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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
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0 replies
jlacosta
Mar 2, 2025
0 replies
H not needed
dchenmathcounts   44
N 17 minutes ago by Ilikeminecraft
Source: USEMO 2019/1
Let $ABCD$ be a cyclic quadrilateral. A circle centered at $O$ passes through $B$ and $D$ and meets lines $BA$ and $BC$ again at points $E$ and $F$ (distinct from $A,B,C$). Let $H$ denote the orthocenter of triangle $DEF.$ Prove that if lines $AC,$ $DO,$ $EF$ are concurrent, then triangle $ABC$ and $EHF$ are similar.

Robin Son
44 replies
dchenmathcounts
May 23, 2020
Ilikeminecraft
17 minutes ago
IZHO 2017 Functional equations
user01   51
N 38 minutes ago by lksb
Source: IZHO 2017 Day 1 Problem 2
Find all functions $f:R \rightarrow R$ such that $$(x+y^2)f(yf(x))=xyf(y^2+f(x))$$, where $x,y \in \mathbb{R}$
51 replies
user01
Jan 14, 2017
lksb
38 minutes ago
chat gpt
fuv870   2
N 39 minutes ago by fuv870
The chat gpt alreadly knows how to solve the problem of IMO USAMO and AMC?
2 replies
fuv870
an hour ago
fuv870
39 minutes ago
Inequality with wx + xy + yz + zw = 1
Fermat -Euler   23
N 41 minutes ago by hgomamogh
Source: IMO ShortList 1990, Problem 24 (THA 2)
Let $ w, x, y, z$ are non-negative reals such that $ wx + xy + yz + zw = 1$.
Show that $ \frac {w^3}{x + y + z} + \frac {x^3}{w + y + z} + \frac {y^3}{w + x + z} + \frac {z^3}{w + x + y}\geq \frac {1}{3}$.
23 replies
Fermat -Euler
Nov 2, 2005
hgomamogh
41 minutes ago
No more topics!
A periodic function
Rushil   6
N May 12, 2022 by parthvartak
Source: IMO 1968 B2
Let $f$ be a real-valued function defined for all real numbers, such that for some $a>0$ we have \[ f(x+a)={1\over2}+\sqrt{f(x)-f(x)^2} \] for all $x$.
Prove that $f$ is periodic, and give an example of such a non-constant $f$ for $a=1$.
6 replies
Rushil
Nov 4, 2005
parthvartak
May 12, 2022
A periodic function
G H J
Source: IMO 1968 B2
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Rushil
1592 posts
#1 • 4 Y
Y by Adventure10, ImSh95, Mango247, Mango247
Let $f$ be a real-valued function defined for all real numbers, such that for some $a>0$ we have \[ f(x+a)={1\over2}+\sqrt{f(x)-f(x)^2} \] for all $x$.
Prove that $f$ is periodic, and give an example of such a non-constant $f$ for $a=1$.
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spider_boy
210 posts
#2 • 4 Y
Y by Adventure10, Adventure10, ImSh95, Mango247
i think you can solve by plugging x->x+a,x->x+2a etc.
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Davron
484 posts
#3 • 3 Y
Y by Adventure10, ImSh95, Mango247
Directly from the equality given: f(x+a) ≥ 1/2 for all x, and hence f(x) ≥ 1/2 for all x.

So f(x+2a) = 1/2 + √( f(x+a) - f(x+a)2 ) = 1/2 + √f(x+a) √(1 - f(x+a)) = 1/2 + √(1/4 - f(x) + f(x)2) = 1/2 + (f(x) - 1/2) = f(x). So f is periodic with period 2a.

We may take f(x) to be arbitrary in the interval [0,1). For example, let f(x) = 1 for 0 ≤ x < 1, f(x) = 1/2 for 1 ≤ x < 2. Then use f(x+2) = f(x) to define f(x) for all other values of x.

kalva
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ivanbart-15
296 posts
#4 • 2 Y
Y by ImSh95, Adventure10
Since $\sqrt{f(x)-(f(x))^{2}}\geq 0$, $f(x+a)\geq \frac{1}{2}$ and $f(x)\geq \frac{1}{2}$.
Squaring the first equation, we obtain:
$(f(x+a))^{2}-f(x+a)+\frac{1}{4}=f(x)-(f(x))^{2}$, which is equivalent to: $|f(x)-\frac{1}{2}|=\sqrt{f(x+a)-(f(x+a))^{2}}$. (*)

Plugging $x+a$ instead of $x$ into the original equation, we get $f(x+2a)-\frac{1}{2}=\sqrt{f(x+a)-(f(x+a))^{2}}$. Backing this into (*) and applying the condition $f(x)\geq \frac{1}{2}$, we get: $ f(x)-\frac{1}{2}=f(x+2a)-\frac{1}{2} $, or equivalently $f(x)=f(x+2a)$.
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DottedCaculator
7303 posts
#5 • 2 Y
Y by centslordm, ImSh95
Solution
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ZETA_in_olympiad
2211 posts
#6 • 1 Y
Y by ImSh95
The functional equation is equivalent to $f(x+a)=\sqrt{f(x)(1-f(x))}+1/2 \implies (f(x+a)-1/2)^2=1/4-(f(x)-1/2)^2 \implies f(x+2a)=f(x).$ Hence $f$ is periodic.

For example, $f(x)=\frac{1}{2}(1+|\cos \frac{\pi x}{2}|),$ which is seen to work. And thus we're done.
This post has been edited 1 time. Last edited by ZETA_in_olympiad, May 12, 2022, 12:34 PM
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parthvartak
12 posts
#7 • 1 Y
Y by ImSh95
A way of aproaching this problem is to simply substitute $x=x+a$.
$\therefore f(x+2a)=\frac{1}{2}+\sqrt{f(x+a)-f(x+a)^2}$
Now substitute $ f(x+a)={1\over2}+\sqrt{f(x)-f(x)^2}$
$\therefore f(x+a)-f(x+a)^2={1\over2}+\sqrt{f(x)-f(x)^2} - ({1\over2}+\sqrt{f(x)-f(x)^2})^{2}={1\over4}-f(x)+f(x)^2=(f(x)-\frac{1}{2})^2.$
$\therefore f(x+2a)=\frac{1}{2}+\sqrt{(f(x)-\frac{1}{2})^2}=\frac{1}{2}+|f(x)-\frac{1}{2}|.$
Note that $\sqrt{a^2}=|a|\neq a$ for negative $a$.
So we must prove $f(x)\geq\frac{1}{2} \forall x\in \mathbb{R}$ to complete our proof.
Luckily, this is easy as $\sqrt a\geq 0 \forall a\in\mathbb{R} \Rightarrow f(x+a)\geq\frac{1}{2} \Rightarrow f(x)\geq \frac{1}{2} \forall x\in \mathbb{R}$.
$\therefore f(x+2a)=f(x) \forall x\in \mathbb{R}$ and we are done!
So, $f(x)$ is periodic with period $2a$.

Finding an example is pretty elusive, but the periodicity should give us the hint of using trignometry.
But $f(x)\geq\frac{1}{2} \forall x\in \mathbb{R}.$ and there is no such condition on trignometric functions, we may also have to use the modulus function.
Combining these two along with the bound on $f(x)$ we get that the function may be of the form$f(x)=\frac{1}{2}+|g(x)|$ where $g(x)$ may be a trignometric function which contains $x$.
After fiddling around with it a little bit we see that $f()=\frac{1}{2}+\frac{1}{2}|cos(\frac{\pi x}{2})|$ satisfies the condition, completing the 2nd part of the problem!
Hope you liked it!!! :thumbup:
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