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)

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          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!