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.)

Comments remain open...

A 7TH GRADE PROBLEM FROM NMO // O PROBLEMA DE CLASA A 7-A DE LA ONM

                     We will solve the following problem (It is Problem 3 from NMO 2026, proposed for grade 7 page 5):

               Determine the pairs  $(a,b)$ of nonzero natural numbers for which the

 numbers  $\frac{a^2+b}{a+b-1}$  and  $\frac{b^2+a}{a+b}$  are natural.

{author} Lucian DRAGOMIR, Oțelu Roșu

Vocabulary  :  NMO = National Mathematics Olympiad

ANSWER CiP

$$a=1,\;\;\;b=1$$

                         Solution CiP

               Let be       $\frac{a^2+b}{a+b-1}=m\in\mathbb{N},\;\;\;\frac{b^2+a}{a+b}=n\in\mathbb{N}\;\;\;\;\;\;\;\;\;(1)$

From (1)  we obtain :

$m(a+b-1)=a^2+b\;,\;\;\;n(a+b)=b^2+a \tag{2}$

          Subtracting the equations (2) we obtain :

$(a+b-1)\cdot m-(a+b)\cdot n=(a-b)(a+b-1) \tag{3}$

          We view  (3)  as a linear Diophantine equation with the unknowns  $m\; and\; n$. A particular, almost obvious solution of  (3)  is  $m_0=a-b\;,\;\;n_0=0.$

 Then, all integer solutions of the equation  (3)  will be :

$m_k=a-b+k\cdot (a+b)\;,\;\;\;n_k=k\cdot (a+b-1)\;\;,\;\;\;k\in\mathbb{Z} \tag{4}$

However, we are only interested in natural number solutions  $(m_k,n_k)\in\mathbb{N}^*\times \mathbb{N}^*$  so  $k>0.$

          The second equation in  (2)  is written

$k\cdot (a+b-1)\cdot (a+b)=b^2+a$

 from which we obtain :

$k=\frac{b^2+a}{(a+b)(a+b-1)} \tag{5}$

                    This is where I got stuck, although the intention was to see from  (5)  what the possible values ​​for  $k$  are.

     Examining the "First Solution" from the official Olympiad materials (see page 15, Problem 3) I found that I was on the right track. Moreover, the expression in  (5)  appears there with the notation  $"p"$  (which in Greek is written  $\pi$). “Diophantus would denote this number using the letter π, hence my choice to include it in the title of the post.”(Idea first shared with COPILOT.) That's how I chose, a little bit cool, sly, the title of the Post there.

"Returning to our sheep," as the proverb would say, we see that :

$k< 1\;\Leftrightarrow\;b^2+a< a^2+2ab+b^2-a-b\;\Leftrightarrow$

$\Leftrightarrow\;a(a-2)+b(2a-1)>0$

for  $a\geqslant 2$. So in the case  $a\geqslant 2$  there are no solutions.

          For  $a=1,\;\;m=\frac{1+b}{b}=1+\frac{1}{b}\;,\;\;n=\frac{b^2+1}{b+1}=b-1+\frac{2}{b+1}$. and  $m\;,n$  are integers if and only if  $b=1$. 

          We got the answer.

$\blacksquare$