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Iran Team Selection Test 2016
MRF2017   9
N 26 minutes ago by SimplisticFormulas
Source: TST3,day1,P2
Let $ABC$ be an arbitrary triangle and $O$ is the circumcenter of $\triangle {ABC}$.Points $X,Y$ lie on $AB,AC$,respectively such that the reflection of $BC$ WRT $XY$ is tangent to circumcircle of $\triangle {AXY}$.Prove that the circumcircle of triangle $AXY$ is tangent to circumcircle of triangle $BOC$.
9 replies
MRF2017
Jul 15, 2016
SimplisticFormulas
26 minutes ago
non-symmetric ineq (for girls)
easternlatincup   36
N 2 hours ago by Tony_stark0094
Source: Chinese Girl's MO 2007
For $ a,b,c\geq 0$ with $ a+b+c=1$, prove that

$ \sqrt{a+\frac{(b-c)^2}{4}}+\sqrt{b}+\sqrt{c}\leq \sqrt{3}$
36 replies
easternlatincup
Dec 30, 2007
Tony_stark0094
2 hours ago
Turbo's en route to visit each cell of the board
Lukaluce   20
N 2 hours ago by Mathgloggers
Source: EGMO 2025 P5
Let $n > 1$ be an integer. In a configuration of an $n \times n$ board, each of the $n^2$ cells contains an arrow, either pointing up, down, left, or right. Given a starting configuration, Turbo the snail starts in one of the cells of the board and travels from cell to cell. In each move, Turbo moves one square unit in the direction indicated by the arrow in her cell (possibly leaving the board). After each move, the arrows in all of the cells rotate $90^{\circ}$ counterclockwise. We call a cell good if, starting from that cell, Turbo visits each cell of the board exactly once, without leaving the board, and returns to her initial cell at the end. Determine, in terms of $n$, the maximum number of good cells over all possible starting configurations.

Proposed by Melek Güngör, Turkey
20 replies
Lukaluce
Apr 14, 2025
Mathgloggers
2 hours ago
Divisibility on 101 integers
BR1F1SZ   3
N 2 hours ago by ClassyPeach
Source: Argentina Cono Sur TST 2024 P2
There are $101$ positive integers $a_1, a_2, \ldots, a_{101}$ such that for every index $i$, with $1 \leqslant i \leqslant 101$, $a_i+1$ is a multiple of $a_{i+1}$. Determine the greatest possible value of the largest of the $101$ numbers.
3 replies
BR1F1SZ
Aug 9, 2024
ClassyPeach
2 hours ago
BMO 2021 problem 3
VicKmath7   19
N 2 hours ago by NuMBeRaToRiC
Source: Balkan MO 2021 P3
Let $a, b$ and $c$ be positive integers satisfying the equation $(a, b) + [a, b]=2021^c$. If $|a-b|$ is a prime number, prove that the number $(a+b)^2+4$ is composite.

Proposed by Serbia
19 replies
VicKmath7
Sep 8, 2021
NuMBeRaToRiC
2 hours ago
USAMO 2002 Problem 4
MithsApprentice   89
N 3 hours ago by blueprimes
Let $\mathbb{R}$ be the set of real numbers. Determine all functions $f: \mathbb{R} \to \mathbb{R}$ such that \[ f(x^2 - y^2) = x f(x) - y f(y)  \] for all pairs of real numbers $x$ and $y$.
89 replies
MithsApprentice
Sep 30, 2005
blueprimes
3 hours ago
pqr/uvw convert
Nguyenhuyen_AG   8
N 3 hours ago by Victoria_Discalceata1
Source: https://github.com/nguyenhuyenag/pqr_convert
Hi everyone,
As we know, the pqr/uvw method is a powerful and useful tool for proving inequalities. However, transforming an expression $f(a,b,c)$ into $f(p,q,r)$ or $f(u,v,w)$ can sometimes be quite complex. That's why I’ve written a program to assist with this process.
I hope you’ll find it helpful!

Download: pqr_convert

Screenshot:
IMAGE
IMAGE
8 replies
Nguyenhuyen_AG
Apr 19, 2025
Victoria_Discalceata1
3 hours ago
Inspired by hlminh
sqing   2
N 3 hours ago by SPQ
Source: Own
Let $ a,b,c $ be real numbers such that $ a^2+b^2+c^2=1. $ Prove that $$ |a-kb|+|b-kc|+|c-ka|\leq \sqrt{3k^2+2k+3}$$Where $ k\geq 0 . $
2 replies
sqing
Yesterday at 4:43 AM
SPQ
3 hours ago
A cyclic inequality
KhuongTrang   3
N 3 hours ago by KhuongTrang
Source: own-CRUX
IMAGE
https://cms.math.ca/.../uploads/2025/04/Wholeissue_51_4.pdf
3 replies
KhuongTrang
Monday at 4:18 PM
KhuongTrang
3 hours ago
Tiling rectangle with smaller rectangles.
MarkBcc168   60
N 4 hours ago by cursed_tangent1434
Source: IMO Shortlist 2017 C1
A rectangle $\mathcal{R}$ with odd integer side lengths is divided into small rectangles with integer side lengths. Prove that there is at least one among the small rectangles whose distances from the four sides of $\mathcal{R}$ are either all odd or all even.

Proposed by Jeck Lim, Singapore
60 replies
MarkBcc168
Jul 10, 2018
cursed_tangent1434
4 hours ago
ALGEBRA INEQUALITY
Tony_stark0094   2
N 4 hours ago by Sedro
$a,b,c > 0$ Prove that $$\frac{a^2+bc}{b+c} + \frac{b^2+ac}{a+c} + \frac {c^2 + ab}{a+b} \geq a+b+c$$
2 replies
Tony_stark0094
5 hours ago
Sedro
4 hours ago
Point that is orthocenter, incenter, and centroid
EmersonSoriano   1
N Apr 5, 2025 by hukilau17
Source: 2017 Peru Southern Cone TST P9
Let $BXC$ be a triangle and $A_1, A_2, A_3$ points in the same plane such that $X$ is the orthocenter of triangle $A_1BC$, $X$ is the incenter of triangle $A_2BC$, and $X$ is the centroid of triangle $A_3BC$. If line $A_1A_3$ is parallel to $BC$, prove that $A_2$ is the midpoint of segment $A_1A_3$.
1 reply
EmersonSoriano
Apr 5, 2025
hukilau17
Apr 5, 2025
Point that is orthocenter, incenter, and centroid
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Source: 2017 Peru Southern Cone TST P9
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EmersonSoriano
44 posts
#1 • 1 Y
Y by PikaPika999
Let $BXC$ be a triangle and $A_1, A_2, A_3$ points in the same plane such that $X$ is the orthocenter of triangle $A_1BC$, $X$ is the incenter of triangle $A_2BC$, and $X$ is the centroid of triangle $A_3BC$. If line $A_1A_3$ is parallel to $BC$, prove that $A_2$ is the midpoint of segment $A_1A_3$.
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hukilau17
283 posts
#2 • 1 Y
Y by PikaPika999
Not bad to complex bash. Let $\triangle BXC$ be inscribed in the unit circle so that
$$|b|=|c|=|x|=1$$Since $A_1$ is the orthocenter of $\triangle BXC$, we have
$$a_1 = b+c+x$$Since $X$ is the centroid of $\triangle A_3BC$, we have $x = \frac{a_3+b+c}3$ and so
$$a_3 = 3x-b-c$$And since $BX$ bisects $\angle A_2BC$ and $CX$ bisects $\angle A_2CB$, we have
$$a_2 = \frac{b\left(\frac{x^2}c\right)\left(c+\frac{x^2}b\right) - c\left(\frac{x^2}b\right)\left(b+\frac{x^2}c\right)}{b\left(\frac{x^2}c\right) - c\left(\frac{x^2}b\right)} = \frac{bc+x^2}{b+c}$$Since $A_1A_3\parallel BC$, we have
$$\frac{a_1-a_3}{b-c} \in \mathbb{R}$$$$\frac{2b+2c-2x}{b-c} = \frac{2bx+2cx-2bc}{cx-bx}$$$$bx+cx-x^2 = bc-bx-cx$$$$2bx+2cx = bc+x^2$$Then rearranging gives
$$a_2 = \frac{bc+x^2}{b+c} = 2x = \frac{(b+c+x) + (3x-b-c)}2 = \frac{a_1+a_3}2$$as desired. $\blacksquare$
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