[A sphere touching the edges or sides of a pyramid] [Extending a tetrahedron to a parallelepiped]

All edges of the triangular pyramid ABCD touch a certain sphere. The three segments connecting the midpoints of the skew edges $AB$ and $CD, AC$ and $BD, AD$ and $BC$ are equal to each other, $\angle ABC=100^{\circ}$. Find the ratio of the heights dropped from vertices A and B.
Let each of the segments connecting the midpoints of opposite edges of the tetrahedron $ABCD$ be equal to $a$. Extend the given tetrahedron to the parallelepiped $AKBLNDMC$ ($AN \| KD \| BM \| LC$), by drawing through its opposite edges pairs of parallel planes (Fig.1). The segments connecting the midpoints of opposite edges of the tetrahedron $ABCD$ are respectively equal and parallel to the edges of the parallelepiped $AKBLNDMC$, hence, all faces of this parallelepiped are rhombuses with all sides equal to $a$. By the theorem on the sum of the squares of the diagonals of a parallelogram,

$$
AB^2 + CD^2 = 4a^2, \quad AC^2 + BD^2 = 4a^2, \quad AD^2 + BC^2 = 4a^2
$$

Therefore,

$$
AB^2 + CD^2 = AC^2 + BD^2 = AD^2 + BC^2
$$

Let the specified sphere touch the edges $AB, CD, AD$, and $BC$ of the tetrahedron $ABCD$ at points $E, F, G$, and $H$ respectively (Fig.2). Then

$$
\begin{gathered}
AB + CD = (AE + BE) + (CF + DF) = (AG + BH) + (CH + DG) = \\
= (AG + DG) + (CH + BH) = AD + BC.
\end{gathered}
$$

Similarly, $AC + BD = AB + CD$ and $AD + BC = AB + CD$. Consider the equalities

$$
AB^2 + CD^2 = AC^2 + BD^2, \quad AB + CD = AC + BD.
$$

Square both sides of the second equality and subtract the first from the result. We get

$$
AB^2 + CD^2 = AC^2 + BD^2, \quad 2AB \cdot CD = 2AC \cdot BD.
$$

Thus, $(AB - CD)^2 = (AC - BD)^2$, from which

$$
AB - CD = AC - BD \quad \text{or} \quad AB - CD = BD - AC,
$$

and since $AB + CD = AC + BD$, in the first case $AB = AC$ and $CD = BD$, which is impossible, since $\angle ABC = 100^\circ > 90^\circ$, and in the second case $-AB = BD$ and $CD = AC$. Similarly, from the equalities

$$
AB^2 + CD^2 = AD^2 + BC^2, \quad AB + CD = AD + BC
$$

we get that $AB = AD$ and $CD = BC$ (this case is impossible, since then from the equalities $CD = AC$ and $CD = BC$ it follows that $AC = BC$) or $AB = BC$ and $CD = AD$. Thus, we have proved that

$$
AB = BD, \quad CD = AC, \quad AB = BC, \quad CD = AD
$$

therefore,

$$
AB = BD = BC, \quad AC = AD = CD,
$$

i.e., $ABCD$ is a regular triangular pyramid with vertex $B$ (Fig.3). Let $h$ be its height dropped from vertex $A$ to the base of the lateral face $BCD$, and $H$ be the height dropped from vertex $B$ to the base $ACD$; $K$ is the midpoint of $CD$. Then

![](https://cdn.mathpix.com/cropped/2024_05_06_6ba7643fcbc3b9c222e2g-17.jpg?height=91&width=666&top_left_y=1984&top_left_x=696)

## Answer

$\sqrt{3} \operatorname{tg} 50^\circ$.