Difference between revisions of "2021 AMC 10B Problems/Problem 7"
MRENTHUSIASM (talk | contribs) m (→Solution) |
MRENTHUSIASM (talk | contribs) m (→Video Solution by Interstigation) |
||
(9 intermediate revisions by 4 users not shown) | |||
Line 8: | Line 8: | ||
<ol style="margin-left: 1.5em;"> | <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>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 | + | <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. 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 */ | ||
Line 19: | Line 21: | ||
filldraw(circle((0,-3),3),white); | filldraw(circle((0,-3),3),white); | ||
filldraw(circle((0,-1),1),white); | filldraw(circle((0,-1),1),white); | ||
− | |||
dot((0,0)); | dot((0,0)); | ||
− | + | label("$A$",(0,0),(0,1.5)); | |
− | + | label("$\ell$",(10,0),(1.5,0)); | |
− | label("$\ell$",(10,0), | ||
</asy> | </asy> | ||
− | |||
− | |||
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> | ||
Line 44: | Line 42: | ||
~Interstigation | ~Interstigation | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/3jC_yOKA7xE | ||
+ | |||
+ | ~Education, the Study of Everything | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2021|ab=B|num-b=6|num-a=8}} | {{AMC10 box|year=2021|ab=B|num-b=6|num-a=8}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 17:22, 16 August 2022
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
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 region we draw the circle with radius south to
- 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
Video Solution
~Education, the Study of Everything
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.