Soubgroups of multiplicative group Z_17

by cip1703, May 12, 2018, 3:00 PM

$\begin{tabular}{ c c c } cell1 & cell2 & cell3 \\ cell4 & cell5 & cell6 \\ cell7 & cell8 & cell9 \end{tabular}$

This post has been edited 1 time. Last edited by cip1703, May 12, 2018, 3:03 PM
Reason: correct

A Problem concerning Areas

by cip1703, Feb 24, 2018, 1:56 PM

LEMA 1
Among the areas marked on the figure we have the relationship
$S_{1}\cdot S_{3}=S_{2}\cdot S_{4}$

http://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvOS8yLzEzNDY2NDUyMTkxNDMwMTZmMThiN2JlMGJiYmQ5NDgzNTFhYjEyLmpwZw==&rn=UmVsYXRpaSBBcmlpLkpQRw==

Proof
Triangles $\Delta DAE,\Delta DCE$ have the same height, so $\frac{area(DAE)}{area(DCE)}=\frac{AE}{AC}$ . Once again, triangles $\Delta BAE,\Delta BCE$ have the same height, so $\frac{area(BAE)}{area(BCE)}=\frac{AE}{AC}$ . It follow $\frac {S_{1}}{S_{4}}=\frac{S_{2}}{S_{3}}$ from where the conclusion.

PROBLEM 1
Find x in the next figure.

http://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvZi9mLzdmMDVkMGVjY2MwNjM4ZWVhOTg2ZTdlZDllZTUyNTM1MmU2NTgzLmpwZw==&rn=Q2FsY3VsdWwgdW5laSBhcmlpLkpQRw==

Answer
$x=\frac{S_{1}\cdot S_{2}(S_{1}+S_{2}+2S_{3})}{S_{3}^{2}-S_{1}\cdot S_{2}}$ Solution
We were doing what he meant in https://artofproblemsolving.com/community/u44261h1597097p9921734.
In the following figure we have $area(FEG)=\frac{S_{1}\cdot S_{2}}{S_{3}}$ .
Denoting with $y=area(AEF)$ ,

http://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvOC8yL2RkNjY2NGZhZWE4NDM0YmYwMWUwY2E2YjYxOTg0YzBkYThkMWRhLmpwZw==&rn=Q2FsY3VsdWwgYWx0ZWkgYXJpaS5KUEc=

triangles $\Delta FAE,\Delta FCE$ have the same height, so $\frac{area(FAE)}{area(FCE)}=\frac{AE}{EC}$ and once again, triangles $\Delta BAE,\Delta BCE$ have the same height, so $\frac{area(BAE)}{area(BCE)}=\frac{AE}{EC}$ .
We write that the two fractions are equal, and we get the equation
$\frac{y}{\frac{S_{1}S_{2}}{S_{3}}+S_{2}}=\frac{y+S_{1}+ \frac{S_{1}S_{2}}{S_{3}}}{S_{2}+S_{3}}$ where the value of $y$ results. Further $x=\frac{S_{1}S_{2}}{S_{3}}+y$ , etc.

Here another such a problem, with the answer also a whole number
http://artofproblemsolving.com/wiki/index.php?title=File:2006amc10b23.gif#filehistory

Attachments:

This post has been edited 13 times. Last edited by cip1703, Feb 24, 2018, 5:30 PM
Reason: end edit

Simultaneous equations with floor functions

by cip1703, Jan 20, 2018, 10:49 AM

==Problem==
Let $\lfloor x \rfloor$ be the greatest integer less than or equal to $x$ and let $\{x\}=x-\lfloor x \rfloor$ . Solve

$\left\{\begin{array}{cc} x+\lfloor y \rfloor + \{z\}=a & (1)\\\\ \{x\}+ y+\lfloor z \rfloor=b & (2)\\\\ \lfloor x \rfloor+\{y\}+z=c & (3)\end{array}\right\|$

where $a$ , $b$ , $c$ are given real numbers.

==Solution==
.........ANSWER: $x=\lfloor \frac{a-b+c}{2} \rfloor+ \{\frac{a+b-c}{2}\}$ , $y=\lfloor \frac{a+b-c}{2} \rfloor+ \{\frac{-a+b+c}{2}\}$ , $z=\lfloor \frac{-a+b+c}{2}\rfloor+ \{\frac{a-b+c}{2}\}.$

...........For any $t\in\mathbb R$ we have
$t=\lfloor t \rfloor + \{t\}$ (4)
and conversely, if $t=k+\alpha, t\in\mathbb Z, \alpha \in [0;1)$ it follow that $\lfloor t \rfloor =k$ and $\{t\}=\alpha.$
Adding relations (1),(2),(3) we get $x+\lfloor y \rfloor + \{z\}+\{x\}+ y+\lfloor z \rfloor+\lfloor x \rfloor+\{y\}+z=a+b+c$
$\iff x+y+z=\frac{a+b+c}{2}$ (5).
Writing (5) as $x+\lfloor y \rfloor +\{z\}+\lfloor z \rfloor+\{y\}=\frac{a+b+c}{2}$ and taking account of (1) we get

$\lfloor z \rfloor+\{y\}=\frac{a+b+c}{2}-a=\frac{-a+b+c}{2}$ , and so $\lfloor z \rfloor=\lfloor\frac{-a+b+c}{2}\rfloor$ , and $\{y\}=\{\frac{-a+b+c}{2}\}$ .
In the same way we obtain $\lfloor x \rfloor+\{z\}=\frac{a-b+c}{2}$ , $\lfloor y \rfloor+\{z\}=\frac{a+b-c}{2}$
Taking account of (1) we have $x=\lfloor x \rfloor +\{x\}=...$ and we get the answer.

See also https://artofproblemsolving.com/texer/boixabdc

This post has been edited 12 times. Last edited by cip1703, Jan 20, 2018, 4:17 PM
Reason: correction

PROBLEM 27463, GMB 12/2017, p. 605

by cip1703, Jan 6, 2018, 1:59 PM

Determine strictly increasing functions $f:\mathbb{N}\rightarrow \mathbb{N}$ such that $\frac{f(x)+f(y)}{1+f(x+y)}$ is not null natural number, for any $x,y\in \mathbb{N}$ .

(Lucian Dragomir, O\c telu-Ro\c su \c si Nicolae St\u aniloiu, Boc\ sa)

Solution Cip

Answer: $f(x)=ax+1,$ $a\in \mathbb{N}, a\geq 1$

Solution:
For $x=y=0$ we have $\frac{2f(0)}{1+f(0)}\in \mathbb{N}^{\bigstar }$ $\Leftrightarrow 2-\frac{2}{1+f(0)}\in \mathbb{N}^{\bigstar }\Rightarrow f(0)=1.$ The fonction $f$ being strictly increasing, we have $f(x+y)>f(x),f(y)$ for all $x,y\geq 1$ , so $2f(x+y)>f(x)+f(y)$ , then
$0<\frac{f(x)+f(y)}{1+f(x+y)}<\frac{2f(x+y)}{1+f(x+y)}=2-\frac{2}{1+f(x+y)}<2.$ It follows that $\frac{f(x)+f(y)}{1+f(x+y)}=1$ , so
$f(x)+f(y)=f(x+y)+1,$ for all $x,y\geq 1$ .
For $y=1$ we get $f(x+1)-f(x)=f(1)-1$ , and noting $a=f(1)-1$ we have $a> f(0)-1=0$ . From $f(x+1)=f(x)+a$ we get by mathematical induction that $f(x)=ax+1,$ for all $x \in \mathbb{N}$ .
Conversely, all such fonctions satisfies the terms of the statement.
$\raggedleft \blacksquare$

This post has been edited 3 times. Last edited by cip1703, Jan 17, 2018, 6:01 AM
Reason: typo mistake

CIP1703