Maths Olympiad Prep

Track / Stage 5 / 15 of 400 #615 of 1964

Problem 615

AIME late
Geometry Difficulty 5.0 Find the answer

Example 1 In the tetrahedron ABCDA B C D, it is known that
AB=AC=AD=DB=5,BC=3,CD=4 A B=A C=A D=D B=5, B C=3, C D=4 \text {. }

Then the volume of the tetrahedron is \qquad .

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

Official solution

Given that BCD=90\angle B C D=90^{\circ}.
As shown in Figure 1, take the midpoint EE of BDB D and connect AEA E and CEC E.
By the properties of a right-angled triangle, we have
BE=CE=DE B E=C E=D E \text {. }
Since AB=AC=AD=DB=5A B=A C=A D=D B=5, we have
ABEACEADE\triangle A B E \cong \triangle A C E \cong \triangle A D E.
Thus, AEBD,AEECA E \perp B D, A E \perp E C.
Therefore, AEA E \perp plane BCDB C D, which means AEA E is the height of plane BCDB C D.
The calculation shows that
Vtetrahedron ABCD=13SBCDAE=13×6×532=53. \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 and solution reproduced as published; topic, difficulty and ordering added by this site.