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Given a rectangle $A$ of sides $L$ and $W$, such that $L$ is less than $W$, what is the average area of all circles entirely contained in $A$? To be more explicit, what is the average area of all circles of all radii with centers located inside of $A$ such that each circle intersects any given side of A at most once?

I used calculus to find the solution to this problem and got that this average area is equal to

$$(4W-3L)(\pi L^2)(72W)^{-1} .$$

However, I was wondering if there exists a simpler, geometric interpretation to this problem.

EDIT: Thanks to joriki, the 36 was corrected to be 72.

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    Well, the random variable X representing the area of a randomly-chosen circle within the rectangle is a continuous random variable, so $E(X)$ would involve computing an integral. The only way around it that I can think of would be if you could show that the distribution of $X$ is a well-known distribution with a given set of parameters with the mean depending on one or more of these parameters (eg normal, exponential, beta, etc).2011-11-08
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    You haven't specified a distribution for the circles; the question doesn't make any sense without one.2011-11-08
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    How is the random choice of circle made? The details could very well make a difference. That is a common phenomenon in geometric probability.2011-11-08
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    @joriki Please explain. The circles' centers may be located at any point inside of the rectangle. Furthermore, all possible radii of these circles are considered such that the circle intersects any given side of the rectangle at most once.2011-11-08
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    @AndréNicolas Could you please elaborate? I am interested in the mean area out of all the circles.2011-11-08
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    @analysisjb: That describes the possible values of the centres and the radii, but not their distribution. Are the radii uniformly distributed? Or the areas of the circles? Are the centres uniformly distributed, or in proportion to the length of the interval of radii they allow? There is no such thing as "the mean area of all the circles" without specifying a distribution.2011-11-08
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    @joriki The centers are uniformly distributed. For any given center, the radii are uniformly distributed obtaining a maximum when the circle intersects a side once.2011-11-08
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    In that case I believe you've got a factor of $2$ wrong. For $W=L$, you can take the average over the octant $0\le y\le x\le L/2$. Then the average over the uniformly distributed radii for point $(x,y)$ is $\frac13\pi y^2$, so the average area is $$\frac{\int_0^{L/2}\int_0^x\frac13\pi y^2\mathrm dy\mathrm dx}{\int_0^{L/2}\int_0^x\mathrm dy\mathrm dx}=\frac{\pi}{72}L^2\;.$$2011-11-09
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    @joriki Wouldn't the average for W=L be given by that integral except shouldn't the limits, instead of being 0 to x, be given by 0 to L/2, as you are dealing with the octant of length L/2, and instead of integrating with respect to x, integrating with respect to L? This would account for the factor of 2.2011-11-09
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    What would integrating with respect to $L$ mean? I thought $L$ is the fixed length of the rectangle? The integration limit is $x$ because the octant is defined by the inequalities $0\le x\le L/2$ and $0\le y\le x$. If you integrate to $L/2$, you're integrating over a quadrant, not an octant, but the limiting side switches within that quadrant, so you can't average over it in a single integral.2011-11-09
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    Yes sorry that was a mistake on my part. This makes sense. I will change the solution to be over 72W, not 36W.2011-11-09
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    There still seems to be a mistake. When $W\gg L$, it should just be the linear average along $L$, which is $$\frac{\int_0^{L/2}\frac13\pi y^2\mathrm dy}{\int_0^{L/2}\mathrm dy}=\frac{\frac19\pi\left(\frac L2\right)^3}{\frac L2}=\frac{\pi}{36}L^2\;,$$ whereas your solution yields $\pi L^2/18$ for this case.2011-11-09
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    But at a distance of L/2 from the sides of length L, the average becomes dependent on W and the maximum radius of any given circle is dependent on it's distance from the side of length W. Thus, the average should be dependent on both L and W.2011-11-09
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    In fact we should be able to build the correct solution from these two cases by splicing a central part into two halves of a square. The averages are $\pi L^2/36$ and $\pi L^2/72$ for the central part and the two halves of the square, respectively, so the overall average should be the weighted average of the two: $$\pi L^2\left(\frac{L^2}{LW}\frac1{72}+\frac{LW-L^2}{LW}\frac1{36}\right)=(2W-L)(\pi L^2)/(72W)\;.$$2011-11-09
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    In response to your last comment: I was considering the case $W\gg L$ (that means $W$ is very large compared to $L$, that is, $L$ is negligible compared to $W$), and in that case the contributions from the outer parts are negligible.2011-11-09

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