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Alexander Bogomolny
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Alexander Bogomolny
@CutTheKnotMath
Freelance mathematician and educational web developer. I am a proud Israeli American, with love and respect for both countries.
NJ, USA Katılım Aralık 2009
342 Takip Edilen20.6K Takipçiler
Lorian Saceanu's Inequality for All Triangles cut-the-knot.org/triangle/Loria… #FigureThat #math #geometry #inequality
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Alexander Bogomolny retweetledi
@KriteeshP What this shows is that "P interior to KLM" => "P interior to the medial traingle" But this is trivial because the former is inscribed in the latter.
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@CutTheKnotMath For P to be interior of KLM.
AP > AK, implies AP > AA'/2, implies AP/AA' >1/2
Hence AP/AA' , BP/BB', CP/CC' all must be more than 1/2 together.
If AP/AA' < 1/2, the area covered by all possible P could be found by plotting AP/AA' = 1/2, which gives a line parallel to base BC.
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@KriteeshP Not certain how it all applies that P is interior to KLM
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@CutTheKnotMath 1/4. Could be redefined as CP > CM, which is redefined as P divides AA' in ratio k : 1, where k > 1. Limiting case is the ratio 1:1.
In case of 1:1 we get locus of P as a line parallel to the opposite side BC Applying Thales theorem and similarity, area is 1/4 for each vertex.
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@KriteeshP I've nothing against (I am all for) the Indians' pride in their culture and history. But, for crying out loud, did anybody hear the word probability c. 600 BCE.
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@CutTheKnotMath Also, have to mention that Baudhayana (600 BCE) gave this theorem in India.
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Seeking problems with 4/7 probability being involved. E.g., Diminishing Hopes II #FigureThat #math #probability cut-the-knot.org/Probability/Se…
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@CutTheKnotMath @dasanil @gcfr20 Now we can use the Bogomolny lower bound on the variance as a function of median.
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Let {x_k} be uniformly distributed on say, [-1,1] with the mean=0. Is that true that the variance is at least the square of the median? #FigureThat #math #probability
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@nntaleb @CutTheKnotMath @gcfr20 I confess the implication of the distribution escapes me, since the actual inequality doesn’t depend on the distribution at all.
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@CutTheKnotMath @nntaleb @gcfr20 yes, xn^2 - xn^2 = 0 so all the left side value of the inequation must be positive, it is!
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@CutTheKnotMath @nntaleb @gcfr20 Minimum variance is achieved when you can bunch the value together as close as possible is a probabilistic interpretation, no?
twitter.com/dasanil/status…
Anil Das@dasanil
@nntaleb @CutTheKnotMath @gcfr20 My approach is (not rigorous, but can be made so) is this: WLOG, assume x_n > 0. Variance is minimized when all positive x = x_n, and all negative x also equal. i.e. x_i = x_n for i >= n, and for j < n, x_j = - (x_n) * (n/n-1). Thus LHS becomes 2n/(2n-1) * x_n^2. QED
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@CutTheKnotMath @nntaleb @gcfr20 if you square the values they're all positive, just pass the xn^2 term of the sum to the right side of the inequation
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@CutTheKnotMath @gcfr20 Sorry.
We can generalize by having S_1,m,S_2 where S_1 and S_2 are sums and m is the median.
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