Problem 752

The height of the given pyramid passes through the center of the circle circumscribed around the base triangle.

## Solution

Let $D H$ be the height of the triangular pyramid $A B C D$, where $A D=B D=C D=269 / 32, A B=13, B C=14, A C=15$. Since $D H$ is perpendicular to the plane $A B C$, the segments $A H, B H$, and $C H$ are the projections of the equal oblique segments $A H, B H$, and $C H$ onto the plane $A B C$. Therefore, $A H=B H=C H$, meaning $H$ is the center of the circle circumscribed around triangle $A B C$. The area $S$ of triangle $A B C$ is found using Heron's formula:

$$S=\sqrt{21 \cdot(21-13)(21-14)(21-15)}=\sqrt{21 \cdot 8 \cdot 7 \cdot 6}=7 \cdot 3 \cdot 4=84$$

Next, we find the radius $R$ of the circle circumscribed around triangle $A B C$:

$$R=a b c /(4 S)=13 \cdot 14 \cdot 15 /(4 \cdot 84)=65 / 8$$

Using the Pythagorean theorem in the right triangle $A D H$, we find the height $D H$ of the pyramid:

$$\begin{gathered} D H=\sqrt{A D^{2}-A H^{2}}=\sqrt{A D^{2}-R^{2}}=\sqrt{(269 / 32)^{2}-(65 / 8)^{2}}= \\ =\sqrt{(269 / 32)^{2}-(260 / 32)^{2}}=\sqrt{269^{2}-260^{2}} / 32= \\ =\sqrt{(269-260)(269+260)} / 32=\sqrt{9 \cdot 529} / 32=3 \cdot 23 / 32=69 / 32 \end{gathered}$$

Therefore,

$$V(A B C D)=\frac{1}{3} S \cdot D H=\frac{1}{3} 84 \cdot 69 / 32=84 \cdot 23 / 32=21 \cdot 23 / 8=483 / 8$$