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old and easy imo inequality
Valentin Vornicu   215
N 2 hours ago by cubres
Source: IMO 2000, Problem 2, IMO Shortlist 2000, A1
Let $ a, b, c$ be positive real numbers so that $ abc = 1$. Prove that
\[ \left( a - 1 + \frac 1b \right) \left( b - 1 + \frac 1c \right) \left( c - 1 + \frac 1a \right) \leq 1. \]
215 replies
Valentin Vornicu
Oct 24, 2005
cubres
2 hours ago
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x^2-x divides by n for some n/\omega(n)+1>x>1
NO_SQUARES   1
N 2 hours ago by a_507_bc
Source: 239 MO 2025 8-9 p6
Let a positive integer number $n$ has $k$ different prime divisors. Prove that there exists a positive integer number $x \in \left(1, \frac{n}{k}+1 \right)$ such that $x^2-x$ divides by $n$.
1 reply
NO_SQUARES
4 hours ago
a_507_bc
2 hours ago
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IMO Genre Predictions
ohiorizzler1434   46
N 2 hours ago by Mrcuberoot
Everybody, with IMO upcoming, what are you predictions for the problem genres?


Personally I predict: predict
46 replies
ohiorizzler1434
May 3, 2025
Mrcuberoot
2 hours ago
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4 wise men and 100 hats. 3 must guess their numbers
NO_SQUARES   1
N 3 hours ago by noemiemath
Source: 239 MO 2025 10-11 p5
There are four wise men in a row, each sees only those following him in the row, i.e. the $1$st sees the other three, the $2$nd sees the $3$rd and $4$th, and the $3$rd sees only the $4$th. The devil has $100$ hats, numbered from $1$ to $100$, he puts one hat on each wise man, and hides the extra $96$ hats. After that, each wise man (in turn: first the first, then the second, etc.) loudly calls a number, trying to guess the number of his hat. The numbers mentioned should not be repeated. When all the wise men have spoken, they take off their hats and check which one of them has guessed. Can the sages to act in such a way that at least three of them knowingly guessed?
1 reply
NO_SQUARES
3 hours ago
noemiemath
3 hours ago
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IMO Shortlist 2011, G4
WakeUp   126
N 3 hours ago by NuMBeRaToRiC
Source: IMO Shortlist 2011, G4
Let $ABC$ be an acute triangle with circumcircle $\Omega$. Let $B_0$ be the midpoint of $AC$ and let $C_0$ be the midpoint of $AB$. Let $D$ be the foot of the altitude from $A$ and let $G$ be the centroid of the triangle $ABC$. Let $\omega$ be a circle through $B_0$ and $C_0$ that is tangent to the circle $\Omega$ at a point $X\not= A$. Prove that the points $D,G$ and $X$ are collinear.

Proposed by Ismail Isaev and Mikhail Isaev, Russia
126 replies
WakeUp
Jul 13, 2012
NuMBeRaToRiC
3 hours ago
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<DPA+ <AQD =< QIP wanted, incircle circumcircle related
parmenides51   42
N 3 hours ago by AR17296174
Source: IMo 2019 SL G6
Let $I$ be the incentre of acute-angled triangle $ABC$. Let the incircle meet $BC, CA$, and $AB$ at $D, E$, and $F,$ respectively. Let line $EF$ intersect the circumcircle of the triangle at $P$ and $Q$, such that $F$ lies between $E$ and $P$. Prove that $\angle DPA + \angle AQD =\angle QIP$.

(Slovakia)
42 replies
1 viewing
parmenides51
Sep 22, 2020
AR17296174
3 hours ago
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Help my diagram has too many points
MarkBcc168   28
N 3 hours ago by AR17296174
Source: IMO Shortlist 2023 G6
Let $ABC$ be an acute-angled triangle with circumcircle $\omega$. A circle $\Gamma$ is internally tangent to $\omega$ at $A$ and also tangent to $BC$ at $D$. Let $AB$ and $AC$ intersect $\Gamma$ at $P$ and $Q$ respectively. Let $M$ and $N$ be points on line $BC$ such that $B$ is the midpoint of $DM$ and $C$ is the midpoint of $DN$. Lines $MP$ and $NQ$ meet at $K$ and intersect $\Gamma$ again at $I$ and $J$ respectively. The ray $KA$ meets the circumcircle of triangle $IJK$ again at $X\neq K$.

Prove that $\angle BXP = \angle CXQ$.

Kian Moshiri, United Kingdom
28 replies
MarkBcc168
Jul 17, 2024
AR17296174
3 hours ago
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A lot of circles
ryan17   8
N 3 hours ago by AR17296174
Source: 2019 Polish MO Finals
Denote by $\Omega$ the circumcircle of the acute triangle $ABC$. Point $D$ is the midpoint of the arc $BC$ of $\Omega$ not containing $A$. Circle $\omega$ centered at $D$ is tangent to the segment $BC$ at point $E$. Tangents to the circle $\omega$ passing through point $A$ intersect line $BC$ at points $K$ and $L$ such that points $B, K, L, C$ lie on the line $BC$ in that order. Circle $\gamma_1$ is tangent to the segments $AL$ and $BL$ and to the circle $\Omega$ at point $M$. Circle $\gamma_2$ is tangent to the segments $AK$ and $CK$ and to the circle $\Omega$ at point $N$. Lines $KN$ and $LM$ intersect at point $P$. Prove that $\sphericalangle KAP = \sphericalangle EAL$.
8 replies
ryan17
Jul 9, 2019
AR17296174
3 hours ago
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NT FE from Taiwan TST
Kitayama_Yuji   13
N 3 hours ago by bin_sherlo
Source: 2024 Taiwan TST Round 2 Mock P3
Let $\mathbb{N}$ be the set of all positive integers. Find all functions $f\colon \mathbb{N}\to \mathbb{N}$ such that $mf(m)+(f(f(m))+n)^2$ divides $4m^4+n^2f(f(n))^2$ for all positive integers $m$ and $n$.
13 replies
Kitayama_Yuji
Mar 29, 2024
bin_sherlo
3 hours ago
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Yet another domino problem
juckter   15
N 3 hours ago by lksb
Source: EGMO 2019 Problem 2
Let $n$ be a positive integer. Dominoes are placed on a $2n \times 2n$ board in such a way that every cell of the board is adjacent to exactly one cell covered by a domino. For each $n$, determine the largest number of dominoes that can be placed in this way.
(A domino is a tile of size $2 \times 1$ or $1 \times 2$. Dominoes are placed on the board in such a way that each domino covers exactly two cells of the board, and dominoes do not overlap. Two cells are said to be adjacent if they are different and share a common side.)
15 replies
juckter
Apr 9, 2019
lksb
3 hours ago
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J
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Nice geometry problem
henderson   5
N Jun 15, 2015 by sunken rock
Source: JBMO Shortlist 2013 G4
Let $I$ be the incenter and $AB$ the shortest side of a triangle $ABC$. The circle with center $I$ and passing through $C$ intersects the ray $AB$ at the point $P$ and the ray $BA$ at the point $Q$. Let $D$ be the point where the excircle of the triangle $ABC$ belonging to angle $A$ touches the side $BC$, and let $E$ be the symmetric of the point $C$ with respect to $D$. Show that the lines $PE$ and $CQ$ are perpendicular.
5 replies
henderson
Jun 3, 2015
sunken rock
Jun 15, 2015
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Nice geometry problem
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Reveal topic tags
geometry
incenter
JBMO
2013
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G H BBookmark kLocked kLocked NReply
Source: JBMO Shortlist 2013 G4
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henderson
312 posts
henderson
#1 h  Jun 3, 2015, 6:57 PM • 2 Y
Y by Adventure10, Mango247
Let $I$ be the incenter and $AB$ the shortest side of a triangle $ABC$. The circle with center $I$ and passing through $C$ intersects the ray $AB$ at the point $P$ and the ray $BA$ at the point $Q$. Let $D$ be the point where the excircle of the triangle $ABC$ belonging to angle $A$ touches the side $BC$, and let $E$ be the symmetric of the point $C$ with respect to $D$. Show that the lines $PE$ and $CQ$ are perpendicular.
This post has been edited 1 time. Last edited by henderson, Jun 3, 2015, 7:12 PM
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colinhy
751 posts
colinhy
#2 h  Jun 3, 2015, 7:35 PM • 2 Y
Y by Adventure10, Mango247
Solution:
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henderson
312 posts
henderson
#3 h  Jun 8, 2015, 6:51 PM • 2 Y
Y by Adventure10, Mango247
Can you explain why $BX = DC$?
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colinhy
751 posts
colinhy
#4 h  Jun 8, 2015, 7:20 PM • 2 Y
Y by Adventure10, Mango247
I think it is common knowledge that $BX = DC$ (you can check out Yufei's handouts for a proof, but it's just length chasing using the fact that tangents from a point are of equal length). Thus, if $M$ is the midpoint of $BC$, then $XM = MD$. Then, since $EC = 2 \cdot DC$ and $BC = 2 \cdot MC$, $BE = 2 \cdot MD = XD$.
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aleksam
101 posts
aleksam
#6 h  Jun 13, 2015, 4:15 PM • 2 Y
Y by Adventure10, Mango247
Let $X$ and $Y$ be the intersection of $CA$ and $CB$, respectivly, with the circle $k(I, CI)$ and $A'$, $B'$ and $C'$ are the projection of $I$ to sides $BC$, $CA$ and $AB$ respectivly. From the fact that $B'$ is the midpoint of $CX$, we have that $AX=XB'-B'A=(s-AB)-(s-BC)=BC-AB$, where s is the semiperimeter of the triangle $ABC$. Now, it is easy to see that triangles $IC'Q$ and $IA'C$ are congruent($IC'=IA', IQ=IC, <IC'Q=<IA'C$) and so $C'Q=A'C$, and we easily deduce $AQ=BC-AB=AX$. So, after short angle chasnig we find $<PCY=(<ABC-<ACB)/2$ and similarly $<XCQ=(<BAC-<ACB)/2$. Now, as in colinhy's solution we can get $<PEB=<ABC/2$, and as $<QCB=90-<ABC/2$ the result follows.
This post has been edited 1 time. Last edited by aleksam, Jun 13, 2015, 4:16 PM
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sunken rock
4391 posts
sunken rock
#7 h  Jun 15, 2015, 1:12 PM • 2 Y
Y by Adventure10, Mango247
From second post, after having seen $BE=BP$, i.e. $\angle BPE=\frac{\hat B}2\ (\ 1\ )$ we can continue with $BQ=BC$, so $\angle BQC=\frac{\hat A+\hat C}2\ (\ 2\ )$, and adding $(1)$ and $(2)$ we get $PE\bot CQ$.

Best regards,
sunken rock
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