, 2025

Solution 1: Translate $\mathcal{P}$ so that it contains the center. The intersection of the translated plane with the rectangular prism is a centrally symmetric hexagon. Let its side lengths be $a$, $b$, $c$, $a$, $b$, and $c$, in that order. Then, for some $t_{a}$, $t_{b}$, and $t_{c}$, the side lengths of the hexagon before the translation were
$$(a - t_{a}, b + t_{b}, c - t_{c}, a + t_{a}, b - t_{b}, c + t_{c}) = (45, 66, 63, 55, 54, 77),$$
from which it follows that $t_{a} = 5$, $t_{b} = 6$, and $t_{c} = 7$.
Now, a translation of a plane can be written in one of three equivalent forms: it can be viewed as a translation in the $x$ direction by a distance $d_{x}$, a translation in the $y$ direction by a distance $d_{y}$, or a translation in the $z$ direction by a distance $d_{z}$ (with coordinate axes chosen as shown below).

As shown above, we can express $t_{a}$, $t_{b}$, and $t_{c}$ in terms of $d_{x}$, $d_{y}$, and $d_{z}$ using the Pythagorean theorem, which yields $t_{a} = \sqrt{d_{y}^{2} + d_{z}^{2}}$, $t_{b} = \sqrt{d_{z}^{2} + d_{x}^{2}}$, and $t_{c} = \sqrt{d_{x}^{2} + d_{y}^{2}}$. Hence,
$$(d_{x}^{2}, d_{y}^{2}, d_{z}^{2}) = \left(\frac{5^{2} + 6^{2} - 7^{2}}{2}, \frac{6^{2} + 7^{2} - 5^{2}}{2}, \frac{7^{2} + 5^{2} - 6^{2}}{2}\right) = (6, 30, 19).$$
We can draw a right pyramid with legs $d_{x}$, $d_{y}$, and $d_{z}$ which has the center of the prism as a vertex with opposite face on $\mathcal{P}$. Then, the height of this pyramid, or the distance from the center to $\mathcal{P}$, is
$$\sqrt{\frac{1}{d_{x}^{-2} + d_{y}^{-2} + d_{z}^{-2}}} = \sqrt{\frac{1}{6^{-1} + 30^{-1} + 19^{-1}}} = \sqrt{\frac{95}{24}};$$

Solution 2: Let the vertices of the hexagon be $ABCDEF$, where $AB = 45$, $BC = 66$, etc. Note that $AB \parallel DE$, $BC \parallel EF$, and $CD \parallel FA$. Let $O$ be the center of the prism, and let $M$, $N$, and $P$ be the midpoints of $AD$, $BE$, and $CF$, respectively.

The key observation is that $MN$ is the midline between $AB$ and $DE$. Hence, plane $OMN$ is the midplane between the faces of the prism containing sides $AB$ and $DE$. Similarly, planes $OMP$ and $ONP$ are the other two midplanes of the prism. Thus, $OM$, $ON$, and $OP$ are mutually orthogonal.

Observe
$$MN = \frac{|AB - DE|}{2} = 5, \quad NP = \frac{|BC - EF|}{2} = 6, \quad \text{and} \quad PM = \frac{|CD - FA|}{2} = 7,$$
so by Heron's formula, we can compute the area of $MNP$ to be $\sqrt{9(9 - 5)(9 - 6)(9 - 7)} = 6\sqrt{6}$.

Moreover, if $x = OM$, $y = ON$, and $z = OP$, then,
$$x^{2} + y^{2} = 5^{2}, \quad y^{2} + z^{2} = 6^{2}, \quad \text{and} \quad z^{2} + x^{2} = 7^{2}.$$
Solving this system of equations gives $x = \sqrt{19}$, $y = \sqrt{6}$, and $z = \sqrt{30}$. Therefore, if $d$ is the distance from $O$ to plane $MNP$ (i.e., the answer), the volume of tetrahedron $OMNP$ can be written as
$$\frac{1}{6} \cdot \sqrt{19} \cdot \sqrt{6} \cdot \sqrt{30} = \frac{1}{3} \cdot (6\sqrt{6}) \cdot d,$$
SO
$$d = \frac{\sqrt{19 \cdot 6 \cdot 30}}{2 \cdot 6 \sqrt{6}} = \sqrt{\frac{95}{24}}.$$