On the regular nonagon

Given the regular nonagon below

prove that

    \[AB^2 + AC \cdot AD = 2AB \cdot AE\]


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An equivalence relation in a triangle

Prove that in any triangle ABC the following equivalence relation holds:

    \[a^2=ab+c^2 \Leftrightarrow \hat{A}= 90^\circ + \frac{\hat{C}}{2}\]


We are working on the following shape.

Let I be the incenter of the triangle. Thus,

    \[\widehat A = {90^0} + \frac{{\widehat C}}{2} \Leftrightarrow \widehat A = \widehat{AIB} \Leftrightarrow \widehat{BAI} = \widehat{AEB} = \frac{\hat{A}}{2}\]

meaning that AB is tangent to the circumcircle of the triangle AIE. Thus,

(1)   \begin{equation*}c^2 = BI \cdot BE \end{equation*}


    \begin{align*} \frac{BI}{BE} = \frac{a+c}{a+b+c} &\overset{(1)}{\Leftrightarrow } c^2 = \frac{a+c}{a+b+c} \cdot BE^2 = \frac{a+c}{a+b+c} \left ( ac - \frac{b^2}{\left ( a+c \right )^2} \right ) \\ &\Leftrightarrow c^2 = \frac{ac\left ( a+c-b \right )}{a+c} \\ &\Leftrightarrow a^2 = ab + c^2 \end{align*}

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On the heptagon

Given a heptagon of side a and diagonals b, c such that b<c ,

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prove that:

    \[\frac{b^2}{a^2}+ \frac{c^2}{b^2} + \frac{a^2}{c^2} =5\]


Let a  be the side of the heptagon and b, c be its diagonals respectively. It holds that

(1)   \begin{equation*} b^2-a^2=  ac \end{equation*}

(2)   \begin{equation*} c^2-b^2  = ab \end{equation*}

(3)   \begin{equation*} a^2-c^2 = -bc \end{equation*}

(4)   \begin{equation*} \frac{c}{a} + \frac{a}{b} -\frac{b}{c} = 2 \end{equation*}

Equation (4) comes naturally from Vieta’s formulae since \frac{c}{a}\; , \; \frac{a}{b}\;, \; -\frac{b}{c} are the roots of the equation t^3-2t^2-t+1=0. Thus,

\begin{aligned} \frac{b^2}{a^2} + \frac{c^2}{b^2} + \frac{a^2}{c^2} &=\frac{b^2-a^2+a^2}{a^2}+ \frac{c^2-b^2+b^2}{b^2}+ \frac{a^2-c^2+c^2}{c^2} \\ &=\left ( \frac{b^2-a^2}{a^2} + \frac{a^2}{a^2} \right ) + \left (\frac{c^2-b^2}{b^2}+\frac{b^2}{b^2} \right ) + \left ( \frac{a^2-c^2}{c^2}+ \frac{c^2}{c^2} \right ) \\ &= \left (\frac{ac}{a^2}+ 1 \right ) + \left ( \frac{ab}{b^2} + 1 \right ) + \left ( -\frac{bc}{c^2}+ 1 \right ) \\ &= \frac{c}{a} + \frac{a}{b} -\frac{b}{c}+ 3 \\ &=2 + 3 \\ &=5 \end{aligned}

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Limit of a sequence

Let f:\mathbb{R}\rightarrow \mathbb{R} be a function such that f(0)=0 and f is differentiable at 0. Let us set

    \[a_n =\sum_{k=1}^{n} f \left( \frac{k}{n^2} \right)\]

Evaluate the limit \lim \limits_{n \rightarrow +\infty} a_n.


Since f is differentiable at 0 , there is some \varepsilon: x \mapsto \varepsilon(x) such that

    \[f(x) = f(0) + x f'(0) + x \varepsilon(x) \quad , \quad \varepsilon(0) =0\]

and \varepsilon is of course continuous.


    \[\sum_{k=1}^{n} f\left ( \frac{k}{n^2} \right ) = \frac{f'(0)}{n} \sum_{k=1}^{n} \frac{k}{n} + \sum_{k=1}^{n} \frac{k}{n^2} \varepsilon \left ( \frac{k}{n^2} \right )\]

Let \epsilon>0. There exists \delta>0 such that |x| \leq \delta which in return means that |\varepsilon(x)| \leq \epsilon. Hence , for n larger than \frac{1}{\delta}+1 it holds that

    \[\left| \sum_{k=1}^n \frac{k}{n^2}\varepsilon \left( \frac{k}{n^2} \right) \right|\leq \epsilon \frac {n}{n}  = \epsilon\]

On the other hand , the sum \displaystyle \frac{1}{n} \sum_{k=1}^{n} \frac{k}{n} is a Riemann sum and converges to \frac{1}{2}.

In conclusion,

    \[\lim_{n \rightarrow +\infty} a_n = \frac{1}{2}\]


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Integral of Jacobi Theta function

Let \vartheta_4(z;q) denote one of the Jacobi Theta functions. Prove that

    \[\int_{0}^{1}\vartheta_4\left ( 0;q \right ) \, \mathrm{d}q = \frac{\pi}{\sinh \pi}\]


We have successively,

    \begin{align*} \int_{0}^{1} \vartheta_4\left ( 0;q \right )\, \mathrm{d}q &= \int_{0}^{1} \sum_{n=-\infty}^{\infty} (-1)^n q^{n^2} \, \mathrm{d}q \\ &= \sum_{n=-\infty}^{\infty} (-1)^n \int_{0}^{1} q^{n^2} \, \mathrm{d}q\\ &=\sum_{n=-\infty}^{\infty} \frac{(-1)^n}{n^2+1} \\ &= \frac{\pi}{\sinh \pi} \end{align*}

The sum is evaluated as follows. Consider the function

    \[f(z) = \frac{\pi \csc \pi z}{z^2+1}\]

and integrate it around a square \Gamma_N with vertices \left ( N+\frac{1}{2} \right )\left ( \pm 1\pm i \right ). The function f has poles at every integer z=n with residue \frac{(-1)^n}{n^2+1} as well as at z=\pm i with residues -\frac{\pi}{2 \sinh \pi}. We also note that as N \rightarrow +\infty the contour integral of f tends to 0. Thus,

    \begin{align*} \frac{1}{2\pi i}\oint \limits_{\Gamma_N} f(z) \, \mathrm{d}z &= \sum_{n=-N}^{N} \mathfrak{Res}_{z=n} \frac{\pi \csc \pi z}{z^2+1} + \mathfrak{Res}\left ( f ; i \right ) + \mathfrak{Res}\left ( f;-i \right ) \\ &= \sum_{n=-N}^{N} \frac{(-1)^n}{n^2+1} - \frac{\pi}{2\sinh \pi} -\frac{\pi}{2 \sinh \pi} \\ &= \sum_{n=-N}^{N} \frac{(-1)^n}{n^2+1} - \frac{\pi}{\sinh \pi} \end{align*}


    \begin{align*} 0 &=\lim_{N \rightarrow +\infty} \oint \limits_{\Gamma_N} f(z) \, \mathrm{d}z \\ &= \lim_{N \rightarrow +\infty} \left ( \sum_{n=-N}^{N} \frac{(-1)^n}{n^2+1} - \frac{\pi}{\sinh \pi} \right )\\ &= \sum_{n=-\infty}^{\infty} \frac{(-1)^n}{n^2+1} - \frac{\pi}{ \sinh \pi} \end{align*}

and the exercise is complete.

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