AIME 2016 I · 第 4 题

Solution 1

Let $B$ and $C$ be the vertices adjacent to $A$ on the same base as $A$, and let $D$ be the last vertex of the triangular pyramid. Then $\angle CAB = 120^\circ$. Let $X$ be the foot of the altitude from $A$ to $\overline{BC}$. Then since $\triangle ABX$ is a $30-60-90$ triangle, $AX = 6$. Since the dihedral angle between $\triangle ABC$ and $\triangle BCD$ is $60^\circ$, $\triangle AXD$ is a $30-60-90$ triangle and $AD = 6\sqrt{3} = h$. Thus $h^2 = \boxed{108}$.

~gundraja

Solution 2

Let $B$ and $C$ be the vertices adjacent to $A$ on the same base as $A$, and let $D$ be the last vertex of the triangular pyramid. Notice that we can already find some lengths. We have $AB=AC=12$ (given) and $BC=BD=\sqrt{144+h^2}$ by the Pythagorean Theorem. Let $M$ be the midpoint of $BC$. Then, we have $AM=6$ ($30-60-90$) triangles and $DM=\sqrt{36+h^2}$ by the Pythagorean Theorem. Applying the Law of Cosines, since $\angle AMD=60^{\circ}$, we get

$$h^2=36+h^2+36-\frac12 \cdot 12 \sqrt{36+h^2} \implies h^2=\boxed{108},$$ as desired.

-A1001