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

({{dup}})
(Standardized answer choices, some minor edits)
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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 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 <math>x</math>. Further, the mats are positioned so that the inner corners each touch an inner corner of an adjacent mat. What is <math>x</math>?
 
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 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 <math>x</math>. Further, the mats are positioned so that the inner corners each touch an inner corner of an adjacent mat. What is <math>x</math>?
  
−
<asy>
+
<asy>unitsize(4mm);
−
unitsize(4mm);
 
 
defaultpen(linewidth(.8)+fontsize(8));
 
defaultpen(linewidth(.8)+fontsize(8));
 
draw(Circle((0,0),4));
 
draw(Circle((0,0),4));
Line 15: Line 14:
 
draw(rotate(300)*mat);
 
draw(rotate(300)*mat);
 
label("\(x\)",(-1.55,2.1),E);
 
label("\(x\)",(-1.55,2.1),E);
−
label("\(1\)",(-0.5,3.8),S);
+
label("\(1\)",(-0.5,3.8),S);</asy>
−
</asy>
 
  
−
<math>\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}</math>
+
<math>\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}</math>
  
 
==Solution==
 
==Solution==
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Let one of the mats be <math>ABCD</math>, and the center be <math>O</math> as shown:  
 
Let one of the mats be <math>ABCD</math>, and the center be <math>O</math> as shown:  
  
−
<asy>
+
<asy>unitsize(8mm);
−
unitsize(8mm);
 
 
defaultpen(linewidth(.8)+fontsize(8));
 
defaultpen(linewidth(.8)+fontsize(8));
 
draw(Circle((0,0),4));
 
draw(Circle((0,0),4));
Line 46: Line 43:
 
draw(Line(0,0)--(0,3.103));
 
draw(Line(0,0)--(0,3.103));
 
draw(Line(0,0)--(-2.687,1.5513));
 
draw(Line(0,0)--(-2.687,1.5513));
−
draw(Line(0,0)--(-0.5,3.9686));
+
draw(Line(0,0)--(-0.5,3.9686));</asy>
−
</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>.  
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==See Also==
 
==See Also==
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{{AMC12 box|year=2008|num-b=21|num-a=23|ab=A}}
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{{AMC12 box|year=2008|ab=A|num-b=21|num-a=23}}
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{{AMC10 box|year=2008|num-b=24|after=Last question|ab=A}}
+
{{AMC10 box|year=2008|ab=A|num-b=24|after=Last Question}}
  
 
[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Trigonometry Problems]]
 
[[Category:Introductory Trigonometry Problems]]

Revision as of 00:32, 26 April 2008

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(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, $x = \frac{3\sqrt{7}-\sqrt{3}}{2} \Rightarrow C$.

Solution 2 (without trigonometry)

This problem needs a solution. If you have a solution for it, please help us out by adding it.

See Also

2008 AMC 12A (Problems • Answer Key • Resources)
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 (Problems • Answer Key • Resources)
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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