Inequality
by giangtruong13, Apr 21, 2025, 8:01 AM
Let 
such that: 
. Prove that: 
such that: 
. Prove that: 
be real numbers.Prove that
Orthoincentre mixup in rmo mock
by Project_Donkey_into_M4, Apr 20, 2025, 6:26 PM
Let 
be a triangle with circumcircle 
, 
be the foot of altitudes from 
onto the opposite sides respectively and 
the orthocentre. Reflect across the line 
to obtain 
. Suppose there exists points 
such that 
is the incentre of 
. If 
and 
be the midpoints of 
and 
respectively, then show that 
are collinear.
be a triangle with circumcircle 
, 
be the foot of altitudes from 
onto the opposite sides respectively and 
the orthocentre. Reflect across the line 
to obtain 
. Suppose there exists points 
such that 
is the incentre of 
. If 
and 
be the midpoints of 
and 
respectively, then show that 
are collinear.Test from Côte d'Ivoire Diophantine equation
by MENELAUSS, Apr 19, 2025, 3:05 PM
determine all triplets 
of natural numbers such that
of natural numbers such that
too many equality cases
by Scilyse, Jul 17, 2024, 12:12 PM
Let be a positive integer, and consider an 
grid. A right-down path is a sequence of grid cells such that each cell is either one cell to the right of or one cell below the previous cell in the sequence. A right-up path is a sequence of grid cells such that each cell is either one cell to the right of or one cell above the previous cell in the sequence.
Prove that the cells of the grid cannot be partitioned into less than right-down or right-up paths. For example, the following partition of the 
grid uses 
paths.
![[asy]
size(4cm);
draw((5,-1)--(0,-1)--(0,-2)--(5,-2)--(5,-3)--(0,-3)--(0,-4)--(5,-4),gray+linewidth(0.5)+miterjoin);
draw((1,-5)--(1,0)--(2,0)--(2,-5)--(3,-5)--(3,0)--(4,0)--(4,-5),gray+linewidth(0.5)+miterjoin);
draw((0,0)--(5,0)--(5,-5)--(0,-5)--cycle,black+linewidth(2.5)+miterjoin);
draw((0,-1)--(3,-1)--(3,-2)--(1,-2)--(1,-4)--(4,-4)--(4,-3)--(2,-3)--(2,-2),black+linewidth(2.5)+miterjoin);
draw((3,0)--(3,-1),black+linewidth(2.5)+miterjoin);
draw((1,-4)--(1,-5),black+linewidth(2.5)+miterjoin);
draw((4,-3)--(4,-1)--(5,-1),black+linewidth(2.5)+miterjoin);
[/asy]](//latex.artofproblemsolving.com/4/7/7/477145f5dd40b84a4daff1c2fdd46ae90f013fb7.png)
Proposed by Zixiang Zhou, Canada
be a positive integer, and consider an 
grid. A right-down path is a sequence of grid cells such that each cell is either one cell to the right of or one cell below the previous cell in the sequence. A right-up path is a sequence of grid cells such that each cell is either one cell to the right of or one cell above the previous cell in the sequence.Prove that the cells of the
grid cannot be partitioned into less than right-down or right-up paths. For example, the following partition of the 
grid uses 
paths.![[asy]
size(4cm);
draw((5,-1)--(0,-1)--(0,-2)--(5,-2)--(5,-3)--(0,-3)--(0,-4)--(5,-4),gray+linewidth(0.5)+miterjoin);
draw((1,-5)--(1,0)--(2,0)--(2,-5)--(3,-5)--(3,0)--(4,0)--(4,-5),gray+linewidth(0.5)+miterjoin);
draw((0,0)--(5,0)--(5,-5)--(0,-5)--cycle,black+linewidth(2.5)+miterjoin);
draw((0,-1)--(3,-1)--(3,-2)--(1,-2)--(1,-4)--(4,-4)--(4,-3)--(2,-3)--(2,-2),black+linewidth(2.5)+miterjoin);
draw((3,0)--(3,-1),black+linewidth(2.5)+miterjoin);
draw((1,-4)--(1,-5),black+linewidth(2.5)+miterjoin);
draw((4,-3)--(4,-1)--(5,-1),black+linewidth(2.5)+miterjoin);
[/asy]](http://latex.artofproblemsolving.com/4/7/7/477145f5dd40b84a4daff1c2fdd46ae90f013fb7.png)
Proposed by Zixiang Zhou, Canada
This post has been edited 8 times. Last edited by Scilyse, Jan 23, 2025, 12:41 PM
Reason: add "proposed by"
Reason: add "proposed by"
The special Miquel's point from a familiar problem
by danil_e, Jul 23, 2023, 2:52 PM
Problem. Let 
be an acute-angled triangle with 
, let 
be its circumcentre. The line through 
perpendicular to intersects circle 
again at 
. The tangents at 
and 
of intersect at 
. 
intersects at 
. 
intersects 
at 
. Let be the midpoint of 
.
Prove that
are concyclic.
be an acute-angled triangle with 
, let 
be its circumcentre. The line through 
perpendicular to intersects circle 
again at 
. The tangents at 
and 
of intersect at 
. 
intersects at 
. 
intersects 
at 
. Let be the midpoint of 
.Prove that
are concyclic.
FE over \mathbb{R}
by megarnie, Nov 13, 2021, 5:29 PM
Find all functions from the reals to the reals so that ![\[f(xy)+f(xf(x^2y))=f(x^2)+f(y^2)+f(f(xy^2))+x \]](//latex.artofproblemsolving.com/d/7/b/d7bf030b047d49477eba887e8bb78c6e5b126fe7.png)
holds for all 
.
![\[f(xy)+f(xf(x^2y))=f(x^2)+f(y^2)+f(f(xy^2))+x \]](http://latex.artofproblemsolving.com/d/7/b/d7bf030b047d49477eba887e8bb78c6e5b126fe7.png)
holds for all 
.This post has been edited 1 time. Last edited by megarnie, Nov 13, 2021, 5:29 PM
\frac{a}{(b-c)^2}+\frac{b}{(a-c)^2}+\frac{c}{(b-a)^2}
by Jjesus, Oct 12, 2020, 2:40 PM
Let 
be real numbers. Knowing that there are positive real numbers 
, different two by two, such that
calculate the value of
in terms of .
be real numbers. Knowing that there are positive real numbers 
, different two by two, such that
calculate the value of
in terms of .This post has been edited 1 time. Last edited by Jjesus, Oct 23, 2020, 9:23 PM
16th ibmo - uruguay 2001/q3.
by carlosbr, Apr 15, 2006, 12:52 AM
Let be a set of 
elements and 
are subsets of (
), such that every one of them has at least 
elements.
Show that there exists
and 
, with 
, such that the number of common elements of 
and 
is greater or equal to: 
be a set of 
elements and 
are subsets of (
), such that every one of them has at least 
elements.Show that there exists
and 
, with 
, such that the number of common elements of 
and 
is greater or equal to: 
The oldest, shortest words — "yes" and "no" — are those which require the most thought.
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