# 2012 AMC 12B Problems/Problem 21

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## 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 1

We can, $\textsc{wlog}$, assume $Y$ coincides with $D$ and $CD\parallel AF$ as before. In which case, we will have $BC=EF=41(\sqrt{3}-1)$. So we have square $AXDZ$ inscribed in equiangular hexagon $ABCDEF$ with $X$ on $\overline{BC}$ and $Z$ on $\overline{EF}$. $[asy] size(200); defaultpen(fontsize(10)+linewidth(1)); pair A=origin,B=(2.5,0),C=B+2.5*dir(60), D=C+1.75*dir(120),E=D-(3.19,0),F=E-1.8*dir(60); pair X=waypoint(B--C,0.345),Z=rotate(90,A)*X,Y=rotate(90,Z)*A; pair Cp=extension(B,C,Y,Y+dir(-60)); draw(A--B--Cp--Y--E--F--cycle); draw(A--X--Y--Z--cycle,linewidth(0.9)+linetype("2 2")); dot("A",A,W,linewidth(4)); dot("B",B,dir(0),linewidth(4)); dot("C",Cp,dir(0),linewidth(4)); dot("E",E,dir(100),linewidth(4)); dot("F",F,W,linewidth(4)); dot("X",X,dir(0),linewidth(4)); dot("D",Y,N,linewidth(4)); dot("Z",Z,W,linewidth(4)); label("u", B--X, SE);label("v", X--Cp, SE); label("40", A--B, S); label("s", A--X, NW); label("s", Y--X, SW); [/asy]$ Let $\angle BXA = \theta$; then $\angle BAX=60^\circ -\theta$. Let $BX=u$. In $\triangle ABX$ we have \begin{align} \frac{2s}{\sqrt{3}}=\frac{u}{\sin(60^\circ-\theta)}=\frac {40}{\sin\theta} \end{align} We also have $\angle CXD=90^\circ - \theta$ and $\angle CDX = \theta-30^\circ$. Let $CX=v$. In $\triangle CDX$ we have \begin{align}\tag{2} \frac{2s}{\sqrt{3}}=\frac{v}{\sin(\theta-30^\circ)}=\frac {CD}{\cos\theta} \end{align} Now $BC=u+v=41(\sqrt{3}-1)$. From $(1)$ and $(2)$ we get\begin{align*} 41(\sqrt{3}-1) &= \frac{2s}{\sqrt{3}}\left(\sin(60^\circ-\theta)+\sin(\theta-30^\circ)\right) \\ &= \frac{2s}{\sqrt{3}} \cdot \frac{\sqrt{3}-1}2\cdot (\sin\theta + \cos\theta) \end{align*} From $(1)$ we get $s\sin\theta = 20\sqrt{3}$ and therefore $s\cos\theta = \sqrt{s^2-3\cdot 20^2}$. Thus $$41(\sqrt{3}-1) = \frac{\sqrt{3}-1}{\sqrt{3}}(20\sqrt{3}+\sqrt{s^2-3\cdot 20^2})$$which simplifies to$$3\cdot 21^2 = s^2-3\cdot 20^2.$$Since $(20, 21, 29)$ is a Pythagorean triple, we get $s=29\sqrt{3}$, i.e. $\framebox{A}$.

## Solution 2

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 3

First, we want to angle chase. Set $ 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

## Solution 4

Let $EZ = x$, $\angle XAB = \alpha$

$\angle BAX = \angle EYZ$, $AX = YZ$, $\angle ZEY = \angle XBA$, $\triangle BAX \cong \triangle EYZ$ by $AAS$, $BX = EZ = x$

$\angle AXB = 180^\circ - 120^\circ - \alpha = 60^\circ - \alpha$

$\frac{XB}{\sin \angle XAB} = \frac{AX}{\sin \angle ABX} = \frac{AB}{\angle AXB}$, $\frac{x}{\sin \alpha} = \frac{AX}{\sin 120^\circ} = \frac{40}{\sin (60^\circ - \alpha)}$

$\angle ZAF = 120^\circ - 90^\circ - \alpha = 30^\circ - \alpha$

$ZF = 41(\sqrt{3} - 1) - x$, $\frac{ZF}{\sin \angle ZAF} = \frac{AZ}{ \sin \angle ZFA}$, $\frac{41(\sqrt{3} - 1) - x}{\sin (30^\circ - \alpha)} = \frac{AZ}{ \sin 120^\circ}$

$\frac{x}{\sin \alpha} = \frac{41(\sqrt{3} - 1) - x}{\sin (30^\circ - \alpha)}= \frac{40}{\sin(60^\circ - \alpha)}$

$40 \cdot \sin \alpha = x(\sin 60^\circ \cos \alpha - \cos 60^\circ \sin \alpha)$ $(1)$

$x(\sin 30^\circ \cos \alpha - \cos 30^\circ \sin \alpha) = [41(\sqrt{3} - 1) - x] \sin \alpha$ $(2)$

By simplifying $(1)$ we get, $\frac{\sqrt{3}}{2} \cdot x \cdot \cos \alpha - \frac{\sin \alpha}{2} \cdot x = 40 \cdot \sin \alpha$ $(3)$

By simplifying $(2)$ we get, $\frac{\cos \alpha}{2} \cdot x + \frac{2 - \sqrt{3}}{2} \cdot \sin \alpha \cdot x = 41(\sqrt{3} - 1)\sin \alpha$ $(4)$

By $\sqrt{3} \cdot (4) - (3)$ we get, $\frac{2\sqrt{3} - 3 +1}{2} \cdot \sin \alpha \cdot x = [41(3-\sqrt{3}) - 40] \sin \alpha$

$(\sqrt{3} - 1)x = 83 - 41\sqrt{3}$, $x = 21 \sqrt{3} - 20$

By the law of cosine $AX = \sqrt{(21 \sqrt{3} - 20)^2 + 40^2 - 2 \cdot (21 \sqrt{3} - 20) 40 \cdot \cos 120^\circ} = \boxed{\textbf{(A) } 29 \sqrt{3}}$

~dolphin7