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ChillOut
 one year ago
Reasking a question that I made 2 years ago and couldn't get to an answer. Maybe now someone can reach it!
Two circles, C and Cr intersect each other. C's radius is 1, while Cr's radius is "r". Their tangent lines form an internal angle of 120° at each intersection point. 1) Calculate the distance "d", which is the distance between the centers of both circles, in terms of r. 2) Find where d has a maximum or a minimum. 3) Find the area of the intersection in terms of π
ChillOut
 one year ago
Reasking a question that I made 2 years ago and couldn't get to an answer. Maybe now someone can reach it! Two circles, C and Cr intersect each other. C's radius is 1, while Cr's radius is "r". Their tangent lines form an internal angle of 120° at each intersection point. 1) Calculate the distance "d", which is the distance between the centers of both circles, in terms of r. 2) Find where d has a maximum or a minimum. 3) Find the area of the intersection in terms of π

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ChillOut
 one year ago
Best ResponseYou've already chosen the best response.2dw:1437616135891:dw

ChillOut
 one year ago
Best ResponseYou've already chosen the best response.2dw:1437616351195:dw

ChillOut
 one year ago
Best ResponseYou've already chosen the best response.2I'm starting to think that this is unsolvable.

ChillOut
 one year ago
Best ResponseYou've already chosen the best response.2All right, so I found this: http://www.had2know.com/academics/intersectionangletwocircles.html Is this of any help?

ChillOut
 one year ago
Best ResponseYou've already chosen the best response.2I swear if I get an answer to this I'll become a math teacher.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.5Cutting out 30 degrees from either ends of the tangent lines, it is easy to see that the angle made by radii segments at the intersection point is 60 degrees : dw:1437618536980:dw

ChillOut
 one year ago
Best ResponseYou've already chosen the best response.2How do I prove that it's 60°?

ChillOut
 one year ago
Best ResponseYou've already chosen the best response.2If it's right, I can just go on with implicit differentiation, yeah?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.5Easy : draw a better diagram that actually represents 120 degrees between the tangents, then argue that the perpendicular to tangent line at the point of tangency passes through the center

ChillOut
 one year ago
Best ResponseYou've already chosen the best response.2Oh, yeah. I see it now. Now I think I can finish it.

ChillOut
 one year ago
Best ResponseYou've already chosen the best response.2I was failing to notice how the radii and the tangents connected to form a triangle.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.5We don't really need calculus for second part, we may use the fact that \(d\) is positive and the max/min value of \(d\) occurs at the same point as \(d^2\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.5dw:1437619117612:dw

ChillOut
 one year ago
Best ResponseYou've already chosen the best response.2Yeah, I did the whole d'(r) and found 1/2 as well.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.5*min value for \(d\) occurs at \(r = 1/2\)

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1I'm getting d^2 = 1 + r^2 + r

ChillOut
 one year ago
Best ResponseYou've already chosen the best response.2Just for future reference, leaving the derivative here: \(\frac{dd}{dr}=(2r1)*(2(r²r+1)^{0.5}\))

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1Law of Cosines c^2 = a^2 + b^2  2*a*b*cos(C) d^2 = 1^2 + r^2  2*1*r*cos(120) d^2 = 1^2 + r^2  2*1*r*(0.5) d^2 = 1^2 + r^2 + r

ChillOut
 one year ago
Best ResponseYou've already chosen the best response.2@jim_thompson5910 , why is the angle between the radii 120º in your picture?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.5@jim_thompson5910 looks 120 degrees has to be the other angle : dw:1437619755534:dw

ChillOut
 one year ago
Best ResponseYou've already chosen the best response.2Now I am confused, If they are orthogonal to the tangent lines, they should have the same angle between them, right?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1look at the attached pic I'm posting angle CAF = 90 degrees since the circle is tangent at point A angle GAD = 90 degrees (same reason) angle DAF = 30 degrees because it's the solution so 90+x = 120 angle DAC = (angle DAF) + (angle CAF) angle DAC = (30) + (90) angle DAC = 120

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1dw:1437619928573:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.5again, the construction is incorrect as 120 degrees has to be the internal angle as per the question

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1dw:1437619979044:dw

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1oh did I flip the angles?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.5I think so, after flipping the angles, ur construction also gives the same expression \(d^2 = 1+r^2r\)

ChillOut
 one year ago
Best ResponseYou've already chosen the best response.2dw:1437620217558:dw This one is supposed to have 120°

ChillOut
 one year ago
Best ResponseYou've already chosen the best response.2So, by using the law of cosines and assuming it's 60° it gives @ganeshie8's expression

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.5It is easy to prove using that new better picture, lets do it

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.5dw:1437620436117:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.5dw:1437620527012:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.5Angle between radius and tangent = 90 therefore angle between radius and other circle's tangent = 12090 = 30 similar reasoning yields the other angle also 30

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.5dw:1437620634836:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.5so the angle between radii segments = 1203030 = 60

ChillOut
 one year ago
Best ResponseYou've already chosen the best response.2Yes, pretty nice! We already have the critical point for the radii. Now I need to remember how to answer c) without resorting to calculus

ChillOut
 one year ago
Best ResponseYou've already chosen the best response.2Just find the area of both sectors segments and add them?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.5calculus is always there if we forget geometry :) lets see if this can be done smartly w/o calc

ChillOut
 one year ago
Best ResponseYou've already chosen the best response.2thing is... I need to find Beta.

ChillOut
 one year ago
Best ResponseYou've already chosen the best response.2Well, I remember the area of a circle's sector. It's given by \(A=r²*arccos(d/2r)\) Where d is the distance between the centers and r the radius of the circle in question.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.5you want to find the area for general case, in terms of \(r\), right ?

ChillOut
 one year ago
Best ResponseYou've already chosen the best response.2This should be ideal.

ChillOut
 one year ago
Best ResponseYou've already chosen the best response.2Well, I'm a lackluster in geometry, as you can see.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.5your method works, start by finding this central angle, \(x\) : dw:1437623030778:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.5once you have \(x\), you can find the area of sector and subtract the area of triangle from it to get : dw:1437623216203:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.5you can do the same for the other piece also! looks good!

ChillOut
 one year ago
Best ResponseYou've already chosen the best response.2The line that cuts through both intersections also divides the angles equally, right?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.5use base angles of isosceles triangle theorem

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.5dw:1437623442647:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.5applying cosine law in that triangle with sides \(1, r, d\) : \[r^2 = 1^2+d^22d\cos(x/2)\] you can find \(x\)

ChillOut
 one year ago
Best ResponseYou've already chosen the best response.2Wouldn't I be able to solve for x only i I had a particular solution for r, which isn't the case?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.5you just want to find the overlapping area for \(r=1/2\) is it ?

ChillOut
 one year ago
Best ResponseYou've already chosen the best response.2Yeah. I don't see any way to find a general solution as d is defined by r.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.5applying cosine law in that triangle with sides \(1, r, d\) : \[r^2 = 1^2+d^22d\cos(x/2)\] \[r^2 = 1+(1+r^2r)2d\cos(x/2) \implies x = 2\arccos\left(\dfrac{2r}{2d}\right)\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.5\(d\) is just a number as you can see, we can always replace \(d^2\) by \(1+r^2r\)

ChillOut
 one year ago
Best ResponseYou've already chosen the best response.2All right, I need to calculate the sector's area.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.5Area of shaded region in circle of radius \(1\) : \[\pi*\dfrac{x}{2\pi} = \arccos\left(\dfrac{2r}{2d}\right)\]

ChillOut
 one year ago
Best ResponseYou've already chosen the best response.2Just subtract the triangle area and we have the segment, right?

ChillOut
 one year ago
Best ResponseYou've already chosen the best response.2So, \(A_{segment}=arccos(\frac{2r}{2d})\frac{1*1*sin(x)}{2}\)

ChillOut
 one year ago
Best ResponseYou've already chosen the best response.2which in turn is \(arccos(\frac{2r}{2d})sin(2arccos(\frac{2r}{2d})) \)

ChillOut
 one year ago
Best ResponseYou've already chosen the best response.2Oops! Missed /2 in the second part.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.5Looks good to me! Area of shaded region in circle of radius \(1\) : \(\large \arccos(\dfrac{2r}{2d})\dfrac{\sin(2\arccos(\frac{2r}{2d}))}{2} \) dw:1437625381250:dw

ChillOut
 one year ago
Best ResponseYou've already chosen the best response.2Now for the other sector... Which will be in terms of "r"

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.5dw:1437625666331:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.5do we get \[y=2\arccos\left(\dfrac{2r^2r }{ 2dr}\right)\] ?

ChillOut
 one year ago
Best ResponseYou've already chosen the best response.2Wait... Let me take a second look.

ChillOut
 one year ago
Best ResponseYou've already chosen the best response.2Yeah, that's what I got.

ChillOut
 one year ago
Best ResponseYou've already chosen the best response.2\(A_{segment}=r^{2}arccos(\frac{2r^{2}r}{2dr})\frac{r^{2}sin(2arccos(\frac{2r^{2}r}{2dr}))}{2}\)

ChillOut
 one year ago
Best ResponseYou've already chosen the best response.2For the circle of radius r. Sorry for the small TeX text

ChillOut
 one year ago
Best ResponseYou've already chosen the best response.2Then, adding the segments' areas, we have \(arccos(\frac{2r}{2d})\frac{sin(arccos(\frac{2r}{2d})}{2}+r²arccos(\frac{2r²r}{2dr})\frac{r²sin(2arccos(\frac{2r²r}{2dr}))}{2}\)

ChillOut
 one year ago
Best ResponseYou've already chosen the best response.2P.S: Missed a "2" before the first arc cosine.
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