How to divide a circle with two perpendicular chords to minimize (and maximize) the following expression

Question

Consider a circle with two perpendicular chords, dividing the circle into four regions $X, Y, Z, W$(labeled):

What is the maximum and minimum possible value of

$$\frac{A(X) + A(Z)}{A(W) + A(Y)}$$

where $A(I)$ denotes the area of $I$?

I know (instinctively) that the value will be maximum when the two chords will be the diameters of the circle, in that case, the area of the four regions will be equal and the value of the expression will be $1$.

I don't know how to rigorously prove this, however. And I have absolutely no idea about minimizing the expression.

Answer 1

Assume that the line $BD$ between areas $X\cup Y$ and $W\cup Z$ is always horizontal, so the other chord will always be vertical. You may assume your circle to be the unit circle (since the radius will cancel out of the expression in any case). You can parametrize your whole setup by the coordinates of the intersection between these two chords, namely the point $D=(x,y)$ with $x^2+y^2\le1$. Depending on these coordinates, you can compute the areas described in your question, using circular segments and triangles. I did this using sage, and the ugly result looks like this:

-(2*x*y - 2*sqrt(-x^2 + 1)*sqrt(-y^2 + 1) + sqrt(2*sqrt(-y^2 + 1)*x -
2*sqrt(-x^2 + 1)*y + 2)*sqrt(-1/2*sqrt(-y^2 + 1)*x + 1/2*sqrt(-x^2 +
1)*y + 1/2) + sqrt(1/2*sqrt(-y^2 + 1)*x - 1/2*sqrt(-x^2 + 1)*y +
1/2)*sqrt(-2*sqrt(-y^2 + 1)*x + 2*sqrt(-x^2 + 1)*y + 2) -
2*arcsin(1/2*sqrt(2*sqrt(-y^2 + 1)*x - 2*sqrt(-x^2 + 1)*y + 2)) -
2*arcsin(1/2*sqrt(-2*sqrt(-y^2 + 1)*x + 2*sqrt(-x^2 + 1)*y + 2)))/(2*x*y
+ 2*sqrt(-x^2 + 1)*sqrt(-y^2 + 1) - sqrt(2*sqrt(-y^2 + 1)*x +
2*sqrt(-x^2 + 1)*y + 2)*sqrt(-1/2*sqrt(-y^2 + 1)*x - 1/2*sqrt(-x^2 +
1)*y + 1/2) - sqrt(1/2*sqrt(-y^2 + 1)*x + 1/2*sqrt(-x^2 + 1)*y +
1/2)*sqrt(-2*sqrt(-y^2 + 1)*x - 2*sqrt(-x^2 + 1)*y + 2) +
2*arcsin(1/2*sqrt(2*sqrt(-y^2 + 1)*x + 2*sqrt(-x^2 + 1)*y + 2)) +
2*arcsin(1/2*sqrt(-2*sqrt(-y^2 + 1)*x - 2*sqrt(-x^2 + 1)*y + 2)))

If you had to do things manually, you might want to spend time simplifying this beast, but since all I care about at the moment is visualizing this, I'm fine as long as my computer can deal with it.

You can polot the result. For $x,y\ge0$ the result looks like this:

In the bottom plane you see your quarter circle of all possible locations for $D$, and in the vertical direction you see the value of your fraction. In this image, the maximal value of $1$ can indeed be observed for $D=(0,0)$. But any point on a horizontal or vertical line through the origin will yield the same value. So it is sufficient that one of the two lines passes through the origin. Which makes sense due to the symmetric way the areas are distributed to numerator and denominator in this case.

The minimal value is “obviously” (although this is no proof!) at $x=y=\frac12\sqrt2$, i.e. half way between these two and at the very rim of the circle. There you get a value of

$$\frac{\pi-2}{\pi+2}\approx 0.222$$

Note that this is the same value David H already gave in a comment.

But as comments already pointed out, it is far from obvious that $D$ has to lie in the first quadrant. In other words, if you don't always associate $W$ with the area that contains the origin, then the maximal value will neccessarily be the reciprocal of your minimal value, i.e.

$$\frac{\pi+2}{\pi-2}\approx 4.504$$

To visualize this case with all quadrants included, you can extend the above plot to the following one:

Since the scales of various areas are so very different, the overall shape might be clearer if you take the logarithms of the fractions you gave. Then you get the following symmetrical result:

As you are considering ways to proove these facts, looking at the plots might suggests possible approaches. For example, you might be able to argue that looking for a minimum in a quarter circle is enough, since all other cases can be reduced to that one. You might want to use polar coordinates, as the mesh in the above plots suggests. You could try to demonstrate that increasing the radius will always decrese the function value, so that it is sufficient to look at configurations which have $D$ on the circle itself. Then you have a much simpler 1d problem, which should be open to common techniques from calculus.

Answer 2

When the considered quotient is not constant it has a maximum value $\mu>1$, and the minimum value is then ${1\over \mu}$. I claim that $$\mu={\pi+2\over\pi-2}\doteq4.504\ ,\tag{1}$$ as conjectured by MvG.

Proof. I'm referring to the above figure. Since the sum of the four areas is constant the quotient in question is maximal when $$S(\alpha,p):={\rm area}(X)+{\rm area}(Z)\qquad\left(-{\pi\over2}\leq\alpha\leq {\pi\over2}, \quad 0\leq p\leq1\right)$$ is maximal.

Turning the turnstile around the point $(p,0)$ and looking at the "infinitesimal sectors" so arising we see that $$\eqalign{{\partial S\over\partial\alpha}&={1\over2}\bigl((r_1^2-r_4^2)+(r_3^2-r_2^2)\bigr)\cr &={1\over2}\bigl((r_1+r_3)^2-(r_4+r_2)^2-\bigr)+(r_2r_4-r_1r_3)\cr &=2\bigl((1-p^2\sin^2\alpha)-(1-p^2\cos^2\alpha)\bigr)+0\cr &=2p^2(\cos^2\alpha-\sin^2\alpha)\ . \cr}$$ From this we conclude the following: Starting at $\alpha=-{\pi\over2}$ we have $S={\pi\over2}$, then $S$ decreases for $-{\pi\over2}<\alpha<-{\pi\over4}$, from then on increases until $\alpha={\pi\over4}$, and finally decreases again to $S={\pi\over2}$ at $\alpha={\pi\over2}$. It follows that for the given $p\geq0$ the area $S$ is maximal at $\alpha={\pi\over4}$.

We now fix $\alpha={\pi\over4}$ and move the center $(p,0)$ of the turnstile from $(0,0)$ to $(1,0)$. Instead of "infinitesimal sectors" we now have "infinitesimal trapezoids" and obtain $${\partial S\over\partial p}={1\over\sqrt{2}}((r_2+r_3)-(r_1+r_4)\bigr)>0\qquad(0<p\leq1)\ .$$ It follows that $S$ is maximal when $\alpha={\pi\over4}$ and $p=1$. In this case one has $S=1+{\pi\over2}$, which leads to the $\mu$ given in $(1)$.