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jlacosta   0
May 1, 2025
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0 replies
jlacosta
May 1, 2025
0 replies
My Unsolved Problem
ZeltaQN2008   0
2 minutes ago
Let \(f:[0,+\infty)\to\mathbb{R}\) be a function which is differentiable on \([0,+\infty)\) and satisfies
\[
\lim_{x\to+\infty}\bigl(f'(x)/e^x\bigr)=0.
\]Prove that
\[
\lim_{x\to+\infty}\bigl(f(x)/e^x\bigr)0.
\]
0 replies
ZeltaQN2008
2 minutes ago
0 replies
polonomials
Ducksohappi   0
14 minutes ago
$P\in \mathbb{R}[x] $ with even-degree
Prove that there is a non-negative integer k such that
$Q_k(x)=P(x)+P(x+1)+...+P(x+k)$
has no real root
0 replies
Ducksohappi
14 minutes ago
0 replies
Four tangent lines concur on the circumcircle
v_Enhance   35
N an hour ago by bin_sherlo
Source: USA TSTST 2018 Problem 3
Let $ABC$ be an acute triangle with incenter $I$, circumcenter $O$, and circumcircle $\Gamma$. Let $M$ be the midpoint of $\overline{AB}$. Ray $AI$ meets $\overline{BC}$ at $D$. Denote by $\omega$ and $\gamma$ the circumcircles of $\triangle BIC$ and $\triangle BAD$, respectively. Line $MO$ meets $\omega$ at $X$ and $Y$, while line $CO$ meets $\omega$ at $C$ and $Q$. Assume that $Q$ lies inside $\triangle ABC$ and $\angle AQM = \angle ACB$.

Consider the tangents to $\omega$ at $X$ and $Y$ and the tangents to $\gamma$ at $A$ and $D$. Given that $\angle BAC \neq 60^{\circ}$, prove that these four lines are concurrent on $\Gamma$.

Evan Chen and Yannick Yao
35 replies
v_Enhance
Jun 26, 2018
bin_sherlo
an hour ago
tangent and tangent again
ItzsleepyXD   0
2 hours ago
Source: holder send me this
Given an acute non-isosceles triangle $ABC$ with circumcircle $\Gamma$. $M$ is the midpoint of segment $BC$ and $N$ is the midpoint of $\overarc{BC}$ of $\Gamma$ (the one that doesn't contain $A$). $X$ and $Y$ are points on $\Gamma$ such that $BX \parallel CY \parallel AM$. Assume there exists point $Z$ on segment $BC$ such that circumcircle of triangle $XYZ$ is tangent to $BC$. Let $\omega$ be the circumcircle of triangle $ZMN$. Line $AM$ meets $\omega$ for the second time at $P$. Let $K$ be a point on $\omega$ such that $KN \parallel AM, \omega_b$ be a circle thaat passes through $B$, $X$ and tangents to $BC$ and $\omega_C$ be a circle that passes through $C,Y$ and tangents to $BC$. Prove that circle with center $K$ and radias $KP$ is tangent to circles $\omega_B,\omega_C$ and $\Gamma$.
0 replies
ItzsleepyXD
2 hours ago
0 replies
No more topics!
Symmetric FE
Phorphyrion   8
N Apr 17, 2025 by jasperE3
Source: 2023 Israel TST Test 7 P1
Find all functions $f:\mathbb{R}\to \mathbb{R}$ such that for all $x, y\in \mathbb{R}$ the following holds:
\[f(x)+f(y)=f(xy)+f(f(x)+f(y))\]
8 replies
Phorphyrion
May 9, 2023
jasperE3
Apr 17, 2025
Symmetric FE
G H J
G H BBookmark kLocked kLocked NReply
Source: 2023 Israel TST Test 7 P1
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Phorphyrion
396 posts
#1 • 2 Y
Y by tiendung2006, dxd29070501
Find all functions $f:\mathbb{R}\to \mathbb{R}$ such that for all $x, y\in \mathbb{R}$ the following holds:
\[f(x)+f(y)=f(xy)+f(f(x)+f(y))\]
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Prod55
127 posts
#2 • 3 Y
Y by tiendung2006, Hoanglahoang, ItsBesi
Very nice!

Let $P(x,y)$ be the given assertion,.

Clearly $f\equiv c$ works, so let's now assume that $f$ is non-constant.

$P(0,0): f(0)=f(2f(0))$
$P(x,0): f(x)=f(f(x)+f(0))$
$P(x,2f(0)): f(x)+f(2f(0))=f(2f(0)x)+f(f(x)+f(2f(0))\Leftrightarrow f(2f(0)x)=f(0)$
so $f(0)=0$ otherwise $f$ must be constant, $P(x,0)$ becomes $f(f(x))=f(x)$.
Observe that $f(a)=f(0)\Leftrightarrow a=0$ since $P(x,a)$ gives $f(ax)=f(0)=0$ and $f$ is not constant.
Now lets define the set $A=\left \{k|  f(x)=f(kx)\forall x \in \mathbb{R} \right \}$
If $f(a)=f(b)$ for some $a\neq b$ then from $P(x,a),P(x,b)$ we get that $f(xa)=f(xb)$ and thus $a/b=k\in A$
Also $f(f(1))=f(1)$ so $f(1)\in A$ ( note that $f(1)\neq 0$ since otherwise
$P(x,1): f(1)=f(f(x)+f(1))\Rightarrow f(f(x))=0$ but $f(f(x))=f(x)$ and hence $f\equiv 0$ a contradiction.)
Now $P(1,1): f(1)=f(2f(1))$ so $2f(1)\in A$ as well and hence for every $x$ we have that
$f(2f(1)x)=f(x)=f(xf(1))\Rightarrow f(x)=f(2x)$ for all $x$, i.e $2\in A$.
$P(x,x): 2f(x)=f(x^2)+f(2f(x))=f(x^2)+f(f(x))=f(x^2)+f(x)\Leftrightarrow f(x)=f(x^2)\Rightarrow x\in A,\forall x\neq 0$.
Now pick any $x,y\neq 0$, then $x/y\in A$ and thus $f(y)=f(y\cdot x/y)=f(x)$.
So all the solution are:
$f(x)\equiv c,\forall x$
$f(0)=0,f(x)=c\neq 0,\forall x\neq 0$
and they clearly work.
This post has been edited 1 time. Last edited by Prod55, May 9, 2023, 8:05 PM
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Mathlover_1
295 posts
#3 • 1 Y
Y by Math.1234
Prod55 wrote:
Very nice!

Let $P(x,y)$ be the given assertion,.

Clearly $f\equiv c$ works, so let's now assume that $f$ is non-constant.

$P(0,0): f(0)=f(2f(0))$
$P(x,0): f(x)=f(f(x)+f(0))$
$P(x,2f(0)): f(x)+f(2f(0))=f(2f(0)x)+f(f(x)+f(2f(0))\Leftrightarrow f(2f(0)x)=f(0)$
so $f(0)=0$ otherwise $f$ must be constant, $P(x,0)$ becomes $f(f(x))=f(x)$.
Observe that $f(a)=f(0)\Leftrightarrow a=0$ since $P(x,a)$ gives $f(ax)=f(0)=0$ and $f$ is not constant.
Now lets define the set $A=\left \{k|  f(x)=f(kx)\forall x \in \mathbb{R} \right \}$
If $f(a)=f(b)$ for some $a\neq b$ then from $P(x,a),P(x,b)$ we get that $f(xa)=f(xb)$ and thus $a/b=k\in A$
Also $f(f(1))=f(1)$ so $f(1)\in A$ ( note that $f(1)\neq 0$ since otherwise
$P(x,1): f(1)=f(f(x)+f(1))\Rightarrow f(f(x))=0$ but $f(f(x))=f(x)$ and hence $f\equiv 0$ a contradiction.)
Now $P(1,1): f(1)=f(2f(1))$ so $2f(1)\in A$ as well and hence for every $x$ we have that
$f(2f(1)x)=f(x)=f(xf(1))\Rightarrow f(x)=f(2x)$ for all $x$, i.e $2\in A$.
$P(x,x): 2f(x)=f(x^2)+f(2f(x))=f(x^2)+f(f(x))=f(x^2)+f(x)\Leftrightarrow f(x)=f(x^2)\Rightarrow x\in A,\forall x\neq 0$.
Now pick any $x,y\neq 0$, then $x/y\in A$ and thus $f(y)=f(y\cdot x/y)=f(x)$.
So all the solution are:
$f(x)\equiv c,\forall x$
$f(0)=0,f(x)=c\neq 0,\forall x\neq 0$
and they clearly work.
l hope that my solution is true
Let $P(x,y)$ be the given assertion
1)$f(0)$$\neq$$0$
$P(x,0)$ $\rightarrow$ $f(x)=f(f(x)+f(0))$ $(1)$
$P(x,1)$ $\rightarrow$ $f(f(x)+f(1))=f(1)$ $(2)$
From $(1)$ we get $f(0)=f(2f(0))$
$P(\frac{1}{2f(0)},2f(0))$ $\rightarrow$ $f(0)=f(1)$ and from $(1)$ and $(2)$ we get f(x)=f(1)=c for all $x$$\in$$R$ $c$$\neq$$0$
2)$f(0)=0$
$P(x,0)$ $\rightarrow$ $f(x)=f(f(x))$ $(1)$
$P(x,1)$ $\rightarrow$ $f(f(x)+f(1))=f(1)$ $(2)$
From $(2)$ we get $f(2f(1)$$=$$f(1)$ then $P(2,f(1))$ $\rightarrow$ $f(2)=f(1)$ $(3)$
From $P(f(x),y)$$=G(x,y)$ and $(1)$ we get $f(f(x)y)=f(xy)$ from $G(1,x)$,$G(2,x)$ and $(3)$ we get $f(x)=f(2x)$ from here we get
$f(x)=f(\frac{x}{2^{\infty}})=f(0)=0$
Then for all $x$$\in$$R$ $f(x)=0$
This post has been edited 1 time. Last edited by Mathlover_1, May 9, 2023, 10:48 PM
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MathLuis
1524 posts
#4 • 1 Y
Y by bin_sherlo
Neat F.E., let $P(x,y)$ the assertion as usual, assume $f$ non-constant as the other case works.
Main Weapon: If $f(a)=f(b)$, from $P(x,a)-P(x,b)$ we get $f(ax)=f(bx)$ for all $x \in \mathbb R$ call this $\star$
Now $P(0,0)$ gives $f(0)=f(2f(0))$ so from $\star$ we have $f(2f(0)x)=f(0)$ so if $f(0) \ne 0$ then $f$ is constant so we get that $f(0)=0$, now clearly $f$ is injective at $0$ or else $f(cx)=0$ for $c \ne 0$ such that $f(c)=0$ which again gives $f$ constant so nope.
Now $P(1,1)$ gives $f(1)=f(2f(1))$ and $P(x,0)$ gives $f(x)=f(f(x))$ so by $\star$ we get $f(2f(1)x)=f(x)=f(f(1)x)$ so $f(x)=f(2x)$ and now by $P(x,x)$ we get that $f(x)=f(x^2)$ so by $\star$ we get that $f(xy)=f(xy^2)$, now pick $x,y \ne 0$ and let $x=\frac{a}{y}$ and $y=\frac{b}{a}$ for any reals $a,b \ne 0$ and we get $f(a)=f(b)$ so $f$ constant for all values but $0$.
Hence the set of solutions is $f(x)=c$ for all reals $x$ and $f(x)=c \ne 0$ for all $x \ne 0$ and $f(0)=0$, thus we are done :D.
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Amitai
9 posts
#5
Y by
Let P(x,y) be the assertion: f(x)+f(y)=f(xy)+f(f(x)+f(y))
assume f(a)=f(b). If we substitute y=a or b, the only term that can change is f(ax). Therfore f(ax)=f(bx) for all x. P(0,0): f(2f(0))=f(0) => f(2f(0)x)=f(0). So either f is constant (which works) or f(0)=0.
P(x,0): f(x)=f(f(x)) => f(2x)=f(2f(x))
(for x,y≠0) f(P(x,y)): f(f(x)+f(y))=f(f(xy)+f(f(x)+f(y)))=
=f(f(x)+f(y))+f(xy)-f(xy(f(x)+f(y))) => f(f(x)+f(y))=f(1).
For x=y≠0 we have f(1)=f(2f(x))=f(2x).
So f is constant except 0 which it is 0 at (it works).
But what if f was from R+ to R+, you might ask.

Well: the equation f(f(x)+f(y))=f(1) still holds, so if we can rewrite the equation as f(x)+f(y)=f(xy)+f(1).
Denote m=Inf(Im(f)). Suppose f(x)<m+ε for arbitrarily small ε>0.
P(x,x): f(1)=2f(x)-f(x^2)<m+2ε. so f(1)=m.
P(x,1/x): f(x)+f(1/x)=2m => f(x)=m for all x. So the only solution is a positive constant function.
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PotatoTheMathematician
14 posts
#6
Y by
Mathlover_1 wrote:
Prod55 wrote:
Very nice!

Let $P(x,y)$ be the given assertion,.

Clearly $f\equiv c$ works, so let's now assume that $f$ is non-constant.

$P(0,0): f(0)=f(2f(0))$
$P(x,0): f(x)=f(f(x)+f(0))$
$P(x,2f(0)): f(x)+f(2f(0))=f(2f(0)x)+f(f(x)+f(2f(0))\Leftrightarrow f(2f(0)x)=f(0)$
so $f(0)=0$ otherwise $f$ must be constant, $P(x,0)$ becomes $f(f(x))=f(x)$.
Observe that $f(a)=f(0)\Leftrightarrow a=0$ since $P(x,a)$ gives $f(ax)=f(0)=0$ and $f$ is not constant.
Now lets define the set $A=\left \{k|  f(x)=f(kx)\forall x \in \mathbb{R} \right \}$
If $f(a)=f(b)$ for some $a\neq b$ then from $P(x,a),P(x,b)$ we get that $f(xa)=f(xb)$ and thus $a/b=k\in A$
Also $f(f(1))=f(1)$ so $f(1)\in A$ ( note that $f(1)\neq 0$ since otherwise
$P(x,1): f(1)=f(f(x)+f(1))\Rightarrow f(f(x))=0$ but $f(f(x))=f(x)$ and hence $f\equiv 0$ a contradiction.)
Now $P(1,1): f(1)=f(2f(1))$ so $2f(1)\in A$ as well and hence for every $x$ we have that
$f(2f(1)x)=f(x)=f(xf(1))\Rightarrow f(x)=f(2x)$ for all $x$, i.e $2\in A$.
$P(x,x): 2f(x)=f(x^2)+f(2f(x))=f(x^2)+f(f(x))=f(x^2)+f(x)\Leftrightarrow f(x)=f(x^2)\Rightarrow x\in A,\forall x\neq 0$.
Now pick any $x,y\neq 0$, then $x/y\in A$ and thus $f(y)=f(y\cdot x/y)=f(x)$.
So all the solution are:
$f(x)\equiv c,\forall x$
$f(0)=0,f(x)=c\neq 0,\forall x\neq 0$
and they clearly work.
l hope that my solution is true
Let $P(x,y)$ be the given assertion
1)$f(0)$$\neq$$0$
$P(x,0)$ $\rightarrow$ $f(x)=f(f(x)+f(0))$ $(1)$
$P(x,1)$ $\rightarrow$ $f(f(x)+f(1))=f(1)$ $(2)$
From $(1)$ we get $f(0)=f(2f(0))$
$P(\frac{1}{2f(0)},2f(0))$ $\rightarrow$ $f(0)=f(1)$ and from $(1)$ and $(2)$ we get f(x)=f(1)=c for all $x$$\in$$R$ $c$$\neq$$0$
2)$f(0)=0$
$P(x,0)$ $\rightarrow$ $f(x)=f(f(x))$ $(1)$
$P(x,1)$ $\rightarrow$ $f(f(x)+f(1))=f(1)$ $(2)$
From $(2)$ we get $f(2f(1)$$=$$f(1)$ then $P(2,f(1))$ $\rightarrow$ $f(2)=f(1)$ $(3)$
From $P(f(x),y)$$=G(x,y)$ and $(1)$ we get $f(f(x)y)=f(xy)$ from $G(1,x)$,$G(2,x)$ and $(3)$ we get $f(x)=f(2x)$ from here we get
$f(x)=f(\frac{x}{2^{\infty}})=f(0)=0$
Then for all $x$$\in$$R$ $f(x)=0$
the last step is wrong, $f(\frac{x}{2^{\infty}})=f(0)$ works if $f$ is continuous at 0 but it's not always true.
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cursed_tangent1434
623 posts
#8 • 2 Y
Y by dolphinday, Ali_Vafa
Solved with hints. Really nice and instructive problem. I think I learnt a lot from this one. The answers are $f(x) = c$ for all $x \in \mathbb{R}$ for a fixed constant $c \in \mathbb{R}$ and
\[f(x) = \begin{cases}
0 & x=0\\
c & x\neq 0
\end{cases}\]for a fixed constant $c \in \mathbb{R}$. It’s easy to see that these functions satisfy the given conditions. We now show these are the only solutions. Let $P(x,y)$ denote the assertion that $f(x)+f(y)=f(xy)+f(f(x)+f(y))$ for real numbers $x$ and $y$. We first start off with fixing certain values of $f$.

Claim : $f(0)=0$ or $f$ is constant.
Proof : Say there exists some $\alpha \neq 0 \in \mathbb{R}$ such that $f(\alpha)=f(0)$. Then, comparing $P(x,0)$ and $P(x,\alpha)$ yields,
\[f(\alpha x)=f(x)+f(\alpha)+f(f(x)+f(\alpha))=f(x)+f(0)-f(f(x)+f(0))=f(0)\]Thus, letting $x= \frac{t}{\alpha}$ implies that $f(t)=f(0)$ for all real numbers $t$. Thus, if $f$ is non-constant there does not exist such $\alpha \neq 0$ such that $f(x)=f(\alpha)$.

Now, $P(0,0)$ implies $f(0)=f(2f(0))$ which according to our previous observation requires, $2f(0)=0$ so we have $f(0)=0$ indeed as claimed.

In what proceed, we assume that $f$ is not constant. First of all, note that $P(x,0)$ implies,
\[f(x)=f(x)+f(0)=f(0)+f(f(x)+f(0))=f(f(x))\]and $P(x,1)$ implies
\[f(1)=f(f(x)+f(1))\]for all real numbers $x$. This will come in useful later. Note that $P(f(x),2)$ gives,
\[f(2f(x)) = f(f(x))+f(2)-f(f(f(x))+f(2))=f(x)+f(2)-f(f(x)+f(2))\]and $P(x,x)$ also gives
\[f(2f(x))=2f(x)-f(x^2)\]Thus, setting these two expressions for $f(2f(x))$ equal we have
\[f(x^2) + f(2) = f(x)+ f(f(x)+f(2))\]and further, when $x=1$ in the above result,
\[f(2)=f(f(1)+f(2))=f(1)\]Thus, $f(2)=f(1)$. Now, returning to $P(x,2)$ we have,
\[f(x)+f(1)=f(x)+f(2)=f(2x)+f(f(x)+f(2)) = f(2x)+f(f(x)+f(1))=f(2x)+f(1)\]which implies that $f(2x)=f(x)$ for all real numbers $x$. Then, applying this result to $P(x,x)$ gives
\[2f(x)=f(x^2)+f(2f(x))=f(x^2)+f(f(x))=f(x^2)+f(x)\]so, $f(x^2)=f(x)$ for all reals $x$. Now, from this based on our previous observation, it follows that $f(xy)=f(xy^2)$ for all real numbers $x$ and $y$. Considering $x= \frac{b}{a}$ and $y= \frac{a^2}{b}$ for any two non-zero $a\neq b $ implies that $f$ is indeed constant over the non-zero reals. Thus, all solutions are indeed as claimed.
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megarnie
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#9
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The only solutions are $f\equiv c$ for some constant $c$, and also $f(x) = c \forall x \ne 0$ and $f(0) = 0$. These clearly work. Now we assume that $f$ is not constant. It suffices to show that there is a constant $c$ for which $f(x) = c \forall x \ne 0$.

Let $P(x,y)$ be the given assertion.

$P(x,1): f(f(x) + f(1)) = f(1)$.

$P(x,0): f(x) = f(f(x) + f(0))$.

So $P(0,0): f(2f(0)) = f(0)$.

$P(x, 2f(0)): f(x) + f(0) = f(2x f(0)) + f(f(x) + f(0))$.

Thus, $f(2xf(0)) = f(0)$. Since $f$ is nonconstant, $f(0) = 0$.

$P(x,0)$ now gives $f(f(x)) = f(x)$.

$P(x,y)$ with $P(x,f(y))$ gives $f(xy) = f(xf(y))$, so also $f(x) = f(xf(1))$.

We have $f(f(x) + f(y) - f(xy) + f(1)) = f(1)$.

$P(x, 2f(1)): f(x) + f(1) = f(2x) + f(f(x) + f(1))$, so $f(x) = f(2x)$. In this way, we get $f(x) = f(2f(x))$ as well, so $P(x,x)$ gives that $f(x) = f(x^2)$. Now comparing $P(x,y)$ and $P(x,y^2)$ gives $f(xy) = f(xy^2)$. For $y \ne 0$, setting $x = \frac 1y$ gives $f(y) = f(1)$, as desired.
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jasperE3
11305 posts
#10 • 1 Y
Y by AlexCenteno2007
Phorphyrion wrote:
Find all functions $f:\mathbb{R}\to \mathbb{R}$ such that for all $x, y\in \mathbb{R}$ the following holds:
\[f(x)+f(y)=f(xy)+f(f(x)+f(y))\]

Let $P(x,y)$ be the assertion $f(x)+f(y)=f(xy)+f(f(x)+f(y))$.
Note that all functions of the form $\boxed{f(x)=c}$, $c\in\mathbb R$ work, otherwise suppose $f$ is nonconstant.
$P(0,0)\Rightarrow f(0)=f(2f(0))$
$P(x,0)\Rightarrow f(x)=f(f(x)+f(0))$
$P(x,2f(0))\Rightarrow f(2xf(0))=f(0)$
Comparing the last two we get $f(0)=0$ since $f$ is nonconstant, so $f(f(x))=f(x)$. Repeat the same process with $1$ instead:
$P(1,1)\Rightarrow f(1)=f(2f(1))$
$P(x,1)\Rightarrow f(1)=f(f(x)+f(1))$
And now we can compare:
$P(x,f(1))\Rightarrow f(x)=f(xf(1))$
$P(x,2f(1))\Rightarrow f(x)=f(2xf(1))$
If $f(1)=0$ then $f$ is constant, so $f(1)\ne0$ and we get $f(x)=f(2x)$.
$P(x,x)\Rightarrow f\left(x^2\right)=f(x)$
$P\left(y,x^2\right)\Rightarrow f\left(x^2y\right)=f(xy)$
So setting $y=\frac1x$, we get $\boxed{f(x)=\begin{cases}0&\text{if }x=0\\c&\text{if }x\ne0\end{cases}}$ where $c\in\mathbb R_{\ne0}$ (since we had $c=f(1)$ here), which satisfies the equation (by casework on whether $x$ or $y$ is $0$).
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