Difference between revisions of "2021 AMC 12B Problems/Problem 15"
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==Problem== | ==Problem== | ||
The figure is constructed from <math>11</math> line segments, each of which has length <math>2</math>. The area of pentagon <math>ABCDE</math> can be written is <math>\sqrt{m} + \sqrt{n}</math>, where <math>m</math> and <math>n</math> are positive integers. What is <math>m + n ?</math> | The figure is constructed from <math>11</math> line segments, each of which has length <math>2</math>. The area of pentagon <math>ABCDE</math> can be written is <math>\sqrt{m} + \sqrt{n}</math>, where <math>m</math> and <math>n</math> are positive integers. What is <math>m + n ?</math> |
Revision as of 08:55, 2 March 2021
- The following problem is from both the 2021 AMC 10B #20 and 2021 AMC 12B #15, so both problems redirect to this page.
Contents
Problem
The figure is constructed from line segments, each of which has length . The area of pentagon can be written is , where and are positive integers. What is
Solution 1
Let be the midpoint of . Noting that and are triangles because of the equilateral triangles, . Also, and so .
Solution 2
Draw diagonals and to split the pentagon into three parts. We can compute the area for each triangle and sum them up at the end. For triangles and , they each have area . For triangle , we can see that and . Using Pythagorean Theorem, the altitude for this triangle is , so the area is . Adding each part up, we get .
Video Solution by OmegaLearn (Extending Lines, Angle Chasing, Trig Area)
~ pi_is_3.14
Video Solution by Hawk Math
https://www.youtube.com/watch?v=p4iCAZRUESs
Video Solution by TheBeautyofMath
https://youtu.be/FV9AnyERgJQ?t=1226
~IceMatrix
See Also
2021 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 14 |
Followed by Problem 16 |
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 |
2021 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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.