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I thought of this question whilst doing some physics.

There is a circle $C_1$ with radius $r$ and a circle $C_2$ with radius $\lambda$. The centre of $C_2$ is on the circumference of $C_1$. What value of $\lambda$ in terms of $r$ will give the area $A$ equal to the area $B$?

Diagram

My attempt: Taking the centre of $C_2$ as the origin and thinking of them as circles on a graph gives an equation for each circle.

C1: $ (x+r)^2+y^2=r^2 $

C2: $x^2+y^2=\lambda^2$

If solve it as a simultaneous equation you can find that their point of intersection is $-\lambda^2/2r$
Then you can do $\int_{-r}^{-\lambda^2/2r} \sqrt{\lambda^2-x^2} dx$ to find area $D$ and $\int_{-\lambda^2/2r}^0 \sqrt{-x^2-2xr} dx$ to find area $E$. Adding these areas together and doubling them should give half the circle, which is $0.5\pi r^2$. Then you should be able to solve the equation for $\lambda$. However when I attempted this (after slogging through about two pages of algebra) I put the equation in Wolfram Alpha and got values for $\lambda$ which were either complex or smaller than $r$. I would like to know if my method is correct and I just made a mistake, or if there is another method to this problem.

enter image description here

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This is an old grazing puzzle. A goat is tethered on the circumference of a circular field. The goat should be able to graze half of the grass in the field. Find the length of the rope. http://datagenetics.com/blog/november42016/index.html

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I think you make a mistake when calculating the area. I don't know how you solved the integrals, so probably there is something going wrong there. Anyway, there is an easier method to compute the area of two overlapping circles. I can explain it here, but it is already very nicely explained on Wolfram.

It will give you the followning equation for $A$:

\begin{align} A(x,r,\lambda) = r^2& \arccos\left(\frac{x^2+r^2-\lambda^2}{2r\lambda}\right)+\lambda^2\arccos\left(\frac{x^2+\lambda^2-r^2}{2r\lambda}\right)-\\ &\frac12\sqrt{(r+\lambda-x)(x+r-\lambda)(x-r+\lambda)(x+r+\lambda)}. \end{align}

In your case, $x=r$, so we can write $$ A(r,\lambda) = r^2 \arccos\left(\frac{2r^2-\lambda^2}{2r\lambda}\right)+\lambda^2\arccos\left(\frac{\lambda}{2r}\right)-\frac12\sqrt{\lambda^2(2r-\lambda)(2r+\lambda)}. $$

Now you need to solve $A(r,\lambda)=\frac12\pi r^2$. I don't see an easy way to solve this; I don't know if this can be algebraically solved. So one way to go is to set $r$ to some real values and find numerically $\lambda$.

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Starting from EdG's answer, considering $$\frac12\pi r^2 = r^2 \arccos\left(\frac{2r^2-\lambda^2}{2r\lambda}\right)+\lambda^2\arccos\left(\frac{\lambda}{2r}\right)-\frac12\sqrt{\lambda^2(2r-\lambda)(2r+\lambda)}$$ let $\lambda=x r$ we just need to find the zero of $$f(x)=\sqrt{x^2 \left(4-x^2\right)}-2 x^2 \cos ^{-1}\left(\frac{x}{2}\right)-2 \cos ^{-1}\left(\frac{1}{x}-\frac{x}{2}\right)+\pi$$ which, for sure, does not show explicit solution.

Using Newton method with $x_0=1$, we obtain the following iterates $$\left( \begin{array}{cc} n & x_n \\ 0 & 1.00000 \\ 1 & 1.12817 \\ 2 & 1.13811 \\ 3 & 1.13816 \end{array} \right)$$ which is the solution for six significant figures.

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