## Task A-3.4.

Let $A B C D$ be a tetrahedron where $\varangle B A C=\varangle C A D=\varangle D A B=90^{\circ}$, $|A D|=2 \sqrt{2}$, and $|A B|=|A C|=3$. Determine the radius of the sphere inscribed in this tetrahedron.
## Solution.

Let $P(X Y Z)$ denote the area of triangle $X Y Z$.

Let $V$ be the volume of the tetrahedron, $O$ be the surface area, and $r$ be the radius of the inscribed sphere.

We will use the known formula $V=\frac{O r}{3}$.

Let $P$ be the midpoint of segment $\overline{B C}$. Since triangle $A B C$ is an isosceles right triangle, we have $|B C|=3 \sqrt{2}$ and $|A P|=\frac{3 \sqrt{2}}{2}$. By the Pythagorean theorem in the right triangle $A P D$,

$$
|D P|=\sqrt{|A D|^{2}+|A P|^{2}}=\sqrt{8+\frac{9}{2}}=\frac{5 \sqrt{2}}{2}
$$

It follows that $P(B C D)=\frac{1}{2} \cdot|B C| \cdot|D P|=\frac{1}{2} \cdot 3 \sqrt{2} \cdot \frac{5 \sqrt{2}}{2}=\frac{15}{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_30_5bba13a74780d91c8bdbg-19.jpg?height=737&width=791&top_left_y=337&top_left_x=567)

Since $P(A B C)=P(A C D)=3 \sqrt{2}$ and $P(A B C)=\frac{9}{2}$, and we have calculated that $P(B C D)=\frac{15}{2}$, it follows that the surface area of the tetrahedron is $O=6 \sqrt{2}+12$.

The volume of the tetrahedron is $V=\frac{1}{3} \cdot 3 \cdot 3 \cdot 2 \sqrt{2}=6 \sqrt{2}$.

From the formula $V=\frac{O r}{3}$, it follows that $r=\frac{3 V}{O}=\frac{18 \sqrt{2}}{6 \sqrt{2}+12}=\frac{3 \sqrt{2}-3}{2}$.

Note: The length $|D P|$ can also be calculated as follows. By the Pythagorean theorem in triangle $A C D$, we have $|C D|=\sqrt{|A C|^{2}+|A D|^{2}}=\sqrt{9+8}=\sqrt{17}$.

By the Pythagorean theorem in triangle $P C D$,

$$
|D P|=\sqrt{|C D|^{2}-|P C|^{2}}=\sqrt{17-\frac{9}{2}}=\frac{5 \sqrt{2}}{2}
$$

Note: If the student does not know the formula $V=\frac{O r}{3}$, it can be derived as follows.

Let $I$ be the center of the inscribed sphere of the tetrahedron $A B C D$. The tetrahedron $A B C D$ can be decomposed into tetrahedra $A B C I, A B D I, B C D I$, and $A C D I$. Each of these 4 tetrahedra has a base that is a face of the tetrahedron $A B C D$, and the height is $r$. Therefore, it follows that

$$
\begin{aligned}
V(A B C D) & =V(A B C I)+V(A B D I)+V(B C D I)+V(A C D I) \\
& =\frac{1}{3} r P(A B C)+\frac{1}{3} r P(A B D)+\frac{1}{3} r P(B C D)+\frac{1}{3} r P(A C D)=\frac{1}{3} r O
\end{aligned}
$$