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

(Solution 2)
Line 14: Line 14:
 
<cmath>a^2+b^2-c^2=ab</cmath>
 
<cmath>a^2+b^2-c^2=ab</cmath>
 
<cmath>c^2=a^2+b^2-ab</cmath>
 
<cmath>c^2=a^2+b^2-ab</cmath>
This looks a lot like Law of Cosines, which is <math>c^2=a^2+b^2-2ab\cosc</math>
+
This looks a lot like Law of Cosines, which is <math>c^2=a^2+b^2-2ab\cos{c}</math>.
<cmath>c^2=a^2+b^2-ab=a^2+b^2-2ab\cosc</cmath>
+
<cmath>c^2=a^2+b^2-ab=a^2+b^2-2ab\cos{c}</cmath>
<cmath>ab=2ab\cosc</cmath>
+
<cmath>ab=2ab\cos{c}</cmath>
<cmath>\frac{1}{2}=\cosc</cmath>
+
<cmath>\frac{1}{2}=\cos{c}</cmath>
<math>\cosc</math> is <math>\frac{1}{2}, so the angle opposite side </math>c<math> is </math>\boxed{60^\circ}$
+
<math>\cos{c}</math> is <math>\frac{1}{2}</math>, so the angle opposite side <math>c</math> is <math>\boxed{60^\circ}</math>
  
 
-aopspandy
 
-aopspandy

Revision as of 19:20, 18 June 2021

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

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

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