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All faces of the triangular pyramid are equal isosceles triangles, and the height of the pyramid coincides with the height of one of its lateral faces. Find the volume of the pyramid if the distance between the farthest opposite edges is 1.
Let the height $D H$ of the pyramid $A B C D$ with vertex $D$ be the height of the lateral face $A D C$. Suppose (Fig.1) that $A C=A D=b, C D=a(a \neq b$, since otherwise $A B C D$ is a regular tetrahedron). Then, from the equality of all faces of the pyramid, it follows that $B D=B C=b$ and the height of the isosceles triangle $B C D$, dropped to the lateral side $B C$, is equal to the height $D H$ of the isosceles triangle $A C D$, dropped to the lateral side $A C$, which is impossible, since $D H$ is perpendicular to the plane of the base of the pyramid, while $D P$ is inclined to this plane. Thus, $D H$ is the height of the isosceles triangle $A D C$, dropped to the base. Let $D A=D C=a, A C=b$ (Fig.2). Then $A B=B C=a$ and $B D=A C=b$. By the Pythagorean theorem,

$$
B H=D H=\sqrt{A D^{2}-A H^{2}}=\sqrt{a^{2}-\frac{b^{2}}{1}}
$$

and since $D H \perp B H$, then

$$
b=B D=B H \cdot \sqrt{2}=\sqrt{a^{2}-\frac{b^{2}}{1}} \cdot \sqrt{2}=\sqrt{2 a^{2}-\frac{b^{2}}{2}} .
$$

From the equality $b=\sqrt{2 a^{2}-\frac{b^{\frac{2}{2}}}{2}}$ we find that $b 2=\frac{\frac{1}{3}}{} a 2$. Therefore, $b>a$, meaning the distance between the largest opposite edges is the distance between $A C$ and $B D$.

Let $M$ be the midpoint of $B D$. Then $H M$ is the height and median of the isosceles triangle $B H D$, and since $H M$ lies in a plane perpendicular to $A C$, $H M$ is the common perpendicular of the lines $B D$ and $A C, H M=\frac{1}{2} B D=\frac{1}{2} b$. According to the problem, $H M=1$, so $b=2$. Then

$$
a=\frac{\sqrt{3}}{2} b=\sqrt{3}, D H=B H=M H \sqrt{2}=\sqrt{2}
$$

Therefore,

$$
V_{A B C D}=\frac{1}{3} S_{\triangle A B C} \cdot D H=\frac{1}{3} \cdot \frac{1}{2} A C \cdot B H \cdot D H=\frac{1}{6} \cdot \frac{1}{2} \cdot 2 \cdot \sqrt{2} \cdot \sqrt{2}=\frac{2}{3}
$$

Answer

## $\frac{2}{3} \cdot$