Difference between revisions of "2002 AMC 12B Problems/Problem 5"
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\qquad\mathrm{(E)}\ 120</math> | \qquad\mathrm{(E)}\ 120</math> | ||
| − | == Solution == | + | == Solution 1 == |
The sum of the degree measures of the angles of a pentagon (as a pentagon can be split into <math>5- 2 = 3</math> triangles) is <math>3 \cdot 180 = 540^{\circ}</math>. If we let <math>v = x - 2d, w = x - d, y = x + d, z = x+2d</math>, it follows that | The sum of the degree measures of the angles of a pentagon (as a pentagon can be split into <math>5- 2 = 3</math> triangles) is <math>3 \cdot 180 = 540^{\circ}</math>. If we let <math>v = x - 2d, w = x - d, y = x + d, z = x+2d</math>, it follows that | ||
<cmath>(x-2d)+(x-d)+x+(x+d)+(x+2d) = 5x = 540 \Longrightarrow x = 108 \ \mathrm{(D)}</cmath> | <cmath>(x-2d)+(x-d)+x+(x+d)+(x+2d) = 5x = 540 \Longrightarrow x = 108 \ \mathrm{(D)}</cmath> | ||
| − | Note that since <math>x</math> is the middle term of an arithmetic sequence with an odd number of terms, it is simply the average of the sequence. | + | Note that since <math>x</math> is the middle term of an arithmetic sequence with an odd number of terms, it is simply the average of the sequence. |
| + | |||
| + | == Solution 2 == | ||
| + | Assume that the pentagon is regular, with the angles is an arithmetic sequence with common difference <math>0</math>. Then <math>x</math>, and any other angle for that matter, will be <math>108</math>. | ||
== See also == | == See also == | ||
Revision as of 22:30, 13 May 2018
Contents
Problem
Let 
and 
be the degree measures of the five angles of a pentagon. Suppose that 
and and form an arithmetic sequence. Find the value of 
.
Solution 1
The sum of the degree measures of the angles of a pentagon (as a pentagon can be split into 
triangles) is 
. If we let 
, it follows that
![\[(x-2d)+(x-d)+x+(x+d)+(x+2d) = 5x = 540 \Longrightarrow x = 108 \ \mathrm{(D)}\]](http://latex.artofproblemsolving.com/2/a/0/2a03fa9f8fc2b8f56860f65f81e93bfc3b933f99.png)
Note that since is the middle term of an arithmetic sequence with an odd number of terms, it is simply the average of the sequence.
Solution 2
Assume that the pentagon is regular, with the angles is an arithmetic sequence with common difference 
. Then , and any other angle for that matter, will be 
.
See also
| 2002 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 4 |
Followed by Problem 6 |
| 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 | |
| All AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
