Circular Geometry
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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?
(source: https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_10B_Problems/Problem_25 )
This is my work can someone please tell me what I did wrong
Call the center of the circle with radius 1 o. Call the intersection point between the 2 circles c and d. Now let oc equal h. Since ac equals 2 we use pythag and get
h^2+1=4
so h=sqrt(3)
Now we know that the desired area is the difference of AOC(the ellipse sector) and the quarter circle. Now we do arithmetic and get sqrt(3)pi-pi
contest-math euclidean-geometry
edited Nov 22 at 6:33
darij grinberg
10.2k33061
asked Nov 22 at 6:04
user501887
133
add a comment |
up vote
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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?
(source: https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_10B_Problems/Problem_25 )
This is my work can someone please tell me what I did wrong
Call the center of the circle with radius 1 o. Call the intersection point between the 2 circles c and d. Now let oc equal h. Since ac equals 2 we use pythag and get
h^2+1=4
so h=sqrt(3)
Now we know that the desired area is the difference of AOC(the ellipse sector) and the quarter circle. Now we do arithmetic and get sqrt(3)pi-pi
contest-math euclidean-geometry
edited Nov 22 at 6:33
darij grinberg
10.2k33061
asked Nov 22 at 6:04
user501887
133
You need to provide a specific drawing for your work (or on top of existing diagram) In general - calculate the intersection between the two large circles and than subtract the small circle. Is this what you tried to do?
– Moti
Nov 22 at 6:42
You used the formula for the area of an ellipse, but that is NOT an ellipse.
– Aretino
Nov 22 at 14:24
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
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?
(source: https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_10B_Problems/Problem_25 )
This is my work can someone please tell me what I did wrong
Call the center of the circle with radius 1 o. Call the intersection point between the 2 circles c and d. Now let oc equal h. Since ac equals 2 we use pythag and get
h^2+1=4
so h=sqrt(3)
Now we know that the desired area is the difference of AOC(the ellipse sector) and the quarter circle. Now we do arithmetic and get sqrt(3)pi-pi
contest-math euclidean-geometry
edited Nov 22 at 6:33
darij grinberg
10.2k33061
asked Nov 22 at 6:04
user501887
133
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?
(source: https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_10B_Problems/Problem_25 )
This is my work can someone please tell me what I did wrong
Call the center of the circle with radius 1 o. Call the intersection point between the 2 circles c and d. Now let oc equal h. Since ac equals 2 we use pythag and get
h^2+1=4
so h=sqrt(3)
Now we know that the desired area is the difference of AOC(the ellipse sector) and the quarter circle. Now we do arithmetic and get sqrt(3)pi-pi
contest-math euclidean-geometry
contest-math euclidean-geometry
edited Nov 22 at 6:33
darij grinberg
10.2k33061
asked Nov 22 at 6:04
user501887
133
edited Nov 22 at 6:33
darij grinberg
10.2k33061
asked Nov 22 at 6:04
user501887
133
edited Nov 22 at 6:33
darij grinberg
10.2k33061
edited Nov 22 at 6:33
darij grinberg
10.2k33061
edited Nov 22 at 6:33
darij grinberg
10.2k33061
10.2k33061
asked Nov 22 at 6:04
user501887
133
asked Nov 22 at 6:04
user501887
133
asked Nov 22 at 6:04
user501887
133
133
You need to provide a specific drawing for your work (or on top of existing diagram) In general - calculate the intersection between the two large circles and than subtract the small circle. Is this what you tried to do?
– Moti
Nov 22 at 6:42
You used the formula for the area of an ellipse, but that is NOT an ellipse.
– Aretino
Nov 22 at 14:24
add a comment |
You need to provide a specific drawing for your work (or on top of existing diagram) In general - calculate the intersection between the two large circles and than subtract the small circle. Is this what you tried to do?
– Moti
Nov 22 at 6:42
You used the formula for the area of an ellipse, but that is NOT an ellipse.
– Aretino
Nov 22 at 14:24
You need to provide a specific drawing for your work (or on top of existing diagram) In general - calculate the intersection between the two large circles and than subtract the small circle. Is this what you tried to do?
– Moti
Nov 22 at 6:42
You need to provide a specific drawing for your work (or on top of existing diagram) In general - calculate the intersection between the two large circles and than subtract the small circle. Is this what you tried to do?
– Moti
Nov 22 at 6:42
You used the formula for the area of an ellipse, but that is NOT an ellipse.
– Aretino
Nov 22 at 14:24
You used the formula for the area of an ellipse, but that is NOT an ellipse.
– Aretino
Nov 22 at 14:24
add a comment |
1 Answer
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Label the intersections $C$ and $D$.
Obviously, $triangle CAB$ and $triangle DAB$ are equilateral. Meaning sector $BCD$ and $ACD$ has an angle of $120^circ$. The area of the shaded region plus the area of the circle in the middle is twice the area of the sector less the area of the triangle:
$$begin{align}
A&=2(A_{text{sector } ACD}-A_{triangle ACD})\
&=2left(frac{4pi}3-sqrt3right)
end{align}$$
Finally, subtract the area of the circle in the middle:
$$A_{text{shaded}}=frac{8pi}3-2sqrt3-pi=frac{5pi}3-2sqrt3$$
answered Nov 22 at 6:46
John Glenn
1,869424
add a comment |
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1 Answer
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up vote
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Label the intersections $C$ and $D$.
Obviously, $triangle CAB$ and $triangle DAB$ are equilateral. Meaning sector $BCD$ and $ACD$ has an angle of $120^circ$. The area of the shaded region plus the area of the circle in the middle is twice the area of the sector less the area of the triangle:
$$begin{align}
A&=2(A_{text{sector } ACD}-A_{triangle ACD})\
&=2left(frac{4pi}3-sqrt3right)
end{align}$$
Finally, subtract the area of the circle in the middle:
$$A_{text{shaded}}=frac{8pi}3-2sqrt3-pi=frac{5pi}3-2sqrt3$$
answered Nov 22 at 6:46
John Glenn
1,869424
add a comment |
up vote
0
down vote
Label the intersections $C$ and $D$.
Obviously, $triangle CAB$ and $triangle DAB$ are equilateral. Meaning sector $BCD$ and $ACD$ has an angle of $120^circ$. The area of the shaded region plus the area of the circle in the middle is twice the area of the sector less the area of the triangle:
$$begin{align}
A&=2(A_{text{sector } ACD}-A_{triangle ACD})\
&=2left(frac{4pi}3-sqrt3right)
end{align}$$
Finally, subtract the area of the circle in the middle:
$$A_{text{shaded}}=frac{8pi}3-2sqrt3-pi=frac{5pi}3-2sqrt3$$
answered Nov 22 at 6:46
John Glenn
1,869424
add a comment |
up vote
0
down vote
up vote
0
down vote
Label the intersections $C$ and $D$.
Obviously, $triangle CAB$ and $triangle DAB$ are equilateral. Meaning sector $BCD$ and $ACD$ has an angle of $120^circ$. The area of the shaded region plus the area of the circle in the middle is twice the area of the sector less the area of the triangle:
$$begin{align}
A&=2(A_{text{sector } ACD}-A_{triangle ACD})\
&=2left(frac{4pi}3-sqrt3right)
end{align}$$
Finally, subtract the area of the circle in the middle:
$$A_{text{shaded}}=frac{8pi}3-2sqrt3-pi=frac{5pi}3-2sqrt3$$
answered Nov 22 at 6:46
John Glenn
1,869424
Label the intersections $C$ and $D$.
Obviously, $triangle CAB$ and $triangle DAB$ are equilateral. Meaning sector $BCD$ and $ACD$ has an angle of $120^circ$. The area of the shaded region plus the area of the circle in the middle is twice the area of the sector less the area of the triangle:
$$begin{align}
A&=2(A_{text{sector } ACD}-A_{triangle ACD})\
&=2left(frac{4pi}3-sqrt3right)
end{align}$$
Finally, subtract the area of the circle in the middle:
$$A_{text{shaded}}=frac{8pi}3-2sqrt3-pi=frac{5pi}3-2sqrt3$$
answered Nov 22 at 6:46
John Glenn
1,869424
answered Nov 22 at 6:46
John Glenn
1,869424
answered Nov 22 at 6:46
John Glenn
1,869424
answered Nov 22 at 6:46
John Glenn
1,869424
1,869424
add a comment |
add a comment |
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You need to provide a specific drawing for your work (or on top of existing diagram) In general - calculate the intersection between the two large circles and than subtract the small circle. Is this what you tried to do?
– Moti
Nov 22 at 6:42
You used the formula for the area of an ellipse, but that is NOT an ellipse.
– Aretino
Nov 22 at 14:24