Difference between revisions of "2012 AMC 12B Problems/Problem 21"
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− | ==Problem== | + | ==Problem 21== |
Square <math>AXYZ</math> is inscribed in equiangular hexagon <math>ABCDEF</math> with <math>X</math> on <math>\overline{BC}</math>, <math>Y</math> on <math>\overline{DE}</math>, and <math>Z</math> on <math>\overline{EF}</math>. Suppose that <math>AB=40</math>, and <math>EF=41(\sqrt{3}-1)</math>. What is the side-length of the square? | Square <math>AXYZ</math> is inscribed in equiangular hexagon <math>ABCDEF</math> with <math>X</math> on <math>\overline{BC}</math>, <math>Y</math> on <math>\overline{DE}</math>, and <math>Z</math> on <math>\overline{EF}</math>. Suppose that <math>AB=40</math>, and <math>EF=41(\sqrt{3}-1)</math>. What is the side-length of the square? | ||
− | <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</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 by djmathman) | |
− | Extend <math>AF</math> and <math>YE</math> so that they meet at <math>G</math>. Then <math>\angle FEG=\angle GFE=60^{\circ}</math>, so <math>\angle FGE=60^{\circ}</math> and | + | ==Solution== |
+ | |||
+ | Extend <math>AF</math> and <math>YE</math> so that they meet at <math>G</math>. Then <math>\angle FEG=\angle GFE=60^{\circ}</math>, so <math>\angle FGE=60^{\circ}</math> and because <math>AB</math> is parallel to <math>YE</math>. Also, since <math>AX</math> is parallel and equal to <math>YZ</math>, we get <math>\angle BAX = \angle ZYE</math>, hence <math>\triangle ABX</math> is congruent to <math>\triangle YEZ</math>. We now get <math>YE=AB=40</math>. | ||
Let <math>a_1=EY=40</math>, <math>a_2=AF</math>, and <math>a_3=EF</math>. | Let <math>a_1=EY=40</math>, <math>a_2=AF</math>, and <math>a_3=EF</math>. | ||
Line 35: | Line 52: | ||
Therefore <math>AZ = 29\sqrt{3} ... \framebox{A}</math> | Therefore <math>AZ = 29\sqrt{3} ... \framebox{A}</math> | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | First, we want to angle chase. Set <math><YXC</math> equal to <math>a</math> degrees. | ||
+ | |||
+ | Now the key idea is that you want to relate the numbers that you have. You know <math>\overline{AB} = 40</math> and that <math>\overline{EZ} + \overline{ZF} = 41(\sqrt{3}-1)</math>. We proceed with the Law of Sines. | ||
+ | |||
+ | Call the side length of the square x. Then we are going to set a constant k equal to <math>\frac{\sin 120^{\circ}}{x}</math>, and this is consistent for every triangle in the diagram because all the angles of the hexagon are equiangular (and so they are all <math>120^{\circ}</math>). | ||
+ | |||
+ | Then we get the following process: | ||
+ | <cmath>\frac{\sin(90-a)}{40} = k</cmath> | ||
+ | <cmath>\cos a = 40k</cmath> | ||
+ | |||
+ | <cmath>\frac{\sin(a-30)}{\overline{EZ}} = k</cmath> | ||
+ | <cmath>\sin(a-30) = \overline{EZ}\cdot k</cmath> | ||
+ | <cmath>\frac{\sin(60-a)}{\overline{ZF}} = k</cmath> | ||
+ | <cmath>\sin(60-a) = \overline{ZF}\cdot k</cmath> | ||
+ | <cmath>\sin(a-30) + \sin(60-a) = k\cdot 41(\sqrt{3}-1)</cmath> | ||
+ | |||
+ | And now expanding using our trig formulas, we get: | ||
+ | <cmath>(\sin a + \cos a)(\frac{\sqrt{3}-1}{2} = k\cdot 41(\sqrt{3}-1)</cmath> | ||
+ | <cmath>\sin a + \cos a = 82k</cmath> | ||
+ | <cmath>\sin a = 42k</cmath> | ||
+ | |||
+ | And so now we have a triangle where <math>\cos a = 40k</math> and <math>\sin a = 42k</math>. Put them in a triangle where the hypotenuse is 1. Then, by the Pythagorean Theorem, we get: | ||
+ | <cmath>\sqrt{(40k)^2 + (42k)^2} = 1</cmath> | ||
+ | <cmath>3364k^2 = 1</cmath> | ||
+ | <cmath>k = \frac{1}{58}</cmath> | ||
+ | |||
+ | And since <math>k = \frac{\sin(120^{\circ})}{x}</math>, then: | ||
+ | <cmath>x = \frac{\sqrt{3}}{2}\cdot58</cmath> | ||
+ | <cmath>x = \boxed{29\sqrt{3}}</cmath> | ||
+ | |||
+ | Solution by IronicNinja | ||
+ | |||
+ | ==Video Solution by Richard Rusczyk== | ||
+ | https://artofproblemsolving.com/videos/amc/2012amc12b/273 | ||
+ | |||
+ | ~dolphin7 | ||
== See Also == | == See Also == |
Latest revision as of 16:32, 6 September 2021
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 by djmathman)
Solution
Extend and so that they meet at . Then , so and because 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
Solution 2
First, we want to angle chase. Set equal to degrees.
Now the key idea is that you want to relate the numbers that you have. You know and that . We proceed with the Law of Sines.
Call the side length of the square x. Then we are going to set a constant k equal to , and this is consistent for every triangle in the diagram because all the angles of the hexagon are equiangular (and so they are all ).
Then we get the following process:
And now expanding using our trig formulas, we get:
And so now we have a triangle where and . Put them in a triangle where the hypotenuse is 1. Then, by the Pythagorean Theorem, we get:
And since , then:
Solution by IronicNinja
Video Solution by Richard Rusczyk
https://artofproblemsolving.com/videos/amc/2012amc12b/273
~dolphin7
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 |
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