where 
and![\[(x-y)[f(x)+f(y)]\leqslant f(x^2-y^2), \forall x,y \in \mathbb{R}\]](http://latex.artofproblemsolving.com/8/0/b/80bb1d7a992c9683a0c2a07b2febe6271f4e60f5.png)
such that 
Y by byron-aj-tom
Prove that for all real numbers 
and 
verifying 
. 
and 
verifying 
. 
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D1019 : Dominoes 2*1
Dattier 5
N
by polishedhardwoodtable
I have a 9*9 grid like this one:
IMAGE
We choose 5 white squares on the lower triangle, 5 black squares on the upper triangle and one on the diagonal, which we remove from the grid.
Like for example here:
IMAGE
Can we completely cover the grid remove from these 11 squares with 2*1 dominoes like this one:
IMAGE
IMAGE
We choose 5 white squares on the lower triangle, 5 black squares on the upper triangle and one on the diagonal, which we remove from the grid.
Like for example here:
IMAGE
Can we completely cover the grid remove from these 11 squares with 2*1 dominoes like this one:
IMAGE
5 replies
Goes through fixed points
CheshireOrb 5
N
by HoRI_DA_GRe8
Source: Vietnam TST 2021 P5
Given a fixed circle 
and two fixed points 
on that circle, let 
be a moving point on such that 
is acute and scalene. Let 
be the midpoint of 
and let 
be the three heights of . In two rays 
, we pick respectively 
such that 
. Let 
be the intersection of 
and 
, and let 
be the second intersection of 
and 
.
a) Show that the circle
always goes through a fixed point.
b) Let
intersects at 
. In the tangent line through 
of 
, we pick 
such that 
. Let 
be the center of 
. Show that 
always goes through a fixed point.
and two fixed points 
on that circle, let 
be a moving point on such that 
is acute and scalene. Let 
be the midpoint of 
and let 
be the three heights of . In two rays 
, we pick respectively 
such that 
. Let 
be the intersection of 
and 
, and let 
be the second intersection of 
and 
.a) Show that the circle
always goes through a fixed point.b) Let
intersects at 
. In the tangent line through 
of 
, we pick 
such that 
. Let 
be the center of 
. Show that 
always goes through a fixed point.
5 replies
n=y^2+108
Havu 6
N
by ektorasmiliotis
Given the positive integer 
where 
.
Prove that
cannot be a perfect cube of a positive integer.
where 
.Prove that
cannot be a perfect cube of a positive integer.
6 replies
Valuable subsets of segments in [1;n]
NO_SQUARES 0
Source: Russian May TST to IMO 2023; group of candidates P6; group of non-candidates P8
The integer 
is given. Let be set of all 
segments of real line of type ![$[i, j]$](//latex.artofproblemsolving.com/a/6/4/a644edd67caa59eaded765c417b6d9822f782ec2.png)
, where 
and 
are integers, 
. A subset 
is said to be valuable if the intersection of any two segments from 
is either empty, or is a segment of nonzero length belonging to . Find the number of valuable subsets of set .
is given. Let be set of all 
segments of real line of type ![$[i, j]$](http://latex.artofproblemsolving.com/a/6/4/a644edd67caa59eaded765c417b6d9822f782ec2.png)
, where 
and 
are integers, 
. A subset 
is said to be valuable if the intersection of any two segments from 
is either empty, or is a segment of nonzero length belonging to . Find the number of valuable subsets of set .
0 replies
Fneqn or Realpoly?
Mathandski 2
N
by jasperE3
Source: India, not sure which year. Found in OTIS pset
Find all polynomials 
with real coefficients obeying
![\[P(x) P(x+1) = P(x^2 + x + 1)\]](//latex.artofproblemsolving.com/b/e/f/beff3420fa8394f4f98c031d81eea6a5a3b8d28c.png)
for all real numbers 
.
with real coefficients obeying![\[P(x) P(x+1) = P(x^2 + x + 1)\]](http://latex.artofproblemsolving.com/b/e/f/beff3420fa8394f4f98c031d81eea6a5a3b8d28c.png)
for all real numbers 
.
2 replies
D1018 : Can you do that ?
Dattier 1
N
by Dattier
Source: les dattes à Dattier
We can find 
, such that 
and 
.
For example :
Can you find such that 
is prime, 
with 
and ?
, such that 
and 
.For example :
Can you find
such that 
is prime, 
with 
and ?
1 reply
D1010 : How it is possible ?
Dattier 14
N
by ehuseyinyigit
Source: les dattes à Dattier
Is it true that
?
A=1728400904217815186787639216753921417860004366580219212750904
024377969478249664644267971025952530803647043121025959018172048
336953969062151534282052863307398281681465366665810775710867856
720572225880311472925624694183944650261079955759251769111321319
421445397848518597584590900951222557860592579005088853698315463
815905425095325508106272375728975
B=2275643401548081847207782760491442295266487354750527085289354
965376765188468052271190172787064418854789322484305145310707614
546573398182642923893780527037224143380886260467760991228567577
953725945090125797351518670892779468968705801340068681556238850
340398780828104506916965606659768601942798676554332768254089685
307970609932846902
?A=1728400904217815186787639216753921417860004366580219212750904
024377969478249664644267971025952530803647043121025959018172048
336953969062151534282052863307398281681465366665810775710867856
720572225880311472925624694183944650261079955759251769111321319
421445397848518597584590900951222557860592579005088853698315463
815905425095325508106272375728975
B=2275643401548081847207782760491442295266487354750527085289354
965376765188468052271190172787064418854789322484305145310707614
546573398182642923893780527037224143380886260467760991228567577
953725945090125797351518670892779468968705801340068681556238850
340398780828104506916965606659768601942798676554332768254089685
307970609932846902
14 replies
Cool one
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Y by
Very easy , Just use the geometric Serie Expansion
(a^(n+1) - b^(n+1)) / (a-b) = a^n + b * a^(n-1) + b^2 * a^(n-2) + ... + a * b^(n-1) + b^n
Since a > b, and Both are positive, this is > (n+1) * b^n and < (n+1) * a^n, And we are done ! Easy Peasy Lemon Squeasy as they say in Kentucky xD
(a^(n+1) - b^(n+1)) / (a-b) = a^n + b * a^(n-1) + b^2 * a^(n-2) + ... + a * b^(n-1) + b^n
Since a > b, and Both are positive, this is > (n+1) * b^n and < (n+1) * a^n, And we are done ! Easy Peasy Lemon Squeasy as they say in Kentucky xD
This post has been edited 1 time. Last edited by vsarg, Mar 15, 2025, 10:12 PM
Reason: Many mistake for me. I edit to ffix it
Reason: Many mistake for me. I edit to ffix it
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#3
Mar 15, 2025, 10:32 PM
Y by
vsarg wrote:
Very easy , Just use the geometric Serie Expansion
(a^(n+1) - b^(n+1)) / (a-b) = a^n + b * a^(n-1) + b^2 * a^(n-2) + ... + a * b^(n-1) + b^n
Since a > b, and Both are positive, this is > (n+1) * b^n and < (n+1) * a^n, And we are done ! Easy Peasy Lemon Squeasy as they say in Kentucky xD
(a^(n+1) - b^(n+1)) / (a-b) = a^n + b * a^(n-1) + b^2 * a^(n-2) + ... + a * b^(n-1) + b^n
Since a > b, and Both are positive, this is > (n+1) * b^n and < (n+1) * a^n, And we are done ! Easy Peasy Lemon Squeasy as they say in Kentucky xD
Not valid you didnt prove.
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Y by
vsarg wrote:
Very easy , Just use the geometric Serie Expansion
(a^(n+1) - b^(n+1)) / (a-b) = a^n + b * a^(n-1) + b^2 * a^(n-2) + ... + a * b^(n-1) + b^n
Since a > b, and Both are positive, this is > (n+1) * b^n and < (n+1) * a^n, And we are done ! Easy Peasy Lemon Squeasy as they say in Kentucky xD
(a^(n+1) - b^(n+1)) / (a-b) = a^n + b * a^(n-1) + b^2 * a^(n-2) + ... + a * b^(n-1) + b^n
Since a > b, and Both are positive, this is > (n+1) * b^n and < (n+1) * a^n, And we are done ! Easy Peasy Lemon Squeasy as they say in Kentucky xD
EXACTLY WHY IS IT GREATER THAT
?This post has been edited 1 time. Last edited by MTA_2024, Mar 15, 2025, 10:42 PM
Reason: Miswriting
Reason: Miswriting
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#5
Mar 15, 2025, 11:55 PM
Y by
EaZ_Shadow wrote:
vsarg wrote:
Very easy , Just use the geometric Serie Expansion
(a^(n+1) - b^(n+1)) / (a-b) = a^n + b * a^(n-1) + b^2 * a^(n-2) + ... + a * b^(n-1) + b^n
Since a > b, and Both are positive, this is > (n+1) * b^n and < (n+1) * a^n, And we are done ! Easy Peasy Lemon Squeasy as they say in Kentucky xD
(a^(n+1) - b^(n+1)) / (a-b) = a^n + b * a^(n-1) + b^2 * a^(n-2) + ... + a * b^(n-1) + b^n
Since a > b, and Both are positive, this is > (n+1) * b^n and < (n+1) * a^n, And we are done ! Easy Peasy Lemon Squeasy as they say in Kentucky xD
Not valid you didnt prove.
His proof is valid (and imo explained thoroughly enough).
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#6
Mar 15, 2025, 11:59 PM
Y by
MTA_2024 wrote:
Prove that for all real numbers and verifying .
and verifying . bro! the middle term is a^n + a^{n-1}b + ... + b^n, so it has 11 terms made up of mixed powers of a and b. Because a>b, then the middle term is larger than (n+1)b^n. Because b<a, the middle term is less than (n+1)a^n. Easy Peasy Lemon Squeezy! Now that's rizz!
I agree! vsarg's proof is thorough and complete! Anyone who says the opposite is just a hater unwilling to embrace the fundamental property of multiplication and factorisations!
This post has been edited 1 time. Last edited by ohiorizzler1434, Mar 15, 2025, 11:59 PM
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#8
Mar 20, 2025, 8:52 AM
Y by
This is the natural solution above! I have an idea that is not only valid for natural numbers ! Let's see! Consider the function 
on the interval ![$[a;b]$](//latex.artofproblemsolving.com/0/0/6/0066f4b2714f88cdaf883b00f877aa1355b7f9ef.png)
. If we apply Lagrange's finite growth theorem, then there is a 
in the interval 
for which 
But since 
we already got the problem. And we didn't have to use the abbreviated calculation formula anywhere, which is only valid for natural numbers !
! Let's see! Consider the function 
on the interval ![$[a;b]$](http://latex.artofproblemsolving.com/0/0/6/0066f4b2714f88cdaf883b00f877aa1355b7f9ef.png)
. If we apply Lagrange's finite growth theorem, then there is a 
in the interval 
for which 
But since 
we already got the problem. And we didn't have to use the abbreviated calculation formula anywhere, which is only valid for natural numbers !This post has been edited 1 time. Last edited by invisibleman, Mar 20, 2025, 8:53 AM
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#9
Mar 20, 2025, 9:58 AM
Y by
what the sigam? what is lagrange finite growth theorem? it does not appear in google saerch.
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#10
Mar 20, 2025, 10:56 AM
Y by
ohiorizzler1434 wrote:
what the sigam? what is lagrange finite growth theorem? it does not appear in google saerch.
Dear friend! Maybe I didn't express myself clearly. I meant this:
https://en.wikipedia.org/wiki/Mean_value_theorem
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Y by
ohiorizzler1434 wrote:
MTA_2024 wrote:
Prove that for all real numbers and verifying .
and verifying . bro! the middle term is a^n + a^{n-1}b + ... + b^n, so it has 11 terms made up of mixed powers of a and b. Because a>b, then the middle term is larger than (n+1)b^n. Because b<a, the middle term is less than (n+1)a^n. Easy Peasy Lemon Squeezy! Now that's rizz!
I agree! vsarg's proof is thorough and complete! Anyone who says the opposite is just a hater unwilling to embrace the fundamental property of multiplication and factorisations!
Very correct Ohifrizzler thank yod for baking me up from meanies like Eazy. In Kentucky we would
horse race them to seeee whos the better one.
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MTA_2024 wrote:
Prove that for all real numbers and verifying .
and verifying . https://math.stackexchange.com/questions/3890494/how-could-someone-conceive-of-using-this-inequality-for-this-proof?noredirect=1
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