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miercuri, 20 octombrie 2021

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Дамашняя работа

Punë me shkrim në Matematikë në Semestrin e I -rë, klasën e 7 -të

Eine bedingte algebraische Identität : Ein Problem der 1985 Österreich-Polen Mathematischer Wettbewerb // Uwarunkowana tożsamość algebraiczna : Problem austriacko-polskiego konkursu matematycznego 1985

See the book

KUCZMA Marcin Emil

144 Problems of the Austrian-Polish Mathematics Competition 1978-1993
The ACADEMIC DISTRIBUTION CENTER, Freeland, Maryland, 1994

at Problem 8.1, page 15. Presented also in CRUX_V12n02_Feb(1986), page 19 (The Olympiad Corner:72).

The statement of the problem

"If $\;a\;,b\;,c\;$ are distinct real numbers whose sum is zero, prove that

$\left [\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c} \right ]\cdot \left [ \frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\right ]=9."\tag{1}$

SOLUTION CiP

Let's make the notations:

$A:=\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c},$

$B:=\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}.$

When we bring to the same denominator, the numerator of the expression $A$ is

$A_1=bc(b-c)+ca(c-a)+ab(a-b).$ But from the condition $a+b+c=0$ we have

$$c=-a-b,\;\;-c=a+b. \tag{2}$$

Replacing $c$ in $A_1$ we get

$A_1\overset{(2)}{=}b\cdot (-a-b)(2b+a)+(-a-b)\cdot a(-2a-b)+ab(a-b)=$

$=(-2b^3-3ab^2-a^2b)+(2a^3+3a ^2b+ab^2)+(a^2b-ab^2)=$

$=(2a^3-2b^3)+(3a^2b-3ab^2)=2(a-b)(a^2+ab+b^2)+3ab(a-b)=$

$=(a-b)(2a^2+5ab+2b^2).$

Let's also note that $2a^2+5ab+2b^2=(2a+b)(a+2b)=(a+a+b)(a+b+b)\overset{(2)}{=}(a-c)(b-c).\tag{*}$

In conclusion we have

$$A=\frac{(a-b)(b-c)(a-c)}{abc}.\tag{3}$$

When we bring to the same denominator, the numerator of the expression $B$ is

$B_1=a(c-a)(a-b)+b(a-b)(b-c)+c(b-c)(c-a)\;\underset{(*)}{\overset{(2)}{=}}\;a(-2a-b)(a-b)+$

$+(ab-b^2)(a+2b)+(-a-b)(-2a^2-5ab-2b^2)=(-2a^3+a^2b+ab^2)+(a^2b+$

$+ab^2-2b^3)+(2a^3+7a^2b+7ab^2+2b^3)=9a^2b+9ab^2=9ab(a+b)\underset{(2)}{=}-9abc.$

In conclusion we have

$$B=\frac{-9abc}{(b-c)(c-a)(a-b)}.\tag{4}$$

Multiplying the relations $(3)$ and $(4)$ we obtain

$$[A]\cdot [B]=\frac{(a-b)(b-c)(a-c)}{abc} \cdot \frac{-9abc}{(b-c)(c-a)(a-b)}=9.$$

$\blacksquare$

REMARKS CiP

joi, 15 iulie 2021

PROBLEM MA127 CRUX MATHEMATICORUM V47 No 6

Pag 275 - En

Pag 276 - Fr

ANSWER CiP

$\log_5 12=\frac{2\cdot a+b}{1-a}$

Solution CiP

We have $\log_5 12 =\log_5(2^2 \cdot 3)=\log_5 2^2+\log_5 3$, so

$\log_5 12 =2\cdot \log_5 2 + \log_5 3. \tag{1}$

First, $\frac{1}{a}=\log_2 10=\log_2 (2\cdot 5)=\log_2 2 +\log_2 5=1+\frac{1}{\log_5 2}$ so

$\frac{1}{\log_5 2}=\frac {1}{a}-1$, whence it results

$\log_5 2 =\frac {a}{1-a}. \tag{2}$

Secondly, $\frac{1}{b}=\log_3 10 =\log_3 2 +\log_3 5=\frac{\log_{10} 2}{\log_{10} 3}+\frac{1}{\log_5 3}$, so $\frac{1}{\log_5 3}=\frac {1}{b}-\frac {a}{b}$, whence it results

$\log_5 3=\frac{b}{1-a}. \tag{3}$

Replacing formulas (2) and (3) in formula (1) we get the answer.

$\blacksquare$

REMARK CiP

More briefly $\log_5 12=\frac{\log_{10} 12}{\log_{10} 5}=\frac{\log_{10} 2^2 \cdot 3}{\log_{10} \frac{10}{2}}=\frac{\log_{10} 2^2 +\log_{10} 3}{\log_{10} 10-\log_{10} 2}=\frac{2\cdot \log_{10} 2 +\log_{10} 3}{1-\log_{10} 2}$ and we get the same answer.

$\blacksquare \blacksquare$

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către eu

miercuri, 7 iulie 2021

PROBLEM OC531 - CRUX MATHEMATICORUM, V47n5

Pag 238 - En

Pag 239 - Fr

ANSWER CiP

If $7 \mid k$ then solutions are $(k,0),\;(k,k),\;(\frac{3k}{7},-\frac{k}{7})$,

and, if any, the pairs $(x_1,y_1),\;(k-x_1+y_1,y_1).$

If $7 \nmid k$ then the solution $(\frac{3k}{7},-\frac{k}{7})$(although rational),

is not made up of integers; the others remain valid.

Examples

If $k=-4$, the equation has solutions

$(-4,0),\;(-4,-4),\;(-1,-3),\;(-6,-3).$

If $k=7$, the equation has solutions

$(7,0),\;(7,7),\;(3,-1),\;(3,6),\;(10,6).$

Solution CiP

There are two pairs $(x,y)$ that obviously check the equation: $(k,0)$ and $(k,k)$.

The given equation is equivalent to

$$x^2-xy+2y^2-kx-ky=0\; ;\; x+y \neq 0. \tag{1} \label{eq1}$$

Let $(x_1,y_1)$ be a solution of equation (1), with the condition $x_1+y_1 \neq 0$. This means that the equation with unknow $x$

$$x^2-(k+y_1)x+2y_1^2-ky_1=0 \tag{2}$$

has the solution $x=x_1$. Being of the second degree for $x$, she has another solution $x_2$ which, according to Viete's Relations, she checks $x_1+x_2=k+y_1$, so $x_2=k-x_1+y_1.$

For the pair $(x_2,y_1)$ to fit the given problem it must be $x_2+y_1 \neq 0$. But $x_2+y_1=0 \; \Leftrightarrow \;k-x_1+2y_1=0 \; \Leftrightarrow \; x_1=k+2y_1$, and writing that $k+2y_1$ check equation (2) we immediately get $y_1=0$, then $x_1=k$.

So for any solution $(x_1,y_1) \neq (k,0)$ of the equation (1) we get another solution by the transformation

$$(x_1,y_1)\; \mapsto \; (k-x_1+y_1,y_1). \tag{3}$$

We have $(k,k) \overset{(3)}{\mapsto} (k-k+k,k)=(k,k)$, so by transforming (3) we do not get a NEW solution. Let's look for all the solutions that remain invariant through this transformation, so $k-x_1+y_1=x_1$, whence $y_1=2x_1-k$. If we write that $(x_1,2x_1-k)$ check equation (1)

$$x_1^2-x_1(2x_1-k)+2(2x_1-k)^2-kx_1-k(2x_1-k)=0$$

$\Leftrightarrow \;7x_1^2-10kx_1+3k^2=0$ we get $x_1=k$ or $x_1=\frac{3k}{7}$. In the secon case $y_1=-\frac{k}{7}$. So the only invariant pairs by transformation (3) are $(k,k)$ and $(\frac{3k}{7}, -\frac{k}{7})$.

So, if $k \;\vdots \; 7$ (means "k is divisible by 7"), then the equation (1) has three solutions $(k,0),\; (k,k),\;(\frac{3k}{7},-\frac{k}{7})$ to which, if there are others $(x_1,y_1)$, an even number of solutions $(x_1,y_1)$ and $(k-x_1+y_1,y_1)$ is added. In total an odd number.

Otherwise ($7 \nmid k$) the solution $(\frac{3k}{7},-\frac{k}{7})$ does not consits of integers, and the number of solutions is even.

$\blacksquare$

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către eu

marți, 16 martie 2021

PROBLEM 4611 CRUX MATHEMATICORUM , V47n02

Page 97 - En

Page 99 - Fr

ANSWER CiP

$\frac{1}{\sin^{4}\frac{\pi}{14}}+\frac{1}{\sin^{4}\frac{3\pi}{14}}+\frac{1}{\sin^{4}\frac{5\pi}{14}}=416$

Solution CiP

As we showed in a previous post, the numbers

$x_{1}=sin\frac{\pi}{14},\;x_{2}=-sin\frac{3\pi}{14},\;x_{3}=sin\frac{5\pi}{14}$

are the roots of the polynomial $8x^{3}-4x^{2}-4x+1$. Then the numbers $y_{k}=\frac{1}{x_{k}}\;,\;k=1,\;2,\;3$ are the roots of the "reverse" equation

$y^{3}-4\cdot y^{2}-4 \cdot y+8=0 \tag{1} \label{eq1}$

The expression to be calculated is equal to $\sum_{k=1}^{3}y_{k}^{4}$.

We write Vieta's formulas for the equation (1)

\begin{cases}y_{1}+y_{2}+y_{3}=4\;\;\; \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(2)\\y_{1}\cdot y_{2}+y_{2}\cdot y_{3}+y_{3} \cdot y_{1}=-4\;\;(3)\\y_{1} \cdot y_{2} \cdot y_{3}=-8 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(4)\end{cases}

From \eqref{eq1} we deduce $y_{k}^{3}=4\cdot y_{k}^{2}+4 \cdot y_{k}-8$ $\Rightarrow$

$y_{k}^{4}=y_{k} \cdot y_{k}^{3}=4\cdot y_{k}^{3}+4 \cdot y_{k}^{2}-8 \cdot y_{k}=$

$=4(4y_{k}^{2}+4y_{k}-8)+4y_{k}^{2}-8y_{k}=20 y_{k}^{2}+8y_{k}-32.$

Then we get

$\sum _{k=1}^{3}y_{k}^{4}=\sum (20 \cdot y_{k}^{2}+8 \cdot y_{k}-32)=20 \cdot \sum_{k=1}^{3} y_{k}^{2}+8 \cdot \sum y_{k} -96 \overset{(2)}{=}$

$=20 \cdot [(\sum y_{k})^{2}-2 \cdot \underset{cycl}{\sum} y_{1}\cdot y_{2}]+8 \cdot 4-96\;\; \underset{(2),(3)}{==}\;\;20 \cdot [4^{2}-2(-4)]-64=416.$

$\blacksquare$

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Tracking Number: 11849
Received: Wed Mar 17 15:14:14 2021

From: Petre Ciobanu
Scoala Gimnaziala "Samuil Micu" SADU
Sibiu, Romania
Email: ptr.ciobanu@gmail.com

Type: Solve a Crux (Numbered) Problem
(problem 4611)

Files:
Pdf_4611.pdf
Sursa_Latex_4611.doc

Comments:
A close problem that the idea of solving suggested to me is Problem 28, page 115 from
E.J. BARBEAU - Polynomials (Springer-Verlag, 1989). I dedicated an article to this problem on my blog

https://ogeometrie-cip.blogspot.com/2021/02/an-equation-that-surprising-is-not.html

I also put on the blog the solution of this problem from you

https://ogeometrie-cip.blogspot.com/2021/03/problem-4611-crux-mathematicorum-v47n02.html