A triangular pyramid is bounded by 4 congruent triangles. What is the volume of the pyramid if the sides of each triangle are $5 \mathrm{~cm}, 6 \mathrm{~cm}$, and $7 \mathrm{~cm}$?
I. solution. From the apex $D$ of the pyramid, we draw the height $A D_{1}=m$, then we drop perpendiculars to the base sides $b=6$ and $c=7$, which cut off segments $A E=x_{1}$ and $A F=x_{2}$ from the common vertex $A$. Since

$$
49-\left(6-x_{1}\right)^{2}=25-x_{1}^{2} \quad \text { and } \quad 36-\left(7-x_{2}\right)^{2}=25-x_{2}^{2}
$$

it follows that

$$
x_{1}=1 \mathrm{~cm} \quad \text { and } \quad x_{2}=\frac{19}{7} \mathrm{~cm}
$$

If $D_{1} E=y_{1} \quad D_{1} F=y_{2} \quad$ and $A D_{1}=n$,

then

$$
x_{1}^{2}+y_{1}^{2}=x_{2}^{2}+y_{2}^{2}=n^{2}
$$

Using Carnot's theorem to calculate the cosine of the angle $\angle BAC=\alpha$, we get

$$
\cos \alpha=\frac{5}{7}
$$

Or if

$$
\cos \alpha=\cos \alpha_{1} \cos \alpha_{2}-\sin \alpha_{1} \sin \alpha_{2}
$$

then

$$
\frac{5}{7}=\frac{x_{1} x_{2}-y_{1} y_{2}}{n^{2}}
$$

From equations (1) and (2), we get

$$
y_{1}^{2}=\frac{49}{6}, y_{2}^{2}=\frac{529}{294}, n^{2}=\frac{55}{6}
$$

Thus

$$
{\overline{D D_{1}}}^{2}=m^{2}=5^{2}-n^{2}
$$

from which

$$
m=\sqrt{\frac{95}{6}}
$$

and

$$
V=\frac{b c \sin \alpha}{2} \cdot \frac{m}{3}=\frac{6 \cdot 7 \cdot \frac{2}{7} \cdot \sqrt{6} \cdot \sqrt{\frac{95}{6}}}{2 \cdot 3}=2 \sqrt{95}=19.4936 \mathrm{~cm}^{3}
$$

(Ernó Kiss, Budapest.)

II. solution. If the edges of the pyramid are $a, b, c$ and the angles of the individual triangles are $\alpha, \beta, \gamma$, then the volume of the pyramid is: (Ábel, Solid Geometry, 4th edition, 124.)

$$
V=\frac{a b c}{3} \sqrt{\sin \sigma \sin (\sigma-\alpha) \sin (\sigma-\beta) \sin (\sigma-\gamma)}
$$

or

$$
V=\frac{a b c}{3} \sqrt{\cos \alpha \cos \beta \cos \gamma}
$$

Substituting the values, we again get:

$$
V=19.4936 \mathrm{~cm}^{3}
$$

(Nándor Ehrenfeld, Nyitra.)