Difference between revisions of "2012 AMC 12B Problems/Problem 21"
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<math> \textbf{(A)}\ 29\sqrt{3} \qquad\textbf{(B)}\ \frac{21}{2}\sqrt{2}+\frac{41}{2}\sqrt{3} \qquad\textbf{(C)}\ \ 20\sqrt{3}+16 \qquad\textbf{(D)}\ 20\sqrt{2}+13 \sqrt{3} \qquad\textbf{(E)}\ 21\sqrt{6}</math> | <math> \textbf{(A)}\ 29\sqrt{3} \qquad\textbf{(B)}\ \frac{21}{2}\sqrt{2}+\frac{41}{2}\sqrt{3} \qquad\textbf{(C)}\ \ 20\sqrt{3}+16 \qquad\textbf{(D)}\ 20\sqrt{2}+13 \sqrt{3} \qquad\textbf{(E)}\ 21\sqrt{6}</math> | ||
+ | |||
+ | <asy> | ||
+ | size(200); | ||
+ | defaultpen(linewidth(1)); | ||
+ | pair A=origin,B=(2.5,0),C=B+2.5*dir(60), D=C+1.75*dir(120),E=D-(3.19,0),F=E-1.8*dir(60); | ||
+ | pair X=waypoint(B--C,0.345),Z=rotate(90,A)*X,Y=rotate(90,Z)*A; | ||
+ | draw(A--B--C--D--E--F--cycle); | ||
+ | draw(A--X--Y--Z--cycle,linewidth(0.9)+linetype("2 2")); | ||
+ | dot("$A$",A,W,linewidth(4)); | ||
+ | dot("$B$",B,dir(0),linewidth(4)); | ||
+ | dot("$C$",C,dir(0),linewidth(4)); | ||
+ | dot("$D$",D,dir(20),linewidth(4)); | ||
+ | dot("$E$",E,dir(100),linewidth(4)); | ||
+ | dot("$F$",F,W,linewidth(4)); | ||
+ | dot("$X$",X,dir(0),linewidth(4)); | ||
+ | dot("$Y$",Y,N,linewidth(4)); | ||
+ | dot("$Z$",Z,W,linewidth(4)); | ||
+ | </asy> | ||
+ | |||
+ | (diagram from AoPS forum) | ||
==Solution== | ==Solution== |
Revision as of 15:55, 30 December 2016
Problem 21
Square is inscribed in equiangular hexagon with on , on , and on . Suppose that , and . What is the side-length of the square?
(diagram from AoPS forum)
Solution
Extend and so that they meet at . Then , so and therefore is parallel to . Also, since is parallel and equal to , we get , hence is congruent to . We now get .
Let , , and .
Drop a perpendicular line from to the line of that meets line at , and a perpendicular line from to the line of that meets at , then is congruent to since is complementary to . Then we have the following equations:
The sum of these two yields that
So, we can now use the law of cosines in :
Therefore
See Also
2012 AMC 12B (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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.