1962 AHSME Problems/Problem 20

Revision as of 22:18, 3 October 2014 by Timneh (talk | contribs) (Solution)

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.)


See Also

1962 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
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