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A circle and two perpendicular lines enclose four regions. Can the regions have distinct rational areas? (I stipulate distinct to eliminate trivial cases in which one of both of the perpendicular lines go through the center of the circle.)

My approach was to let the circle be $x^2$+$y^2$=$1/\pi$ (so the area of the circle is one), then let the perpendicular lines be $y=a$ and $x=b$, then express the four areas in terms of $a$ and $b$ and try to show that the areas cannot all be rational. But I quickly got lost in very complicated expressions for the areas.

I've tried to work with this, but I have come up with nothing.

Here is another question about four regions in a circle, which has not received any answer yet; I'm not sure if it would help answer my question.

This resolved question makes me suspect that the answer to my question is no.

Dan
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  • You may assume that the circle is of the form $x^2 + y^2 = R^2$, and the lines are, as you say, $x = a$ and $y = b$. Do you by any chance know closed-form expressions for the four resulting areas in terms of $a$, $b$ and $R$? – Jeroen van der Meer Dec 01 '21 at 12:13
  • @JeroenvanderMeer Yes, but the expressions are overwhelmingly complicated (and they do not seem to simplify), such that I have no hope that those expressions would allow me to answer the question. – Dan Dec 01 '21 at 12:49
  • These painful expressions might become quite nice (yielding stuff onlly involving terms like $\sqrt{3}$, $\sqrt{5}$ and $\pi$) if you take say $R = 1$, $x = 1/2$ and $y = 1/3$. After that you can try to scale by an appropriate factor to make all these numbers rational. This is just a guess, though... – Jeroen van der Meer Dec 01 '21 at 12:51

1 Answers1

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This isn't a complete answer, but maybe someone can build off of it.

I'm going to slightly re-parametrize your variables. Let the circle still be $x^2+y^2=\frac{1}{\pi}$, but let the lines be $x=\frac{a}{\sqrt{\pi}}$ and $y=\frac{b}{\sqrt{\pi}}$ (this is just to make the later calculations look nicer). Then also assume that $0 < a < b < 1$.

Then the area of the top-right section is $$\int_{\frac{a}{\sqrt{\pi}}}^{\frac{\sqrt{1-b^2}}{\sqrt{\pi}}}\left(\sqrt{\frac{1}{\pi}-x^2}-\frac{b}{\sqrt{\pi}}\right)dx$$ which is $$t_1=\frac{ab}{\pi}+\frac{\arccos(b)-b\sqrt{1-b^2}}{2\pi}-\frac{\arcsin(a)+a\sqrt{1-a^2}}{2\pi}$$

The area of the bottom-left section is then $t_3=\frac{1}{2}+\frac{2ab}{\pi}-t_1$ (using the linked question). Similarly, the top-left section is $$t_2=\frac{\arccos(b)-b\sqrt{1-b^2}}{\pi}-t_1$$

and the bottom-right section is $t_4=\frac{1}{2}-\frac{2ab}{\pi}-t_2$

If $t_1$ is rational, then for $t_3$ to be rational, $r_1=\frac{ab}{\pi}$ must be rational. Also for $t_2$ to be rational, $r_2=\frac{\arccos(b)-b\sqrt{1-b^2}}{\pi}$ must be rational.

So using that $r_1$ and $r_2$ must be rational, that then also means (since $t_1$ must be rational) that $r_3=\frac{\arcsin(a)+a\sqrt{1-a^2}}{\pi}$ must be rational. All the sections can in fact be represented in terms of $r_1, r_2, r_3$. Specifically, $t_1=r_1+\frac{r_2}{2}-\frac{r_3}{2}$, $t_2=-r_1+\frac{r_2}{2}+\frac{r_3}{2}$, $t_3=\frac{1}{2}+r_1-\frac{r_2}{2}+\frac{r_3}{2}$, and $t_4=\frac{1}{2}-r_1-\frac{r_2}{2}-\frac{r_3}{2}$. So it is necessary and sufficient for $r_1, r_2, r_3$ to be rational.

That's as far as I've gotten, and I've tried adding, multiplying and dividing $r_i$ with each other, but with little headway. I think it's true that neither $a$ nor $b$ can be algebraic (nor even algebraic multiples of $\pi$), but I'm not sure about this.

Varun Vejalla
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