Circles' intersection area is π/2, smaller circle's radius is 1; larger circle's radius=?

[Wikh]

There are two circles like on the picture above. The smaller circles's radius is 1 and its
center point is 1 unit away from the bigger circle's center point. The intersection area of the two circles is equal to π/2.
The radius of the bigger circle is the question.

With the help of the computer I can find that the radius is 1.1587285.
I can write an equation but that is too complex to solve it without computer.
My father told me that somebody was able to calculate it without using computer approximately 40 years ago.
I would be grateful if anybody could help me.

 

[fresh_42]

Say [imath] a [/imath] is the area of the greater circle, [imath] b=\pi [/imath] is the area of the smaller one, and [imath] c=\pi/2 [/imath] is the intersection area. We know b and c and [imath] a=\pi \cdot R^2 [/imath]. All we need is the distance [imath] \overline{CD} [/imath] in dependency of [imath] R, [/imath] the radius of the larger circle. The distance [imath] \overline{CD} [/imath] creates two circle segments, see https://en.wikipedia.org/wiki/Circular_segment.

I haven't solved it, yet, and I don't want to spoil your problem, but that is how I would approach it.

 

[Dr.Peterson]

There are two circles like on the picture above. The smaller circles's radius is 1 and its
center point is 1 unit away from the bigger circle's center point. The intersection area of the two circles is equal to π/2.
The radius of the bigger circle is the question.

With the help of the computer I can find that the radius is 1.1587285.
I can write an equation but that is too complex to solve it without computer.
My father told me that somebody was able to calculate it without using computer approximately 40 years ago.
I would be grateful if anybody could help me.

My guess would be that they used one of the various numerical approximation algorithms, by hand. They existed before computers ...

If you want to show us your equation, we can discuss whether it can be approximated reasonably, or perhaps needs some simplification.

 

[fresh_42]

I am a bit confused by the image. Do you mean the quadrilateral as the intersection area or the actual entire intersection area?
I assume it is the latter, both circle segments. If so, then consider only the first quadrant of the coordinate system. Then the bigger circle intersects the smaller half circle into two areas of exactly [imath] \pi/4. [/imath] All we need is to determine the parameter [imath] R, [/imath] the radius of the bigger circle. This can be done by integrating the corresponding curves. A bit of work to do, but it should be possible.

 

[Wikh]

I am a bit confused by the image. Do you mean the quadrilateral as the intersection area or the actual entire intersection area?
I assume it is the latter, both circle segments. If so, then consider only the first quadrant of the coordinate system. Then the bigger circle intersects the smaller half circle into two areas of exactly [imath] \pi/4. [/imath] All we need is to determine the parameter [imath] R, [/imath] the radius of the bigger circle. This can be done by integrating the corresponding curves. A bit of work to do, but it should be possible.

Your approach is correct, the entire intersection area is equal to pi/2. The parameter R is the question. I have tried to solve it with integration but the equation was also too complex to solve it without computer.

 

[Wikh]

My guess would be that they used one of the various numerical approximation algorithms, by hand. They existed before computers ...

If you want to show us your equation, we can discuss whether it can be approximated reasonably, or perhaps needs some simplification.

I managed to determine the center angles of the circles where the only variable was the radius of the bigger circle. I used the law of cosine.
My equation is:

 

[fresh_42]

We have [imath] (x-1)^2+y^2=1 \wedge x^2+y^2= R^2 \wedge x,y>0 [/imath] to determine the coordinates of the point [imath] P [/imath] depending only on [imath] R. [/imath]

[imath] B= G+C = \pi/4[/imath]
[imath] A+G+C = A + \pi/4 = R^2\pi / 4 [/imath]
[imath] P=(x,y) \text{ solves } (x-1)^2+y^2=1\wedge x^2+y^2=R^2\wedge x,y>0\Longrightarrow x=R^2/2 [/imath]
[imath] A+G =\int_0^{R^2/2} \sqrt{R^2-x^2}\,dx[/imath]
[imath] G =\int_{0}^{R^2/2} \sqrt{1-x^2}\,dx[/imath]
[imath] C =\int_{R^2/2}^R \sqrt{R^2-x^2}\,dx[/imath]

This means we have four variables [imath] A,G,C,R [/imath] and five equations (one equation in hand). There are three integrals with [imath] \arcsin [/imath] terms. I think (hope) we can reduce our system such that we have two equations with [imath] R [/imath] and [imath] \arcsin f(R) [/imath] so that we can end up with one equation in [imath] R. [/imath]

 

[Dr.Peterson]

View attachment 36822
We have [imath] (x-1)^2+y^2=1 \wedge x^2+y^2= R^2 \wedge x,y>0 [/imath] to determine the coordinates of the point [imath] P [/imath] depending only on [imath] R. [/imath]

[imath] B= G+C = \pi/4[/imath]
[imath] A+G+C = A + \pi/4 = R^2\pi / 4 [/imath]
[imath] P=(x,y) \text{ solves } (x-1)^2+y^2=1\wedge x^2+y^2=R^2\wedge x,y>0\Longrightarrow x=R^2/2 [/imath]
[imath] A+G =\int_0^{R^2/2} \sqrt{R^2-x^2}\,dx[/imath]
[imath] G =\int_{0}^{R^2/2} \sqrt{1-x^2}\,dx[/imath]
[imath] C =\int_{R^2/2}^R \sqrt{R^2-x^2}\,dx[/imath]

This means we have four variables [imath] A,G,C,R [/imath] and five equations (one equation in hand). There are three integrals with [imath] \arcsin [/imath] terms. I think (hope) we can reduce our system such that we have two equations with [imath] R [/imath] and [imath] \arcsin f(R) [/imath] so that we can end up with one equation in [imath] R. [/imath]

I assume your work will end up producing an equation similar to the OP's:

I managed to determine the center angles of the circles where the only variable was the radius of the bigger circle. I used the law of cosine.
My equation is: View attachment 36818

(This has an extra variable, x. I don't know what that represents.)

The main question is not, as I understand it, how to get an equation containing R, but how to solve for R. That will require a numerical method, because R is found both inside and outside trig functions.

Has either of you tried doing that? (without using a computer - but presumably using trig tables or the like)

By the way, this is the classic "goat grazing problem", found here, for example, and here. (Yours is the "interior grazing problem".) I believe you can get a somewhat simpler equation if you let the variable be the angle BAC in the OP; solve for that angle, and then use it to find R.

 

[fresh_42]

I assume your work will end up producing an equation similar to the OP's:

Yes, it seems so. I had hoped for an additional equation for the area C by the extra integral and eliminate the arctan terms with it.

 

[fresh_42]

I ended up with [imath] 0=-\dfrac{\pi}{4}+ \dfrac{\pi R^2}{4} -\dfrac{1}{4}R\sqrt{4-R^2}+\arcsin\left(\dfrac{R}{2}\right)-\dfrac{1}{2}R^2 \arctan\left(\dfrac{R}{\sqrt{4-R^2}}\right) [/imath] where inspection as well as WA says: numerical solution only: [imath] R \approx 1.15872847301812...[/imath]

FWIW: [imath] 1/\sin 60° \approx 1.1547005383792515... \approx R [/imath] (A020832 in OEIS).

 

[BobP]

I arrived at an equation by geometrical means, simply the sum of two circle sectors minus the area of a triangle, (to calculate the area of just the top half of the required area).
The equation is
[math]\arcsin (R/2) + (R^{2}/2)\arccos(R/2) - (R/2)\sqrt{1- R^{2}/4}=\pi/4.[/math]Solving, using the Newton-Raphson method (with a pocket calculator) and a starting value of R = 1.0 produces the sequence
1.0,
1.1635,
1.158731,
1.15872847,
1.15872847.

 

[Wikh]

I arrived at an equation by geometrical means, simply the sum of two circle sectors minus the area of a triangle, (to calculate the area of just the top half of the required area).
The equation is
[math]\arcsin (R/2) + (R^{2}/2)\arccos(R/2) - (R/2)\sqrt{1- R^{2}/4}=\pi/4.[/math]Solving, using the Newton-Raphson method (with a pocket calculator) and a starting value of R = 1.0 produces the sequence
1.0,
1.1635,
1.158731,
1.15872847,
1.15872847.

Thank you very much. I have never learnt iteration methods, I am just an economist. Your solution is very helpful.

 

[khansaheb]

I have never learnt iteration methods, I am just an economist.

Most Nobel prizes in economy is won for mathematical achievements. Any kind of forecasting that is done by the economists - will use iterations.