8.1. Points $K, L, M, N$ are the centers of the circles inscribed in the faces $S A B, S A C$, $S B C$ and $A B C$ of the tetrahedron $S A B C$. It is known that $A B=S C=5, A C=S B=7, B C=S A=8$. Find the volume of the tetrahedron $K L M N$. Round your answer to the nearest hundredth if necessary.
Answer: 0.66 (exact value: $\frac{\sqrt{11}}{5}$).

Solution. We will solve the problem in a general form. Let $BC = SA = a$, $AC = SB = b$, $AB = SC = c$. The faces of the tetrahedron $SABC$ are equal triangles (such a tetrahedron is called equifacial). Draw a plane through each edge of the tetrahedron, parallel to the opposite edge (see Fig. 2). These six planes uniquely determine a parallelepiped, called the accompanying parallelepiped for the tetrahedron $SABC$. The edges of the tetrahedron are the diagonals of the faces of the parallelepiped, and the edges of the parallelepiped are equal to the segments connecting the centers of the opposite faces of the parallelepiped (the midpoints of the skew edges of the tetrahedron). Since the opposite edges of the tetrahedron are equal, the diagonals of the parallel faces of the accompanying parallelepiped are also equal, so its faces are rectangles, and the parallelepiped itself is rectangular. Denoting the lengths of the edges of the parallelepiped by $x, y, z$, we get $x^2 + y^2 = a^2$, $y^2 + z^2 = b^2$, $z^2 + x^2 = c^2$. Adding these three equations, we find $x^2 + y^2 + z^2 = \frac{a^2 + b^2 + c^2}{2}$, from which $x^2 = \frac{a^2 + c^2 - b^2}{2}$, $y^2 = \frac{b^2 + c^2 - a^2}{2}$, $z^2 = \frac{a^2 + b^2 - c^2}{2}$. Therefore, the volume of the accompanying parallelepiped is $V = xyz = \frac{\sqrt{2}}{4} \sqrt{(a^2 + c^2 - b^2)(b^2 + c^2 - a^2)(a^2 + b^2 - c^2)}$, and the volume of the original equifacial tetrahedron is $V_{SABC} = \frac{1}{3} V = \frac{\sqrt{2}}{12} \sqrt{(a^2 + c^2 - b^2)(b^2 + c^2 - a^2)(a^2 + b^2 - c^2)}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_72d2011c4e3388928cb1g-5.jpg?height=455&width=554&top_left_y=683&top_left_x=134)

Fig. 1

![](https://cdn.mathpix.com/cropped/2024_05_06_72d2011c4e3388928cb1g-5.jpg?height=494&width=560&top_left_y=644&top_left_x=751)

Fig. 2

![](https://cdn.mathpix.com/cropped/2024_05_06_72d2011c4e3388928cb1g-5.jpg?height=462&width=537&top_left_y=674&top_left_x=1388)

Fig. 3

From the problem statement, it follows that the tetrahedron $KLMN$ is also equifacial. Express the lengths of the edges of the parallelepiped accompanying the tetrahedron $KLMN$ in terms of the lengths $x, y, z$ of the edges of the parallelepiped accompanying the tetrahedron $SABC$. Denote (see Fig. 1) by $Q$ and $R$ the midpoints of the edges $SC$ and $AB$, and by $F$ and $G$ the feet of the angle bisectors of the faces $SAB$ and $CAB$, respectively. Let $l$ be the line passing through points $Q$ and $R$. Since $l$ is perpendicular to the faces of the accompanying parallelepiped (see Fig. 2), $l \perp SC$ and $l \perp AB$. Rotate the tetrahedron $SABC$ by $180^\circ$ around the line $l$. In this case, the tetrahedra $SABC$ and $KLMN$ (and the parallelepiped accompanying $SABC$) will map onto themselves. Since the vertices $M$ and $L$ will map to each other, the segment $ML$ is perpendicular to the line $l$ and intersects it at its midpoint - point $D$. Similarly, we get $KN \perp l$, $E = KN \cap l$, $KE = EN$. Also note that the edges $SC$ and $AB$ are used equivalently in the reasoning, so $QD = RE$.

Let $\Pi$ be the plane passing through point $Q$ and perpendicular to the edge $SC$. Consider the orthogonal projection of the tetrahedron $SABC$ onto the plane $\Pi$ (see Fig. 3). Points $A', B', F', G', K', N'$ are the orthogonal projections of points $A, B, F, G, K, N$ onto the plane $\Pi$, point $Q$ is the orthogonal projection of points $S$ and $C$, the segment $QR$ lies in the plane $\Pi$, and since $R$ is the midpoint of $AB$, $R$ is the midpoint of $A'B'$, and similarly, $E$ is the midpoint of $K'N'$, and $R$ is the midpoint of $F'G'$.

Point $K$, being the center of the circle inscribed in triangle $SAB$, divides the bisector $SF$ in the ratio $\frac{FK}{FS} = \frac{c}{a+b+c}$. Since the ratio of lengths is preserved under orthogonal projection, $\frac{FK}{FS} = \frac{F'K'}{F'S'}$, and since $KN \perp QR$ and $AB \perp QR$, $K'N' \perp QR$ and $A'B' \perp QR$. Therefore, $K'N' \| A'B'$ and $\frac{RE}{RQ} = \frac{F'K'}{F'S'} = \frac{FK}{FS} = \frac{c}{a+b+c}$. Therefore, the edge $DE$ of the parallelepiped accompanying the tetrahedron $KLMN$ is $DE = RQ - RE - DQ = RQ - 2RE = z\left(1 - \frac{2c}{a+b+c}\right) = z \frac{a+b-c}{a+b+c}$. Similarly, considering the pairs of edges $SA, BC$ and $SB, AC$, we get that the other edges of the parallelepiped accompanying the tetrahedron $KLMN$ are $\frac{b+c-a}{a+b+c}$ and $x \frac{a+c-b}{a+b+c}$, respectively. Therefore, the volume of the tetrahedron $KLMN$ is $V_{KLMN} = \frac{(b+c-a)(a+c-b)(a+b-c)}{(a+b+c)^3} V_{SABC} = \frac{\sqrt{2}}{12} \cdot \frac{(b+c-a)(a+c-b)(a+b-c)}{(a+b+c)^3} \sqrt{(a^2 + c^2 - b^2)(b^2 + c^2 - a^2)(a^2 + b^2 - c^2)}$. For $a = 8$, $b = 7$, $c = 5$, we get $V_{KLMN} = \frac{\sqrt{11}}{5} \approx 0.66$.

Remark. The problem can be solved in other ways as well. For example, to find the lengths of the edges of the tetrahedron $KLMN$, one can proceed as follows: find the lengths of the bisectors $SA_1$ and $SC_1$ of triangles $BSC$ and $ASB$, respectively, as well as the lengths of the segments $BA_1$ and $BC_1$. Then find the angle $B$ in triangle $ABC$, and then the segment $A_1C_1$ in triangle $BA_1C_1$. In triangle $SA_1C_1$, use the cosine theorem to find the angle $S$ and, knowing in what ratio points $K$ and $L$ divide the bisectors $SA_1$ and $SC_1$, find the segments $SK$ and $SL$. Finally, from triangle $SKL$, find the length $KL$. The other edges of the tetrahedron $KLMN$ are found similarly.