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2008 AMC 12A Problems/Problem 22

Revision as of 23:10, 19 February 2008 by Xantos C. Guin (talk | contribs) (New page: ==Problem== A round table has radius <math>4</math>. Six rectangular place mats are placed on the table. Each place mat has width <math>1</math> and length <math>x</math> as shown. They ar...)
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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("<math>x</math>",(-1.55,2.1),E); label("<math>1</math>",(-0.5,3.8),S); [/asy]

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

Solution

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("<math>x</math>",(-1.55,2.1),E); label("<math>x</math>",(0.03,1.5),E); label("<math>A</math>",(-3.6,2.5513),E); label("<math>B</math>",(-3.15,1.35),E); label("<math>C</math>",(0.05,3.20),E); label("<math>D</math>",(-0.75,4.15),E); label("<math>O</math>",(0.00,-0.10),E); label("<math>1</math>",(-0.1,3.8),S); label("<math>4</math>",(-0.4,2.2),S); draw(Line(0,0)--(0,3.103)); draw(Line(0,0)--(-2.687,1.5513)); draw(Line(0,0)--(-0.5,3.9686)); [/asy]

Since there are $6$ mats, $\Delta BOC$ is equilateral. 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, $x = \frac{3\sqrt{7}-\sqrt{3}}{2} \Rightarrow C$.

See Also

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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