Difference between revisions of "2022 AMC 12B Problems/Problem 10"
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==Solution== | ==Solution== | ||
− | Consider triangle <math>AFG</math>. <math> | + | Consider triangle <math>AFG</math>. Note that <math>AF = 2</math>, <math>AG = 1</math>, and <math>\angle GAF = 120 ^{\circ}</math> because it is an interior angle of a regular hexagon.<sup>See note for details.</sup> |
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+ | By the [[Law of Cosines]], we have: | ||
<cmath>\begin{align*} | <cmath>\begin{align*} |
Revision as of 18:31, 17 November 2022
Problem
Regular hexagon has side length . Let be the midpoint of , and let be the midpoint of . What is the perimeter of ?
Solution
Consider triangle . Note that , , and because it is an interior angle of a regular hexagon.See note for details.
By the Law of Cosines, we have:
By SAS Congruence, triangles , , , and are congruent, and by CPCTC, quadrilateral is a rhombus. Therefore, its perimeter is .
Note: The sum of the interior angles of any polygon with sides is given by . Therefore, the sum of the interior angles of a hexagon is , and each interior angle of a regular hexagon measures .
See Also
2022 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 9 |
Followed by Problem 11 |
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 |
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