# Very HARD geomerty question I invented... Challenge!!!!! Watch

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**Two circles, each with radius r and centres A and B, are drawn on the surface of a sphere with radius 3r such that the area of the overlap of the circles on the surface of the sphere is a tenth of the total surface area of the sphere. Find the length of the straight line AB.**

This has GOT to be hard... I don't have a clue how to do it myself, just made it up and thought... crikey... I don't even know where to begin.

Maybe some of the university level mathematicians/phsyicists can do this.

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#5

(Original post by

sweet... you just confirmed it being beautifully hard...

**El Stevo**)sweet... you just confirmed it being beautifully hard...

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#6

(Original post by

I think El Stevo means are you taking A as a point of the sphere and measuring radii along great circles of the sphere, or is A within the sphere and the radius measured directly to the points of the circle.

**RichE**)I think El Stevo means are you taking A as a point of the sphere and measuring radii along great circles of the sphere, or is A within the sphere and the radius measured directly to the points of the circle.

from what mik said, i think the circle is a 'lens', and the radius he has given is from the highest point of the lens to the rim....

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#7

(Original post by

This has GOT to be hard... I don't have a clue how to do it myself, just made it up and thought... crikey... I don't even know where to begin.

Maybe some of the university level mathematicians/phsyicists can do this.

**mik1w**)**Two circles, each with radius r and centres A and B, are drawn on the surface of a sphere with radius 3r such that the area of the overlap of the circles on the surface of the sphere is a tenth of the total surface area of the sphere. Find the length of the straight line AB.**This has GOT to be hard... I don't have a clue how to do it myself, just made it up and thought... crikey... I don't even know where to begin.

Maybe some of the university level mathematicians/phsyicists can do this.

Area of Circle 1 + Area of Circle 2 - [ 2 * Area of Overlap ]

=> pi(r^2) + pi(r^2) - [ 2 * A Tenth of Surface area of Sphere with radius 3r]

=> 2pi(r^2) - 2.([1/10].4pi(3r)^2)

=> 2pi(r^2) - (36/5).pi(r^2)

=> (-26/5)pi(r^2)

So the entire area of overlapping of the two circles is negative! (i.e. in practical terms this is impossible).

However I imagine that the distance between A and B is imaginary?

Euclid.

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yeah. if you imagine the perfect sphere is Earth then drawing a circle on the ground will represent that circle. I think it's non-Euclidean geometry..!

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#9

(Original post by

Assume the two circles overlap and are placed firmly on the sphere. Then the area of the entire two circles is:

Area of Circle 1 + Area of Circle 2 - [ 2 * Area of Overlap ]

=> pi(r^2) + pi(r^2) - [ 2 * A Tenth of Surface area of Sphere with radius 3r]

=> 2pi(r^2) - 2.([1/10].4pi(3r)^2)

=> 2pi(r^2) - (36/5).pi(r^2)

=> (-26/5)pi(r^2)

So the entire area of overlapping of the two circles is negative! (i.e. in practical terms this is impossible).

However I imagine that the distance between A and B is imaginary?

Euclid.

**Euclid**)Assume the two circles overlap and are placed firmly on the sphere. Then the area of the entire two circles is:

Area of Circle 1 + Area of Circle 2 - [ 2 * Area of Overlap ]

=> pi(r^2) + pi(r^2) - [ 2 * A Tenth of Surface area of Sphere with radius 3r]

=> 2pi(r^2) - 2.([1/10].4pi(3r)^2)

=> 2pi(r^2) - (36/5).pi(r^2)

=> (-26/5)pi(r^2)

So the entire area of overlapping of the two circles is negative! (i.e. in practical terms this is impossible).

However I imagine that the distance between A and B is imaginary?

Euclid.

edit: after an beautiful el stevo botch, i may be gettign somewhere...

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#10

i have the area of the two overlapping spherical caps to be: {18.Pi.(r^2)}{(4/5)-([sqrt3]/2)}

edit: no i don't ... schoolboy error at start...

edit: no i don't ... schoolboy error at start...

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#13

I'm going to pass this onto my maths teacher to see what she comes up with I'm too busy with exams to even attempt this Plus, I don't think we have even been taught how to even attempt to tackle this question :P

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#14

ok... seen as you just "made it up" as you said... how do you know thats its actually possible to work out!! it could be impossible to work out (not enough data given in the whestion) or soemthing like that. although i havnt done M2 yet (doing M1 now) so im not too sure about the formulas. just wondering how you all think its possible when noones got the answer?

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#15

(Original post by

I'm going to pass this onto my maths teacher to see what she comes up with I'm too busy with exams to even attempt this Plus, I don't think we have even been taught how to even attempt to tackle this question :P

**Jedi Ezekiel**)I'm going to pass this onto my maths teacher to see what she comes up with I'm too busy with exams to even attempt this Plus, I don't think we have even been taught how to even attempt to tackle this question :P

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(Original post by

ok... seen as you just "made it up" as you said... how do you know thats its actually possible to work out!! it could be impossible to work out (not enough data given in the whestion) or soemthing like that. although i havnt done M2 yet (doing M1 now) so im not too sure about the formulas. just wondering how you all think its possible when noones got the answer?

**kevinb456**)ok... seen as you just "made it up" as you said... how do you know thats its actually possible to work out!! it could be impossible to work out (not enough data given in the whestion) or soemthing like that. although i havnt done M2 yet (doing M1 now) so im not too sure about the formulas. just wondering how you all think its possible when noones got the answer?

the only possible error was in making the overlap area of the two circles as a fraction of the whole circle too large so that of the distance AB=0 the overlap area would still be too small.

despite this there should still be an imaginary answer to the problem if 1/10 is too large.

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#17

yeah there is an imaginary answer... of wait let me just think of it... yep.. erm its... 3. there ya go, imaginary answer (no, i havnt done complex/imaginary numbers yet :P )

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#18

(Original post by

its not pir^2 were dealing with though... the circles are in fact spherical caps, whose surface area can be calculated with the formula 2Pirh, when h is the height between the top of the cap and the base of the cap, and r is the radius of the base... i think i could do it if i knew how to work out r from the curved radius, but i can't find a way...

edit: after an beautiful el stevo botch, i may be gettign somewhere...

**El Stevo**)its not pir^2 were dealing with though... the circles are in fact spherical caps, whose surface area can be calculated with the formula 2Pirh, when h is the height between the top of the cap and the base of the cap, and r is the radius of the base... i think i could do it if i knew how to work out r from the curved radius, but i can't find a way...

edit: after an beautiful el stevo botch, i may be gettign somewhere...

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#19

my answer i posted for the area of the two lenses is wrong... just the server was down so much i couldn't fix it......

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#20

well... i had a go again this morning, starting with the spherical caps being adjacent instead of overlapping...

Circumference of the sphere = 2.Pi.(3r) = 6.Pi.r

Curved Diameter of Spherical Cap = 2.r

Two spherical caps, adjacent, give curved length 4.r. With arc length 4.r, and circumference of 6.Pi.r, angle subtended is 2.Pi.{(4.r)/(6.Pi.r)} = 4/3 radians.

If the caps are adjacent, the curved distance between A and B is 2r, and the angle at the centre is 2/3 radians. With sphere radius 3r, and angle of 2/3 rads. Linear AB can be found using the cosine rule, and is:

r(3sqrt2)(sqrt(1-cos{2/3}) which can be simplified to 6r.sin(1/3)

thats as far as i have got... i now need to work out the total width of overlapping caps, from which i can calculate the angle. i can then take 2/3 rads off this angle (removing the outer radius of each cap, leaving curved length AB) and then use the cosine rule to work out the linear length AB. i can work out the area of the overlapping caps, but not sure how to go on from there...

so.. i have 6r.sin(@-2/3) where @ fits (@[email protected]) = 3Pi/{10Cos(1/3)}

================================ =======================

Working for Linear[AB]

(a^2)=(b^2)+(c^2)-2bcCos%

(a^2)=(3r)^2+(3r)^2-2(3r)(3r)Cos(2/3)

(a^2)=18(r^2)- 18(r^2)Cos(2/3)

a = sqrt {18(r^2)- 18(r^2)Cos(2/3)}

a = r(3sqrt2)sqrt(1-cos(2/3)

this looked more plicompicated than it need be, so i worked it again using sine rule...

2c = distance ab

c/sin(1/3) = 3r/sin(pi/2)

c = 3r[sin(1/3)]

Linear AB = 6r[sin(1/3)

================================ ========================

Using the area of a segment of a circle was (1/2)(r^2)@ i deduced the area of a segment of a spherical cap to be [email protected] Similarly, using the area of a sector of a circle was (r^2)[email protected], I deduced the area of a sector of a spherical cap to be [email protected] The area which equals 1/10 of the surface area of the sphere is two lots of the difference of segment and sector. So I connived the following.

rh(@)[email protected] = (1/2){(18/5)Pi(r^2)}

rh(@[email protected]) = (9/5)(r^2)Pi

h(@[email protected]) = (9/5)(r)Pi

(@[email protected]) = (9/5h)(r)Pi

i worked out h using pythagoras where the hypotenuse was 3r, and the base was 3rSin(1/3)..

(3r)^2 = (x^2) + {3rsin(1/3)}^2

(x^2) = 9(r^2) - 9(r^2)({sin^2}{1/3})

(x^2) = 9(r^2)(1-{sin^2}{1/3})

(x^2) = 9(r^2)(cos^2}{1/3})

x = 3rCos(1/3)

h = 3r - x

h = 3r{1-cos(1/3)}

if h = 3r{1-cos(1/3)} then

(@[email protected]) = (9/5h)(r)Pi ===> (@[email protected]) = (9/{15r{1-cos(1/3)}})(r)Pi

(@[email protected]) = (9/{15{1-cos(1/3)}})Pi

(@[email protected]) = (3/{5{1-cos(1/3)}})Pi

(@[email protected]) = 3Pi/{5{1-cos(1/3)}}

Thanks to Galois, @ is 16.44813889 rads. There being only 2Pi rads around the centre of a sphere shows the overlap must be bigger than the spherical cap itself, thus the length AB is imaginary.

However, to summarise the linear length AB is 6r.sin(@-2/3) when @ is 16.44813889 rads (8 dp) and r is the 1/3 the radius of the sphere.

Circumference of the sphere = 2.Pi.(3r) = 6.Pi.r

Curved Diameter of Spherical Cap = 2.r

Two spherical caps, adjacent, give curved length 4.r. With arc length 4.r, and circumference of 6.Pi.r, angle subtended is 2.Pi.{(4.r)/(6.Pi.r)} = 4/3 radians.

If the caps are adjacent, the curved distance between A and B is 2r, and the angle at the centre is 2/3 radians. With sphere radius 3r, and angle of 2/3 rads. Linear AB can be found using the cosine rule, and is:

r(3sqrt2)(sqrt(1-cos{2/3}) which can be simplified to 6r.sin(1/3)

thats as far as i have got... i now need to work out the total width of overlapping caps, from which i can calculate the angle. i can then take 2/3 rads off this angle (removing the outer radius of each cap, leaving curved length AB) and then use the cosine rule to work out the linear length AB. i can work out the area of the overlapping caps, but not sure how to go on from there...

so.. i have 6r.sin(@-2/3) where @ fits (@[email protected]) = 3Pi/{10Cos(1/3)}

================================ =======================

Working for Linear[AB]

(a^2)=(b^2)+(c^2)-2bcCos%

(a^2)=(3r)^2+(3r)^2-2(3r)(3r)Cos(2/3)

(a^2)=18(r^2)- 18(r^2)Cos(2/3)

a = sqrt {18(r^2)- 18(r^2)Cos(2/3)}

a = r(3sqrt2)sqrt(1-cos(2/3)

this looked more plicompicated than it need be, so i worked it again using sine rule...

2c = distance ab

c/sin(1/3) = 3r/sin(pi/2)

c = 3r[sin(1/3)]

Linear AB = 6r[sin(1/3)

================================ ========================

Using the area of a segment of a circle was (1/2)(r^2)@ i deduced the area of a segment of a spherical cap to be [email protected] Similarly, using the area of a sector of a circle was (r^2)[email protected], I deduced the area of a sector of a spherical cap to be [email protected] The area which equals 1/10 of the surface area of the sphere is two lots of the difference of segment and sector. So I connived the following.

rh(@)[email protected] = (1/2){(18/5)Pi(r^2)}

rh(@[email protected]) = (9/5)(r^2)Pi

h(@[email protected]) = (9/5)(r)Pi

(@[email protected]) = (9/5h)(r)Pi

i worked out h using pythagoras where the hypotenuse was 3r, and the base was 3rSin(1/3)..

(3r)^2 = (x^2) + {3rsin(1/3)}^2

(x^2) = 9(r^2) - 9(r^2)({sin^2}{1/3})

(x^2) = 9(r^2)(1-{sin^2}{1/3})

(x^2) = 9(r^2)(cos^2}{1/3})

x = 3rCos(1/3)

h = 3r - x

h = 3r{1-cos(1/3)}

if h = 3r{1-cos(1/3)} then

(@[email protected]) = (9/5h)(r)Pi ===> (@[email protected]) = (9/{15r{1-cos(1/3)}})(r)Pi

(@[email protected]) = (9/{15{1-cos(1/3)}})Pi

(@[email protected]) = (3/{5{1-cos(1/3)}})Pi

(@[email protected]) = 3Pi/{5{1-cos(1/3)}}

Thanks to Galois, @ is 16.44813889 rads. There being only 2Pi rads around the centre of a sphere shows the overlap must be bigger than the spherical cap itself, thus the length AB is imaginary.

However, to summarise the linear length AB is 6r.sin(@-2/3) when @ is 16.44813889 rads (8 dp) and r is the 1/3 the radius of the sphere.

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