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Triangular Numbers in action
integrated_JRC   29
N an hour ago by Aiden-1089
Source: RMO 2018 P5
Find all natural numbers $n$ such that $1+[\sqrt{2n}]~$ divides $2n$.

( For any real number $x$ , $[x]$ denotes the largest integer not exceeding $x$. )
29 replies
integrated_JRC
Oct 7, 2018
Aiden-1089
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A+b+c=0
Xixas   18
N Apr 17, 2025 by sqing
Source: Lithuanian Mathematical Olympiad 2006
Show that if $a+b+c=0$ then $(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b})(\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c})=9$.
18 replies
Xixas
Apr 12, 2006
sqing
Apr 17, 2025
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A+b+c=0
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Source: Lithuanian Mathematical Olympiad 2006
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Xixas
297 posts
Xixas
#1 h  Apr 12, 2006, 2:31 PM • 2 Y
Y by Adventure10, Mango247
Show that if $a+b+c=0$ then $(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b})(\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c})=9$.
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Michael
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Michael
#2 h  Apr 12, 2006, 3:15 PM • 3 Y
Y by Mathskidd, Adventure10, Mango247
Xixas wrote:
Show that if $a+b+c=0$ then $(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b})(\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c})=9$.

The following identity is easily verified:
\[ {{y-z}\over{x}}+{{z-x}\over{y}}+{{x-y}\over{z}}=-{{(y-z)(z-x)(x-y)}\over{xyz}}\quad\hbox{for}\quad x,y,z\neq 0.\leqno (1) \]
Set $x=b-c,y=c-a,z=a-b$ in (1). In view of $a+b+c=0$ we have $y-z=b+c-2a=-3a,z-x=-3b,x-y=-3c$, and (1) becomes
\[ {{-3a}\over{b-c}}+{{-3b}\over{c-a}}+{{-3c}\over{a-b}}=-{{(-3a)(-3b)(-3c)}\over{(b-c)(c-a)(a-b)}}, \]
i.e.,
\[ {{a}\over{b-c}}+{{b}\over{c-a}}+{{c}\over{a-b}}=-{{9abc}\over{(b-c)(c-a)(a-b)}}. \]
Now set $x=a,y=b,z=c$ in (1):
\[ {{b-c}\over{a}}+{{c-a}\over{b}}+{{a-b}\over{c}}=-{{(b-c)(c-a)(a-b)}\over{abc}}. \]
Multiplying the last two equalities we get the claimed one.
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stergiu
1648 posts
stergiu
#3 h  Apr 12, 2006, 5:53 PM • 2 Y
Y by Adventure10, Mango247
Nice work !

The problem comes from the Austrian- Polish competition.I have recently read a very similar - almost the same - solution in an older issue of Crux.

Babis
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indybar
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indybar
#4 h  Apr 13, 2006, 9:12 AM • 2 Y
Y by Adventure10, Mango247
Indeed :D
http://www.kalva.demon.co.uk/aus-pol/ap85.html
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silouan
3952 posts
silouan
#5 h  Apr 13, 2006, 4:49 PM • 2 Y
Y by Adventure10, Mango247
The problem indeed is classic and quite well-known (at least in Greece ).
Another solution is . Let $S$ the given product

Notice that $\frac{a}{b-c}(\frac{c-a}{b}+\frac{a-b}{c})=\frac{2a^2}{bc}$

Similar for the other two .

We expand the given product and we use he above .so
$S=3+\frac{2(a^3+b^3+c^3)}{abc}$ .But $a^3+b^3+c^3=3abc$ when $a+b+c=0$

so $S=9$ and we are done
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ArcticMonkey
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ArcticMonkey
#6 h  Apr 22, 2006, 12:19 PM • 2 Y
Y by Adventure10, Mango247
Another solution..

Click to reveal hidden text
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Xixas
297 posts
Xixas
#7 h  Apr 22, 2006, 3:42 PM • 3 Y
Y by Kunihiko_Chikaya, Adventure10, Mango247
Ghm :wacko: Can we apply AM-HM to potentialy negative $x, y, z$?
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ArcticMonkey
46 posts
ArcticMonkey
#8 h  Apr 22, 2006, 4:18 PM • 2 Y
Y by Adventure10, Mango247
Hm I have to admit I really didn't consider that restriction. Isn't it only for the inequalities with geometric mean? :? Actually it is also possible to proof this inequality using Cauchy-Schwarz: (Though it seems that this time we really have a restriction because of those d$a$mn sqrt{} things. :))

$(x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})\geq (\sqrt{x}\cdot\sqrt{\frac{1}{x}} +\sqrt{y}\cdot\sqrt{\frac{1}{y}}+ \sqrt{z}\cdot\sqrt{\frac{1}{z}})^2=9$

[EDIT]

Ok, we just have to try $(-10, 1, 1)$ and see that my solution sucks. :oops:
This post has been edited 1 time. Last edited by ArcticMonkey, Apr 22, 2006, 4:23 PM
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Xixas
297 posts
Xixas
#9 h  Apr 22, 2006, 4:20 PM • 2 Y
Y by Adventure10, Mango247
That's the reason that prevented me from using Cauchy-Schwarz on the contest.
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dgreenb801
1896 posts
dgreenb801
#10 h  Mar 29, 2009, 8:36 PM • 2 Y
Y by Adventure10, Mango247
Since $ a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$
It follows that if $ a + b + c = 0$, then $ a^3 + b^3 + c^3 = 3abc$
The numerator of the left factor is
$ - \sum (a^3 - a^2b - a^2c + abc) = - \sum (a^3 - a^2(b + c) + abc) = - \sum (2a^3 + abc)$
Since $ 3abc = a^3 + b^3 + c^3$, the left factor is
$ \frac { - 3 \sum a^3}{(a - b)(b - c)(c - a)}$
The right factor is
$ \frac {b^2c - bc^2 + c^2a - ca^2 + a^2b - ab^2}{abc}$
But the denominator, $ abc = \frac {1}{3} \sum a^3$. Everything cancels, and we're left with 9.
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CrazyBoy
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CrazyBoy
#11 h  Feb 12, 2012, 5:15 PM • 2 Y
Y by Adventure10, Mango247
silouan wrote:
Notice that $\frac{a}{b-c}(\frac{c-a}{b}+\frac{a-b}{c})=\frac{2a^2}{bc}$
How did you come up with that ?
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sqing
42194 posts
sqing
#12 h  Aug 9, 2014, 7:34 AM • 3 Y
Y by boss_1998, Adventure10, Mango247
Xixas wrote:
Show that if $a+b+c=0$ then $(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b})(\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c})=9$.
$\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}=-\frac{9abc}{(a-b)(b-c)(c-a)},$

$\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c}=-\frac{(a-b)(b-c)(c-a)}{abc}.$
here
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Kunihiko_Chikaya
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Kunihiko_Chikaya
#13 h  Aug 3, 2015, 3:29 AM • 2 Y
Y by Adventure10, Mango247
The problem is very classic. In Japanese univesity entrance exam, Nara University entrance exam/Medicine has posed te same problem in early 1960 or later 1950.

I would like to know the original problem or articles regarding the problem.

Are there anyone having the information ？

Any information would be appreciated.

Thanks in advance.

kunny
This post has been edited 2 times. Last edited by Kunihiko_Chikaya, Aug 3, 2015, 8:17 AM
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arqady
30252 posts
arqady
#14 h  Aug 3, 2015, 5:41 AM • 3 Y
Y by Kunihiko_Chikaya, Adventure10, Mango247
I have found this problem in the following book.
В.Кречмар, "Задачник по алгебре", (1937 year, page 15, problem 18).
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soupynoodles
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soupynoodles
#15 h  Aug 6, 2015, 4:34 PM • 2 Y
Y by Adventure10, Mango247
Since a+b+c=0, c=-a-b
Therefore, substituting this in you get the attached image.
Take out 9 from the numerator, cancel 2a^4 + 5(a^3)b - 5a(b^3) - 2b^4 from the top and bottom... and you're done!
Attachments:
This post has been edited 1 time. Last edited by soupynoodles, Aug 6, 2015, 4:36 PM
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sqing
42194 posts
sqing
#16 h  Mar 13, 2017, 10:41 PM • 3 Y
Y by Adventure10, Mango247, alice211
Xixas wrote:
Show that if $a+b+c=0$ then $(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b})(\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c})=9$.
Kyrgyzstan national 2016
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sqing
42194 posts
sqing
#17 h  Sep 4, 2017, 12:27 AM • 3 Y
Y by Adventure10, Mango247, AlexCenteno2007
Let $a,b,c$ be real numbers such that $a+b+c=0$ .Prove that$$\frac{a^2+b^2+c^2}{2}\cdot \frac{a^5+b^5+c^5}{5}=\frac{a^7+b^7+c^7}{7}$$
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AlexCenteno2007
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AlexCenteno2007
#18 h  Apr 17, 2025, 12:06 AM
Y by
sqing wrote:
Let $a,b,c$ be real numbers such that $a+b+c=0$ .Prove that$$\frac{a^2+b^2+c^2}{2}\cdot \frac{a^5+b^5+c^5}{5}=\frac{a^7+b^7+c^7}{7}$$

Solution:
By elementary symmetric polynomials we have S1=$a+b+c=0$
S2=-2$\sigma(2)$ and
S3=3$\sigma(3)$
By substituting in a direct way we have to
$$\frac{S2}{2}\cdot \frac{S2}{2}\cdot\frac{S3}{3}=\frac{a^7+b^7+c^7}{7}$$But I know that S7=$7\sigma(3)\cdot \sigma(2)^2$
From this the result is immediate.
This post has been edited 3 times. Last edited by AlexCenteno2007, Apr 17, 2025, 12:08 AM
Reason: Error
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sqing
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sqing
#19 h  Apr 17, 2025, 2:18 AM
Y by
Thanks.
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