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Ingham shoed (Mean value theorems in the theory of the Riemann Zeta function, P.L.M.S., (2), 27, (1926), 273-300) $$\int_{1}^{T}\left|\zeta\left(\frac{1}{2}+it\right)\right|^{4}dt=\frac{T\log^{4}\left(T\right)}{2\pi^{2}}+O\left(T\log^{2}\left(T\right)\right) $$ as $T\rightarrow\infty $ and ...


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What is triangle $\triangle ABC$ called in relation to triangle $\triangle DEF$? According to Wikipedia, the inner triangle is called the Gergonne triangle, contact triangle or intouch triangle of the outer. What is triangle $\triangle DEF$ called in relation to triangle $\triangle ABC$? I don't know an answer to this yet, but searching the web ...


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Different aproach: we can just assume that the circle you want to circumscribe is centererd at the centre of the disk of the Beltrami Klein Model. and that the segments of the n-gon are just lines in this model. Then from Euclidean geometry we now from Regular n-gon between two concentric circles we learn $n \geq \pi/\cos^{-1} r/R $ with $ r$ and $R$ ...


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In the picture (Klein model) below it is clearly shown that the half of the central angle $\Phi(=\frac{2\pi}{n})$ equals the the angle of parallelism belonging to $\rho$, the radius of the circle. That is, $$\Pi(\rho)\ge \frac{\pi}{n}$$ or equivalently $$n\ge \frac{\pi}{\Pi(\rho)}$$is the condition that the tangent lines meet within the plane or exactly on ...


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$ AB,BC, CA$ are sides and parts of semicircles centered on x-axis with centers respectively at Infinity, Q and P... Angle between two lines also equals angle between their perpendiculars. The semi-circles are geodesics in ( one of two) Poincare's hyperbolic geometry models.


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Note that $\measuredangle BAC$ is the angle centered at $A$ (and not, e.g. $B$). The geodesic $AC$ is orthogonal to the geodesic $AP$ (why?), implying the first relation. The second is similar, you only have to reflect the angle $\measuredangle 0BQ$ about the tangent to $BC$ in $B$. The last equation in your question ("if you notice") is true, cause $OPA$ ...


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In Euclidean geometry, the radius does not matter — one can superscribe a regular $n$-gon around a circle of radius $r$, for every $n\ge 3$ and every $r>0$. But in hyperbolic geometry, $n$ must grow with $r$. The reason is the divergence of geodesics from one another, a hallmark of hyperbolic geometry. When trying to circumscribe an $n$-gon, we ...


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Use the isometry obtained from the $n \times n$ identity matrix by replacing the upper left $2 \times 2$ submatrix with $\pmatrix{5/4 & 3/4 \\ 3/4 & 5/4}$. This matrix preserves the line which is the intersection of the hyperboloid with the $x_1,x_2$-coordinate plane, and it translates along that line. This matrix is the one which has a "light ray" ...


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Here is an outline of how I would attack this problem. I am going to describe it for an equilateral triangle where $p=q=r$, but I guess that the method extends with little trouble. The hyperbolic equilateral triangle $\Delta_H$ has a kind of "barycentric subdivision", where each side is subdivided at its midpoint, and the barycenter of $\Delta_H$ is the ...


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(I'm assuming $h$ means hyperbolic distance). You can certainly have a situation like A--B \/ /\ C--D where $|AB|=|CD|$ and $|BC|=|AD|$. Being in the hyperbolic plane doesn't prevent that. For example you could construct it by drawing a circle with center $X$ and two diameters $AD$ and $BC$. Then triangles $AXB$ and $CXD$ are congruent (by ...


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I think pringles are shaped because they are designed to fit perfectly in your mouth .the roof of your mouth holds that shape and when you began to crunch your tongue fits perfectly under it to give you that awesome full taste of your chip



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