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Difference between revisions of "2019 AMC 10A Problems/Problem 16"

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Education, the Study of Everything
 
Education, the Study of Everything
  
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==Video Solution 2 by SpreadTheMathLove==
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https://www.youtube.com/watch?v=q9L66I9f9nI
  
 
==Video Solution==
 
==Video Solution==

Revision as of 00:06, 21 January 2023

The following problem is from both the 2019 AMC 10A #16 and 2019 AMC 12A #10, so both problems redirect to this page.

Problem

The figure below shows $13$ circles of radius $1$ within a larger circle. All the intersections occur at points of tangency. What is the area of the region, shaded in the figure, inside the larger circle but outside all the circles of radius $1 ?$

[asy]unitsize(20);filldraw(circle((0,0),2*sqrt(3)+1),rgb(0.5,0.5,0.5));filldraw(circle((-2,0),1),white);filldraw(circle((0,0),1),white);filldraw(circle((2,0),1),white);filldraw(circle((1,sqrt(3)),1),white);filldraw(circle((3,sqrt(3)),1),white);filldraw(circle((-1,sqrt(3)),1),white);filldraw(circle((-3,sqrt(3)),1),white);filldraw(circle((1,-1*sqrt(3)),1),white);filldraw(circle((3,-1*sqrt(3)),1),white);filldraw(circle((-1,-1*sqrt(3)),1),white);filldraw(circle((-3,-1*sqrt(3)),1),white);filldraw(circle((0,2*sqrt(3)),1),white);filldraw(circle((0,-2*sqrt(3)),1),white);[/asy]

$\textbf{(A) } 4 \pi \sqrt{3} \qquad\textbf{(B) } 7 \pi \qquad\textbf{(C) } \pi\left(3\sqrt{3} +2\right) \qquad\textbf{(D) } 10 \pi \left(\sqrt{3} - 1\right) \qquad\textbf{(E) } \pi\left(\sqrt{3} + 6\right)$

Solution 1

[asy] unitsize(20);filldraw(circle((0,0),2*sqrt(3)+1),rgb(0.5,0.5,0.5));filldraw(circle((-2,0),1),white);filldraw(circle((0,0),1),white);filldraw(circle((2,0),1),white);filldraw(circle((1,sqrt(3)),1),white);filldraw(circle((3,sqrt(3)),1),white);filldraw(circle((-1,sqrt(3)),1),white);filldraw(circle((-3,sqrt(3)),1),white);filldraw(circle((1,-1*sqrt(3)),1),white);filldraw(circle((3,-1*sqrt(3)),1),white);filldraw(circle((-1,-1*sqrt(3)),1),white);filldraw(circle((-3,-1*sqrt(3)),1),white);filldraw(circle((0,2*sqrt(3)),1),white);filldraw(circle((0,-2*sqrt(3)),1),white); pair O,A,B,C; O=(0,0); A=(-1,sqrt(3)); B=(1,sqrt(3)); C=(0,sqrt(3)*2); draw(O--A); draw(A--C); draw(B--C); draw(O--B); draw(A--B); draw(O--C); dot(A); dot(B); dot(C); dot(O); label("A",A, W); label("O",O,S); label("B",B,E); label("C",C, N); [/asy]

In the diagram above, notice that triangle $OAB$ and triangle $ABC$ are congruent and equilateral with side length $2$. We can see the radius of the larger circle is two times the altitude of $OAB$ plus $1$ (the distance from point $C$ to the edge of the circle). Using $30^{\circ}-60^{\circ}-90^{\circ}$ triangles, we know the altitude is $\sqrt{3}$. Therefore, the radius of the larger circle is $2\sqrt{3}+1$.

The area of the larger circle is thus $\left(2\sqrt{3}+1\right)^2 \pi = \left(13+4\sqrt{3}\right)\pi$, and the sum of the areas of the smaller circles is $13\pi$, so the area of the dark region is $\left(13+4\sqrt{3}\right)\pi-13\pi = \boxed{\textbf{(A) } 4 \pi \sqrt{3}}$.

Solution 2

We can form an equilateral triangle with side length $6$ from the centers of three of the unit circles tangent to the outer circle. The radius of the outer circle is the circumradius of the triangle plus $1$. By using $R = \frac{abc}{4K}$ or $R=\frac{a}{2\sin{A}}$, we get the radius as $\frac{6}{\sqrt{3}}+1$.

The shaded area is thus $\pi((\frac{6}{\sqrt{3}}+1)^2-13) = \boxed{\textbf{(A) } 4 \pi \sqrt{3}}$.

Solution 3

Like in Solution 2, we can form an equilateral triangle with side length $6$ from the centers of three of the unit circles tangent to the outer circle. We can find the height of this triangle to be $3\sqrt{3}$. Then, we can form another equilateral triangle from the centers of the second and third circles in the third row and the center of the bottom circle with side length $2$. The height of this triangle is clearly $\sqrt{3}$. Therefore the diameter of the large circle is $4\sqrt{3} + 2$ and the radius is $\frac{4\sqrt{3}+2}{2} = 2\sqrt{3} + 1$. The area of the large circle is thus $\pi\left(2\sqrt{3} + 1\right)^{2} = \pi \cdot \left(13 + 4\sqrt{3}\right) = 13\pi + 4\pi\sqrt{3}$. The total area of the $13$ smaller circles is $13\pi$, so the shaded area is $\left(13\pi + 4\pi\sqrt{3}\right) - 13\pi = \boxed{\textbf{(A) } 4 \pi \sqrt{3}}$.

Solution 4

[asy] unitsize(20);filldraw(circle((0,0),2*sqrt(3)+1),rgb(0.5,0.5,0.5));filldraw(circle((-2,0),1),white);filldraw(circle((0,0),1),white);filldraw(circle((2,0),1),white);filldraw(circle((1,sqrt(3)),1),white);filldraw(circle((3,sqrt(3)),1),white);filldraw(circle((-1,sqrt(3)),1),white);filldraw(circle((-3,sqrt(3)),1),white);filldraw(circle((1,-1*sqrt(3)),1),white);filldraw(circle((3,-1*sqrt(3)),1),white);filldraw(circle((-1,-1*sqrt(3)),1),white);filldraw(circle((-3,-1*sqrt(3)),1),white);filldraw(circle((0,2*sqrt(3)),1),white);filldraw(circle((0,-2*sqrt(3)),1),white); pair A,B,C; A=(0,sqrt(3)*2); B=(-2,0); C=(0,0); draw(A--B); draw(A--C); draw(B--C); dot(A); dot(B); dot(C); label("A",A, N); label("B",B, W); label("C",C, S); [/asy]

In the diagram above, $AB=4$ and $BC=2$, so $AC=\sqrt{4^2-2^2}=2\sqrt{3}$. The larger circle's radius is $AC+1=2\sqrt{3}+1$, so the larger circle's area is $\pi\left(2\sqrt{3}+1\right)^2=\pi\left(13+4\sqrt{3}\right)=13\pi+4\pi\sqrt{3}$. Now, subtracting the combined area of the smaller circles gives $13\pi+4\pi\sqrt{3}-13\pi=\boxed{\textbf{(A) } 4 \pi \sqrt{3}}$.

Video Solution1

https://youtu.be/BBoWwpToBZ8

Education, the Study of Everything

Video Solution 2 by SpreadTheMathLove

https://www.youtube.com/watch?v=q9L66I9f9nI

Video Solution

https://youtu.be/WOz6CpF-6mI

~savannahsolver

Video Solution

https://youtu.be/NsQbhYfGh1Q?t=1793

~ pi_is_3.14

See Also

2019 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 17
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
2019 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
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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