2021 AMC 12B Problems/Problem 15
- The following problem is from both the 2021 AMC 10B #20 and 2021 AMC 12B #15, so both problems redirect to this page.
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Video Solution by OmegaLearn (Extending Lines, Angle Chasing, Trig Area)
- 5 Video Solution by Hawk Math
- 6 Video Solution by Mathematical Dexterity (Basic Area Formulas)
- 7 Video Solution by TheBeautyofMath
- 8 Video Solution by Interstigation (Ignore Useless Segments)
- 9 Video Solution by The Power of Logic
- 10 See Also
Problem
The figure is constructed from line segments, each of which has length . The area of pentagon can be written as , 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, (or split into two 30-60-90 triangles) 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 Mathematical Dexterity (Basic Area Formulas)
https://www.youtube.com/watch?v=7kDTlVixsD0
Video Solution by TheBeautyofMath
https://youtu.be/FV9AnyERgJQ?t=1226
~IceMatrix
Video Solution by Interstigation (Ignore Useless Segments)
~Interstigation
Video Solution by The Power of Logic
https://www.youtube.com/watch?v=f8L5K2yIXUc
~The Power of Logic
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.