2022 AMC 12B Problems/Problem 10
Contents
Problem
Regular hexagon has side length . Let be the midpoint of , and let be the midpoint of . What is the perimeter of ?
Diagram
~MRENTHUSIASM
Solution 1
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 .
~mathboy100
Solution 2
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 .
Solution 3
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 .
~sirswagger21
Note: the last part of this solution could have been simplified by noting that
Solution 4
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)
Solution 5 (Answer Choices + Pythagorean Theorem Extension)
Like the previous solutions, note that and are all congruent by SAS. It follows that quadrilateral is a rhombus.
Recall the Pythagorean Theorem, which states for all right triangles, where is the hypotenuse of the triangle. However, by drawing a quick diagram of an obtuse triangle, we can see that , in any given obtuse triangle.
Since is a regular hexagon, all of its angles are obtuse. It follows that is an obtuse triangle. Using the extended Pythagorean Theorem for obtuse triangles, we have:
Since is a rhombus, the perimeter is . This eliminates all answer choices but and , since in all of those options . Lastly, is eliminated due to the triangle inequality, as is not greater than .
Hence, the answer is .
~SwordOfJustice
Video Solution 1 by mop 2024
https://youtu.be/ezGvZgBLe8k&t=0s
~r00tsOfUnity
Video Solution 2 (Under 1 min!)
~Education, the Study of Everything
Video Solution 3 by SpreadTheMathLove
https://www.youtube.com/watch?v=6I3ZNpI7qwE
Video Solution(1-16)
~~Hayabusa1
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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