Joy of Mathematics

Arcosine function

September 13, 2020 / Leave a comment

Given a function $f:[-1, 1] \rightarrow [0, \pi]$ such that

(1) $\begin{equation*} \cos f(x) = x \quad \text{for all} \;\; x \in [-1 , 1] \end{equation*}$

Evaluate $f(0)$ .
Prove that $f$ is one to one.
Prove that $f \left ( \cos x \right ) = x$ for all $x \in [0, \pi]$ .
Find the range of $f$ .
Sketch the graph of $f$ .

Solution

Setting $x=0$ at $(1)$ we have that
$\cos f(0) =0 \Leftrightarrow x = \frac{\pi}{2} + \kappa \pi \; , \; \kappa \in \mathbb{Z}$

However,

$\begin{align*} 0 \leq f(0) \leq \pi &\Rightarrow 0 \leq \kappa \pi + \frac{\pi}{2} \leq \pi \\ &\Rightarrow - \frac{\pi}{2} \leq \kappa \pi \leq \frac{\pi}{2} \\ &\Rightarrow -\frac{1}{2} \leq \kappa \leq \frac{1}{2} \\ &\!\!\!\!\!\!\overset{\kappa \in \mathbb{Z}}{=\! =\! =\!\Rightarrow } \kappa =0 \end{align*}$

Thus $f(0)=\frac{\pi}{2}$ .
Let $x_1, x_2 \in [-1, 1]$ such that $f(x_1) = f(x_2)$ . Thus,
$\begin{align*} f(x_1) = f(x_2) &\Leftrightarrow \cos f(x_1) = \cos f(x_2) \\ &\Leftrightarrow x_1 = x_2 \end{align*}$

Hence $f$ is $1-1$ .
Setting $x = \cos x$ we have
$\begin{align*} \cos f(x) =x &\Leftrightarrow \cos f \left ( \cos x \right ) = \cos x \\ &\Leftrightarrow f \left ( \cos x \right ) = x \end{align*}$

since $\cos$ is strictly decreasing in $[0, \pi]$ .
$f \left( [-1, 1] \right) = [ 0, \pi]$ because $f^{-1}(x) = \cos x \; , \; x \in [0, \pi]$ .
The graph is seen at the following figure:

Trigonometric inequality

September 13, 2020 / Leave a comment

Let $0 \leq \alpha < \frac{\pi}{2}$ . Prove that

$2 \sin \alpha + \tan \alpha \geq 3\alpha$

Hyperbolic Triangle

September 13, 2020 / 1 Comment on Hyperbolic Triangle

The vertices of a triangle lie on the hyperbola $y=\frac{1}{x}$ . Prove that its orthocentre also lies on the hyperbola.

Solution

We are working on the following figure

Rendered by QuickLaTeX.com

Let $\mathrm{A} \left( p, \frac{1}{p} \right)$ , $\mathrm{B} \left( q, \frac{1}{q} \right)$ and $\Gamma \left( r , \frac{1}{r} \right)$ . Let us denote as $\mathrm{H}$ its orthocentre. We have that:

$\lambda_{\mathrm{B}\Gamma} = \frac{\frac{1}{r} - \frac{1}{q}}{r-q} = - \frac{1}{qr}$

Hence, the slope of the altitude $\mathrm{AZ}$ is $qr$ . Similarly,

$\lambda_{\mathrm{AB}} = \frac{\frac{1}{q} - \frac{1}{p}}{q-p} = - \frac{1}{pq}$

Hence, the slope of the altitude $\Gamma \mathrm{E}$ is $pq$ . Hence,

$\left ( \varepsilon \right )_{\mathrm{B} \Gamma} : y - \frac{1}{p} = qr \left ( x - p \right )$

and

$\left ( \varepsilon \right )_{\Gamma \mathrm{E}} : y - \frac{1}{r} = pq \left ( x - r \right )$

Solving this linear system we have

$\begin{align*} \left ( \varepsilon \right )_{\mathrm{B} \Gamma} = \left ( \varepsilon \right )_{\Gamma \mathrm{E}} &\Leftrightarrow qr \left ( x- p \right ) + \frac{1}{p} = \frac{1}{r} + pq \left ( x-r \right ) \\ &\Leftrightarrow x = - \frac{1}{pqr} \end{align*}$

and finally $y=-pqr$ . So, $\mathrm{H} \left( - \frac{1}{pqr} , -pqr \right)$ . This proves the claim.

Limit of radius of inscribed circle

September 3, 2020 / Leave a comment

Consider the points $A(0, 0)$ , $B(1, 0) , C(x, 1)$ with $x>0$ . Let $r(x)$ be the radius of the inscribed circle of the triangle $ABC$ .

Prove that

$\lim_{x \rightarrow +\infty} \rho(x)=0$

Solution

Since $\tan A = \frac{1}{x}$ we deduce that $\hat{A} \rightarrow 0$ as $x \rightarrow +\infty$ . Since the incenter $I$ lies on the bisector of $A$ , it follows that if $D$ is the projection of $I$ on the $x'x$ axis