GAZETA MATEMATICĂ Seria B N0 2/2020

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A PROBLEM WITH A SQUARE

          Seeing the post on facebook

https://www.facebook.com/angel.silvapalacios.1/posts/509305606696302

 

I will write a solution using complex numbers.

     At each point $P$ of the plan corresponds a complex number $z_{P}$. In our figure we choose the origin so that $z_{M}=0$, and the real axis BC and let $z_{A}=\alpha$. Since the point $M$ is the middle

 of the segment $[BC]$ we can assign $z_{B}=-\beta$, $z_{C}=\beta$. In the given figure we have $\beta =m$. Each vector $\overrightarrow{PQ}$ corresponds to the complex number $z_{Q}-z_{P}$. This correspondence between vectors and complex numbers is a linear application. I will also use the fact that, if the vector $\overrightarrow{PQ}$ rotate anticlockwise around the point $P$ by $\theta$ degrees, obtaining the result $\overrightarrow{PR}$, then

$z_{R}-z_{P}=(z_{Q}-z_{P}) \cdot (\cos \theta + \imath \sin \theta)$, or $=(z_{Q}-z_{P})\cdot \exp(\imath \theta)$.

          Because $\overrightarrow{AL}=2\cdot \overrightarrow{LB}$, according to a well-known formula we have

$\overrightarrow{ML}=\frac{\overrightarrow{MA}+2\cdot \overrightarrow{MB}}{1+2}.$

 (It is obtained from $\overrightarrow{AL}=2\cdot \overrightarrow{LB} \Leftrightarrow \overrightarrow{ML}- \overrightarrow{MA}=2(\overrightarrow{MB}- \overrightarrow{ML}) \Leftrightarrow (1+2) \overrightarrow{ML}=\overrightarrow{MA}+2\overrightarrow{MB}$, etc.). So to the vector $\overrightarrow{ML}$ corresponds the complex number $\frac{\alpha -2 \beta}{3}$. Ad-hoc we note this by

$\overrightarrow{ML}$   $\mapsto$   $\frac{\alpha-2\beta}{3}.$

 In the same way, from $\overrightarrow{AN}=3\cdot \overrightarrow{NC}$ we deduce

$\overrightarrow{MN}=\frac{\overrightarrow{MA}+3\cdot \overrightarrow{MC}}{1+3}$

 so

$\overrightarrow{MN}$  $\mapsto$  $\frac{\alpha +3 \beta}{4}$.

 But we have that $\overrightarrow{ML}$ is obtained from $\overrightarrow{ML}$ by a $90^{\circ}$ anticlockwise rotation around the point $M$, so

$\frac{\alpha +3\beta}{4}\cdot (\cos 90^{\circ}+\imath \sin 90^{\circ})=\frac{\alpha-2\beta}{3}$

$\Leftrightarrow$   $ \frac{\alpha+3\beta}{4}\cdot \imath=\frac{\alpha-2\beta}{3}$

$\Leftrightarrow$   $3\alpha \imath +9\beta \imath=4\alpha-8\beta$.

 We deduce $(4-3\imath)\cdot\alpha=(8+9\imath)\cdot\beta$, $\alpha=\frac{8+9\imath)\cdot \beta}{4-3\imath}=\beta\cdot \frac{(8+9\imath)(4+3\imath)}{16+9}$, so

$(1)$          $\alpha=\frac{1+12\imath}{5}\cdot \beta$,    $\beta=\frac{1-12\imath}{29}\cdot\alpha$.

      Using (1) for the vectors below we calculate the corresponding complex numbers

$\overrightarrow{AB}$   $\mapsto$   $-\beta-\alpha=-6\beta \frac{1+2\imath}{5}$,

$\overrightarrow{AC}$   $\mapsto$   $\beta-\alpha=4\beta\frac{1-3\imath}{5}$.

     By rotating the vector $\overrightarrow{AB}$ with the angle $\theta$, counterclockwise around point $A$ is obtained a vector  to which corresponds the complex number z

$(2)$             $-6\beta\cdot \frac{1+2\imath}{5}\cdot(\cos \theta +\imath \sin \theta)$.

 At the same time, the result of this operation is a vector collinear  with $\overrightarrow{AC}$,
so it is represented in form $\lambda \cdot \overrightarrow{AC}$ (with $\lambda $ -positive real number) and to which corresponds the complex number 

$(3)$                      $\lambda \cdot  4\beta \cdot \frac{1-3\imath}{5}$.

 If we equal the two numbers in (2), (3)

$-6\beta \cdot \frac{1+2\imath}{5}(\cos \theta +\imath \sin \theta)=\lambda \cdot 4\beta\cdot \frac{1-3\imath}{5}$

 $\Rightarrow$  $\cos \theta +\imath \sin \theta=-\lambda \frac{2}{3}\cdot \frac{1-3\imath}{1+2\imath}=-\lambda \cdot \frac{2}{3}\cdot \frac{(1-3\imath)(1-2\imath)}{1+4}=-\lambda\cdot \frac{2}{3}\cdot \frac{-5-5\imath}{5}$, so

$(4)$                   $\cos \theta+\imath \sin \theta=\lambda \cdot \frac{2}{3}(1+\imath)$.

The number to the left of the equation (4) has module 1 so it has to be $\left | \lambda \cdot \frac{2}{3}(1+\imath)\right |=1$ and we obtain $\lambda =\frac{3}{2\sqrt{2}}$. Then the formula (4) is written

$\cos \theta +\imath \sin \theta=\frac{1}{\sqrt{2}}(1+\imath)$

 $\Leftrightarrow$     $\cos \theta +\imath \sin \theta =\cos 45^{\circ}+\imath \sin 45^{\circ}$

whence it results $\theta=45^{\circ}$.

$\blacksquare$

 

          REMARKS

      1)  At points $L,\;N,\;K$ correspond the complex numbers

$z_{L}=\frac{-3+4\imath}{5}\beta$,  $z_{N}=\frac{4+3\imath}{5}\beta$,   $z_{K}=\frac{1+7\imath}{5}\beta$

(the latter is obtained by rotating the vector $\overrightarrow{NM}$ by $90^{\circ}$, clockwise, arround point $N$; we obtain for $\overrightarrow{NK}=-\frac{4+3\imath}{5}\beta \cdot (-\imath)$ and $z_{K}=z_{N}+\frac{-3+4\imath}{5}$).

A graphical representation using GeoGebra can be seen in the following figure (I chose $\beta=5$). Note that point $K$ is vertically from $A$ to $BC$.

  

     2) If we do not specify the origin or the real axis then

$z_{A}=\alpha$, $z_{B}=\beta$, $z_{C}=\gamma$

 and the calculations would have given $z_{M}=\frac{\beta+\gamma}{2}$

$\overrightarrow{ML}$     $\mapsto$     $\frac{(\alpha-\frac{\beta+\gamma}{2})+2\cdot(\beta-\frac{\beta+\gamma}{2})}{1+2}=\frac{2\alpha+\beta-3\gamma}{6}$

$\overrightarrow{MN}$     $\mapsto$     $\frac{\alpha-\frac{\beta+\gamma}{2}+3\cdot (\gamma-\frac{\beta+\gamma}{2})}{1+3}=\frac{\alpha-2\beta+\gamma}{4}$.

Because the vector $\overrightarrow{ML}$ is obtained by rotating counterclockwise, around the point $M$ of the vector $\overrightarrow{MN}$ by $90^{\circ}$, we have

$\overrightarrow{ML}$  $\mapsto$  $\imath \cdot \frac{\alpha-2\beta+\gamma}{4}$

and from here

$\frac{2\alpha+\beta-3\gamma}{6}=\frac{\alpha\imath-2\beta\imath+\gamma\imath}{4}$ 

$\Leftrightarrow$  $\alpha \cdot (4-3\imath)=(-2-6\imath)\beta+(6+3\imath)\gamma$

$\Leftrightarrow$  $\alpha=\frac{2-6\imath}{5}\cdot\beta+\frac{3+6\imath}{5}\cdot\gamma$.

Also, because rotating the vector $\overrightarrow{AB}$ with the $\theta$ angle we have a collinear vector with $\overrightarrow{AL}$, we have the equation

$(\beta-\alpha)\cdot (\cos \theta+\imath \sin \theta)=\lambda \cdot (\gamma-\alpha)$  $(\lambda>0)$

so 

$\cos \theta+\imath \sin=\lambda \cdot \frac{\gamma-\alpha}{\beta-\alpha}=\lambda \cdot \frac{\frac{2-6\imath}{5}\cdot (\gamma-\beta)}{\frac{3+6\imath}{5}\cdot(\beta-\gamma)}=\lambda \cdot(-\frac{2-6\imath}{3+6\imath})=\frac{30(1+\imath}{45}\cdot \lambda=\frac{2\lambda}{3}(1+\imath)$

and we find equation (4), etc.

DIDACTICA MATEMATICĂ N0 1/2020

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GAZETA MATEMATICĂ Seria B N0 1/2020

 

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GAZETA MATEMATICĂ Seria B N0 1/2019

 

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Dedicated to CONSTANTIN STELESCU on its 60th anniversary //Problem 111 Crux Mathematicorum v2n2 (Feb 1976)

      In the magazine

in Volume 2 Number 2 (February, 1976) page 25

     
     Two solutions of the exercise appear in Number 5 (May, 1976 page 95-96)

        Solution CiP

          The numbers $a,b,c$ can be considered as the roots of a third degree equation

$$x^{3}-px^{2}+qx-r=0.$$

 With the numbers $x_{1}=b-c,\;x_{2}=c-a,\;x_{3}=a-b$ we will form a third degree equation that they check. We will first calculate 

$$x_{1}+x_{2}+x_{3}=0$$

$$x_{1}\cdot x_{2}+x_{1}\cdot x_{3}+x_{2} \cdot x_{3}=$$

  $=(b-c)(c-a)+(c-a)(a-b)+(a-b)(b-c)=bc+ca+ab-a^{2}-b^{2}-c^{2}=$

$$=3\cdot q-p^{2}\overset{\underset{\mathrm{def}}{}}{=}Q$$

the last equality being in accordance with the Vieta's formulas,

$$x_{1} \cdot x_{2} \cdot x_{3}=(b-c)(c-a)(a-b)\overset{\underset{\mathrm{def}}{}}{=}R.$$

 Again with the Viète's formulas (for those who prefer the original French name François Viète instead of the Latinised form of his name, "Franciscus Vieta"), the numbers $b-c,\;c-a,\;a-b$ will be the roots of the equation

$$x^{3}-0\cdot x^{2}+Q\cdot x-R=0.$$

 But then $\frac{1}{x_{1}},\;\frac{1}{x_{2}},\;\frac{1}{x_{3}}$ will be the roots of the "written backwards" equation

$$R\cdot x^{3}-Q\cdot x^{2}-1=0$$

 and the Vieta's formulas give

$\sum_{cycl}\frac{1}{x_{1}} =\frac{Q}{R}$,            $\sum_{cycl} \frac{1}{x_{1}}\cdot \frac{1}{x_{2}}=0$.

     Now $\frac{1}{(b-c)^{2}}+\frac{1}{(c-a)^{2}}+\frac{1}{(a-b)^{2}}=\sum \frac {1}{x_{1}^{2}}=(\sum \frac {1}{x_{1}})^{2}-2 \cdot \sum \frac {1}{x_{1}\cdot x_{2}}=(\frac{Q}{R})^{2}-2\cdot 0=$

$$=(\sum_{cycl} \frac {1}{b-a})^{2}$$

that is, a square number.

$\blacksquare$