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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
Yesterday at 11:16 PM
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

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[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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0 replies
jlacosta
Yesterday at 11:16 PM
0 replies
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
positive integers forming a perfect square
cielblue   0
30 minutes ago
Find all positive integers $n$ such that $2^n-n^2+1$ is a perfect square.
0 replies
cielblue
30 minutes ago
0 replies
Function equation
LeDuonggg   6
N 41 minutes ago by MathLuis
Find all functions $f: \mathbb{R^+} \rightarrow \mathbb{R^+}$ , such that for all $x,y>0$:
\[ f(x+f(y))=\dfrac{f(x)}{1+f(xy)}\]
6 replies
LeDuonggg
Yesterday at 2:59 PM
MathLuis
41 minutes ago
A nice and easy gem off of StackExchange
NamelyOrange   0
42 minutes ago
Source: https://math.stackexchange.com/questions/3818796/
Define $S$ as the set of all numbers of the form $2^i5^j$ for some nonnegative $i$ and $j$. Find (with proof) all pairs $(m,n)$ such that $m,n\in S$ and $m-n=1$.
0 replies
NamelyOrange
42 minutes ago
0 replies
at everystep a, b, c are replaced by a+\gcd(b,c), b+\gcd(a,c), c+\gcd(a,b)
NJAX   8
N an hour ago by Assassino9931
Source: 2nd Al-Khwarizmi International Junior Mathematical Olympiad 2024, Day2, Problem 8
Three positive integers are written on the board. In every minute, instead of the numbers $a, b, c$, Elbek writes $a+\gcd(b,c), b+\gcd(a,c), c+\gcd(a,b)$ . Prove that there will be two numbers on the board after some minutes, such that one is divisible by the other.
Note. $\gcd(x,y)$ - Greatest common divisor of numbers $x$ and $y$

Proposed by Sergey Berlov, Russia
8 replies
NJAX
May 31, 2024
Assassino9931
an hour ago
Increments and Decrements in Square Grid
ike.chen   23
N an hour ago by Andyexists
Source: ISL 2022/C3
In each square of a garden shaped like a $2022 \times 2022$ board, there is initially a tree of height $0$. A gardener and a lumberjack alternate turns playing the following game, with the gardener taking the first turn:
[list]
[*] The gardener chooses a square in the garden. Each tree on that square and all the surrounding squares (of which there are at most eight) then becomes one unit taller.
[*] The lumberjack then chooses four different squares on the board. Each tree of positive height on those squares then becomes one unit shorter.
[/list]
We say that a tree is majestic if its height is at least $10^6$. Determine the largest $K$ such that the gardener can ensure there are eventually $K$ majestic trees on the board, no matter how the lumberjack plays.
23 replies
ike.chen
Jul 9, 2023
Andyexists
an hour ago
4-var inequality
RainbowNeos   5
N 2 hours ago by RainbowNeos
Given $a,b,c,d>0$, show that
\[\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}\geq 4+\frac{8(a-c)^2}{(a+b+c+d)^2}.\]
5 replies
RainbowNeos
Yesterday at 9:31 AM
RainbowNeos
2 hours ago
Hard diophant equation
MuradSafarli   2
N 2 hours ago by MuradSafarli
Find all positive integers $x, y, z, t$ such that the equation

$$
2017^x + 6^y + 2^z = 2025^t
$$
is satisfied.
2 replies
MuradSafarli
3 hours ago
MuradSafarli
2 hours ago
An almost identity polynomial
nAalniaOMliO   6
N 2 hours ago by Primeniyazidayi
Source: Belarusian National Olympiad 2025
Let $n$ be a positive integer and $P(x)$ be a polynomial with integer coefficients such that $P(1)=1,P(2)=2,\ldots,P(n)=n$.
Prove that $P(0)$ is divisible by $2 \cdot 3 \cdot \ldots \cdot n$.
6 replies
nAalniaOMliO
Mar 28, 2025
Primeniyazidayi
2 hours ago
Euler's function
luutrongphuc   2
N 3 hours ago by KevinYang2.71
Find all real numbers \(\alpha\) such that for every positive real \(c\), there exists an integer \(n>1\) satisfying
\[
\frac{\varphi(n!)}{n^\alpha\,(n-1)!} \;>\; c.
\]
2 replies
luutrongphuc
5 hours ago
KevinYang2.71
3 hours ago
Wot n' Minimization
y-is-the-best-_   25
N 3 hours ago by john0512
Source: IMO SL 2019 A3
Let $n \geqslant 3$ be a positive integer and let $\left(a_{1}, a_{2}, \ldots, a_{n}\right)$ be a strictly increasing sequence of $n$ positive real numbers with sum equal to 2. Let $X$ be a subset of $\{1,2, \ldots, n\}$ such that the value of
\[
\left|1-\sum_{i \in X} a_{i}\right|
\]is minimised. Prove that there exists a strictly increasing sequence of $n$ positive real numbers $\left(b_{1}, b_{2}, \ldots, b_{n}\right)$ with sum equal to 2 such that
\[
\sum_{i \in X} b_{i}=1.
\]
25 replies
y-is-the-best-_
Sep 23, 2020
john0512
3 hours ago
Line AT passes through either S_1 or S_2
v_Enhance   88
N 3 hours ago by bjump
Source: USA December TST for 57th IMO 2016, Problem 2
Let $ABC$ be a scalene triangle with circumcircle $\Omega$, and suppose the incircle of $ABC$ touches $BC$ at $D$. The angle bisector of $\angle A$ meets $BC$ and $\Omega$ at $E$ and $F$. The circumcircle of $\triangle DEF$ intersects the $A$-excircle at $S_1$, $S_2$, and $\Omega$ at $T \neq F$. Prove that line $AT$ passes through either $S_1$ or $S_2$.

Proposed by Evan Chen
88 replies
v_Enhance
Dec 21, 2015
bjump
3 hours ago
Inequality with a,b,c
GeoMorocco   4
N 3 hours ago by Natrium
Source: Morocco Training
Let $   a,b,c   $ be positive real numbers such that : $   ab+bc+ca=3   $ . Prove that : $$\frac{\sqrt{1+a^2}}{1+ab}+\frac{\sqrt{1+b^2}}{1+bc}+\frac{\sqrt{1+c^2}}{1+ca}\ge \sqrt{\frac{3(a+b+c)}{2}}$$
4 replies
GeoMorocco
Apr 11, 2025
Natrium
3 hours ago
China Northern MO 2009 p4 CNMO
parkjungmin   1
N 3 hours ago by WallyWalrus
Source: China Northern MO 2009 p4 CNMO P4
The problem is too difficult.
1 reply
parkjungmin
Apr 30, 2025
WallyWalrus
3 hours ago
Polynomial Squares
zacchro   26
N 3 hours ago by Mathandski
Source: USA December TST for IMO 2017, Problem 3, by Alison Miller
Let $P, Q \in \mathbb{R}[x]$ be relatively prime nonconstant polynomials. Show that there can be at most three real numbers $\lambda$ such that $P + \lambda Q$ is the square of a polynomial.

Alison Miller
26 replies
zacchro
Dec 11, 2016
Mathandski
3 hours ago
Convex hexagon with AD=BC+EF, BE=AF+CD, CF=DE+AB
Valentin Vornicu   16
N Dec 22, 2013 by dizzy
Source: Kazakhstan international contest 2006, Problem 6
Let $ ABCDEF$ be a convex hexagon such that $ AD = BC + EF$, $ BE = AF + CD$, $ CF = DE + AB$. Prove that:
\[ \frac {AB}{DE} = \frac {CD}{AF} = \frac {EF}{BC}.
\]
16 replies
Valentin Vornicu
Jan 22, 2006
dizzy
Dec 22, 2013
Convex hexagon with AD=BC+EF, BE=AF+CD, CF=DE+AB
G H J
Source: Kazakhstan international contest 2006, Problem 6
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Valentin Vornicu
7301 posts
#1 • 2 Y
Y by Adventure10, Mango247
Let $ ABCDEF$ be a convex hexagon such that $ AD = BC + EF$, $ BE = AF + CD$, $ CF = DE + AB$. Prove that:
\[ \frac {AB}{DE} = \frac {CD}{AF} = \frac {EF}{BC}.
\]
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Sung-yoon Kim
324 posts
#2 • 2 Y
Y by Adventure10 and 1 other user
It can be proved by using vectors. Actually the same idea was used in IMO 2003 #3.
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perfect_radio
2607 posts
#3 • 5 Y
Y by BSJL, Adventure10, Crazy4Hitman, pokpokben, and 1 other user
I had a similar idea, but I was very DUMB not to see what happened if I went on this path.

I got the INEQUALITIES, but I thought that adding them would never give something useful:

$\left| \overrightarrow{AD} \right| = \left| \overrightarrow{BC} \right| + \left| \overrightarrow{EF} \right| \geq \left| \overrightarrow{FC} + \overrightarrow{BE} \right|$. (I don't know why, but this inequality seemed too natural for me to yield anything; really don't know why)

Squaring and adding: $\left( \overrightarrow{AD}+\overrightarrow{EB}+\overrightarrow{CF} \right)^2 \leq 0$. Thus, $BC \| EF$ etc. Also, $BC \| DA$. The conclusion is now obvious.

Thanks, Sung-yoon Kim.
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Sung-yoon Kim
324 posts
#4 • 2 Y
Y by Adventure10 and 1 other user
Yeah I think you're on the right way..
Quote:
$\left| \overrightarrow{AD} \right| = \left| \overrightarrow{BC} \right| + \left| \overrightarrow{EF} \right| \geq \left| \overrightarrow{FC} + \overrightarrow{BE} \right|$
Consider a parellelogram BCFX. Then $BC+EF=XF+FE \geq XE=|\overrightarrow{XB}+\overrightarrow{BE}|=|\overrightarrow{CF}+\overrightarrow{BE}|$. The rest is same as yours.
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Tiks
1144 posts
#5 • 1 Y
Y by Adventure10
Guys I think we shoulld find sinthetical solution for this one,I havn't do it still :( ,but I will try again ;).
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Davron
484 posts
#6 • 2 Y
Y by Adventure10, Mango247
Yes it is a good idea to find a synthetic proof for this problem.

Davron
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jin
383 posts
#7 • 1 Y
Y by Adventure10
That's really hard to find a such solution.
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Tiks
1144 posts
#8 • 2 Y
Y by Adventure10, Mango247
Yes,it is realy hard job,if that can help you I know that ther are sintetic solution,I know a man who has found it,but I want find it from misalf,so I havn't ask him about that solution. :)
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barasawala
124 posts
#9 • 2 Y
Y by Adventure10, Mango247
perfect_radio wrote:
The conclusion is now obvious.

Well, after proving all these parallels, I still can't see why it's obvious...
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perfect_radio
2607 posts
#10 • 3 Y
Y by Adventure10, Crazy4Hitman, Mango247
Let $AD \cap CF = \left\{ T \right\}$. The following statements hold:
- $ABCT$ is a parallelogram;
- $DEFT$ is a parallelogram;
- $\triangle ATF \sim \triangle DTC$;
- the conclusion.
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maggot
72 posts
#11 • 1 Y
Y by Adventure10
perfect_radio wrote:
Also, $BC \| DA$.

And why is this?
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perfect_radio
2607 posts
#12 • 2 Y
Y by Adventure10 and 1 other user
$\left( \overrightarrow{AD}+\overrightarrow{EB}+\overrightarrow{CF}\right)^{2}\leq 0$ implies $\overrightarrow{AD}+\overrightarrow{EB}+\overrightarrow{CF}= \overrightarrow 0$. Also note that $\overrightarrow{EB}+\overrightarrow{CF}= \overrightarrow{CB}+\overrightarrow{EF}$.
Hence,
\[\overrightarrow{AD}= \overrightarrow{BC}+\overrightarrow{FE}.\]
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vittasko
1327 posts
#13 • 1 Y
Y by Adventure10
A simple construction of the configuration, as the problem states, is as follows:

A triangle $\bigtriangleup KLM$ is given and let $A$ be, a fixed point on the extension of the sideline $KL$ $($ $K,$ between $A,$ $L$ $).$

The line through the $A$ and parallel to $LM,$ intersects the sideline $KM$ at a point, so be it $B.$

The line through the $B$ and parallel to $KL,$ intersects the sideline $LM$ at a point, so be it $C.$

The line through the $C$ and parallel to $KM,$ intersects the sideline $KL$ at a point, so be it $D.$

The line through the $D$ and parallel to $LM,$ intersects the sideline $KM$ at a point, so be it $E.$

Through $A,$ $E$ now, we draw to lines parallel to $KM\equiv BE$ and $KL\equiv AD$ respectively, which intersect the sideline $LM,$ at points $F,$ $F'$ and we will prove that $F'\equiv F.$

It is easy to show that $CF = CM+MF = DE+AB$ $,(1)$ and $CF' = CL+LF' = AB+DE$ $,(2)$

From $(1),$ $(2)$ $\Longrightarrow$ $CF' = CF$ $\Longrightarrow$ $F'\equiv F.$

So, it has already been constructed the configuration as the problem states.

$\bullet$ From $AB\parallel DE$ $\Longrightarrow$ $\frac{AB}{DE}= \frac{KA}{KD}= \frac{KB}{KE}$ $,(3)$

But, $\frac{KA}{KD}= \frac{EF}{BC}$ $,(4)$ and $\frac{KB}{KE}= \frac{CD}{AF}$ $,(5)$

From $(3),)$ $(4),$ $(5)$ $\Longrightarrow$ $\frac{AB}{DE}= \frac{CD}{AF}= \frac{EF}{BC}$ and the proof is completed.

Kostas Vittas.
Attachments:
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prowler
312 posts
#14 • 2 Y
Y by Adventure10, Mango247
Lemma: Let $ABCD$ be arbitrary quadrilateral. Denote
$a,b,c,d$ it's sides and $e,f$ diagonals, then $a^{2}+c^{2}+2bd\geq e^{2}+f^{2}$.



Proof: Denote $M, N, P$ midpoints of $BC, AD, BD$, from
triangle inequality $MN\leq MP+NP$ or $2MN\leq b+d$. We will
remember Euler theorem for quadrilateral and midpoint of opposite
sides or diagonals: $b^{2}+d^{2}+e^{2}+f^{2}=a^{2}+c^{2}+4MN^{2}\leq a^{2}+c^{2}+(b+d)^{2}$. Equality is hold when $BC \parallel AD$ and our lemma is
proved.


Now we will solve the problem.

Denote $AB=a,BC=b,CD=c, DE=d, EF=e, AF=f, AC=x_{1}, BD=x_{2}, CE=x_{3}, DF=x_{4}, EA=x_{5}, FB=x_{6}, AD=l_{1}, BE=l_{2}, CF=l_{3}$.

We have
$l_{1}=b+e, l_{2}= c+f, l_{3}=a+f$. Apply our theorem for $ABCD$ and
$DEFA$ we have $a^{2}+c^{2}+2bl_{1}\geq x_{1}^{2}+x_{2}^{2}$ and
$d^{2}+f^{2}+2el_{1}\geq x_{4}^{2}+x_{5}^{2}$. Adding all of them:
\[a^{2}+c^{2}+d^{2}+f^{2}+2l_{1}^{2}\geq x_{1}^{2}+x_{2}^{2}+x_{4}^{2}+x_{5}^{2}\]
Summing all such quadrilaterals we get
\[2(a^{2}+b^{2}+c^{2}+d^{2}+e^{2}+f^{2})+2(l_{1}^{2}+l_{2}^{2}+l_{3}^{2})\geq 2(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}+x_{5}^{2}+x_{6}^{2}) \]
Next step is to apply our lemma for quadrilateral $BCEF$:
$x_{3}^{2}+x_{6}^{2}+2be\geq l_{2}^{2}+l_{3}^{2}$. Summing with inequalities for
$CDAF$, $DEAB$:
\[(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}+x_{5}^{2}+x_{6}^{2})+2ad+2be+2cf\geq 2(l_{1}^{2}+l_{2}^{2}+l_{3}^{2}) \]
Multiplying second inequality by $2$ and adding with first one
we get
\[2(a^{2}+b^{2}+c^{2}+d^{2}+e^{2}+f^{2})+4(ad+be+cf)\geq 2(l_{1}^{2}+l_{2}^{2}+l_{3}^{2}) \]
or just $2(a+d)^{2}+2(b+e)^{2}+2(c+f)^{2}\geq 2(l_{1}^{2}+l_{2}^{2}+l_{3}^{2})$

As we see the equality is hold and thus from lemma's case
of equality we get $BC \parallel AD \parallel EF$ and similarly other lines are
parallel. Let a line through $C$ parallel to $AB$ intersects $AD$ at
$P$, then $AP=b$ and thus $PD=e$ and $PD\parallel EF$. Hence $PDEF$ is
parallelogram and $C,P,F$ are collinear as are $A,P,D$. So triangles
$APF$ and $DPC$ are similar and from here
\[\frac{AP}{DP}=\frac{PF}{PC}=\frac{AF}{CD}\]
Rewriting in hexagon's sides:
\[\frac{BC}{EF}=\frac{DE}{AB}=\frac{AF}{CD}\]
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prowler
312 posts
#15 • 2 Y
Y by Adventure10, Mango247
This was problem was proposed for Mathematical Reflections 2 in 2006,
by Nairi Sedrakyan (Armenia).

Mathematical Reflections: http://www.awesomemath.org
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ISLAMIDIN
5 posts
#16 • 1 Y
Y by Adventure10
in a convex hexagon with AD=BC+EF, CF=DE+AB,BE=AF+CD.THERE WE CAN FIND EASILY.A simple construction of the configuration, as the problem states, is as follows:

A triangle \bigtriangleup KLM is given and let A be, a fixed point on the extension of the sideline KL ( K, between A, L ).

The line through the A and parallel to LM, intersects the sideline KM at a point, so be it B.

The line through the B and parallel to KL, intersects the sideline LM at a point, so be it C.

The line through the C and parallel to KM, intersects the sideline KL at a point, so be it D.

The line through the D and parallel to LM, intersects the sideline KM at a point, so be it E.

Through A, E now, we draw to lines parallel to KM\equiv BE and KL\equiv AD respectively, which intersect the sideline LM, at points F, F' and we will prove that F'\equiv F.

It is easy to show that CF = CM+MF = DE+AB ,(1) and CF' = CL+LF' = AB+DE ,(2)

From (1), (2) \Longrightarrow CF' = CF \Longrightarrow F'\equiv F.

So, it has already been constructed the configuration as the problem states.

\bullet From AB\parallel DE \Longrightarrow \frac{AB}{DE}= \frac{KA}{KD}= \frac{KB}{KE} ,(3)

But, \frac{KA}{KD}= \frac{EF}{BC} ,(4) and \frac{KB}{KE}= \frac{CD}{AF} ,(5)

From (3),) (4), (5) \Longrightarrow \frac{AB}{DE}= \frac{CD}{AF}= \frac{EF}{BC} and the proof is completed.
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dizzy
181 posts
#17 • 2 Y
Y by Adventure10, Mango247
prowler wrote:
Lemma: Let $ABCD$ be arbitrary quadrilateral. Denote
$a,b,c,d$ it's sides and $e,f$ diagonals, then $a^{2}+c^{2}+2bd\geq e^{2}+f^{2}$.



Proof: Denote $M, N, P$ midpoints of $BC, AD, BD$, from
triangle inequality $MN\leq MP+NP$ or $2MN\leq b+d$. We will
remember Euler theorem for quadrilateral and midpoint of opposite
sides or diagonals: $b^{2}+d^{2}+e^{2}+f^{2}=a^{2}+c^{2}+4MN^{2}\leq a^{2}+c^{2}+(b+d)^{2}$. Equality is hold when $BC \parallel AD$ and our lemma is
proved.

Sorry for reviving the old topic, but your lemma is not true, and therefore the whole solution as well. From Euler theorem we have not as you wrote $b^{2}+d^{2}+e^{2}+f^{2}=a^{2}+c^{2}+4MN^{2}$ but $ a^{2}+b^{2}+c^{2}+d^{2}=e^{2}+f^{2}+4MN^{2} $. So your lemma rewrites as $ e^2+f^2+2bd\geq a^2+c^2 $. :wink:
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