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

(Solution 2 (without trigonometry))
(Solution 1 (trigonometry))
Line 41: Line 41:
 
label("\(1\)",(-0.1,3.8),S);
 
label("\(1\)",(-0.1,3.8),S);
 
label("\(4\)",(-0.4,2.2),S);
 
label("\(4\)",(-0.4,2.2),S);
draw(Line(0,0)--(0,3.103));
+
draw((0,0)--(0,3.103));
draw(Line(0,0)--(-2.687,1.5513));
+
draw((0,0)--(-2.687,1.5513));
draw(Line(0,0)--(-0.5,3.9686));</asy>
+
draw((0,0)--(-0.5,3.9686));</asy>
  
 
Since there are <math>6</math> mats, <math>\Delta BOC</math> is [[equilateral]]. So, <math>BC=CO=x</math>. Also, <math>\angle OCD = \angle OCB + \angle BCD = 60^\circ+90^\circ=150^\circ</math>.  
 
Since there are <math>6</math> mats, <math>\Delta BOC</math> is [[equilateral]]. So, <math>BC=CO=x</math>. Also, <math>\angle OCD = \angle OCB + \angle BCD = 60^\circ+90^\circ=150^\circ</math>.  

Revision as of 00:35, 25 January 2011

The following problem is from both the 2008 AMC 12A #22 and 2004 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

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. 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)

See Math Jam

See Also

2008 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 21
Followed by
Problem 23
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
2008 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 24
Followed by
Last Question
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 10 Problems and Solutions
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