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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 $s_1$. –  Henning Makholm Jun 20 '12 at 23:50
    
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. –  Ali Jun 20 '12 at 23:59
    
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 $A_{02}<A_{22}<A_{12}$. –  Henning Makholm Jun 21 '12 at 0:03
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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. –  Gerry Myerson Jun 21 '12 at 2:06
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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. –  Gerry Myerson Jun 21 '12 at 5:36
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1 Answer 1

The shape of $A_1$ is always the same, so we can calculate its area as the sum of two circular segments: $A_1 = r^2(2\pi/3 - \sqrt{3}/2)$.

Let's first let $t$ be the origin (why the hell would you name the origin of a ray $t$?!). Let's set $r=1$, keeping in mind all areas will be scaled by $r^2$ later. The coordinates of the vertices of $A_{11}$ are $(1/8,\pm \sqrt{15}/8)$, and the subtended angles are $2\tan^{-1} \sqrt{15}$ and $2\tan^{-1} \sqrt{15}/7$. This then gives $A_{11} \approx 0.350767 (r^2)$, $A_{1} \approx 1.22837$, and $A_{12} \approx 0.877603$. Note that all these numbers can be made precise; they're just huge ugly expressions, and remember they are multiplied by $r^2$.

Now let $s_2$ be located at coordinates $(R,0)$ where $R > 1$ according to our assumptions. By similar reasoning, the vertices of $A_2$ are $(\frac{2R^2-1}{2R},\pm \sqrt{1-\frac{1}{4R^2}})$. Similarly now, we can compute $A_{22}$. The expression for $A_{22}/A_{12}$ is horrendously large, so I will just have Mathematica plot it as a function of $R$:

enter image description here

The limit according to Mathematica is $1.09003$.

Edit: I have corrected a number of mistakes. Now, the area ratio is always greater than unity for $R>1$.

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I should just like to add that numerically evaluating the areas is extremely simple as the sum of circular segments, and can even be done in a numerically robust way when the segments are thin slivers. For C code, see the function CircularSectorArea in this file –  Victor Liu Jun 21 '12 at 5:22
    
Nice. Is it true that $A_{22}\gt A_{12}$ for all values of $R$? –  Gerry Myerson Jun 21 '12 at 5:44
    
No. The plot clearly shows that it crosses over the area of $A_{12}$. Mathematica says this happens around $R=9.23574$. –  Victor Liu Jun 21 '12 at 5:48
    
Thanks. The plot doesn't show this, unless you put in a horizontal line, labeled, at $A_{12}$ and the plot extends below this line. –  Gerry Myerson Jun 21 '12 at 7:33
    
@VictorLiu: How can that be? If we set $R=1$, then $A_{22}$ and $A_{12}$ coincide. Are there two $R$ values that give the same $A_{22}$ area? –  Henning Makholm Jun 21 '12 at 10:23
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