4,0,8): sqrt(16 + 64) = sqrt(80). Good.

AD: from (0,0,0) to (0,5,8): sqrt(25 +64)=sqrt(89). Good.

Now, what about the other edges?

BC: between (4,5,0) and (4,0,8): sqrt( (0)^2 + (5)^2 + ( -8)^2 ) = sqrt(25 +64)=sqrt(89). Good, since BC should be sqrt89.

BD: between (4,5,0) and (0,5,8): sqrt( (4)^2 +0 + ( -8)^2 )=sqrt(16 +64)=sqrt(80). Good, BD is sqrt80.

CD: between (4,0,8) and (0,5,8): sqrt( (4)^2 + ( -5)^2 +0 )=sqrt(16+25)=sqrt(41). Perfect, CD is sqrt41.

Perfect! So this coordinate system works. So the four vertices are:

A = (0,0,0),

B = (4,5,0),

C = (4,0,8),

D = (0,5,8).

Great, now that I have coordinates, I can compute the areas of the faces easily using vectors or the cross product.

First, let's compute the area of face ABC. The points are A(0,0,0), B(4,5,0), C(4,0,8).

Vectors AB = (4,5,0), AC = (4,0,8).

The cross product AB × AC is determinant:

i  j  k

4  5  0

4  0  8

Calculates to:

i*(5*8 -0*0) - j*(4*8 -0*4) + k*(4*0 -5*4)

= i*(40) - j*(32) + k*(-20)

So the cross product vector is (40, -32, -20).

The magnitude of this cross product is sqrt(40² + (-32)^2 + (-20)^2) = sqrt(1600 + 1024 + 400) = sqrt(3024).

Simplify sqrt(3024):

3024 ÷ 16 = 189, so sqrt(16*189) = 4*sqrt(189). 189 = 9*21, so sqrt(189)=3*sqrt(21). Thus, magnitude is 4*3*sqrt21 = 12√21.

Therefore, the area of triangle ABC is half that: 6√21.

Since all four faces are congruent, each has area 6√21. Therefore, total surface area A = 4 *6√21 =24√21.

Wait, let me confirm with another face. Let's compute face ABD: points A(0,0,0), B(4,5,0), D(0,5,8).

Vectors AB = (4,5,0), AD = (0,5,8).

Cross product AB × AD:

i  j  k

4  5  0

0  5  8

= i*(5*8 -0*5) - j*(4*8 -0*0) + k*(4*5 -5*0)

= i*(40) - j*(32) + k*(20)

So cross product is (40, -32, 20). The magnitude is sqrt(40² + (-32)^2 +20²) = same as before sqrt(1600 +1024 +400)=sqrt(3024)= same as before, so area is also 6√21. Good.

Similarly, other faces will have the same area. So total surface area is indeed 24√21.

Therefore, the inradius r = 3V / A.

We have Volume V = 160/3,

So 3V = 160,

A =24√21,

Thus, r = 160 / (24√21) = (160)/(24√21) = (20)/(3√21).

Rationalizing the denominator:

(20)/(3√21) * (√21/√21) = (20√21)/(3*21) = (20√21)/63.

Simplify numerator and denominator: 20 and 63 are coprime (since 63=7*3^2, 20=2^2*5, no common factors). Also, 21 factors into 3*7, which is square-free. So the expression is (20√21)/63, so m=20, n=21, p=63.

Wait, but 20 and 63 have a common factor? 20 is 2^2*5, 63 is 7*3^2. No common factors, so they are coprime. And 21 is square-free. So the answer is 20 +21 +63= 104.

Wait, but let me check my steps again to be sure.

First, coordinates were set correctly, and the cross product calculations for the area of the face gave 6√21 per face, so total area 24√21. Volume was 160/3. Then inradius is 3*(160/3)/(24√21) = 160/(24√21) = 20/(3√21) = 20√21/(3*21) = 20√21/63. So yes, that's correct.

Therefore, m=20, n=21, p=63. Sum is 20+21+63=104.

Wait, but hold on, the problem states "the distance can be written in the form m√n/p where m and p are relatively prime, and n is not divisible by the square of any prime." Here, 20 and 63 are coprime, and 21 factors into 3*7, which is square-free. So that's okay. So the answer is 104.

But let me check once more the cross product calculation for area.

For face ABC: vectors AB=(4,5,0), AC=(4,0,8). Cross product:

i component: (5)(8) - (0)(0) =40,

j component: - [ (4)(8) - (0)(4) ] = -32,

k component: (4)(0) - (5)(4) = -20.

So cross product (40, -32, -20). The magnitude squared is 40² + (-32)^2 + (-20)^2 = 1600 + 1024 + 400 = 3024. sqrt(3024). Let me factor 3024:

Divide by 16: 3024 /16 = 189. So sqrt(16*189)=4*sqrt(189). Then 189=9*21, so sqrt(189)=3*sqrt(21). So total sqrt(3024)=4*3*sqrt21=12√21. So area is 6√21. Correct.

Volume via box: the tetrahedron is formed by four vertices of the box, such that it's a disphenoid. The volume formula for such a tetrahedron is indeed (a*b*c)/3 where a,b,c are the box dimensions. Here, 4*5*8 /3 = 160/3. Correct.

Thus, inradius is 3V/A = (3*(160/3))/(24√21) = 160/(24√21) simplifies to 20/(3√21) then rationalized to 20√21/63. So yes, the answer is 20+21+63=104.

Wait, but I recall that sometimes in these problems, the answer might require simplifying the fraction further or checking if there's a miscalculation. Let me check once again:

Wait, 3V/A: 3*(160/3) is 160. Divided by A=24√21 gives 160/(24√21). 160 divided by 24 is 20/3. So 20/(3√21). Multiply numerator and denominator by √21: 20√21/(3*21) = 20√21/63. Since 20 and 63 have no common factors, that's reduced. So yes, 20√21/63. So m=20, n=21, p=63. Sum is 104.

Hmm, but I have a feeling that maybe I made a mistake in the surface area. Let me check another face, say BCD.

Points B(4,5,0), C(4,0,8), D(0,5,8).

Vectors BC = (0, -5,8), BD = (-4,0,8).

Cross product BC × BD:

i   j   k

0  -5   8

-4  0   8

Calculates to:

i*(-5*8 -8*0) - j*(0*8 -8*(-4)) + k*(0*0 - (-5)*(-4))

= i*(-40) - j*(0 +32) + k*(0 -20)

= (-40, -32, -20)

The magnitude is same as before sqrt( (-40)^2 + (-32)^2 + (-20)^2 ) = sqrt(1600 +1024 +400)=sqrt(3024)=12√21. Area is 6√21. So yes, same as others. So total area is indeed 24√21.

Therefore, I think the answer is 104. But let me check if the problem says "distance from I to each of the faces are all equal". So inradius is correctly calculated as 20√21 /63. So m+n+p=20+21+63=104.

Wait, but I just recalled that sometimes in some tetrahedrons, the inradius formula is different? Wait no, the formula r = 3V/A is standard for tetrahedrons, analogous to 2D's A = r*s.

Yes, in 3D, the formula for the inradius is indeed r = 3V / A, where A is the total surface area. So that's correct.

Therefore, I think the answer is 104. However, since I might have made an error in the cross product or calculations, but all steps seem consistent. Let me see if 20√21 divided by 63 can be simplified further? 20 and 63 have no common factors, and 21 is square-free. So yes, that's the simplest form.

Therefore, the final answer is 20 + 21 + 63 = 104.

**Final Answer**
\boxed{104}
</think>

Given a tetrahedron \(ABCD\) with edge lengths \(AB = CD = \sqrt{41}\), \(AC = BD = \sqrt{80}\), and \(BC = AD = \sqrt{89}\), we need to find the distance from the incenter to each face, expressed in the form \(\frac{m \sqrt{n}}{p}\), and compute \(m + n + p\).

First, we recognize the tetrahedron is a disphenoid, which can be inscribed in a rectangular box with dimensions \(4 \times 5 \times 8\). The volume \(V\) of the tetrahedron is \(\frac{1}{3}\) of the box's volume:
\[
V = \frac{1}{3} \times 4 \times 5 \times 8 = \frac{160}{3}
\]

Next, we calculate the surface area. Each face of the tetrahedron is a triangle with sides \(\sqrt{41}\), \(\sqrt{80}\), and \(\sqrt{89}\). Using coordinates for vertices \(A(0,0,0)\), \(B(4,5,0)\), \(C(4,0,8)\), and \(D(0,5,8)\), we compute the area of one face using the cross product of vectors. For face \(ABC\):
- Vectors \(\overrightarrow{AB} = (4, 5, 0)\) and \(\overrightarrow{AC} = (4, 0, 8)\)
- Cross product \(\overrightarrow{AB} \times \overrightarrow{AC} = (40, -32, -20)\)
- Magnitude of cross product: \(\sqrt{40^2 + (-32)^2 + (-20)^2} = \sqrt{3024} = 12\sqrt{21}\)
- Area of one face: \(\frac{1}{2} \times 12\sqrt{21} = 6\sqrt{21}\)

Total surface area \(A\) is four times this area:
\[
A = 4 \times 6\sqrt{21} = 24\sqrt{21}
\]

The inradius \(r\) is given by \(r = \frac{3V}{A}\):
\[
r = \frac{3 \times \frac{160}{3}}{24\sqrt{21}} = \frac{160}{24\sqrt{21}} = \frac{20}{3\sqrt{21}} = \frac{20\sqrt{21}}{63}
\]

Thus, \(m = 20\), \(n = 21\), and \(p = 63\). Summing these values:
\[
m + n + p = 20 + 21 + 63 = 104
\]

\[
\boxed{104}
\]


Solution:
Observe that the "worst" possible box is one of the maximum possible length.
By symmetry, the height and the width are the same in this antioptimal box. (If the height and width weren't the same, the extra difference between them could be used to make the length longer.) Thus, let the width and height be of length $a$ and the length be $L$.
We're given that the volume is $23$; thus, $a^2L=23$. We're also given that the surface area is $54=2\cdot27$; thus, $a^2+2aL=27$.
From the first equation, we can get $L=\dfrac{23}{a^2}$. We do a bunch of algebra:
\begin{align*}
L&=\dfrac{23}{a^2} \\
27&=a^2+2aL \\
&=a^2+2a\left(\dfrac{23}{a^2}\right) \\
&=a^2+\dfrac{46}a \\
27a&=a^3+46 \\
a^3-27a+46&=0. \\
\end{align*}
We can use the Rational Root Theorem and test a few values. It turns out that $a=2$ works. We use synthetic division to divide by $a-2$:
[Asdf.png](https://artofproblemsolving.com/wiki/index.php/File:Asdf.png)
As we expect, the remainder is $0$, and we are left with the polynomial $x^2+2x-23$. We can now simply use the quadratic formula and find that the remaining roots are $\dfrac{-2\pm\sqrt{4-4(-23)}}2=\dfrac{-2\pm\sqrt{96}}2=\dfrac{-2\pm4\sqrt{6}}2=-1\pm2\sqrt6$. We want the smallest $a$ to maximize $L$, and it turns out that $a=2$ is in fact the smallest root. Thus, we let $a=2$. Substituting this into $L=\dfrac{23}{a^2}$, we find that $L=\dfrac{23}4$. However, this is not our answer! This is simply the length of the box; we want the radius of the sphere enclosing it. We know that the diameter of the sphere is the diagonal of the box, and the 3D Pythagorean Theorem can give us the space diagonal. Applying it, we find that the diagonal has length $\sqrt{2^2+2^2+\left(\dfrac{23}4\right)^2}=\sqrt{8+\dfrac{529}{16}}=\sqrt{\dfrac{128+529}{16}}=\dfrac{\sqrt{657}}4$. This is the diameter; we halve it to find the radius, $\dfrac{\sqrt{657}}8$. We then square this and end up with $\dfrac{657}{64}$, giving us an answer of $657+64=\boxed{721}$.
~Technodoggo
Denote by $x$, $y$, $z$ the length, width, and height of a rectangular box.
We have
\begin{align*}
xy + yz + zx & = \frac{54}{2}  \hspace{1cm} (1) \\
xyz & = 23 \hspace{1cm} (2)
\end{align*}
We have
\begin{align*}
4 r^2 & = x^2 + y^2 + z^2 \\
& = \left( x + y + z \right)^2 - 2 \cdot \left( xy + yz + zx \right) \\
& = \left( x + y + z \right)^2 - 54 .
\end{align*}
Therefore, we solve the following constrained optimization problem:
\begin{align*}
\max_{x,y,z} \ & x + y + z \\
\mbox{subject to } & (1), (2)
\end{align*}
First, we prove that an optimal solution must have at least two out of $x$, $y$, $z$ that are the same.
Denote by $\lambda$ and $\eta$ lagrangian multipliers of constraints (1) and (2), respectively.
Consider the following Lagrangian:
\begin{align*}
\max_{x,y,z, \lambda, \eta} & x + y + z + \lambda \left( xy + yz + zx - 27 \right)
+ \eta \left( xyz - 23 \right) .
\end{align*}
Taking first-order-condition with respect to $x$, $y$, $z$, respectively, we get
\begin{align*}
1 + \lambda \left( y + z \right) + \eta yz & = 0 \hspace{1cm} (3) \\
1 + \lambda \left( z + x \right) + \eta zx & = 0 \hspace{1cm} (4) \\
1 + \lambda \left( x + y \right) + \eta xy & = 0 \hspace{1cm} (5)
\end{align*}
Suppose there is an optimal solution with $x$, $y$, $z$ that are all distinct.
Taking $(4)-(3)$, we get
\[ \left( x - y \right) \left( \lambda + \eta z \right) = 0 . \]
Because $x \neq y$, we have
\[ \lambda + \eta z = 0 \hspace{1cm} (6) \]
Analogously, we have
\begin{align*}
\lambda + \eta x & = 0 \hspace{1cm} (7)
\end{align*}
Taking $(6) - (7)$, we get $\eta \left( z - x \right) = 0$.
Because $z \neq x$, we have $\eta = 0$. Plugging this into (6), we get $\lambda = 0$.
However, the solution that $\lambda = \eta = 0$ is a contradiction with (3).
Therefore, in an optimal solution, we cannot have $x$, $y$, and $z$ to be all distinct.
W.L.O.G, in our remaining analysis, we assume an optimal solution satisfies $y = z$.
Therefore, we need to solve the following two-variable optimization problem:
\begin{align*}
\max_{x,y} \ & x + 2y \\
\mbox{subject to } & 2 xy + y^2 = 27 \\
& xy^2 = 23
\end{align*}
Replacing $x$ with $y$ by using the constraint $xy^2 = 23$, we solve the following single-variable optimization problem:
\begin{align*}
\max_y \ &  \frac{23}{y^2} + 2y \hspace{1cm} (8) \\
\mbox{subject to } & \frac{46}{y} + y^2 = 27 \hspace{1cm} (9)
\end{align*}
By solving (9), we get $y = 2$ and $-1 + 2 \sqrt{6}$.
Plugging $y = 2$ into (8), we get $\frac{23}{y^2} + 2y = \frac{39}{4}$.
Plugging $y = -1 + 2 \sqrt{6}$ into (8), we get $\frac{23}{y^2} + 2y = \frac{96 \sqrt{6} - 21}{23}$.
We have $\frac{96 \sqrt{6} - 21}{23} < \frac{39}{4}$.
Therefore, the maximum value of $x + y + z$ is $\frac{39}{4}$.
Therefore,
\begin{align*}
r^2 & = \frac{1}{4} \left( \left( x + y + z \right)^2 - 54 \right) \\
& = \frac{1}{4} \left( \left( \frac{39}{4} \right)^2 - 54 \right) \\
& = \frac{657}{64} .
\end{align*}
Therefore, the answer is
$657 + 64 = \boxed{\textbf{(721) }}$.
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
First, let's list the conditions:
Denote by $l$, $w$, $h$ the length, width, and height of a rectangular box.
\[lwh=23\]
\begin{align*}
2(lw+wh+hl)&=54\\
lw+wh+hl&=27.
\end{align*}
Applying the Pythagorean theorem, we can establish that
\begin{align*}
(2r)^2&=(l^2+w^2+h^2)\\
4r^2&=(l^2+w^2+h^2)\\
4r^2&=(l+w+h)^2-2(lw+wh+hl)\\
4r^2&=(l+w+h)^2-54.
\end{align*}
We can spot Vieta's formula hidden inside this equation and call this $m$. Now we have three equations:
\[lwh=23\]
\[lw+wh+hl)=27\]
\[l+w+h=m\]
Let there be a cubic equation. $x^3+bx^2+cx+d=0$. Its roots are $l$, $w$ and $h$. We can use our formulas from before to derive $c$ and $d$.
\[-b=l+w+h=m\]
\[c=lw+wh+lh=27\]
\[-d=lwh=23\]
We can now rewrite the equation from before:
$x^3-mx^2+27x-23=0$
To find the maximum $r$ we need the maximum $m$. This only occurs when this equation has double roots illustrated with graph below.

[AIME 2024 I P15 Pic1.PNG](https://artofproblemsolving.com/wiki/index.php/File:AIME_2024_I_P15_Pic1.PNG) [](https://artofproblemsolving.com/wiki/index.php/File:AIME_2024_I_P15_Pic1.PNG)
WLOG we can set $h=w$.
Thus:
\[lw+w^2+wl=27\]
\[lw^2=23\]
We can substitute $l$ and form a depressed cubic equation with $w$.
\begin{align*}
lw^2&=23\\
l&=\frac{23}{w^2}\\
2\left(\frac{23}{w^2}\right)w+w^2&=27\\
\frac{46}{w}+w^2&=27\\
w^2+\frac{46}{w}-27&=0\\
w^3 -27w+46&=0.
\end{align*}
Based on Rational Root Theorem the possible rational roots are $\pm1, \pm2, \pm23$
A quick test reveals that $2$ is a root of the equation. Comparing coefficients we can factorize the equation into:
$(w-2)(w^2+2w-23)=0$
Besides $2$, we derive another positive root using the quadratic formula, $2\sqrt{6}-1$
But to maximize the $m$ we need to pick the smaller $w$, which is $2$.
Substituting this into $l=\frac{23}{w^2}$, we find that $l=\dfrac{23}4$.
Applying it to our equation above:
\begin{align*}
4r^2&=(l+w+h)^2-54\\
4r^2&=(l+2w)^2-54\\
4r^2&=\left(\dfrac{23}4+2\cdot2\right)^2-54\\
4r^2&=\left(\dfrac{39}4\right)^2-54\\
4r^2&=\left(\dfrac{1521}{16}\right)-54\\
4r^2&=\left(\dfrac{657}{16}\right)\\
r^2&=\left(\dfrac{657}{64}\right).
\end{align*}
$657+64=\boxed{721}$.
~[luckuso](https://artofproblemsolving.comhttps://artofproblemsolving.com/wiki/index.php/User:Cyantist)
to find the maximum m for $x^3-mx^2+27x-23=0$
rewrite m as function of x and calculate derivatives to get maximum value, \[m(x) =-x + 27x^{-1} - 23x^{-2}\]
\[m'(x) = -1 - 27x^{-2} -46x^{-3} = 0\]
\[x^3 -27x+46=0\]
\[(x-2)(x^2+2x-23)=0\]
when x = 2, \[m= 2 + \frac{27}{2}  - \frac{23}{4}  = \frac{39}{4}\]
the rest is similar to solution 3
~[luckuso](https://artofproblemsolving.comhttps://artofproblemsolving.com/wiki/index.php/User:Cyantist)

Answer: 721

<｜begin▁of▁sentence｜><｜User｜>Let $\mathcal{B}$ be the set of rectangular boxes with surface area $54$ and volume $23$. Let $r$ be the radius of the smallest sphere that can contain each of the rectangular boxes that are elements of $\mathcal{B}$. The value of $r^2$ can be written as $\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$.<｜Assistant｜><think>

Okay, so I need to find the radius squared of the smallest sphere that can contain any rectangular box from the set 𝓑, where each box has a surface area of 54 and a volume of 23. Then, express r² as a fraction p/q and find p + q. Hmm, let's start by understanding the problem step by step.

First, a rectangular box has three dimensions: length, width, and height. Let me denote them as x, y, z. The surface area of a rectangular box is given by 2(xy + yz + zx) = 54, and the volume is xyz = 23. The sphere that contains the box must have a diameter at least as long as the space diagonal of the box. The space diagonal d of a box with sides x, y, z is sqrt(x² + y² + z²). Therefore, the radius r of the sphere must be at least d/2, so r = d/2. Therefore, to minimize the radius, we need to minimize the space diagonal d, given the constraints of surface area and volume.

Wait, but actually, the problem says "the smallest sphere that can contain each of the rectangular boxes that are elements of 𝓑". So, the sphere must work for all such boxes in 𝓑. Therefore, we need the maximum possible space diagonal among all boxes in 𝓑, because the sphere must contain every box, so its radius must be at least half the maximum space diagonal. Therefore, r is half the maximum of sqrt(x² + y² + z²) over all boxes in 𝓑. Therefore, we need to find the maximum value of sqrt(x² + y² + z²) given that 2(xy + yz + zx) = 54 and xyz = 23. Then, r² would be (maximum(x² + y² + z²))/4, so we need to compute that maximum and divide by 4.

Therefore, the key is to maximize x² + y² + z² given the constraints 2(xy + yz + zx) = 54 and xyz = 23. Let me note that x, y, z are positive real numbers.

Hmm, so how do I approach this optimization problem with constraints? Maybe using Lagrange multipliers? Or perhaps some inequality?

Alternatively, since we have two constraints, maybe we can express variables in terms of each other. Let me think.

Let me denote S = x + y + z, P = xy + yz + zx, and Q = xyz. Wait, but in our case, we know that 2P = 54, so P = 27, and Q = 23. We need to find the maximum of x² + y² + z². But x² + y² + z² = (x + y + z)^2 - 2(xy + yz + zx) = S² - 2P. Since P is fixed at 27, then x² + y² + z² = S² - 54. Therefore, to maximize x² + y² + z², we need to maximize S², so maximize S. Alternatively, since S is x + y + z, so maximizing S would maximize S², but we need to see if there's a relationship between S and the given constraints.

Alternatively, perhaps we can use the method of Lagrange multipliers here. Let me set up the problem.

We need to maximize f(x, y, z) = x² + y² + z²

subject to the constraints:

g(x, y, z) = 2(xy + yz + zx) - 54 = 0,

h(x, y, z) = xyz - 23 = 0.

So, using Lagrange multipliers with two constraints, we can set up the gradients:

∇f = λ∇g + μ∇h,

where λ and μ are the Lagrange multipliers.

Calculating the gradients:

∇f = (2x, 2y, 2z),

∇g = (2(y + z), 2(x + z), 2(x + y)),

∇h = (yz, xz, xy).

Therefore, the equations are:

2x = λ * 2(y + z) + μ * yz,

2y = λ * 2(x + z) + μ * xz,

2z = λ * 2(x + y) + μ * xy.

Hmm, that's a system of three equations. Let me see if I can find a symmetric solution here. Maybe assuming two variables are equal? Let me suppose that two variables are equal, say x = y. Let me see if that works.

Let me suppose x = y. Then, let me denote x = y = a, and z = b. Then, the constraints become:

Surface area: 2(a² + 2ab) = 54 ⇒ a² + 2ab = 27,

Volume: a²b = 23.

We need to maximize x² + y² + z² = 2a² + b².

Let me see if this assumption leads to a feasible solution. Let me try to solve for a and b.

From the volume equation: b = 23/(a²). Substitute into the surface area equation:

a² + 2a*(23/(a²)) = 27 ⇒ a² + (46)/a = 27.

Multiply both sides by a:

a³ + 46 = 27a ⇒ a³ -27a +46 = 0.

Hmm, solving this cubic equation. Let me see if there is an integer root. Trying a=2: 8 -54 +46 = 0. 8 -54 is -46 +46 = 0. Hey, a=2 is a root!

Therefore, (a - 2)(a² + 2a -23) = 0. So, the other roots are from a² + 2a -23 =0, which would be a = [-2 ± sqrt(4 +92)]/2 = [-2 ± sqrt(96)]/2 = [-2 ± 4*sqrt(6)]/2 = -1 ± 2*sqrt(6). Since a must be positive, so a = -1 + 2√6 ≈ -1 + 4.898 ≈ 3.898, or the other root is negative. So, possible a=2 or a≈3.898.

Wait, so when a=2, then b =23/(2²)=23/4=5.75. Let me check the surface area:

a² + 2ab = 4 + 2*2*(23/4) = 4 + (4*(23/4)*2?) Wait, let me compute again:

Wait, the surface area equation when x=y=a and z=b is 2(a² + 2ab) = 54, so a² + 2ab =27. Plugging a=2 and b=23/4:

2² + 2*2*(23/4) =4 + (4*(23/4)) =4 +23=27. Correct. So that works.

Similarly, for the other root a = -1 + 2√6 ≈3.898, let's compute b=23/a². Let me compute a²:

a = -1 + 2√6, so a +1 = 2√6, so (a +1)^2 = 24, so a² + 2a +1 =24 ⇒ a² =23 -2a. Wait, maybe better to compute numerically:

a ≈3.898, so a²≈15.19, so b≈23/15.19≈1.515. Then check surface area:

a² + 2ab ≈15.19 +2*3.898*1.515 ≈15.19 + 11.81≈27. So that works too.

Therefore, there are two critical points when x=y. So, under the assumption x=y, we can have two possible boxes. Let me compute x² + y² + z² for both cases.

First case: a=2, b=23/4:

2a² + b² = 2*(4) + (23/4)^2 =8 + (529/16)= (128/16)+(529/16)=657/16≈41.0625.

Second case: a≈3.898, b≈1.515:

2a² + b²≈2*(15.19)+(2.295)≈30.38 +2.295≈32.675. Wait, but that's less than the first case. Wait, but we are trying to maximize x² + y² + z², so the first case gives a higher value. Hmm, so perhaps the maximum occurs at a=2, but maybe there are other cases where variables are not equal?

Alternatively, maybe the maximum occurs when two variables are equal, but perhaps another case where all variables are different?

Alternatively, perhaps the maximum is achieved when one variable approaches zero? Wait, but the volume is fixed at 23, so if one variable approaches zero, the product would approach zero, which contradicts the volume being 23. So variables can't be too small.

Alternatively, perhaps the maximum occurs when one variable is as large as possible, but constrained by the surface area and volume.

Alternatively, maybe the case when two variables are equal gives the maximum, so perhaps the first case with a=2 gives the maximum x² + y² + z² of 657/16. But let's check another case where variables are different.

Alternatively, let me see if the Lagrange equations can be solved more generally.

We had the equations from the Lagrange multipliers:

2x = λ * 2(y + z) + μ * yz,

2y = λ * 2(x + z) + μ * xz,

2z = λ * 2(x + y) + μ * xy.

Let me denote these equations as (1), (2), (3).

Let me try to subtract equation (1) and (2):

2x - 2y = 2λ(y + z - x - z) + μ(yz - xz)

Simplify:

2(x - y) = 2λ(y - x) + μ z(y - x)

Factor out (x - y):

2(x - y) = -2λ(x - y) + μ z (y - x)

Wait, let me re-express:

Left side: 2(x - y)

Right side: 2λ[2(y + z) - 2(x + z)]? Wait, no, perhaps better to re-express step by step.

Wait, equation (1) minus equation (2):

2x - 2y = 2λ[(y + z) - (x + z)] + μ(yz - xz)

Simplify the terms inside the brackets:

(y + z) - (x + z) = y - x

Similarly, yz - xz = z(y - x)

Therefore,

2(x - y) = 2λ(y - x) + μ z (y - x)

Factor out (y - x) on the right:

2(x - y) = (y - x)(2λ + μ z)

Bring all terms to left:

2(x - y) + (x - y)(2λ + μ z) = 0

Factor (x - y):

(x - y)[2 + 2λ + μ z] = 0

Therefore, either x = y, or [2 + 2λ + μ z] =0.

Similarly, subtracting equation (1) - (3):

2x - 2z = 2λ[(y + z) - (x + y)] + μ(yz - xy)

Simplify:

Left: 2(x - z)

Right: 2λ(z - x) + μ y(z - x)

Therefore,

2(x - z) = (z - x)(2λ + μ y)

Bring terms together:

2(x - z) + (x - z)(2λ + μ y) =0

Factor (x - z):

(x - z)[2 + 2λ + μ y] =0

Thus, either x = z or [2 + 2λ + μ y] =0.

Similarly, subtracting equation (2) - (3):

2y - 2z = 2λ[(x + z) - (x + y)] + μ(xz - xy)

Simplify:

Left: 2(y - z)

Right: 2λ(z - y) + μ x(z - y)

Thus,

2(y - z) = (z - y)(2λ + μ x)

Bring together:

2(y - z) + (y - z)(2λ + μ x)=0

Factor (y - z):

(y - z)[2 + 2λ + μ x] =0

Therefore, either y = z or [2 + 2λ + μ x] =0.

So, from these three differences, we have possibilities:

Either two variables are equal, or the terms in brackets are zero.

Case 1: All variables are equal. Then x = y = z. Let's check if this is possible.

If x = y = z, then surface area is 2(3x²)=6x²=54 ⇒ x²=9 ⇒x=3. Volume would be x³=27, but the volume is supposed to be 23. Since 27≠23, this is impossible. So variables can't all be equal.

Case 2: Two variables equal, say x = y ≠ z. Then, as we considered earlier, leading to the two solutions. Alternatively, other pairs.

Alternatively, suppose that none of the variables are equal, so then the other factors must be zero. That is,

From (x - y)[2 + 2λ + μ z] =0,

since x ≠ y, then 2 + 2λ + μ z =0,

Similarly, from (x - z)[2 + 2λ + μ y] =0,

since x ≠ z, then 2 + 2λ + μ y =0,

Similarly, from (y - z)[2 + 2λ + μ x] =0,

since y ≠ z, then 2 + 2λ + μ x =0.

Therefore, we have the three equations:

2 + 2λ + μ z =0,

2 + 2λ + μ y =0,

2 + 2λ + μ x =0.

Subtracting the first equation from the second:

(2 + 2λ + μ y) - (2 + 2λ + μ z) =0 ⇒ μ(y - z)=0.

Since we are in the case where y ≠ z, this implies μ=0.

Similarly, subtracting first and third equations:

μ(x - z)=0, so again μ=0.

If μ=0, then from the first equation, 2 + 2λ =0 ⇒ λ = -1.

So, substituting back into the original Lagrange equations:

∇f = λ∇g + μ∇h,

with μ=0 and λ = -1,

So,

2x = -1 * 2(y + z) + 0,

2y = -1 * 2(x + z) +0,

2z = -1 * 2(x + y) +0.

Therefore,

2x = -2(y + z) ⇒ x = - (y + z),

Similarly,

2y = -2(x + z) ⇒ y = - (x + z),

2z = -2(x + y) ⇒ z = - (x + y).

But x, y, z are lengths, so they must be positive. However, the equations above would require x, y, z to be negative sums of positive numbers, which is impossible. Therefore, this case leads to a contradiction. Therefore, the only possible case is when two variables are equal.

Therefore, the critical points occur when two variables are equal. So, we can proceed under the assumption that two variables are equal, which we already considered.

Earlier, when we set x = y, we found two solutions: one with a=2, b=23/4, and another with a≈3.898, b≈1.515. The first gives a higher x² + y² + z². Let me check if there are other cases where different pairs are equal, like x = z or y = z, but due to symmetry, they should give the same results.

Therefore, the maximum of x² + y² + z² under the given constraints is 657/16, which would give r² = (657/16)/4 = 657/64. Wait, but hold on, let me confirm.

Wait, the space diagonal squared is x² + y² + z², so the radius is half the space diagonal, so r² is (x² + y² + z²)/4. Therefore, if the maximum x² + y² + z² is 657/16, then r² would be (657/16)/4 = 657/64. But let me check if that is indeed the maximum.

Wait, but when I considered the case where x = y ≈3.898, then x² + y² + z² was approximately 32.675, which is about 32.675, whereas 657/16 is 41.0625, so that is indeed larger. So that case gives a larger value. So that is the maximum?

Wait, but perhaps there are other critical points when different variables are set equal? Let me check another case where, say, x = z. Let me try that.

Let me suppose x = z = a, and y = b. Then, similar to before:

Surface area: 2(a² + 2ab) =54 ⇒ a² + 2ab =27,

Volume: a²b =23.

Same equations as when we set x=y. So, the solutions would be the same, just with variables renamed. So, the maximum x² + y² + z² would still be 2a² + b², same as before. So, same results. Similarly for y = z.

Therefore, the maximum occurs when two variables are equal to 2, and the third is 23/4, giving x² + y² + z² = 2*(2)^2 + (23/4)^2 = 8 + (529/16) = (128 + 529)/16 = 657/16. Therefore, r² is 657/(16*4) = 657/64. But wait, 657 and 64, are they coprime?

Let me check: 657 divided by 3 is 219, so 657 = 3*219 = 3*3*73. 64 is 2^6. So, no common factors. Therefore, p =657, q=64, so p+q=657+64=721.

Wait, but before I conclude, I need to make sure that this is indeed the maximum. Is there a possibility that when variables are all different, we can get a higher value?

Alternatively, perhaps the maximum occurs at the critical point we found, and the other critical point is a minimum?

Wait, let me see. Let me consider the two cases when x=y:

First case: a=2, b=23/4≈5.75. Then, the dimensions are 2,2,5.75. The space diagonal squared is 2² +2² + (23/4)^2 = 4 +4 + (529/16)= 8 + 33.0625=41.0625.

Second case: a≈3.898, b≈1.515. The space diagonal squared is approx 3.898² +3.898² +1.515²≈15.19+15.19+2.295≈32.675. So that is indeed smaller. So the first case gives a larger value. So that is the maximum.

Alternatively, perhaps another critical point where variables are not equal? But according to the Lagrange multipliers, the only possible critical points are when two variables are equal, so that should be the maximum.

Alternatively, let me see if there are other possibilities. Let me suppose that variables are different. Let me try some numbers.

Suppose, for example, that one side is very large, but then the other sides must be small to keep the volume at 23. But the surface area is fixed at 54. Let me see.

Suppose x is large, then y and z must be small. Let me see if that can lead to a larger space diagonal.

Suppose x approaches infinity, but then y and z must approach zero to keep the volume fixed at 23. However, the surface area would be 2(xy + yz + zx). If x is very large and y,z very small, then terms like xy and xz would dominate the surface area. But since surface area is fixed at 54, if x increases, then y and z must decrease such that xy and xz don't make the surface area too big. Hmm, perhaps there's a balance here.

Alternatively, maybe the maximum occurs at the critical point we found. Let me see.

Alternatively, let me consider using the method of substitution. Let me denote variables as x, y, z with constraints:

2(xy + yz + zx)=54 ⇒ xy + yz + zx =27,

and xyz=23.

We need to maximize x² + y² + z².

Let me consider that for fixed P = xy + yz + zx and Q = xyz, the expression x² + y² + z² is related to (x + y + z)^2. Since x² + y² + z² = S² - 2P, where S =x + y + z. So to maximize x² + y² + z², we need to maximize S². Therefore, we need to maximize S =x + y + z.

Therefore, the problem reduces to maximizing S =x + y + z given that xy + yz + zx =27 and xyz=23.

Hmm, so perhaps we can use the method of Lagrange multipliers again but for maximizing S with those constraints. Alternatively, use inequalities.

Alternatively, perhaps using the AM ≥ GM inequality? Not sure.

Alternatively, let me consider that for three variables with fixed product and fixed sum of products, how to maximize the sum.

Alternatively, let me consider substituting variables. Let me suppose variables are a, b, c. Let me think of them as roots of a cubic equation. Let me consider the cubic equation t³ - St² + Pt - Q =0, so t³ - St² +27t -23=0.

The roots of this equation are x, y, z. To have real positive roots, the cubic must have three real positive roots. The maximum S would correspond to the case when two roots are equal? Because often extrema occur at symmetric points.

Wait, in optimization under constraints, the extremum is often achieved when variables are equal or when some variables are equal. Since in our previous case, when two variables are equal, we get a higher S. Let me compute S in both cases.

First case: x=y=2, z=23/4=5.75. Then S =2+2+5.75=9.75.

Second case: x=y≈3.898, z≈1.515. Then S≈3.898+3.898+1.515≈9.311.

So indeed, the first case gives a higher S, so that is the maximum. Therefore, the maximum S is 9.75=39/4, so S²=(39/4)^2=1521/16. Then x² + y² + z² = S² - 2P = 1521/16 - 54 = 1521/16 - 864/16 = (1521 - 864)/16 = 657/16, which matches the previous result. Therefore, that is indeed the maximum.

Therefore, the minimal sphere radius squared is (657/16)/4 = 657/64. Since 657 and 64 are coprime (as 657 factors into 3×3×73 and 64 is 2^6), so p=657 and q=64, so p+q=657+64=721.

Wait, but before I finish, let me just confirm once more that there isn't a case where variables are all different that gives a higher value. Suppose, for instance, take x=1, then y*z=23, and the surface area would be 2(y + z + yz). Wait, let me see:

Wait, if x=1, then the surface area equation becomes 2( y + z + yz ) =54 ⇒ y + z + yz =27. But since y*z=23, so let me set y*z=23, so z=23/y. Then,

y + 23/y +23 =27 ⇒ y +23/y =4.

Multiply by y: y² +23 =4y ⇒ y² -4y +23=0. The discriminant is 16 -92 = -76 <0, so no real solution. So x cannot be 1.

Alternatively, try x=3. Then y*z=23/3≈7.6667. The surface area equation:

2(3y +3z + yz)=54 ⇒3y +3z + yz=27.

Let me set z=23/(3y). Then,

3y + 3*(23/(3y)) + y*(23/(3y)) =27 ⇒3y + 23/y +23/3 =27.

Multiply all terms by 3y:

9y² +69 +23y =81y ⇒9y² +23y +69 -81y=0 ⇒9y² -58y +69=0.

Discriminant: 58² -4*9*69 =3364 -2484=880. sqrt(880)=~29.66. So solutions:

y=(58 ±29.66)/18. Let me compute:

58 +29.66≈87.66/18≈4.87,

58-29.66≈28.34/18≈1.574.

So y≈4.87, z≈23/(3*4.87)≈23/14.6≈1.575,

or y≈1.574, z≈23/(3*1.574)≈23/4.72≈4.87.

So the dimensions would be x=3, y≈4.87, z≈1.575. Then compute x² + y² + z²≈9 +23.7 +2.48≈35.18, which is less than 657/16≈41.06. So indeed smaller.

Alternatively, trying another value, say x=4. Let's see:

x=4, then y*z=23/4=5.75,

Surface area equation: 2(4y +4z + yz)=54 ⇒4y +4z + yz=27.

Let z=5.75/y,

Then 4y +4*(5.75/y) + y*(5.75/y)=27 ⇒4y +23/y +5.75=27 ⇒4y +23/y =21.25.

Multiply by y:4y² +23 =21.25 y ⇒4y² -21.25 y +23=0.

Multiply all terms by 4 to eliminate decimals: 16y² -85y +92=0.

Discriminant: 85² -4*16*92=7225 - 5888=1337. sqrt(1337)≈36.57.

Solutions: y=(85 ±36.57)/32.

First solution: (85+36.57)/32≈121.57/32≈3.799,

Second: (85-36.57)/32≈48.43/32≈1.513.

Thus, y≈3.8, z≈5.75/3.8≈1.513,

or y≈1.513, z≈3.8.

Thus, dimensions 4, 3.8, 1.513. Then x² + y² + z²≈16 +14.44 +2.29≈32.73, still less than 41.

Therefore, it seems that the case with two sides 2 and one side 23/4 gives the maximum. Therefore, the answer should be 657/64, so 657+64=721.

Wait, but let me check if there is another critical point when variables are not two equal. Wait, according to the Lagrange multipliers, the only possible critical points are when two variables are equal, so that should cover all possibilities. Therefore, I think this is correct.

Therefore, the answer is 721.

**Final Answer**
\boxed{721}
</think>

Given a set \(\mathcal{B}\) of rectangular boxes with surface area 54 and volume 23, we need to find the radius \(r\) of the smallest sphere that can contain each of these boxes. The value of \(r^2\) is to be expressed as \(\frac{p}{q}\) where \(p\) and \(q\) are coprime, and we need to find \(p + q\).

1. **Constraints and Objective**:
   - Surface area: \(2(xy + yz + zx) = 54 \implies xy + yz + zx = 27\)
   - Volume: \(xyz = 23\)
   - We need to maximize the space diagonal \( \sqrt{x^2 + y^2 + z^2} \), which requires maximizing \(x^2 + y^2 + z^2\).

2. **Using Lagrange Multipliers**:
   - We set up the Lagrangian function and derive the system of equations. After analysis, we find that the critical points occur when two variables are equal.

3. **Case Analysis**:
   - Assume \(x = y = a\) and \(z = b\):
     - Surface area: \(a^2 + 2ab = 27\)
     - Volume: \(a^2b = 23\)
     - Solving these, we find two solutions:
       - \(a = 2\), \(b = \frac{23}{4}\)
       - \(a \approx 3.898\), \(b \approx 1.515\)

4. **Maximizing \(x^2 + y^2 + z^2\)**:
   - For \(a = 2\) and \(b = \frac{23}{4}\):
     - \(x^2 + y^2 + z^2 = 2(2^2) + \left(\frac{23}{4}\right)^2 = 8 + \frac{529}{16} = \frac{657}{16}\)
   - This is the maximum value as other cases yield smaller results.

5. **Calculating \(r^2\)**:
   - The space diagonal squared is \(\frac{657}{16}\), so \(r^2 = \frac{657}{64}\).
   - \(p = 657\) and \(q = 64\), which are coprime.

6. **Final Result**:
   - \(p + q = 657 + 64 = 721\).

\boxed{721}