Difference between revisions of "2021 AMC 10A Problems/Problem 21"
MRENTHUSIASM (talk | contribs) m (→Diagram) |
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so we get <math>PQ=16\sqrt3</math> and <math>YZ=36,</math> respectively. | so we get <math>PQ=16\sqrt3</math> and <math>YZ=36,</math> respectively. | ||
− | By the equilateral triangles | + | By the equilateral triangles, we find the perimeter of hexagon <math>ABCDEF:</math> |
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
AB+BC+CD+DE+EF+FA&=AZ+PC+CD+DQ+YF+FA \\ | AB+BC+CD+DE+EF+FA&=AZ+PC+CD+DQ+YF+FA \\ |
Revision as of 10:45, 5 October 2021
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
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 the equilateral triangles, we find the perimeter of hexagon Finally, the answer is
~sugar_rush (Fundamental Logic)
~MRENTHUSIASM (Reconstruction)
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.