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

Revision as of 13:14, 3 January 2020

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

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)

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 $\Delta OEC$ 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

$[asy]unitsize(8mm); defaultpen(linewidth(.8)+fontsize(8)); draw(Circle((0,0),4)); path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle; draw(mat); draw(rotate(60)*mat); draw(rotate(120)*mat); draw(rotate(180)*mat); draw(rotate(240)*mat); draw(rotate(300)*mat); label("$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, $MD = MG = \frac{\sqrt{3}x}{2} \Rightarrow DG = \sqrt{3}x$ . $AE = 2 + \sqrt{3}x$

Use the Pythagorean theorem on 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 pythagorean theorem 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

@@ Line 128: / Line 128: @@
 ==See Also==
-{{AMC10 box|year=2008|ab=A|num-b=24|num-a=Last question}}
+{{AMC10 box|year=2008|ab=A|num-b=24|after=Last question}}
 {{AMC12 box|year=2008|ab=A|num-b=21|num-a=23}}
 {{MAA Notice}}

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