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Difference between revisions of "2012 AMC 12B Problems/Problem 21"

(→‎Problem 21)
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Therefore <math>AZ = 29\sqrt{3} ... \framebox{A}</math>
 
Therefore <math>AZ = 29\sqrt{3} ... \framebox{A}</math>
  
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==Solution 2==
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First, we want to angle chase. Set <math><YXC</math> equal to <math>a</math> degrees.
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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>).
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Then we get the following process:
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<cmath>\frac{\sin(90-a)}{40} = k</cmath>
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<cmath>\cos a = 40k</cmath>
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<cmath>\frac{\sin(a-30)}{\overline{EZ}} = k</cmath>
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<cmath>\sin(a-30) = \overline{EZ}\cdot k</cmath>
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<cmath>\frac{\sin(60-a)}{\overline{ZF}} = k</cmath>
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<cmath>\sin(60-a) = \overline{ZF}\cdot k</cmath>
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<cmath>\sin(a-30) + \sin(60-a) = k\cdot 41(\sqrt{3}-1)</cmath>
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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>
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<cmath>\sin a + \cos a = 82k</cmath>
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<cmath>\sin a = 42k</cmath>
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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:
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<cmath>\sqrt{(40k)^2 + (42k)^2} = 1</cmath>
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<cmath>3364k^2 = 1</cmath>
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<cmath>k = \frac{1}{58}</cmath>
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And since <math>k = \frac{\sin(120^{\circ})}{x}</math>, then:
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<cmath>x = \frac{\sqrt{3}}{2}\cdot\frac{1}{58}</cmath>
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<cmath>x = \boxed{29\sqrt{3}}</cmath>
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Solution by IronicNinja
 
== See Also ==
 
== See Also ==
  

Revision as of 13:47, 25 November 2019

Problem 21

Square $AXYZ$ is inscribed in equiangular hexagon $ABCDEF$ with $X$ on $\overline{BC}$, $Y$ on $\overline{DE}$, and $Z$ on $\overline{EF}$. Suppose that $AB=40$, and $EF=41(\sqrt{3}-1)$. What is the side-length of the square?

$\textbf{(A)}\ 29\sqrt{3} \qquad\textbf{(B)}\ \frac{21}{2}\sqrt{2}+\frac{41}{2}\sqrt{3} \qquad\textbf{(C)}\ \ 20\sqrt{3}+16 \qquad\textbf{(D)}\ 20\sqrt{2}+13 \sqrt{3} \qquad\textbf{(E)}\ 21\sqrt{6}$

[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)

Solution

Extend $AF$ and $YE$ so that they meet at $G$. Then $\angle FEG=\angle GFE=60^{\circ}$, so $\angle FGE=60^{\circ}$ and therefore $AB$ is parallel to $YE$. Also, since $AX$ is parallel and equal to $YZ$, we get $\angle BAX = \angle ZYE$, hence $\triangle ABX$ is congruent to $\triangle YEZ$. We now get $YE=AB=40$.

Let $a_1=EY=40$, $a_2=AF$, and $a_3=EF$.

Drop a perpendicular line from $A$ to the line of $EF$ that meets line $EF$ at $K$, and a perpendicular line from $Y$ to the line of $EF$ that meets $EF$ at $L$, then $\triangle AKZ$ is congruent to $\triangle ZLY$ since $\angle YZL$ is complementary to $\angle KZA$. Then we have the following equations:

\[\frac{\sqrt{3}}{2}a_2 = AK=ZL = ZE+\frac{1}{2} a_1\] \[\frac{\sqrt{3}}{2}a_1 = YL =ZK = ZF+\frac{1}{2} a_2\]

The sum of these two yields that

\[\frac{\sqrt{3}}{2}(a_1+a_2) = \frac{1}{2} (a_1+a_2) + ZE+ZF = \frac{1}{2} (a_1+a_2) + EF\] \[\frac{\sqrt{3}-1}{2}(a_1+a_2) = 41(\sqrt{3}-1)\] \[a_1+a_2=82\] \[a_2=82-40=42.\]

So, we can now use the law of cosines in $\triangle AGY$:

\[2AZ^2 = AY^2 = AG^2 + YG^2 - 2AG\cdot YG \cdot \cos 60^{\circ}\] \[= (a_2+a_3)^2 + (a_1+a_3)^2 - (a_2+a_3)(a_1+a_3)\] \[= (41\sqrt{3}+1)^2 + (41\sqrt{3}-1)^2 - (41\sqrt{3}+1)(41\sqrt{3}-1)\] \[= 6 \cdot 41^2 + 2 - 3 \cdot 41^2 + 1 = 3 (41^2 + 1) = 3\cdot 1682\] \[AZ^2 = 3 \cdot 841 = 3 \cdot 29^2\]

Therefore $AZ = 29\sqrt{3} ... \framebox{A}$


Solution 2

First, we want to angle chase. Set $<YXC$ equal to $a$ degrees.

Now the key idea is that you want to relate the numbers that you have. You know $\overline{AB} = 40$ and that $\overline{EZ} + \overline{ZF} = 41(\sqrt{3}-1)$. 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 $\frac{\sin 120^{\circ}}{x}$, and this is consistent for every triangle in the diagram because all the angles of the hexagon are equiangular (and so they are all $120^{\circ}$).

Then we get the following process: \[\frac{\sin(90-a)}{40} = k\] \[\cos a = 40k\]

\[\frac{\sin(a-30)}{\overline{EZ}} = k\] \[\sin(a-30) = \overline{EZ}\cdot k\] \[\frac{\sin(60-a)}{\overline{ZF}} = k\] \[\sin(60-a) = \overline{ZF}\cdot k\] \[\sin(a-30) + \sin(60-a) = k\cdot 41(\sqrt{3}-1)\]

And now expanding using our trig formulas, we get: \[(\sin a + \cos a)(\frac{\sqrt{3}-1}{2} = k\cdot 41(\sqrt{3}-1)\] \[\sin a + \cos a = 82k\] \[\sin a = 42k\]

And so now we have a triangle where $\cos a = 40k$ and $\sin a = 42k$. Put them in a triangle where the hypotenuse is 1. Then, by the Pythagorean Theorem, we get: \[\sqrt{(40k)^2 + (42k)^2} = 1\] \[3364k^2 = 1\] \[k = \frac{1}{58}\]

And since $k = \frac{\sin(120^{\circ})}{x}$, then: \[x = \frac{\sqrt{3}}{2}\cdot\frac{1}{58}\] \[x = \boxed{29\sqrt{3}}\]

Solution by IronicNinja

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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