7. Let $A B C D E$ be a square pyramid of height $\frac{1}{2}$ with square base $A B C D$ of side length $A B=12$ (so $E$ is the vertex of the pyramid, and the foot of the altitude from $E$ to $A B C D$ is the center of square $A B C D)$. The faces $A D E$ and $C D E$ meet at an acute angle of measure $\alpha$ (so that $0^{\circ}<\alpha<90^{\circ}$ ). Find $\tan \alpha$.
Answer: $\frac{17}{144}$ Let $X$ be the projection of $A$ onto $D E$. Let $b=A B=12$.
The key fact in this computation is that if $Y$ is the projection of $A$ onto face $C D E$, then the projection of $Y$ onto line $D E$ coincides with the projection of $A$ onto line $D E$ (i.e. $X$ as defined above). We compute $A Y=\frac{b}{\sqrt{b^{2}+1}}$ 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 $A X=2[A E D] / E D=\frac{b \sqrt{b^{2}+1} / 2}{\sqrt{2 b^{2}+1} / 2}$.
But $\alpha=\angle A X Y$, so from $\left(b^{2}+1\right)^{2}-\left(\sqrt{2 b^{2}+1}\right)^{2}=\left(b^{2}\right)^{2}$, it easily follows that $\tan \alpha=\frac{\sqrt{2 b^{2}+1}}{b^{2}}=\frac{17}{144}$.