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

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#1
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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14 years ago
#2
you mean curved circles, or rings pressed onto the sphere until they stop?
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#3
curved circles
as in, the circumference < 2pi*r
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14 years ago
#4
sweet... you just confirmed it being beautifully hard...
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14 years ago
#5
(Original post by El Stevo)
sweet... you just confirmed it being beautifully hard...
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.
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14 years ago
#6
(Original post by 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.
i knew there was a better way of putting it...

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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14 years ago
#7
(Original post by 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.
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.
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#8
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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14 years ago
#9
(Original post by 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.
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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14 years ago
#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...
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14 years ago
#11
(Original post by mik1w)
think it's non-Euclidean geometry..!
hahahahaha
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14 years ago
#12
make it:

{18.Pi.(r^2)}{sqrt[1-cos(2/3)]-(1/2)+cos(2/3)}
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14 years ago
#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 years ago
#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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14 years ago
#15
(Original post by 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
At least you actually have math teachers who can do the harder A-Level math
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#16
(Original post by 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?
I can picture it in my head. I just don't know how to do the maths of it

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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14 years ago
#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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14 years ago
#18
(Original post by 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...
Ah right, I didn't read the bit where they're supposed to be lenses, I assumed they were rings.
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14 years ago
#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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14 years ago
#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.
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