Joy of Mathematics

Equality in a triangle

April 4, 2022 / Leave a comment

Prove that in any triangle $ABC$ it holds that:

$16\sum m_a^2 m_b^2 = 9 \left ( s^2 + r^2 + 4Rr \right )^2 - 144 s^2 Rr$

Consider two parallel planes that have a distance $d \leq 2r$ and intersect a sphere of radius $r$ . (both planes intersect the sphere.) Prove that the area of the surface of the sphere enclosed by both planes is only dependant by $r, d$ and not by the position of the two planes with respect to the sphere.

Solution

WLOG we assume that the two planes which intersect the sphere of radius $r$ are perpedicular to $x$ -axis with equations $x=a$ and $x=b$ respectively.

(In the scheme the points $A$ and $B$ are points of intersection of a meridian cycle with the planes $x=a$ and $x=b$ respectively and the points $A'(a,0,0)$ and $B'(b,0,0)$ are the projections of the points $A$ and $B$ respectively, on $x$ -axis.)
The area in question is the area of the surface by revolution of the meridian’s arc $\overset{\frown}{AB}$ which lies between the planes, rotating it about the $x$ -axis. The arc $\overset{\frown}{AB}$ is the graph of the function $y(x)=\sqrt{r^2-x^2}\,,\; x\in[a,b]$ , and the area in question is

$\begin{align*} A&=2\pi\int_{a}^{b}y(x)\,\sqrt{1+(y'(x))^2}\;dx \\ &=2\pi\int_{a}^{b}\sqrt{r^2-x^2}\sqrt{1+\frac{x^2}{r^2-x^2}}\;dx \\ &=2\pi\int_{a}^{b}\sqrt{r^2-x^2}\sqrt{\frac{r^2}{r^2-x^2}}\;dx \\ &=2\pi\,r\int_{a}^{b}dx \\ &=2\pi\,r\,(b-a)\\ &=2\pi\,r\,d\,. \end{align*}$

Factorization

January 27, 2022 / Leave a comment

Let $\mathrm{ABC}$ be a right triangle at $\mathrm{A}$ . Factor $a^3+b^3+c^3$ .

Solution

The following equations hold:

(1) $\begin{equation*} a^2 = b^2 + c^2 \end{equation*}$

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

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

We have successively:

$\begin{align*} a^3+b^3+c^3 &=a \cdot a^2+ b^3+c^3 \\ &=a(b^2+c^2)+b^3+c^3 \\ &=ab^2+ac^2+b^3+c^3 \\ &=b^2(a+b)+c^2(a+c) \\ &=(a^2-c^2)(a+b)+(a^2-b^2)(a+c)= \\ &=\cdots \\ &=(a+b)(a+c)(2a-b-c) \end{align*}$

Factorial series

January 1, 2022 / Leave a comment

Prove that

$\sum_{n=0}^{\infty} \frac{(4n)!}{(4n+4)!} = \frac{\ln 2}{4} - \frac{\pi}{24}$

Solution

We have successively:

$\begin{align*} \sum_{n =0}^{\infty} \frac{(4n)!}{(4n+4)!} & =\sum_{n=0}^{\infty} \frac{\Gamma(4n+1)}{\Gamma(4n+5)} \\ &= \frac{1}{6} \sum_{n=0}^{\infty} \mathrm{B}(4n+1,4) \\ & = \frac{1}{6}\sum_{n=0}^{\infty} \int_{0}^{1}x^{4n}(1-x)^3\, \mathrm{d}x \\ & = \frac{1}{6}\int_{0}^{1} \frac{(1-x)^2}{(1+x^2)(1+x)} \, \mathrm{d}x \\ &= \frac{1}{6} \int_{0}^{1} \left ( \frac{2}{x+1} - \frac{x+1}{x^2+1} \right )\, \mathrm{d}x \\ &= \frac{1}{6} \int_{0}^{1} \left ( \frac{2}{x+1} - \frac{x}{x^2+1} - \frac{1}{x^2+1} \right )\, \mathrm{d}x \\ & = \frac{\ln 2}{4}-\frac{\pi}{24} \end{align*}$

The composition is a metric

December 18, 2021 / Leave a comment

Let $d$ be a metric , $f$ be a strictly increasing function and concave on $[0, +\infty)$ such that $f(0)=0$ . Prove that $f \circ d$ is a metric.

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Equality in a triangle

Planes and sphere

Factorization

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The composition is a metric

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