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 "2019 AMC 10A Problems/Problem 16"

m (Added name)
(Fixed grammar)
Line 34: Line 34:
 
</asy>
 
</asy>
  
−
In the diagram above, notice that triangle <math>OAB</math> and triangle <math>ABC</math> are congruent and equilateral with side-length 2. We can see the radius of the larger circle is two times the altitude of <math>OAB</math> plus 1 (distance point C to the edge of the circle). Using <math>30-60-90</math> triangles, we know the altitude is <math>\sqrt{3}</math>. Therefore, the radius of the larger circle is <math>2\sqrt{3}+1</math>.
+
In the diagram above, notice that triangle <math>OAB</math> and triangle <math>ABC</math> are congruent and equilateral with side length <math>2</math>. We can see the radius of the larger circle is two times the altitude of <math>OAB</math> plus <math>1</math> (the distance from point <math>C</math> to the edge of the circle). Using <math>30^{\circ}-60^{\circ}-90^{\circ}</math> triangles, we know the altitude is <math>\sqrt{3}</math>. Therefore, the radius of the larger circle is <math>2\sqrt{3}+1</math>.
  
−
The area of the larger circle is thus, <math>(2\sqrt{3}+1)^2 \pi = (13+4\sqrt{3})\pi</math>, and the sum of the areas of the smaller circles is <math>13\pi</math>, so the area of the dark region is <math>(13+4\sqrt{3})\pi-13\pi = 4\sqrt{3}\pi</math>, which implies <math>\implies \boxed{A}</math>
+
The area of the larger circle is thus <math>(2\sqrt{3}+1)^2 \pi = (13+4\sqrt{3})\pi</math>, and the sum of the areas of the smaller circles is <math>13\pi</math>, so the area of the dark region is <math>(13+4\sqrt{3})\pi-13\pi = 4\sqrt{3}\pi</math>, which implies the answer is <math>\boxed{A}</math>
  
 
-eric2020
 
-eric2020
Line 43: Line 43:
  
 
==Solution 2==
 
==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 <math>6</math> 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 <math>1</math>. 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>.
  
 
The shaded region is the area of the outer circle is <math>\pi((\frac{6}{\sqrt{3}}+1)^2-13) = \boxed{4\pi\sqrt{3} \implies A}</math>
 
The shaded region is the area of the outer circle is <math>\pi((\frac{6}{\sqrt{3}}+1)^2-13) = \boxed{4\pi\sqrt{3} \implies A}</math>
  
 
==Solution 3==
 
==Solution 3==
−
Like 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. We can find the height of this triangle to be <math>3\sqrt{3}</math>. Then, we can form another equilateral triangle from the centers of the 2nd and 3rd circles in the third row and the center of the bottom circle with length 2. The height of this triangle is clearly <math>\sqrt{3}</math>. Therefore the diameter of the large circle is <math>4\sqrt{3} + 2</math> and the radius is <math>\frac{4\sqrt{3}+2}{2} = 2\sqrt{3} + 1</math>. Finding the area is: <math>A = \pi r^{2} = \pi (2\sqrt{3} + 1)^{2} = \pi \cdot (13 + 4\sqrt{3}) = (13\pi + 4\pi\sqrt{3})</math> The area of the 13 circles are <math>13\pi</math> and so the shaded area is <math>(13\pi + 4\pi\sqrt{3}) - 13\pi = \boxed{4\pi\sqrt{3} \implies A}</math>
+
Like in Solution 2, we can form an equilateral triangle with side length <math>6</math> from the centers of three of the unit circles tangent to the outer circle. We can find the height of this triangle to be <math>3\sqrt{3}</math>. Then, we can form another equilateral triangle from the centers of the 2nd and 3rd circles in the third row and the center of the bottom circle with side length <math>2</math>. The height of this triangle is clearly <math>\sqrt{3}</math>. Therefore the diameter of the large circle is <math>4\sqrt{3} + 2</math> and the radius is <math>\frac{4\sqrt{3}+2}{2} = 2\sqrt{3} + 1</math>. The area is thus <math>A = \pi r^{2} = \pi (2\sqrt{3} + 1)^{2} = \pi \cdot (13 + 4\sqrt{3}) = (13\pi + 4\pi\sqrt{3})</math>. The total area of the 13 circles is <math>13\pi</math> and so the shaded area is <math>(13\pi + 4\pi\sqrt{3}) - 13\pi = \boxed{4\pi\sqrt{3} \implies A}</math>
  
 
<math>\linebreak</math>
 
<math>\linebreak</math>

Revision as of 18:50, 10 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 $1$ (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 $(2\sqrt{3}+1)^2 \pi = (13+4\sqrt{3})\pi$, 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 the answer is $\boxed{A}$

-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 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 $R = \frac{abc}{4A}$ 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 $\pi((\frac{6}{\sqrt{3}}+1)^2-13) = \boxed{4\pi\sqrt{3} \implies A}$

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 2nd and 3rd 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 is thus $A = \pi r^{2} = \pi (2\sqrt{3} + 1)^{2} = \pi \cdot (13 + 4\sqrt{3}) = (13\pi + 4\pi\sqrt{3})$. The total area of the 13 circles is $13\pi$ and so the shaded area is $(13\pi + 4\pi\sqrt{3}) - 13\pi = \boxed{4\pi\sqrt{3} \implies A}$

$\linebreak$

~DBlack2021

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10A_Problems/Problem_16&oldid=101757"