Difference between revisions of "1982 AHSME Problems/Problem 23"
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MRENTHUSIASM (talk | contribs) m (→Solution 1 (Law of Sines and Law of Cosines)) |
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\textbf{(E)}\ \text{none of these} </math> | \textbf{(E)}\ \text{none of these} </math> | ||
− | == Solution 1 == | + | == Solution 1 (Law of Sines and Law of Cosines) == |
In <math>\triangle ABC,</math> let <math>a=n,b=n+1,c=n+2,</math> and <math>\angle A=\theta</math> for some positive integer <math>n.</math> We are given that <math>\angle C=2\theta,</math> and we need <math>\cos\theta.</math> | In <math>\triangle ABC,</math> let <math>a=n,b=n+1,c=n+2,</math> and <math>\angle A=\theta</math> for some positive integer <math>n.</math> We are given that <math>\angle C=2\theta,</math> and we need <math>\cos\theta.</math> | ||
− | We apply the Law of Cosines to solve for <math>\cos\ | + | We apply the Law of Cosines to solve for <math>\cos\theta:</math> <cmath>\cos\theta=\frac{b^2+c^2-a^2}{2bc}=\frac{n+5}{2(n+2)}.</cmath> |
− | + | We apply the Law of Sines to <math>\angle A</math> and <math>\angle C:</math> <cmath>\frac{\sin\theta}{a}=\frac{\sin(2\theta)}{c}.</cmath> | |
− | + | By the Double-Angle Formula <math>\sin(2\theta)=2\sin\theta\cos\theta,</math> we simplify and rearrange to solve for <math>\cos\theta:</math> <cmath>\cos\theta=\frac{c}{2a}=\frac{n+2}{2n}.</cmath> | |
− | + | We equate the expressions for <math>\cos\theta:</math> <cmath>\frac{n+5}{2(n+2)}=\frac{n+2}{2n},</cmath> | |
− | |||
from which <math>n=4.</math> By substitution, the answer is <math>\cos\theta=\boxed{\textbf{(A)}\ \frac{3}{4}}.</math> | from which <math>n=4.</math> By substitution, the answer is <math>\cos\theta=\boxed{\textbf{(A)}\ \frac{3}{4}}.</math> | ||
~MRENTHUSIASM | ~MRENTHUSIASM | ||
− | == Solution 2 == | + | == Solution 2 (Law of Cosines Only) == |
This solution uses the same variable definitions as Solution 1 does. Moreover, we conclude that <math>\cos\theta=\frac{n+5}{2(n+2)}</math> from the second paragraph of Solution 1. | This solution uses the same variable definitions as Solution 1 does. Moreover, we conclude that <math>\cos\theta=\frac{n+5}{2(n+2)}</math> from the second paragraph of Solution 1. | ||
− | We apply the Law of Cosines to solve for <math>\cos\ | + | We apply the Law of Cosines to solve for <math>\cos(2\theta):</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 | + | By the Double-Angle Formula <math>\cos(2\theta)=2\cos^2\theta-1,</math> we set up an equation for <math>n:</math> <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 |
Latest revision as of 23:29, 15 September 2021
Contents
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 (Law of Sines and Law of Cosines)
In let
and
for some positive integer
We are given that
and we need
We apply the Law of Cosines to solve for
We apply the Law of Sines to
and
By the Double-Angle Formula
we simplify and rearrange to solve for
We equate the expressions for
from which
By substitution, the answer is
~MRENTHUSIASM
Solution 2 (Law of 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 set up an equation for
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.