Difference between revisions of "2002 AMC 12B Problems/Problem 5"
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== Solution == | == Solution == | ||
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+ | === Solution 1 === | ||
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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 | ||
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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. | ||
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+ | You can always assume the values are the same so <math>\frac{540}{5}=108</math> | ||
+ | |||
+ | === Solution 2 === | ||
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+ | Let <math>v</math>, <math>w</math>, <math>x</math>, <math>y</math>, <math>z</math> be <math>v</math>, <math>v + d</math>, <math>v+2d</math>, <math>v+3d</math>, <math>v+4d</math>, respectively. Then we have <cmath>v + w + x + y + z = 5v + 10d = 180^{\circ} (5 - 2) = 540^{\circ}</cmath> | ||
+ | Dividing the equation by <math>5</math>, we have <cmath>v + 2d = x = 108^{\circ} \mathrm {(D)}</cmath> | ||
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+ | ~ Nafer | ||
== See also == | == See also == |
Latest revision as of 05:58, 8 August 2023
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
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
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.
You can always assume the values are the same so
Solution 2
Let ,
,
,
,
be
,
,
,
,
, respectively. Then we have
Dividing the equation by
, we have
~ Nafer
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.