Difference between revisions of "2021 AMC 12B Problems/Problem 15"
m (→Solution 2: Added a diagram) |
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==Solution 2== | ==Solution 2== | ||
+ | |||
+ | <asy> | ||
+ | /* Made by samrocksnature, adapted by Tucker */ | ||
+ | pair A=(-2.4638,4.10658); | ||
+ | pair B=(-4,2.6567453480756127); | ||
+ | pair C=(-3.47132,0.6335248637894945); | ||
+ | pair D=(-1.464483379039766,0.6335248637894945); | ||
+ | pair E=(-0.956630463955801,2.6567453480756127); | ||
+ | pair F=(-2,2); | ||
+ | pair G=(-3,2); | ||
+ | draw(A--B--C--D--E--A); | ||
+ | draw(A--C--A--D); | ||
+ | label("A",A,N); | ||
+ | label("B",B,W); | ||
+ | label("C",C,S); | ||
+ | label("D",D,S); | ||
+ | label("E",E,dir(0)); | ||
+ | dot(A^^B^^C^^D^^E); | ||
+ | dot(F^^G, gray); | ||
+ | draw(A--G--A--F, gray); | ||
+ | draw(B--F--C, gray); | ||
+ | draw(E--G--D, gray); | ||
+ | </asy> | ||
Draw diagonals <math>AC</math> and <math>AD</math> to split the pentagon into three parts. We can compute the area for each triangle and sum them up at the end. For triangles <math>ABC</math> and <math>ADE</math>, they each have area <math>2\cdot\frac{1}{2}\cdot\frac{4\sqrt{3}}{4}=\sqrt{3}</math>. For triangle <math>ACD</math>, we can see that <math>AC=AD=2\sqrt{3}</math> and <math>CD=2</math>. Using Pythagorean Theorem, the altitude for this triangle is <math>\sqrt{11}</math>, so the area is <math>\sqrt{11}</math>. Adding each part up, we get <math>2\sqrt{3}+\sqrt{11}=\sqrt{12}+\sqrt{11} \implies \boxed{\textbf{(D)} ~23}</math>. | Draw diagonals <math>AC</math> and <math>AD</math> to split the pentagon into three parts. We can compute the area for each triangle and sum them up at the end. For triangles <math>ABC</math> and <math>ADE</math>, they each have area <math>2\cdot\frac{1}{2}\cdot\frac{4\sqrt{3}}{4}=\sqrt{3}</math>. For triangle <math>ACD</math>, we can see that <math>AC=AD=2\sqrt{3}</math> and <math>CD=2</math>. Using Pythagorean Theorem, the altitude for this triangle is <math>\sqrt{11}</math>, so the area is <math>\sqrt{11}</math>. Adding each part up, we get <math>2\sqrt{3}+\sqrt{11}=\sqrt{12}+\sqrt{11} \implies \boxed{\textbf{(D)} ~23}</math>. |
Revision as of 22:05, 12 February 2021
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 .
~Lcz
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
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.