2018 AMC 10B Problems/Problem 7


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 $\bold{\boxed{\text{(D)19}}}$.


Note: This solution is time-consuming and is not recommended.

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: 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





~Education, the Study of Everything

Video Solution


See Also

2018 AMC 10B (ProblemsAnswer KeyResources)
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. AMC logo.png