Joy of Mathematics

A complex inequality

January 5, 2023 Tolaso

Let $z \in \mathbb{C}$ such that $|z+1| \leq 1$ and $|z^2+1|\leq 1$ . Prove that $|z|\leq 1$ .

Solution

We have successively:

$\begin{align*} \left | 2z \right | &= \left | \left ( z +1 \right )^2 - \left ( z^2+1 \right ) \right | \\ &\leq \left | z +1 \right |^2 + \left | z^2+1 \right | \\ &=1^2 + 1 \\ &=2 \end{align*}$

Floor of square root

December 21, 2022 Tolaso

Let $x \in \mathbb{R}$ be a positive real number. Prove that

$\left \lfloor \sqrt{\left \lfloor x \right \rfloor} \right \rfloor = \left \lfloor \sqrt{x} \right \rfloor$

Solution

We have successively:

$\begin{align*} n = \left \lfloor \sqrt{x} \right \rfloor & \Rightarrow n \leq \sqrt{x} < n+1 \\ &\Rightarrow n^2 \leq x < (n+1)^2 \\ &\Rightarrow n^2 \leq \left \lfloor x \right \rfloor < (n+1) ^2\\ &\Rightarrow n \leq \sqrt{\left \lfloor x \right \rfloor} < n+1 \\ &\Rightarrow n = \left \lfloor \sqrt{\left \lfloor x \right \rfloor} \right \rfloor \end{align*}$

Generalization of Pythagoras’ theorem

November 24, 2022 Tolaso

In the following figure the triagle $\mathrm{AB} \Gamma$ is right angled. Using the sides of the triangle we draw equilateral triangles as shown in the picture.

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Prove that

$\mathrm{E}_1 = \mathrm{E}_2 + \mathrm{E}_3$

Solution

Since $\mathrm{AB} \Gamma$ is right angled , we know that Pythagoras’ theorem applies; hence:

$\mathrm{AB}^2 + \mathrm{A} \Gamma^2 = \mathrm{B}\Gamma^2$

On the other hand the area of an equilateral triangle of side $\alpha$ is given by the formula $\displaystyle \mathrm{E} = \frac{\alpha^2 \sqrt{3}}{4}$ . Hence,

$\mathrm{AB}^2 + \mathrm{A} \Gamma^2 = \mathrm{B}\Gamma^2 \Leftrightarrow \frac{\sqrt{3}\mathrm{AB}^2}{4} + \frac{\sqrt{3} \mathrm{A} \Gamma^2}{4} = \frac{\sqrt{3} \mathrm{B}\Gamma^2}{4} \Leftrightarrow \mathrm{E}_3 + \mathrm{E}_2 = \mathrm{E}_1$

and the result follows.

A binomial sum

October 22, 2022 Tolaso

Evaluate the sum

$\sum_{k=0}^n \binom{n}{k}\frac{k!}{(n+1+k)!}$

Solution

We have successively

$\begin{align*} \sum_{k=0}^n \binom{n}{k}\frac{k!}{(n+1+k)!} &= \sum_{k=0}^{n} \frac{n!}{\left ( n-k \right )! \left ( n+k+1 \right )!} \\ &=\frac{n!}{\left ( 2n+1 \right )!} \sum_{k=0}^{n} \binom{2n+1}{n-k} \\ &= \frac{n!}{\left ( 2n+1 \right )!} \sum_{k=0}^{n} \binom{2n+1}{k} \\ &=\frac{n!}{2 \cdot \left ( 2n+1 \right )!} \sum_{k=0}^{2n+1} \binom{2n+1}{k} \\ &= \frac{n! \cdot 2^{2n}}{\left ( 2n+1 \right )!} \end{align*}$

Equality

July 24, 2022 Tolaso

Prove that in any triangle $ABC$ it holds that

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

Solution

We have successively:

$\begin{align*} 16 \sum m_a^2 m_b^2 &= 16 \sum \frac{2 \left ( b^2 + c^2 \right ) - a^2}{4} \cdot \frac{2 \left ( a^2 + c^2 \right ) - b^2}{4} \\ &= \sum \left ( 2b^2 + 2c^2 - a^2 \right ) \left ( 2a^2 + 2c^2 - b^2 \right ) \\ &= \sum \left ( 5a^2 b^2 + 2b^2 c^2 + 2a^2 c^2 - 2a^4 - 2b^4 + 4c^4 \right ) \\ &= 9 \left ( a^2 b^2 + b^2 c^2 + c^2 a^2 \right ) \\ &= 9 \left ( ab + bc + ca \right )^2 - 8abc \left ( a + b +c \right ) \\ &= 9 \left ( s^2 + r^2 + 4Rr \right )^2 - 144 Rrs^2 \end{align*}$