Gheorghe ANDREI - COMPLEX NUMBERS (part II) - Some Problems -- 5 / Gheorghe ANDREI - NUMERE COMPLEXE (partea II) - Câteva Probleme -- 5 / Gheorghe ANDREI – LICZBY ZŁOŻONE (część II) – Niektóre problemy – 5

 I wrote about the book here. 

It's not for nothing that I wrote somewhere that THIS BOOK IS FULL OF SCRATCHES. (Maybe that's why the author withdrew Part I from sale)

                    We are now debating Problem 62***, page 157 (solved on pages 516-519).

       

                     " 62***.  Calculate  $P_n\; Q_n\;and\; R_n$, where  $n\in \mathbb{N},\;n\geqslant 3,\;a\in\mathbb{R}^*\;:$

 $$a)\;\;\;P_n=\prod_{k=0}^{n-1} \cos \left (a+\frac{k\pi}{n}\right )\;;$$

$$b)\;\;Q_n=\prod_{k=0}^{n-1}\sin \left (a+\frac{k\pi}{n}\right )\;;$$

$$c)\;\;R_n=\prod_{k=0}^{n-1}\tan \left (a+\frac{k\pi}{n}\right )."$$

ANSWER

$a)\;\;P_{2n}=\frac{(-1)^n\sin 2na}{2^{2n-1}}\;\;;\;\;P_{2n+1}=\frac{(-1)^n\cos (2n+1)a}{2^{2n}}$

$b)\;\;Q_n=\frac{\sin na}{2^{n-1}}$

$c)\;\;R_{2n}=-1\;\;;\;\;R_{2n+1}=(-1)^n\tan (2n+1)a$

Solution CiP

               We will start with b). A solution that, at some point, requires a square root extraction is found in TURTOIU Fanică - Probleme de Trigonometrie, Ed. Tehnică, București, 1979 (pages 86-87, Problem 2.38 - received by the kindness of Liviu PODGORNEI). This is a personal attempt.

          Let  $z_k=\cos\left (a+\frac{k\pi}{n}\right )+\imath \sin \left (a+\frac{k\pi}{n}\right )\;,\;\;0\leqslant k\leqslant n-1.$  We have

$|z_k|=1=z_k\cdot \bar z_k$  so  $\bar z_k=\cos \left (a+\frac{k\pi}{n}\right )-\imath \sin \left (a+\frac{k\pi}{n}\right )=\frac{1}{z_k}.$  In addition

$z_k^{2n}=\cos 2n \left (a+\frac{k\pi}{n}\right )+\imath \sin 2n\left (a+\frac{k\pi}{n}\right )=\cos 2na+\imath \sin 2na \tag{1}$

Then

$\sin \left (a+\frac{k\pi}{n}\right )=\frac{z_k-\bar z_k}{2\imath}=\frac{z_k-\frac{1}{z_k}}{2\imath}=\frac{z_k^2-1}{2\imath z_k} \tag{2}$

          From  (1)  it follows that the equation

$z^{2n}=\cos 2na+\imath \sin 2na \tag{3}$

has the roots  $\pm z_k\;,\;\;k\in \{0,\;1,\dots ,\;n-1 \}.$  Denoting

$f(z)=z^{2n}-(\cos na +\imath \sin na) \tag{4}$

we have  $f(z)=(z-z_0)(z-z_1)\dots (z-z_{n-1})(z+z_0)(z+z_1)\dots (z+z_{n-1})\;$,  or

$$f(z)=\prod_{k=0}^{n-1}(z^2-z_k^2) \tag{5}$$

From here we go further  $$\prod_{k=0}^{n-1}(z_k^2-1)=(-1)^n\prod_{k=0}^{n-1}(1-z_k^2)=(-1)^nf(1)\overset{(4)}{=}(-1)^n(1-\cos 2na-\sin 2na)=$$

$$=(-1)^n(2\sin^2na-2\imath \sin na \cos na)=(-1)^n(-2\imath^2\sin^2 na-2\imath \sin na \cos na)$$

so

$$\prod_{k=0}^{n-1}(z_k^2-1)=-2\imath (-1)^n\sin na(\cos na+\imath \sin na) \tag{6}$$

          On the other hand

$$\prod_{k=0}^{n-1}z_k=\prod_{k=0}^{n-1}\left [(\cos a+\imath \sin a)\left (\cos \frac{k\pi}{n}+\imath \sin\frac{k\pi}{n}\right )\right]=$$

$$=(\cos a+\imath \sin a)^n\prod_{k=0}^{n-1}\left (\cos \frac{k\pi}{n}+\imath \sin \frac{k\pi}{n}\right )=$$

$$=(\cos na+\imath \sin na)\left [\cos \left (\sum_{k=0}^{n-1}k\right )\frac{\pi}{n}+\imath \sin \left (\sum_{k=0}^{n-1}k\right )\frac{\pi}{n}\right ]=$$

$=(\cos na+\imath \sin na)\left [\cos (n-1)\frac{\pi}{2}+\imath \sin (n-1)\frac{\pi}{2}\right ]$

because  $\sum_{k=0}^{n-1}k=\frac{(n-1)n}{2}.$  Analyzing the cases  $n=4m,\;4m+1,\;4m+2,\;4m+3$  it is immediately seen that  $\cos (n-1)\frac{\pi}{2}+\imath \sin (n-1)\frac{\pi}{2}=-\imath^{n+1}$ , and then the previous calculation gives us

$$\prod_{k=0}^{n-1}z_k=-\imath^{n+1}(\cos na+\imath\sin na) \tag{7}$$

          Now we can finish

$$Q_n=\prod_{k=0}^{n-1}\sin \left (a+\frac{k\pi}{n}\right )\overset{(2)}{=}\frac{\prod_{k=0}^{n-1}(z_k^2-1)}{2^n\imath^n\prod_{k=0}^{n-1}z_k}\overset{(6)}{\underset{(7)}{=}}\frac{-2\imath(-1)^n\sin na(\cos na+\imath\sin na)}{2^n\imath^n(-\imath^{n+1})(\cos na+\imath \sin na)}=$$

$=\frac{\imath \cdot (-1)^n\sin na}{2^{n-1}\imath^{2n}\cdot \imath}=\frac{(-1)^n \sin na}{2^{n-1}(-1)^n}=\frac{\sin na}{2^{n-1}}.$  We got the answer of  b).

     Remark CiP   On page 518 the following stupid answer is given :

$$Q_n=\begin{cases}\frac{(-1)^n\cdot \sin na}{2^{n-1}}\;\;\;if\;\;n-even\\\frac{(-1)^{n+1}\cdot \sin na}{2^{n-1}}\;\;\;if\;\;n-odd\end{cases} \tag{Q}$$

............................................................................................................................

                    a)  With the same notations as in point b), first we have

$\cos\left (a+\frac{k\pi}{n}\right )=\frac{z_k+\bar z_k}{2}=\frac{z_k^2+1}{2z_k} \tag{8}$

          Then, starting from  (5)  we see that  

$$f(\imath)=\prod_{k=0}^{n-1}(-1-z_k^2)=(-1)^n\prod_{k=0}^{n-1}(z_k^2+1)$$

so

$$\prod_{k=0}^{n-1}(z_k^2+1)=(-1)^nf(\imath) \tag{9}$$

          Then we can write

$$P_n=\prod_{k=0}^{n-1}\cos \left (a+\frac{k\pi}{n}\right )\overset{(8)}{=}\frac{\prod_{k=0}^{n-1}(z_k^2+1)}{2^n\prod_{k=0}^{n-1}z_k}\overset{(7)}{\underset{(9)}{=}}\frac{(-1)^nf(\imath)}{2^n(-\imath^{n+1})(\cos na+\imath \sin na)} \tag{10}$$

But  $f(\imath)\underset{(4)}{=}(-1)^n-\cos 2na-\imath \sin 2na$  and we will treat the  $n-even$  and  $n-odd$  cases separately.

          $\fbox{$n$-even}$

          $f(\imath)=1-\cos na-\imath \sin na=2\sin^2 na-2\imath \sin na \cos na=$

$=-2\imath^2\sin^2 na-2\imath \sin na \cos na=-2\imath \sin na(\cos na+\imath \sin na)$

In  (10)  also appears  $-\imath^{n+1}=-\imath^n\cdot \imath=(-\imath)(-1)^{\frac{n}{2}}$  and then we continue like this

$P_n=\frac{-2\imath \sin na(\cos na+\imath \sin na)}{2^n(-\imath)(-1)^{\frac{n}{2}}(\cos na+\imath \sin na)}=\frac{(-1)^{\frac{n}{2}}\sin na}{2^{n-1}}.$

          $\fbox{$n$-odd}$

          $(-1)^nf(\imath)=-(-1-\cos 2na-\imath \sin 2na)=2\cos^2 na+2\imath \sin na \cos na=$

$=2\cos na(\cos na+\imath \sin na)$

and because  $-\imath^{n+1}=-(-1)^{\frac{n+1}{2}}=-(-1)^{\frac{n-1}{2}+1}=-(-1)^{\frac{n-1}{2}}(-1)=(-1)^{\frac{n-1}{2}}$

we have  $P_n\underset{(10)}{=}\frac{2\cos na(\cos na+\imath \sin na)}{2^n(-1)^{\frac{n-1}{2}}(\cos na+\imath \sin na)}=\frac{(-1)^{\frac{n-1}{2}}\cos na}{2^{n-1}}.$  We got the answer of  a).

      Remark CiP   Although in the solution I used COPILOT, who first told me incorrectly that the minus sign always appears in the formula, then came back and corrected the mistake, we must admit that in the book, on page 517, the correct answer is given. Marius DRĂGAN and Sorin RĂDULESCU are cited as the authors of the solution.

.............................................................................................................................

                    c)  $R_{2n}=\frac{Q_{2n}}{P_{2n}}=\frac{\frac{\sin 2na}{2^{2n-1}}}{\frac{(-1)^n\sin 2na}{2^{2n-1}}}=(-1)^n\;;$

                         $R_{2n+1}=\frac{Q_{2n+1}}{P_{2n+1}}=\frac{\frac{\sin (2n+1)a}{2^{2n}}}{\frac{(-1)^n\cos (2n+1)a}{2^{2n}}}=(-1)^n\tan(2n+1)a.$

$\blacksquare\;\blacksquare\;\blacksquare$

An article about SOME TRIGONOMETRIC IDENTITIES // hg

 See the full article in GMB 7/2005, pages 289-293.

First identity :

(is formula (5) , page 290, in the cited article)

$$\prod_{k=0}^{n-1}\cos \frac{x+2k\pi}{n}=\frac{1}{2^{n-1}}\left [\cos\frac{n\pi}{2}+(-1)^{n+1}\cdot \cos x\right ] \tag{1}$$

           For  $x=\pi$  we obtain :

$\cos \frac{\pi}{n}\cdot \cos \frac{3\pi}{n} \dots  \cos\frac{(2n-1)\pi}{n}=\frac{1}{2^{n-1}}\left [\cos \frac{n\pi}{2}+(-1)^{n+1}\cdot \cos \pi \right]=\frac{1}{2^{n-1}}\left [\cos \frac{n\pi}{2}+(-1)^n\right ]$

i.e. Exercise 9a).

          For  $x=0$  we obtain :

$\cos \frac{2\pi}{n}\cdot \cos \frac{4\pi}{n}\dots \cos \frac{2(n-1)\pi}{n}=\frac{1}{2^{n-1}} \left [\cos \frac{n\pi}{2}+(-1)^{n+1}\cdot \cos 0\right ]=\frac{1}{2^{n-1}}\left [ \cos \frac{n\pi}{2}-(-1)^n\right ]$

i.e. Exercise 9b).

In particular we have the equalities :

$\cos \frac{\pi}{5}\cdot \cos \frac{3\pi}{5}\cdot \cos \frac{5 \pi}{5}\cdot \cos \frac{7\pi}{5}\cdot \cos \frac{9\pi}{5}=\frac{1}{2^4}\left [ \cos\frac{5\pi}{2}-1\right ]=-\frac{1}{16}\;;$

$\cos \frac{2\pi}{5}\cdot \cos \frac{4\pi}{5}\cdot \cos \frac{6\pi}{5}\cdot \cos\frac{8\pi}{5}=\frac{1}{2^4}\left [\cos \frac{5\pi}{2}+1\right ]=\frac{1}{16}.$

(in construction)

OLIMPIADA NAȚIONALĂ de MATEMATICĂ - cioburi // NATIONAL MATHEMATICS OLYMPIAD - some shards // NATIONALE MATHEMATIK-OLYMPIADE - Scherben

 (O încercare de regăsire a profesorului meu din Liceu, Vasile BIVOLARU. A plecat in GERMANIA, sotia, Ecaterina, a murit acolo  -  mi-a spus fiul lor, Cristian, pe Facebook)

          Scanning, little by little, my collection of  "Gazeta Matematica" magazines (aka GMB), I reached issue 6 from 1978 where the Topics from the 1977 Final Stage are published, the only one I participated in, in 10th grade. 

(in consrtuction)

C : 2816 - A concrete trigonometric equality // კონკრეტული ტრიგონომეტრიული თანასწორობა

The problem was published in GMB 12 / 2004; I don't think it's here. It was solved in GMB 6 / 2005.

    "C:2816.  Prove the equality  $\frac{\sin 110^{\circ}}{\sin 30^{\circ}}-\frac{\sin 60^{\circ}}{\sin 80^{\circ}}=1."$

{authors :  Romanța GHIȚĂ and Ioan GHIȚĂ, Blaj

               Solution CiP - taken from the cited source

          The given equality is successively equivalent to :

$\sin 110^{\circ}\cdot \sin 80^{\circ}=\sin 30^{\circ}\cdot \sin 60^{\circ}+\sin 30^{\circ}\cdot \sin 80^{\circ}\;\Leftrightarrow$

$\Leftrightarrow\;\;2\sin 70^{\circ}\cdot \sin 80^{\circ}=\sin 60^{\circ}+\sin 80^{\circ}\;\Leftrightarrow$

$\Leftrightarrow\;\;\cos 10^{\circ}-\cos 150^{\circ}=\sin 60^{\circ}+\sin 80^{\circ}$

which is obvious, because  $\cos 10^{\circ}=\sin 80^{\circ}\;\;and\;\; -\cos 150^{\circ}=\sin 60^{\circ}=\frac{\sqrt{3}}{2}.$

$\blacksquare$

          REMARK CiP   It would be interesting to study some trigonometric equations starting from the present equality, as I did in the post of September/7/ 2025  A forgotten trigonometric equation.

A problem with Symmetric Polynomials // Um problema com polinómios simétricos

 It is problem 3 016, 2/1977 page 59 from the "Revista Matematica a Elevilor din Timisoara" (Mathematical Magazine of Students from Timisoara).

         "3 016.  If  $a,\;b,\;c$  are nonzero real numbers, such that 

 $a+b+c=0$  and  $a^3+b^3+c^3=a^5+b^5+c^5$ , 

then  $a^2+b^2+c^2=\frac{6}{5}.$

{author : } Titu ANDREESCU, student, Timisoara"

Solution CiP

                    We will use the following algebraic identities :

$(a+b+c)^2-a^2-b^2-c^2=2(ab+bc+ca) \tag{1}$

$(a+b+c)^3-a^3-b^3-c^3=3(a+b)(b+c)(c+a) \tag{2}$

$(a+b+c)^5-a^5-b^5-c^5=$

$=5(a+b)(b+c)(c+a)(a^2+b^2+c^2+ab+bc+ca) \tag{3}$

          (1)  is obtained from the identity :

$(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca$.

          For  (2)  let  $P(a,b,c)=(a+b+c)^3-a^3-b^3-c^3$. If  $b=-a$  then  $Pa,-a,c)=c^3-a^3+a^3-c^3=0$  so  $P$  has the factor  $a+b$. From symmetry it will also have the factors  $b+c,\;c+a$  so, being of degree 3, it results 

$P(a,b.c)=(a+b)(b+c)(c+a)\cdot Q(a,b,c)\;,\;\;deg(Q)=0$

hence  $Q(a,b,c)$  is a constant. As  $P(1,1,0)=2^3-1-1=6$, from  $6=(1+1)(1+0)(0+1)\cdot Q(1,1,0)$  it foloow   $Q=3.$

          For  (3)  let  $P(a,b,c)=(a+b+c)^5-a^5-b^5-c^5$.  $P(a,-a,c)=c^5-a^5+a^5-c^5=0$  so

$P(a,b,c)=(a+b)(b+c)(c+a)\cdot Q(a,b,c)\;,\;\deg(Q)=2 \tag{4}$

So  $ Q$  has the form  $Q(a,b,c)=m\cdot (a^2+b^2+c^2)+n\cdot (ab+bc+ca)$

If we put in  (4)  first  $a=b=1,\;c=0$,  and second  $a=b=c=1$  we get 

$P(1,1,0)=2^5-1-1=30=2(m\cdot 2+n),\;$

$\;\;P(1,1,1)=3^5-1-1-1=240=8(m\cdot 3+n\cdot 3)$

hence  $m=n=5$.

               Let's move on to solving the problem. In the condition  $a+b+c=0$ , we have :

 - from  (1)  

$\sum a^2=-2\sum ab \tag{5}$

  - from  (2)   $0-\sum a^3\;\underset{a+b=-c\;etc}{=}\;\;\;3(-c)(-a)(-b)$ , so

$\sum a^3=3abc \tag{6}$

  - from  (3)  $0-\sum a^5=5(-c)(-a)(-b)(\sum a^2+\sum ab)\overset{(5)}{=}-5abc(\sum a^2-\frac{1}{2}\sum a^2$c , so

$\sum a^5=\frac{5}{2}abc\sum a^2 \tag{7}$

But we also have  $\sum a^3=\sum a^5$ , so from  (6)  and  (7)  it results 

 $3abc=\frac{5}{2}abc \sum a^2$

hence the answer.

$\blacksquare$

 

A RIGHT TRIANGLE and ... one more // Un TRIUNGHI DREPTUNGHIC și ... încă unul // Правоаголен триаголник и ... уште еден

 It's a fairly basic geometry problem (proposed for sixth grade). 

    "E :  6209*.  In triangle  $ABC$  with  $\sphericalangle A=90^{\circ}$ the bisector of angle  $C$ intersects the perpendicular bisector of side  $AB$  in  $D$. Find  $\sphericalangle  CBD$.

 Ion MACREA, professor Brașov"

                    Figure of the problem (CiP) : 

               REMARK CiP  The order of important lines in a triangle is known to be: altitude -  - - bisectors - median - perpendicular bisectors. Since CF is the median and CA the altitude, line CD is between them, so point D is on the extension of segment EF, beyond F and not elsewhere.

ANSWER CiP

$\sphericalangle CDB=90^{\circ}$

                    Solution CiP 

               We rely on the following characterization of right triangles :

                           In a right triangle, the median to the hypotenuse

                          is equal to one half of the hypotenuse.

              Conversely, if in a triangle the median drawn from a vertex

                      has length equal to one half of the side to which it is drawn,

                       then the angle at that vertex is a right angle.

A complete figure is below. Point  $F$  is the midpoint of the leg  $[AB],\;\;DE\perp AB$.

(Basically,  $EF$  is the perpendicular bisector of side  $[AB]$.) 

          $FE,\;AC \perp AB \Rightarrow FE \parallel AC$  and then, 

once

 E is the midpoint of the segment BC                         (1)

 and the second time we have the equalities of angles

$\sphericalangle EDC=\sphericalangle ECD \;\;\;\overset{CD\; is\; the}{\underset{bisector\; of \;\sphericalangle ACB}{=}}\;\;\;\sphericalangle ACD\tag{2}$

             $(2)\Rightarrow\;\;CDE$  is isosceles triangle, with base  $ CD$  so

$DE=EC\underset{(1)}{=}EB=\frac{BC}{2}$

but this says that, in triangle  $BCD$  

    the median  $DE$  drawn from the vertex  $D$  has length equal to one half of the side  $BC$  to which it is drawn,

so  the angle  $\sphericalangle BDC$  at the vertex  $D$  is a right angle.

$\blacksquare$

Two High School Quizzes I Still Remember (9th Grade / Year 1) // Doua Extemporale de pe vremea Liceului (clasa a 9-a/anul I) // Дзве школьныя віктарыны, якія я дагэтуль памятаю (9 клас / 1 год)

          I found two Quizzes in my 1st year high school ALGEBRA textbook.(On the first page is my signature and a personal stamp. I had several math teachers at the beginning of that school year. 

         I will talk about the first of them, Vasile BIVOLARU, who only stayed for three weeks, being incorporated for a 6-month military internship, another time. 

I spent the first two years of high school with Professor Gheorghe DUMITRU, the one who gave me my grades.

First Quiz, with the last teacher

Second quiz

I also found among the pages of the book an Essay on Political Science. (The writing in this essay does not appear to be mine.)

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