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Difference between revisions of "2012 AMC 10A Problems/Problem 18"

(→‎Solution 2 (Looking at the answer choices))
(→‎Solution 3 (Areas))
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Notice that we have three sectors at each ends of the fidget spinner. Realize that the central angles of a circle always sum up to 360. Since we know one of the central angles is equal to 120, then we subtract 360 to 120 which gives us 240. Knowing that, each sector is <math>\frac{2}{3}</math> of a circle and since there is 3 of them, <math>\frac{2}{3}*3=2</math> circles. To find the area of those circles, we have to use <math>\pi r^2</math> again, but since there are 2 circles, then it is <math>2\pi r^2</math>, which gives us <math>2\pi</math>.  
+
Notice that we have three sectors exterior to the hexagon. Realize that the central angles of a circle always sum up to 360. Since we know one of the central angles is equal to 120, then we subtract 360 to 120 which gives us 240. Knowing that, each sector is <math>\frac{2}{3}</math> of a circle and since there is 3 of them, <math>\frac{2}{3}*3=2</math> circles. To find the area of those circles, we have to use <math>\pi r^2</math> again, but since there are 2 circles, then it is <math>2\pi r^2</math>, which gives us <math>2\pi</math>.  
  
 
Now we have enough information to find the total area,  
 
Now we have enough information to find the total area,  

Revision as of 01:13, 29 November 2022

The following problem is from both the 2012 AMC 12A #14 and 2012 AMC 10A #18, so both problems redirect to this page.

Problem 14

The closed curve in the figure is made up of 9 congruent circular arcs each of length $\frac{2\pi}{3}$, where each of the centers of the corresponding circles is among the vertices of a regular hexagon of side 2. What is the area enclosed by the curve?

[asy] defaultpen(fontsize(6pt)); dotfactor=4; label("$\circ$",(0,1)); label("$\circ$",(0.865,0.5)); label("$\circ$",(-0.865,0.5)); label("$\circ$",(0.865,-0.5)); label("$\circ$",(-0.865,-0.5)); label("$\circ$",(0,-1)); dot((0,1.5)); dot((-0.4325,0.75)); dot((0.4325,0.75)); dot((-0.4325,-0.75)); dot((0.4325,-0.75)); dot((-0.865,0)); dot((0.865,0)); dot((-1.2975,-0.75)); dot((1.2975,-0.75)); draw(Arc((0,1),0.5,210,-30)); draw(Arc((0.865,0.5),0.5,150,270)); draw(Arc((0.865,-0.5),0.5,90,-150)); draw(Arc((0.865,-0.5),0.5,90,-150)); draw(Arc((0,-1),0.5,30,150)); draw(Arc((-0.865,-0.5),0.5,330,90)); draw(Arc((-0.865,0.5),0.5,-90,30)); [/asy]

$\textbf{(A)}\ 2\pi+6\qquad\textbf{(B)}\ 2\pi+4\sqrt{3}\qquad\textbf{(C)}\ 3\pi+4\qquad\textbf{(D)}\ 2\pi+3\sqrt{3}+2\qquad\textbf{(E)}\ \pi+6\sqrt{3}$

Solution 1

Draw the hexagon between the centers of the circles, and compute its area $(6)(0.5)(2\sqrt{3})=6\sqrt{3}$. Then add the areas of the three sectors outside the hexagon ($2\pi$) and subtract the areas of the three sectors inside the hexagon but outside the figure($\pi$) to get the area enclosed in the curved figure $(\pi+6\sqrt{3})$, which is $\boxed{\textbf{(E)}\ \pi+6\sqrt{3}}$.


Solution 2 (Looking at the answer choices)

We see that after forming the hexagon using the sectors outside the hexagon, there will be three sectors left. Each sector has an area of $\frac{\pi}{3},$ so the three combined make $\pi.$ Since the side length of the hexagon is $2,$ it's area doesn't have $\pi$ in it, so we know that the final answer will be $\pi + \text{(area of hexagon)}.$ Looking at the answer choices, the only answer with only one $\pi$ is $\boxed{\textbf{(E)}}.$

Solution 3 (Areas)

As you can see, this diagram looks like a fidget spinner ;). Fidget spinners aside, we need to add stuff to our diagram to make it look easier. In the directions, they were talking about the centers of each arc create a hexagon, so let's add that to our diagram.

[asy] defaultpen(fontsize(6pt)); dotfactor=4; label("$\circ$",(0,1)); label("$\circ$",(0.865,0.5)); label("$\circ$",(-0.865,0.5)); label("$\circ$",(0.865,-0.5)); label("$\circ$",(-0.865,-0.5)); label("$\circ$",(0,-1)); dot((0,1.5)); dot((-0.432,0.75)); dot((0.4325,0.75)); dot((-0.4325,-0.75)); dot((0.4325,-0.75)); dot((-0.865,0)); dot((0.865,0)); dot((-1.2975,-0.75)); dot((1.2975,-0.75)); draw(Arc((0,1),0.5,210,-30)); draw(Arc((0.865,0.5),0.5,150,270)); draw(Arc((0.865,-0.5),0.5,90,-150)); draw(Arc((0.865,-0.5),0.5,90,-150)); draw(Arc((0,-1),0.5,30,150)); draw(Arc((-0.865,-0.5),0.5,330,90)); draw(Arc((-0.865,0.5),0.5,-90,30)); draw((0,1)--(0.865, 0.5)--(0.865,-0.5)--(0,-1)--(-0.865,-0.5)--(-0.865,0.5)--(0,1)); [/asy]

The side length of the hexagon is 2 and if we plug it in to the area of a regular hexagon formula $\frac{3\sqrt 3}{2}s^2$ we get $6\sqrt 3$.

Note that the interior angles of a regular hexagon is 120 because of the formula $\frac{180(n-2)}{n}$ where n is the number of sides. Knowing that, each sector is $\frac{1}{3}$ of a circle. This would mean the three sectors inside the hexagon altogether equal a full circle. Knowing that the radius is 1, we can use the area of a circle $\pi r^2$ and subtract it to $6\sqrt 3$. Thus we get the total area of $6\sqrt 3 - \pi$.


Notice that we have three sectors exterior to the hexagon. Realize that the central angles of a circle always sum up to 360. Since we know one of the central angles is equal to 120, then we subtract 360 to 120 which gives us 240. Knowing that, each sector is $\frac{2}{3}$ of a circle and since there is 3 of them, $\frac{2}{3}*3=2$ circles. To find the area of those circles, we have to use $\pi r^2$ again, but since there are 2 circles, then it is $2\pi r^2$, which gives us $2\pi$.

Now we have enough information to find the total area,

$(6\sqrt 3 -\pi+2\pi)=\textbf{(E)}\ \pi+6\sqrt{3}$

~ghfhgvghj10

See Also

2012 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
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
2012 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
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

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