A tetrahedron has two faces that are equilateral triangles with unit side length, and two faces that are isosceles right triangles. What is the volume of the tetrahedron?
Let's construct the network of the test. It is easy to see that the hypotenuse of a right-angled triangle cannot be a unit length (1. figure), because then the length of the leg would be $\frac{\sqrt{2}}{2}$, and a regular triangle with a unit side cannot fit this. Therefore, the sought solid consists of $2$ regular and $2$ right-angled triangles with unit legs (2. figure).

Let the base be the right-angled triangle $A B C$. The projection of the fourth vertex $D$ onto the base plane is $D^{\prime}$ (3. figure). The right-angled triangles $D D^{\prime} A, D D^{\prime} B, D D^{\prime} C$ are congruent, as their hypotenuse is exactly the height of the solid; thus, $D^{\prime} A=D^{\prime} B=D^{\prime} C$. The point $D^{\prime}$ is equidistant from the points $A, B, C$, and since $A B C$ is a right-angled triangle, $D^{\prime}$ is the center of the Thales circle circumscribed around the triangle $A B C$. Therefore, $D^{\prime} A=D^{\prime} B=D^{\prime} C=\frac{\sqrt{2}}{2}$, and the right-angled triangle $D D^{\prime} B$ is also isosceles. Hence, the height of the tetrahedron is $\frac{\sqrt{2}}{2}$.

The volume of the tetrahedron: $V=\frac{A_{t} \cdot m}{3}=\frac{1}{3} \cdot \frac{1}{2} \cdot \frac{\sqrt{2}}{2}=\frac{\sqrt{2}}{12}$ cubic units.

Remark. Several people noticed that the given tetrahedron is exactly one quarter of a regular unit-sided octahedron, and thus its volume can be easily written down.
![](https://cdn.mathpix.com/cropped/2024_05_02_a5a3ec234a842547f5d4g-1.jpg?height=1596&width=624&top_left_y=800&top_left_x=772)

![](https://cdn.mathpix.com/cropped/2024_05_02_a5a3ec234a842547f5d4g-2.jpg?height=491&width=507&top_left_y=134&top_left_x=801)