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 x² = 41 - y² = 41 - 25 = 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, from (0,0,0) to (x,0,0), (0,y,0), (0,0,z). Then, the four vertices of the tetrahedron can be (0,0,0), (x,y,0), (x,0,z), (0,y,z), or something like that? Wait, maybe not exactly. Let me think again.

Alternatively, the vertices can be placed at (a, b, c), (-a, -b, c), (-a, b, -c), (a, -b, -c). Let me see if that works. Let me try to set coordinates such that the centroid is at the origin. Let me suppose the coordinates are as follows:

Let me denote the four vertices as follows:

A = (x, y, z),

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

C = (-x, y, -z),

D = (x, -y, -z).

This way, the midpoints of the edges AB, AC, AD would be at the origin? Wait, let me check midpoint of AB: [(x + (-x))/2, (y + (-y))/2, (z + z)/2] = (0, 0, z). Hmm, not the origin. Maybe this isn't the right coordinate system.

Alternatively, perhaps the standard way to place a disphenoid in a box is to have the vertices at (a, b, c), (-a, -b, -c), (a, -b, -c), (-a, b, c). Wait, maybe that's overcomplicating. Alternatively, let me think of the box with edges 2a, 2b, 2c, so that the coordinates can be symmetric around the origin. Let me try this approach.

Suppose the box has edges of length 2x, 2y, 2z, so that the eight vertices are all combinations of (±x, ±y, ±z). Then, choosing four vertices such that each pair of opposite edges of the tetrahedron correspond to the face diagonals. Let me see.

Alternatively, perhaps the vertices of the tetrahedron can be (x, y, z), (-x, -y, z), (-x, y, -z), (x, -y, -z). Let me check the distances between these points.

Compute AB: between (x,y,z) and (-x,-y,z). The distance squared would be (2x)^2 + (2y)^2 + 0 = 4x² + 4y². Similarly, CD is between (-x,y,-z) and (x,-y,-z). The distance squared is (2x)^2 + (2y)^2 + 0 = same as AB. So that's good, so AB and CD have length 2√(x² + y²). Similarly, AC is between (x,y,z) and (-x,y,-z). The distance squared is (2x)^2 + 0 + (2z)^2 = 4x² + 4z². BD is between (-x,-y,z) and (x,-y,-z). Distance squared is (2x)^2 + 0 + (2z)^2, same as AC. Then AD is between (x,y,z) and (x,-y,-z). Distance squared is 0 + (2y)^2 + (2z)^2 = 4y² + 4z². Similarly, BC is between (-x,-y,z) and (-x,y,-z). Distance squared is 0 + (2y)^2 + (2z)^2, same as AD. So that works. So in this coordinate system, the edges are as required if we set 2√(x² + y²) = sqrt(41), so 4(x² + y²) = 41. Wait, but in the problem, AB is sqrt(41). Wait, in my calculation above, AB squared is 4x² + 4y², so that would be (AB)^2 = 4(x² + y²). So setting that equal to 41, so x² + y² = 41/4. Similarly, AC squared is 4x² + 4z² = 80, so x² + z² = 20. And AD squared is 4y² + 4z² = 89, so y² + z² = 89/4.

Wait, but earlier when I considered the box edges as x, y, z (not doubled), I had equations x² + y² = 41, x² + z² = 80, y² + z² = 89, leading to x=4, y=5, z=8. But here, in this coordinate system, the edges of the box would be 2x, 2y, 2z, so perhaps there is a scaling factor here. Let me see.

Alternatively, maybe I need to adjust variables. Let me denote the box edges as a, b, c, so that the coordinates are (a, b, c), etc. Then, following the same logic, the edges of the tetrahedron would be sqrt{(2a)^2 + (2b)^2}, etc. Wait, perhaps my initial approach was better. Let me go back.

Earlier, I considered the box with edges x, y, z (so from 0 to x, 0 to y, 0 to z), and then the face diagonals would be sqrt(x² + y²), sqrt(x² + z²), sqrt(y² + z²). Then, the edges of the tetrahedron are those face diagonals. So in that case, the edges of the tetrahedron are the face diagonals of the box, so each edge of the tetrahedron corresponds to a face diagonal of the box. Therefore, the tetrahedron can be thought of as having vertices at four alternate vertices of the box, such that each edge of the tetrahedron is a face diagonal. So for example, the vertices could be (0,0,0), (x,y,0), (x,0,z), (0,y,z). Let me check the distances between these points.

Distance between (0,0,0) and (x,y,0) is sqrt(x² + y²), which would be one edge. Between (0,0,0) and (x,0,z) is sqrt(x² + z²). Between (0,0,0) and (0,y,z) is sqrt(y² + z²). Then between (x,y,0) and (x,0,z) is sqrt( (0)^2 + y² + z² ). Wait, let me compute that: the coordinates difference is (0, -y, z), so squared distance is y² + z². Similarly between (x,y,0) and (0,y,z): sqrt(x² + 0 + z²). And between (x,0,z) and (0,y,z): sqrt(x² + y² + 0). So indeed, each pair of vertices among these four points have distances equal to the face diagonals. So in this case, the edges of the tetrahedron are exactly the face diagonals of the box. Therefore, the edges of the tetrahedron are sqrt(x² + y²), sqrt(x² + z²), sqrt(y² + z²), each appearing twice. So that's perfect because in the problem, the edges are given as AB=CD= sqrt(41), AC=BD= sqrt(80), AD=BC= sqrt(89). Therefore, if we set:

sqrt(x² + y²) = sqrt(41),

sqrt(x² + z²) = sqrt(80),

sqrt(y² + z²) = sqrt(89).

Which is exactly the same as the equations I had before, leading to x=4, y=5, z=8. Therefore, the box has dimensions 4,5,8. Therefore, the tetrahedron can be embedded in a box with dimensions 4,5,8, with vertices at (0,0,0), (4,5,0), (4,0,8), (0,5,8). Let me confirm the distances here.

Compute AB: between (0,0,0) and (4,5,0): sqrt( (4)^2 +5^2 +0 ) = sqrt(16+25)=sqrt(41). Good.

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

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

Then BC: between (4,5,0) and (4,0,8): sqrt(0 +25 +64)=sqrt(89). Good.

BD: between (4,5,0) and (0,5,8): sqrt(16 +0 +64)=sqrt(80). Good.

CD: between (4,0,8) and (0,5,8): sqrt(16 +25 +0)=sqrt(41). Perfect. So that works. So the coordinates are set as follows:

Let me assign the vertices as:

A = (0, 0, 0),

B = (4, 5, 0),

C = (4, 0, 8),

D = (0, 5, 8).

Alright, so now that I have coordinates for all four vertices, I can compute the volume and the surface area.

First, let's compute the volume of the tetrahedron. Since it's embedded in a box, maybe there's an easy way. The volume of a tetrahedron in a box like this is 1/3 * base area * height? Wait, but perhaps more straightforward with coordinates.

Alternatively, since the tetrahedron is formed by four vertices of a rectangular box, the volume can be calculated as (abc)/3, where a, b, c are the edges of the box? Wait, let me check that. Wait, actually, in a box with edges x, y, z, the volume of the tetrahedron formed by three edges from a vertex is (x y z)/6. But in this case, the tetrahedron is using four vertices such that each edge is a face diagonal. Hmm, maybe the formula is different.

Alternatively, let me use the scalar triple product. Since I have coordinates for the four points, I can compute vectors from one vertex to the other three and then compute the scalar triple product divided by 6.

Let me take point A as (0,0,0). Then vectors AB, AC, AD are:

AB = B - A = (4,5,0),

AC = C - A = (4,0,8),

AD = D - A = (0,5,8).

The volume is 1/6 times the absolute value of the scalar triple product of AB, AC, AD.

Compute the scalar triple product AB ⋅ (AC × AD).

First compute AC × AD:

AC = (4, 0, 8),

AD = (0,5,8).

Cross product:

|i   j   k |

|4   0   8|

|0   5   8|

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

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

= (-40, -32, 20)

Then AB ⋅ (AC × AD) = (4,5,0) ⋅ (-40, -32, 20) = 4*(-40) + 5*(-32) + 0*20 = -160 -160 + 0 = -320.

The absolute value is 320, so the volume is 320 / 6 = 160/3 ≈ 53.333...

Wait, let me confirm that cross product again:

First component (j and k components of AC and AD):

i component: (AC_y * AD_z - AC_z * AD_y) = 0*8 - 8*5 = 0 - 40 = -40,

j component: -(AC_x * AD_z - AC_z * AD_x) = -(4*8 - 8*0) = -(32 - 0) = -32,

k component: (AC_x * AD_y - AC_y * AD_x) = 4*5 - 0*0 = 20 - 0 = 20.

Yes, so cross product is (-40, -32, 20). Then dot with AB (4,5,0):

4*(-40) = -160,

5*(-32) = -160,

0*20 = 0,

Total: -320. Absolute value 320. Volume is 320 / 6 = 160/3. Okay, so Volume V = 160/3.

Now, need to compute the total surface area. The tetrahedron has four triangular faces. Each face is a triangle with known side lengths. Since all edges are given, perhaps I can compute the area of each triangle using Heron's formula.

Alternatively, since we have coordinates, maybe compute the area via vectors or coordinates. Let me see which is easier.

First, let's note the four faces:

Face ABC: formed by points A, B, C.

Face ABD: points A, B, D.

Face ACD: points A, C, D.

Face BCD: points B, C, D.

Alternatively, since all edges are given, maybe Heron's formula is straightforward. Let me try that.

First, for triangle ABC:

Sides AB = sqrt(41), AC = sqrt(80), BC = sqrt(89). Wait, hold on, BC is between B(4,5,0) and C(4,0,8). Let me confirm BC's length again. The coordinates difference is (0, -5, 8). So squared length is 0 + 25 + 64 = 89, so sqrt(89). So yes, that's correct.

So triangle ABC has sides sqrt(41), sqrt(80), sqrt(89). Let me compute its area using Heron's formula.

First, compute the semi-perimeter: s = (sqrt(41) + sqrt(80) + sqrt(89))/2. Hmm, that might get messy. Alternatively, since I have coordinates, maybe compute the area via the cross product.

For triangle ABC, vectors AB and AC can be used. The area is 1/2 the magnitude of AB × AC.

AB is (4,5,0), AC is (4,0,8). Cross product:

|i   j   k |

|4   5   0|

|4   0   8|

= 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 squared is 40² + (-32)^2 + (-20)^2 = 1600 + 1024 + 400 = 3024. So the magnitude is sqrt(3024). Let me compute sqrt(3024):

3024 divided by 16 is 189, so sqrt(16*189) = 4*sqrt(189). 189 is 9*21, so sqrt(189)=3*sqrt(21). Thus sqrt(3024)=4*3*sqrt(21)=12√21. Therefore, the magnitude is 12√21, so the area is 1/2 * 12√21 = 6√21.

Okay, so area of ABC is 6√21.

Similarly, let's compute area of ABD. Points A, B, D.

AB = sqrt(41), AD = sqrt(89), BD = sqrt(80). Wait, BD is between B(4,5,0) and D(0,5,8). The coordinates difference is (-4,0,8), so squared length is 16 + 0 + 64 = 80, so sqrt(80). So sides are sqrt(41), sqrt(89), sqrt(80). Same as triangle ABC, so maybe the area is the same?

Alternatively, compute via vectors. Vectors AB and AD.

AB is (4,5,0), AD is (0,5,8). Cross product:

|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 squared is same as before: 40² + (-32)^2 +20² = 1600 + 1024 + 400 = 3024. So same magnitude, 12√21, so area is 6√21 again. So ABD also has area 6√21.

Now, moving on to triangle ACD: points A, C, D.

AC = sqrt(80), AD = sqrt(89), CD = sqrt(41). So sides sqrt(80), sqrt(89), sqrt(41). Same as the other triangles, so perhaps same area?

Let me confirm with vectors. Vectors AC and AD.

AC is (4,0,8), AD is (0,5,8). Cross product:

|i   j   k |

|4   0   8|

|0   5   8|

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

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

So cross product is (-40, -32, 20). The magnitude squared is same as before: (-40)^2 + (-32)^2 +20^2 = 1600 + 1024 + 400 = 3024. So again, magnitude 12√21, area 6√21. So ACD is also 6√21.

Finally, the last face, BCD: points B, C, D.

BC = sqrt(89), CD = sqrt(41), BD = sqrt(80). So sides sqrt(89), sqrt(41), sqrt(80). Again, same as the others. Let me check with vectors.

Take vectors BC and BD.

Coordinates:

Point B is (4,5,0), C is (4,0,8), D is (0,5,8).

Vectors BC = C - B = (0, -5, 8),

BD = D - B = (-4, 0, 8).

Compute cross product BC × BD:

|i   j   k |

|0  -5   8|

|-4  0   8|

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

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

= (-40, -32, -20)

The magnitude squared is (-40)^2 + (-32)^2 + (-20)^2 = same 1600 + 1024 + 400 = 3024. So magnitude 12√21, area 6√21. So all four faces have the same area! That's nice, so the total surface area A = 4 * 6√21 = 24√21.

Therefore, the inradius r = 3V / A = 3*(160/3) / (24√21) = (160) / (24√21) = (160 ÷ 8) / (24 ÷ 8 * √21) = 20 / (3√21). Wait, let me compute step by step:

3V is 3*(160/3) = 160.

A is 24√21.

So r = 160 / (24√21) = (160 ÷ 8) / (24 ÷ 8 * √21) = 20 / (3√21). Then, rationalizing the denominator:

20/(3√21) = (20√21)/(3*21) = (20√21)/63. So that would be the inradius. Wait, but let me check again:

Wait, 160 divided by 24 is 160/24 = 20/3. So 20/3 divided by √21 is (20)/(3√21). Then, rationalizing, multiply numerator and denominator by √21:

(20√21)/(3*21) = (20√21)/63. Since 20 and 63 have a common factor? 20 is 2²×5, 63 is 7×3². So they are co-prime? Yes, 20 and 63 share no common prime factors. And 21 is 3×7, so √21 is simplified. So the distance is (20√21)/63, so m=20, n=21, p=63. Wait, but 20 and 63 are coprime, and 21 is square-free. So then m + n + p is 20 +21 +63= 104. Hmm, but let me confirm all steps again to be sure.

Wait, first, volume was 160/3. Correct? Let me check the scalar triple product again. The cross product AC × AD was (-40, -32, 20). Then AB ⋅ (AC × AD) was (4,5,0) ⋅ (-40, -32, 20) = 4*(-40) +5*(-32) +0*20 = -160 -160 = -320. Absolute value 320, so volume is 320/6 = 160/3. That's correct.

Total surface area: each face is 6√21, four faces so 24√21. Correct.

Thus, inradius r = 3*(160/3) / (24√21) = 160/(24√21) = 20/(3√21) = (20√21)/63. So yes, that's the distance. So m=20, n=21, p=63. Then 20+21+63=104. Hmm, but I need to check if the problem states "distance can be written in the form m√n / p where m and p are coprime, and n is square-free". Here, 20 and 63 are coprime (since 20 factors into 2²×5, 63 is 7×3²; no common factors). And 21 is 3×7, which is square-free. So that should be correct. Wait, but let me check if maybe I made a mistake in the cross product or area calculations.

Wait, let me check the cross product for triangle BCD again. Vectors BC = (0, -5,8) and BD = (-4,0,8). The cross product components:

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

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

k component: 0*0 - (-5)(-4) = 0 -20 = -20,

So cross product is (-40, -32, -20). The magnitude squared is (-40)^2 + (-32)^2 + (-20)^2 = 1600 + 1024 + 400 = 3024. sqrt(3024) = sqrt(16*189) = 4*sqrt(189). Wait, 189 is 9*21, so sqrt(189)=3*sqrt(21). So sqrt(3024)=4*3*sqrt(21)=12√21. So the area is 1/2 * 12√21 = 6√21. Correct. So all faces are indeed 6√21. So total area 24√21. So that's correct.

Therefore, the inradius is indeed (20√21)/63, so the answer is 20 +21 +63=104. Wait, but before I get too confident, let me see if there's another way to compute the inradius, maybe using the formula for a disphenoid?

Alternatively, in a disphenoid, the inradius can be calculated as 3V/A, which is exactly what I did. Alternatively, maybe there is a formula in terms of the edges?

Alternatively, since the tetrahedron is formed in a box, perhaps there's a formula for the inradius in terms of the box dimensions?

Let me think. The box has dimensions x=4, y=5, z=8. The tetrahedron is formed by four vertices such that each edge is a face diagonal. The inradius is the distance from the center of the box to each face? Wait, in a disphenoid, is the incenter at the centroid of the box?

Wait, the centroid of the tetrahedron is the average of its vertices. Let me compute that. The coordinates of the four vertices are:

A: (0,0,0),

B: (4,5,0),

C: (4,0,8),

D: (0,5,8).

The centroid (geometric center) would be the average of the coordinates:

x: (0 +4 +4 +0)/4 = 8/4 = 2,

y: (0 +5 +0 +5)/4 = 10/4 = 2.5,

z: (0 +0 +8 +8)/4 = 16/4 =4.

So centroid is at (2, 2.5, 4). Which is exactly the center of the box, since the box goes from 0 to 4 in x, 0 to5 in y, 0 to8 in z. So the box's center is (2, 2.5, 4). So the centroid of the tetrahedron coincides with the center of the box. Now, in a disphenoid, is the incenter the same as the centroid? Hmm, in a disphenoid, which is a tetrahedron with congruent opposite edges, it is also called a "rhombohedron" but wait, no, a rhombohedron is a different shape. Wait, but in any case, in a disphenoid, the incenter and centroid might coincide? Because of symmetry.

Since the tetrahedron is symmetric with respect to the center of the box, so the incenter should be at the centroid. Therefore, the distance from the centroid to each face is equal. Therefore, perhaps I can compute the distance from the centroid (2, 2.5, 4) to one of the faces, and that should be the inradius. Let me try that as a check.

Let me pick a face, say face ABC. The equation of the plane ABC can be found, then compute the distance from (2, 2.5, 4) to that plane.

First, find the equation of plane ABC. Points A(0,0,0), B(4,5,0), C(4,0,8).

Let me compute the normal vector using vectors AB and AC.

Vectors AB = (4,5,0), AC = (4,0,8). The cross product AB × AC is as before:

(4,5,0) × (4,0,8) = determinant:

i (5*8 -0*0) - j (4*8 -0*4) + k (4*0 -5*4) = (40, -32, -20). So the normal vector is (40, -32, -20). We can simplify this by dividing by 4: (10, -8, -5). So the plane equation is 10x -8y -5z = d. Since point A (0,0,0) is on the plane, d=0. So the equation is 10x -8y -5z =0.

The distance from point (2, 2.5,4) to this plane is |10*2 -8*2.5 -5*4| / sqrt(10² + (-8)^2 + (-5)^2).

Compute numerator: 20 - 20 -20 = (20 -20) -20 = 0 -20 = -20. Absolute value is 20.

Denominator: sqrt(100 +64 +25) = sqrt(189). sqrt(189)=sqrt(9*21)=3√21. So distance is 20/(3√21). Which is the same as before, which rationalized is 20√21 /63. So that's consistent. Therefore, this method also gives the same result. So that's good.

Therefore, the answer should be 20 +21 +63=104. Wait, but before I conclude, let me check if the problem specifies that the tetrahedron is ABCD with those edge lengths, and I assumed the coordinates correctly. Let me just confirm once again the edge lengths with the coordinates:

AB: between (0,0,0) and (4,5,0): sqrt(16+25)=sqrt(41). Correct.

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

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

BC: (4,5,0) to (4,0,8): sqrt(0+25+64)=sqrt(89). Correct.

BD: (4,5,0) to (0,5,8): sqrt(16+0+64)=sqrt(80). Correct.

CD: (4,0,8) to (0,5,8): sqrt(16+25+0)=sqrt(41). Correct.

All edges match. So coordinates are correct. Calculations seem correct. So the answer is 104. Hmm, but let me check if maybe I made a mistake in the problem's question. Wait, the problem says "the distances from I to each of the faces of the tetrahedron are all equal". So that point is the incenter, and the formula I used is correct. So I think this is right. Maybe 104 is the answer. Wait, but let me check if 20 and 63 can be simplified? 20 is 2^2*5, 63 is 7*3^2. No common factors, so 20/63 is reduced. And 21 is square-free. So yes, the problem's conditions are satisfied. So I think that's the answer.

**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 as a disphenoid, which can be inscribed in a rectangular box with dimensions \(4 \times 5 \times 8\). The vertices of the tetrahedron are placed at the vertices of the box such that each edge of the tetrahedron is a face diagonal of the box. The coordinates of the vertices are:
- \(A = (0, 0, 0)\)
- \(B = (4, 5, 0)\)
- \(C = (4, 0, 8)\)
- \(D = (0, 5, 8)\)

**Volume Calculation:**
Using the scalar triple product of vectors from vertex \(A\):
- Vectors: \(\overrightarrow{AB} = (4, 5, 0)\), \(\overrightarrow{AC} = (4, 0, 8)\), \(\overrightarrow{AD} = (0, 5, 8)\)
- Cross product \(\overrightarrow{AC} \times \overrightarrow{AD} = (-40, -32, 20)\)
- Dot product \(\overrightarrow{AB} \cdot (\overrightarrow{AC} \times \overrightarrow{AD}) = -320\)
- Volume \(V = \frac{1}{6} | -320 | = \frac{160}{3}\)

**Surface Area Calculation:**
Each face is a triangle with area calculated using the cross product:
- For face \(ABC\), cross product magnitude \(12\sqrt{21}\), area \(6\sqrt{21}\)
- All four faces have the same area, total surface area \(A = 4 \times 6\sqrt{21} = 24\sqrt{21}\)

**Inradius Calculation:**
Using the formula \(r = \frac{3V}{A}\):
- \(r = \frac{3 \times \frac{160}{3}}{24\sqrt{21}} = \frac{160}{24\sqrt{21}} = \frac{20}{3\sqrt{21}} = \frac{20\sqrt{21}}{63}\)

The distance is \(\frac{20\sqrt{21}}{63}\), leading to \(m = 20\), \(n = 21\), \(p = 63\). Thus, \(m + n + p = 20 + 21 + 63 = 104\).

\[
\boxed{104}
\]


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}


Solution:
By properties of logarithms, we can simplify the given equation to $x\log_xy=4y\log_yx=10$. Let us break this into two separate equations:
\[x\log_xy=10\]
\[4y\log_yx=10.\]
We multiply the two equations to get:
\[4xy\left(\log_xy\log_yx\right)=100.\]
Also by properties of logarithms, we know that $\log_ab\cdot\log_ba=1$; thus, $\log_xy\cdot\log_yx=1$. Therefore, our equation simplifies to:
\[4xy=100\implies xy=\boxed{025}.\]
~Technodoggo
Convert the two equations into exponents:
\[x^{10}=y^x~(1)\]
\[y^{10}=x^{4y}~(2).\]
Take $(1)$ to the power of $\frac{1}{x}$:
\[x^{\frac{10}{x}}=y.\]
Plug this into $(2)$:
\[x^{(\frac{10}{x})(10)}=x^{4(x^{\frac{10}{x}})}\]
\[{\frac{100}{x}}={4x^{\frac{10}{x}}}\]
\[{\frac{25}{x}}={x^{\frac{10}{x}}}=y,\]
So $xy=\boxed{025}$
~alexanderruan
Similar to solution 2, we have:
$x^{10}=y^x$ and $y^{10}=x^{4y}$
Take the tenth root of the first equation to get
$x=y^{\frac{x}{10}}$
Substitute into the second equation to get
$y^{10}=y^{\frac{4xy}{10}}$
This means that $10=\frac{4xy}{10}$, or $100=4xy$, meaning that $xy=\boxed{25}$.
~MC413551

Answer: 025

<｜begin▁of▁sentence｜><｜User｜>There exist real numbers $x$ and $y$, both greater than 1, such that $\log_x\left(y^x\right)=\log_y\left(x^{4y}\right)=10$. Find $xy$.<｜Assistant｜><think>

Okay, so I have this problem here: There exist real numbers x and y, both greater than 1, such that log base x of y^x equals 10, and log base y of x^{4y} also equals 10. I need to find the product xy. Hmm, logarithms with variables as bases and exponents. Let me try to recall logarithm properties and see how to approach this.

First, let me write down the given equations. The first one is log_x(y^x) = 10. The second one is log_y(x^{4y}) = 10. Since both equations are equal to 10, maybe I can set them equal to each other or find a way to relate x and y through these equations.

Starting with the first equation: log_x(y^x) = 10. I remember that log_b(a^c) = c * log_b(a). So applying that property here, this becomes x * log_x(y) = 10. Hmm, so x times the logarithm of y with base x equals 10. Maybe I can rewrite the logarithm in terms of natural logs or something else?

Alternatively, maybe it's better to convert the logarithmic equations into exponential form. Let me try that. For the first equation, log_x(y^x) = 10 means that x^10 = y^x. Because log_b(a) = c is equivalent to b^c = a. So here, the base is x, the result is 10, so x^10 = y^x. Got that equation: x^{10} = y^{x}.

Similarly, the second equation: log_y(x^{4y}) = 10. Converting that, we get y^{10} = x^{4y}. So the second equation is y^{10} = x^{4y}.

Now, so we have two equations:

1) x^{10} = y^{x}

2) y^{10} = x^{4y}

Hmm, so we need to solve for x and y here. Since both equations involve exponents with x and y in both the base and exponent, this might be a bit tricky. Maybe taking logarithms on both sides would help? Let me try that.

Starting with the first equation: x^{10} = y^{x}. Taking natural logarithm of both sides: ln(x^{10}) = ln(y^x). Using the power rule again, that becomes 10 ln x = x ln y. Similarly, for the second equation: y^{10} = x^{4y}, take natural logs: 10 ln y = 4y ln x.

So now we have two equations:

1) 10 ln x = x ln y

2) 10 ln y = 4y ln x

Hmm, so maybe we can denote some variables here to simplify. Let me let a = ln x and b = ln y. Then, since x and y are greater than 1, a and b are positive real numbers.

Substituting, the first equation becomes 10a = x b. Wait, but x is e^{a}, since a = ln x. Similarly, y is e^{b}. So substituting x = e^{a} and y = e^{b}, then:

First equation: 10a = e^{a} * b

Second equation: 10b = 4 e^{b} * a

Hmm, so now we have:

1) 10a = e^{a} * b

2) 10b = 4 e^{b} * a

Hmm, this seems a bit complicated, but maybe we can solve for one variable in terms of the other and substitute.

From the first equation, let's solve for b: b = (10a)/e^{a}

Then plug this into the second equation:

10*(10a / e^{a}) = 4 e^{b} * a

Simplify left side: 100a / e^{a} = 4 a e^{b}

We can divide both sides by a, since a is positive (because x >1, so ln x ≠0), so:

100 / e^{a} = 4 e^{b}

Then, divide both sides by 4:

25 / e^{a} = e^{b}

So e^{b} = 25 e^{-a}

Take natural log of both sides: b = ln(25) - a

But from earlier, we had b = (10a)/e^{a}

Therefore, setting those equal:

(10a)/e^{a} = ln(25) - a

Hmm, so now we have an equation with just a. Let me write that:

(10a)/e^{a} + a = ln(25)

Factor out a:

a (10/e^{a} + 1) = ln(25)

Hmm, this is a transcendental equation, which might not have an algebraic solution. Maybe we can find a substitution or guess a solution?

Alternatively, perhaps there is a clever substitution or relationship between x and y. Let me think again about the original equations.

We have x^{10} = y^{x} and y^{10} = x^{4y}

Maybe we can express y in terms of x from the first equation and substitute into the second?

From the first equation: y^x = x^{10}, so y = (x^{10})^{1/x} = x^{10/x}

Similarly, from the second equation: x^{4y} = y^{10}, so substituting y from above:

x^{4y} = (x^{10/x})^{10} = x^{100/x}

Therefore, x^{4y} = x^{100/x}

Since the bases are the same and x >1, the exponents must be equal:

4y = 100/x

So 4y = 100/x => y = 25/x

Ah! That's a useful relationship: y = 25/x. So y is 25 over x. So now we can substitute this into one of the previous equations.

We had y = x^{10/x} from the first equation, and also y =25/x from the second equation. Therefore, setting them equal:

x^{10/x} = 25/x

Hmm, so x^{10/x} * x =25

Which is x^{10/x +1} =25

Hmm, that's still a bit complicated. Let me write that again:

x^{10/x} = 25/x

Multiply both sides by x:

x^{10/x} * x =25

Which is x^{10/x +1} =25

Alternatively, perhaps take natural logs of both sides of the equation x^{10/x} =25/x.

Taking ln of both sides:

(10/x) ln x = ln(25/x) = ln25 - ln x

So:

(10/x) ln x + ln x = ln25

Factor out ln x:

ln x (10/x +1) = ln25

Hmm, this is similar to the equation I had earlier with a substitution. Let me let a = ln x again, then:

a (10/e^{a} +1 ) = ln25

Wait, but maybe with the substitution y =25/x, perhaps we can find x and y more directly?

We have y =25/x, so xy =25. Wait, is that possible? If y =25/x, then xy=25. So is the answer 25? Wait, but let me check that.

Wait, hold on. If we have y =25/x, then indeed xy=25. But let me confirm if that's consistent with the equations.

Wait, but let me see. Let me suppose that xy=25, so y=25/x. Let me check if this holds with the original equations.

Wait, but let me see if that's the case. Let me suppose that xy=25, so y=25/x. Let me plug this into the first equation: x^{10}= y^x = (25/x)^x. So x^{10} = (25^x)/(x^x). Multiply both sides by x^x:

x^{10} * x^x =25^x

x^{x +10} =25^x

Take both sides to the power 1/x:

x^{(x +10)/x} =25

x^{1 +10/x} =25

Hmm, so x^{1 +10/x} =25. Let me see if this can be true for some x.

Alternatively, let me suppose that x is 5. Let me test x=5. Then y=25/5=5. So x=y=5. Let me check the original equations.

First equation: log_5(5^5)= log_5(3125)=5, but the problem says it should be 10. So that's not working. Hmm, so x=5 and y=5 is not a solution.

Hmm, so maybe my assumption that xy=25 is incorrect? Wait, but we derived y=25/x from the equations. Wait, let me check that step again.

From the second equation, after substituting y from the first equation:

We had from first equation: y =x^{10/x}

Then substituted into the second equation: x^{4y}= y^{10}

So x^{4y}= (x^{10/x})^{10}=x^{100/x}

Therefore, exponents must be equal: 4y =100/x, so y=25/x. That seems correct. So that relationship must hold. Therefore, y=25/x, so xy=25. But when I tested with x=5, it didn't work. So maybe there is another x and y where xy=25, but x and y are not 5?

Wait, perhaps x and y are different numbers such that their product is 25. Let me try x= something else. Let me suppose x=25, then y=1. But y must be greater than 1, so that's invalid. How about x= sqrt(25)=5, but that gives y=5 again. Hmm.

Alternatively, maybe x= something like 25^{1/ something}. Let me think. Let me try to proceed with the equation x^{1 +10/x}=25.

Let me set t = x. So the equation is t^{1 +10/t}=25. Let me take natural logs:

(1 +10/t) ln t = ln25.

Hmm, this equation might require some trial and error. Let me see if I can find a t that satisfies this.

Suppose t=5. Then left side: (1 +10/5) ln5 = (1+2) ln5 =3 ln5 ≈3*1.609≈4.828. But ln25≈3.218. Not equal.

Hmm, too big. Let me try t= something larger. Let me try t=10.

Left side: (1 +10/10) ln10 =2 * ln10≈2*2.302≈4.605. Still bigger than ln25≈3.218.

Hmm, need to make the left side smaller. Let me try t=20.

(1 +10/20) ln20 = (1.5)*ln20≈1.5*2.995≈4.493. Still too big.

Hmm, maybe t=25?

(1 +10/25) ln25 = (1.4)*3.218≈4.505. Still too big.

Hmm, going the other way. Maybe t= something less than 5?

Wait, but x has to be greater than 1. Let me try t=2.

(1 +10/2) ln2 =6 *0.693≈4.158. Still bigger than 3.218.

Hmm, t=1. Let me see, but x must be greater than1. Let me try t approaching 1 from the right. If t approaches 1, then (1 +10/t) approaches 11, and ln t approaches 0. So the product approaches 0. Not helpful.

Hmm, maybe I made a mistake in the derivation. Let me check again.

Starting over:

First equation: log_x(y^x)=10 ⇒ x^{10}=y^x.

Second equation: log_y(x^{4y})=10 ⇒ y^{10}=x^{4y}

From first equation: y = x^{10/x}

Substitute into second equation: (x^{10/x})^{10} =x^{4y}

So x^{100/x}=x^{4y} ⇒ 100/x =4y ⇒ y=25/x. So that's correct.

Therefore, y=25/x. So substituting back into first equation, x^{10}=(25/x)^x.

So x^{10} = (25^x)/(x^x). Multiply both sides by x^x:

x^{10} *x^x =25^x ⇒x^{x +10}=25^x.

Divide both sides by x^{x}:

x^{10}= (25^x)/(x^x)= (25/x)^x.

Wait, so x^{10}=(25/x)^x. Hmm, so x^{10}=(25/x)^x. Let me write this as (25/x)^x =x^{10}

Take both sides to the power 1/x:

25/x =x^{10/x}

Multiply both sides by x:

25 =x^{10/x +1}

Wait, so 25 =x^{1 +10/x}

Which is the same as before. Hmm. Let me set z = x. So 25 = z^{1 +10/z}

Take natural logs: ln25 = (1 +10/z) ln z

Let me denote f(z) = (1 +10/z) ln z. We need to find z>1 such that f(z)=ln25≈3.218.

Let me try z=5: f(5)= (1+2) ln5≈3*1.609≈4.828>3.218

z=10: (1+1) ln10≈2*2.302≈4.604>3.218

z=15: (1 +10/15) ln15≈(1.6667)*2.708≈4.413>3.218

z=20: (1+0.5) ln20≈1.5*2.995≈4.493>3.218

Hmm, still too big. Wait, maybe z=25?

(1 +10/25) ln25≈1.4*3.218≈4.505>3.218

Hmm, still higher. Wait, so as z increases, 10/z decreases, so 1 +10/z approaches 1, and ln z increases. So the product might have a minimum?

Wait, perhaps I need to try a larger z? Let me try z=50.

(1 +10/50) ln50 =1.2 * 3.912≈4.694. Still higher.

Hmm, z=100: (1 +0.1) ln100≈1.1*4.605≈5.065. Hmm, even higher. Wait, so as z increases, the function first increases? Wait, maybe I need to check derivative.

Alternatively, maybe trying smaller z?

Wait, z= sqrt(25)=5, but that was too big. Wait, let me try z= something like 3.

z=3: (1 +10/3) ln3≈(4.333)*1.098≈4.766>3.218

Hmm, z=4:

(1 +10/4) ln4= (1+2.5)*1.386≈3.5*1.386≈4.851>3.218

Hmm, z=1. Let me see z approaching 1 from the right. Let z=1.1:

(1 +10/1.1) ln1.1≈(1+9.0909)*0.0953≈10.0909*0.0953≈0.963. That's less than 3.218.

Ah! So at z=1.1, f(z)=0.963, which is less than 3.218, and at z=5, it's 4.828. So the function increases from z=1 to some point?

Wait, so between z=1.1 and z=5, the function increases from ~1 to ~4.8, so it must cross 3.218 somewhere between z=1.1 and z=5. Let me try z=2.

Wait, z=2 gives f(z)= (1+5)*ln2≈6*0.693≈4.158>3.218

Hmm, so between z=1.1 and z=2, the function goes from 0.96 to 4.15. So crosses 3.218 somewhere between z=1.5?

Wait, let me try z=1.5:

(1 +10/1.5) ln1.5≈(1+6.6667)*0.4055≈7.6667*0.4055≈3.111. Close to 3.218.

Hmm, 3.111 vs 3.218. So a bit lower. Let me try z=1.6:

(1 +10/1.6) ln1.6≈(1 +6.25)*0.4700≈7.25*0.4700≈3.4025. That's over 3.218.

So between z=1.5 and z=1.6, f(z) crosses 3.218.

Let me try z=1.55:

10/1.55≈6.4516, so 1+6.4516≈7.4516

ln(1.55)≈0.4383

Multiply: 7.4516 *0.4383≈3.266. Hmm, that's close to 3.218.

Hmm, z=1.55 gives ~3.266, which is a bit higher than 3.218.

Try z=1.53:

10/1.53≈6.5359, so 1+6.5359≈7.5359

ln(1.53)≈0.4245

Multiply: 7.5359 *0.4245≈3.200. That's very close to 3.218.

Hmm, 3.200 vs 3.218. Close. Let me try z=1.535:

10/1.535≈6.517, so 1+6.517≈7.517

ln(1.535)≈0.4273

Multiply: 7.517 *0.4273≈3.208. Hmm, getting closer.

z=1.54:

10/1.54≈6.4935, so total 7.4935

ln(1.54)= approx 0.4333

Multiply: 7.4935 *0.4333≈3.246. Hmm, over again.

Hmm, so between 1.53 and 1.54. Let me try z=1.535:

Wait, already did that. Maybe z=1.532:

10/1.532≈6.529, so total 7.529

ln(1.532)= approx 0.426

7.529 *0.426≈3.206. Hmm.

Hmm, this is getting tedious. Maybe there's an exact solution?

Wait, perhaps the equation can be manipulated to find an exact value?

We have x^{1 +10/x}=25. Let me suppose that 1 +10/x is an exponent that can be expressed as a logarithm. Alternatively, maybe x is 25^{something}?

Alternatively, let me suppose that x=5^k for some k. Let me try substituting x=5^k. Then y=25/x=25/(5^k)=5^{2 -k}

Let me see if this substitution helps.

First equation: x^{10}=y^x ⇒ (5^k)^{10}= (5^{2 -k})^{5^k}

Simplify left side:5^{10k}

Right side:5^{(2 -k)*5^k}

Therefore, exponents must be equal:

10k = (2 -k)*5^k

Hmm, so 10k = (2 -k)5^k. Let me see if this equation can be solved for k.

Trying k=1: 10*1= (2-1)*5^1 ⇒10=5. Not equal.

k=2: 10*2= (2-2)*5^2 ⇒20=0. Nope.

k=0.5: 10*(0.5)= (2 -0.5)*5^{0.5} ⇒5=1.5*sqrt(5)≈1.5*2.236≈3.354. Not equal.

k= something else. Let me try k= ?

Hmm, maybe k= ?

Alternatively, let me try k= log something. Not sure.

Alternatively, perhaps k=2?

Wait, tried that. Hmm.

Alternatively, maybe k= - something? But x must be greater than1, so 5^k>1 ⇒k>0.

Hmm, maybe k= 1/2?

Wait tried that. Hmm.

Alternatively, let me try k= log_5 (something). Hmm, not sure.

Alternatively, perhaps this approach isn't helpful. Maybe there's another way.

Wait, going back to the equations:

We have two equations:

1) x^{10} = y^x

2) y^{10}=x^{4y}

And we have y=25/x. Let me substitute y=25/x into the first equation:

x^{10} = (25/x)^x ⇒x^{10} =25^x /x^x ⇒x^{10}*x^x =25^x ⇒x^{x +10}=25^x

Let me write this as (x^{(x +10)/x}) =25 ⇒x^{1 +10/x}=25

Let me set t =10/x. Then, 10/x =t ⇒x=10/t. Substitute into the equation:

(10/t)^{1 + t} =25

Hmm, so (10/t)^{1 + t}=25. Let me see if this can be solved for t.

Take natural logs: (1 + t) ln(10/t)=ln25

Hmm, still complicated. Maybe guess a value for t?

Suppose t=2, so x=5. Then left side: (1+2) ln(10/2)=3 ln5≈3*1.609≈4.828 vs ln25≈3.218. Not equal.

t=4, so x=2.5. Then (1+4) ln(10/4)=5 ln(2.5)≈5*0.916≈4.58. Still too big.

t=5, x=2. (1+5) ln(10/5)=6 ln2≈6*0.693≈4.158. Still too big.

t=6, x≈1.6667. (1+6) ln(10/6)=7 ln(1.6667)≈7*0.5108≈3.576. Closer to 3.218.

t=7, x≈1.4286. (8) ln(10/7)=8*ln(1.4286)≈8*0.3567≈2.853. Now it's less than 3.218.

Hmm, so between t=6 and t=7. Let me try t=6.5, x≈10/6.5≈1.538.

Then (1+6.5)=7.5, ln(10/6.5)=ln(1.5385)≈0.432.

Multiply:7.5*0.432≈3.24. Close to 3.218.

Hmm, so t≈6.5 gives 3.24, which is close. Let me try t=6.6:

x=10/6.6≈1.5151

Then (1+6.6)=7.6, ln(10/6.6)=ln(1.5151)≈0.4155

Multiply:7.6*0.4155≈3.16. Hmm, that's under.

Hmm, so between t=6.5 and t=6.6. Let me try t=6.55:

x≈10/6.55≈1.5267

(1+6.55)=7.55, ln(10/6.55)=ln(1.5267)≈0.422

Multiply:7.55*0.422≈3.20. Close to 3.218.

Hmm, t=6.53:

x≈10/6.53≈1.5314

(1+6.53)=7.53, ln(10/6.53)=ln(1.5314)≈0.426

Multiply:7.53*0.426≈3.206. Hmm, very close.

Hmm, so t≈6.53, so x≈1.5314, then y=25/x≈25/1.5314≈16.33.

Wait, but this is getting into approximate values. The problem says "real numbers", so maybe the answer is 25? But when I tested x=5 and y=5, it didn't work, but maybe there's another solution where xy=25?

Wait, but according to the equations, we derived that y=25/x, so their product must be 25. Therefore, even if the individual values of x and y are not integers, their product is 25. Therefore, the answer is 25?

Wait, but in the problem statement, it says "there exist real numbers x and y, both greater than 1", so even if we can't find exact values for x and y, their product is fixed at 25. Therefore, maybe the answer is 25?

Wait, but let me confirm with the equations. Since we derived that y=25/x from the two equations, so regardless of the values of x and y, their product must be 25. Therefore, the answer is 25. So maybe I overcomplicated it by trying to find x and y?

Wait, let me see again:

We had the two equations:

1) x^{10}=y^x

2) y^{10}=x^{4y}

From equation 1, express y in terms of x: y =x^{10/x}

Substitute into equation 2:

(y)^10 = (x^{10/x})^{10}=x^{100/x}

And equation 2 says that equals x^{4y}, so:

x^{100/x}=x^{4y} ⇒100/x=4y ⇒ y=25/x.

Therefore, y=25/x, so xy=25. Therefore, regardless of the values of x and y, their product must be 25. Therefore, the answer is 25. So even if x and y are not integers, their product is 25. Therefore, the problem is designed so that you don't need to find x and y individually, just their product, which is 25.

Therefore, the answer is 25. I think I overcomplicated it by trying to solve for x and y. The key was to realize that substituting one equation into the other gives y=25/x, so product is 25.

Wait, but let me check with the approximate values I had before. If x≈1.53 and y≈16.33, then xy≈1.53*16.33≈25, which checks out. So even though x and y are not nice numbers, their product is exactly 25. Therefore, the problem is designed so that you can find xy without solving for x and y explicitly. So the answer is 25.

**Final Answer**
\boxed{25}
</think>

Given the equations \(\log_x(y^x) = 10\) and \(\log_y(x^{4y}) = 10\), we start by converting these logarithmic equations into their exponential forms.

For the first equation:
\[
\log_x(y^x) = 10 \implies x^{10} = y^x
\]

For the second equation:
\[
\log_y(x^{4y}) = 10 \implies y^{10} = x^{4y}
\]

We now have two equations:
1. \(x^{10} = y^x\)
2. \(y^{10} = x^{4y}\)

From the first equation, we express \(y\) in terms of \(x\):
\[
y = x^{10/x}
\]

Substituting this expression for \(y\) into the second equation:
\[
(x^{10/x})^{10} = x^{4y}
\]
\[
x^{100/x} = x^{4y}
\]

Since the bases are the same, we equate the exponents:
\[
\frac{100}{x} = 4y \implies y = \frac{25}{x}
\]

Thus, we find that \(y = \frac{25}{x}\). Therefore, the product \(xy\) is:
\[
xy = x \cdot \frac{25}{x} = 25
\]

Hence, the final answer is:
\[
\boxed{25}
\]