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jlacosta   0
Today at 3:57 PM
Congratulations to all the mathletes who competed at National MATHCOUNTS! If you missed the exciting Countdown Round, you can watch the video at this link. Are you interested in training for MATHCOUNTS or AMC 10 contests? How would you like to train for these math competitions in half the time? We have accelerated sections which meet twice per week instead of once starting on July 8th (7:30pm ET). These sections fill quickly so enroll today!

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0 replies
1 viewing
jlacosta
Today at 3:57 PM
0 replies
Number Theory
TheMathBob   4
N a few seconds ago by fe.
Source: Polish Math Olympiad 2021 2nd round p3 day 1
Positive integers $a,b,z$ satisfy the equation $ab=z^2+1$. Prove that there exist positive integers $x,y$ such that
$$\frac{a}{b}=\frac{x^2+1}{y^2+1}$$
4 replies
TheMathBob
Feb 13, 2021
fe.
a few seconds ago
IMO Shortlist 2014 N4
hajimbrak   74
N 17 minutes ago by LenaEnjoyer
Let $n > 1$ be a given integer. Prove that infinitely many terms of the sequence $(a_k )_{k\ge 1}$, defined by \[a_k=\left\lfloor\frac{n^k}{k}\right\rfloor,\] are odd. (For a real number $x$, $\lfloor x\rfloor$ denotes the largest integer not exceeding $x$.)

Proposed by Hong Kong
74 replies
hajimbrak
Jul 11, 2015
LenaEnjoyer
17 minutes ago
USAMO 2003 Problem 1
MithsApprentice   71
N an hour ago by cubres
Prove that for every positive integer $n$ there exists an $n$-digit number divisible by $5^n$ all of whose digits are odd.
71 replies
MithsApprentice
Sep 27, 2005
cubres
an hour ago
IMO Genre Predictions
ohiorizzler1434   76
N 2 hours ago by aidan0626
Everybody, with IMO upcoming, what are you predictions for the problem genres?


Personally I predict: predict
76 replies
ohiorizzler1434
May 3, 2025
aidan0626
2 hours ago
No more topics!
Functional Geometry
GreekIdiot   2
N Apr 30, 2025 by Double07
Source: BMO 2024 SL G7
Let $f: \pi \to \mathbb R$ be a function from the Euclidean plane to the real numbers such that $f(A)+f(B)+f(C)=f(O)+f(G)+f(H)$ for any acute triangle $\Delta ABC$ with circumcenter $O$, centroid $G$ and orthocenter $H$. Prove that $f$ is constant.
2 replies
GreekIdiot
Apr 27, 2025
Double07
Apr 30, 2025
Functional Geometry
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G H BBookmark kLocked kLocked NReply
Source: BMO 2024 SL G7
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GreekIdiot
277 posts
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Let $f: \pi \to \mathbb R$ be a function from the Euclidean plane to the real numbers such that $f(A)+f(B)+f(C)=f(O)+f(G)+f(H)$ for any acute triangle $\Delta ABC$ with circumcenter $O$, centroid $G$ and orthocenter $H$. Prove that $f$ is constant.
This post has been edited 1 time. Last edited by GreekIdiot, Apr 27, 2025, 1:08 PM
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ItzsleepyXD
151 posts
#2
Y by
Consider $\triangle ABC$ with circumcenter $O$, centroid $G$ and orthocenter $H$.
$D,E,F,M_{OG},M_{OH},M_{HG}$ is midpoint of $BC,CA,AB,OG,OH,HG$
Define $O_A,G_A,H_A$ be circumcenter, centroid and orthocenter of $\triangle AEF$.
and $H_B,O_B,G_B,H_C,O_C,G_C$ analogously
by $\triangle ABC, \triangle AFE , \triangle BFD , \triangle CDE , \triangle DEF$ and some calculation
known that $f(O)+f(G)=f(M_{OG})+f(M_{HG})$
Let $M'$ be reflection of $M_{HG}$ wrt point $M_{OG}$ notice that $f(M_{HG}=f(M')$
so f is constant. done $\square$
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Double07
94 posts
#3
Y by
Use complex numbers:

For a point $A(a)$ we use $f(a)=f(A)$.

WLOG suppose $f(0)=0$.

Let $\omega=\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}$.

We have $f(a)+f(\omega a)+f(\omega^2 a)=3\cdot f(0)=0, \forall a\in \mathbb{C}$.

We also have $f(x)+f(\omega a)+f(\omega^2 a)=0+f\left(\dfrac{x-a}{3}\right)+f(x-a), \forall a,x\in \mathbb{C}$ with $|a|=|x|>|a-x|$ (so that the triangle is acute).

So $f(x)-f(a)=f\left(\dfrac{x-a}{3}\right)+f(x-a)$.

Because $f(x)-f(y)$ depends on just $x-y$ (for $x,y$ complex with $|x|=|y|$), we can consider function $g$ such that $g(x-y)=f(x)-f(y)$.

So $g(x-a)=f\left(\dfrac{x-a}{3}\right)+f(x-a)$.

Notice that $g$ is additive: $g(x-y)+g(y-z)=f(x)-f(y)+f(y)-f(z)=f(x)-f(z)=g(x-z)$.

So $g(nx)=ng(x)$, $\forall n\in \mathbb{Z}$ and $x\in\mathbb{C}$.

So $g(x)+g(-x)=0\implies g(a-x)+g((-a)-(-x))=0\implies f(a)-f(x)+f(-a)-f(-x)=0\implies$
$\implies f(x)+f(-x)=f(a)+f(-a)$, so $f(x)+f(-x)$ is a constant $c$.

But $f(1)+f(\omega)+f(\omega^2)=0=-(f(-1)+f(-\omega)+f(-\omega^2))\implies 3c=0\implies c=0\implies f$ is odd.

$f(x)-f(y)=g(x-y)\implies f(x)+f(-y)=g(x-y)\implies f(x)+f(y)=g(x+y)$, where $|x|=|y|$.

Choose $x=\omega a$ and $y=\omega^2 a$ in the above to get $-f(a)=f(\omega a)+f(\omega^2 a)=g(-a)=-g(a)\implies f\equiv g$.

But we know that $g(x)=f(x)+f\left(\frac{x}{3}\right)\implies f\left(\frac{x}{3}\right)=0, \forall x\in \mathbb{C}$, so we're done.
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