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

Solution 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 with side length $4$ . $\triangle AFE$ is a right triangle with $\overline{AF}=\frac12$ . The length of $\overline{EF}=x+\frac{\sqrt{3}}{2}$ . 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 $x = \frac{3\sqrt{7}-\sqrt{3}}{2}\ \boxed{\text{C}}$

Solution 6 (Estimation)

We will treat the area between two adjacent placemats as an equilateral triangle. Because the inside hexagon is regular, the distance from the center to the closest vertex of an equilateral triangle $x.$ The altitude of an equilateral triangle with side length 1 is $\frac{\sqrt{3}}{2},$ which means $x$ is roughly $4-\frac{\sqrt{3}}{2}.$ Estimating, $\frac{3\sqrt{7}}{2}$ ( $\sqrt{7}$ is about $2.65$ which can be easily estimated by comparing $26^2=676$ and $27^2=729$ ) is about $4$ (a little less, which is what we'd expect), so we circle $\boxed{\text{(C) }\frac{3\sqrt{7}-\sqrt{3}}{2}}$ and are happy that we just solved an $AMC$ $10$ problem $25$ without having to do any actual thinking.

2008 AMC 10A (Problems • Answer Key • Resources)
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All AMC 10 Problems and Solutions

2008 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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All AMC 12 Problems and Solutions

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