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Difference between revisions of "2004 AMC 10B Problems/Problem 25"

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== Solution ==
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== Solution 1 ==
  
 
The area of the small circle is <math>\pi</math>. We can add it to the shaded region, compute the area of the new region, and then subtract the area of the small circle from the result.
 
The area of the small circle is <math>\pi</math>. We can add it to the shaded region, compute the area of the new region, and then subtract the area of the small circle from the result.
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<math>\left( \frac 83 \pi - 2\sqrt 3 \right) - \pi = \boxed{\mathrm{(B)\ } \frac 53 \pi - 2\sqrt 3 }</math>.
 
<math>\left( \frac 83 \pi - 2\sqrt 3 \right) - \pi = \boxed{\mathrm{(B)\ } \frac 53 \pi - 2\sqrt 3 }</math>.
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== Solution 2 (Cheap - Out Of Time) ==
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Notice that answer choices (C), (D), and (E) are exactly <math>\pi</math> more than answer choices (A), (B). This suggests that answer choices (C), (D), and (E) are meant to trick you into forgetting to subtract the circle with radius <math>1</math> (area <math>\pi</math>) From this, (A) or (B) are probably the answer choices (since you do need to subtract <math>\pi</math> at the end!) Since the angle measures of the sectors are <math>120^{\circ},</math> and we are working with the areas of some equilateral triangles, it only makes sense for the answer to have a <math>\sqrt 3</math> and not a <math>\sqrt 2</math> in them (There's no <math>45^{\circ}</math> angle either). So, our answer is <math>\boxed{\mathrm{(B)\ } \frac 53 \pi - 2\sqrt 3 }</math>.
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~lpieleanu (minor editing and extra details)
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==Video Solution 1 by SpreadTheMathLove=
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https://www.youtube.com/watch?v=ajApE9bdUJQ
  
 
== See also ==
 
== See also ==

Latest revision as of 17:24, 26 June 2024

Problem

A circle of radius $1$ is internally tangent to two circles of radius $2$ at points $A$ and $B$, where $AB$ is a diameter of the smaller circle. What is the area of the region, shaded in the picture, that is outside the smaller circle and inside each of the two larger circles?

$\mathrm{(A) \ } \frac{5}{3} \pi - 3\sqrt 2 \qquad \mathrm{(B) \ } \frac{5}{3} \pi - 2\sqrt 3 \qquad \mathrm{(C) \ } \frac{8}{3} \pi - 3\sqrt 3 \qquad \mathrm{(D) \ } \frac{8}{3} \pi - 3\sqrt 2 \qquad \mathrm{(E) \ } \frac{8}{3} \pi - 2\sqrt 3$


[asy] unitsize(1cm); defaultpen(0.8); pair O=(0,0), A=(0,1), B=(0,-1); path bigc1 = Circle(A,2), bigc2 = Circle(B,2), smallc = Circle(O,1); pair[] P = intersectionpoints(bigc1, bigc2); filldraw( arc(A,P[0],P[1])--arc(B,P[1],P[0])--cycle, lightgray, black ); draw(bigc1); draw(bigc2); unfill(smallc); draw(smallc); dot(O); dot(A); dot(B); label("$A$",A,N); label("$B$",B,S); draw( O--dir(30) ); draw( A--(A+2*dir(30)) ); draw( B--(B+2*dir(210)) ); label("$1$", O--dir(30), N ); label("$2$", A--(A+2*dir(30)), N ); label("$2$", B--(B+2*dir(210)), S ); [/asy]

Solution 1

The area of the small circle is $\pi$. We can add it to the shaded region, compute the area of the new region, and then subtract the area of the small circle from the result.

Let $C$ and $D$ be the intersections of the two large circles. Connect them to $A$ and $B$ to get the picture below:

[asy] unitsize(1.5cm); defaultpen(0.8); pair O=(0,0), A=(0,1), B=(0,-1); path bigc1 = Circle(A,2), bigc2 = Circle(B,2), smallc = Circle(O,1); pair[] P = intersectionpoints(bigc1, bigc2); filldraw( arc(A,P[0],P[1])--arc(B,P[1],P[0])--cycle, lightgray, black ); draw(bigc1); draw(bigc2); dot(A); dot(B); label("$A$",A,N); label("$B$",B,S); /* dot(O); unfill(smallc); draw(smallc); draw( O--dir(30) ); draw( A--(A+2*dir(30)) ); draw( B--(B+2*dir(210)) ); label("$1$", O--dir(30), N ); label("$2$", A--(A+2*dir(30)), N ); label("$2$", B--(B+2*dir(210)), S ); */ label("$C$",P[0],W); label("$D$",P[1],E); draw( P[0]--A--P[1] ); draw( P[0]--B--P[1] ); draw( A--B ); [/asy]

We can see that the triangles $\triangle ABC$ and $\triangle ABD$ are both equilateral with side $2$.

Take a look at the lower circle. The angle $ABC$ is $60^\circ$, thus sector $ABC$ is $1/6$ of the circle. The same is true for sector $ABD$ of the lower circle, and sectors $CAB$ and $BAD$ of the upper circle.

If we now sum the areas of these four sectors, we will almost get the area of the new shaded region - except that each of the two equilateral triangles will be counted twice. The area of an equilateral triangle given its side, $s,$ is $\frac{\sqrt3}4s^2.$

Therefore, the area of the new shaded region is $4\cdot \left( \frac 16 \cdot \pi\cdot 2^2 \right) - 2 \cdot \left( 4 \cdot \frac{\sqrt3}4 \right) = \frac 83 \pi - 2\sqrt 3$. Lastly, we must subtract the area of the circle that we added earlier, $\pi$, and we get

$\left( \frac 83 \pi - 2\sqrt 3 \right) - \pi = \boxed{\mathrm{(B)\ } \frac 53 \pi - 2\sqrt 3 }$.

Solution 2 (Cheap - Out Of Time)

Notice that answer choices (C), (D), and (E) are exactly $\pi$ more than answer choices (A), (B). This suggests that answer choices (C), (D), and (E) are meant to trick you into forgetting to subtract the circle with radius $1$ (area $\pi$) From this, (A) or (B) are probably the answer choices (since you do need to subtract $\pi$ at the end!) Since the angle measures of the sectors are $120^{\circ},$ and we are working with the areas of some equilateral triangles, it only makes sense for the answer to have a $\sqrt 3$ and not a $\sqrt 2$ in them (There's no $45^{\circ}$ angle either). So, our answer is $\boxed{\mathrm{(B)\ } \frac 53 \pi - 2\sqrt 3 }$.

~lpieleanu (minor editing and extra details)

=Video Solution 1 by SpreadTheMathLove

https://www.youtube.com/watch?v=ajApE9bdUJQ

See also

2004 AMC 10B (Problems • Answer Key • Resources)
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