Ce? se întâmplă când are toată familia febră ?? // What happens when the whole family has a fever?

          It seems that the Mathematical Olympiad has begun.

          Let's try to put ourselves in the shoes of a 5th grader. Let's take a list of issues given not too long ago. For example, let's consult the magazine below. 

(Unfortunately I can't show you more than that.)

- The picture is courtesy of SSMR -

          We will choose three 5th grade problems from the first stage of the Olympiad as examples.

          Problem 1 (at page 1) "A test has 20 questions, which are to be solved in 150

                       minutes. The questions are of three types: easy, medium and difficult. 

                       It is expected that each of the 7 easy questions can be solved in 4 

                      minutes and each of the medium questions can be solved in 8 minutes.

                       How many minutes are left for solving difficult questions?"      

                               A. 56            B. 58            C.62            D. 66            E. 70

Answer CiP        D. 66

Solution CiP

               Easy questions and medium questions require a solving time of $7 \cdot 4+7\cdot 8=24$  minutes. So for the difficult questions, $150-84=66$ minutes remain. The answer is D.

$\blacksquare$

               Remark CiP We have 7 easy questions and 7 medium questions, so the number of difficult questions is $20-7-7=6$. So each difficult question is expected to be solved in 11 minutes.

< end Rem>

          In light of the statement in Problem 1, it seems that Problems 1-7 are considered easy, Problems 8-14 are medium, and Problems 15-20 are difficult. I will take one of each. 

           I also mention that the Magazine in the image above does not contain the corresponding answers to these questions. Despite the other qualities: e.g. a clear English that would make even Shakespeare envious. News about the adventures at this Olympics also appeared in the press. That's also where I found out where you can find the Answers. The same lovely SSMR.

          Problem 11 (at page 2) "The largest integer which divided by 2022 gives a

                    quotient smaller than the remainder is :

A. 1048043482         B. 4086461         C. 16185         D. 8091         E. 9         " 

 

Answer CiP    B. 4086461

Solution CiP

               By dividing a number $D$ by $2022$ we obtain the quotient $Q$ and the remainder $R$, according to the long division as below

$$\begin{array}{c|c} D & 2022 \\ \hline \vdots & Q \\ \hline R & \; \end{array}$$

and the Division with remainder Theorem is written

$$D=2022\cdot Q+R\;\; \quad R<2022 \tag{1}$$

The statement also specifies the condition 

$$Q\;<\;R \tag{2}$$

So the first relation (1) gives us $D<2022R+R$ , or $D<2023R$. But from the second relation (1) we  have $D\leqslant 2021$, and so

$$D<2023 \cdot 2021 =4\;088\; 483.$$

So the answer "A."  is excluded, being a larger number.

     We see from (1) that the largest admissible number $D$ occurs when $R$ and $Q$ are the largest, also verifying the condition (2), so 

$$Q=R-1=2020.$$

Hence $D=2022 \cdot 2020+2021=4\;086\;461$, so answer is B. 

$\blacksquare$

           Remark CiP  Answer options C and E also verify the condition (2)

$$16\;185 \;:\;2022 \;=\;8\quad remainder\;9\;\quad \quad 9\;:\;2022\;=\;0\quad remainer\;9$$

but option D does not $8\;091\;;\;2022\;=\;4\quad remainder\;3.$

<end Rem>

          Problem 18 (at page 3) "If $S(n)$ denotes the sum of the digits of the number

                  $n$, then the number of $n-$s so that $\;2021\leqslant n+S(n)\leqslant 2022$ is:

A. 0         B. 1         C. 2         D. 3         E. 4    "

Answer CiP   C. 2

$$1996+S(1996)=1996+25=2021\;\quad \; 2014+S(2014)=2014+7=2021$$

Solution CiP

          $n=0$ does not satisfy the condition.

           If $n\leqslant 1999$ we have $S(n)\leqslant 28$, and then

$$n\geqslant 2021-S(n)\geqslant 2021-28=1993.$$

We have to try the numbers:

$$1993,\quad 1994,\quad\cdots,1999.\quad  \tag{1}$$

After some calculations:

$$1993+S(1993)=1993+22=2015,$$

$$1994+S(1994)=1994+23=2017,$$

$$1995+S(1995)=1995+24=2019,$$

$1996+S(1996)=1996+25=2021$ - this number is convenient

we see that for the other numbers in the list (1) the value $n+S(n)$ exceeds 2023.

          If $n\geqslant 2000$ (anyway $n\leqslant 2022$) we have $S(n)\geqslant 2$ so

 $n\leqslant 2022-S(n)\leqslant 2020$.

Among the numbers   2000, 2001, ..2020 the highest $S(n)$ value is $S(2019)=12$ and then, from $n\geqslant 2021-S(n)$ we get $n\geqslant 2021-12=2009$, so we have to try only the numbers

$$2009,\quad 2010,\;\dots\;2020.\quad \tag{2}$$

Observing the evolution of calculations:

$$2009+S(2009)=2009+11=2020$$

$$2010+S(2010)=2010+3=2013$$

$$2011+S(2011)=2011+4=2015$$

$$2012+S(2012)=2012+5=2017$$

$$2013+S(2013)=2013+6=2019$$

$2014+S(2014)=2014+7=2021$ - this number is convenient

$$2015+S(2015)=2015+8=2023$$

it turns out that from here on out, there can be no more convenient numbers in list (2).

$\blacksquare$

SILBERBERG Gheorghe and a terrifying Problem : S.L.24.220

           The author, currently a teacher at UVT, was an international Olympian in mathematics. Originally from the city of Lugoj, he came with a bus full of students to the Timisoara County Olympics.

          Published in the Magazine(aka REVISTE) "GAZETA MATEMATICĂ : Supliment cu Exerciții", September 2024 at page 10. The statement is:

          "Let $f:\mathbb{R} \rightarrow \mathbb{Z}$ be a monotone and surjective function with properties:

              (i) $f(f(x))=f(x),\;\forall x\in \mathbb{R};$

              (ii) $f(2x)-f(x)=f\left ( x+\frac{1}{2}\right ),\; \forall x \in \mathbb{R}.$

     a) Show that $f\left ( k-\frac{1}{2^n} \right )=k-1$, whatever $k \in \mathbb{Z}$ and $n \in \mathbb{N}.$

     b) Determine the function $f$. "

ANSWER CiP

a) Induction on $n\in \mathbb{N}$

b) f(x)=[x] - the Floor function

                    Solution CiP

               The function $f$ being surjective, for $k\in\mathbb{Z}$ there exists $x_k\in\mathbb{R}$ such that $f(x_k)=k.$ Now $k=f(x_k)\underset{(i)}{=}f(f(x_k))=f(k)$, so
$$f(k)=k,\;\;\forall k\in\mathbb{Z} \tag{1}$$

          We see from (1) that the function $f$ is weakly increasing i.e. 

$$x<y\;\Rightarrow\;f(x)\leqslant f(y). \tag{1#}$$

          a) Equation

$$f\left (k-\frac{1}{2^n}\right )=k-1 \tag{2}$$

is true for $n=0$ according to (1). From (ii) we have

$$f\left (2\left (k-\frac{1}{2}\right)\right )-f\left (k-\frac{1}{2}\right )=f\left (k-\frac{1}{2}+\frac{1}{2}\right )\;\Leftrightarrow$$

$$\Leftrightarrow\;f(2k-1)-f\left (k-\frac{1}{2}\right )=f(k)\;\underset{(1)}{\Leftrightarrow}$$

$$\Leftrightarrow\;(2k-1)-f\left (k-\frac{1}{2}\right )=k\;\Rightarrow\;f\left (k-\frac{1}{2}\right )=k-1$$

so (2) is true for $n=1$.

{ edit nov 23, 2024: It seems that an inductive reasoning of (2) after $n$ has no immediate chance of success. } 

          From $(1)$ and $(1\#)$ we obtain the important estimate

$$k\leqslant x<k+1\;\;\;\Rightarrow\;k\leqslant f(x)\leqslant k+1,\;\;k\in\mathbb{Z}. \tag{3}$$

          Let us assume that for a certain $k_0\in \mathbb{Z}$ we have $\color {Red}{f\left (k_0-\frac{1}{4}\right ) \neq k_0-1}$. Hence, from (1#) follow $f\left ( k_0-\frac{1}{4}\right )=k_0$.

We have equality

$$f(2x+1)-f(2x)=f(x+1)-f(x) ,\;\;\forall x \in \mathbb{R}.\tag{4}$$

Indeed, applying (ii) to the underlined expressions

$$f(2x+1)-f(2x)=\underline{f \left ( 2 \left ( x+\frac{1}{2}\right ) \right )}-f(2x)=$$

$$=f\left (x+\frac{1}{2} \right )+f\left(x+\frac{1}{2}+\frac{1}{2} \right )-f(2x)=f(x+1)-\underline{f(2x)+f\left (x+\frac{1}{2} \right )}=$$

$$=f(x+1)-f(x).$$

Taking in (4) $x=k_0-\frac{1}{4}$ we get $f \left ( 2 \left (k_0-\frac{1}{4} \right )+1 \right )-f \left (2 \left( k_0-\frac{1}{4} \right ) \right )=f \left (k_0-\frac{1}{4}+1\right)-f(\left (k_0-\frac{1}{4} \right )\;\Leftrightarrow$

$$\Leftrightarrow\;f\left (2k_0+1-\frac{1}{2} \right )-f \left (2k_0-\frac{1}{2}\right )=f\left (k_0-\frac{1}{4}+1 \right )-f\left (k_0-\frac{1}{4} \right ) \;\Leftrightarrow$$

$$\overset{(2)\;for\;n=1}{\underset{f(k_0-1/4)=k_0}{\Leftrightarrow}}\;2k_0-(2k_0-1)=f\left (k_0+1-\frac{1}{4}\right )-k_0$$

hence $f\left (k_0+1-\frac{1}{4}\right )=k_0+1$, so applying a inductive reasoning results

$$f\left (k-\frac{1}{4}\right )=k,\;\forall k\geqslant k_0,\;k\in \mathbb{Z}.\tag{5}$$

Applying (ii) again for $k-\frac{1}{4}$ we get $f\left (2\left (k-\frac{1}{4}\right) \right )-f\left (k-\frac{1}{4} \right )=f\left (k-\frac{1}{4}+\frac{1}{2} \right )$

$$\Leftrightarrow\;f\left (2k-\frac{1}{2} \right )-f\left (k-\frac{1}{4} \right )=f\left ( k+\frac{1}{4} \right )\;\overset {(2)\;for\;n=1}{\underset{(5)}{\Leftrightarrow}}$$

$$\Leftrightarrow\;(2k-1)-k=f\left (k+\frac{1}{4} \right )\;\Leftrightarrow\;f\left (k+\frac{1}{4} \right )=k-1.$$

But then we get $k=f(k)\leqslant f \left ( k+\frac{1}{4} \right )=k-1$, FALSE.

 So $f\left (k-\frac{1}{4} \right )=k-1,\; \forall k \in \mathbb{Z}$   hence (2) is true for $n=2$.

{edit nov 27, 2024: This is how we can prove (2) by induction on $n$ }

          Let the predicate depending on the variable $n\in \mathbb{N}$ be

$$P(n)\;:\;\;"\forall k \in \mathbb{Z},\;f\left (k-\frac{1}{2^n} \right )=k-1"$$

     For $n\in \{0,\;1,\;2\}$ we have $P(n)-{\color{Green}{true}}$.

     We now assume $P(n)$-true for some $n$ and we will show that $P(n+1)$ is also true, so according to mathematical induction it results $\forall n\in \mathbb{N}\;P(n)$-true.

     If, by absurdity, $P(n+1)$ is false, it means that there exists $k_0\in \mathbb{Z}$ such that $f\left ( k_0-\frac{1}{2^{n+1}} \right ) \neq k_0-1$. But, because of (3), we must then have 

$$f\left (k_0-\frac{1}{2^{n+1}}\right )=k_0. \tag{6}$$

Applying (ii) for $x=k_0-\frac{1}{2^{n+1}}$ we get

$$f\left (2\left (k_0-\frac{1}{2^{n+1}}\right ) \right )-f\left ( k_0-\frac{1}{2^{n+1}}\right )=f\left ( k_0-\frac{1}{2^{n+1}}+\frac{1}{2} \right )\;\Leftrightarrow$$

$$\Leftrightarrow\;f\left ( 2k_0-\frac{1}{2^n}\right )-f \left (k_0-\frac{1}{2^{n+1}} \right )=f\left (k_0+\frac{1}{2}-\frac{1}{2^{n+1}} \right )\;\Leftrightarrow$$

$$\overset{P(n)-true}{\underset{(6)}{\Leftrightarrow}} \;(2k_0-1)-k_0=f\left (k_0+\frac{1}{2}-\frac{1}{2^{n+1}} \right )\;\;\Rightarrow$$

$$\Rightarrow\;\;\;f\left( k_0+\frac{1}{2}-\frac{1}{2^{n+1}} \right )=k_0-1.$$

But, since $\frac{1}{2}-\frac{1}{2^{n+1}}>0$, we have the inequalities 

$k_0=f(k_0)\leqslant f\left (k_0+\frac{1}{2}-\frac{1}{2^{n+1}} \right )=k_0-1$-FALSE.

          With this  "a)" is demonstrated.

          b) Let's show that for $k\leqslant x <k+1$ we have $f(x)=k$.

        If we choose a $n>-log_2(k+1-x)$ we have

 $-n<log_2(k+1-x)\;\;\Rightarrow\;\;2^{-n}<k+1-x\;\;\Rightarrow\;\;x<k+1-\frac{1}{2^n}.$

Further $k=f(k)\leqslant f(x)\leqslant f\left ( k+1-\frac{1}{2^n}  \right )\underset {(2)}{=}k$, so $f(x)=k$. We got the answer.

$\blacksquare$

A PROBLEM with the ISO_80000 SPECIFICATION

           "Find the real numbers $x$ and $y$, if  $\lg^2\frac{x}{y}=3\cdot \lg\frac{x}{2024}\cdot \lg \frac{2024}{y}$."

[Wikipedia says that, according to the ISO 80000 specification(it costs a lot to read it, and time...), "$\lg$" should be the standard notation for the decimal logarithm $log_{10}.$ ]

***  The sign "***" means that the problem does not have a known author, as it appears in the MAGAZINE(aka REVISTE) Supliment cu Exerciții, September 2024; proposed for the 10th grade on page 9 with number S.L24.211.

ANSWER CiP

$$x=2024,\;\;y=2024$$

Solution CiP

            Instead, we will solve the problem (...more than that, I don't venture to try...)
                    "Find the real numbers $x$ and $y$, if $\lg^2\frac{x}{y}=\lambda \cdot \lg\frac{x}{a} \cdot \lg \frac{a}{y}$"

       $a$ and $\lambda$ being given positive real numbers, $\color {Red}{\lambda <4}$."

                                                         The ANSWER  will be $\underline {x=a\;,\;y=a}$

     The existence conditions of the problem, $\frac{x}{y}>0,\; \frac{x}{a}>0,\;\frac{a}{y}>0$ combined give $x>0$ and $y>0$.

     I will use the inequality 

$$4\cdot u \cdot v \leqslant (u+v)^2 \tag{1}$$

with the sign $"="$ in (1) if and only if $u=v$. Let $A:=\lg\frac{x}{y}$;

$$A^2\underset{eq}{=}\lambda \cdot \lg\frac{x}{a}\cdot \lg\frac{a}{y}=\frac{\lambda}{4}\cdot 4\lg\frac{x}{a}\lg\frac{a}{y}\;\;\;\;\overset{(1)}{\underset{u=\lg\frac{x}{a}\;v=\lg\frac{a}{y}}{\leqslant}}\; \;\;\;\frac{\lambda}{4} \cdot \left ( \lg\frac{x}{a}+\lg \frac{a}{y} \right)^2 =$$

$$=\frac{\lambda}{4}\cdot \left [\lg \left (\frac{x}{a}\cdot \frac{a}{y}\right )\right ]^2=\frac{\lambda}{4}\cdot \lg^2 \frac{x}{y}=\frac{\lambda}{4}\cdot A^2,$$

hence $A^2 \leqslant \frac{\lambda}{4} \cdot A^2\;\Leftrightarrow\;A^2\cdot (1-\frac{\lambda}{4})\leqslant 0$ but which in the given condition $\lambda <4$ implies $A=0.$

     I got $\lg\frac{x}{y}=0$ so $x=y$ and, from the equation, that one of the conditions $\lg \frac{x}{a}=0$ or $\lg \frac{a}{y}=0$ occurs, so $x=a$ or $y=a$, (!)actually both.

$\blacksquare$

PROBLEM E:16993 author Mihaela BERINDEANU, Bucharest

 "Let $a,\;b\in \mathbb{N}^*$ be such that the number $\frac{a+3}{b}+\frac{b+3}{a}$ is an integer.

If $(a,b)$ is the greatest common divisor of the numbers $a$ and $b$, then show that $(a,b) \leqslant \sqrt{3(a+b)}$."

          From the MAGAZINE(aka REVISTE) Gazeta Matematica seria B no.9/2024, page 426, the proposed problem for the 7th grade

          ANSWER CiP

             A case where the equal sign occurs is

$$a=6,\;b=6.$$

          Solution CiP

          Let $k=\frac{a+3}{b}+\frac{b+3}{a}\in \mathbb{Z}$, actually $k\in\mathbb{N}^*$.

After calculations, we have the relationship

$$a^2+b^2-kab+3a+3b=0.\tag{1}$$

If $d=(a,b)$ then $a=d\cdot a_1,\;b=d \cdot b_1$ with $(a_1,b_1)=1.$Replacing these in (1) we have

$$d^2 \cdot a_1^2+d^2 \cdot b_1^2-k\cdot da_1\cdot db_1+3d\cdot a_1+3d\cdot b_1=0\;\Leftrightarrow$$

$$da_1^2+db_1^2-dka_1b_1+\underline{3(a_1+b_1)}=0.$$

In the last equation, the underlined term must be divisible by the number $d$, because all other terms are divisible by it. From here we get, with natural numbers

$$d \mid 3(a_1+b_1)\;\Rightarrow\;3(a_1+b_1)=d \cdot c \geqslant d\;\Rightarrow\;$$

$$\Rightarrow\;3\left ( \frac{a}{d}+\frac{b}{d} \right ) \geqslant d\;\Leftrightarrow \; 3(a+b)\geqslant d^2\;\;\Leftrightarrow\;(a,b)\leqslant \sqrt{3(a+b)}.$$

          For $a=6$ and $b=6$ we have $\frac{a+3}{b}+\frac{b+3}{a}=\frac{9}{6}+\frac{9}{6}=3$, and
$$d=(6,6)=6=\sqrt{36}=\sqrt{3\cdot(6+6)}.$$

 $\blacksquare$

MENELAOS' THEOREM: the affine version

 

          Given a triangle $ABC$ and a straight line $s$ that does not pass through any of its vertices.

The line $s$ should intersect the lines determined by the sides of the triangle at the points $M,\;N,\;P$ as in the figure. (According to the position of the line $s$ in front of

the triangle, we can have one or all three of these intersection points on the extensions of the sides of the triangle.)

We will consider some vectors associated with this configuration : $\overrightarrow{AP},\;\overrightarrow {BM},\;\overrightarrow{CN}.$

          The points $P,\;A,\;B\;$ being collinear, and so are $B,\;M,\;C\;$ as well as $A,\;C,\;N\;$, we can find some scalars $\alpha,\;\beta,\;\gamma\;$ so that (see "Barycentric coordinate on the Straight Line"):

$$\color {Red} {\overrightarrow{AP}=\gamma \cdot \overrightarrow{AB}},\;\;\overrightarrow{BP}=(\gamma-1)\overrightarrow{AB},\;\;\frac{\overline{PA}}{\overline{PB}}=\frac{\gamma}{\gamma-1} \tag{1}$$

$$\color {Red}{\overrightarrow{BM}=\alpha \cdot \overrightarrow{BC}},\;\;\overrightarrow{CM}=(\alpha-1)\overrightarrow{BC},\;\;\frac{\overline{MB}}{\overline{MC}}=\frac{\alpha}{\alpha-1} \tag{2}$$

$$\overrightarrow{CN}=\beta \cdot \overrightarrow {CA},\;\;\overrightarrow{AN}=(\beta-1)\overrightarrow{CA},\;\;\frac{\overline{NC}}{\overline{NA}}=\frac{\beta}{\beta-1} \tag{3}$$

(will continue)

My books from the ELECTRONIC LIBRARY

Scan the following code

One, two, three, four // Come in, please, and shut the door

          It is a  fun poem for children. You can learn English and Mathematics at the same time.

          The following example contains ONLY math... And some gold!

$$\arctan \frac{1}{2}+\arctan \frac{3}{4}=2\arctan \frac{1}{\varphi}$$

Enjoy!

          

Revista de Matematică a Elevilor și Profesorilor din Județul Caraș-Severin (RMCS) // Mathematics Magazine of Students and Teachers from Caraș-Severin County

          I found an interesting problem, for class 5, in RMCS_39--2012.

In translation

             "V. 241   Consider the set $M=\{1,\;8,\;15,\;22,\;29, \dots ,\;134\}$. Show that any 12-element subset of the set M contains two elements whose sum is equal to 142.

Bihor County Olympiad"

              I solve this problem with the help of the Pigeonhole principle. A post on AOPS also helped me.

             First note that since $1=7\cdot 0+1,\;8=7\cdot 1+1,\;15=7\cdot 2+1,\dots ,\;134=7\cdot 19+1$, the set M has 20 elements. Let's arrange the elements of the set M on 10 columns like this

$$\begin{array}{|r|r|}\hline 8&15&22&29&36&43&50&57&64&1 \\ \hline  134&127&120&113&106&99&92&85&78&71 \\ \hline \end{array}$$

The sum of the elements in the first nine columns is 142. Let us choose some set of 12 elements from the elements of the set M. Let's assume that, in the worst case, we also chose elements 1 and 71 (from the last column). However, 12-2=10 elements remain. With only nine columns available, two of the ten elements must be in the same column, and they will have the sum 142.

$\blacksquare$

          Remark CiP  There are sets with 11 elements that do not satisfy the requirement of the problem. For example

$$\{1,\;8,\;15,\;22,\;29,\;36,\;43,\;50,\;57,\;64,\;70\}.$$

          L.E. You can also consult the solution from RMCS here, at page 19.

                   Other issues of the RMCS can be found here.

A problem with ... echo : E: 16 967 from GMB 6-7-$\infty$ /2024

           The first Problem, E:15 002, appeared in GMB_4/2016, at page 212 (the password for open file is  ogeometrie ). I solved it myself in the post "Rezolvarea problemelor din Gazeta Matematica (seria B) Nr 4/2016 pentru clasa a VIII-a". The official solution to the problem appeared in GMB_10/2016, at pages 473-474; I have not yet compared it with my solution...

          After almost 2000 problems, another version, by the same author, appears in GMB_6-7-8/2024, at page 369 (same pasword !!).

 ANSWER CiP  $a=2,\;\;b=2.$

               Solution CiP

          We add  $2+2=4$  to both sides of the equation; so

$$\left ( \frac{2a-1}{b+2}+2 \right )+\left ( \frac{2b-1}{a+2}+2 \right )=\frac{6}{a+b}+4\;\Leftrightarrow$$

$$\Leftrightarrow\;\frac{2a+2b+3}{b+2}+\frac{2b+2a+3}{a+2}=\frac{6+4a+4b}{a+b};$$

since $2a+2b+3 \neq 0$, being an odd number, we can simplify the equation with him, and get

$$\frac{1}{b+2}+\frac{1}{a+2}=\frac{2}{a+b}\Leftrightarrow \frac{1}{b+2}-\frac{1}{a+b}=\frac{1}{a+b}-\frac{1}{a+2}\Leftrightarrow $$

$$\Leftrightarrow \frac{a-2}{(b+2)(a+b)}=\frac{2-b}{(a+b)(a+2)}\;\;\;\underset{a+b \neq 0}{\Leftrightarrow} \;\;\;\frac{a-2}{b+2}=\frac{2-b}{a+2}\Leftrightarrow$$

$$\Leftrightarrow a^2-4=4-b^2\Leftrightarrow a^2+b^2=8;$$

the last equation is possible with integer numbers only if  $a^2=4,\;b^2=4$ and, because $a,b \neq -2$, we have $a=2$ and $b=2.$

$\blacksquare$

GAZETA MATEMATICĂ Seria B N0 6-7-8/2024

           This magazine is for my students, who do not have the possibility to subscribe. The password for opening the file is known only to them.

Issues of the magazine, older or newer, can be found here.

You can find other mathematics magazines here.

Sorry, the links are no longer valid.

Ciąg ... podstępów // Un șir cu ... șmecherie

 From the link https://www.deltami.edu.pl/2024/09/klub-44/

          Statement of the problem

          "The sequence $\;a_0,\;a_1,\;a_2,\;...\;$ is defined by the formulas

           $a_0=3,\;a_{n+1}=a_n^2-2,\;n=0,\;1,\;2,...\;.$ Calculate the limit

$$\lim_{n \to \infty}\frac{1}{a_n}\prod_{i=0}^{n-1}a_i$$

            or demonstrate that this limit does not exist."

The demonstration is from the same cited place .

          ANSWER : The limit is  $\frac{1}{\sqrt{5}}.$

          Proof : Personal preamble CiP

             A first observation is that $\;a_n>2.$ (It would be too pedantic to prove this by induction; however: $a_0>2\;$ and assuming $a_n>2\;$ we get

 $a_{n+1}=a_n^2-2>2^2-2=2.$)

          We have, because $a_n>2,$

$\;\;a_{n+1}-a_n=a_n^2-2-a_n=(a_n+1)(a_n-2)>0$, so $a_{n+1}>a_n$, meaning the sequence $(a_n)_{n\geqslant 0}$ is strictly increasing.

         Then we have  $a_n\geqslant n+3$ (an extremely poor estimate). (Induction is equally trivial: $a_0\geqslant 3$, and assuming $\;a_n \geqslant n+3$, result

 $a_{n+1}\geqslant (n+3)^2-2=n^2+6n+7\geqslant n+4.$)

However, from the above it follows that

$$\lim_{n \to \infty}a_n=\infty \;. \tag{1}$$

I wrote all this to justify the statement "Jasne, że $a_n \to \infty$". Next, we copy the quoted solution.

<end Preamble>

          We will define the sequence $(b_n)_{n \geqslant 0}$ by

 $b_0=1,\;b_n=a_0\cdots a_{n-1},\;n=1,\;2,\dots$ 

 and we have to investigate the limit of the sequence $\left ( \frac{b_n}{a_n}\right )_{n \geqslant 0}.$

          We see that

$$b_{n+1}=a_n \cdot b_n \tag{2}$$

and we will prove by Induction that 

$$a_n^2-5\cdot b_n^2=4. \tag{3}$$

     Obvious $\;a_0^2-5\cdot b_0^2=3^2-5\cdot 1=4$, and assuming (3) true, result

$$a_{n+1}^2-5\cdot b_{n+1}^2 \overset{def}{\underset{(2)}{=}}(a_n^2-2)^2-5 \cdot (a_n\cdot b_n)^2=\underline{a_n^4}-4a_n^2+4-\underline{5a_n^2b_n^2}=\underline{a_n^2}\cdot (a_n^2-5b_n)-4a_n^2+4 \underset{(3)}{=}a_n^2\cdot 4-4a_n^2+4=4$$

so (3) also occurs for $\;n+1.$ This complete the Induction.

     Now we have 

$$\left (\frac{b_n}{a_n} \right )^2=\frac{b_n^2}{a_n^2}\underset{(3)}{=}\frac{\frac{a_n^2-4}{5}}{a_n^2}=\frac{a_n^2-4}{5a_n^2}=\frac{1}{5}-\frac{4}{5}\cdot \frac{1}{a_n^2}\xrightarrow{(1)}\frac{1}{5}-\frac{4}{5}\cdot 0=\frac{1}{5}.$$

Therefore $\;\frac{b_n}{a_n} \to \frac{1}{\sqrt{5}}.$

$\blacksquare$

Doru ISAC ... are? cineva această Carte ??

 

Увод у апстрактну алгебру, аутор Ц.К. ТАИЛОР //An Introduction to Abstract Algebra, author C.K. TAYLOR

 The book can be found at the link https://drive.google.com/file/d/1fWFTtrDG4rAsuaP1qP-PQjrDdCVbNoW0/view?usp=drive_link

 I took it from the link https://bookboon.com/en/statistics-and-mathematics-ebooks 

 I have added a first page with other mathematical publications of interest. 

Don't be afraid to access the links presented. To view some publications, you must enter the password that I indicated.

Књига се налази на линку https://drive.google.com/file/d/1fWFTtrDG4rAsuaP1qP-PQjrDdCVbNoW0/view?usp=drive_link

Узео сам са линка https://bookboon.com/en/statistics-and-mathematics-ebooks

Додао сам прву страницу са другим математичким публикацијама од интереса.

Немојте се плашити да приступите представљеним везама. Да бисте видели неке публикације, морате унети лозинку коју сам навео.

ROMANIAN MATHEMATICAL COMPETITIONS - 2018

           This will be the LAST volume of RMC that I post on the Blog. Click on the image to download.

The password to open the file is  ogeometrie .

You can find other magazines  here .

Romanian Mathematics Competitions - 2000 // Kompetisi Matematika Rumania - 2000

 (indoneziana)

Jika bukan karena Blog ini, saya tidak akan memiliki akses ke beberapa RMC lama:

https://www.m4th-lab.net/

Lihat RMC di tempat ini.

Untuk majalah lain lihat di sini.

Kata sandi untuk membuka akun adalah   ogeometrie

Acad Gheorghe VRĂNCEANU / Academician Gheorghe VRANCEANU / 学者ゲオルゲ・ヴランシアヌ

 Click on the image to download. The password to open the file is ogeometrie.

See also in the Electronic Library

VRĂNCEANU Gheorghe

Geometrie Analitică, Proiectivă și Diferențială

Ed. Didactică și Pedagogică, București, 1962

VRĂNCEANU Gheorghe (Acad)

Lecții de Geometrie Diferențială

Ed. Didactică și Pedagogică, București, vol_1(1976)

VRĂNCEANU G., TELEMAN C.

Geometrie euclidiană, geometrii neeuclidiene, teoria relativității (ed. II)

Ed. Tehnică, București, 1967 , compressed

In haste, part of the edition had the wrong cover.

I will scan and post the other volumes when I have time.

REVISTA MATEMATICĂ (a elevilor) din TIMIȘOARA - nr 1/1974

 Click on the image to download. Password is   ogeometrie .

For other collections, see here.

DIDACTICA MATEMATICĂ N0 1/2015

 Click on the image to download. The password to open the file is  ogeometrie .

Other Mathematics Magazines can be found here.

DECIZIE de Recalculare... vs DECIZIA initiala 227189 / 04.IUN.2021

 Decizia initiala

Talonul de Pensie luna JUL 2024

Decizia ....de Recalculare din 02.08.2024

primita in 26.aug.2024

S-au preluat Numarul Total de Puncte din Decizia anterioara : 35,0738. 

S-au adaugat 6,41117 puncte de stabilitate pentru Stagiul de Cotizare ce depaseste 25 ani, adica 

10 ani, 1 luna, 28 zile.

          Pensia cuvenita incepand cu 01.Sep.2024  (inclusiv diminuarea datorita anticiparii) este 2555 lei, deci fata de actuala pensie de 2182 lei rezulta o crestere bruta cu 373 lei.....

         

In sfarsit si Talonul de Pensie pe luna SEP....

.....Si cu asta...GATA (desi au uitat sa treaca pe Talon si la care? BANCA imi incasez pensia... Ocazie de noi intarzieri)

Dan BRÂNZEI : Inside is Geometry! Those who love it and those who don't are invited in!

 

Evaluarea nationala

          At the National Assessment for VIII-th grade graduates, the students received the following subjects: see the section "Subiecte si bareme" (Subjects and scales).

          Regarding the Mathematics test, the subject is in the

 "ENVIII_Matematica_2024_var_07_LRO.pdf" 

file, and the correction scale in the 

"ENVIII_Matematica_2024_bar_07_LRO.pdf"

file. There was also, like every year, a reserve option : see "Matematică rezervă - 27 iunie 2024" section; see files "ENVIII_Matematica_2024_var_02_LRO.pdf" and "ENVIII_Matematica_2024_bar_02_LRO.pdf".

          Two problems were considered more difficult. Problem III.4 and problem III.5.

In translation: 

                         "The attached figure shows the isosceles triangle $ABC$ with $AB=AC$.

                       The height from the top $A$ intersects the side $BC$ at point $D$ and $AD=BC$. 

                       The height from vertex $B$ intersects side $AC$ at point $E$. Heights $AD$

                        and $BE$ intersect at point $H$.

                          a) Show that angles $DAC$ and $EBC$ have the same measure.

                          b) Prove that $AH=3 \cdot HD$."

In translation:

                         "The attached figure shows the circle with center $O$, in which $CD$ is

                          the diameter. Point $B$ belongs to the circle so that lines $BO$ and $CD$

                          are perpendicular. Point $M$ belongs to the small arc $BC$, lines $DM$

                          and $BO$ intersect at point $N$, $DN=2 \cdot MN$ and $MN=4\;cm$.

                               a)  Show that the measure of the angle $CMD$ is $90^{\circ}$.

                               b)  Calculate the area of ​​the triangle $DON$."

               The full Subject and Scale are in the Images below.

 //////////////////////////

///////////////////////////////////////////////////////////////

                    Solution and comments CiP

                The solution to exercise III.4.a) is standard, noting that the angle $\measuredangle DAC$ is, in the right triangle $ACD$, the complement of the angle $\measuredangle C$, and the angle $\measuredangle EBC$ is, in the right triangle $BCE$, the complement of the same angle $\measuredangle C$.

          In exercise III.4.b) the key to the solution is the consideration of similar triangles $ACD$ and $BDH$. The order of the vertices in the relation of similarity is very important: respectively equal angles are opposed to respectively proportional sides. In more detail,

          first, $BD=DC=\frac{BC}{2}=\frac{AD}{2}$;

          secondly $\measuredangle BDH =90^{\circ}=\measuredangle ADC$ and 

$$\measuredangle HBD \underset{same\;angle}{=}\measuredangle EBC \underset{a)}{=}\measuredangle CAD.$$

Then from the last row above we have based on the "AA" criterion that

$$\Delta \;BDH \sim \Delta \;ADC$$

hence

$$\frac{BD}{AD}=\frac{DH}{DC}=\frac{BH}{AC}. \tag{$\sigma$}$$

But $\frac{BD}{AD}=\frac{\frac{BC}{2}}{BC}=\frac{1}{2}$, so from ($\sigma$) $\frac{HD}{DC}=\frac{1}{2}$. From here $HD=\frac{DC}{2}=\frac{BC}{4}=\frac{AD}{4}$.

          We get $AH=AD-HD=AD-\frac{AD}{4}=3 \cdot \frac{AD}{4}=3 \cdot HD.$

$\square$

               The key to exercise III.5.b) consists in determining the radius of the circle. For this, all the solutions I have come across are based on writing segments ratios. The most common is $\frac{OD}{ND}=\frac{MD}{CD}$. In the official scale, it results from writing the cosine of angle $\measuredangle C$ in two different triangles. Or one could use, less common with our students, the similarity of the triangles $DNO$ and $DCM$.

          But a simple idea can solve the problem without resorting to the most elementary knowledge of geometry. Join points C and M.

          $BO$ is the median of the segment $[CD]$, so the point $N$ on it verifies $NC=ND$. Hence 

$$NC=8.$$

          The right triangle $CMN$ has the leg $MN$ half of the hypotenuse $CN$ therefore it is a "30-60-90" triangle. (I have marked this on the figure with blue lines.) We still have $\measuredangle CND=180^{\circ}-60^{\circ}=120^{\circ}$. But in the isosceles triangle $CDN$, the median $NO$ from the vertex $N$ is also a bisector, so $\measuredangle CNO =\measuredangle DON =120^{\circ}\;:\;2=60^{\circ}$ (marked with red). It follows easily that the triangles $CNO$ and $DNO$ are of the same type. So $NO=4$ and we now have enough data to finish the problem.

$\blacksquare$

Juliusz Brzeziński Galois teori genom övningar // Galois Theory Through Exercises by Juliusz Brzeziński

                A source where you can download this book is here.

 

                I read an earlier version ("2015"), written for students. 

                 An interesting problem caught my attention: on page 102, problem 16.79 (in the "2015" version it is page 87, problem 16.40). But without indications of a solution. I have made posts with this type of problem here, here and here.

  

48-63-69 TRIANGLE : Să trăiești NAȘULE !! // Long live Godmother !!

           I used to follow this geometry blog. 

           I actually contributed to the publication of an problem (#963).

           The title is and is not a joke: I named my blog "ogeometrie" in the hope that those who will access "gogeometry" will also find me.

The figure above is a triangle with angles $\measuredangle {A}=48^{\circ},\;\measuredangle{B}=63^{\circ},\;\measuredangle{C}=69^{\circ}.$

          This triangle is the subject of Problem #1561. Today (2024-06-24) two solutions are already published. GregMarch 3, 2024 at 3:27 AM  and  Sumith PeirisMarch 5, 2024 at 11:48 AM. I also tried the solution, but no matter how much "angle chasing" I did, it was not enough. It is proven from the already published solutions that we must also appeal to a circumscribed circle.

Lemma ni Euler // Euler's lemma

 (filipineza) 

                    The name "EULER's Lemma" is given by me. This property of the circumcenter and the orthocenter is used in the demonstration of the "Circle of the Nine Points Theorem".

                    THEOREM (EULER's Lemma)

                    "We denote by O, G and H respectively the center of the 

                circumscribed circle, the center of gravity and the orthocenter

                 of a triangle ABC.

                     Let A' be the midpoint of side [BC]. We have equality

$$AH=2\cdot OA'. " \tag {e}$$

               Rem CiP

                In reality we have more, the relationship between vectors takes place

$$\overrightarrow{AH}=2 \cdot \overrightarrow{OA'}. \tag {E}$$

                    CiP Proof (following the text from the images at the beginning) 

               First we demonstrate

               Lemma 1 Two triangles with respectively parallel sides are similar.

                   Proof of Lemma_1

                  Let $A_1B_1C_1$ and $A_2B_2C_2$ be two triangles having

$$A_1B_1 \parallel A_2B_2\;,\;\;A_1C_1 \parallel A_2C_2\;,\;\;B_1C_1 \parallel B_2C_2\;. \tag{$\pi$}$$

Since [the angles] $\angle B_1A_1C_1,\;\angle B_2A_2C_2$ have respectively parallel sides, we have

$$\angle A_1 \equiv \angle A_2\;\;\;or \;\;\;\measuredangle A_1+\measuredangle A_2=180^{\circ} \tag {$\lambda_A$}$$

Analogously we have

$$\angle B_1 \equiv \angle B _2\;\;\;or \;\;\; \measuredangle B_1+\measuredangle B_2=180^{\circ} \tag{$\lambda_B$};$$

$$\angle C_1 \equiv \angle C_2\;\;\;or \;\;\; \measuredangle C_1+\measuredangle C_2=180^{\circ} . \tag{$\lambda_C$}$$

If in two of the relations $(\lambda_{A,B,C}$ the case of supplementarity would appear, e.g.

$$\measuredangle A_1+\measuredangle A_2=180^{\circ}=\measuredangle B_1+\measuredangle B_2$$

then

$\measuredangle C_1+\measuredangle C_2=180^{\circ}-(\measuredangle A_1+\measuredangle B_1)+180^{\circ}-(\measuredangle A_2+\measuredangle B_2)=(180^{\circ}-\measuredangle A_1-\measuredangle A_2)+(180^{\circ}-\measuredangle B_1-\measuredangle B_2)=0$

impossible. 

          Therefore, in two of the relations $(\lambda_{A,B,C})$ we have $"\equiv "$ (and implicitly, based on the sum of the angles of a triangle we have $"\equiv "$ in the remaining case as well). The triangles will be similar.  

$\square $<end Proof of L_1>

             Lemma 2 Two triangles with respective perpendicular sides are similar.

               Same justification.

                    Proof of (e) [apud E. RUSU, "Cum gindim si rezolvam 200 de probleme", Ed. Albatros, Buc., 1972, pp 226-227, Probl. 176]

              In [the triangles] $\Delta AHB$ and $\Delta A'OB'$ we have

$$AH \perp BC \perp OA'$$

so $AH \parallel AB$; analogous $BH \parallel OB'$. Then $A'B' \parallel AB$ from the middle theorem. 

        With the Lemma 1 result $\Delta AHB \sim \Delta A'OB'$, having the similarity ratio $\frac{AB}{A'B'}=2.$ So also $\frac{AH}{A'O}=2.$

               Remark CiP The fact that (e) also occurs as vectors  (i.e. (E)) results from the observations:

($\alpha$)   $O,\;H\;\in Int \Delta ABC\;\;\Leftrightarrow\;\widehat{BAC},\;\widehat{ABC},\;\widehat {BCA}\;<90^{\circ}$

($\beta$)   $O,\;H\;\in Ext \Delta ABC\;\;\Leftrightarrow\;\widehat{A}\;>90^{\circ}\;or\;\widehat{B}\;>90^{\circ}\;or\; \widehat{C}\;>90^{\circ}$

($\gamma$)  $H=A,\;O=A'\;\;\Leftrightarrow\;\widehat{A}=90^{\circ}.$

Moreover, in the case $(\beta)$ we can specify:

          $\widehat{BAC}>90^{\circ}\;\Rightarrow\;H\in Int \{opposite\;angle\;at\;the\;apex\;with\; \angle BAC\}$

                                                                                   and $O\in \{the\;half-plane\;determined\;by\;the\; line\;BC,\;which\;does\;\underline{not}\;contain\;the\;vertex\;A \}$

$\blacksquare \blacksquare$