1962 AHSME Problems/Problem 20

Revision as of 20:00, 29 October 2020 by Shadow-18 (talk | contribs) (Solution)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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.)

Solution 2

If we write the five terms as $a$, $a - n$, $a - 2n$, $a + n$ and $a + 2n$, we can see that adding them up, we get $5a = 540$ through this, we can see that $a = 108$, $\fbox{\textbf{(A)}}$

See Also

1962 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
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 31 32 33 34 35 36 37 38 39 40
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS