# Difference between revisions of "2008 AMC 12A Problems/Problem 22"

The following problem is from both the 2008 AMC 12A #22 and 2008 AMC 10A #25, so both problems redirect to this page.

## Problem

A round table has radius $4$. Six rectangular place mats are placed on the table. Each place mat has width $1$ and length $x$ as shown. They are positioned so that each mat has two corners on the edge of the table, these two corners being end points of the same side of length $x$. Further, the mats are positioned so that the inner corners each touch an inner corner of an adjacent mat. What is $x$?

$[asy]unitsize(4mm); defaultpen(linewidth(.8)+fontsize(8)); draw(Circle((0,0),4)); path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle; draw(mat); draw(rotate(60)*mat); draw(rotate(120)*mat); draw(rotate(180)*mat); draw(rotate(240)*mat); draw(rotate(300)*mat); label("$$x$$",(-1.55,2.1),E); label("$$1$$",(-0.5,3.8),S);[/asy]$

$\mathrm{(A)}\ 2\sqrt{5}-\sqrt{3}\qquad\mathrm{(B)}\ 3\qquad\mathrm{(C)}\ \frac{3\sqrt{7}-\sqrt{3}}{2}\qquad\mathrm{(D)}\ 2\sqrt{3}\qquad\mathrm{(E)}\ \frac{5+2\sqrt{3}}{2}$

## Solution 1 (Trigonometry)

Let one of the mats be $ABCD$, and the center be $O$ as shown:

$[asy]unitsize(8mm); defaultpen(linewidth(.8)+fontsize(8)); draw(Circle((0,0),4)); path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle; draw(mat); draw(rotate(60)*mat); draw(rotate(120)*mat); draw(rotate(180)*mat); draw(rotate(240)*mat); draw(rotate(300)*mat); label("$$x$$",(-1.55,2.1),E); label("$$x$$",(0.03,1.5),E); label("$$A$$",(-3.6,2.5513),E); label("$$B$$",(-3.15,1.35),E); label("$$C$$",(0.05,3.20),E); label("$$D$$",(-0.75,4.15),E); label("$$O$$",(0.00,-0.10),E); label("$$1$$",(-0.1,3.8),S); label("$$4$$",(-0.4,2.2),S); draw((0,0)--(0,3.103)); draw((0,0)--(-2.687,1.5513)); draw((0,0)--(-0.5,3.9686));[/asy]$

Since there are $6$ mats, $\Delta BOC$ is equilateral (the hexagon with side length $x$ is regular). So, $BC=CO=x$. Also, $\angle OCD = \angle OCB + \angle BCD = 60^\circ+90^\circ=150^\circ$.

By the Law of Cosines: $4^2=1^2+x^2-2\cdot1\cdot x \cdot \cos(150^\circ) \Rightarrow x^2 + x\sqrt{3} - 15 = 0 \Rightarrow x = \frac{-\sqrt{3}\pm 3\sqrt{7}}{2}$.

Since $x$ must be positive, $x = \frac{3\sqrt{7}-\sqrt{3}}{2} \Rightarrow (C)$.

## Solution 2 (without trigonometry)

Draw $OD$ and $OC$ as in the diagram. Draw the altitude from $O$ to $DC$ and call the intersection $E$.

$[asy]unitsize(8mm); defaultpen(linewidth(.8)+fontsize(8)); draw(Circle((0,0),4)); path mat=((-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle); draw(mat); draw(rotate(60)*mat); draw(rotate(120)*mat); draw(rotate(180)*mat); draw(rotate(240)*mat); draw(rotate(300)*mat); pair D = rotate(300)*(-3.687,1.5513); pair C = rotate(300)*(-2.687,1.5513); pair EE = foot((0.00,0.00),D,C); draw(D--EE--(0,0)); label("$$x$$",(-1.55,2.1),E); label("$$x$$",(0.03,1.5),E); label("$$A$$",(-3.6,2.5513),E); label("$$B$$",(-3.15,1.35),E); label("$$C$$",(0.05,3.20),E); label("$$D$$",(-0.75,4.15),E); label("$$O$$",(0.00,-0.10),E); label("$$1$$",(-0.1,3.8),S); label("$$4$$",(-0.4,2.2),S); draw((0,0)--(0,3.103)); draw((0,0)--(-2.687,1.5513)); draw((0,0)--(-0.5,3.9686)); label("$$E$$", EE,SE); [/asy]$

As proved in the first solution, $\angle OCD = 150^\circ$. That makes $\Delta OCE$ a $30-60-90$ triangle, so $OE = \frac{x}{2}$ and $CE= \frac{x\sqrt 3}{2}$

Since $\Delta ODE$ is a right triangle, $\left({\frac{x}{2}}\right)^2 + \left({\frac{x\sqrt 3}{2} +1}\right)^2 = 4^2 \Rightarrow x^2+x\sqrt3-15 = 0$

Solving for $x$ gives $x =\frac{3\sqrt{7}-\sqrt{3}}{2}\Rightarrow (C)$

## Solution 3 (simply Pythagorean Theorem)

$[asy]unitsize(8mm); defaultpen(linewidth(.8)+fontsize(8)); draw(Circle((0,0),4)); path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle; draw(mat); draw(rotate(60)*mat); draw(rotate(120)*mat); draw(rotate(180)*mat); draw(rotate(240)*mat); draw(rotate(300)*mat); label("$$x$$",(-1.55,2.1),E); label("$$x$$",(0.03,1.5),E); label("$$A$$",(-3.6,2.5513),E); label("$$B$$",(-3.15,1.35),E); label("$$C$$",(0.05,3.20),E); label("$$D$$",(-0.75,4.15),E); label("$$E$$",(0,4.17)); label("$$F$$",(0.75,4.15),W); label("$$O$$",(0.00,-0.10),E); label("$$1$$",(-0.1,3.8),S); label("$$4$$",(-0.4,2.2),S); draw((0,0)--(0,3.103)); draw((0,0)--(-2.687,1.5513)); draw((0,0)--(-0.5,3.9686)); draw((0,0)--(-0.5,3.9686));[/asy]$

By symmetry, $E$ is the midpoint of $DF$ and $OE$ is an extension of $OC$. Thus $\angle OED = 90^\circ$. Since $OD = 4$ and $DE = \frac{1}{2}$, $OE = \sqrt{16-\frac{1}{4}} = \frac{\sqrt{63}}{2} = \frac{3\sqrt{7}}{2}$. Since $\triangle CED$ is $30-60-90$, $CE = \frac{\sqrt{3}}{2}$ (this can also be deduced from Pythagoras on $\triangle CED$).

Thus $OC = \frac{3\sqrt{7}-\sqrt{3}}{2}$. As previous solutions noted, $\triangle BOC$ is equilateral, and thus the desired length is $x = OC \implies (C)$.

## Solution 3

$[asy]unitsize(8mm); defaultpen(linewidth(.8)+fontsize(8)); draw(Circle((0,0),4)); path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle; draw(mat); draw(rotate(60)*mat); draw(rotate(120)*mat); draw(rotate(180)*mat); draw(rotate(240)*mat); draw(rotate(300)*mat); label("$$x$$",(-1.95,3),E); label("$$A$$",(-3.6,2.5513),E); label("$$C$$",(0.05,3.20),E); label("$$E$$",(0.40,-3.60),E); label("$$B$$",(-0.75,4.15),E); label("$$D$$",(-2.62,1.5),E); label("$$F$$",(-2.64,-1.43),E); label("$$G$$",(-0.2,-2.8),E); label("$$\sqrt{3}x$$",(-1.5,-0.5),E); label("$$M$$",(-2,-0.9),E); label("$$O$$",(0.00,-0.10),E); label("$$1$$",(-2.7,2.3),S); label("$$1$$",(0.1,-3.4),S); label("$$8$$",(-0.3,0),S); draw((0,-3.103)--(-2.687,1.5513)); draw((0.5,-3.9686)--(-0.5,3.9686));[/asy]$

Looking at the diagram above, we know that $BE$ is a diameter of circle $O$ due to symmetry. Due to Thales' theorem, triangle $ABE$ is a right triangle with $A = 90 ^\circ$. $AE$ lies on $AD$ and $GE$ because $BAD$ is also a right angle. To find the length of $DG$, notice that if we draw a line from $F$ to $M$, the midpoint of line $DG$, it creates two $30$ - $60$ - $90$ triangles. Therefore, $MD = MG = \frac{\sqrt{3}x}{2} \Rightarrow DG = \sqrt{3}x$. $AE = 2 + \sqrt{3}x$

Applying the Pythagorean theorem to triangle $ABE$, we get $$(2+\sqrt{3}x)^2 + x^2 = 8^2 \Rightarrow 4 + 3x^2 + 4\sqrt{3}x + x^2 = 64 \Rightarrow x^2 + \sqrt{3}x - 15 = 0$$ Using the quadratic formula to solve, we get $$x = \frac{-\sqrt{3} \pm \sqrt{3 -4(1)(-15)}}{2} = \frac{\pm 3\sqrt{7} - \sqrt{3}}{2}$$ $x$ must be positive, therefore $$x = \frac{3\sqrt{7} - \sqrt{3}}{2} \Rightarrow (C)$$

~Zeric Hang

## Soultion 4 (coordinate bashing)

$[asy]unitsize(8mm); defaultpen(linewidth(.8)+fontsize(8)); draw(Circle((0,0),4)); path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle; draw(mat); draw(rotate(60)*mat); draw(rotate(120)*mat); draw(rotate(180)*mat); draw(rotate(240)*mat); draw(rotate(300)*mat); label("$$x$$",(-1.55,2.1),E); label("$$1$$",(-0.5,3.8),S);[/asy]$

We will let $O(0,0)$ be the origin. This way the coordinates of $C$ will be $(0,y)$. By $30-60-90$, the coordinates of $D$ will be $\left(-\frac{1}{2}, y + \frac{\sqrt{3}}{2}\right)$. The distance any point with coordinates $(x, y)$ is from the origin is $\sqrt{x^2 + y^2}$. Therefore, the distance $D$ is from the origin is $4$ and $\frac{1}{4} + x^2 + x\sqrt{3} + \frac{3}{4} = x^2 + x\sqrt{3} + 1 = 16$. We get the quadratic equation mentioned in solution 2. Using the quadratic formula, we get that $x = \frac{3\sqrt{7}-\sqrt{3}}{2} \Rightarrow (C)$

Note: Since $C$ and $D$ are not labeled in the diagram, refer to solution 1 for the location of points $C$ and $D$.

## Solution 5

$[asy]unitsize(8mm); defaultpen(linewidth(.8)+fontsize(8)); draw(Circle((0,0),4)); path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle; draw(mat); draw(rotate(60)*mat); draw(rotate(120)*mat); draw(rotate(180)*mat); draw(rotate(240)*mat); draw(rotate(300)*mat); label("$$x$$",(-1.55,2.1),E); label("$$A$$",(-3.6,2.5513),E); label("$$B$$",(-3.15,1.35),E); label("$$C$$",(0.05,3.20),E); label("$$D$$",(-0.75,4.15),E); label("$$E$$",(0.3,4.15),E); label("$$F$$",(-3.4,1.89),E); draw((0.5,3.9686)--(-3.13,2.45)); draw((0.5,3.9686)--(-2.95,2));[/asy]$

Notice that $\overarc{AE}$ is $\frac16$ the circumference of the circle. Therefore, $\overline{AE}$ is the side length of an inscribed hexagon. The side length of such a hexagon is simply the radius of the circle, or $4$. $\angle AFE$ is a right angle, therefore $\triangle AFE$ is a right triangle. $\overline{AF}$ is half the length of $1$, or $\frac{1}{2}$. The length of $\overline{EF}$ is $x$ plus the altitude length of one of the equilateral triangles, or $x+\frac{\sqrt{3}}{2}$. Using the Pythagorean Theorem, we get

$4^2 = \left(\frac{1}{2}\right)^2 + \left(x+\frac{\sqrt{3}}{2}\right)^2$

Solving for $x$, we get $x = \frac{3\sqrt{7}-\sqrt{3}}{2}\ \boxed{\text{C}}$