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ISI UGB 2025 P7
SomeonecoolLovesMaths   5
N 21 minutes ago by quasar_lord
Source: ISI UGB 2025 P7
Consider a ball that moves inside an acute-angled triangle along a straight line, unit it hits the boundary, which is when it changes direction according to the mirror law, just like a ray of light (angle of incidence = angle of reflection). Prove that there exists a triangular periodic path for the ball, as pictured below.

IMAGE
5 replies
SomeonecoolLovesMaths
2 hours ago
quasar_lord
21 minutes ago
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ISI UGB 2025 P4
SomeonecoolLovesMaths   1
N 24 minutes ago by quasar_lord
Source: ISI UGB 2025 P4
Let $S^1 = \{ z \in \mathbb{C} \mid |z| =1 \}$ be the unit circle in the complex plane. Let $f \colon S^1 \longrightarrow S^2$ be the map given by $f(z) = z^2$. We define $f^{(1)} \colon = f$ and $f^{(k+1)} \colon = f \circ f^{(k)}$ for $k \geq 1$. The smallest positive integer $n$ such that $f^{(n)}(z) = z$ is called the period of $z$. Determine the total number of points in $S^1$ of period $2025$.
(Hint : $2025 = 3^4 \times 5^2$)
1 reply
SomeonecoolLovesMaths
2 hours ago
quasar_lord
24 minutes ago
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Asymmetric FE
sman96   14
N 33 minutes ago by mkultra42
Source: BdMO 2025 Higher Secondary P8
Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that$$f(xf(y)-y) + f(xy-x) + f(x+y) = 2xy$$for all $x, y \in \mathbb{R}$.
14 replies
sman96
Feb 8, 2025
mkultra42
33 minutes ago
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ISI UGB 2025 P2
SomeonecoolLovesMaths   1
N 35 minutes ago by Project_Donkey_into_M4
Source: ISI UGB 2025 P2
If the interior angles of a triangle $ABC$ satisfy the equality, $$\sin ^2 A + \sin ^2 B + \sin^2 C = 2 \left( \cos ^2 A + \cos ^2 B + \cos ^2 C \right),$$prove that the triangle must have a right angle.
1 reply
1 viewing
SomeonecoolLovesMaths
2 hours ago
Project_Donkey_into_M4
35 minutes ago
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hard inequality
moldovan   2
N 36 minutes ago by ririgggg
Source: Austria 1987
Let $ x_1,...,x_n$ be positive real numbers. Prove that:

$ \displaystyle\sum_{k=1}^{n}x_k+\sqrt{\displaystyle\sum_{k=1}^{n}x_k^2} \le \frac{n+\sqrt{n}}{n^2} \left( \displaystyle\sum_{k=1}^{n} \frac{1}{x_k} \right) \left( \displaystyle\sum_{k=1}^{n} x_k^2 \right).$
2 replies
moldovan
Jul 8, 2009
ririgggg
36 minutes ago
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ISI UGB 2025 P5
SomeonecoolLovesMaths   3
N 39 minutes ago by lakshya2009
Source: ISI UGB 2025 P5
Let $a,b,c$ be nonzero real numbers such that $a+b+c \neq 0$. Assume that $$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{a+b+c}$$Show that for any odd integer $k$, $$\frac{1}{a^k} + \frac{1}{b^k} + \frac{1}{c^k} = \frac{1}{a^k+b^k+c^k}.$$
3 replies
SomeonecoolLovesMaths
2 hours ago
lakshya2009
39 minutes ago
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Radiant sets
BR1F1SZ   1
N an hour ago by alexheinis
Source: 2025 Francophone MO Juniors P1
A finite set $\mathcal S$ of distinct positive real numbers is called radiant if it satisfies the following property: if $a$ and $b$ are two distinct elements of $\mathcal S$, then $a^2 + b^2$ is also an element of $\mathcal S$.
[list=a]
[*]Does there exist a radiant set with a size greater than or equal to $4$?
[*]Determine all radiant sets of size $2$ or $3$.
[/list]
1 reply
BR1F1SZ
Yesterday at 11:12 PM
alexheinis
an hour ago
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Knights NOT crowded on the chessboard
mshtand1   1
N an hour ago by sarjinius
Source: Ukrainian Mathematical Olympiad 2025. Day 1, Problem 8.4
What is the maximum number of knights that can be placed on a chessboard of size \(8 \times 8\) such that any knight, after making 1 or 2 arbitrary moves, does not land on a square occupied by another knight?

Proposed by Bogdan Rublov
1 reply
mshtand1
Mar 13, 2025
sarjinius
an hour ago
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Taking antipode on isosceles triangle's circumcenter
Nuran2010   0
2 hours ago
Source: Azerbaijan Al-Khwarizmi IJMO TST 2025
In isosceles triangle, the condition $AB=AC>BC$ is satisfied. Point $D$ is taken on the circumcircle of $ABC$ such that $\angle CAD=90^{\circ}$.A line parallel to $AC$ which passes from $D$ intersects $AB$ and $BC$ respectively at $E$ and $F$.Show that circumcircle of $ADE$ passes from circumcenter of $DFC$.
0 replies
Nuran2010
2 hours ago
0 replies
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R to R, with x+f(xy)=f(1+f(y))x
NicoN9   3
N 2 hours ago by EeEeRUT
Source: Own.
Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that\[ x+f(xy)=f(1+f(y))x \]for all $x, y\in \mathbb{R}$.
3 replies
NicoN9
5 hours ago
EeEeRUT
2 hours ago
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find angle
TBazar   7
N 2 hours ago by TBazar
Given $ABC$ triangle with $AC>BC$. We take $M$, $N$ point on AC, AB respectively such that $AM=BC$, $CM=BN$. $BM$, $AN$ lines intersect at point $K$. If $2\angle AKM=\angle ACB$, find $\angle ACB$
7 replies
TBazar
May 8, 2025
TBazar
2 hours ago
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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
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Functional Geometry
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Reveal topic tags
geometry
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circumcircle
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Source: BMO 2024 SL G7
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GreekIdiot
222 posts
GreekIdiot
#1 h  Apr 27, 2025, 1:08 PM
Y by
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
141 posts
ItzsleepyXD
#2 h  Apr 30, 2025, 3:50 AM
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
85 posts
Double07
#3 h  Apr 30, 2025, 8:29 PM
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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