2021 AMC 10B Problems/Problem 7

Problem

In a plane, four circles with radii $1,3,5,$ and $7$ are tangent to line $\ell$ at the same point $A,$ but they may be on either side of $\ell$ . Region $S$ consists of all the points that lie inside exactly one of the four circles. What is the maximum possible area of region $S$ ?

$\textbf{(A) }24\pi \qquad \textbf{(B) }32\pi \qquad \textbf{(C) }64\pi \qquad \textbf{(D) }65\pi \qquad \textbf{(E) }84\pi$

Solution 1

$[asy] /* diagram made by samrocksnature */ pair A=(10,0); pair B=(-10,0); draw(A--B); draw(circle((0,-1),1)); draw(circle((0,-3),3)); draw(circle((0,-5),5)); draw(circle((0,7),7)); dot((0,7)); draw((0,7)--(0,0)); label("$7$",(0,3.5),E); label("$\ell$",(-9,0),S); [/asy]$ After a bit of wishful thinking and inspection, we find that the above configuration maximizes our area, which is $49 \pi + (25-9) \pi=65 \pi \rightarrow \boxed{\textbf{(D)}}$

~ samrocksnature

Solution 2 (Explains Solution 1 Using Intuition)

Suppose each circle lies north or south to line $\ell.$ We construct the circles one by one:

1. Without the loss of generality, we draw the circle with radius $7$ north to $\ell.$

2. To maximize the area of the desired region, we draw the circle with radius $5$ south to $\ell.$

3. Now, we need to subtract out the circle with radius $3$ at least. The optimal situation is that the circle with radius $3$ encompasses the circle with radius $1,$ so that we do not need to subtract more. That is, the two smallest circles are on the same side of $\ell,$ but can be on either side. The diagram in Solution 1 shows one possible positions of the four circles.

Together, the answer is $7^2\pi+5^2\pi-3^2\pi=\boxed{\textbf{(D) }65\pi}.$

~MRENTHUSIASM

Video Solution by OmegaLearn (Area of Circles and Logic)

https://youtu.be/yPIFmrJvUxM

~ pi_is_3.14

Video Solution by TheBeautyofMath

https://youtu.be/GYpAm8v1h-U?t=206

~IceMatrix

Video Solution by Interstigation

https://youtu.be/DvpN56Ob6Zw?t=555

~Interstigation

2021 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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

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