A Problem from the KANGAROO Contest // Un problème du concours KANGOUROU

 KANGAROO is a widely known mathematics competition. Here we discuss a Percentage Problem, taken from Book, which you will soon be able to find scanned here.

                          <<At "Frank Einstein" College, the number of students decreased by

                                 10% in one year, but the percentage of girls increased from 50%

                                  to 55%. The number of girls...

        A. grew up with 0,5%     B. grew up with 1%     C.  remained the same

        D. decreased by  1%     E.  decreased by 0,5%  >>

(1991: Grades 6-7 Problem #29, page 9

 and Grade 8 Problem #22, page 11)

ANSWER CiP  :  D

A complete answer:

 the number of girls decreased by 1% and the number of boys decreased by 19%

                            Solution CiP 

                       Let's denote by  $b,\;f$  respectively the number of boys and girls at the beginning of the year and by $B,\;F$ their number at the end. From the text of the problem it follows that we have the equations:

$B+F=(1-10\text {%}) \cdot (b+f) \tag{1}$

$f=50\text{%} \cdot (b+f) \tag{2}$

$F=55\text{%} \cdot (B+F) \tag{3}$

From (2) we obtain  $f=b$  (which even grandma would have deduced). So from (1) we can write

$B+F=\frac{90}{100} \cdot 2f \tag{4}$

which substituted into (3) gives us

$F=\frac{55}{100} \cdot \frac{90}{100}\cdot 2f=\frac{99}{100} \cdot f=(1-1\text{%}) \cdot f$

So the number of girls decreased by 1%, hence the answer D.

     Further, from  $B+F\overset{(1)}{=}\frac{90}{100} \cdot 2f\;$, replacing  $f$  with  $b$  and  $F$  with  $\frac{99}{100} \cdot b$  we obtain  $B+\frac{99}{100}\cdot b=\frac{90}{100}\cdot b \Rightarrow $

$\Rightarrow B=\left (\frac{180}{100}-\frac{99}{100} \right )=\frac{81}{100}\cdot b=(1-19\text{%})\cdot b$

which completes the answer.

$\blacksquare$

         We are trying to formulate a more general problem :

    The total number of students decreased by  $t \text{%}$  and

 the percentage of girls increased from  $g\text{%}$  to  $h\text{%}$.      

                                      Determine the percentages by which the number of

                                   girls and boys changed.

  

ANSWER CiP (with the previous notations)

$F=\frac{(100-t)\cdot h}{g\cdot 100} \cdot f \tag{F}$

$B=\frac{(100-t)\cdot (100-h)}{(100-g) \cdot 100} \cdot b \tag {B}$

 

                         Solution CiP

          From the problem statement we write the equations :

$B+F=\left (1-\frac{t}{100} \right ) \cdot (b+f) \tag {i}$

$f=\frac{g}{100} \cdot (b+f) \tag{ii}$

$F=\frac{h}{100} \cdot (B+F) \tag{iii}$

Equations (ii) and (iii) lead to the relations :

$(100-g)\cdot f=g \cdot b \tag{b,f}$

$(100-h) \cdot F=h \cdot B \tag{B,F}$

     Substituting  $B\underset{(B,F)}{=}\frac{100-h}{h}\cdot F$  and  $b\underset{(b,f)}{=}\frac{100-g}{g}\cdot f$  in equation (i) we have

$\frac{100-h}{h}\cdot F +F=\frac{100-t}{100}\cdot \left ( \frac{100-g}{g}\cdot f+f\right )\;\Leftrightarrow$

$\Leftrightarrow\;\frac{100}{h}\cdot F=\frac{100-t}{100}\cdot \frac{100}{g} \cdot f\;\Rightarrow\; F=\frac{(100-t)\cdot h}{g\cdot 100} \cdot f$  that is, equation (F).

     Substituting  $F\underset{(B,F)}{=}\frac{h}{100-h}\cdot B$  and  $f\underset{(b,f)}{=}\frac{g}{100-g}\cdot b$  in equation (i) we have

$B+\frac{h}{100-h} \cdot B=\frac{100-t}{100}\cdot \left (b+\frac{g}{100-g}\cdot b\right )\;\Leftrightarrow$

$\Leftrightarrow\;\frac{100}{100-h}\cdot B=\frac{100-t}{100}\cdot \frac{100}{100-g} \cdot b\;\Rightarrow\; B=\frac{(100-t)\cdot (100-h)}{(100-g)\cdot 100} \cdot b$  that is, equation (B).

$\blacksquare \;\blacksquare$

      In the initial problem :  $t=10,\;g=50,\;h=55$  so

$F=\frac{90\cdot 55}{100\cdot 50}\cdot f=\frac{99}{100}\cdot f=(1-1\text{%})\cdot f$,

  $B=\frac{90\cdot 45}{100\cdot 50}\cdot b=\frac{81}{100}\cdot b=(1-19\text{%})\cdot b.$

Other EXAMPLES :

             - If the total number of students decreases by 12% and the number of girls increases from 44% to 50% then

$F=\frac{(100-12)\cdot 50}{44\cdot 100}\cdot f=\frac{88\cdot 50}{44\cdot 100}\cdot f=f$, 

 so the percentage of girls remains the same, and

$B=\frac{(100-12)\cdot (100-50)}{(100-44)\cdot 100}\cdot b=\frac{44}{56}\cdot b\approx 0,79\cdot b=(1-21\text{%})\cdot  b$

so the percentage of boys decreases by about 21%.

             - If the total number of students decrease by  8%  and the number of girls increases from  46% to 54%  then

$F=\frac{(100-8)\cdot 54}{46\cdot 100}\cdot f=\frac{92\cdot 54}{46\cdot 100}\cdot f=\frac{108}{100}\cdot f=(1+8\text{%})\cdot f$

so the percentage of girls increases by  8%.  ATTENTION !! : it is a coincidence that  54%-46%=8%. 

$B=\frac{(100-8)\cdot (100-54)}{(100-46)\cdot 100}\cdot b=\frac{92\cdot 46}{54 \cdot 100}\cdot b=\frac{529}{675}\cdot b\approx 0,78 \cdot b=(1-22\text{%})\cdot b$

so the percentage of boys decreases by about 22%.

             - If the total number of students decrease by  48% (a bit of a bizarre situation)  and the number of girls increases from  52% to 64%  then

$F=\frac{(100-48)\cdot 64}{52\cdot 100}\cdot f=\frac{52\cdot 64}{52\cdot 100}\cdot f=\frac{64}{100}\cdot f=(1-36\text{%})\cdot f$

so the percentage of girls decreases by  36%;

$B=\frac{(100-48)\cdot (100-64)}{(100-52)\cdot 100}\cdot b=\frac{52\cdot 36}{48\cdot 100}\cdot b=\frac{39}{100}\cdot b=(1-61\text{%})\cdot b$

so the percentage of boys decreases by  61%.

              Remarks CiP

            - If the total number of students increases instead of decreasing then we can juggle with negative percentage values, but the formulas remain the same (as in the second example, the number of girls)

            - We have seen in some examples that the answer obtained is only approximate. It would be interesting to study the relationships between the quantities  $t,\; g,\;h\;$ for which the answer is expressed as an integer percentage, i.e. 

$\frac{(100-t)\cdot h}{g}\;\;,\;\;\;\frac{(100-t)\cdot (100-h)}{100-g}$

are integers.

<end Rem>

A Paradox of Infinite Long Division // Un paradosso della divisione infinita

 Everything started from the aspects provided in the Post "An easy...".(I still haven't found a simple explanation.)

         I will extract from there the equation 

$26^3-24^3-12^3+1^3=(2^3+1)\cdot 225$

which expresses that  $2025=9\cdot 225$. I'm thinking of seeing what a writing by 225 looks like, which maybe I haven't taken into account yet. I force a division for this,  $26^3-24^3-12^3+1\;:\;(2^3+1)$ , imitating the long division of two polynomials.

\begin{array}{ccccccc}\;\;\;26^3&-24^3&\;&-12^3&+1&\vdash&2^3+1\\-26^3&\;&-13^3&&&&13^3-12^3-\left(\frac{13}{2}\right)^3+\left(\frac{13}{4}\right )^3-\left (\frac{13}{8}\right)^3+\left (\frac{13}{16}\right)^3-\left (\frac{13}{32}\right )^3+\cdots\\\hline \setminus&-24^3&-13^3&-12^3&+1\\\;&24^3&\;&12^3\\\hline \;&\setminus&-13^3&\setminus&+1\\\;&\;&13^3&+\left(\frac{13}{2}\right)^3\\\hline \;&\;&\setminus&\left(\frac{13}{2}\right)^3&\;&+1\\\;&\;&\;&-\left(\frac{13}{2}\right)^3&-\left(\frac{13}{4}\right)^3\\\hline \;&\;&\;&\setminus&-\left(\frac{13}{4}\right)^3&+1\\\;&\;&\;&\;&\left(\frac{13}{4}\right)^3+\left (\frac{13}{8}\right)^3\\\hline \;&\;&\;&\;&\left (\frac{13}{8}\right)^3&+1\\\;&\;&\;&\;&-\left(\frac{13}{8}\right)^3-\left(\frac{13}{16}\right)^3\\\hline \;&\;&\;&\;&-\left(\frac{13}{16}\right)^3&+1\\\;&\;&\;&\;&\left(\frac{13}{16}\right)^3+\left(\frac{13}{32}\right)^3\\\hline \;&\;&\;&\;&\left(\frac{13}{32}\right)^3&\color{Red}{+1}\\\:&\;&\;&\;&\dots&\dots\end{array}

The division can continue indefinitely. The quotient of this division is

$13^3-12^3-\left ( \frac{13}{2}\right )^3+\left ( \frac{13}{4}\right )^3-\left ( \frac{13}{8}\right )^3+\left (\frac{13}{16}\right )^3-\left ( \frac{13}{32}\right )^3+ \cdots=$

$=-12^3+13^3 \cdot \left [1-\left (\frac{1}{2}\right )^3+\left (\frac{1}{2}\right )^6-\left (\frac{1}{2} \right )^9+\left (\frac{1}{2} \right )^{12}-\left (\frac{1}{2} \right)^{15}+\cdots \right ]=$

$=13^3\cdot \frac{1}{1+\frac{1}{8}}-12^3=\frac{8}{9}\cdot 13^3-12^3=\frac{26^3}{9}-12^3.$

We would have expected the result to be 225, so 

$\color{Red}{\frac{26^3}{9}-12^3=225}$

But this is NOT true, and in this I considered, mistakenly, that it was a paradox. We must take into account that for each partial remainder of the division there remains a  $\color{Red}{+1}$. The correct result of the division is

$\color{Blue}{\frac{26^3}{9}-12^3+\frac{1}{9}}$

          I asked ChatGPT what other possibilities there were to write the number 225 as a sum of cubes and he gave me the following reasonable results, but with which I made no progress at all :

$6^3+2^3+1^3$

$7^3-4^3-3^3-3^3$

$7^3-6^3+5^3-3^3$

$7^3-5^3+2^3-1^3$

$8^3-7^3+4^3-2^3$

$9^3-8^3+2^3$

$10^3-7^3-6^3-6^3$

An easy problem for... Leonhard EULER // Ein einfaches Problem für... Leonhard Euler

 I've messed up my blogs (petrell-man and/or mathematyka2), and none of them know how to type in LaTeX. And then I made a third one matemattica0, where a shy LaTeX form goes. 

              EULER is famous for elucidating Fermat's number  $F_5=2^{2^5}+1=641 \times 6700417$. He is the one who saw that  $641=2^7\times 5+1=2^4+5^4$.

That's why I thought that only he could help me with the problem:

              "1. (page 51)  We will call a positive integer  $n\;cool$  if it is a perfect square and there exists positive integers  $a,\;b,\;c,\;d$  different from each other, with  $a>b\;and\;c>d$ ,  such that  $n=a^3-b^3+c^3-d^3$.  For instance,  225  is cool because  $225=15^2\;\;and\;\;225=7^3-5^3+2^3-1^3$.  Show that :

a)  2025 is cool ;

b)  there are infinitely many cool numbers.

(Relu CIUPEA)" 

ANSWER CiP

a) $2025=45^2=11^3-3^3+9^3-2^3=16^3-15^3+11^3-3^3$

b)  $225\cdot k^6=(15k^3)^2=(7k^2)^3-(5k^2)^3+(2k^2)^3-(k^2)^3,\;k\in\mathbb{N}^*$ 

Solution CiP

                   a) was solved by ChatGPT. I chose the simplest of several equalities.

                    b) We multiply the example provided in the statement by  $k^6$.

$\blacksquare$

                   An infinite number of remarks can be made, but I still haven't managed to find a simple way, accessible to a junior, to achieve the result.

            At first I thought it was easy, starting from

$2025=9\cdot 225=(2^3+1^3)(7^3-5^3+2^3-1^3)=14^3-10^3+7^3-5^3+4^3-1^3.$

But there is no further path to the result in sight.

             Let's write 225 differently then. We know  $225=1^3+2^3+3^3+4^3+5^3$, and $3^3+4^3+5^3=6^3$, so  $225=1^3+2^3+6^3$  but even that doesn't get us where we want to go.

             ChatGPT gave us a few more representations that didn't quite fit our requirements :

$2025=15^3-11^3-3^3+2^3$

$2025=26^3-24^3-12^3+1^3$

$2025=29^3-26^3-17^3+5^3$

$2025=34^3-31^3-20^3+8^3$

When I asked him how he calculated it, he told me he used a code and the PYTHON language.

<end Rem>

"THE TREEPENNY OPERA" : Identities with sums of three binomials // "DIE DREIGROSCHEN OPERA" : Identitäten mit Summen von drei Binomien

 Inspiration from the title of a Bertolt BRECHT play.

                  We propose to establish algebraic identities of the form :

$x^2+(x+a)^2+(x+b)^2=(x+m)^2+(x+n)^2+(x+p)^2 \tag{1}$

where  $a,\;b,\;m,\;n,\;p\;$ are real numbers.

My greatest achievement is the formula in

 

                    PROPOSITION 1.  Let  $a,\;b\;$ be given numbers. It holds 

                      the algebraic identity :

$x^2+(x+a)^2+x+b)^2=\left (x+\frac{2(a+b)}{3}\right )^2+\left (x+\frac{2b-a}{3}\right )^2+\left ( x+\frac{2a-b}{3}\right )^2 \tag{2}$

          The proof can be done by direct calculation.

$\square$

For example

$x^2+(x+1)^2+(x+5)^2=(x+4)^2+(x+3)^2+(x-1)^2$

Some choices may be unfortunate, e.g.

$x^2+(x+1)^2+(x+2)^2=(x+2)^2+(x+1)^2+x^2$

$x^2+(x+2)^2+(x+4)^2=(x+4)^2+(x+2)^2+x^2$

$x^2+(x+3)^2+(x+6)^2=(x+6)^2+(x+3)^2+x^2$

while others give us equal terms

$x^2+(x+0)^2+(x+3)^2=(x+2)^2+(x+2)^2+(x-1)^2$

$x^2+(x+3)^2+(x+0)^2=(x+2)^2+(x-1)^2+(x+2)^2$

More seriously we have

$x^2+(x-1)^2+(x+4)^2=(x+2)^2+(x+3)^2+(x-2)^2$

but putting  $x-1=y$  we have  $(y+1)^2+y^2+(y+5)^2=(y+3)^2+(y+4)^2+(y-1)^2$  which is precisely the example from the beginning.

Another                                                     $x^2+(x-2)^2+(x+5)^2=(x+2)^2+(x+4)^2+(x-3)^2$

in which if we put  $x-3=y$ , write it backwards, somehow ordering the terms

$y^2+(y+5)^2+(y+7)^2=(y+1)^2+(y+3)^2+(y+8)^2$;

if instead we put  $x-2=z$  we have, ordering now

$z^2+(z+2)^2+(z+7)^2=(z-1)^2+(z+4)^2+(z+6)^2$

and if  $z-1=t$  and write it backwards, we have

$t^2+(t+5)^2+(t+7)^2=(t+1)^2+(t+3)^2+(t+8)^2$

which is the same as the identity in  $"y"$.

Finally

$x^2+(x-1)^2+(x+7)^2=(x+4)^2+(x+5)^2+(x-3)^2$

Ordering we have

$(x-1)^2+x^2+(x+7)^2=(x-3)^2+(x+4)^2+(x+5)^2\;;$

and putting  $x-1=y$  we obtain

$y^2+(y+1)^2+(y+8)^2=(y-2)^2+(y+3)^2+(y+6)^2\;;$

putting  $x-3=z$  in the backwards we obtain

$z^2+(z+7)^2+(z+8)^2=(z+2)^2+(z+3)^2+(z+10)^2$

so we get a bunch of other identities.

          We can write it somehow "in integers" on the (2)

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

where  $a$  and  $c$  are arbitrary numbers.

          Now let's see how we got to (2).

          Identifying in (1) the coefficients of  $x^1$  and  $x^0$  we have the relations

$\begin{cases} m+n+p=a+b\\m^2+n^2+p^2=a^2+b^2\end{cases} \tag{4}$

Eliminating the unknown  $n$  we obtain

  $m^2+(a+b-m-p)^2+p^2=a^2+b^2\Leftrightarrow2m^2+(a+b)^2-2m(a+b)-2p(a+b)+2mp+2p^2=a^2+b^2$

which divided by 2 and ordered by  $p $ gives us

$p^2+p(m-a-b)+m^2-m(a+b)+ab=0 \tag{5}$

The equation (5) in the unknown  $p$  has the discriminant

 $\Delta_p=(m-a-b)^2-4[m^2-m(a+b)+ab]=(a-b)^2+2m(a+b)-3m^2$. 

Assuming that it is a perfect square

$\Delta_p=(a-b)^2+2m(a+b)-3m^3=t^2 \tag{6}$

then the roots of the equation (5) will be

$p_{1,2}=\frac{a+b-m\pm t}{2} \tag{7}$

Writing (6) as an equation in the unknown  $m$

$3m^2-2m(a+b)+t^2-(a-b)^2=0 \tag{8}$

we set the condition that its half-discriminant is a perfect square

$\Delta^{'}_m=(a+b)^2-3[t^2-(a-b)^2]=s^2$

so

$s^2+3t^2=(a+b)^2+3(a-b)^2 \tag{9}$

The values ​​of m are given by

$m_{1,2}=\frac{a+b\pm s}{2} \tag{10}$

          Choosing in (9)  $s=a+b,\;t=a-b$  we obtain

           - from (10) : $m_{1,2}=\frac{a+b \pm (a+b)}{2}\in \left \{0,\;\frac{2(a+b)}{3} \right \}$; we take the second value

           - from (7) : $p_{1,2}=\frac{a+b-\frac{2(a+b)}{3} \pm (a-b)}{2} \in \left \{\frac{2b-a}{3},\;\frac{2a-b}{3}\right \}$.

Then  $n=a+b-m-p$  take the respectively values  $\frac{2a-b}{3},\;\frac{-a+2b}{3}$.

     If  $m=\frac{2(a+b)}{3},\;p_1=\frac{2b-a}{3},\;n=\frac{2a-b}{3}$  we obtain (2). And the values  $p_2=\frac{2a-b}{3},\;n_2=\frac{-a+2b}{3}$ give us the same formula (2) with  $a$  and  $b$  swapped.

          If we notice that in (9)  the right-hand side is  $4a^2-4ab+4b^2=(2a-b)^2+3b^2$  then we can still make the choice $s=2a-b,\;t=b$  but by the end we get the same formula (2) again. More than that, since we already know a solution for the quadratic equation (9) in the unknowns  $s\; and\; t$, we can find a general solution to it depending on two parameters  $\lambda,\;\mu$:

$s=\frac{(3\mu^2-\lambda^2)(a+b)-6\lambda \mu (a-b)}{\lambda^2+3\mu^2},\;\;\;t=\frac{(\lambda^2-3\mu^2)(a-b)-2\lambda \mu (a+b)}{\lambda^2+3\mu^2}$

We are not going in this direction anymore, but what has been said so far has shown that the following Proposition takes place:

                    PROPOSITION 2.  Let  $a,\;b,\;c\;$ be given numbers. We can

                          determine, in terms of  $a,\; b,\; c$ , two numbers $?(a,b,c)$,

                          $\;??(a,b,c)$  such that

$x^2+(x+a)^2+(x+b)^2=(x+c)^2+(x+?)^2+(x+??)^2 \tag{11}$

         Indeed, from (4) we obtain the conditions

$n+p=a+b-c,\;\;n^2+p^2=a^2+b^2-c^2$

which can express  $n$  and  $p$  in terms of  $a,\;b,\;c$ :

$n,p=\frac{a+b-c \pm \sqrt{a^2+b^2-3c^2-2ab+2bc+2ac}}{2}$

Conditions in which these expressions are rational, i.e. without radicals, we leave them to the study of others who are more skilled.

              Examples :        $x^2+(x+1)^2+(x+2)^2=(x+3)^2+(x-\imath \sqrt{2})^2+(x+\imath \sqrt{2})^2$

$x^2+(x+4)^2+(x+5)^2=(x+6)^2+(x+1)^2+(x+2)^2$

$x^2+(x+7)^2+(x+8)^2=(x+9)^2+(x+3-\sqrt{7})^2+(x+3+\sqrt{7})^2$

In a Triangle, a Median is equal to an Altitude : $m_B=h_A$ // In einem Dreieck hat die Mediane die gleiche Länge wie die Höhe

I didn't discover that an altitude and a median in a triangle can be equal.

However, if  $h_A$  is equal to  $m_A$  it also appears in the Problem below. Until then, let's digress a bit.

          Let's take their formulas. Altitude :  $h_A=\frac{2}{a}\cdot \sqrt{s(s-a)(s-b)(s-c)}$

where  $s=\frac{a+b+c}{2}$ (the notations are like here).

  Median :  $m_B=\frac{1}{2}\sqrt{2a^2+2c^2-b^2}$.

We prefer to write  $h_A=\frac{2}{a}\cdot \sqrt{\frac{a+b+c}{2}\cdot \frac{-a+b+c}{2}\cdot \frac{a-b+c}{2}\cdot \frac{a+b-c}{2}}$  or developed

$h_A=\frac{1}{2a}\cdot \sqrt{2a^2b^2+2b^2c^2+2c^2a^2-a^4-b^4-c^4} \tag{1}$

          The equality  $h_A^2=m_A^2$  is written : 

$\frac{1}{4a^2}(2a^2b^2+2b^2c^2+2c^2a^2-a^4-b^4-c^4)=\frac{1}{4}(2a^2+2c^2-b^2)\Leftrightarrow$

$\Leftrightarrow \frac{b^2}{2}+\frac{b^2c^2}{2a^2}+\not{\frac{c^2}{2}}-\frac{a^4}{4}-\frac{b^4}{4a^2}-\frac{c^4}{4a^2}=\frac{a^2}{2}+\not{\frac{c^2}{2}}-\frac{b^2}{4} \Leftrightarrow$

$\Leftrightarrow \frac{3b^2}{4}-\frac{b^4+c^4-2b^2c^2}{4a^2}=\frac{3a^2}{4}$  so in the end we get

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

I mention that we will not use this relationship today.

          I recently saw the Problem  E:17209 in a shop window :

In translation :

               "E:17209.  In triangle ABC, median  $BM$  and altitude  $AD$  are

             concurrent at point  $P$   and have the same length. Show that  $BP = 2PD$.

 Dragoș Ionuț MATEI, Constanța"

I am attaching a figure about this problem.

There are several cases of the figure, depending on the position of point D on line BC. In any of the cases we will prove, which immediately implies the conclusion of the problem :

ANSWER CiP

$\widehat{MBC}=30^{\circ}$

                    Solution CiP (Attention, the problem is proposed for Grade 6)

              The position of point D on line BC allows us to distinguish the cases :

$(i)\;\;D=B\;;\;\;(ii)\;\;D=C\;;\;\;(iii)\;\;B-D-C\;;\;\;(iv)\;\;B-C-D\;;\;\;(v)\;\;D-B-C$

          (i)  In this case triangle  $ABC$ has a right angle at vertex  $B$.

The hypothesis  $BM=AD$  shows that  $ABM$  is an equilateral triangle, so  $CAB$ is a 30 60 90  triangle. Hence  $\widehat{MBC}=30^{\circ}$. In the equation  $BP=2PD$  all segments are zero.

          (ii)  In this case  $ABC$  has a rightangle at vertex  $C$.

From  $AD=BM=2b$(say) it follow  $MC=BM/2$  so  $BMC$ is now a 30-60-90 triangle. Because $D=C,\;P=M$  then  $BM=2MC\Rightarrow BP=2PD$.

          (iii)  The figure looks like this

We will proof this case, but we will see that the proof also applies to the others.

          We draw, both in size and orientation  $DN\overset{=}{\parallel}MC$. Quadrilateral  $CMND$  is a parallelogram.

So is  $ADMN$; in this we have

$MN=AD=BM \tag{1}$

and since  $AD\perp BC$, we also have  $MN\perp CD$. But then  $CMDN$  is a rhombus, so   $CN=ND=CM$. Here too, the line  $DC$ is the perpendicular bisector of the segment  $[MN]$, and then

$BM=BN \tag{2}$

From (1) and (2) it follows that  $BM=MN=BN$, therefore the triangle  $BMN$  is equilateral. In it, $BC$  is the bisector of the  $\measuredangle{MBN}$, therefore  $\widehat{MBC}=30^{\circ}$.

     Triangle  $BPD$  is right-angled with  $\widehat{PBD}=30^{\circ}$, so  $BP=2PD$.

          (iv)  The figure looks like this

Same proof

          (iv)  The figure looks like this

Same proof

$\blacksquare\;\blacksquare$

A Curious Inequality with... Radicals // ... रेडिकलहरूसँगको एउटा जिज्ञासु असमानता

 The page presents the problems from the fifth onwards proposed for the 7th grade (the first four are on the previous page)

As it says on the second cover: "In grades VII-XII, the first 8 problems are intended for deepening the subject and preparing for national exams, and the last 4 are addressed to those who wish to additionally prepare for competitions". 

So Problem S:E25.203 is considered easy. But still... ??

         In translation : 

    S:E25.203.  The positive real numbers  $a$  and  $b$  verify the relation  $a+b+ab=3.$

                a) Verify that, if  $a=2$  then  $a+b>2>\sqrt{a}+\sqrt{b}.$

                b)  Prove that  $a+b\geqslant \sqrt{a}+\sqrt{b}$,  for any  $a$  and  $b.$

ANSWER CiP

$\textbf{b)}\;\;\;  a+b\geqslant 2\geqslant \sqrt{a}+\sqrt{b} \tag{A}$.

 The  $"="$ sign occurs if and only if  $a=b=1$

                         Solution CiP

               a) From the given relation we obtain

$b=\frac{3-a}{1+a} \tag{1}$

and it is observed that the given numbers are within the range of values  $0<a,\;b<3$.

          $a=2\underset{(1)}{\Rightarrow}b=\frac{1}{3}$  and obviously  $a+b>2$.

     After that, we have  $\sqrt{a}+\sqrt{b}=\sqrt{2}+\sqrt{\frac{1}{3}}=\frac{3\sqrt{2}+\sqrt{3}}{3}.$  But

$(\sqrt{a}+\sqrt{b})^2=\frac{(3\sqrt{2}+\sqrt{3})^2}{9}=\frac{18+3+6\sqrt{6}}{9}=\frac{7+2\sqrt{6}}{3}<4$, because of  

$\sqrt{24}<\sqrt{25}\Rightarrow 2\sqrt{6}<5\Rightarrow 7+2\sqrt{6}<7+5 \Rightarrow \frac{7+2\sqrt{6}}{3}<4$.

Interestingly, we don't get a more precise result if we start with the "stronger" inequality  $\sqrt{288}<\sqrt{289}$. So

$2<\frac{289}{144}\Rightarrow \sqrt{2}<\sqrt{\frac{289}{144}}=\frac{17}{12}\Rightarrow 4\sqrt{2}<\frac{17}{3}\Rightarrow 6-4\sqrt{2}>6-\frac{17}{3}=\frac{1}{3}\Rightarrow \sqrt{4-4\sqrt{2}+(\sqrt{2})^2}>\sqrt{\frac{1}{3}}\Rightarrow$

$\Rightarrow \sqrt{(2-\sqrt{2})^2}>\sqrt{\frac{1}{3}}\Rightarrow 2-\sqrt{2}>\sqrt{\frac{1}{3}} \Rightarrow 2>\sqrt{2}+\sqrt{\frac{1}{3}}$.

               b)  Only after numerous numerical simulations was I led to the solution of this case, by introducing the intermediate term  $2$  between the two inequalities.

        In the first part we use

  $x+\frac{4}{x}\geqslant 4,\;for\;x>0 \tag{2}$

 ($\Leftrightarrow (\sqrt{x})^2-2\cdot \sqrt{x} \cdot \frac{2}{\sqrt{x}}+\left (\frac{2}{\sqrt{x}}\right )^2\geqslant 0\Leftrightarrow \left (\sqrt{x}-\frac{2}{\sqrt{x}}\right )^2\geqslant 0;\;"="\;iff\;\;x=2$)

We have  

$a+b\underset{(1)}{=}a+\frac{3-a}{1+a}=(a-1)+\left ( \frac{3-a}{1+a}+1 \right )=(a-1)+\frac{4}{1+a}=$

$=\left ((1+a)+\frac{4}{1+a}\right )-2\;\overset{(2)}{\underset{with\;x=1+a}{\geqslant}}\;4-2=2$

thus left side of (A). Equal sign occurs here iff  $1+a=2$  so  $a=1$  and hence  $b\overset{(1)}{=}\frac{3-1}{1+1}=1$.

For right side of (A),  $(\sqrt{ab}-1)^2\geqslant 0\Rightarrow ab-2\sqrt{ab}+1\geqslant 0 \;\overset{ab=3-a-b}{\Rightarrow}$

$\Rightarrow 3-a-b-2\sqrt{ab}+1\geqslant 0\Rightarrow a+b+2\sqrt{ab}\leqslant 4\Rightarrow (\sqrt{a}+\sqrt{b})^2\leqslant 4$

so  $\sqrt{a}+\sqrt{b}\leqslant 2$. The  $"="$  sign occurs if and only if  $a\cdot b=1$  which together with  $a+b=3-ab=2$  give  $a=b=1$.

$\blacksquare$

Neculai STANCIU (Buzău) in the MATHEMATICAL JOURNAL // Neculai STANCIU (Buzău) în GAZETA MATEMATICĂ

 In the recent issue of the Exercise Supplement (aka SGM) he proposed the Problem S:E25.199. The problem statement is :

                    "Consider  $\Delta ABC$  a triangle such that  $\measuredangle A=2\measuredangle C$.

                      Prove that  $\frac{AB}{BC}=\frac{1}{2\cos C}.$

Neculai STANCIU, Buzău"

Solution CiP

We construct  $AD$ - the bisector of angle  $\widehat{BAC}$. We have 

$$\measuredangle CAD=\measuredangle BAD=\frac{\measuredangle BAC}{2}=\measuredangle C$$

          We now construct  $BE\parallel AD,\;E\in AC$. We have for the angles formed with the secant $AB$  that

$\measuredangle BAD=\measuredangle ABE=\measuredangle C \tag{1}$

and the exterior angle  $\widehat{BAC}$ of triangle  $\Delta ABE$ shows us that

$\measuredangle BAC=\measuredangle ABE+\measuredangle AEB\Leftrightarrow 2\measuredangle C\overset{(1)}{=}\measuredangle C+\measuredangle AEB\Rightarrow \measuredangle AEB=\measuredangle C$

From all the angles equal to  $C$  in the figure, we see that we have the isosceles triangles  $BAE$  and  $CBE$, so

$ AB=AE\;,\;BC=BE\tag{2}$

Constructing the height  $AF$  in the isosceles triangle  $ABE,\;AF\perp BE$, it will also be the median, so  $BF=EF=BE/2$. Then, in triangle  $ABF$

$\cos C\underset{(1)}{=}\cos \widehat{ABF}=\frac{BF}{AB}=\frac{BE/2}{AB}\underset{(2)}{=}\frac{BC/2}{AB}=\frac{BC}{2\cdot AB}$,  which is equivalent to the statement  $\frac{AB}{BC}=\frac{1}{2\cos C}.$

$\blacksquare$

          Remark CiP  The solution does not use any advanced trigonometry formulas, only definitions, otherwise from the Law of Sines in triangle  $ABC$  we could easily obtain

$\frac{AB}{\sin C}=\frac{BC}{\sin A}\Rightarrow \frac{AB}{BC}=\frac{\sin C}{\sin A}=\frac{\sin C}{\sin 2C}=\frac{\sin C}{2\sin C \cdot \cos C}=\frac{1}{2\cos C}$.

Starting from a Trigonometric Identity without variables // Ξεκινώντας από μια Τριγωνομετρική Ταυτότητα χωρίς μεταβλητές

           We have on my Page, at No. 6, the Identity

$\frac{\tan10^{\circ}}{\tan20^{\circ}\cdot \tan40^{\circ}}=\frac{1}{\sqrt{3}}$

It is equivalent to  $\frac{\tan10^{\circ}}{\tan20^{\circ}\cdot \tan40^{\circ}}=\tan30^{\circ}$  or else

$$\tan20^{\circ}\cdot \tan30^{\circ}\cdot \tan40^{\circ}=\tan10^{\circ} \tag{1}$$

               Let's consider the equation

$$ \tan x\cdot \tan\frac{3x}{2}\cdot \tan 2x=\tan \frac{x}{2}\tag{2}$$

(1) shows that  $x=20^{\circ}$  is a solution for (2).

The problem of completely solving this equation remains open for now.

A forgotten trigonometric equation // Egy elfeledett trigonometrikus egyenlet

 We will solve the following trigonometric equation here :

$$\tan(30^{\circ}-x)\cdot \tan(30^{\circ}+x)=\frac{\tan x}{\sqrt{3}} \tag{1}$$

I first mentioned it in the Post here. There we also showed that the equation  (1)  is equivalent to the equation

$$t^3-3\sqrt{3}\cdot t^2-3\cdot t+\sqrt{3}=0,\;\;t=\tan x \tag{2}$$

ANSWER CiP

$$x_k^{\circ}=20^{\circ}+k\cdot 60^{\circ},\;\;k\in\mathbb{Z} \tag{3}$$

                    Solution CiP

              The calculation has already been done, but we are presenting it for convenience.

$(1)\overset{t=\tan x}{\Leftrightarrow} t^3-3\sqrt{3}\cdot t^2-3\cdot t+\sqrt{3}=0\Leftrightarrow \sqrt{3}=\frac{3t-t^3}{1-3t^2}\Leftrightarrow$

$\Leftrightarrow \sqrt{3}=\frac{3\cdot \tan x-\tan^3x}{1-3\cdot \tan^2x}\Leftrightarrow \tan 60^{\circ}=\tan 3x\Leftrightarrow 3x=60^{\circ}+k\cdot 180^{\circ},\;\;k\in \mathbb{Z}$

From here we get the answer  (3)

$\blacksquare$

          Remark CiP   It all started with solving the first equation from No. 6 on Page here. Now, after we have solved the equation  (1) (see also the Post from September 3, 2025), we can write the identities :

$$\frac{\tan10^{\circ}\cdot \tan50^{\circ}}{\tan20^{\circ}}=\frac{1}{\sqrt{3}}\tag{4}$$

$$\frac{\tan10^{\circ}\cdot \tan70^{\circ}}{\tan40^{\circ}}=\frac{1}{\sqrt{3}}\tag{5}$$

$$\frac{\tan50^{\circ}\cdot \tan70^{\circ}}{\tan80^{\circ}}=\frac{1}{\sqrt{3}}\tag{6}$$

          But  (4), considering that  $\frac{1}{\tan20^{\circ}}=\cot20^{\circ}$, is exactly the first identity from No. 6. Seeking to express all quantities as tangents of angles not exceeding  $45^{\circ}$, we further have  $\frac{\tan10^{\circ}\cdot \tan50^{\circ}}{\tan20^{\circ}}=\frac{\tan10^{\circ} \cdot \color{Red}{\cot40^{\circ}}}{\tan20^{\circ}}=\frac{\tan10^{\circ}\cdot \color{Red}{\frac{1}{\tan40^{\circ}}}}{\tan20^{\circ}}=\frac{\tan10^{\circ}}{\tan20^{\circ}\cdot \tan40^{\circ}}=\frac{1}{\sqrt{3}}$

which is the second entity from the aforementioned No. 6.

     Doing the same thing with  (5) : 

$\frac{1}{\sqrt{3}}=\frac{\tan10^{\circ}\cdot \color{Red}{\cot20^{\circ}}}{\tan40^{\circ}}=\frac{\tan10^{\circ}\cdot \color{Red}{\frac{1}{\tan20^{\circ}}}}{\tan40^{\circ}}=\frac{\tan10^{\circ}}{\tan20^{\circ}\cdot \tan{40^{\circ}}}$

and with (6) :

$\frac{1}{\sqrt{3}}=\frac{\color{Red}{\cot40^{\circ}\cdot \cot20^{\circ}}}{\color{Red}{\cot10^{\circ}}}=\frac{\frac{1}{\tan40^{\circ}}\cdot \frac{1}{\tan20^{\circ}}}{\frac{1}{\tan10^{\circ}}}=\frac{\tan10^{\circ}}{\tan20^{\circ}\cdot \tan40^{\circ}}$

So the second of the identities in No. 6 is the most important.

              Let's demonstrate it directly then. We will see that the pattern is the same.

$\frac{\tan10^{\circ}}{\tan20^{\circ}\cdot \tan40^{\circ}}=\frac{\frac{\sin10^{\circ}}{\cos10^{\circ}}}{\frac{\sin20^{\circ}}{\cos20^{\circ}}\cdot \frac{\sin40^{\circ}}{\cos40^{\circ}}}=\frac{8\cdot \sin10^{\circ}\cdot \cos20^{\circ}\cdot \cos40^{\circ}}{8\cdot \cos10^{\circ}\cdot \sin20^{\circ}\cdot \sin40^{\circ}}\tag{7}$

The numerator in (7) is  $4\cdot \frac{\sin20^{\circ}}{\cos10^{\circ}}\cdot \cos20^{\circ}\cdot \cos40^{\circ}=\frac{1}{\cos10^{\circ}}\cdot 2\cdot \sin40^{\circ}\cdot \cos40^{\circ}=\frac{1}{\cos10^{\circ}}\cdot \sin80^{\circ}=1$

The denominator in (7) is  $4\cdot \cos10^{\circ}\cdot(\cos20^{\circ}-\cos60^{\circ})=4\cos10^{\circ}\cdot \cos20^{\circ}-4\cos10^{\circ}\cdot \frac{1}{2}=$

$=2(\cos10^{\circ}+\cos30^{\circ})-2\cos10^{\circ}=2\cos10^{\circ}+2\cdot \frac{\sqrt{3}}{2}-2\cos10^{\circ}=\sqrt{3}$

So the final value in (7) is  $\frac{1}{\sqrt{3}}$.

<end Rem>

The Expressions for Roots of Cubic Polynomial // تعبيرات جذور كثيرة الحدود التكعيبية، مرة أخرى بمساعدة اللاهوت

                We have discussed this issue before. We will discuss the similarities and differences with the current situation at the end.

          Here we consider the problem : 

                   Let  $\tau$  be one of the roots of the equation

$t^3-3\sqrt{3}\cdot t^2-3\cdot t+\sqrt{3}=0 \tag{1}$

                   The other two roots are

$\tau_2=\frac{\sqrt{3}}{2}\cdot \tau^2-5\cdot \tau+\frac{\sqrt{3}}{2}\;,\;\;\tau_3=-\frac{\sqrt{3}}{2}\cdot \tau^2+4\cdot \tau+\frac{5\sqrt{3}}{2} \tag{2}$ 

SOLUTION  CiP

               From the fact that  $\tau$  checks  (1)  we obtain some useful formulas in future calculations

$\tau^3=3\sqrt{3}\cdot \tau^2+3\cdot \tau -\sqrt{3} \tag{3}$

$\tau^4=30\cdot \tau^2+8\sqrt{3} \cdot \tau-9 \tag{4}$

$\frac{1}{\tau}=-\frac{\sqrt{3}}{3}\cdot \tau^2+3\cdot \tau+\sqrt{3} \tag{5}$

          Indeed, $\tau^3-3\sqrt{3}\cdot \tau^2-3\cdot \tau +\sqrt{3}=0 \Rightarrow \tau^3=3\sqrt{3} \cdot \tau^2+3\cdot \tau-\sqrt{3}$,  i.e. (3). Then  $\tau^4=\tau \cdot \tau^3\overset{(3)}{=}$

$=\tau (3\sqrt{3}\cdot \tau^2+3\cdot \tau-\sqrt{3})=3\sqrt{3}\cdot \tau^3+3\cdot \tau^2-\sqrt{3}\cdot \tau\overset{(3)}{=}3\sqrt{3}(3\sqrt{3}\cdot \tau^2+3\cdot \tau-\sqrt{3})+3\cdot \tau^2-\sqrt{3}\cdot \tau=30\cdot \tau^2+8\sqrt{3}\cdot \tau-9$ i.e. (4). Finally, $\sqrt{3}=-\tau^3+3\sqrt{3}\cdot \tau^2+3\cdot \tau=\tau \cdot (-\tau^2+3\sqrt{3}\cdot \tau+3)\Rightarrow\frac{1}{\tau}=\frac{-\tau^2+3\sqrt{3}\cdot \tau+3}{\sqrt{3}}=-\frac{1}{\sqrt{3}}\cdot \tau^2+3\cdot \tau+\sqrt{3}$, i.e. (5).

          From the first formula of Vieta  $\tau_1+\tau_2+\tau_3=3\sqrt{3}$  we get

$$\tau_2+\tau_3=3\sqrt{3}-\tau \tag{6}$$

and from the third  $\tau_1\cdot \tau_2\cdot \tau_3=-\sqrt{3}$  we deduce

$$\tau_2\cdot \tau_3=-\frac{\sqrt{3}}{\tau} \tag{7}$$

(6) and (7) show that  $\tau_{2,3}$  are the roots of the quadratic equation

$t^2-(3\sqrt{3}-\tau) \cdot t-\frac{\sqrt{3}}{\tau}=0 \tag{8}$

          Let's calculate its discriminant:  $\Delta_2=(3\sqrt{3}-\tau)^2+\frac{4\sqrt{3}}{\tau}=$

$\overset{\color{Red}{!!!}}{=}27-6\sqrt{3}\cdot \tau+\tau^2+\frac{4\sqrt{3}\cdot \tau}{\color{Red}{\tau^2}}=\frac{\tau^4-6\sqrt{3}\cdot \tau^3+27\cdot \tau^2+4\sqrt{3}\cdot \tau}{\color{Red}{\tau^2}}=$

$\overset{(4)}{\underset{(3)}{=}}\frac{(30\cdot \tau^2+8\sqrt{3}\cdot \tau-9)-6\sqrt{3}(3\sqrt{3}\cdot \tau^2+3\cdot \tau-\sqrt{3})+27 \tau^2+4\sqrt{3}\cdot \tau}{\color{Red}{\tau^2}}=\frac{3\cdot \tau^2-6\sqrt{3}\cdot \tau+9}{\color{Red}{\tau^2}}=\frac{(\sqrt{3}\cdot \tau-3)^{\color{Red}2}}{\color{Red}{\tau^2}}$

<  Hallelujah, I got a perfect square  !!!  > 

Let's do a little more calculation to make writing easier :  $\pm\sqrt{\Delta_2}=\frac{\sqrt{3}\cdot \tau-3}{\tau}=$

$=(\sqrt{3}\cdot \tau-3)\cdot \frac{1}{\tau}\overset{(5)}{=}(\sqrt{3}\cdot \tau-3)\left (-\frac{\sqrt{3}}{3}\cdot \tau^2+3\cdot \tau+\sqrt{3}\right )=$

$=-\tau^3+\sqrt{3}\cdot \tau^2+3\sqrt{3}\cdot \tau^2-9\cdot \tau+3\cdot \tau-3\sqrt{3}=$

$\overset{(3)}{=}-(3\sqrt{3}\cdot \tau^2+3\cdot \tau-\sqrt{3})+4\sqrt{3}\cdot \tau^2-6\cdot \tau-3\sqrt{3}=\color{Blue}{\sqrt{3}\cdot \tau^2-9\cdot \tau-2\sqrt{3}}$

 !!! Remember this expression !!!

With these, the equation  (8)  has the roots

$\tau_{2,3}=\frac{3\sqrt{3}-\tau \pm \sqrt{\Delta_2}}{2}=\frac{3\sqrt{3}-\tau \pm (\sqrt{3}\cdot \tau^2-9\cdot \tau-2\sqrt{3})}{2} \tag{9}$

from where we obtain the expressions  (2).

$\blacksquare$

               REMARKS CiP

                    1.  In the calculation of  $\Delta_2$  that follows immediately after the equation  (8), I made a trick so that at least the denominator would be a perfect square. Then I was lucky enough !!  Compare with the Post  << Not only God does not help, but also Allah does not "put you in trouble">>.

If in (7) we had used (5), obtaining  $\tau_2\cdot \tau_3=\tau^2-3\sqrt{3}\cdot \tau-3$, then for the equation (8) we would have had  $\Delta_2=(3\sqrt{3}-\tau)^2-4(\tau^2-3\sqrt{3}\cdot \tau-3)=39+6\sqrt{3}\cdot \tau -3\cdot \tau^2.$  Seeing this expression of  $\Delta_2$  we wondered, as we did in the Post How? can we recognize a perfect square??, "how the hell" can we notice - compare with (9) - that

$$\color{Blue}{39+6\sqrt{3}\cdot \tau-3\cdot \tau^2=(\sqrt{3}\cdot \tau^2-9\cdot \tau-2\sqrt{3})^2}$$

Problem NOT solved yet.

                    2.  The theoretical part about when and how such a statement is possible was exposed in  the  Post "The Rational Expressions for Roots of Cubic Polynomial". There are some differences, but only apparent.

          First of all, that the polynomial  $f=X^3-3\sqrt{3}\cdot X^2-3\cdot X+\sqrt{3}$  is not in  $\mathbb{Q}[X]$  but in  $\mathbb{Q}(\sqrt{3})[X].$

          Then we need to make sure that the polynomial $f$  is irreducible. Maybe this is not that important, and I don't have an immediate justification at hand. 

          Third, its discriminant must be a perfect square (see Proposition in the Post). From  $f$  we obtain the depressed cubic by substitution  $t=u+\sqrt{3}$, obtainig the polynomial  $u^3-12\cdot u-8\sqrt{3}$. Its discriminant, the same as for  $f$, is  $-4p^3-27q^2=4\cdot12^3-27\cdot(64\cdot3)=1\;728$  so equal to $3\cdot 576=(24\sqrt{3})^2$, hence perfect square in  $\mathbb{Q}(\sqrt{3})$.

<end REM's>

Phương trình đại số có nghiệm lượng giác //Algebraic Equation with Trigonometric Solution

 We are showing here what we have left from a post from yesterday. I was helped for this in the AOPS Forum.

               The roots of the equation

$$t^3-3\sqrt{3}\cdot t^2-3\cdot t+\sqrt{3}=0 \tag{E}  $$

             are  $\tan20^{\circ},\;\;\tan80^{\circ},\;\;\tan140^{\circ}$

Adapted Solution by CiP

$(E)\Leftrightarrow \sqrt{3}\cdot (1-3t^2)=3t-t^3 \Leftrightarrow \sqrt{3}=\frac{3t-t^3}{1-3t^2}\;\underset{t=\tan x}{=}\frac{3\cdot \tan x-\tan^3x}{1-3\cdot \tan^2x}=\tan (3x)$

so, taking the equality of the extreme terms,  $3\cdot x=60^{\circ}+k\cdot 180^{\circ},\;\;k\in \mathbb{Z}$, hence  $x_k=20^{\circ}+k\cdot 60^{\circ}$. Returning to the variable  $t$,  we have the roots

$t_1=\tan( 20^{\circ}+3k\cdot 60^{\circ})=\tan (20^{\circ}+k\cdot 180^{\circ})=\tan20^{\circ},$

$t_2=\tan(20^{\circ}+(3k+1)\cdot 60^{\circ})=\tan(80^{\circ}+k\cdot 180^{\circ})=\tan80^{\circ}$

$t_3=\tan(20^{\circ}+(3k+2)\cdot 60^{\circ})=\tan(140^{\circ}+k\cdot 180^{\circ})=\tan140^{\circ}$.

$\blacksquare$

          Remark Cip  Vieta's formulas for the roots of the equation (E) lead to the equalities

$\tan20^{\circ}-\tan40^{\circ}+\tan80^{\circ}=3\sqrt{3}$

$\tan20^{\circ}\cdot \tan 40^{\circ}+\tan 40^{\circ}\cdot \tan 80^{\circ}-\tan 20^{\circ}\cdot \tan 80^{\circ}=3$

$\tan20^{\circ}\cdot \tan 40^{\circ}\cdot \tan80^{\circ}=\sqrt{3}$

<end Rem>

A trigonometric identity without variables

 We will prove the trigonometric identity

$$\tan10^{\circ}\cdot \tan50^{\circ}\cdot \cot20^{\circ}=\frac{1}{\sqrt{3}} \tag{1}$$

The problem appears at Solutions in GMB 3/2005, number O:1063, page 132. You can find more about this magazine here and here.

Official Solution

      Using that we have

 $\sin50^{\circ}=\cos(90^{\circ}-50^{\circ})=\cos40^{\circ},\;\;\cos10^{\circ}=\sin(90^{\circ}-10^{\circ})=\sin80^{\circ}$

and  $\cos50^{\circ}=\sin(90^{\circ}-50^{\circ})=\sin40^{\circ}$  we obtain

$\tan10^{\circ}\cdot \tan50^{\circ}\cdot \cot20^{\circ}=\frac{\sin10^{\circ}}{\cos10^{\circ}} \cdot \frac{\sin50^{\circ}}{\cos50^{\circ}} \cdot \frac{\cos 20^{\circ}}{\sin20^{\circ}}=\frac{\sin10^{\circ}\cdot \cos40^{\circ}\cdot \cos20^{\circ}}{\sin80^{\circ}\cdot \sin40^{\circ}\cdot \sin20^{\circ} } \tag{2}$

          Up in (2) we have, from  $\sin20^{\circ}=2\cdot \sin10^{\circ}\cdot \cos10^{\circ}$  if we replace  $2\cdot \sin10^{\circ}$ :

$8\cdot \sin10^{\circ}\cdot \cos20^{\circ}\cdot \cos40^{\circ}=\frac{4\cdot \sin20^{\circ}\cdot \cos20^{\circ}\cos40^{\circ}}{ \cos10^{\circ}}=\frac{2\cdot \sin40^{\circ}\cdot \cos40^{\circ}}{\cos10^{\circ}}=\frac{\sin80^{\circ}}{\cos10^{\circ}}=1.$

          Down in (2) we have, using  $2\sin u\sin v=\cos(v-u)-\cos(v+u)$, replacing  $\sin80^{\circ}=\cos10^{\circ}$, and using  $2\cos u \cos v=\cos(v-u)+\cos(v+u)$

$8\cdot \sin20^{\circ}\cdot \sin40^{\circ}\cdot \sin80^{\circ}=4(\cos20^{\circ}-\cos60^{\circ})\cdot \cos10^{\circ}=$

$=4\cdot \cos20^{\circ}\cdot \cos10^{\circ}-4\cdot \frac{1}{2}\cdot \cos10^{\circ}=2(\cos10^{\circ}+\cos30^{\circ})-2\cdot \cos10^{\circ}=$

$=2\cdot \cos10^{\circ}+2\cdot \frac{\sqrt{3}}{2}-2\cdot \cos10^{\circ}=\sqrt{3}$

These values ​​entered in  (2) give the answer.

$\blacksquare$

               Just as "After the battle, everyone is a general", we will also provide a solution.

                    Solution CiP

                Lemma CiP      $t_1=\tan20^{\circ}$  is  one of the roots of the equation

$$t^3-3\sqrt{3}\cdot t^2-3\cdot t+\sqrt{3}=0 \tag{3}$$

               Proof of Lemma

          In the triple angle formula  $\tan 3\theta=\frac{3\tan \theta-\tan^3 \theta}{1-3\tan^2 \theta}$  we substitute

$\theta=20^{\circ},\;\;t=\tan \theta$  and because  $\tan3\theta=\tan 60^{\circ}=\sqrt{3}$, we have

$\frac{3t-t^3}{1-3t^2}=\sqrt{3}\Leftrightarrow\;3t-t^3=\sqrt{3}-3\sqrt{3}\cdot t^2\Leftrightarrow\;t^3-3\sqrt{3}\cdot t^2-3\cdot t+\sqrt{3}=0$

that is (3).

qed Lemma$\square$

          To prove (1), we successively transcribe it (1)$\;\Leftrightarrow$

$\Leftrightarrow\;\frac{\tan(30^{\circ}-20^{\circ})\cdot \tan(30^{\circ}+20^{\circ})}{\tan20^{\circ}}=\frac{1}{\sqrt{3}}\Leftrightarrow\;\tan(30^{\circ}-20^{\circ})\cdot \tan(30^{\circ}+20^{\circ})=\frac{\tan20^{\circ}}{\sqrt{3}}$

and we have the equivalent problem: 

              Verify that the equation 

$$\tan(30^{\circ}-x)\cdot \tan (30^{\circ}+x)=\frac{\tan x}{\sqrt{3}} \tag{4}$$

               is satisfied by  $x=20^{\circ}$.

The left side of  (4)  is written :

$$\frac{\sin(30^{\circ}-x) \cdot \sin(30^{\circ}+x)}{\cos(30^{\circ}-x)\cdot \cos(30^{\circ}+x)}\;\;\overset{2\sin u\sin v=\cos((v-u)-\cos(v+u)}{\underset{2\cos u\cos v=\cos(v-u)+\cos(v+u)}{=}}$$

$=\frac{\cos2x-\cos60^{\circ}}{\cos2x+\cos60^{\circ}}\;\;\underset{\tan x=t}{=}\;\;\frac{\frac{1-t^2}{1+t^2}-\frac{1}{2}}{\frac{1-t^2}{1+t^2}+\frac{1}{2}}=\frac{2-2t^2-1-t^2}{2-2t^2+1+t^2}=\frac{1-3t^2}{3-t^2}$

so we need to show that the equation  $\frac{1-3t^2}{3-t^2}=\frac{t}{\sqrt{3}}$  is verified by  $t=\tan20^{\circ}$. But the equation with the unknown  $t$ is written equivalently

$\sqrt{3}-3\sqrt{3}\cdot t^2=3\cdot t-t^3\;\Leftrightarrow\;t^3-3\sqrt{3}\cdot t^2-3\cdot t+\sqrt{3}=0$

which the Lemma shows has the solution  $t=\tan20^{\circ}$.

$\blacksquare$

          Remark CiP  By this we have NOT completely solved the equation  (4) nor what are the other two roots of the equation  (3).

<end Rem>

BUBBLE's algebra // Algebra lui BULĂ

           Bubble is a comic character. In Romanian, his name also has a trivial connotation.

          Everyone knows that multiplication is distributive over addition...

$$a\cdot (b+c)=a\cdot b+a\cdot c \tag{$\cdot$ / +}$$

Here we imply a certain "priority" of multiplication over addition, so as not to write excessively pedantically   $a\cdot (b+c)=(a\cdot b)+(a\cdot c)$.

          What if addition were distributive to multiplication ?

A very narrow-minded math teacher would compare a student who thinks like this to Bubble. This would be written

$$a+b\cdot c=(a+b)\cdot (a+c) \tag{+/$\cdot$}$$

or, excessively pedantically, like this    $a+(b\cdot c)=(a+b)\cdot (a+c).$

          Let's not forget that it is possible for two operations to be mutually distributive of each other e.g. in the case of Boolean algebras.

          I thought about a similar situation when I was developing initial assessment tests. I gave a problem like this to 8th graders :

                 Let  $a,\;b,\;c$  be real numbers that satisfy 

$a+b+c=1 \tag{1}$

                  Prove that  $a+b\cdot c=(a+b)(a+c) \tag{2}$

         There are many ways to demonstrate this, from a complete lack of reaction to the statement, to the most complicated calculations. Which? would be the shortest, and therefore most elegant, demonstration?? I would say that this would be :

$a+b\cdot c=a\cdot 1+bc \underset{(1)}{=}a(a+b+c)+bc=a^2+ab+ac+bc=a(a+b)+c(a+b)=(a+b)(a+c)$

 

          The problem in the image is related to the same context.

(I don't think there's any need for translation of E : 6134*.) "I. SAFTA, Pitești" is a well-known name in the field of problem-solving.

                                                       Solution CiP

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

$=\frac{a^2+2ab+b^2}{2}+\frac{b^2+2bc+c^2}{2}+\frac{c^2+2ca+a^2}{2}=$

$=\underline{\frac{(a+b)^2}{2}+\frac{(b+c)^2}{2}+\frac{(c+a)^2}{2}}=$

$\overset{a+b=1-c}{\underset{and \;the\; others}{=}}\frac{(1-c)^2}{2}+\frac{(1-a)^2}{2}+\frac{(1-b)^2}{2}=\underline{\left (\frac{1-c}{\sqrt{2}}\right )^2+\left ( \frac{1-a}{\sqrt{2}}\right )^2+\left( \frac{1-b}{\sqrt{2}}\right )^2}$

$\blacksquare$

Regarding the Problem 16 667

 Here, page 47, the solution was published. I haven't found the Statement yet.

     We select the property in the photo.

 ANSWER CiP

We will show that one of the numbers  $a_k$  is equal to 1, 

and the others are zero. Then  $b=k$.

                              Solution CiP

               We will treat the case  $n=4$  in detail, so for

$a_1,\;\;a_2,\;\;a_3,\;\;a_4\geqslant 0 \tag{1}$

we have equations

\begin{cases}a_1+a_2+a_3+a_4=1\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(2)\\1\cdot a_1+2\cdot a_2+3\cdot a_3+4\cdot a_4=b\;\;\;\;\;\;\;\;\;\;\;\;(3)\\1^2\cdot a_1+2^2\cdot a_2+3^2\cdot a_3+4^2\cdot a_4=b^2\;\;\;\;\;(4)\end{cases}

Subtracting (2) from (3) we have  $b-1=a_2+2\cdot a_3+3\cdot a_4\geqslant 0$,  so  $b\geqslant 1.$

          $\textbf{If}\;\;\bf{b\geqslant 4}$  then 

$b^2-4\cdot b\;\underset{(4)-4\times (3)}{=}(a_1+4a_2+9a_3+16a_4)-4\cdot (a_1+2a_2+3a_3+4a_4)$,  so

$0\leqslant b^2-4b=-3a_1-4a_2-3a_3\overset{(1)}{\leqslant} 0 \tag{5}$

Therefore, in (5) we have everywhere the sign  $"="$. In particular

$b=4,\;\;a_1=a_2=a_3=0,\;\;and\;\;a_4\overset{(2)}{=}1 \tag{6}$

        $\textbf{Let}\;\;\bf{b\in[i,i+1)}$  for some  $i\in\{1,\;2,\;3\}$. We calculate first

$b^2-b\cdot i\;\overset{(4)}{\underset{i\times (3)}{=}}(1a_1+4a_2+9a_3+16a_4)-(i\cdot a_1+2i\cdot a_2+3i\cdot a_3+4i\cdot a_4)$  so

$b^2-bi=(1-i)a_1+2(2-i)a_2+3(3-i)a_3+4(4-i)a_4 \tag{7}$

On the other hand

$b-i\cdot 1\overset{(3)}{\underset{i\times (2)}{=}}(a_1+2a_2+3a_3+4a_4)-(i\cdot a_1+i\cdot a_2+i\cdot a_3+i\cdot a_4)=$

$=(1-i)a_1+(2-i)a_2+(3-i)a_3+(4-i)a_4$

which, multiplied by  $b$  gives us

$b^2-bi=b(1-i)a_1+b(2-i)a_2+b(3-i)a_3+b(4-i)a_4 \tag{8}$

Equating (7) and (8)

$(1-i)a_1+2(2-i)a_2+3(3-i)a_3+4(4-i)a_4=b(1-i)a_1+b(2-i)a_2+b(3-i)a_3+b(4-i)a_4$

we obtain

$(b-1)(i-1)a_1+(b-2)(i-2)a_2+(b-3)(i-3)a_3+(b-4)(i-4)a_4=0 \tag{9}$

               For  $i=1$, (so  $1\leqslant b<2$), the first term cancels out, and all coefficients of   $a_2,\;a_3,\;a_4$  are  $>0$.

Therefore  $a_2=a_3=a_4=0$,  so (cf. (2))  $a_1=1$  and then  $b\overset{(3)}{=}1$.

                For  $i=2$, (so  $2\leqslant b<3$), then the second term cancels out, and all coefficients of  $a_1,\;a_3,\;a_4$  are  $>0$.

Therefore  $a_1=a_3=a_4=0$,  so (cf.(2))  $a_2=1$  and then  $b\overset{(3)}{=}2$.

                For  $i=3$, (so  $3\leqslant b <4$), then the third term cancels out, and all coefficients of  $a_1,\;a_2,\;a_4$  are  $>0$.

Therefore  $a_1=a_2=a_4=0$, so (cf.(2))  $a_3=1$  and then  $b\overset{(3)}{=}3$.

The last three conclusions, together with (6), validate the answer in this particular case.

               The general case extends what is shown in case  $n=4$. So let be the numbers

$a_1,\;a_2,\;\dots,a_n\;\geqslant 0 \tag{11}$

that verify the relations

$$a_1+a_2+\dots+a_n=\sum_{k=1}^na_k=1 \tag{12}$$

$$a_1+2\cdot a_2+\dots+n\cdot a_n=\sum_{k=1}^nk\cdot a_k=b \tag{13}$$

$$a_1+4\cdot a_2+\dots +n^2\cdot a_n=\sum_{k=1}^nk^2\cdot a_k=b^2 \tag{14}$$

We have

$$b-1\overset{(13),(12)}{=}\sum_{k=1}^nk\cdot a_k-\sum_{k=1}^na_k=\sum_{k=1}^n(k-1)\cdot a_k=\sum_{k=2}^n(k-1)a_k\geqslant 0$$

so  $b\geqslant 1$.

          If  $b\geqslant n$  then

$$0\leqslant b^2-n\cdot b\overset{(14),(13)}{=}\sum_{k=1}^nk^2\cdot a_k-n \sum_{k=1}^nk\cdot a_k=\sum_{k=1}^n(k^2-nk)a_k=\sum_{k=1}^{n-1}k(k-n)a_k\leqslant 0$$

and according to (11) it result  $a_1=\dots=a_{n-1}=0,\;b=n,\;a_n\overset{(12)}{=}1$.

          If  $b\in[i,i+1)$  for some  $i\in\{1,\dots,n-1\}$, then, on the one hand

$$b^2-i\cdot b\;\overset{(14),(13)}{=}\sum_{k=1}^nk^2a_k-i\cdot \sum_{k=1}^nka_k=\sum_{k=1}^nk(k-i)a_k \tag{15}$$

On the other hand

$$b-i\;\overset{(13),(12)}{=}\sum_{k=1}^nka_k-i\cdot \sum_{k=1}^na_k=\sum_{k=1}^n(k-i)a_k$$

which, multiplied by  $b$  gives us

$$b^2-ib=\sum_{k=1}^nb(k-i)a_k \tag{16}$$

Equating (15) and (16)  we obtain

$$\sum_{k=1}^n(b-k)(i-k)a_k=0$$

or, excluding the term  $a_i$ wich has coefficient zero

$(b-1)(i-1)a_1+\dots+(b-i+1)a_{i-1}+(b-i-1)(-1)a_{i+1}+\dots (b-n)(i-n)a_n=0$.

Let's analyze the coefficients of the numbers  $a_k$  in the equation above :

          for  $k<i\;\Rightarrow\;(b-k)(i-k)\;\overset{k<i\leqslant b}{>}0;$

          for  $k>i\;\Rightarrow\;(b-k)(i-k)=(k-b)(k-i)\overset{k\geqslant i+1>b}{>}0$

Then it turns out that  $a_1=\dots=a_{i-1}=0=a_{i+1}=\dots =a_n$,  so  $a_i\overset{(12)}{=}1$  and  $b\underset{(13)}{=}i.$

$\blacksquare$

PLM inseamna o trivialitate.... PLV este prescurtarea oficiala

 Am postat in Aug/01/2024 Decizia de Recalculare a pensiei.

Un comentator anonim, cam sugubat, s-a interesat de Decizia de trecere la Pensia pentru Limita de Varsta. Prescurtat PLV. I-am raspuns ... la fel de trivial. Scuze cititorilor mai sensibili...

Aici sunt cele 3 pagini din Decizie...

Talonul de pensie pe luna AUGUST 2025, insemnat cu compararile din "Marea Recalculare"

Am incasat si niste diferente pntru lunile MAI-JUL, cum se vede mai jos

Voi posta si Talonul pe SEPtembrie, care asa va fi de aici incolo....cat voi trai.

Edit Sep 8, 2025 : Iata Talonul ....final: