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

Latest revision as of 17:22, 16 August 2022

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:

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

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

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: $[asy] /* diagram made by samrocksnature, edited by MRENTHUSIASM */ pair A=(10,0); pair B=(-10,0); draw(A--B); filldraw(circle((0,7),7),yellow); filldraw(circle((0,-5),5),yellow); filldraw(circle((0,-3),3),white); filldraw(circle((0,-1),1),white); dot((0,0)); label("$A$",(0,0),(0,1.5)); label("$\ell$",(10,0),(1.5,0)); [/asy]$ 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

Video Solution

https://youtu.be/3jC_yOKA7xE

~Education, the Study of Everything

@@ Line 3: / Line 3: @@
 <math>\textbf{(A) }24\pi \qquad \textbf{(B) }32\pi \qquad \textbf{(C) }64\pi \qquad \textbf{(D) }65\pi \qquad \textbf{(E) }84\pi</math>
-==Solution 1==
+==Solution==
+Suppose that line <math>\ell</math> is horizontal, and each circle lies either north or south to <math>\ell.</math> We construct the circles one by one:
+<ol style="margin-left: 1.5em;">
+  <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 region <math>S,</math> we draw the circle with radius <math>5</math> south to <math>\ell.</math></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. </li>
+</ol>
+The diagram below shows one possible configuration of the four circles:
 <asy>
-/* diagram made by samrocksnature */
+/* diagram made by samrocksnature, edited by MRENTHUSIASM */
 pair A=(10,0);
 pair B=(-10,0);
 draw(A--B);
-draw(circle((0,-1),1));
+filldraw(circle((0,7),7),yellow);
-draw(circle((0,-3),3));
+filldraw(circle((0,-5),5),yellow);
-draw(circle((0,-5),5));
+filldraw(circle((0,-3),3),white);
-draw(circle((0,7),7));
+filldraw(circle((0,-1),1),white);
-dot((0,7));
+dot((0,0));
-draw((0,7)--(0,0));
+label("$A$",(0,0),(0,1.5));
-label("$7$",(0,3.5),E);
+label("$\ell$",(10,0),(1.5,0));
-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 <math>49 \pi + (25-9) \pi=65 \pi \rightarrow \boxed{\textbf{(D)}}</math>
+Together, the answer is <math>\pi\cdot7^2+\pi\cdot5^2-\pi\cdot3^2=\boxed{\textbf{(D) }65\pi}.</math>
-~ samrocksnature
+~samrocksnature ~MRENTHUSIASM
-==Solution 2 (Explains Solution 1 Using Intuition)==
-Suppose each circle lies north or south to line <math>\ell.</math> We construct the circles one by one:
-. Without the loss of generality, we draw the circle with radius <math>7</math> north to <math>\ell.</math>
-. To maximize the area of the desired region, we draw the circle with radius <math>5</math> south to <math>\ell.</math>
-. Now, we need to subtract out the circle with radius <math>3</math> <b>at least</b>. The optimal situation is that the circle with radius <math>3</math> encompasses the circle with radius <math>1,</math> so that 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 in Solution 1 shows one possible positions of the four circles.
-Together, the answer is <math>7^2\pi+5^2\pi-3^2\pi=\boxed{\textbf{(D) }65\pi}.</math>
-~MRENTHUSIASM
 == Video Solution by OmegaLearn (Area of Circles and Logic) ==
@@ Line 48: / Line 42: @@
 ~Interstigation
+==Video Solution==
+https://youtu.be/3jC_yOKA7xE
+~Education, the Study of Everything
 ==See Also==
 {{AMC10 box|year=2021|ab=B|num-b=6|num-a=8}}
 {{MAA Notice}}

2021 AMC 10B (Problems • Answer Key • Resources)
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