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Difference between revisions of "2018 AMC 10B Problems/Problem 7"

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<math>\textbf{(A) } 16 \qquad \textbf{(B) } 17 \qquad \textbf{(C) } 18 \qquad \textbf{(D) } 19 \qquad \textbf{(E) } 36</math>
 
<math>\textbf{(A) } 16 \qquad \textbf{(B) } 17 \qquad \textbf{(C) } 18 \qquad \textbf{(D) } 19 \qquad \textbf{(E) } 36</math>
  
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==Solution==
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==Solution 1==
  
 
Use the answer choices and calculate them. The one that works is D
 
Use the answer choices and calculate them. The one that works is D
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 +
==Solution 2==
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Let the radius of the smaller semicircles be <math>r</math>. This means that the radius of the larger semicircle is <math>nr</math>. The sum of the areas of the smaller semicircles is <math>\frac{\pinr^2}{2}</math>, while the area of the larger semicircle subtracted by the sum of the areas of the smaller semicircle is <math>\frac{\pin^2r^2}{2}-\frac{\pinr^2}{2}=\frac{\pir^2(n^2-n)}{2}</math>. Taking the ratio of the areas of the smaller semicircles to this value, we get <math>\frac{n}{n^2-n}=\frac{1}{18}</math>. Solving, we get <math>n=19</math>, or <math>\boxed{\textbf{(D) } 19}</math>.
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Solution by ElectroVortex
  
 
==See Also==
 
==See Also==

Revision as of 15:52, 16 February 2018

In the figure below, $N$ congruent semicircles lie on the diameter of a large semicircle, with their diameters covering the diameter of the large semicircle with no overlap. Let $A$ be the combined area of the small semicircles and $B$ be the area of the region inside the large semicircle but outside the semicircles. The ratio $A:B$ is $1:18$. What is $N$?


[asy] draw((0,0)--(18,0)); draw(arc((9,0),9,0,180)); filldraw(arc((1,0),1,0,180)--cycle,gray(0.8)); filldraw(arc((3,0),1,0,180)--cycle,gray(0.8)); filldraw(arc((5,0),1,0,180)--cycle,gray(0.8)); filldraw(arc((7,0),1,0,180)--cycle,gray(0.8)); label("...",(9,0.5)); filldraw(arc((11,0),1,0,180)--cycle,gray(0.8)); filldraw(arc((13,0),1,0,180)--cycle,gray(0.8)); filldraw(arc((15,0),1,0,180)--cycle,gray(0.8)); filldraw(arc((17,0),1,0,180)--cycle,gray(0.8)); [/asy]


$\textbf{(A) } 16 \qquad \textbf{(B) } 17 \qquad \textbf{(C) } 18 \qquad \textbf{(D) } 19 \qquad \textbf{(E) } 36$

Solution 1

Use the answer choices and calculate them. The one that works is D

Solution 2

Let the radius of the smaller semicircles be $r$. This means that the radius of the larger semicircle is $nr$. The sum of the areas of the smaller semicircles is $\frac{\pinr^2}{2}$ (Error compiling LaTeX. Unknown error_msg), while the area of the larger semicircle subtracted by the sum of the areas of the smaller semicircle is $\frac{\pin^2r^2}{2}-\frac{\pinr^2}{2}=\frac{\pir^2(n^2-n)}{2}$ (Error compiling LaTeX. Unknown error_msg). Taking the ratio of the areas of the smaller semicircles to this value, we get $\frac{n}{n^2-n}=\frac{1}{18}$. Solving, we get $n=19$, or $\boxed{\textbf{(D) } 19}$.

Solution by ElectroVortex

See Also

2018 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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