1982 AMC 12 Problem 18

Question

In the adjoining figure of a rectangular solid,

\angle DHG=45^\circ

and

\angle FHB=60^\circ.

Find the cosine of

\angle BHD.

\frac{\sqrt3}{6}

\frac{\sqrt2}{6}

\frac{\sqrt6}{3}

\frac{\sqrt6}{4}

\frac{\sqrt6-\sqrt2}{4}

Assign coordinates at

H

along the three mutually perpendicular edges

Translate the two given angles into relationships among the three edge lengths, then use a dot product

Let

H=(0,0,0),

F=(u,0,0),

G=(0,v,0),

and

C=(0,0,w).

Then

B=(u,0,w)

and

D=(0,v,w).

The

60^\circ

condition gives

w=\sqrt3u,

while the

45^\circ

condition gives

v=w.

Hence

\begin{aligned} \cos\angle BHD &=\frac{w^2} {\sqrt{u^2+w^2}\sqrt{2w^2}}\\ &=\frac{\sqrt6}{4}. \end{aligned}

Therefore, the correct answer is D.

Problem 18 in other years