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Difference between revisions of "2021 AMC 10B Problems/Problem 7"

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   <li>Without the loss of generality, we draw the circle with radius <math>7</math> north to <math>\ell.</math></li><p>
 
   <li>Without the loss of generality, we draw the circle with radius <math>7</math> north to <math>\ell.</math></li><p>
 
   <li>To maximize the area of the desired region, we draw the circle with radius <math>5</math> south to <math>\ell</math> by intuition.</li><p>
 
   <li>To maximize the area of the desired region, we draw the circle with radius <math>5</math> south to <math>\ell</math> by intuition.</li><p>
−
   <li>Now, we need to subtract the circle with radius <math>3</math> <i><b>at least</b></i>. The optimal situation is that the circle with radius <math>3</math> encompasses the circle with radius <math>1,</math> in which we do not need to subtract more. That is, the two smallest circles are on the same side of <math>\ell,</math> but can be on either side. The diagram below shows one possible configuration of the four circles:
+
   <li>Now, we need to subtract the circle with radius <math>3</math> <i><b>at least</b></i>. The optimal situation is that the circle with radius <math>3</math> encompasses the circle with radius <math>1,</math> in which we do not need to subtract more. That is, the two smallest circles are on the same side of <math>\ell,</math> but can be on either side. </li>
 +
</ol>
 +
The diagram below shows one possible configuration of the four circles:
 
<asy>
 
<asy>
 
/* diagram made by samrocksnature, edited by MRENTHUSIASM */
 
/* diagram made by samrocksnature, edited by MRENTHUSIASM */
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label("$\ell$",(10,0),E);
 
label("$\ell$",(10,0),E);
 
</asy>
 
</asy>
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</li>
 
−
</ol>
 
 
Together, the answer is <math>\pi\cdot7^2+\pi\cdot5^2-\pi\cdot3^2=\boxed{\textbf{(D) }65\pi}.</math>
 
Together, the answer is <math>\pi\cdot7^2+\pi\cdot5^2-\pi\cdot3^2=\boxed{\textbf{(D) }65\pi}.</math>
  

Revision as of 06:49, 4 September 2021

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

Suppose that line $\ell$ is horizontal, and each circle lies either north or south to $\ell.$ 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 $\ell$ by intuition.

  3. Now, we need to subtract the circle with radius $3$ at least. The optimal situation is that the circle with radius $3$ encompasses the circle with radius $1,$ in which 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 below shows one possible configuration of the four circles: [asy] /* diagram made by samrocksnature, edited by MRENTHUSIASM */ pair A=(10,0); pair B=(-10,0); draw(A--B); filldraw(circle((0,7),7),yellow); filldraw(circle((0,-5),5),yellow); filldraw(circle((0,-3),3),white); filldraw(circle((0,-1),1),white); dot((0,7)); dot((0,0)); draw((0,7)--(0,0)); label("$7$",(0,3.5),E); label("$\ell$",(10,0),E); [/asy] Together, the answer is $\pi\cdot7^2+\pi\cdot5^2-\pi\cdot3^2=\boxed{\textbf{(D) }65\pi}.$

~samrocksnature ~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

See Also

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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