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

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~Flame_Thrower
 
~Flame_Thrower
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Note: This solution is time-consuming and practically useless.
 
  
 
==Solution 2 (More Algebraic Approach)==
 
==Solution 2 (More Algebraic Approach)==
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Solution by: Archimedes15
 
Solution by: Archimedes15
  
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Note: you could have just set <math>r</math> as <math>1</math> in the beginning.
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Note: using WLOG you could have just set <math>r</math> as <math>1</math> in the beginning.
  
 
==Solution 3==
 
==Solution 3==
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Each small semicircle is <math>\frac{1}{N^2}</math> of the large semicircle. Since <math>N</math> small semicircles make <math>\frac{1}{19}</math> of the large one, <math>\frac{N}{N^2} = \frac1{19}</math>. Solving this, we get <math>\boxed{19}</math>. ...
+
Each small semicircle is <math>\frac{1}{N^2}</math> of the large semicircle. Since <math>N</math> small semicircles make <math>\frac{1}{19}</math> of the large one, <math>\frac{N}{N^2} = \frac1{19}</math>. Solving this, we get <math>\boxed{19}</math>.  
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 +
... Don't think this one works
  
 
==Video Solution==
 
==Video Solution==
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~savannahsolver
 
~savannahsolver
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==Video Solution (HOW TO THINK CREATIVELY!!!)==
 +
https://youtu.be/qdyL6-Vmrrk
 +
 +
~Education, the Study of Everything
 +
 +
 +
 +
== Video Solution ==
 +
https://youtu.be/NsQbhYfGh1Q?t=1120
  
 
==See Also==
 
==See Also==

Latest revision as of 20:55, 24 October 2024

Problem

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 (Work using Answer Choices)

Use the answer choices and calculate them. The one that works is $\bold{\boxed{\text{(D)19}}}$.

~Flame_Thrower

Solution 2 (More Algebraic Approach)

Let the number of semicircles be $n$ and let the radius of each semicircle be $r$. To find the total area of all of the small semicircles, we have $n \cdot \frac{\pi \cdot r^2}{2}$.

Next, we have to find the area of the larger semicircle. The radius of the large semicircle can be deduced to be $n \cdot r$. So, the area of the larger semicircle is $\frac{\pi \cdot n^2 \cdot r^2}{2}$.

Now that we have found the area of both A and B, we can find the ratio. $\frac{A}{B}=\frac{1}{18}$, so part-to-whole ratio is $1:19$. When we divide the area of the small semicircles combined by the area of the larger semicircles, we get $\frac{1}{n}$. This is equal to $\frac{1}{19}$. By setting them equal, we find that $n = 19$. This is our answer, which corresponds to choice $\bold{\boxed{\text{(D)19}}}$.

Solution by: Archimedes15


Note: using WLOG you could have just set $r$ as $1$ in the beginning.

Solution 3

Each small semicircle is $\frac{1}{N^2}$ of the large semicircle. Since $N$ small semicircles make $\frac{1}{19}$ of the large one, $\frac{N}{N^2} = \frac1{19}$. Solving this, we get $\boxed{19}$.

... Don't think this one works

Video Solution

https://youtu.be/4KT0AtJZQzU

~savannahsolver

Video Solution (HOW TO THINK CREATIVELY!!!)

https://youtu.be/qdyL6-Vmrrk

~Education, the Study of Everything


Video Solution

https://youtu.be/NsQbhYfGh1Q?t=1120

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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