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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
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
May 1, 2025
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
BMO 2024 SL A4
MuradSafarli   3
N 14 minutes ago by sqing
A4.
Let \(a \geq b \geq c \geq 0\) be real numbers such that \(ab + bc + ca = 3\).
Prove that:
\[
3 + (2 - \sqrt{3}) \cdot \frac{(b-c)^2}{b+(\sqrt{3}-1)c} \leq a+b+c
\]and determine all the cases when the equality occurs.
3 replies
MuradSafarli
Apr 27, 2025
sqing
14 minutes ago
an exponential inequality with two variables
teresafang   5
N 15 minutes ago by teresafang
x and y are positive real numbers.prove that [(x^y)/y]^(1/2)+[(y^x)/x]^(1/2)>=2.
sorry.I’m not good at English.Also I don’t know how to use Letax.
5 replies
teresafang
May 4, 2025
teresafang
15 minutes ago
Inspired by Austria 2025
sqing   6
N 16 minutes ago by sqing
Source: Own
Let $ a,b\geq 0 ,a,b\neq 1$ and $  a^2+b^2=1. $ Prove that$$   (a + b ) \left( \frac{a}{(b -1)^2} + \frac{b}{(a - 1)^2} \right) \geq 12+8\sqrt 2$$
6 replies
sqing
Today at 2:01 AM
sqing
16 minutes ago
Geometry
gggzul   4
N 20 minutes ago by gggzul
In trapezoid $ABCD$ segments $AB$ and $CD$ are parallel. Angle bisectors of $\angle A$ and $\angle C$ meet at $P$. Angle bisectors of $\angle B$ and $\angle D$ meet at $Q$. Prove that $ABPQ$ is cyclic
4 replies
gggzul
Today at 8:22 AM
gggzul
20 minutes ago
Incentre-excentre geometry
oVlad   1
N an hour ago by mashumaro
Source: Romania Junior TST 2025 Day 2 P2
Consider a scalene triangle $ABC$ with incentre $I$ and excentres $I_a,I_b,$ and $I_c$, opposite the vertices $A,B,$ and $C$ respectively. The incircle touches $BC,CA,$ and $AB$ at $E,F,$ and $G$ respectively. Prove that the circles $IEI_a,IFI_b,$ and $IGI_c$ have a common point other than $I$.
1 reply
oVlad
2 hours ago
mashumaro
an hour ago
Two equal angles
jayme   5
N an hour ago by Captainscrubz
Dear Mathlinkers,

1. ABCD a square
2. I the midpoint of AB
3. 1 the circle center at A passing through B
4. Q the point of intersection of 1 with the segment IC
5. X the foot of the perpendicular to BC from Q
6. Y the point of intersection of 1 with the segment AX
7. M the point of intersection of CY and AB.

Prove : <ACI = <IYM.

Sincerely
Jean-Louis
5 replies
jayme
May 2, 2025
Captainscrubz
an hour ago
Geometry
Lukariman   1
N an hour ago by Lukariman
Given circle (O) and point P outside (O). From P draw tangents PA and PB to (O) with contact points A, B. On the opposite ray of ray BP, take point M. The circle circumscribing triangle APM intersects (O) at the second point D. Let H be the projection of B on AM. Prove that <HDM = 2∠AMP.
1 reply
Lukariman
2 hours ago
Lukariman
an hour ago
Inequality involving square root cube root and 8th root
bamboozled   3
N 2 hours ago by Jackson0423
If $a,b,c,d,e,f,g,h,k\in R^+$ and $a+b+c=d+e+f=g+h+k=8$, then find the maximum value of $\sqrt{ad^3 g^4} +\sqrt[3]{be^3 h^4} + \sqrt[8]{cf^3 k^4}$
3 replies
bamboozled
Today at 4:46 AM
Jackson0423
2 hours ago
Geo metry
TUAN2k8   1
N 3 hours ago by SimplisticFormulas
Help me plss!
Given an acute triangle $ABC$. Points $D$ and $E$ lie on segments $AB$ and $AC$, respectively. Lines $BD$ and $CE$ intersect at point $F$. The circumcircles of triangles $BDF$ and $CEF$ intersect at a second point $P$. The circumcircles of triangles $ABC$ and $ADE$ intersect at a second point $Q$. Point $K$ lies on segment $AP$ such that $KQ \perp AQ$. Prove that triangles $\triangle BKD$ and $\triangle CKE$ are similar.
1 reply
TUAN2k8
4 hours ago
SimplisticFormulas
3 hours ago
Nordic 2025 P3
anirbanbz   9
N Today at 8:11 AM by Tsikaloudakis
Source: Nordic 2025
Let $ABC$ be an acute triangle with orthocenter $H$ and circumcenter $O$. Let $E$ and $F$ be points on the line segments $AC$ and $AB$ respectively such that $AEHF$ is a parallelogram. Prove that $\vert OE \vert = \vert OF \vert$.
9 replies
anirbanbz
Mar 25, 2025
Tsikaloudakis
Today at 8:11 AM
Aime type Geo
ehuseyinyigit   1
N Today at 7:44 AM by ehuseyinyigit
Source: Turkish First Round 2024
In a scalene triangle $ABC$, let $M$ be the midpoint of side $BC$. Let the line perpendicular to $AC$ at point $C$ intersect $AM$ at $N$. If $(BMN)$ is tangent to $AB$ at $B$, find $AB/MA$.
1 reply
ehuseyinyigit
Yesterday at 9:04 PM
ehuseyinyigit
Today at 7:44 AM
Arbitrary point on BC and its relation with orthocenter
falantrng   34
N Today at 7:41 AM by Mamadi
Source: Balkan MO 2025 P2
In an acute-angled triangle \(ABC\), \(H\) be the orthocenter of it and \(D\) be any point on the side \(BC\). The points \(E, F\) are on the segments \(AB, AC\), respectively, such that the points \(A, B, D, F\) and \(A, C, D, E\) are cyclic. The segments \(BF\) and \(CE\) intersect at \(P.\) \(L\) is a point on \(HA\) such that \(LC\) is tangent to the circumcircle of triangle \(PBC\) at \(C.\) \(BH\) and \(CP\) intersect at \(X\). Prove that the points \(D, X, \) and \(L\) lie on the same line.

Proposed by Theoklitos Parayiou, Cyprus
34 replies
falantrng
Apr 27, 2025
Mamadi
Today at 7:41 AM
Parallelograms and concyclicity
Lukaluce   31
N Today at 4:15 AM by Ihatecombin
Source: EGMO 2025 P4
Let $ABC$ be an acute triangle with incentre $I$ and $AB \neq AC$. Let lines $BI$ and $CI$ intersect the circumcircle of $ABC$ at $P \neq B$ and $Q \neq C$, respectively. Consider points $R$ and $S$ such that $AQRB$ and $ACSP$ are parallelograms (with $AQ \parallel RB, AB \parallel QR, AC \parallel SP$, and $AP \parallel CS$). Let $T$ be the point of intersection of lines $RB$ and $SC$. Prove that points $R, S, T$, and $I$ are concyclic.
31 replies
Lukaluce
Apr 14, 2025
Ihatecombin
Today at 4:15 AM
Concurrency from isogonal Mittenpunkt configuration
MarkBcc168   17
N Today at 12:25 AM by Ilikeminecraft
Source: Fake USAMO 2020 P3
Let $\triangle ABC$ be a scalene triangle with circumcenter $O$, incenter $I$, and incircle $\omega$. Let $\omega$ touch the sides $\overline{BC}$, $\overline{CA}$, and $\overline{AB}$ at points $D$, $E$, and $F$ respectively. Let $T$ be the projection of $D$ to $\overline{EF}$. The line $AT$ intersects the circumcircle of $\triangle ABC$ again at point $X\ne A$. The circumcircles of $\triangle AEX$ and $\triangle AFX$ intersect $\omega$ again at points $P\ne E$ and $Q\ne F$ respectively. Prove that the lines $EQ$, $FP$, and $OI$ are concurrent.

Proposed by MarkBcc168.
17 replies
MarkBcc168
Apr 28, 2020
Ilikeminecraft
Today at 12:25 AM
IMO 2009, Problem 5
orl   90
N Apr 26, 2025 by mkultra42
Source: IMO 2009, Problem 5
Determine all functions $ f$ from the set of positive integers to the set of positive integers such that, for all positive integers $ a$ and $ b$, there exists a non-degenerate triangle with sides of lengths
\[ a, f(b) \text{ and } f(b + f(a) - 1).\]
(A triangle is non-degenerate if its vertices are not collinear.)

Proposed by Bruno Le Floch, France
90 replies
orl
Jul 16, 2009
mkultra42
Apr 26, 2025
IMO 2009, Problem 5
G H J
Source: IMO 2009, Problem 5
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bobthegod78
2982 posts
#82
Y by
Very cool problem! Let $P(a,b)$ denote the assertion.

Initially, we prove $f$ is unbounded. If $f$ was bounded, then take a sufficiently large $a$, and obviously the statement is not true.

We claim $f(1)=1$. Consider $P(1,b)$, then obviously $f(b) = f(b + f(1)-1)$.
If indeed $f(1)>1$, then the sequence would be periodic every $f(1)-1$ terms, implying that the image of $f$ is $f(1), f(2), \dots, f(f(1)-1)$. But this implies $f$ is bounded, contradiction.

Then taking $P(a,1)$, we have $f(f(a))=a$, so $f$ is an involution and therefore bijective.

We prove that $f((k-1)f(2) - (k-2)) = k$ for $k\geq 2$ through induction. The base case, $k=2$, is obvious.
Now assume this is true for $k=2, 3, \dots, m$. Consider $P(2,(m-1)f(2) - (m-2))$. We have $2, m, f(m f(2) - (m-1))$ are the sides of a triangle. Since $f$ is bijective, in fact, we must have $f(mf(2)-(m-1))=m+1$. This proves the claim.

But since $f((k-1)f(2) - (k-2))=k$, the set $S = \{(k-1)f(2) - (k-2) \mid k\geq 2, k \in \mathbb N \}$ must contain all the positive integers at least 2, which implies $f(2)=2$. It is easy to see this finishes.
This post has been edited 1 time. Last edited by bobthegod78, Nov 17, 2023, 11:19 PM
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abeot
125 posts
#83 • 2 Y
Y by megarnie, centslordm
Denote the assertion as $P(a, b)$.

First,
\[ P(1, b) \implies f(b) = f(b+f(1) - 1) \]Now, if $f(1) \neq 1$ then $f$ is periodic. If we make $a$ very large then we find that this case fails. Thus, $f(1) = 1$. Then,
\[ P(a, 1) \implies f(f(a)) = a \]This means that $f$ is both injective and surjective. Plug
\[ P(2, b) \implies f(b) - 1 \leq f(b + f(2) - 1) \leq f(b) + 1 \]Note that in particular since $f(b+f(2) - 1) \neq f(b)$ by injectivity, we have that either $f(b+f(2)-1) = f(b)+1$ or $f(b+f(2) - 1) = f(b) - 1$.

Now, suppose that $f(b+f(2)-1) = f(b) - 1$ for some $b$. Denote $m = f(2) - 1$. Then $f(b+2m) = f(b+m) - 1 = f(b) - 2$. By induction, then
\[ f(b+(k+1)m) = f(b+km) - 1 = f(b) - k - 1 \]for all positive integers $k$. But then we eventually get a nonnegative output, contradiction.

So we must have $f(b+f(2) - 1) = f(b) + 1$. If $m = f(2) - 1$, then by induction,
\[ f(b + (k+1)m) = f(b+km) + 1 = f(b) + k + 1 \]If $f(2) - 1 > 1$ then take some $c$ such that $c \not \equiv 1 \pmod{m}$. Since $f(c) \neq f(f(2)) = 2$, then there exists nonnegative integers $k$ and $j$ such that
\[ f(c+km) = f(c) + k = 2+j = f(1+jm) \]Then this contradicts injectivity.

Thus, $f(2) - 1 = 1$, and so we have $f(b+1) = f(b) + 1$. This implies that $f(n) = n$, which obviously works. $\blacksquare$
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amazingtheorem
17 posts
#84
Y by
The only solution is $f(n)=n$ for all $n \in \mathbb{N}$. It's easy to check that this function satisfies the problem condition.

Now, let $P(a,b)$ be the assertation of there is a triangle whose sides are $a,f(b)$, and $f(b+f(a)-1)$. This would later implies $$a+f(b)\ge f(b+f(a)-1)+1, a+f(b+f(a)-1)\ge f(b)+1,\text{and } f(b)+f(b+f(a)-1)\ge a+1.$$
Claim 1. $f(1)=1.$
Proof.

Assume to the contrary, we have $f(1)=c\ge 2$. Observe that $P(a,1)$ gives $f(b)\ge f(b+c-1)$ for all $b \in \mathbb{N}$. This implies the chain of inequality: $$f(b)\ge f(b+c-1)\ge f(b+2c-2)\ge f(b+3c-3)\ge \cdots $$It means that for every $i\in \{1,2,3,...,c-1 \}$, there is $M_i\in \mathbb{N}$ such that the sequence $f(i+M_i(c-1)), f(i+(M_i+1)(c-1)), f(i+(M_i+2)(c-1)), \cdots$ is a constant sequence. We now let every terms of that sequence be $N_i$.
Choose $r=\max(M_1,M_2,...,M_{c-1})+2$. Notice that the sequence $f(1+(c-1)r),f(2+(c-1)r),f(3+(c-1)r), \cdots$ is periodic in which the period is at most $c-1$.
Now, choose $a=2max(N_1,N_2,...,N_{c-1})$. Notice $P(a,1+(c-1)r)$ implies $$f(1+(c-1)r)+f(1+(c-1)r+f(a)-1)\ge a+1.$$However, $f(1+(c-1)r)+f(1+(c-1)r+f(a)-1)\le a$ by the definition of $a$. This is a contradiction. $\blacksquare$

Now, we have $f(1)=1$. Notice that $P(a,1)$ tells us $$f(f(a))+1\ge a+1 \text{and } a+1\ge f(f(a))+1. $$This implies $f(f(a))=a$ for every $a\in \mathbb{N}$, which gives us the bijectivity of $f$.
Let $k\ge 2$ be a natural number such that $f(k)=2$. We also have $k=f(2)$.

Claim 2. $f(1+(k-1)l)=l+1$ for all $l\in \mathbb{N}_0$
Proof.
We will use induction. For $l=0$ and $l=1$, it is clear. Assume that we now have $f(1+(k-1)d)=d+1$ for every $d=0,1,2,...,l$. For $d=l+1$, by the bijectivity of $f$, we have $f(1+(k-1)(l+1))\ge l+1$. However, from $P(f(a),k)$, we have $f(a)+1\ge f(a+k-1)$ for all $a \in \mathbb{N}$. Substituting $a=1+(k-1)l$, we get $l+2\ge f(1+(k-1)(l+1))$. This forces $f(1+(k-1)(l+1))=l+2$. This completes the induction. $\blacksquare$

We now have $f(1+(k-1)l)=l+1$ for all $l\in \mathbb{N}_0$. This implies $f(l)=1+(k-1)(l-1)$ for every $l \in \mathbb{N}$. Now, the bijectivity of $f$ forces $k=2$ (If not, then there is no $q\in \mathbb{N}$ such that $f(q)=2$). This brings us to $f(l)=l$ for every $l\in \mathbb{N}$.

To sum up, the only function that satisfies the problem condition is $f(n)=n$ for every $n \in \mathbb{N}$. It is easy to check that this solution indeed satisfies the property given.
This post has been edited 1 time. Last edited by amazingtheorem, Jan 23, 2024, 6:47 PM
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Sedro
5845 posts
#85
Y by
We claim that the only solution is $\boxed{f(x)=x}$. It is easy to see that it satisfies the condition in the problem statement; we now prove there are no other solutions. Before we begin, we note all inequalities below are obtained from the triangle inequality unless specified otherwise.

Claim: $f(1)=1$.

Proof: Let $a=1$, and we have $f(b)+1> f(b+f(1)-1)$ and $f(b+f(1)-1)+1> f(b)$. Together, these imply $f(b) = f(b+f(1)-1)$. For the sake of contradiction, assume $f(1)>1$. Then, there are arbitrarily large positive integers $d$ such that $f(d)=f(1)$. Let $a=d$ and fix $b$ to obtain $f(b)+f(b+f(d)-1) = f(b)+f(b+f(1)-1) > d$. Taking $d$ large enough, we have a contradiction, and hence $f(1)=1$.

Claim: $f$ is an involution.

Proof: Let $b=1$, and we have $a+1 > f(f(a))$ and $f(f(a))+1 > a$, which together imply $f(f(a))=a$, as desired.

Claim: $f(2)=2$.

Proof: Assume for the sake of contradiction that $f(2)=r>2$, and hence $f(r)=2$. We prove by strong induction that for any nonnegative integer $k$, we have $f(r+k(r-1))=k+2$. The base case, $k=0$, is trivial, so we proceed to the inductive step.

Assume that this claim holds for $0,1,\dots,k$; we show it holds for $k+1$. Let $a=2$ and $b=r+k(r-1)$; then, we have $a+f(b) = k+4 > f(r+(k+1)(r-1))$. Combining this with the facts $f(1)=1$, $f(r+k(r-1)) = k+2$ for $k=0,1,\dots, k$, and $f$ is injective, we must have $f(r+(k+1)(r-1))=k+3$, which completes the inductive step. But this implies that $f(r+(r-2)(r-1)) = r$, so in order not to violate the injectivity of $f$, we must have $r+(r-2)(r-1)=r$. However, this implies $r\in \{1,2\}$, which is a contradiction. Hence, $f(2)=2$.

Claim: $f$ is the identity function.

Proof: we will show that $f(x)=x$ using strong induction. We already know that $f(1)=1$ and $f(2)=2$; these are our base cases. For the inductive step, we assume $f(x)=x$ for all positive integers less than or equal to $x$, and prove $f(x+1)=x+1$. Note that to preserve the injectivity of $f$, we must have $f(x+1)>x$. Then, let $a=2$, $b=x$, and we must have $x+2 > f(x+1)$. This implies $f(x+1)=x+1$, which completes the inductive step. We are now done. $\blacksquare$
This post has been edited 1 time. Last edited by Sedro, Jun 17, 2024, 2:56 PM
Reason: Cleaned up proof
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Bryan0224
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#86
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Guys I really don’t know latex
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ihatemath123
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#87
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The only solution is $f(x)=x$, which is easy to check.

Taking $P(1,a)$ gives us $f(a) = f(a+f(1)-1)$. If $f(1) \neq 1$, it follows that $f(x)$ is periodic with period $f(1)-1$; then, in our original equation, we can repeatedly increase $a$ by $f(1)-1$ to get a contradiction. So, $f(1) = 1$.

Taking $P(a,1)$ gives us $a = f(f(a))$, so in particular, $f$ is injective.

Claim: We have $f(x) \geq x$.
Proof: If $f(a+1) \leq a$, then taking $P(f(a+1),b)$ for any integer $b$ gives us \[f(a+b) \leq a-1+f(b),\]contradicting injectivity.

Combining the above claim with $f(f(x))$ gives us $f(x)=x$, as desired.
This post has been edited 2 times. Last edited by ihatemath123, Aug 11, 2024, 3:25 PM
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cursed_tangent1434
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#88
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Solved with reni_wee. A bit long winded, but this was the solution that we found. Really fun and different kind of problem.

The answer is $f(n) = n$ for all $n\in \mathbb{Z}_{>0}$. It’s easy to see that these functions satisfy the given equation. We now show these are the only solutions. Let $P(a,b)$ be the assertion that $a$ , $f(b)$ and $f(b+f(a)-1)$ are the sides of a non-degenerate triangle. We start off by proving the following characteristic of $f$.

Claim : The function $f$ is an involution, i.e $f(f(a))=a$ for all positive integers $a$.
Proof : Note that from $(1,b)$ we have,
\[f(b)-1 < f(b+f(1)-1)< f(b)+1\]which implies $f(b+f(1)-1)=f(b)$ for all positive integers $b$. Now, if $f(1)>1$ this implies that $f$ is periodic with period $T = |f(1)-1|$. Then, since $f$ is a periodic function over the positive integers it must be bounded both above and below. Say $f(m)=M$ is the maximum of this function. Then, $P(2M,m)$ implies,
\[2M < f(m)+f(m+f(M)-1)\le 2M\]which is a clear contradiction. Thus, $f$ cannot be periodic and hence $f(1)=1$. Then, $P(a,1)$ implies,
\[a-1=a -f(1) < f(f(a))<a+f(1)a+1\]and thus, $f(f(a))=a$ for all positive integers $a$ as desired.

Note that now in fact $f$ is both injective and surjective. Next we look at $P(2,f(b))$. This tells us that
\[2 , f(f(b)) \text{ and } f(f(b)+f(2)-1)\]are the sides of a non-degenerate triangle. Thus, $2 , f(b)$ and $f(f(b)+f(2)-1)$ are sides of a non-degenerate triangle. This implies,
\[b-2 < f(f(b)+f(2)-1)<b+2\]We now have 3 cases to explore.

Case 1 : $f(f(b)+f(2)-1) = b$. Thus,
\[f(f(b)+f(2)-1)=b = f(f(b))\]which since $f$ is injective implies $f(b)+f(2)-1=f(b)$ or $f(2)=1$ which is a clear contradiction.

Case 2 : $f(f(b)+f(2)-1)=b-1$. Thus,
\[f(f(b)+f(2)-1)=b-1=f(f(b-1))\]which since $f$ is injective implies $f(b)+f(2)-1=f(b-1)$ But for $f=2$ this rewrites to $2f(2)-1=f(1)=1$ which implies $f(2)=1$ which is again a clear contradiction.

Case 3 : $f(f(b)+f(2)-1)=b+1$. Thus,
\[f(f(b)+f(2)-1)=b-1=f(f(b+1))\]which since $f$ is injective implies $f(b)+f(2)-1 = f(b+1)$. Thus,
\[f(b+1)=f(b)+f(2)-1 \ge f(b)+1>f(b)\]implies that $f$ is strictly increasing. Combining this with our previous observation that $f$ is injective and surjective, it follows that $f$ is indeed the identity function, as claimed.
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HamstPan38825
8859 posts
#89
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This solution is very clear and motivated, making the problem nice but not particularly difficult.

From letting $a = 1$, we have $f(b) = f(b+f(1) - 1)$ for all $b \in \mathbb N$. If $f(1) - 1 > 0$, then $f$ is eventually periodic, thus taking sufficiently large $a$ in the original assertion yields a contradiction.

Thus $f(1) = 1$. Setting $b = 1$, it follows that $f(f(a)) = a$, hence $f$ is bijective.

Claim: [Key Claim] For all $n \in \mathbb N$, $f(n+1) = f(n) + f(2) - 1$.

Proof: We induct on $n$, with the $n=1$ case clear. First, observe that
\[n-1 \leq f(f(n) + f(2) - 1) \leq n+1\]by the condition.

First Case: If $f(f(n) + f(2) - 1) = n-1$, applying $f$ to both sides yields \[f(n) = f(n-1) - (f(2) - 1).\]But the inductive hypothesis forces $f(2) = 1$, which contradicts $f$ bijective.

Second Case: If $f(f(n) + f(2) - 1) = n$, applying $f$ to both sides yields $f(n) = f(n) - (f(2) - 1)$, hence once again $f(2) = 1$, contradiction.

Third Case: If $f(f(n) + f(2) - 1) = n+1$, applying $f$ to both sides yields $f(n) + f(2) - 1 = f(n+1)$, which is the desired conclusion. $\blacksquare$

So $f$ is bijective and linear, implying $f \equiv n$, which works.
This post has been edited 2 times. Last edited by HamstPan38825, Nov 5, 2024, 4:28 PM
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bin_sherlo
719 posts
#90 • 1 Y
Y by tiny_brain123
Answer is $f(n)=n$ for each positive integer.
Claim: $f(1)=1$.
Proof:$a=1$ implies $f(b)=f(b+f(1)-1)$. If $f(1)\neq 1$, then $f$ is periodic. Denote by $T$ the period. \[f(b)+f(b+f(a)-1)=f(b)+f(b+f(a+nT)-1)\geq a+nT\]Which is impossible for sufficiently large $n$.$\square$
$b=1$ gives $f(f(a))=a$ hence $f$ is an involution. This yields $f$ is bijective.
Claim: $f(2)-1\geq |f(a+1)-f(a)|$.
Proof: $a,b\rightarrow f(a),2$ implies $f(2)+f(a+1)\geq f(a)+1\iff f(2)-1\geq f(a)-f(a+1)$. Similarily $f(a)+f(2)\geq f(a+1)+1\iff f(2)-1\geq f(a+1)-f(a)$ which gives the conclusion.$\square$
Claim: $f(2)=2$.
Proof: Let $f(c)=2\iff f(2)=c$ by injectivity. Note that $c\neq 1$. $f(a),c,f(a+1)$ are sides hence $a=c-1$ yields $c+1\geq f(c-1)\geq c-1$. Now we split into $3$ cases.
Suppose that $f(c-1)=c$. Then, $c=3$ by injectivity and $f(a),2,f(a+2)$ are sides which implies $f(a+2)\leq f(a)+1$. Since $f(1)=1,f(3)=2$, induction with utilising injectivity gives $f(2k-1)=k$. However, $4k-3=f(f(4k-3))=f(2k-1)=k$ does not hold for $k>1$ which results in a contradiction.
Assume that $f(c-1)=c+1\iff f(c+1)=c-1$. We observe that $f(a),f(c+1),f(a+c)$ are sides of a triangle thus, $f(a+c)-f(a)\leq c-2$. Let $m_i$ be the maximum among $\{f((i-1)c+1),\dots,f(ic)\}$. $m_i$ increases at most $c-2$ hence \[\max\{f(1),\dots,f(nc)\}\leq m_1+(n-1)(c-1)=m_1+cn-n-c+1\]Since $f$ is injective, $\max\{f(1),\dots,f(nc)\}\geq nc$ which implies $m_1+cn-c-n+1\geq cn$ or $m_1-c\geq n$ which is impossible for sufficiently large $n$.
So we get $f(c-1)=c-1$. Pick $a=c,b=c-1$ which yields $f(c),f(c-1),f(2c-2)$ are sides or $2,c-1,f(2c-2)$ are sides. Thus, $f(2c-2)\leq c$. Choose $b=2c-2$ to get $f(a),f(2c-2),f(a+2c-3)$ are sides. So we conclude that
\[f(a+2c-3)\leq f(a)+f(2c-2)-1\leq f(a)+c-1\implies f(a+2c-3)-f(a)\leq c-1\]Let $m_i$ be the maximum among $\{f((i-1)(2c-3)+1),\dots,f(i(2c-3))\}$. $m_i$ increases at most $c-1$ thus,
\[\max\{f(1),\dots,f(n(2c-3))\}\leq m_1+(c-1)(n-1)=m_1+cn-c-n+1\]Since $f$ is injective, $\max\{f(1),\dots,f(n(2c-3))\}\geq n(2c-3)$ so $2cn-3n\leq m_1+cn-c-n+1$ which is equavilent to $(c-2)n\leq m_1-c+1$. Since $c\neq 1$ and $c$ cannot be larger than $2$, $c=2$.$\square$
We have $|f(a+1)-f(a)|\leq 1$ and $f$ is injective hence $f(n)=n$ for all positive integers as desired.$\blacksquare$
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Mathandski
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#91
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Subjective Rating (MOHs) $       $
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Ilikeminecraft
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#92
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The answer is $f\equiv x.$
Pick $a = 1$ first, and we have $f(b) = f(b + f(1) - 1).$ If $f(1)\neq1,$ then $f$ has period $f(1) - 1 > 1.$ However, if we fix $b,$ and pick an arbitrarily large $a \equiv1\pmod{f(1)-1}$ such that $a > f(b) + f(b + f(1) - 1),$ then this is a contradiction. Thus, $f(1) = 1.$

Take $b = 1$ and we get that $f$ is an involution.
Hence, $f$ is injective and surjective.

Plug in $b = f(b)$ and we get the new equation $a, b, f(f(a) + f(b) - 1)$ forms a triangle.

We now prove $f\equiv x$ using strong induction.
Assume $f(s) = 2.$ Take $2, 2$ and we get $4 > f(3).$ Hence, either $f(3) = 2$ or $f(3) = 3.$
In the first case, we have $f(3) = 2, f(2) = 3.$ Take $2, 2$ and so $4 > f(5),$ contradiction.
To finish, easy induction suffices. Assume $f(k) = k$ for all $k < n.$
Take $n-1, 2$ and we get $f(n) < n + 1,$ so $f\equiv n.$
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asdf334
7585 posts
#93
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hey that was kind of fun lol
Okay so we need to brainstorm. The first piece of information we can get comes from plugging in $a=1$; we get $f(b)=f(b+f(1)-1)$. In that case if $f(1)\ne 1$ then we ultimately find that $f$ is periodic and thus bounded, a contradiction to the original triplet given as we can set $a$ to be large.

Thus $f(1)=1$, hence we obtain from $b=1$ that $a=f(f(a))$. Now write $a\to f(a)$ to get the nicer triplet $f(a)$, $f(b)$, and $f(a+b-1)$.

Turns out we are almost done. Suppose that $f(2)=S$ and $f(S)=2$. What information can we get from this? Well, we can set $a+b-1=S$, but this restricts our options and doesn't work as well.

Instead write $a=S$ to obtain that $(2,f(b),f(b+S-1))$ form the side lengths of a triangle. Note that $f$ is injective and surjective; if $S>2$ then we must have $|f(b)-f(b+S-1)|=1$. But now consider the set $\{b+0(S-1),b+1(S-1),b+2(S-1),\dots\}$. When adjacent values in the set are plugged into $f$, they yield consecutive outputs. Furthermore, all outputs are distinct. As a result, the outputs are simply $\{f(b)+0,f(b)+1,f(b)+2,\dots\}$.

But if $S>2$ then there are still an infinite number of inputs outside this set, and only a finite number of outputs left (less than $f(b)$). That is a contradiction, so $S=f(2)=2$.

Hence $f(n)\le n$ for $n\in \{1,2\}$. Now we can induct: if $f(n)\le n$ and $n\ge 2$, write
\[n+2\ge f(n)+f(2)>f(n+1)\]thus proving that $f(n+1)\le n+1$ and so $f(n)\le n$ always. As $f$ is injective we get $f(n)=n$ always, done. $\blacksquare$
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fearsum_fyz
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#94
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Note that the given condition can be rewritten as
$$|a - f(b)| < f(b + f(a) - 1) < a + f(b)$$
Claim 1: $f(1) = 1$.
Proof. Substituting $a = 1$ gives
$f(b) - 1 < f(b + f(1) - 1) < f(b) + 1$
$\implies f(b + f(1) - 1) = f(b)$
Now assume, for the sake of contradiction, that $d = f(1) - 1 > 0$. We have $f(1 + nd) = f(1) = d + 1$ for any $n \in \mathbb{N}$.
Then substituting $a = 1 + nd, b = 1$ into $a < f(b) + f(b + f(a) - 1)$ gives
$1 + nd < f(1) + f(1 + f(1 + nd) - 1) = f(1) + f(1 + d) = 2f(1) = 2d + 2$ for all $n \in \mathbb{N}$, a contradiction. $\square$

Claim 2: $f$ is an involution.
Proof. Simply plug in $b = 1$, giving $a - 1 < f(f(a)) < a + 1$, forcing $f(f(a)) = a$. $\square$

We can now try to replicate the 'sandwiching' we did, but with $a = 2$ instead of $1$, leading to the following claim:

Claim 3: Let $f(2) - 2 = d$. Then $f(b + (d + 1)) = f(b) \pm 1$ for all $b$.
Proof. Note that $f(b) - 2 < f(b + d + 1) < f(b) + 2$, but $f(b + d + 1) \neq f(b)$ by Claim 2. Hence, $f(b + d + 1) = f(b) \pm 1$. $\square$

Using Claim 2 again, this can be further rewritten as:

Claim π: For all $b$, either $f(b + 1) = f(b) + (d + 1)$ or $f(b - 1) = f(b) + (d + 1)$.

Now we are at the home stretch. We can show, by induction, that we'll always have the first case and never the second one.

Finish: As the base case, we have $f(1) = 1$ and $f(2) = d + 2$.
Assume, as the inductive hypothesis, that $f(b + 1) = f(b) + (d + 1)$ for all $b \leq n - 1$. We'd like to show that $f(n + 1) = f(n) + (d + 1)$.
Assume for the sake of contradiction that $f(n + 1) \neq f(n) + (d + 1)$. Then $f(n - 1) = f(n) + (d + 1)$ by Claim π.
However, by the inductive hypothesis, $f(n - 1) = f(n) - (d + 1)$. This is a contradiction.

We are done.
This post has been edited 1 time. Last edited by fearsum_fyz, Apr 26, 2025, 11:19 AM
Reason: clarity
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lelouchvigeo
181 posts
#95 • 1 Y
Y by alexanderhamilton124
Solution
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mkultra42
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#96
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Setting \(a=1\) gives \(f(b)=f(b+f(1)-1)\) which implies \(f(b+k(f(1)-1))=f(b)\).

If \(f(1)-1>0\), then for a fixed \(b\) and \(a=a+k(f(1)-1)\), for \(k\) large enough, gives a contradiction.

Thus \(f(1)=1\).

Now for \(b=1\) we get \(a=f(f(a))\). In particular, the function is bijective.

For \(b=f(2)\) we get \(f(a+(f(2)-1))= f(a) \pm 1\).

Assume that the sign is not always \(+\), and note that it cannot be \(-\) indefinitely, so there must be \(k\) such that:

\[f(a+k(f(2)-1))=f(a+(k-1)(f(2)-1))-1,\quad f(a+(k+1)(f(2)-1))=f(a+k(f(2)-1))+1=f(a+(k-1)(f(2)-1))\]
We get a contradiction by injectivity.

Thus, \(f(a+k(f(2)-1))=f(a)+k\).

For \(a=1\) we get \(f(1+k(f(2)-1))=k+1=f(f(k+1)) \implies f(k+1)=1+k(f(2)-1))\).

Surjectivity tells us that we must have \(f(2)-1=1\) and thus \(f(n)=n, \forall n\).
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