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what is the probablity of random pick up three points inside a regular triangle which form a triangle and contain the center of the regualr triangle

the three points are randomly picked within the regular triangle and then form a new triangle and the new triangle have to contain the center of the original regular triangle what is the probability

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    Good question. The answer is 1/4 for a circular arena and 33/128 for a square one, so probably very close to that for an equilateral triangle.2011-10-16
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    Correction! It is exactly 1/4 for a square arena too. This strongly suggest that it's also 1/4 for the equilateral triangle, but the best rigorous bound I have so far is $\frac{8}{36}.2011-10-16
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    Potential calculation sketch: Denote the triangle $T$ with center $C$ (which kind?). For two given points $x,y$ inside the triangle, find the region of points $z\in T$ for which $C\in\Delta xyz$ by drawing lines $xC,yC$ and looking at the subtriangle opposite $C$ from $x,y$; let $f(x,y)$ be its area. Then $$P=\frac{1}{|T|^3}\iint_{T\times T}f(x,y)dA_1dA_2.$$ We can change to polar coordinates centered at $C$ and split $T$ into 3 regions demarcated by $x,y,z$; fix $x$ in one region and split $\iint$ with $y$ in one or the other 2 regions, derive $f$ w/ geometry in each case, evaluate. I think..2011-10-16
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    Rather, the 3 regions are demarcated by the vertices of the triangle, or sectioned off as $CU,CV,CW$.2011-10-16
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    @Henning: The $1/4$ guess seemed quite plausible, I was hoping for a nice symmetry argument that would obviate the complicated integration, but numerically the answer seems to be around $0.2453$, with the error probably only in the last digit.2011-10-16
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    @joriki, I've added an answer with some symmetry-based arguments giving a lower bound agreeing with your numerical results.2011-10-16
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    The circle case is related to an old mathoverflow question of mine: http://mathoverflow.net/questions/2014/if-you-break-a-stick-at-two-points-chosen-uniformly-the-probability-the-three-re2011-11-02

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