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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
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0 replies
jlacosta
May 1, 2025
0 replies
IMO Shortlist 2009 - Problem G3
April   48
N 5 minutes ago by Markas
Let $ABC$ be a triangle. The incircle of $ABC$ touches the sides $AB$ and $AC$ at the points $Z$ and $Y$, respectively. Let $G$ be the point where the lines $BY$ and $CZ$ meet, and let $R$ and $S$ be points such that the two quadrilaterals $BCYR$ and $BCSZ$ are parallelogram.
Prove that $GR=GS$.

Proposed by Hossein Karke Abadi, Iran
48 replies
April
Jul 5, 2010
Markas
5 minutes ago
Mount Inequality erupts in all directions!
BR1F1SZ   3
N 6 minutes ago by NumberzAndStuff
Source: Austria National MO Part 1 Problem 1
Let $a$, $b$ and $c$ be pairwise distinct nonnegative real numbers. Prove that
\[
(a + b + c) \left( \frac{a}{(b - c)^2} + \frac{b}{(c - a)^2} + \frac{c}{(a - b)^2} \right) > 4.
\](Karl Czakler)
3 replies
BR1F1SZ
May 5, 2025
NumberzAndStuff
6 minutes ago
Show that (DEN) passes through the midpoint of BC
v_Enhance   24
N 6 minutes ago by Markas
Source: Sharygin First Round 2013, Problem 21
Chords $BC$ and $DE$ of circle $\omega$ meet at point $A$. The line through $D$ parallel to $BC$ meets $\omega$ again at $F$, and $FA$ meets $\omega$ again at $T$. Let $M = ET \cap BC$ and let $N$ be the reflection of $A$ over $M$. Show that $(DEN)$ passes through the midpoint of $BC$.
24 replies
v_Enhance
Apr 7, 2013
Markas
6 minutes ago
2020 EGMO P5: P is the incentre of CDE
alifenix-   49
N 7 minutes ago by Markas
Source: 2020 EGMO P5
Consider the triangle $ABC$ with $\angle BCA > 90^{\circ}$. The circumcircle $\Gamma$ of $ABC$ has radius $R$. There is a point $P$ in the interior of the line segment $AB$ such that $PB = PC$ and the length of $PA$ is $R$. The perpendicular bisector of $PB$ intersects $\Gamma$ at the points $D$ and $E$.

Prove $P$ is the incentre of triangle $CDE$.
49 replies
alifenix-
Apr 18, 2020
Markas
7 minutes ago
No more topics!
FE over R
jasperE3   7
N Apr 27, 2025 by jasperE3
Source: Evan Chen
Find all functions $f:\mathbb R\to\mathbb R$ such that $f(x+f(y))+f(xy)=f(x+1)f(y+1)-1\forall x,y\in\mathbb R$.
7 replies
jasperE3
Apr 27, 2021
jasperE3
Apr 27, 2025
FE over R
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G H BBookmark kLocked kLocked NReply
Source: Evan Chen
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jasperE3
11321 posts
#1 • 3 Y
Y by HWenslawski, oVlad, nathantareep
Find all functions $f:\mathbb R\to\mathbb R$ such that $f(x+f(y))+f(xy)=f(x+1)f(y+1)-1\forall x,y\in\mathbb R$.
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v_Enhance
6877 posts
#2 • 4 Y
Y by HamstPan38825, MathLuis, Mango247, Mango247
For context, this is a made-up functional equation mentioned at the end of section 6 of https://web.evanchen.cc/handouts/FuncEq-Intro/FuncEq-Intro.pdf, where it's shown that $f$ injective would let you finish. However, $f$ is not necessarily injective, e.g. if $f$ is constant. I make no promise that there is a nice solution to this equation. :D
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gghx
1072 posts
#3 • 1 Y
Y by Mango247
Can't solve the entire thing. But add the additional condition $f(f(0))=-f(0)$ and it is solvable, and relatively hard too. Try it!
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jasperE3
11321 posts
#4
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What is the solution with $f(f(0))=-f(0)$?
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gghx
1072 posts
#7
Y by
$P(x,y): f(x+f(y))+f(xy)=f(x+1)f(y+1)-1$.
$P(0,0): f(f(0))+f(0)=f(1)^2-1$

Case 1: $f(1)=1$.
Note $f(f(x)+1)=f(x+f(1))=f(x+1)$.
$P(f(x),y),P(f(y),x): f(xf(y))=f(yf(x))$. Taking $x=0,f(0)=f(yf(0))$.
If $f(0)\ne 0$, $f$ is constant. Easy to find solutions.
If $f(0)=0$, $P(x,0): f(x+1)=f(x)+1$.
Original equation becomes $f(x+f(y))+f(xy)=f(x)f(y)+f(x)+f(y)$.
Putting $y\rightarrow y+1$, $f(x+f(y))+f(xy+x)=f(x)f(y)+2f(x)+f(y)$.
Hence, $f(xy+x)=f(x)f(y)+f(x)$ which is a well known FE. (Cauchy+multiplicative).
Only solution $f(x)=x$ (or $0$ which dosent fit).

Case 2: $f(1)=-1$.
$f(f(x)+1)=f(x+f(1))=f(x-1)$.
$P(x,1): f(x-1)+f(x)=f(x+1)f(2)-1$ or $f(f(x)+1)+f(x)=f(x+1)f(2)-1$.
This means that if $f(a)=f(b)$ then $f(a+1)=f(b+1)$.
If $f(a)=f(b)$,
$P(x,a),P(x,b): f(x)=f(cx)$ for some $c$.
$P(x,yc): f(y+1)=f(yc+1)=f(y+1/c)$.
This means that $f(x)=f(x+d)$ for some $d$. $P(x,y),P(x,y+d)$ gives $f$ constant.

Else $f$ is injective and it is easy to finish.


I can't remember if this is what I thought of back in April but it should be correct.

Furthermore, it is interesting to see if anyone can remove the condition / use a weaker one. I probably tried this back then and failed.
This post has been edited 2 times. Last edited by gghx, Sep 21, 2021, 12:24 AM
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Dennis1208
21 posts
#8
Y by
gghx wrote:
$P(x,y): f(x+f(y))+f(xy)=f(x+1)f(y+1)-1$.
$P(0,0): f(f(0))+f(0)=f(1)^2-1$

Case 1: $f(1)=1$.
Note $f(f(x)+1)=f(x+f(1))=f(x+1)$.
$P(f(x),y),P(f(y),x): f(xf(y))=f(yf(x))$. Taking $x=0,f(0)=f(yf(0))$.
If $f(0)\ne 0$, $f$ is constant. Easy to find solutions.
If $f(0)=0$, $P(x,0): f(x+1)=f(x)+1$.
Original equation becomes $f(x+f(y))+f(xy)=f(x)f(y)+f(x)+f(y)$.
Putting $y\rightarrow y+1$, $f(x+f(y))+f(xy+x)=f(x)f(y)+2f(x)+f(y)$.
Hence, $f(xy+x)=f(x)f(y)+f(x)$ which is a well known FE. (Cauchy+multiplicative).
Only solution $f(x)=x$ (or $0$ which dosent fit).

Case 2: $f(1)=-1$.
$f(f(x)+1)=f(x+f(1))=f(x-1)$.
$P(x,1): f(x-1)+f(x)=f(x+1)f(2)-1$ or $f(f(x)+1)+f(x)=f(x+1)f(2)-1$.
This means that if $f(a)=f(b)$ then $f(a+1)=f(b+1)$.
If $f(a)=f(b)$,
$P(x,a),P(x,b): f(x)=f(cx)$ for some $c$.
$P(x,yc): f(y+1)=f(yc+1)=f(y+1/c)$.
This means that $f(x)=f(x+d)$ for some $d$. $P(x,y),P(x,y+d)$ gives $f$ constant.

Else $f$ is injective and it is easy to finish.


I can't remember if this is what I thought of back in April but it should be correct.

Furthermore, it is interesting to see if anyone can remove the condition / use a weaker one. I probably tried this back then and failed.
Can you help me how to prove that there are only 2 cases for $f(1)$?
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megarnie
5606 posts
#9
Y by
Dennis1208 wrote:
Can you help me how to prove that there are only 2 cases for $f(1)$?
gghx is assuming $f(f(0)) = -f(0)$
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jasperE3
11321 posts
#10
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jasperE3 wrote:
Find all functions $f:\mathbb R\to\mathbb R$ such that $f(x+f(y))+f(xy)=f(x+1)f(y+1)-1\forall x,y\in\mathbb R$.

Let $P(x,y)$ be the assertion $f(x+f(y))+f(xy)=f(x+1)f(y+1)-1$.

If $f(1)=f(2)=0$ we have:
$P(1,1)\Rightarrow 0=-1$, contradiction.
So $f(1)\ne0$ or $f(2)\ne0$.

Claim: $f$ is injective (barring constant solutions)
Suppose $f(a)=f(b)$ for some $a\ne b$.
$P(0,a)\Rightarrow f(f(a))+f(0)=f(1)f(a+1)-1$
$P(0,b)\Rightarrow f(f(a))+f(0)=f(1)f(b+1)-1$
Comparing, we get $f(1)(f(a+1)-f(b+1))=0$.
$P(1,a)\Rightarrow f(1+f(a))+f(a)=f(2)f(a+1)-1$
$P(1,b)\Rightarrow f(1+f(a))+f(a)=f(2)f(b+1)-1$
Comparing, we get $f(2)(f(a+1)-f(b+1))=0$.
So $f(a+1)=f(b+1)$, whether $f(1)\ne0$ or $f(2)\ne0$.
$P(x,a)\Rightarrow f(x+f(a))+f(xa)=f(x+1)f(a+1)-1$
$P(x,b)\Rightarrow f(x+f(a))+f(xb)=f(x+1)f(a+1)-1$
Comparing, we get $f(xa)=f(xb)$. If one of $a,b$ is zero, then the other isn't, so $f$ is constant. The only constant solutions are $\boxed{f(x)=\frac{\sqrt5-1}2}$ and $\boxed{f(x)=\frac{-\sqrt5-1}2}$, so assume from now on that $ab\ne0$, we get $f(x)=f\left(\frac abx\right)$.
$P\left(y,\frac abx\right)\Rightarrow f(y+f(x))+f(xy)=f\left(\frac abx+1\right)f(y+1)-1$
$P(y,x)\Rightarrow f(y+f(x))+f(xy)=f(x+1)f(y+1)-1$
Comparing, we get $f(y+1)\left(f\left(\frac abx+1\right)-f(x+1)\right)=0$. Setting $y\in\{0,1\}$ such that $f(y+1)\ne0$, this becomes $f\left(\frac abx+1\right)=f(x+1)$. Then:
$$f(x)=f\left(\frac abx-\frac ab+1\right)=f\left(x-1+\frac ba\right).$$Let $p=\frac ba-1\ne0$, this is just $f(x)=f(x+p)$.
$P(x,y+p)\Rightarrow f(x+f(y))+f(xy+px)=f(x+1)f(y+1)-1$
$P(x,y)\Rightarrow f(x+f(y))+f(xy)=f(x+1)f(y+1)-1$
Comparing, we get $f(xy+px)=f(xy)$. Setting $y=0$, we get $f(px)=f(0)$, so $f$ is a constant function whose solutions we already encountered.

Swapping $x,y$ we get $f(x+f(y))=f(y+f(x))$, so using injectivity and setting $y=0$, we get $f(x)=x+f(0)$, testing gives $\boxed{f(x)=x}$.
This post has been edited 1 time. Last edited by jasperE3, Apr 27, 2025, 8:34 PM
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