2012 AMC 12B Problems/Problem 21

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 because $AB$ 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: $x = \frac{\sqrt{3}}{2}\cdot58$ $x = \boxed{29\sqrt{3}}$

Solution by IronicNinja

Video Solution by Richard Rusczyk

https://artofproblemsolving.com/videos/amc/2012amc12b/273

~dolphin7

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

Page

Toolbox

Search

2012 AMC 12B Problems/Problem 21

Contents

Problem 21

Solution

Solution 2

Video Solution by Richard Rusczyk

See Also