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

Revision as of 00:20, 5 December 2019

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

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 $\boxed{19}$ .

Revision as of 13:29, 15 August 2019 (view source) Xyab (talk \| contribs) m (→‎Solution 2 (More Algebraic Approach)) ← Older edit		Revision as of 00:20, 5 December 2019 (view source) Anmol04 (talk \| contribs) m (added "Problem" on the top) Newer edit →
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		+	== Problem ==
		+
	In the figure below, <math>N</math> congruent semicircles lie on the diameter of a large semicircle, with their diameters covering the diameter of the large semicircle with no overlap. Let <math>A</math> be the combined area of the small semicircles and <math>B</math> be the area of the region inside the large semicircle but outside the semicircles. The ratio <math>A:B</math> is <math>1:18</math>. What is <math>N</math>?		In the figure below, <math>N</math> congruent semicircles lie on the diameter of a large semicircle, with their diameters covering the diameter of the large semicircle with no overlap. Let <math>A</math> be the combined area of the small semicircles and <math>B</math> be the area of the region inside the large semicircle but outside the semicircles. The ratio <math>A:B</math> is <math>1:18</math>. What is <math>N</math>?

2018 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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