and 
Find the minimum 
Y by
Out of all the nondegenerate triangles with positive integer sides and perimeter 
, find the one with the smallest area.
, find the one with the smallest area.G
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Geometric inequality with Fermat point
Assassino9931 6
N
by arqady
Source: Balkan MO Shortlist 2024 G2
Let 
be an acute triangle and let 
be an interior point for it such that 
. Prove that
When does equality hold?
be an acute triangle and let 
be an interior point for it such that 
. Prove that
When does equality hold?
6 replies
Straight-edge is a social construct
anantmudgal09 27
N
by cj13609517288
Source: INMO 2023 P6
Euclid has a tool called cyclos which allows him to do the following:
[list]
[*] Given three non-collinear marked points, draw the circle passing through them.
[*] Given two marked points, draw the circle with them as endpoints of a diameter.
[*] Mark any intersection points of two drawn circles or mark a new point on a drawn circle.
[/list]
Show that given two marked points, Euclid can draw a circle centered at one of them and passing through the other, using only the cyclos.
Proposed by Rohan Goyal, Anant Mudgal, and Daniel Hu
[list]
[*] Given three non-collinear marked points, draw the circle passing through them.
[*] Given two marked points, draw the circle with them as endpoints of a diameter.
[*] Mark any intersection points of two drawn circles or mark a new point on a drawn circle.
[/list]
Show that given two marked points, Euclid can draw a circle centered at one of them and passing through the other, using only the cyclos.
Proposed by Rohan Goyal, Anant Mudgal, and Daniel Hu
27 replies
+1 wWeighted Blocks
ilovemath04 51
N
by Maximilian113
Source: ISL 2019 C2
You are given a set of 
blocks, each weighing at least 
; their total weight is 
. Prove that for every real number 
with 
you can choose a subset of the blocks whose total weight is at least but at most 
.
blocks, each weighing at least 
; their total weight is 
. Prove that for every real number 
with 
you can choose a subset of the blocks whose total weight is at least but at most 
.
51 replies
Very easy NT
GreekIdiot 7
N
by Primeniyazidayi
Prove that there exists no natural number 
such that 
.
such that 
.
7 replies
Azer and Babek playing a game on a chessboard
Nuran2010 1
N
by Diamond-jumper76
Source: Azerbaijan Al-Khwarizmi IJMO TST
Azer and Babek have a 
chessboard. Initially, Azer colors all cells of this chessboard with some colors. Then, Babek takes 
rows and columns and looks at the 
cells in the intersection. Babek wants to have all these cells in a same color, but Azer doesn't. With at least how many colors, Azer can reach his goal?
chessboard. Initially, Azer colors all cells of this chessboard with some colors. Then, Babek takes 
rows and columns and looks at the 
cells in the intersection. Babek wants to have all these cells in a same color, but Azer doesn't. With at least how many colors, Azer can reach his goal?
1 reply
Something nice
KhuongTrang 27
N
by arqady
Source: own
Problem. Given 
be non-negative real numbers such that 
Prove that
be non-negative real numbers such that 
Prove that
27 replies
1 viewing
Pairwise distance-one products
y-is-the-best-_ 28
N
by john0512
Source: IMO 2019 SL A4
Let 
be a positive integer and 
be real numbers such that ![\[a_1+a_2+\dots+a_n=0.\]](//latex.artofproblemsolving.com/0/f/d/0fde8d75cebc35103ae9acf8ebe393653d336d60.png)
Define the set 
by
![\[A=\left\{(i, j)\,|\,1 \leqslant i<j \leqslant n,\left|a_{i}-a_{j}\right| \geqslant 1\right\}\]](//latex.artofproblemsolving.com/6/9/7/6974ebf20c44bcb4d52d85adc0805e28aafa5d6e.png)
Prove that, if is not empty, then
![\[\sum_{(i, j) \in A} a_{i} a_{j}<0.\]](//latex.artofproblemsolving.com/b/e/3/be3e65a633dc2ac092a264968f1cf8514d3d6124.png)
be a positive integer and 
be real numbers such that ![\[a_1+a_2+\dots+a_n=0.\]](http://latex.artofproblemsolving.com/0/f/d/0fde8d75cebc35103ae9acf8ebe393653d336d60.png)
Define the set 
by![\[A=\left\{(i, j)\,|\,1 \leqslant i<j \leqslant n,\left|a_{i}-a_{j}\right| \geqslant 1\right\}\]](http://latex.artofproblemsolving.com/6/9/7/6974ebf20c44bcb4d52d85adc0805e28aafa5d6e.png)
Prove that, if is not empty, then![\[\sum_{(i, j) \in A} a_{i} a_{j}<0.\]](http://latex.artofproblemsolving.com/b/e/3/be3e65a633dc2ac092a264968f1cf8514d3d6124.png)
28 replies
2^x+3^x = yx^2
truongphatt2668 8
N
by Tamam
Prove that the following equation has infinite integer solutions:
8 replies
Fibonacci...?
Jackson0423 1
N
by KAME06
The sequence 
is defined by 
and for all positive integers 
, 
. Show that for every positive integer 
, there exists a suitable positive integer 
such that 
is a multiple of .
is defined by 
and for all positive integers 
, 
. Show that for every positive integer 
, there exists a suitable positive integer 
such that 
is a multiple of .1 reply
4 concyclic points
buzzychaoz 18
N
by bjump
Source: Japan Mathematical Olympiad Finals 2015 Q4
Scalene triangle has circumcircle 
and incenter 
. The incircle of triangle touches side 
at 
respectively. Circumcircle of triangle 
intersects again at distinct from 
, circumcircle of triangle 
intersects again at 
distinct from 
. Prove that the points 
are concyclic.
has circumcircle 
and incenter 
. The incircle of triangle touches side 
at 
respectively. Circumcircle of triangle 
intersects again at distinct from 
, circumcircle of triangle 
intersects again at 
distinct from 
. Prove that the points 
are concyclic.
18 replies
angles in triangle
AndrewTom 33
N
by zuat.e
Source: BrMO 2012/13 Round 2
The point lies inside triangle so that 
. The point is such that 
is a parallelogram. Prove that 
.
lies inside triangle so that 
. The point is such that 
is a parallelogram. Prove that 
.
33 replies
Gheorghe Țițeica 2025 Grade 7 P3
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Source: Gheorghe Țițeica 2025
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#2
Mar 29, 2025, 3:19 PM
Y by
If degenerate triangles were allowed, the optimal choice would be a collinear triangle. For a non-degenerate triangle we want to come as close as possible to collinearity; triangles with sidelengths 25, 26, 49 or 2, 49, 49 come to mind. Let us try to produce a rigorous argument for the claim that the latter sidelengths 2, 49, 49 do indeed produce the smallest possible area.
We compare the areas of a triangle with sidelengths
and a triangle with sidelengths 
, assuming that 
and 
. Using Heron's formula, we see that it suffices to compare the terms
for the first triangle and
for the second triangle. As the first two factors are the same in both terms, we can reduce the comparison to the terms
and
Subtracting the last term from the one before it, we get 
, which is positive due to 
. This shows that the area of the second triangle is smaller.
Now, if we have a non-degenerate triangle with sidelengths and and 
and with minimal area, the above step from sidelengths to sidelengths must be barred for some reason (otherwise the area would not be minimal). The only possible reason for that is that a non-degenerate triangle with sidelengths does not exist due to a violation of the triangle inequalities. As the triangle inequalities for the 
-
-
-triangle are fulfilled, the only triangle inequality which can be violated for the hypothetical triangle with the new sidelengths is 
, meaning that in fact we have
As the triangle inequalities for the ---triangle are fulfilled (due to its existence), we have 
and therefore
So, taking these inequalities together, we have
and, adding 1, we get
Consequently, 
is equal either to 
or to 
. Assuming the first case, 
, we get
which is not possible for an integer value of . So the second case must be true: 
. This leads us to
which means
and 
We are almost done. We still need to find out what and are. As we already know that 
, the Heronian formula tells us that the area of our minimal triangle only depends on the term
which is equivalent to 
As is already fixed (
), the only variable here is the product 
. Now
For the positive integer value of , we need to get as far away from 
as possible to minimalize . As 
, we must investigate the lower end of the spectrum for . The value 
is forbidden, as would be 
then, violating 
and the triangle inequality 
. The best acceptable choice is obviously 
. So we finally have
We compare the areas of a triangle with sidelengths
and a triangle with sidelengths 
, assuming that 
and 
. Using Heron's formula, we see that it suffices to compare the terms
for the first triangle and
for the second triangle. As the first two factors are the same in both terms, we can reduce the comparison to the terms
and
Subtracting the last term from the one before it, we get 
, which is positive due to 
. This shows that the area of the second triangle is smaller.Now, if we have a non-degenerate triangle with sidelengths
and and 
and with minimal area, the above step from sidelengths to sidelengths must be barred for some reason (otherwise the area would not be minimal). The only possible reason for that is that a non-degenerate triangle with sidelengths does not exist due to a violation of the triangle inequalities. As the triangle inequalities for the 
-
-
-triangle are fulfilled, the only triangle inequality which can be violated for the hypothetical triangle with the new sidelengths is 
, meaning that in fact we have
As the triangle inequalities for the ---triangle are fulfilled (due to its existence), we have 
and therefore
So, taking these inequalities together, we have
and, adding 1, we get
Consequently, 
is equal either to 
or to 
. Assuming the first case, 
, we get
which is not possible for an integer value of . So the second case must be true: 
. This leads us to
which means
and 
We are almost done. We still need to find out what and are. As we already know that 
, the Heronian formula tells us that the area of our minimal triangle only depends on the term
which is equivalent to 
As is already fixed (
), the only variable here is the product 
. Now
For the positive integer value of , we need to get as far away from 
as possible to minimalize . As 
, we must investigate the lower end of the spectrum for . The value 
is forbidden, as would be 
then, violating 
and the triangle inequality 
. The best acceptable choice is obviously 
. So we finally have
This post has been edited 1 time. Last edited by Rainbow1971, Mar 29, 2025, 3:33 PM
Reason: Correction of minor error in the first paragraph
Reason: Correction of minor error in the first paragraph
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