Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution 2 (without trigonometry))
(→‎Solution)
Line 91: Line 91:
  
 
Solving for <math>x</math> gives <math> x =\frac{3\sqrt{7}-\sqrt{3}}{2}\Rightarrow C </math>
 
Solving for <math>x</math> gives <math> x =\frac{3\sqrt{7}-\sqrt{3}}{2}\Rightarrow C </math>
 +
 +
=== Solution 2 (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("\(E\)",(0,4.19));
 +
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, <math>E</math> is the midpoint of <math>DF</math> and <math>OE</math> is an extension of <math>CE</math>. Thus <math>\angle OED = 90^\circ</math>. Since <math>OD = 4</math> and <math>DE = \frac{1}{2}</math>, <math>DE = \sqrt{16-\frac{1}{4}} = \frac{\sqrt{63}}{2} = \frac{3\sqrt{7}}{2}</math>. Since <math>\triangle CED</math> is <math>30-60-90</math>, <math>CE = \frac{\sqrt{3}}{2}</math> (or this can also be deduced from Pythagoras on <math>\triangle CED</math>). Thus <math>OC = \frac{3\sqrt{7}-\sqrt{3}}{2}</math>. As previous solutions noted, <math>\triangle BOC</math> is equilateral, and thus the desired length is <math>x = OC \implies C</math>.
  
 
==Solution 3==
 
==Solution 3==

Revision as of 17:51, 25 October 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 $\Delta ODE$ 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 2 (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("\(E\)",(0,4.19)); 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 $CE$. Thus $\angle OED = 90^\circ$. Since $OD = 4$ and $DE = \frac{1}{2}$, $DE = \sqrt{16-\frac{1}{4}} = \frac{\sqrt{63}}{2} = \frac{3\sqrt{7}}{2}$. Since $\triangle CED$ is $30-60-90$, $CE = \frac{\sqrt{3}}{2}$ (or 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, $MD = MG = \frac{\sqrt{3}x}{2} \Rightarrow DG = \sqrt{3}x$. $AE = 2 + \sqrt{3}x$

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 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

Soultion 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 would be $(0,x)$. By 30-60-90, the coordinates of D would be $(\sqrt{-1}{2}, x + \frac{\sqrt{3}}{2})$. The distance $(x, y)$ is from the origin is just $\sqrt{x^2 + y^2}$. Therefore, the distance D is from the origin is both 4 and $\frac{1}{4} + x^2 + x\sqrt{3} + \frac{3}{4} = x^2 + x\sqrt{3} + 1 = 16$. We get the equation mentioned in all the previous solution, using the quadratic formula, we get that $x = \frac{3\sqrt{7}-\sqrt{3}}{2} \Rightarrow C$

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 one-sixth the circumference of the circle. Therefore, $\overline{AE}$ is the side length of an inscribed hexagon. The side length of such a hexagon is simply the radius, or $4$. $\angle AFE$ is a right angle, therefore $\triangle AFE$ is a right triangle. $\overline{AF}$ is half the length of $1$, or $\frac{1}{2}$. The length of $\overline{EF}$ is $x$ plus the altitude length of one of the equilateral triangles, or $x+\frac{\sqrt{3}}{2}$. 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 $x =$

$\frac{3\sqrt{7}-\sqrt{3}}{2}$ , or $\boxed{\text{C}}$.


See Also

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
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_12A_Problems/Problem_22&oldid=135808"