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The following is a geometry problem that I came across with in the course of a research project.

Consider a ray starting at some initial point $t$. Place point $s_1$ at distance $r$ from $t$ on the ray and draw a circle centered at $s_1$ that passes through $t$. Likewise, centered at $t$, an arc with radius $r$ goes through $s_1$. Let $\mathcal{A_1}$ be the area enclosed between the intersecting arcs.

Next, arbitrarily place another point somewhere on the free end of the ray and call it $s_2$ such that $|s_1 - t| < |s_2 - t|$, where $|.|$ denotes the Euclidean distance. A circle with radius $r$ is centered at $s_2$ and another arc centered at $t$ goes through $s_2$. The area enclosed between these intersecting arcs we call $\mathcal{A}_2$. It is easy to show that $\mathcal{A}_1 < \mathcal{A}_2 < \lim_{|s_2 - t| \to \infty} \mathcal{A}_2 = \frac{1}{2} \pi r^2$.

Now, assume that we mark the segments of the ray within the enclosed areas in the middle and arcs centered at $t$ pass through the marks segmenting $\mathcal{A}_1$ and $\mathcal{A}_2$. We call these segmented areas $\mathcal{A}_{11}$ and $\mathcal{A}_{12}$ and $\mathcal{A}_{21}$ and $\mathcal{A}_{22}$ as depicted below (dashed lines are the arcs centered at $t$).

enter image description here

Question: How does $\mathcal{A}_{22}$ change as $s_2$ gets farther from $t$? (i.e., does it increase or decrease?) What can we say about $\mathcal{A}_{22}$ in comparison with $\mathcal{A}_{12}$?

Any idea or comment is much appreciated.

EDIT: The question has been edited in a way that makes the comments incomprehensible. Please see the edit history if you want to make sense of the comments.

EDIT: Here is the link to the same question at mathoverflow.net

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    Do you have numeric evidence that $A_{12}>A_{22}$. Intuitively, I would reason: If we move $s_1$ a bit farther right, $A_{11}$ would shrink to a point and $A_{12}$ would be a circle of radius $\frac r2$, whose area is definitely less than that of $A_{22}$. It looks unintuitive to me that the area of the $A_{\cdot 2}$ would be increasing from $s_2$ to $s_1$ only suddenly decrease to the right of $s_1$.2012-06-20
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    Thanks for the comment, @HenningMakholm. The fact is that $s_1$ is a fixed point at distance $r$ from $t$ and $s_2$ is the only point that we arbitrarily choose. In other words, we are not allowed to move $s_1$. Regarding your question on the numerical evidence, I should say no. In fact, I do not know any way to numerically evaluate these regions. Whatever I said is just based on my intuition and of course I am not sure of its correctness.2012-06-20
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    Who says we're not allowed to imagine $s_1$ being somewhere else? But if that bothers you, just put an $s_0$ half a unit to the right of $s_1$ instead, and consider how unlikely it seems that $A_{02}.2012-06-21
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    Well, if you want to move $s_1$ anywhere, you are also changing the radius $r$. In other words, the circle centered at $s_1$ must always pass through $t$. This is, in fact, part of the problem definition.2012-06-21
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    Again, if that gives you problems, just imagine that the point midways between the original $s_1$ and $t$ is called some thing else than $s_1$. I suggested $s_0$, but you may also call is $s_7$ or $q$.2012-06-21
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    If you let $s_2\to\infty$, then it's easy to calculate $A_{21}$ as it's a third of the circle minus a 120-30-30 triangle with short side $r$. Then you can get $A_{22}$. So if you can find a way to calculat, or estimate, $A_{11}$, then you can get $A_{12}$ (since $A_1$ is easy), and compare.2012-06-21
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    @HenningMakholm: I am not concerned with naming. My point is that by moving $s_1$ to the right, you also shrink radius $r$. Note that both circles centered at $s_1$ and $s_2$ have the same radius $r$. What you are trying to do by moving $s_1$ to the right is similar to down scaling the picture and does not change the problem.2012-06-21
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    Well, freaking _don't move it_, then. Leave $s_1$ exactly where it is, and instead put a _new_ point to the right of $s_1$, and draw the same circles centered at $t$ that you do for $s_1$ and $s_2$, just intersecting your _new_ point instead of the $s_1$ and $s_2$. The conclusion is exactly the same no matter how you describe the operation.2012-06-21
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    @HenningMakholm: I think I am not getting you quite clearly. With that _new_ point, say $s_0$, are we getting rid of $A_{01}$? Would that _new_ circle centered at $s_0$ contain only $A_{02}$ with which we compare $A_{12}$ and $A_{22}$?2012-06-21
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    @GerryMyerson: Thanks for your comment. I am not sure though, given that $A_{22} < A_{12}$ when $s_2 \to \infty$, can we claim that $A_{22} < A_{12}$ for any $s_2$ between $s_1$ and $\infty$?2012-06-21
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    Before you decide on the implications, better check the facts. If $A_{22}\gt A_{12}$ as $s_2\to\infty$, then that will prove that there is some finite place where $A_{22}\gt A_{12}$. In other words, this is an attempt to prove that the conjecture is *wrong*.2012-06-21
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    Ali, please edit in a link to the identical question posted at MathOverflow, and please edit in a link to this question over there.2012-06-21
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    @Ali: Yes, with the new point being put at a distance $\frac R2$ from $T$, $A_{01}$ has shrunk to a circle of radius $0$, so $A_{02}$ is the entire circle of radios $\frac R2$.2012-06-21

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