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Can someone explain how to do this purely geometrically?
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Cubes and squares
y-is-the-best-_ 61
N
by ezpotd
Source: IMO 2019 SL N2
Find all triples 
of positive integers such that 
.
of positive integers such that 
.
61 replies
Chess game challenge
adihaya 21
N
by Mr.Sharkman
Source: 2014 BAMO-12 #5
A chess tournament took place between 
players. Every player played every other player once, with no draws. In addition, each player had a numerical rating before the tournament began, with no two players having equal ratings. It turns out there were exactly 
games in which the lower-rated player beat the higher-rated player. Prove that there is some player who won no less than 
and no more than 
games.
players. Every player played every other player once, with no draws. In addition, each player had a numerical rating before the tournament began, with no two players having equal ratings. It turns out there were exactly 
games in which the lower-rated player beat the higher-rated player. Prove that there is some player who won no less than 
and no more than 
games.
21 replies
[ELMO2] The Multiplication Table
v_Enhance 27
N
by Mr.Sharkman
Source: ELMO 2015, Problem 2 (Shortlist N1)
Let 
, 
, and 
be positive integers. Prove that ![\[ \sum_{i = 1}^n \min\left(\left\lfloor \frac{x}{i} \right\rfloor, m \right) = \sum_{i = 1}^m \min\left(\left\lfloor \frac{x}{i} \right\rfloor, n \right). \]](//latex.artofproblemsolving.com/b/7/a/b7aebfe11f646d604d2e8cb9d5bd70ec796949e8.png)
Proposed by Yang Liu
, 
, and 
be positive integers. Prove that ![\[ \sum_{i = 1}^n \min\left(\left\lfloor \frac{x}{i} \right\rfloor, m \right) = \sum_{i = 1}^m \min\left(\left\lfloor \frac{x}{i} \right\rfloor, n \right). \]](http://latex.artofproblemsolving.com/b/7/a/b7aebfe11f646d604d2e8cb9d5bd70ec796949e8.png)
Proposed by Yang Liu
27 replies
Problem 1
randomusername 74
N
by Mr.Sharkman
Source: IMO 2015, Problem 1
We say that a finite set 
of points in the plane is balanced if, for any two different points 
and 
in , there is a point 
in such that 
. We say that is centre-free if for any three different points , and in , there is no points 
in such that 
.
(a) Show that for all integers
, there exists a balanced set consisting of points.
(b) Determine all integers for which there exists a balanced centre-free set consisting of points.
Proposed by Netherlands
of points in the plane is balanced if, for any two different points 
and 
in , there is a point 
in such that 
. We say that is centre-free if for any three different points , and in , there is no points 
in such that 
.(a) Show that for all integers
, there exists a balanced set consisting of points.(b) Determine all integers
for which there exists a balanced centre-free set consisting of points.Proposed by Netherlands
74 replies
Find Triples of Integers
termas 41
N
by ilikemath247365
Source: IMO 2015 problem 2
Find all positive integers 
such that
are all powers of 
.
Proposed by Serbia
such that
are all powers of 
.Proposed by Serbia
41 replies
DO NOT OVERSLEEP JOHN MACKEY’S CLASS
ike.chen 31
N
by Mr.Sharkman
Source: USA TSTST 2023/4
Let be an integer and let 
be the complete graph on vertices. Each edge of is colored either red, green, or blue. Let denote the number of triangles in with all edges of the same color, and let denote the number of triangles in with all edges of different colors. Prove
![\[ B\le 2A+\frac{n(n-1)}{3}.\]](//latex.artofproblemsolving.com/6/5/9/659d0199c3d685d43f1ced3704dfa59b02b586b3.png)
(The complete graph on vertices is the graph on vertices with 
edges, with exactly one edge joining every pair of vertices. A triangle consists of the set of 
edges between 
of these vertices.)
Proposed by Ankan Bhattacharya
be an integer and let 
be the complete graph on vertices. Each edge of is colored either red, green, or blue. Let denote the number of triangles in with all edges of the same color, and let denote the number of triangles in with all edges of different colors. Prove![\[ B\le 2A+\frac{n(n-1)}{3}.\]](http://latex.artofproblemsolving.com/6/5/9/659d0199c3d685d43f1ced3704dfa59b02b586b3.png)
(The complete graph on vertices is the graph on vertices with 
edges, with exactly one edge joining every pair of vertices. A triangle consists of the set of 
edges between 
of these vertices.)Proposed by Ankan Bhattacharya
31 replies
Grade IX - Problem I
icx 23
N
by shendrew7
Source: Romanian National Mathematical Olympiad 2007
Let 
such that the equation ![\[x^{2}-(a^{2}+b^{2}+c^{2}+d^{2}+1)x+ab+bc+cd+da=0 \]](//latex.artofproblemsolving.com/9/0/4/904d1b23d5c744118a7d68df1e2e94bddac8c1cd.png)
has an integer solution. Prove that the other solution is integer too and both solutions are perfect squares.
such that the equation ![\[x^{2}-(a^{2}+b^{2}+c^{2}+d^{2}+1)x+ab+bc+cd+da=0 \]](http://latex.artofproblemsolving.com/9/0/4/904d1b23d5c744118a7d68df1e2e94bddac8c1cd.png)
has an integer solution. Prove that the other solution is integer too and both solutions are perfect squares.
23 replies
USAMO 2002 Problem 2
MithsApprentice 35
N
by sami1618
Let 
be a triangle such that
![\[ \left( \cot \dfrac{A}{2} \right)^2 + \left( 2\cot \dfrac{B}{2} \right)^2 + \left( 3\cot \dfrac{C}{2} \right)^2 = \left( \dfrac{6s}{7r} \right)^2, \]](//latex.artofproblemsolving.com/6/9/b/69b1964be5f92eb44a36b0b8604bf473fe27e210.png)
where
and 
denote its semiperimeter and its inradius, respectively. Prove that triangle is similar to a triangle 
whose side lengths are all positive integers with no common divisors and determine these integers.
be a triangle such that![\[ \left( \cot \dfrac{A}{2} \right)^2 + \left( 2\cot \dfrac{B}{2} \right)^2 + \left( 3\cot \dfrac{C}{2} \right)^2 = \left( \dfrac{6s}{7r} \right)^2, \]](http://latex.artofproblemsolving.com/6/9/b/69b1964be5f92eb44a36b0b8604bf473fe27e210.png)
where
and 
denote its semiperimeter and its inradius, respectively. Prove that triangle is similar to a triangle 
whose side lengths are all positive integers with no common divisors and determine these integers.
35 replies
1 viewing
Center lies on altitude
plagueis 17
N
by bin_sherlo
Source: Mexico National Olympiad 2018 Problem 6
Let be an acute-angled triangle with circumference 
. Let the angle bisectors of 
and 
intersect again at 
and 
. Let 
be the intersection point of these angle bisectors. Let 
and 
be the respective reflections of and in 
and 
. Prove that the center of the circle passing through , , lies on the altitude of triangle from .
Proposed by Victor Domínguez and Ariel García
be an acute-angled triangle with circumference 
. Let the angle bisectors of 
and 
intersect again at 
and 
. Let 
be the intersection point of these angle bisectors. Let 
and 
be the respective reflections of and in 
and 
. Prove that the center of the circle passing through , , lies on the altitude of triangle from .Proposed by Victor Domínguez and Ariel García
17 replies
IMO Shortlist 2014 C6
hajimbrak 22
N
by awesomeming327.
We are given an infinite deck of cards, each with a real number on it. For every real number , there is exactly one card in the deck that has written on it. Now two players draw disjoint sets and of 
cards each from this deck. We would like to define a rule that declares one of them a winner. This rule should satisfy the following conditions:
1. The winner only depends on the relative order of the
cards: if the cards are laid down in increasing order face down and we are told which card belongs to which player, but not what numbers are written on them, we can still decide the winner.
2. If we write the elements of both sets in increasing order as
and 
, and 
for all 
, then beats .
3. If three players draw three disjoint sets
from the deck, beats and beats then also beats .
How many ways are there to define such a rule? Here, we consider two rules as different if there exist two sets and such that beats according to one rule, but beats according to the other.
Proposed by Ilya Bogdanov, Russia
, there is exactly one card in the deck that has written on it. Now two players draw disjoint sets and of 
cards each from this deck. We would like to define a rule that declares one of them a winner. This rule should satisfy the following conditions:1. The winner only depends on the relative order of the
cards: if the cards are laid down in increasing order face down and we are told which card belongs to which player, but not what numbers are written on them, we can still decide the winner.2. If we write the elements of both sets in increasing order as
and 
, and 
for all 
, then beats .3. If three players draw three disjoint sets
from the deck, beats and beats then also beats .How many ways are there to define such a rule? Here, we consider two rules as different if there exist two sets
and such that beats according to one rule, but beats according to the other.Proposed by Ilya Bogdanov, Russia
22 replies
Radical Axes and circles
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#3
Apr 22, 2025, 2:53 AM
Y by
SirAppel wrote:
is the radical axis, so then the intersection with 
must have equal powers with respect to both circles, which only occurs when the segment is bisectedthanks
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#4
Apr 22, 2025, 4:33 AM
Y by
egmo grind 
This post has been edited 1 time. Last edited by martianrunner, Apr 22, 2025, 4:33 AM
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#5
Apr 22, 2025, 7:53 AM
Y by
Let AP intersect BC at M
By power of point we get
Hence,
By power of point we get
Hence,
This post has been edited 1 time. Last edited by spiderman0, Apr 22, 2025, 7:53 AM
Reason: Typo
Reason: Typo
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