Difference between revisions of "2021 AMC 10B Problems/Problem 7"
MRENTHUSIASM (talk | contribs) m (Colored the diagram so it makes things clearer.) |
MRENTHUSIASM (talk | contribs) (Cleaned up the page by combining solutions. Also, I credited samrocksnature as the primary author.) |
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<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 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 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: | ||
<asy> | <asy> | ||
/* diagram made by samrocksnature, edited by MRENTHUSIASM */ | /* diagram made by samrocksnature, edited by MRENTHUSIASM */ | ||
Line 15: | Line 20: | ||
filldraw(circle((0,-1),1),white); | filldraw(circle((0,-1),1),white); | ||
dot((0,7)); | dot((0,7)); | ||
+ | dot((0,0)); | ||
draw((0,7)--(0,0)); | draw((0,7)--(0,0)); | ||
label("$7$",(0,3.5),E); | label("$7$",(0,3.5),E); | ||
label("$\ell$",(-9,0),S); | label("$\ell$",(-9,0),S); | ||
</asy> | </asy> | ||
− | + | </li> | |
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</ol> | </ol> | ||
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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> | ||
− | ~MRENTHUSIASM | + | ~samrocksnature ~MRENTHUSIASM |
== Video Solution by OmegaLearn (Area of Circles and Logic) == | == Video Solution by OmegaLearn (Area of Circles and Logic) == |
Revision as of 06:39, 4 September 2021
Contents
[hide]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
Suppose that line is horizontal, and each circle lies either north or south to We construct the circles one by one:
- Without the loss of generality, we draw the circle with radius north to
- To maximize the area of the desired region, we draw the circle with radius south to by intuition.
- Now, we need to subtract the circle with radius at least. The optimal situation is that the circle with radius encompasses the circle with radius in which 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. The diagram below shows one possible configuration of the four circles:
Together, the answer is
~samrocksnature ~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.