On the regular nonagon

April 9, 2019 / Leave a comment

Given the regular nonagon below

prove that

$AB^2 + AC \cdot AD = 2AB \cdot AE$

Solution

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An equivalence relation in a triangle

April 8, 2019 / Leave a comment

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

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

Solution

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

Hence,

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

On the heptagon

April 8, 2019 / Leave a comment

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

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prove that:

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

Solution

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,

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

Limit of a sequence

April 6, 2019 / Leave a comment

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 $\lim \limits_{n \rightarrow +\infty} a_n$ .

Solution

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.

Thus,

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

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,

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

Integral of Jacobi Theta function

March 25, 2019 / 2 Comments on Integral of Jacobi Theta function

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

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

Solution

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,

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

Hence,

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

and the exercise is complete.

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