1982 AHSME Problems/Problem 23

Problem

The lengths of the sides of a triangle are consecutive integers, and the largest angle is twice the smallest angle. The cosine of the smallest angle is

$\textbf{(A)}\ \frac{3}{4}\qquad \textbf{(B)}\ \frac{7}{10}\qquad \textbf{(C)}\ \frac{2}{3}\qquad \textbf{(D)}\ \frac{9}{14}\qquad \textbf{(E)}\ \text{none of these}$

Solution 1

In $\triangle ABC,$ let $a=n,b=n+1,c=n+2,$ and $\angle A=\theta$ for some positive integer $n.$ We are given that $\angle C=2\theta,$ and we need $\cos\theta.$

We apply the Law of Cosines to solve for $\cos\angle A:$ $\cos\theta=\frac{b^2+c^2-a^2}{2bc}=\frac{n+5}{2(n+2)}.$ We apply the Law of Cosines to solve for $\cos\angle C:$ $\cos(2\theta)=\frac{a^2+b^2-c^2}{2ab}=\frac{n-3}{2n}.$ By the Double-Angle Formula $\cos(2\theta)=2\cos^2\theta-1,$ we have $2\left(\frac{n+5}{2(n+2)}\right)^2-1=\frac{n-3}{2n},$ from which $n=-3,-\frac12,4.$ Recall that $n$ is a positive integer, so $n=4.$ By substitution, the answer is $\boxed{\textbf{(A)}\ \frac{3}{4}}.$

~MRENTHUSIASM

Solution 2

1982 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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1982 AHSME Problems/Problem 23

Contents

Problem

Solution 1

Solution 2

See Also