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

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==Solution 2==
 
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We can form an equilateral triangle of side length <math>4</math> with three unit circles and their radii included in each side, one going through the center and the other two connecting to a unit circle tangent to the larger circle. Therefore, based off <math>30 - 60 - 90</math> triangles, the altitude of the triangle from the center of the middle unit circle to the center of the outer unit circle is <math>2 \sqrt{3}</math>. Therefore, adding the radius of the last unit circle to the altitude gives of the radius of the large circle, <math>1 + 2 \sqrt{3}</math>. Finding the area of the large circle gives us <math>13 \pi + 4 \pi \sqrt{3}</math>. The combined area of the unit circles is <math>13 \pi</math>. Thus, the area of the shaded region is the area of the larger circle minus the areas of the unit circles:
 
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<math>13 \pi + 4 \pi \sqrt{3} - 13 \pi = \boxed {4 \pi \sqrt{3} \implies (A)} </math>
 
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==See Also==
 
==See Also==

Revision as of 18:45, 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

[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

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

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

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