2021 AMC 10A Problem 13

Test the three faces that meet at $A$ with the converse of the Pythagorean theorem:
$$AB^2 + AC^2 = 4 + 9 = 13 = BC^2,\quad AB^2 + AD^2 = 4 + 16 = 20 = BD^2,\quad AC^2 + AD^2 = 9 + 16 = 25 = CD^2.$$
So the angles $\angle BAC$, $\angle BAD$, $\angle CAD$ are all right angles: the edges $AB$, $AC$, $AD$ are mutually perpendicular, like three edges of a box meeting at a corner.

Take face $ABC$ as the base; it is a right triangle of area $\frac12 \cdot 2 \cdot 3 = 3$. Since $AD$ is perpendicular to both $AB$ and $AC$, it is perpendicular to the base plane, so the height is $AD = 4$. Then
$$V = \frac13 \cdot 3 \cdot 4 = 4.$$

The answer is $\boxed{\textbf{(C)}\ 4}$.

Whenever a tetrahedron's six edges are given, first look for right angles: a "corner of a box" tetrahedron with perpendicular edges $a, b, c$ at one vertex has volume $\frac{abc}{6}$. The given lengths $\sqrt{13}$, $2\sqrt5 = \sqrt{20}$, and $5 = \sqrt{25}$ are the hypotenuses of the legs $(2,3)$, $(2,4)$, $(3,4)$ respectively.

Place $A$ at the origin, $B = (2,0,0)$, $C = (0,3,0)$, $D = (0,0,4)$; all six distances match the given ones, and the volume of a corner tetrahedron is $\frac{2 \cdot 3 \cdot 4}{6} = 4$.

• Using (1/3)(base area)(height) with the wrong height, or forgetting the 1/6 factor for a corner tetrahedron (answer 24 or 12).
• Not recognizing the right angles and attempting a Heron-style or Cayley-Menger computation, which wastes most of the available time.