[Perpendicularity of lines and planes] [Maximum/minimum distance]

In tetrahedron $ABCD$, the dihedral angles at edges $AB$, $AC$, and $BD$ are right angles. One of the segments connecting the midpoints of opposite edges of the tetrahedron has a length of $a$, and the other has a length of $a \sqrt{6}$. Find the length of the longest edge of the tetrahedron.
Planes of faces $ABD$ and $ACD$ are perpendicular to the plane of face $ABC$ and intersect along the line $AD$, which means that the edge $AD$ is perpendicular to the plane of face $ABC$. Therefore, $\angle BAD = \angle CAD = 90^\circ$. Planes of faces $ABC$ and $BCD$ are perpendicular to the plane of face $ABD$ and intersect along the line $BC$, which means that the edge $BC$ is perpendicular to the plane of face $ABD$. Therefore, $\angle CBD = \angle ABC = 90^\circ$. The segment $CD$ is the common hypotenuse of the right triangles $ACD$ and $BCD$, so $CD > AC, CD > AC, CD > BD, CD > BC$, and since $AC$ is the hypotenuse of the right triangle $ABC$, then $CB > AC > AB$. Consequently, $CD$ is the largest edge of the tetrahedron $ABCD$. Let $K, L, M$, and $N$ be the midpoints of the edges $BD, CD, AC$, and $AB$ respectively. Segments $KL$ and $MN$ are the midlines of triangles $BCD$ and $ABC$, so $KL \parallel BC, KL = \frac{1}{2} BC, MN \parallel BC, MN = \frac{1}{2} BC$. Therefore, quadrilateral $KLMN$ is a parallelogram, and since $AD \perp BC$ and $KN \parallel AD$, then $KN \perp KL$. Therefore, $KLMN$ is a rectangle. Consequently, its diagonals $KM$ and $LN$ are equal. Let $P$ and $Q$ be the midpoints of the edges $AD$ and $BC$ respectively. Denote $BC = x, AD = y$. Then $KL = \frac{x}{2}, KN = \frac{y}{2}$. Suppose $KM = NL = a \sqrt{6}$. Then $PQ = a$

$$
\begin{gathered}
. \text { From the right triangle } KLN \text { we find that } \frac{x^2}{4} + \frac{y^2}{4} = 6a^2 \cdot \text { Therefore, } \\
a^2 = PQ^2 = AP^2 + AQ^2 = AP^2 + BQ^2 + AB^2 = \frac{y^2}{4} + \frac{x^2}{4} + AB^2 = 6a^2 + AB^2 > a^2, \\
\text { which is impossible. Therefore, } KM = LN = a \text { and } PQ = a \sqrt{6}. \text { Then }
\end{gathered}
$$

$$
\frac{x^2}{4} + \frac{y^2}{4} = a^2, x^2 + y^2 = 4a^2
$$

$$
\begin{gathered}
6a^2 = PQ^2 = AP^2 + AQ^2 = AP^2 + (BQ^2 + AB^2) = \frac{y^2}{4} + \frac{x^2}{4} + AB^2 = a^2 + AB^2, \\
AB^2 = 6a^2 - a^2 = 5a^2. \\
\text { Therefore, } \\
CD^2 = AD^2 + AC^2 = AD^2 + (BC^2 + AB^2) = y^2 + x^2 + 5a^2 = 4a^2 + 5a^2 = 9a^2.
\end{gathered}
$$

Therefore, $CD = 3a$.

## Answer

$3a$.

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