Difference between revisions of "1982 AHSME Problems/Problem 23"
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We apply the Law of Cosines to solve for <math>\cos\angle C:</math> <cmath>\cos(2\theta)=\frac{a^2+b^2-c^2}{2ab}=\frac{n-3}{2n}.</cmath> | We apply the Law of Cosines to solve for <math>\cos\angle C:</math> <cmath>\cos(2\theta)=\frac{a^2+b^2-c^2}{2ab}=\frac{n-3}{2n}.</cmath> | ||
− | By the Double-Angle Formula <math>\cos(2\theta)=2\cos^2\theta-1,</math> we have <cmath>2\left(\frac{n+5}{2(n+2)}\right)^2-1 | + | By the Double-Angle Formula <math>\cos(2\theta)=2\cos^2\theta-1,</math> we have <cmath>\frac{n-3}{2n}=2\left(\frac{n+5}{2(n+2)}\right)^2-1</cmath> from which <math>n=-3,-\frac12,4.</math> Recall that <math>n</math> is a positive integer, so <math>n=4.</math> By substitution, the answer is <math>\cos\theta=\boxed{\textbf{(A)}\ \frac{3}{4}}.</math> |
~MRENTHUSIASM | ~MRENTHUSIASM |
Revision as of 04:52, 15 September 2021
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
Solution 1 (Sines and Cosines)
In let
and
for some positive integer
We are given that
and we need
We apply the Law of Cosines to solve for
Let the brackets denote areas. We write in terms of
and
respectively:
Recall that
holds for all
Equating the last two expressions and then simplifying, we have
Equating the expressions for
we get
from which
By substitution, the answer is
~MRENTHUSIASM
Solution 2 (Cosines Only)
This solution uses the same variable definitions as Solution 1 does. Moreover, we conclude that from the second paragraph of Solution 1.
We apply the Law of Cosines to solve for
By the Double-Angle Formula
we have
from which
Recall that
is a positive integer, so
By substitution, the answer is
~MRENTHUSIASM
See Also
1982 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
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