A necessary correction: it seems that I was wrong in my calculations, not only in life

 The answer to the Problem in the post here is WRONG. It was brought to my attention by a commenter, unfortunately anonymous.

          The answer will be replaced there with this one :

Corrected   ANSWER CiP

The statement is NOT true for polynomials of degree 4 :

If  $P(x)=x^4\;,\;\;Q(x)=x^4+x^2\;,\;\; R(x)=2x^4+x^2\;\;$  then  $P(x)\leqslant Q(x)\leqslant R(x)$

but  $\lambda \cdot P(x)+(1-\lambda)\cdot R(x)=(2-\lambda)\cdot x^4+(1-\lambda)\cdot x^2\neq Q(x)$

$\square$

Three Nice Polynomials of Degree III // Tre pæne polynomier af grad III

 We will solve Problem 7 (page 38, in the magazine from yesterday's Post) from Team Selection Test(TSTs) 1 for IMO.

" Problem 7.  Consider three polynomials of degree 3 with real 

coefficients $P,\;Q,\;R$  such that  $P(x)\leqslant Q(x)\leqslant R(x)$  for any real  $x$

  and, in addition, there exists a real number  $a$  such that  $P(a)=R(a).$

  Show that there exists a constant  $\lambda \in [0,\;1]$  with the property

  $Q=\lambda P+(1-\lambda)R.$  Does the statement remain true in the case 

                         when  $P,\;Q,\;R$  have degree 4?

{authors : } I. Cuculescu and L. Panaitopol, Bucharest"

ANSWER CiP (see correction)

It is not entirely true for polynomials of degree 4. For example, if 

$P(x)=x^4\;,\;\;Q(x)=x^4+x^2\;,\;\;R(x)=x^4+2x^2$  then  $P(x)\leqslant Q(x)\leqslant R(x)$

  and we have $Q(x)=\lambda \cdot P(x)+(1-\lambda)\cdot R(x)$  but with  $\lambda =-1 \not \in [0,\;1]$

Solution CiP

                   Obviously  $P(a)=Q(a)=R(a).$  Let's define polynomials

$S(x)=R(x)-Q(x)\;\geqslant 0\;,\;\forall x\;\;;\;\;T(x)=R(x)-P(x)\;\geqslant 0\;,\;\forall x \tag{1}$

          We have  $S(a)=0$  therefore  $S(x)=(x-a)S_2(x)$ , for a certain polynomial  $S_2$  of degree 2.

Since when  $x$  passes through the value  $x=a$ , the polynomial  $S$  does not change sign, we must have 

$S(x)=(x-a)^2\cdot S_1(x) \tag{2}$

with  $S_1$-a first degree polynomial. But the polynomial  $S_1$  should have a root, through which if   $x$  passes, the expression in  (2)  changes sign again. Contradiction, so  $S_1$  is a constant.

          In exactly the same way results

$T(x)=(x-a)^2\cdot T_1(x) \tag{3}$

with  $T_1$  a constant.

          Let  $\lambda :=\frac{S(x)}{T(x)}\overset{(2)}{\underset{(3)}{=}}\frac{S_1}{T_1}=constant\;$.  Since  $\lambda=\frac{R(x)-Q(x)}{R(x)-P(x)}$  and  $R(x)-Q(x)\leqslant R(x)-P(x)$  we have  $0\leqslant \lambda \leqslant 1$  and 

 $Q(x)=\lambda \cdot P(x)+(1-\lambda )\cdot R(x).$

$\blacksquare$

Problem E : 6271

 Click on the image and use the password : ogeometrie . The QR code on page 2 contains a link to the Books in my Electronic Library. (Click on the year of publication. Same password if needed.) A collection of important magazines is at the letter G : GAZETA MATEMATICĂ seria B.

On page 292 :

                          " E : 6271*.  Find the maximum and minimum of the sum  $x+y$

                                                   and the product  $xy$ , knowing that  $5x+6y=150$ ,

 and  $x\; and\; y$  are natural numbers different from zero.           

{author : } Gh. MARGHESCU, student, București "

ANSWER CiP

$max\;(x+y)=29$  for  $x=24\;, \;y=5$

$min\;(x+y)=26$   for  $x=6\;,\;y=20$

$max\;(x\cdot y)=180$  for  $x=12\;,\;y=15\;\;or\;\;x=18\;,\;y=10$

$min\;(x\cdot y)=120$  for  $x=6\;,\;y=20\;\;or\;\;x=24\;,\;y=5$

Solution CiP

                  We have the relationships :

$5x+6y=150\;\;,\;\;\;5x=150-6y\;\;\;6y=150-5x \tag{1}$

          From  $6\mid 150\;\;and\;\;6\mid 6y$  and from the second formula (1) it follows that  $6\mid 5x$  and because  $gcd(5,6)=1$  we have  $6\mid x$  so

$x=6\cdot x_1\;,\;\;x_1\in\mathbb{N^*} \tag{2}$

Similarly, we have  $5\mid 150\;\;and\;\;5\mid 5x$ , and from the third formula (1) it results  $5\mid 6y$ , so

$y=5\cdot y_1\;\;\;y_1\in\mathbb{N^*} \tag{3}$

          Then  $5x+6y=150\;\;\overset{(2)}{\underset{(3)}{\Leftrightarrow}}\;\;\;5\cdot 6x_1+6\cdot 5y_1=150\Leftrightarrow$

$\Leftrightarrow\;\;\;\;\;\;\;\;\;\;x_1+y_1=5 \tag{4}$

If  $x_1=n\in\mathbb{N^*}$ , then  $y_1\overset{(4)}{=}5-x_1=5-n$ , so the nonzero natural number solutions for  (4)  are :

$x_1=n\;\;,\;\;\;y_1=5-n\;\;\;\;\;1\leqslant n \leqslant 4 \tag{5}$

But

$x+y\;\;\;\overset{(2)\;(3)}{\underset{(5)}{=}}\;\;6\cdot n+5\cdot (5-n)=n+25 \tag{6}$

and then  $1\leqslant n \leqslant 4 \Leftrightarrow26\leqslant n+25 \leqslant 29\underset{(6)}{\Leftrightarrow}$

$\Leftrightarrow\;\;\;\;\;26 \leqslant x+y \leqslant 29 \tag{7}$

We have in  (7)  $x+y=26$  for  $n=1$ , so  $x_1=1\;,\;y_1=4$  and we get from  (2)  and  (3)  $x=6\;,\;y=20$  for the minimum value.  And  $x+y=29$  for  $n=4$ , so $x_1=4\;,\;y_1=1$  so  $x=24\;,\;y=5$  for the maximum value.

          Since there are a small number of values ​​for  $x\; and\; y$ , we will list all the products  $x\cdot y$  in the table below, from where we will also obtain the answer.

\begin{array}{c|c|c|c|c}n&1&2&3&4\\\hline x&6&12&18&24\\\hline y&20&15&10&5\\\hline x\cdot y&120&180&180&120\\\end{array}

$\blacksquare$

The E : 17461 problem, half easy, half hard // Задача E: 17461, наполовину простая, наполовину сложная

 From GMB (page 159), also proposed for 5th grade.

          " E : 17461.  Let  $N=3+3^2+3^3+\dots+3^{4n+1}\;,\;n\in\mathbb{N}.$

            a)  Show that  $2N+3$  s a perfect square.

            b)  For  $n$  even natural numbers, determine the last two digits of  $N$.

                                                                                  {author : } Marin CHIRCIU, Pitești "

Partial  ANSWER  CiP

a)  $2N+3=(3^{2n+1})^2$

b) The answer is not a unique number. We have :

$3+3^2+3^3+3^4+3^5+3^6+3^7+3^8+3^9=29\;5\color{Yellow}{23}$

$3+3^2+3^3+\dots+3^{17}=193\;710\;2\color{Pink}{43}$

Solution CiP

                    a) $2N+3=\color{Red}{3}+N+N=$

$=\color{Red}{3}+(3+3)+(3^2+3^2)+(3^3+3^3)+\dots+(3^{4n+1}+3^{4n+1})=$

$=\color{Red}{(3+3+3)}+(3^2+3^2)+(3^3+3^3)+\dots+(3^{4n+1}+3^{4n+1}).$

Let's observe that  $3+3+3=3^2$ ...and in general

$3^k+3^k+3^k=3^{k+1} \tag{1}$

Then, the red parenthesis results is $\color{Red}{3^2}$ , which appears as the term in the second parenthesis thus :

$(\color{Red}{3^2}+3^2+3^2)+(3^3+3^3)+\dots +(3^{4n+1}+3^{4n+1}).$

     Continuing this calculation, we finally obtain the result

$2N+3=\color{Red}{3^{4n+1}}+3^{4n+1}+3^{4n+1}\underset{(1)}{=}3^{4n+2}=(3^{2n+1})^2\;,$

i.e. a perfect square.

$\square$

Our Beloved Magazine "GAZETA MATEMATICĂ series B" - GMB 3 / 2026 // Notre Magazine Bien-Aimé "GAZETA MATEMATICĂ série B" - GMB 3 / 2026

  Click on the image and use the password : ogeometrie .The QR code on page 2 contains a link to the Books in my Electronic Library. (Click on the year of publication. Same password if needed.) A collection of important magazines is at the letter G : GAZETA MATEMATICĂ seria B.

          A problem to illustrate the love for numbers, proposed for 5th grade (page 159) :

                    "E : 17457 .   Write the number  $220^{2n+1}$  as the sum of 5 distinct

                      nonzero perfect squares. 

{author : } Marin CHIRCIU, Pitești "

ANSWER CiP

$220=1^2+3^2+4^2+5^2+13^2$  hence

$220^{2n+1}=(220^n)^2+(3\cdot 220^n)^2+(4 ^2\cdot 220^n)^2+(5\cdot 220^n)^2+(13\cdot 220^n)^2$

Solution CiP

               We have  $220^{2n+1}=(220^n)^2\cdot 220^1$ , so let's try to write the number  $220$  as the sum of five distinct nonzero perfect squares. 

We write some sums,  $\color{Green}{highlighting}$  a perfect square term:

$220=\color{Green}{196}+24$

$220=\color{Green}{169}+51$

$220=\color{Green}{144}+76$

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

But any number is written as the sum of four squares, according to Lagrange's Four-Square Theorem, so let's try this with the second term in the writings above.

          Asking a friend I found out that : 

<< Adevărul e că  $24$ are foarte puține reprezentări valide.

Singura reprezentare cu pătrate întregi ne-negative este:

$$\boxed{{\displaystyle 24=4^2+2^2+2^2+0^2}}$$

>>

In translation : <<The truth is that 24 has very few valid representations. The only representation with non-negative integer squares is: ... >> But the squares here are not all nonzero and not all distinct.

Moving on to term $51$ , the same friend gave me the answer you see in the first row above.

The problem is now solved.

$\blacksquare$

The Exercise Supplement of the Mathematical Gazette // Príloha k cvičeniam v Matematickom vestníku

 Click on the image and use the password : ogeometrie .The QR code on page 2 contains a link to the Books in my Electronic Library. (Click on the year of publication. Same password if needed.) A collection of important magazines is at the letter G : GAZETA MATEMATICĂ seria B.

       

                 To see how simple the exercises and problems are, we will take the following problem proposed  for 6th grade:

    

               "S.E26.100.  The degree measures of the angles of an acute-angled triangle

 are expressed as prime numbers. Prove that the triangle is isosceles.

* * * {no author}"

ANSWER CiP

The triangle has angle measures  $2^{\circ}\;,\;\;89^{\circ}\;,\;\;89^{\circ}$

    Solution CiP

               If  $x^{\circ},\;y^{\circ},\;z^{\circ}$  are the angles of this triangle, then :

$x^{\circ}\;,\;y^{\circ},\;z^{\circ}\;<90^{\circ}\;\;and\;\;x^{\circ}+y^{\circ}+z^{\circ}=180^{\circ} \tag{1}$

          Being prime numbers, if they were all odd then their sum would be odd. We contradict (1). So one of the numbers, let's say  $z^{\circ}$  is  $2^{\circ}$.  Then the second condition in (1) becomes :

$x^{\circ}+y^{\circ}=178^{\circ} \tag{2}$

     An obvious solution for  (2) , which meets the requirements of the problem, is  $x^{\circ}=y^{\circ}=89^{\circ}$.  This triangle is isosceles.

     $x^{\circ}\underset{(1)}{<}89^{\circ}\Rightarrow y\underset{(2)}{=}178^{\circ}-x^{\circ}>89^{\circ}$.  And we can still have  $y^{\circ}=90^{\circ}$ ,  which violates all the requirements.

$\blacksquare$

Relationships between Trigonometric Values ​​Deduced from Equations // Relaciones entre valores trigonométricos deducidos de ecuaciones // In Memoriam Professor Constantin BUȘE

           In the GMB 7/2005, mentioned here, the solution to problem 25223 (published in GMB/1/2005, page 40) appears, which is the source of this post. (With nostalgia, I found on page 299 the name of my college colleague Constantin BUȘE...deceased in the meantime)

         " 25223.  To solve in  $\mathbb{C}$   the equation :  $3x^3-3x+1=0.$

[author : ] Gh. Szöllősy, Sighetul Marmației"

ANSWER

$x_1=\frac{2}{\sqrt{3}}\cos\frac{5\pi}{18}\;,\;x_2=\frac{2}{\sqrt{3}}\cos \frac{7\pi}{18}\;,\;x_3=\frac{2}{\sqrt{3}}\cos\frac{17\pi}{18}$

 Solution

               We are looking for solutions of the kind  $x_t=a\cdot \cos t,\;a>0,\;t\in [0,\pi]$. Then

$3x_t^3-3x_t+1=0\Leftrightarrow 3a^3\cos^3x_t-3a\cos x_t+1=0 \tag{1}$

  Because  $\cos 3t=4\cos^3t-3\cos t$  we are looking to determine  $a$  such that  $\frac{3a^3}{4}=\frac{3a}{3}$  which leads to  $a=\frac{2}{\sqrt{3}}.$  Thus we have

$\frac{2}{\sqrt{3}}=\frac{3a^3\cos^3x_t-3a\cos x_t}{4\cos^3 x_t-3\cos x_t}\underset{(1)}{=}\frac{-1}{\cos 3t}$

From  $\cos 3t=-\frac{\sqrt{3}}{2}$  we obtain  $3t \in \left \{\pm \frac{5\pi}{6}+2k\pi\right \}\cap [0,\pi]$, so  $t\in \left \{\frac{5\pi}{18},\;\frac{7\pi}{18},\;\frac{17\pi}{18}\right\}$  hence the Answer.

$\blacksquare$

          REMARK CiP  Putting  $x=\frac{2}{\sqrt{3}}\cdot y$  we see that  $y$  are solutions of the equation

$3\cdot \frac{8}{3\sqrt{3}}y^3-3\cdot \frac{2}{\sqrt{3}}y+1=0\;\Leftrightarrow\;8y^3-6y+\sqrt{3}=0 \tag{2}$

that is  $y_{1-3}\in \left \{\cos\frac{5\pi}{18},\;\cos\frac{7\pi}{18},\;\cos\frac{17\pi}{18}=-\cos\frac{\pi}{18}\right \}.$  We therefore have, from Vieta's Formulas :

$\cos\frac{\pi}{18}-\cos\frac{5\pi}{18}-\cos\frac{7\pi}{18}=0$

$\cos \frac{\pi}{18}\cdot \cos \frac{5\pi}{18}-\cos \frac{5\pi}{18}\cdot \cos \frac {7\pi}{18}+\cos \frac{\pi}{18}\cdot \cos \frac{7\pi}{18}=-\frac{3}{4}$

$\cos \frac{\pi}{18}\cdot \cos \frac{5\pi}{18} \cdot \cos \frac{7\pi}{18}=\frac{\sqrt{3}}{8}$

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