2021 AMC 10A Problems/Problem 21
Contents
Problem
Let be an equiangular hexagon. The lines and determine a triangle with area , and the lines and determine a triangle with area . The perimeter of hexagon can be expressed as , where and are positive integers and is not divisible by the square of any prime. What is ?
Diagram
~MRENTHUSIASM
Solution 1
Let and be the intersections and respectively.
The sum of the interior angles of any hexagon is Since hexagon is equiangular, each of its interior angles is By angle chasing, we conclude that the interior angles of and are all Therefore, these triangles are all equilateral triangles, from which and are both equilateral triangles.
We are given that so we get and respectively.
By equilateral triangles and segment addition, we find the perimeter of hexagon Finally, the answer is
~sugar_rush ~MRENTHUSIASM
Solution 2
Let the length Then, we have We get We want and it follows that Finally, the answer is
~mathboy282
Solution 3
Since it is an equiangular hexagon, each interior angle measures 120 degrees, and by angle chasing you can conclude that all the triangles are equilateral. Since the area formula for an equiangular triangle is s^2 sqrt(3)/4, the side lengths of the triangles are 36 and 16 sqrt(3). In the above diagram the sum of the lengths of AF, AB, and FE is 36 and the sum of the lengths of BC, CD, and DE is 16sqrt(3). Since these are also the side lengths of the hexagon, the answer is 36+16+3=55 (C).
~NamyaB.
Video Solution by OmegaLearn (Angle Chasing and Equilateral Triangles)
~ pi_is_3.14
Video Solution by TheBeautyofMath
~IceMatrix
Video Solution by MRENTHUSIASM (English & Chinese)
https://www.youtube.com/watch?v=0n8EAu2VAiM
~MRENTHUSIASM
See Also
2021 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
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 10 Problems and Solutions |
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