Question

Pyramid $OABCD$ has square base $ABCD,$ congruent edges $\overline{OA},$ $\overline{OB},$ $\overline{OC},$ and $\overline{OD},$ and $\angle AOB=45^\circ.$ Let $\theta$ be the measure of the dihedral angle formed by faces $OAB$ and $OBC.$ Given that $\cos\theta=m+\sqrt n,$ where $m$ and $n$ are integers, find $m+n.$

Answer

Answer: 5

Concepts:3D geometry coordinate geometry vector

Difficulty rating: 2450

Small Hint:

Place the square’s vertices at $(\pm1,\pm1,0)$ and the apex at $(0,0,h)$

Big Hint:

Find $h^2$ from $\angle AOB,$ then take the supplement of the angle between suitable face normals

Solution:

Take adjacent base vertices $A=(1,1,0),$ $B=(-1,1,0),$ $C=(-1,-1,0)$ and $O=(0,0,h).$ From $\angle AOB=45^\circ,$ $\begin{aligned}\frac{h^2}{h^2+2}&=\frac1{\sqrt2},\\h^2&=2+2\sqrt2.\end{aligned}$ Normals to the two faces may be taken as $(0,2h,2)$ and $(-2h,0,2).$ Their acute angle has cosine $\frac{1}{h^2+1}=3-2\sqrt2.$ The interior dihedral angle is its supplement, so $\cos\theta=2\sqrt2-3=-3+\sqrt8.$ Thus $m+n=-3+8=5.$

1995 AIME Problem 12

Answer: 5

Difficulty rating: 2450

Small Hint:

Big Hint:

Solution:

Problem 12 in Other Years