1982 AHSME Problems/Problem 18

Problem 18

In the adjoining figure of a rectangular solid, $\angle DHG=45^\circ$ and $\angle FHB=60^\circ$ . Find the cosine of $\angle BHD$ .

$[asy] import three;defaultpen(linewidth(0.7)+fontsize(10)); currentprojection=orthographic(1/3+1/10,1-1/10,1/3); real r=sqrt(3); triple A=(0,0,r), B=(0,r,r), C=(1,r,r), D=(1,0,r), E=O, F=(0,r,0), G=(1,0,0), H=(1,r,0); draw(D--G--H--D--A--B--C--D--B--F--H--B^^C--H); draw(A--E^^G--E^^F--E, linetype("4 4")); label("$A$", A, N); label("$B$", B, dir(0)); label("$C$", C, N); label("$D$", D, W); label("$E$", E, NW); label("$F$", F, S); label("$G$", G, W); label("$H$", H, S); triple H45=(1,r-0.15,0.1), H60=(1-0.05, r, 0.07); label("$45^\circ$", H45, dir(125), fontsize(8)); label("$60^\circ$", H60, dir(25), fontsize(8));[/asy]$

$\text {(A)} \frac{\sqrt{3}}{6} \qquad \text {(B)} \frac{\sqrt{2}}{6} \qquad \text {(C)} \frac{\sqrt{6}}{3} \qquad \text{(D)}\frac{\sqrt{6}}{4}\qquad \text{(E)}\frac{\sqrt{6}-\sqrt{2}}{4}$

Solution

WLOG, let $CD=1$ .

Looking at square GHDC, we see that $\angle DHC=45$ , which implies that $DC=CH=1$ and $DH=\sqrt{2}$

Taking each cross-section one at a time, we look at square DHFB. We obviously know that CHB is a $30$ degree angle, giving $BH=\frac{2\sqrt{3}}{3}$ , and $BC=\frac{\sqrt{3}}{3}$ .

Looking at square ABCD, we see that $BD^2=1+\frac{1}{3}\implies BD^2=\frac{4}{3}=\frac{2\sqrt{3}}{3}$ .

We now look at triangle DBH, which has side lengths $BD=\frac{2\sqrt{3}}{3}$ , $BH=\frac{2\sqrt{3}}{3}$ and $DH=\sqrt{2}$ . Because $BD$ is opposite angle DHB, by the law of cosines, $\frac{4}{3}=\frac{4}{3}+2-2(\frac{2\sqrt{3}}{3})(\sqrt{2})\cos(\theta)\implies 2=\frac{4\sqrt{6}\cos(\theta)}{3} \implies \frac{2}{\frac{4\sqrt{6}}{3}}\implies \frac{6}{4\sqrt{6}} \implies \frac{\sqrt{6}}{4}$

Thus, the answer is $(\textbf{D})$

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1982 AHSME Problems/Problem 18

Problem 18

Solution