2022 AMC 12B Problems/Problem 10
Regular hexagon has side length . Let be the midpoint of , and let be the midpoint of . What is the perimeter of ?
Let the center of the hexagon be . , , , , , and are all equilateral triangles with side length . Thus, , and . By symmetry, . Thus, by the Pythagorean theorem, . Because and , . Thus, the solution to our problem is .
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, the perimeter of 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 .
We use a coordinates approach. Letting the origin be the center of the hexagon, we can let Then, and
We use the distance formula four times to get Thus, the perimeter of .
Note: the last part of this solution could have been simplified by noting that
Note that triangles and are all congruent, since they have side lengths of and and an included angle of
By the Law of Cosines, Therefore,
-Benedict T (countmath1)
Video Solution 1
~Education, the Study of Everything
Video Solution 2 by SpreadTheMathLove
|2022 AMC 12B (Problems • Answer Key • Resources)|
|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.