Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.0 AIME, harder Find the answer

9. In tetrahedron $\mathrm{ABCD}$ , the length of edge $\mathrm{AB}$ is $3 \mathrm{~cm}$ , the area of face $\mathrm{ABC}$ is $15 \mathrm{~cm}^{2}$ , and the area of face $\mathrm{ABD}$ is $12 \mathrm{~cm}^{2}$ . The angle between these two faces is $30^{\circ}$ . Find the volume of the tetrahedron (in $\mathrm{cm}^{3}$ ).

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

Let $\mathrm{V}$ be the volume of tetrahedron $\mathrm{ABCD}$ , and $\mathrm{h}$ be the height from $\mathrm{D}$ to the base $\mathrm{ABC}$ . Then, $\mathrm{V}=\frac{1}{3} \mathrm{hS} \triangle \mathrm{ABC}$ . To determine $V$ , we only need to determine $h$ . Draw $\mathrm{DK} \perp \mathrm{AB}$ at $\mathrm{K}$ , and connect $\mathrm{KH}$ . By the inverse of the three perpendiculars theorem, $\mathrm{HK} \perp \mathrm{AB}$ , thus $\angle \mathrm{DKH}=30^{\circ}$ .
From $\mathrm{S} \triangle A B D=\frac{1}{2} \mathrm{DK} \cdot \mathrm{AB}$ , we get
$\begin{aligned} \mathrm{DK} & =\frac{2 \mathrm{~S} \triangle A B D}{3}=8 . \\ \therefore \quad \mathrm{h} & =8 \sin 30^{\circ}=4, \\ \mathrm{~V} & =\frac{1}{3} 4 \times 15=20\left(\mathrm{~cm}^{3}\right) . \end{aligned}$

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.