Difference between revisions of "1962 AHSME Problems/Problem 20"

(Solution)
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==Solution==
 
==Solution==
{{solution}}
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If the angles are in an arithmetic progression, they can be expressed as
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<math>a</math>, <math>a+n</math>, <math>a+2n</math>, <math>a+3n</math>, and <math>a+4n</math> for some real numbers <math>a</math> and <math>n</math>.
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Now we know that the sum of the degree measures of the angles of a pentagon is <math>180(5-2)=540</math>.
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Adding our expressions for the five angles together, we get <math>5a+10n=540</math>.
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We now divide by 5 to get <math>a+2n=108</math>. It so happens that <math>a+2n</math> is one of the angles we defined earlier, so that angle must have a measure of <math>\boxed{108\textbf{ (A)}}</math>.
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(In fact, for any arithmetic progression with an odd number of terms,
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the middle term is equal to the average of all the terms.)

Revision as of 16:11, 16 April 2014

Problem

The angles of a pentagon are in arithmetic progression. One of the angles in degrees, must be:

$\textbf{(A)}\ 108\qquad\textbf{(B)}\ 90\qquad\textbf{(C)}\ 72\qquad\textbf{(D)}\ 54\qquad\textbf{(E)}\ 36$

Solution

If the angles are in an arithmetic progression, they can be expressed as $a$, $a+n$, $a+2n$, $a+3n$, and $a+4n$ for some real numbers $a$ and $n$. Now we know that the sum of the degree measures of the angles of a pentagon is $180(5-2)=540$. Adding our expressions for the five angles together, we get $5a+10n=540$. We now divide by 5 to get $a+2n=108$. It so happens that $a+2n$ is one of the angles we defined earlier, so that angle must have a measure of $\boxed{108\textbf{ (A)}}$. (In fact, for any arithmetic progression with an odd number of terms, the middle term is equal to the average of all the terms.)

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