1962 AHSME Problems/Problem 20

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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1962 AHSME Problems/Problem 20

Problem

Solution