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2004 AMC 10A Problems/Problem 21: Difference between revisions

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==Solution 2==
==Solution 2==


As mentioned in Solution #1, we can make an equation for the area of the shaded region in terms of <math>\theta</math>.
As mentioned in Solution #1, we can make an equation for the area of the shaded region in terms of <imath>\theta</imath>.


<math>\implies\dfrac{2\theta}{2\pi} \cdot \pi +\dfrac{2(\pi-\theta)}{2\pi} \cdot (4\pi-\pi)+\dfrac{2\theta}{2\pi}(9\pi-4\pi)=\theta +3\pi-3\theta+5\theta=3\theta+3\pi</math>.
<imath>\implies\dfrac{2\theta}{2\pi} \cdot \pi +\dfrac{2(\pi-\theta)}{2\pi} \cdot (4\pi-\pi)+\dfrac{2\theta}{2\pi}(9\pi-4\pi)=\theta +3\pi-3\theta+5\theta=3\theta+3\pi</imath>.


So, the shaded region is <math>3\theta+3\pi</math>. This means that the unshaded region is <math>9\pi-(3\theta+3\pi)</math>.
So, the shaded region is <imath>3\theta+3\pi</imath>. This means that the unshaded region is <imath>9\pi-(3\theta+3\pi)</imath>.


Also, the shaded region is <math>\frac{8}{13}</math> of the unshaded region. Hence, we can now make an equation and solve for <math>\theta</math>.
Also, the shaded region is <imath>\frac{8}{13}</imath> of the unshaded region. Hence, we can now make an equation and solve for <imath>\theta</imath>.




<math>3\theta+3\pi=\frac{8}{13}(9\pi-(3\theta+3\pi)\implies 39\theta+39\pi=8(6\pi-3\theta)\implies 39\theta+39\pi=48\pi-24\theta</math>.
<imath>3\theta+3\pi=\frac{8}{13}(9\pi-(3\theta+3\pi)\implies 39\theta+39\pi=8(6\pi-3\theta)\implies 39\theta+39\pi=48\pi-24\theta</imath>.


Simplifying, we get <math>63\theta=9\pi\implies \theta=\boxed{\mathrm{(B)}\ \frac{\pi}{7}}</math>
Simplifying, we get <imath>63\theta=9\pi\implies \theta=\boxed{\mathrm{(B)}\ \frac{\pi}{7}}</imath>
 
==Solution 3==
First, we can find the area of the shaded region, which turns out to be 24\pi/7.


== Video Solution by OmegaLearn ==
== Video Solution by OmegaLearn ==

Revision as of 19:45, 10 November 2025

Problem

Two distinct lines pass through the center of three concentric circles of radii 3, 2, and 1. The area of the shaded region in the diagram is $\frac{8}{13}$ of the area of the unshaded region. What is the radian measure of the acute angle formed by the two lines? (Note: $\pi$ radian is $180$ degree.)

[asy] size(85); fill((-30,0)..(-24,18)--(0,0)--(-24,-18)..cycle,gray(0.7)); fill((30,0)..(24,18)--(0,0)--(24,-18)..cycle,gray(0.7)); fill((-20,0)..(0,20)--(0,-20)..cycle,white); fill((20,0)..(0,20)--(0,-20)..cycle,white); fill((0,20)..(-16,12)--(0,0)--(16,12)..cycle,gray(0.7)); fill((0,-20)..(-16,-12)--(0,0)--(16,-12)..cycle,gray(0.7)); fill((0,10)..(-10,0)--(10,0)..cycle,white); fill((0,-10)..(-10,0)--(10,0)..cycle,white); fill((-10,0)..(-8,6)--(0,0)--(-8,-6)..cycle,gray(0.7)); fill((10,0)..(8,6)--(0,0)--(8,-6)..cycle,gray(0.7)); draw(Circle((0,0),10),linewidth(0.7)); draw(Circle((0,0),20),linewidth(0.7)); draw(Circle((0,0),30),linewidth(0.7)); draw((-28,-21)--(28,21),linewidth(0.7)); draw((-28,21)--(28,-21),linewidth(0.7));[/asy]

$\mathrm{(A) \ } \frac{\pi}{8} \qquad \mathrm{(B) \ } \frac{\pi}{7} \qquad \mathrm{(C) \ } \frac{\pi}{6} \qquad \mathrm{(D) \ } \frac{\pi}{5} \qquad \mathrm{(E) \ } \frac{\pi}{4}$

Solution 1

Let the area of the shaded region be $S$, the area of the unshaded region be $U$, and the acute angle that is formed by the two lines be $\theta$. We can set up two equations between $S$ and $U$:

$S+U=9\pi$

$S=\dfrac{8}{13}U$

Thus $\dfrac{21}{13}U=9\pi$, and $U=\dfrac{39\pi}{7}$, and thus $S=\dfrac{8}{13}\cdot \dfrac{39\pi}{7}=\dfrac{24\pi}{7}$.

Now we can make a formula for the area of the shaded region in terms of $\theta$:

$\dfrac{2\theta}{2\pi} \cdot \pi +\dfrac{2(\pi-\theta)}{2\pi} \cdot (4\pi-\pi)+\dfrac{2\theta}{2\pi}(9\pi-4\pi)=\theta +3\pi-3\theta+5\theta=3\theta+3\pi=\dfrac{24\pi}{7}$

Thus $3\theta=\dfrac{3\pi}{7}\Rightarrow \theta=\dfrac{\pi}{7}\Rightarrow\boxed{\mathrm{(B)}\ \frac{\pi}{7}}$

Solution 2

As mentioned in Solution #1, we can make an equation for the area of the shaded region in terms of $\theta$.

$\implies\dfrac{2\theta}{2\pi} \cdot \pi +\dfrac{2(\pi-\theta)}{2\pi} \cdot (4\pi-\pi)+\dfrac{2\theta}{2\pi}(9\pi-4\pi)=\theta +3\pi-3\theta+5\theta=3\theta+3\pi$.

So, the shaded region is $3\theta+3\pi$. This means that the unshaded region is $9\pi-(3\theta+3\pi)$.

Also, the shaded region is $\frac{8}{13}$ of the unshaded region. Hence, we can now make an equation and solve for $\theta$.


$3\theta+3\pi=\frac{8}{13}(9\pi-(3\theta+3\pi)\implies 39\theta+39\pi=8(6\pi-3\theta)\implies 39\theta+39\pi=48\pi-24\theta$.

Simplifying, we get $63\theta=9\pi\implies \theta=\boxed{\mathrm{(B)}\ \frac{\pi}{7}}$

Solution 3

First, we can find the area of the shaded region, which turns out to be 24\pi/7.

Video Solution by OmegaLearn

https://youtu.be/t3EWtMnJu2Y?t=49

~ pi_is_3.14

See also

2004 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
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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