Difference between revisions of "2004 AMC 10B Problems/Problem 25"
m (→Solution) |
|||
(12 intermediate revisions by 8 users not shown) | |||
Line 41: | Line 41: | ||
</asy> | </asy> | ||
− | == Solution == | + | == 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. | ||
Line 85: | Line 85: | ||
Take a look at the lower circle. The angle <math>ABC</math> is <math>60^\circ</math>, thus sector <math>ABC</math> is <math>1/6</math> of the circle. The same is true for sector <math>ABD</math> of the lower circle, and sectors <math>CAB</math> and <math>BAD</math> of the upper circle. | Take a look at the lower circle. The angle <math>ABC</math> is <math>60^\circ</math>, thus sector <math>ABC</math> is <math>1/6</math> of the circle. The same is true for sector <math>ABD</math> of the lower circle, and sectors <math>CAB</math> and <math>BAD</math> 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, <math>s,</math> is <math>\frac{\sqrt3} | + | 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, <math>s,</math> is <math>\frac{\sqrt3}4s^2.</math> |
Therefore, the area of the new shaded region is <math>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</math>. Lastly, we must subtract the area of the circle that we added earlier, <math>\pi</math>, and we get | Therefore, the area of the new shaded region is <math>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</math>. Lastly, we must subtract the area of the circle that we added earlier, <math>\pi</math>, and we get | ||
+ | |||
<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>. | ||
+ | |||
+ | == Solution 2 (Cheap - Out Of Time) == | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | ~lpieleanu (minor editing and extra details) | ||
+ | |||
+ | ==Video Solution 1 by SpreadTheMathLove= | ||
+ | https://www.youtube.com/watch?v=ajApE9bdUJQ | ||
== See also == | == See also == |
Latest revision as of 18:24, 26 June 2024
Contents
Problem
A circle of radius is internally tangent to two circles of radius
at points
and
, where
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?
Solution 1
The area of the small circle is . 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 and
be the intersections of the two large circles. Connect them to
and
to get the picture below:
We can see that the triangles and
are both equilateral with side
.
Take a look at the lower circle. The angle is
, thus sector
is
of the circle. The same is true for sector
of the lower circle, and sectors
and
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, is
Therefore, the area of the new shaded region is . Lastly, we must subtract the area of the circle that we added earlier,
, and we get
.
Solution 2 (Cheap - Out Of Time)
Notice that answer choices (C), (D), and (E) are exactly 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
(area
) From this, (A) or (B) are probably the answer choices (since you do need to subtract
at the end!) Since the angle measures of the sectors are
and we are working with the areas of some equilateral triangles, it only makes sense for the answer to have a
and not a
in them (There's no
angle either). So, our answer is
.
~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) | ||
Preceded by Problem 24 |
Followed by Last Question | |
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.