Difference between revisions of "1962 AHSME Problems/Problem 20"
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| − | {{ | + | If the angles are in an arithmetic progression, they can be expressed as |
| + | <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>. | ||
| + | Now we know that the sum of the degree measures of the angles of a pentagon is <math>180(5-2)=540</math>. | ||
| + | Adding our expressions for the five angles together, we get <math>5a+10n=540</math>. | ||
| + | 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>. | ||
| + | (In fact, for any arithmetic progression with an odd number of terms, | ||
| + | the middle term is equal to the average of all the terms.) | ||
Revision as of 15:11, 16 April 2014
Problem
The angles of a pentagon are in arithmetic progression. One of the angles in degrees, must be:
Solution
If the angles are in an arithmetic progression, they can be expressed as
, 
, 
, 
, and 
for some real numbers and 
.
Now we know that the sum of the degree measures of the angles of a pentagon is 
.
Adding our expressions for the five angles together, we get 
.
We now divide by 5 to get 
. It so happens that is one of the angles we defined earlier, so that angle must have a measure of 
.
(In fact, for any arithmetic progression with an odd number of terms,
the middle term is equal to the average of all the terms.)
