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Difference between revisions of "1981 AHSME Problems/Problem 21"

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<math>\textbf{(A)}\ 15^\circ\qquad\textbf{(B)}\ 30^\circ\qquad\textbf{(C)}\ 45^\circ\qquad\textbf{(D)}\ 60^\circ\qquad\textbf{(E)}\ 150^\circ</math>
 
<math>\textbf{(A)}\ 15^\circ\qquad\textbf{(B)}\ 30^\circ\qquad\textbf{(C)}\ 45^\circ\qquad\textbf{(D)}\ 60^\circ\qquad\textbf{(E)}\ 150^\circ</math>
  
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==Solution==
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==Solution 1==
  
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First notice that exchanging <math>a</math> for <math>b</math> in the original equation must also work. Therefore, <math>a=b</math>. Replacing <math>b</math> for <math>a</math> and expanding/simplifying in the original equation yields <math>4a^2-c^2=3a^2</math>, or <math>a^2=c^2</math>. Since <math>a</math> and <math>c</math> are positive, <math>a=c</math>. Therefore, we have an equilateral triangle and the angle opposite <math>c</math> is just 45^\circ\qquad\textbf{(D)}\.
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We will try to solve for a possible value of the variables. First notice that exchanging <math>a</math> for <math>b</math> in the original equation must also work. Therefore, <math>a=b</math> works. Replacing <math>b</math> for <math>a</math> and expanding/simplifying in the original equation yields <math>4a^2-c^2=3a^2</math>, or <math>a^2=c^2</math>. Since <math>a</math> and <math>c</math> are positive, <math>a=c</math>. Therefore, we have an equilateral triangle and the angle opposite <math>c</math> is just <math>\textbf{(D)}\ 60^\circ\qquad</math>.
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==Solution 2==
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<cmath>(a+b+c)(a+b-c)=3ab</cmath>
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<cmath>a^2+2ab+b^2-c^2=3ab</cmath>
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<cmath>a^2+b^2-c^2=ab</cmath>
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<cmath>c^2=a^2+b^2-ab</cmath>
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This looks a lot like Law of Cosines, which is <math>c^2=a^2+b^2-2ab\cos{c}</math>.
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<cmath>c^2=a^2+b^2-ab=a^2+b^2-2ab\cos{c}</cmath>
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<cmath>ab=2ab\cos{c}</cmath>
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<cmath>\frac{1}{2}=\cos{c}</cmath>
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<math>\cos{c}</math> is <math>\frac{1}{2}</math>, so the angle opposite side <math>c</math> is <math>\boxed{60^\circ}</math>.
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-aopspandy
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==See also==
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{{AHSME box|year=1981|num-b=20|num-a=22}}
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{{MAA Notice}}

Latest revision as of 14:13, 28 June 2025

Problem 21

In a triangle with sides of lengths $a$, $b$, and $c$, $(a+b+c)(a+b-c) = 3ab$. The measure of the angle opposite the side length $c$ is

$\textbf{(A)}\ 15^\circ\qquad\textbf{(B)}\ 30^\circ\qquad\textbf{(C)}\ 45^\circ\qquad\textbf{(D)}\ 60^\circ\qquad\textbf{(E)}\ 150^\circ$

Solution 1

We will try to solve for a possible value of the variables. First notice that exchanging $a$ for $b$ in the original equation must also work. Therefore, $a=b$ works. Replacing $b$ for $a$ and expanding/simplifying in the original equation yields $4a^2-c^2=3a^2$, or $a^2=c^2$. Since $a$ and $c$ are positive, $a=c$. Therefore, we have an equilateral triangle and the angle opposite $c$ is just $\textbf{(D)}\ 60^\circ\qquad$.

Solution 2

\[(a+b+c)(a+b-c)=3ab\] \[a^2+2ab+b^2-c^2=3ab\] \[a^2+b^2-c^2=ab\] \[c^2=a^2+b^2-ab\] This looks a lot like Law of Cosines, which is $c^2=a^2+b^2-2ab\cos{c}$. \[c^2=a^2+b^2-ab=a^2+b^2-2ab\cos{c}\] \[ab=2ab\cos{c}\] \[\frac{1}{2}=\cos{c}\] $\cos{c}$ is $\frac{1}{2}$, so the angle opposite side $c$ is $\boxed{60^\circ}$.

-aopspandy

See also

1981 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
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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