Track / Stage 5 / 27 of 400 #627 of 1964

Problem 627

AIME late

Geometry Difficulty 5.0 Find the answer

In the tetrahedron $A B C D$ , it is known that
$A B=A C=A D=D B=5, B C=3, C D=4 \text {. }$

Then the volume of the tetrahedron is $\qquad$ .

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Official solution

Given that $\angle B C D=90^{\circ}$ .
As shown in Figure 1, take the midpoint $E$ of $B D$ and connect $A E$ and $C E$ .
By the properties of a right-angled triangle, we have
$B E=C E=D E \text {. }$
Since $A B=A C=A D=D B=5$ , we have
$\triangle A B E \cong \triangle A C E \cong \triangle A D E$ .
Thus, $A E \perp B D, A E \perp E C$ .
Therefore, $A E \perp$ plane $B C D$ , which means $A E$ is the height of plane $B C D$ .
The calculation shows that
$\begin{array}{l} V_{\text {tetrahedron }ABCD}=\frac{1}{3} S_{\triangle B C D} \cdot A E \\ =\frac{1}{3} \times 6 \times \frac{5 \sqrt{3}}{2}=5 \sqrt{3} . \end{array}$

Source: NuminaMath-1.5, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.