Difference between revisions of "2021 AMC 10B Problems/Problem 7"
MRENTHUSIASM (talk | contribs) m (Added in See Also.) |
MRENTHUSIASM (talk | contribs) (Added in Solution 2, and changed $l$ to $\ell$ for better writing purposes.) |
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==Problem== | ==Problem== | ||
− | In a plane, four circles with radii <math>1,3,5,</math> and <math>7</math> are tangent to line <math> | + | In a plane, four circles with radii <math>1,3,5,</math> and <math>7</math> are tangent to line <math>\ell</math> at the same point <math>A,</math> but they may be on either side of <math>\ell</math>. Region <math>S</math> consists of all the points that lie inside exactly one of the four circles. What is the maximum possible area of region <math>S</math>? |
<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> | <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== | + | ==Solution 1== |
<asy> | <asy> | ||
/* diagram made by samrocksnature */ | /* diagram made by samrocksnature */ | ||
Line 16: | Line 16: | ||
draw((0,7)--(0,0)); | draw((0,7)--(0,0)); | ||
label("$7$",(0,3.5),E); | label("$7$",(0,3.5),E); | ||
− | label("$ | + | label("$\ell$",(-9,0),S); |
</asy> | </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> | 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> | ||
~ samrocksnature | ~ samrocksnature | ||
+ | |||
+ | ==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: | ||
+ | |||
+ | 1. Without the loss of generality, we draw the circle with radius <math>7</math> north to <math>\ell.</math> | ||
+ | |||
+ | 2. To maximize the area of the desired region, we draw the circle with radius <math>5</math> south to <math>\ell.</math> | ||
+ | |||
+ | 3. 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. | ||
+ | |||
+ | 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) == | == Video Solution by OmegaLearn (Area of Circles and Logic) == |
Revision as of 13:28, 5 March 2021
Contents
Problem
In a plane, four circles with radii and are tangent to line at the same point but they may be on either side of . Region consists of all the points that lie inside exactly one of the four circles. What is the maximum possible area of region ?
Solution 1
After a bit of wishful thinking and inspection, we find that the above configuration maximizes our area, which is
~ samrocksnature
Solution 2 (Explains Solution 1 Using Intuition)
Suppose each circle lies north or south to line We construct the circles one by one:
1. Without the loss of generality, we draw the circle with radius north to
2. To maximize the area of the desired region, we draw the circle with radius south to
3. Now, we need to subtract out the circle with radius at least. The optimal situation is that the circle with radius encompasses the circle with radius so that we do not need to subtract more. That is, the two smallest circles are on the same side of but can be on either side.
Together, the answer is
~MRENTHUSIASM
Video Solution by OmegaLearn (Area of Circles and Logic)
~ 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.