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

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(You also form an equilateral triangle with side length of <math>4</math> that goes through the center)~mn28407
 
(You also form an equilateral triangle with side length of <math>4</math> that goes through the center)~mn28407
  
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==Solution 3==
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==Solution 2==
 
We can form an equilateral triangle with side length 6 from the centers of 3 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 <math>R = \frac{abc}{4A}</math> or <math>R=\frac{a}{2\sin{A}}</math>, we get the radius <math>\frac{6}{\sqrt{3}}+1</math>.
 
We can form an equilateral triangle with side length 6 from the centers of 3 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 <math>R = \frac{abc}{4A}</math> or <math>R=\frac{a}{2\sin{A}}</math>, we get the radius <math>\frac{6}{\sqrt{3}}+1</math>.
  

Revision as of 22:06, 9 February 2019

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]

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

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 (distance point C to the edge of the circle). Using $30-60-90$ 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, $(2\sqrt{3}+1)^2 \pi = (13+4\sqrt{3})\pi$, and the sum of the areas of the smaller circles is $13\pi$, so the area of the dark region is $(13+4\sqrt{3})\pi-13\pi = 4\sqrt{3}\pi$, which implies $\implies \boxed{A}$

-eric2020

(You also form an equilateral triangle with side length of $4$ that goes through the center)~mn28407

Solution 2

We can form an equilateral triangle with side length 6 from the centers of 3 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 $R = \frac{abc}{4A}$ or $R=\frac{a}{2\sin{A}}$, we get the radius $\frac{6}{\sqrt{3}}+1$.

The shaded region is the area of the outer circle is $\pi((\frac{6}{\sqrt{3}}+1)^2-13) = \boxed{4\pi\sqrt{3} \implies A}$

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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