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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
Converse of a classic orthocenter problem
spartacle   43
N 5 minutes ago by ihategeo_1969
Source: USA TSTST 2020 Problem 6
Let $A$, $B$, $C$, $D$ be four points such that no three are collinear and $D$ is not the orthocenter of $ABC$. Let $P$, $Q$, $R$ be the orthocenters of $\triangle BCD$, $\triangle CAD$, $\triangle ABD$, respectively. Suppose that the lines $AP$, $BQ$, $CR$ are pairwise distinct and are concurrent. Show that the four points $A$, $B$, $C$, $D$ lie on a circle.

Andrew Gu
43 replies
spartacle
Dec 14, 2020
ihategeo_1969
5 minutes ago
Symmetric points part 2
CyclicISLscelesTrapezoid   22
N 6 minutes ago by ihategeo_1969
Source: USA TSTST 2022/6
Let $O$ and $H$ be the circumcenter and orthocenter, respectively, of an acute scalene triangle $ABC$. The perpendicular bisector of $\overline{AH}$ intersects $\overline{AB}$ and $\overline{AC}$ at $X_A$ and $Y_A$ respectively. Let $K_A$ denote the intersection of the circumcircles of triangles $OX_AY_A$ and $BOC$ other than $O$.

Define $K_B$ and $K_C$ analogously by repeating this construction two more times. Prove that $K_A$, $K_B$, $K_C$, and $O$ are concyclic.

Hongzhou Lin
22 replies
CyclicISLscelesTrapezoid
Jun 27, 2022
ihategeo_1969
6 minutes ago
Periodicity of factorials
Cats_on_a_computer   0
24 minutes ago
Source: Thrill and challenge of pre-college mathematics
Let a_k denote the first non zero digit of the decimal representation of k!. Does the sequence a_1, a_2, a_3, … eventually become periodic?
0 replies
Cats_on_a_computer
24 minutes ago
0 replies
Cyclic Quad. and Intersections
Thelink_20   11
N 34 minutes ago by americancheeseburger4281
Source: My Problem
Let $ABCD$ be a quadrilateral inscribed in a circle $\Gamma$. Let $AC\cap BD=E$, $AB\cap CD=F$, $(AEF)\cap\Gamma=X$, $(BEF)\cap\Gamma=Y$, $(CEF)\cap\Gamma=Z$, $(DEF)\cap\Gamma=W$, $XZ\cap YW=M$, $XY\cap ZW=N$. Prove that $MN$ lies over $EF$.
11 replies
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Thelink_20
Oct 29, 2024
americancheeseburger4281
34 minutes ago
No more topics!
R+ Functional Equation
Mathdreams   10
N Apr 15, 2025 by TestX01
Source: Nepal TST 2025, Problem 3
Find all functions $f : \mathbb{R}^+ \rightarrow \mathbb{R}^+$ such that \[f(f(x)) + xf(xy) = x + f(y)\]for all positive real numbers $x$ and $y$.

(Andrew Brahms, USA)
10 replies
Mathdreams
Apr 11, 2025
TestX01
Apr 15, 2025
R+ Functional Equation
G H J
G H BBookmark kLocked kLocked NReply
Source: Nepal TST 2025, Problem 3
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Mathdreams
1472 posts
#1 • 1 Y
Y by khan.academy
Find all functions $f : \mathbb{R}^+ \rightarrow \mathbb{R}^+$ such that \[f(f(x)) + xf(xy) = x + f(y)\]for all positive real numbers $x$ and $y$.

(Andrew Brahms, USA)
This post has been edited 1 time. Last edited by Mathdreams, Apr 11, 2025, 1:28 PM
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megarnie
5611 posts
#2 • 4 Y
Y by khan.academy, KevinYang2.71, abrahms, Alex-131
The only solutions are $f(x) = \frac cx$ for some constant $c$ and $f\equiv 1$, which clearly work.

Let $P(x,y)$ be the given assertion. Clearly $1$ is the only constant solution, so assume $f$ is not constant.

Claim: $f$ is injective.
Proof: Suppose $f(a) = f(b)$ for some positive reals $a,b$.

$P(x,a)$ with $P(x,b)$ gives that $f(xa) = f(xb)$ for all $x \in \mathbb R^{+}$.

Now, $P(a,x)$ compared with $P(b,x)$ gives $af(ax) - a = b f(bx) - b$, so $a(f(ax) - 1) = b (f(bx) - 1)$. But, since $f(ax) = f(bx)$, we have \[ a(f(ax) - 1) = b(f(ax) - 1) \]Since $f$ isn't constant, we can choose $x$ where $f(ax) \ne 1$, so $a = b$. $\square$

$P(x, f(x)): f(f(x)) + xf(xf(x)) = x + f(f(x))$, so $xf(xf(x)) = x \implies f(xf(x)) = 1$.

$P(1, y): f(f(1)) = 1$.

Injectivity implies $xf(x) = f(1)$ for all $x$, so $f(x) = \frac{f(1)}{x}$, and setting $c = f(1)$ gives the desired result.
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pco
23515 posts
#3 • 4 Y
Y by khan.academy, Maksat_B, Sedro, abrahms
Mathdreams wrote:
Find all functions $f : \mathbb{R}^+ \rightarrow \mathbb{R}^+$ such that \[f(f(x)) + xf(xy) = x + f(y)\]for all positive real numbers $x$ and $y$.
Let $P(x,y)$ be the assertion $f(f(x))+xf(xy)=x+f(y)$
Let $c=f(1)$ and $d=f(2)$

Subtracting $P(x,1)$ from $P(x,2)$, we get $f(2x)=f(x)+\frac{d-c}x$
Subtracting $P(2,1)$ from $P(2,x)$, we get $f(2x)=d+\frac{f(x)-c}2$

Subtracting : $f(x)=2\frac{c-d}x+2d-c$

Plugging $f(x)=\frac ax+b$ in original equation, we get $(a,b)=(\text{anything},0)$ or $(0,1)$ and solutions :
$\boxed{\text{S1 : }f(x)=1\quad\forall x>0}$, which indeed fits

$\boxed{\text{S2 : }f(x)=\frac ax\quad\forall x>0}$, which indeed fits, whatever is $a>0$
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jasperE3
11395 posts
#4 • 1 Y
Y by AlexCenteno2007
Let $P(x,y)$ be the assertion $f(f(x))+xf(xy)=x+f(y)$.
$P(x,f(x))\Rightarrow f(xf(x))=1$
$P(f(x),x)\Rightarrow f(f(f(x)))=f(x)$
$P(f(x),y)\Rightarrow f(x)f(yf(x))=f(y)$, in particular we have $f(f(x))=\frac{f(1)}{f(x)}$
$P(x,f(y))\Rightarrow xf(x)^2-(xf(y)+f(1))f(x)+f(y)f(1)=0\Rightarrow f(x)\in\left\{f(y),\frac{f(1)}x\right\}$ for each $x,y\in\mathbb R^+$ (we solved this as a quadratic in $f(x)$)
If $f(x)\ne\frac{f(1)}x$ for some $x$ then by varying $y$ over $\mathbb R^+$ we get that $f$ is constant, and testing, the only constant solution is $\boxed{f(x)=1}$ for all $x$.
Otherwise, $\boxed{f(x)=\frac cx}$ which works for any $c\in\mathbb R^+$.
This post has been edited 1 time. Last edited by jasperE3, Apr 11, 2025, 5:24 PM
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Tony_stark0094
69 posts
#6
Y by
if $f \equiv c$ then $c$ must be $1$ further assume $f$ is not constant:
it's easy to get $f(f(1))=1$
and observe that $f$ is injective:
now $P(x,f(x)): f(f(x))+xf(xf(x))=x+f(f(x)) \implies f(xf(x))=1$
from injectivity $xf(x)=f(1) \implies f(x)=\frac {f(1)}{x}$
hence $f(x)=1 \forall x \in R$ and $f(x)=\frac {f(1)}{x} \forall x \in R$ are the only solutions
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jasperE3
11395 posts
#7
Y by
Tony_stark0094 wrote:
it's easy to get $f(f(1))=1$
and observe that $f$ is injective:

how?
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Tony_stark0094
69 posts
#8
Y by
jasperE3 wrote:
Tony_stark0094 wrote:
it's easy to get $f(f(1))=1$
and observe that $f$ is injective:

how?

$P(1,1): f(f(1))+f(1)=1+f(1) \implies f(f(1))=1$
for injectivity assume $f(a)=f(b)$
then subtract $P(a,1)$ from $P(b,1)$
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jasperE3
11395 posts
#9
Y by
Tony_stark0094 wrote:
jasperE3 wrote:
Tony_stark0094 wrote:
it's easy to get $f(f(1))=1$
and observe that $f$ is injective:

how?

$P(1,1): f(f(1))+f(1)=1+f(1) \implies f(f(1))=1$
for injectivity assume $f(a)=f(b)$
then subtract $P(a,1)$ from $P(b,1)$

and what if $f(a)=f(b)=1$ (which is indeed the particular case $f(xf(x))=f(f(1))=1$ that you use injectivity for)
This post has been edited 1 time. Last edited by jasperE3, Apr 12, 2025, 7:35 AM
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ThatApollo777
73 posts
#10
Y by
Claim : the only solutions are $f(x) = 1$ and $f(x) = \frac{c}{x}$.
Pf : Its easy to check these work, we now show these are only solutions.

Claim 1: If $f$ is not injective, its identically $1$.
Let $P(x, y)$ be the assertion. $$P(1,1) \implies f(f(1)) = 1$$Assuming $f(a) = f(b)$ for $a \neq b$. Let $r = \frac{b}{a} \neq 1$. $$P(a, y) - P(b, y) \implies a(f(ay) - 1) = b(f(by)-1)$$Putting $y = \frac{f(1)}{a}$ we can conclude: $$f(rf(1)) = 1$$$$P(\frac{x}{f(1)}, f(1)) - P(\frac{x}{f(1)}, rf(1)) \implies f(x) = f(rx)$$$$P(x, \frac{t}{x}) - P(rx, \frac{t}{x}) \implies f(t) = 1$$Since $r \neq 1$.

Now, assuming $f$ is injective consider $$P(x, \frac{f(1)}{x}) : f(f(x)) = f(\frac{f(1)}{x}) \implies f(x) = \frac{f(1)}{x}$$
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cursed_tangent1434
653 posts
#11
Y by
The answers are $f(x) = 1$ for all $x\in \mathbb{R}^+$ and $f(x)= \frac{c}{x}$ for all $x\in \mathbb{R}^+$ for some fixed constant $c \in \mathbb{R}^+$. It’s easy to see that these functions satisfy the given equation. We now show these are the only solutions. Let $P(x,y)$ be the assertion that $f(f(x))+xf(xy)=x+f(y)$ for positive real numbers $x$ and $y$.

In what follows we assume that $f$ is not constant one. Say there does now exist some $x_0 \ne 1$ such that $f(x_0) \ne 1$ (i.e for all $x_0\ne 1$ we have $f(x_0)=1$). As $f$ is not constant one this indicates that $f(1) \ne 1$. Then, $P\left(x,\frac{1}{x}\right)$ yields,
\begin{align*}
f(f(x)) + xf(1) &= x+f\left(\frac{1}{x}\right)\\
f(1) + xf(1) &= x+1 \\
(x+1)f(1) &= x+1
\end{align*}which is a clear contradiction since $x+1>0$ and $f(1) \ne 1$. Thus, there indeed exists some $x_0 \ne 1$ such that $f(x_0) \ne 1$. Now, $P(1,1)$ implies that
\[f(f(1))+f(1)=1+f(1)\]from which we have $f(f(1))=1$. We now make the following observation.

Claim : The function $f$ is injective.

Proof : We first show that it is injective at all points except 1. For this, note that if there exists $t_1 \ne t_2 $ such that $f(t_1) =f(t_2) \ne 1$. Then, $P(t_1,1)$ and $P(t_2,1)$ yeild,
\[f(f(t_1))+t_1f(t_1)=t_1+f(1)\]\[f(f(t_2))+t_2f(t_2)=t_2+f(1)\]whose difference implies
\[(t_1-t_2)f(t_1)=t_1-t_2\]which since $f(t_1) \ne 1$ implies that $t_1=t_2$ which is a contradiction. Hence, $f$ is indeed injective at all points except 1.

With this observation in hand, consider $x_0 \ne 1$ such that $f(x_0) \ne 1$ and $\alpha$ such that $f(\alpha)=1$. Then, from $P\left(x_0 , \frac{\alpha}{x_0}\right)$ we have
\begin{align*}
f(f(x_0)) + x_0f(\alpha) &= x_0 + f\left(\frac{\alpha}{x_0}\right)\\
f(f(x_0)) &=  f\left(\frac{\alpha}{x_0}\right)
\end{align*}Now,
\[f(f(f(x_0))) = f\left(f\left(\frac{\alpha}{x_0}\right)\right)=f(x_0)\]which since $f(x_0) \ne 1$ implies $f(f(x_0))=x_0 \ne 1$. Thus,
\begin{align*}
f(f(x_0)) &=  f\left(\frac{\alpha}{x_0}\right)\\
f(x_0) &= \frac{\alpha}{x_0}
\end{align*}In particular, if there exists $\alpha_1,\alpha_2 \in \mathbb{R}^+$ such that $f(\alpha_1)=f(\alpha_2)=1$ we have
\[\frac{\alpha_1}{x_0} = f(x_0) = \frac{\alpha_2}{x_0}\]which implies $\alpha_1=\alpha_2$ proving the claim.

Thus, $f$ is injective and
\[f(x) = \frac{f(1)}{x}\]for all $x \ne f(1)$ and $f(f(1))=1$. This implies that indeed $f(x) = \frac{c}{x}$ for some fixed constant $c \in \mathbb{R}^+$ as desired.
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TestX01
341 posts
#12
Y by
cutie patootie problem

We claim either $f$ is one or $\frac{c}{x}$. These clearly work. Let $P(x,y)$ denote the assertion.
Firstly, $P(1,1)$ gives $f(f(1))=1$.
$P(x,1)$ yields $f(f(x))+xf(x)=x+f(1)$. Hence $f(f(x))=xf(1)-xf(x)$. Thus, subbing back, $xf(xy)+f(1)-xf(x)=f(y)$. Now take $x\to f(1)$ so we get
\[f(1)f(f(1)y)=f(y)\]Now, we will take $y\to f(1)y$ in $xf(xy)+f(1)-xf(x)=f(y)$ to get $\frac{xf(xy)}{f(1)}+f(1)-xf(x)=\frac{f(y)}{f(1)}$. Comparing this with $xf(xy)+f(1)-xf(x)=f(y)$, we have
\[xf(xy)\left(\frac{1}{f(1)}-1\right)=f(y)\left(\frac{1}{f(1)}-1\right)\]This gives us two cases. Either $f(1)=1$ or $xf(xy)=f(y)$ for all $x,y$. In the latter case, we would have $xf(x)=f(1)$, and $f(x)=\frac{c}{x}$ which is a solution.

Now, suppose $f(1)=1$. Then, taking $P\left(x,\frac{1}{x}\right)$, we have $f(f(x))+x=x+f\left(\frac{1}{x}\right)$ hence $f(f(x))=f\left(\frac{1}{x}\right)$. Assume that $f$ is not always constant, as if it was constant then we would have $cx=x$ hence $c=1$ as desired. We shall prove that $f$ is injective, which would finish as then cancelling one $f$ we get $f(x)=\frac{1}{x}$.

Let $f(a)=f(b)$ such that WLOG $\frac{b}{a}>1$. Take $y=a,b$, so we have $f(f(x))+xf(ax)=x+f(a)$ and $f(f(x))+xf(bx)=x+f(b)$. Subtracting we have
\[x(f(ax)-f(bx))=f(a)-f(b)=0\]Hence, as $x\neq 0$, we have
\[f(ax)=f(bx)\quad f(x)=f(cx)\]where $c=\frac{b}{a}>1$ by scaling down $x$.

Now, consider $P(cx,y)$ so we get $f(f(x))+cxf(xy)=cx+f(y)$. Comparing with $P(x,y)$ we have
\[xf(xy)(c-1)=x(c-1)\]Yet $c-1>0$. Thus, we have $xf(xy)=x$ or $f(xy)=1$. Taking $y=1$ gives $f(x)=1$ for all $x$, a contradiction as we have dealt with constant $f$.

Thus, $f$ must be injective, and we are done.
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