HMMT February, 2015

Geometry Difficulty 5.8 AIME, harder Prove it United States

Problem:

Let $ABCDE$ be a square pyramid of height $\frac{1}{2}$ with square base $ABCD$ of side length $AB=12$ (so $E$ is the vertex of the pyramid, and the foot of the altitude from $E$ to $ABCD$ is the center of square $ABCD$ ). The faces $ADE$ and $CDE$ meet at an acute angle of measure $\alpha$ (so that $0^{\circ}<\alpha<90^{\circ}$ ). Find $\tan \alpha$ .

Solution

Solution:

$\boxed{\frac{17}{144}}$

Let $X$ be the projection of $A$ onto $DE$ . Let $b=AB=12$ .

The key fact in this computation is that if $Y$ is the projection of $A$ onto face $CDE$ , then the projection of $Y$ onto line $DE$ coincides with the projection of $A$ onto line $DE$ (i.e., $X$ as defined above).

We compute $AY=\frac{b}{\sqrt{b^{2}+1}}$ by looking at the angle formed by the faces and the square base (via $1/2 - b/2 - \sqrt{b^{2}+1}/2$ right triangle).

Now we compute $AX=2[ AED ] / ED = \frac{b \sqrt{b^{2}+1}/2}{\sqrt{2b^{2}+1}/2}$ .

But $\alpha=\angle AXY$ , so from $\left(b^{2}+1\right)^{2}-\left(\sqrt{2b^{2}+1}\right)^{2}=\left(b^{2}\right)^{2}$ , it easily follows that

$tan \alpha = \frac{\sqrt{2b^{2}+1}}{b^{2}} = \frac{17}{144}.$

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.