Joy of Mathematics

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:

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

Factorization

January 27, 2022 Tolaso

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:

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

Equal segments

October 14, 2021 Tolaso

Let $f(x) = \frac{1}{x}$ and let $(\varepsilon)$ be a line that intersects $\mathcal{C}_f$ at the points $\mathrm{A}$ , $\Delta$ and the axis at the points $\mathrm{B}$ and $\Gamma$ .

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Prove that $\mathrm{AB}= \Gamma \Delta$ .

Solution

Let $(\varepsilon):y= \alpha x + \beta$ with $\alpha \neq 0$ . Then,

$\alpha x + \beta = \frac{1}{x} \Leftrightarrow \alpha x^2 + \beta x - 1 = 0 \Leftrightarrow \left\{\begin{matrix} x_1 + x_2 &= & - \dfrac{\beta}{\alpha} \\\\ x_1 x_2 & = & - \dfrac{1}{\alpha} \end{matrix}\right.$

Hence,

$\begin{align*} \overrightarrow{\mathrm{BA}} &= \left ( x_1, \frac{1}{x_1} - \beta \right ) \\ &= \left ( - \frac{\beta}{\alpha} -x_2, - \frac{1}{x_2} \right ) \\ &= \overrightarrow{\Delta \Gamma} \end{align*}$