ogeometrie

marți, 16 februarie 2021

Getting to know a new problem book ...

See the book here

In Cap. 8 (Miscellaneous Problems) at page 236, Problem 47

SOLUTION CiP

The numbers $x,\;y,\;z$ are the roots of a third degree equation, with unknown $w$

(1) $w^{3}-(x+y+z)\cdot w^{2}+(xy+yz+zx)\cdot w -xyz=0$

(according to Vieta's formulas). From the given conditions, and we can write the second one $\frac{yz+zx+xy}{xyz}=0$, the equation (1) is

(2) $w^{3}-x \cdot y \cdot z=0$.

From (2) we get

$w^{4}=xyz \cdot w$, $w^{5}=xyz \cdot w^{2}$ $w^{6}=xyz \cdot w^{3}$.

We put $x,\; y,\; z$ in turn instead of $w$ in the last relation, we sum up the obtained results in and we get

$\sum x^{6}=xyz \cdot \sum x^{3}$.

It follows from here $\frac{x^{6}+y^{6}+z^{6}}{x^{3}+y^{3}+z^{3}}=xyz$.

$\blacksquare$

REMARK CiP

Numbers that check the given conditions must be complex numbers. (Because $\sum x^{2}=(\sum x)^{2}-2 \cdot \sum xy =0$ and $xyz \neq 0$.)

From relationships

\begin{cases}x+y=-z\\\frac{1}{x}+\frac{1}{y}=-\frac{1}{z} \end{cases}

we deduce

$x+y=-z$, $x \cdot y =z^{2}$

so $x$ and $y$ are the roots of the quadratic equation with unknown $t$

$t^{2}+z \cdot t +z^{2}=0$

that is $\frac{-1\pm\imath \sqrt{3}}{2}\cdot z$. So the general family of solutions of the given equations is

$(\frac{-1+\imath \sqrt{3}}{2}\cdot z\;,\; \frac{-1-\imath \sqrt{3}}{2} \cdot z\;,\;z)\;\;\;z \in \mathbb{C}\setminus \{ 0 \}$.

$\square$

luni, 8 februarie 2021

PROBLEM MA 102 - Crux Mathematicorum Vol. 47 No 1

En, pag4

Fr, pag 5

ANSWER CiP $704$ squares

Solution CiP

Another grid of 31 matches has an area of 10 squares:

A grid with $b$ matches horizontally and $a$ matches vertically contains $a \cdot (b+1)+b \cdot (a+1)$ matches. We have the equation

$2\cdot a \cdot b +a+b=337$

$\Leftrightarrow \;4ab+a+b=674$ $\;\Leftrightarrow \;(2a+1)\cdot (2b+1)=675$.

Examining all the decompositions into two factors of the number $675$ we obtain the table below.

\begin{matrix}
&2a+1 &3 &5 &9 &15 &25 \\
&2b+1 &225 &135 &75 &45 &27 \\
&a &1 &2 &4 &7 &12 \\
&b &112 &67 &37 &22 &13 \\
&area &112 &134 &148 &154 &156
\end{matrix}

The total area is $112+134+148+154+156=704.$

$\blacksquare$

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Tracking Number: 11633
Received: Tue Feb 9 13:38:13 2021

From: Petre Ciobanu
Scoala Gimnaziala "Samuil Micu" SADU
Sibiu, Romania
Email: ptr.ciobanu@gmail.com

Type: Solve a MathemAttic Problem
(problem MA102)

Files:
MA102.pdf

Comments:
See my Blog
https://ogeometrie-cip.blogspot.com/2021/02/problem-ma-102-crux-mathematicorum-vol.html

=====

Fisierul Sursa Latex

https://artofproblemsolving.com/texer/rypsefyz

Crux Mathematicorum crux@cms.math.ca

15:39 (acum 1 minut)

către eu

====================================================================

Added Jul 13, 2021

Good answer see V47n06, pag 278

In solving them, they count in more detail the total number of matches:

"There are $(a+1)$ rows of horizontal mathces, each containing $b$ matches. Similarly, there are $(b+1)$ columns of vertical matches, each containing $a$ matches. So the total number of matches is

$(a+1)b+(b+1)a=$...

=end added=

miercuri, 3 februarie 2021

Identities comprising $tg 10^{\circ},\; tg 50^{\circ},\; tg 70^{\circ}$

Prove that

$tg^{2} 10^{\circ}+tg^{2} 50^{\circ}+tg^{2} 70^{\circ}=9$.

Solution CiP

We have $tg 30^{\circ}=\frac{1}{\sqrt{3}}$ and, applying the formula $tg 3\alpha=\frac{3tg \alpha-tg^{3} \alpha}{1-3tg^{2} \alpha}$,

$\Leftrightarrow \; \frac {3tg 10^{\circ}-tg^{3} 10^{\circ}}{1-3 tg^{2} 10^{\circ}}=\frac{1}{\sqrt{3}}\;\Leftrightarrow \; tg^{3} 10^{\circ}-\sqrt{3} tg^{2} 10^{\circ}-3 tg 10^{\circ}+\frac{1}{\sqrt{3}}=0.$

So $tg 10^{\circ}$ is the root of the third degree equation

(1) $z^{3}-\sqrt{3} \cdot z^{2}-3 \cdot z +\frac{1}{\sqrt{3}}=0$.

The same calculations go for $tg(30^{\circ}+180^{\circ})=\frac{1}{\sqrt{3}}$ and $tg(30^{\circ}+2 \cdot 180^{\circ})=\frac {1}{\sqrt{3}}$, so $tg 70^{\circ}$ and $tg 130^{\circ}=-tg 50^{\circ}$ they are also roots of the equation (1). These are all the roots of the equation (1).

Now we write Vièta's relations for equation (1) and its roots $tg10^{\circ},\; -tg 50^{\circ}\; tg 70^{\circ}$ - arranging a little -

\begin{cases} tg10^{\circ}-tg50^{\circ}+tg70^{\circ}=\sqrt{3}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(2)\\tg10^{\circ} \cdot tg50^{\circ}+tg50^{\circ} \cdot tg70^{\circ}-tg70^{\circ} \cdot tg10^{\circ}=3\;\;(3)\\tg10^{\circ} \cdot tg 50^{\circ} \cdot tg 70^{\circ}=\frac{1}{\sqrt{3}}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(4)\end{cases}

(2)-(4) represent the fundamental relations between the three numbers $tg10^{\circ}, tg50^{\circ},tg70^{\circ}$, from where we can obtain as many symmetrical relations as we want.

From algebraic identity

$(A+B+C)^{2}=A^{2}+B^{2}+C^{2}+2(A\cdot B+B \cdot C+C \cdot A)$

we can express

$tg^{2}10^{\circ}+tg^{2}50^{\circ}+tg^{2}70^{\circ}=(tg10^{\circ}-tg50^{\circ}+tg70^{\circ})^{2}+2\cdot (tg10^{\circ} \cdot tg50^{\circ}+tg50^{\circ} \cdot tg70^{\circ}-tg70^{\circ} \cdot tg10^{\circ})=(\sqrt{3})^{2}+2 \cdot 3=9.$

$\blacksquare$

This relationship appeared in another context in

https://www.facebook.com/sanchir.bold.14/posts/3756482811079358

see the problem on page 221.