2021 AMC 12B Problems/Problem 15: Difference between revisions
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== Video Solution by OmegaLearn (Extending Lines, Angle Chasing, Trig Area) == | |||
https://youtu.be/QtSbAKUb1VE | |||
~ pi_is_3.14 | |||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2021|ab=B|num-b=14|num-a=16}} | {{AMC12 box|year=2021|ab=B|num-b=14|num-a=16}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 20:25, 11 February 2021
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 
![[asy] /* Made by samrocksnature */ 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--F--A--G); draw(B--F--C); draw(E--G--D); label("A",A,N); label("B",B,W); label("C",C,W); label("D",D,E); label("E",E,dir(0)); dot(A^^B^^C^^D^^E^^F^^G); [/asy]](http://latex.artofproblemsolving.com/2/9/0/290bd73a3648761d08879fa132a08542027b4484.png)
Solution
Let 
be the midpoint of 
. Noting that 
and 
are 
triangles because of the equilateral triangles, ![$AM=\sqrt{AD^2-MD^2}=\sqrt{12-1}=\sqrt{11} \implies [ACD]=\sqrt{11}$](http://latex.artofproblemsolving.com/8/c/1/8c1ec2de466d0d2c336d8b455f5b950a717543df.png)
. Also, ![$[AED]=2*2*\frac{1}{2}*\sin{120^o}=\sqrt{3}$](http://latex.artofproblemsolving.com/3/8/f/38f52258a27fcca9c04e078a4b4b385d19efca85.png)
and so ![$[ABCDE]=[ACD]+2[AED]=\sqrt{11}+2\sqrt{3}=\sqrt{11}+\sqrt{12} \implies \boxed{(\textbf{D})23}$](http://latex.artofproblemsolving.com/d/9/2/d92ddebc97a72b7708445d46f1b26acb91359545.png)
.
~Lcz
Video Solution by OmegaLearn (Extending Lines, Angle Chasing, Trig Area)
~ pi_is_3.14
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 | |
These problems are copyrighted © by the Mathematical Association of America. 
