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"

(→‎Solution 1)
m (→‎Solution 1)
 
(8 intermediate revisions by 5 users not shown)
Line 8: Line 8:
  
 
<math>\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)</math>
 
<math>\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)</math>
−
 
−
==Video Solution 1==
 
−
https://youtu.be/BBoWwpToBZ8
 
−
 
−
Education, the Study of Everything
 
−
 
  
 
==Solution 1==
 
==Solution 1==
Line 19: Line 13:
 
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);
 
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;
+
pair O,A,B,C,H;
 
O=(0,0);
 
O=(0,0);
 
A=(-1,sqrt(3));
 
A=(-1,sqrt(3));
 
B=(1,sqrt(3));
 
B=(1,sqrt(3));
 
C=(0,sqrt(3)*2);
 
C=(0,sqrt(3)*2);
 +
H=(0,sqrt(3));
 
draw(O--A);
 
draw(O--A);
 
draw(A--C);
 
draw(A--C);
Line 38: Line 33:
 
label("B",B,E);
 
label("B",B,E);
 
label("C",C, N);
 
label("C",C, N);
 +
label("H",H, NE);
 
</asy>
 
</asy>
  
−
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>.
+
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 <math>2\overline{OH} + 1</math>. Using <math>30^{\circ}-60^{\circ}-90^{\circ}</math> triangles, we know <math>\overline{OH} = \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>\left(2\sqrt{3}+1\right)^2 \pi = \left(13+4\sqrt{3}\right)\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>\left(13+4\sqrt{3}\right)\pi-13\pi = \boxed{\textbf{(A) } 4 \pi \sqrt{3}}</math>.
 
The area of the larger circle is thus <math>\left(2\sqrt{3}+1\right)^2 \pi = \left(13+4\sqrt{3}\right)\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>\left(13+4\sqrt{3}\right)\pi-13\pi = \boxed{\textbf{(A) } 4 \pi \sqrt{3}}</math>.
Line 46: Line 42:
 
==Solution 2==
 
==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. 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 as <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}{4K}</math> or <math>R=\frac{a}{2\sin{A}}</math>, we get the radius as <math>\frac{6}{\sqrt{3}}+1</math>.
  
 
The shaded area is thus <math>\pi((\frac{6}{\sqrt{3}}+1)^2-13) = \boxed{\textbf{(A) } 4 \pi \sqrt{3}}</math>.
 
The shaded area is thus <math>\pi((\frac{6}{\sqrt{3}}+1)^2-13) = \boxed{\textbf{(A) } 4 \pi \sqrt{3}}</math>.
Line 74: Line 70:
  
 
In the diagram above, <math>AB=4</math> and <math>BC=2</math>, so <math>AC=\sqrt{4^2-2^2}=2\sqrt{3}</math>. The larger circle's radius is <math>AC+1=2\sqrt{3}+1</math>, so the larger circle's area is <math>\pi\left(2\sqrt{3}+1\right)^2=\pi\left(13+4\sqrt{3}\right)=13\pi+4\pi\sqrt{3}</math>. Now, subtracting the combined area of the smaller circles gives <math>13\pi+4\pi\sqrt{3}-13\pi=\boxed{\textbf{(A) } 4 \pi \sqrt{3}}</math>.
 
In the diagram above, <math>AB=4</math> and <math>BC=2</math>, so <math>AC=\sqrt{4^2-2^2}=2\sqrt{3}</math>. The larger circle's radius is <math>AC+1=2\sqrt{3}+1</math>, so the larger circle's area is <math>\pi\left(2\sqrt{3}+1\right)^2=\pi\left(13+4\sqrt{3}\right)=13\pi+4\pi\sqrt{3}</math>. Now, subtracting the combined area of the smaller circles gives <math>13\pi+4\pi\sqrt{3}-13\pi=\boxed{\textbf{(A) } 4 \pi \sqrt{3}}</math>.
 +
 +
==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==
 
==Video Solution==
Line 79: Line 83:
  
 
~savannahsolver
 
~savannahsolver
 +
 +
== Video Solution by OmegaLearn ==
 +
https://youtu.be/NsQbhYfGh1Q?t=1793
 +
 +
~ pi_is_3.14
  
 
==See Also==
 
==See Also==

Latest revision as of 07:20, 20 November 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,H; O=(0,0); A=(-1,sqrt(3)); B=(1,sqrt(3)); C=(0,sqrt(3)*2); H=(0,sqrt(3)); 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); label("H",H, NE); [/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 $2\overline{OH} + 1$. Using $30^{\circ}-60^{\circ}-90^{\circ}$ triangles, we know $\overline{OH} = \sqrt{3}$. 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 by OmegaLearn

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

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=204803"