such that 
.
Where 
.
such that for all 
Y by
Prove that if a convex quadrilateral 
satisfies the equation
![\[
(AB + CD)^2 + (AD + BC)^2 = (AC + BD)^2,
\]](//latex.artofproblemsolving.com/7/d/a/7daf2d6ac00609b5252b2b8b05d96a5ca0952a94.png)
then must be a rectangle.
satisfies the equation![\[
(AB + CD)^2 + (AD + BC)^2 = (AC + BD)^2,
\]](http://latex.artofproblemsolving.com/7/d/a/7daf2d6ac00609b5252b2b8b05d96a5ca0952a94.png)
then must be a rectangle.G
Topic
First Poster
Last Poster
Orthoincentre mixup in rmo mock
Project_Donkey_into_M4 2
N
by Physicsknight
Source: Mock RMO 2018,TDP and Kayak P5
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.
2 replies
1 viewing
The special Miquel's point from a familiar problem
danil_e 8
N
by anantmudgal09
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.
8 replies
Inspired by old results
sqing 1
N
by lbh_qys
Source: Own
Let 
be real numbers.Prove that
be real numbers.Prove that
1 reply
1 viewing
too many equality cases
Scilyse 17
N
by Confident-man
Source: 2023 ISL C6
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.
IMAGE
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.IMAGE
Proposed by Zixiang Zhou, Canada
17 replies
FE over \mathbb{R}
megarnie 6
N
by jasperE3
Source: Own
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 
.
6 replies
Looks a Mumbai candy
Physicsknight 0
Source: Shourya
Let 
be 
real numbers such that 
for all 
and such that 
Find the maximum possible value of the expression
be 
real numbers such that 
for all 
and such that 
Find the maximum possible value of the expression
0 replies
>=512 different isosceles triangles whose vertices have the same color
parmenides51 3
N
by AlexCenteno2007
Source: Mathematics Regional Olympiad of Mexico West 2016 P6
The vertices of a regular polygon with 
sides are colored gold or silver. Prove that there are at least 
different isosceles triangles whose vertices have the same color.
sides are colored gold or silver. Prove that there are at least 
different isosceles triangles whose vertices have the same color.
3 replies
Fourth power ineq
Project_Donkey_into_M4 1
N
by sqing
Source: 2018 Mock RMO tdp and kayak P1
Let 
such that 
. Prove that![\[\sqrt[4]{(1-a^4)(1-b^4)(1-c^4)(1-d^4)}\geq 255\cdot abcd.\]](//latex.artofproblemsolving.com/d/e/6/de61611f628ea796c706d05941d7ee9b3fc22916.png)
such that 
. Prove that![\[\sqrt[4]{(1-a^4)(1-b^4)(1-c^4)(1-d^4)}\geq 255\cdot abcd.\]](http://latex.artofproblemsolving.com/d/e/6/de61611f628ea796c706d05941d7ee9b3fc22916.png)
1 reply
1 viewing
Dear Sqing: So Many Inequalities...
hashtagmath 36
N
by sqing
I have noticed thousands upon thousands of inequalities that you have posted to HSO and was wondering where you get the inspiration, imagination, and even the validation that such inequalities are true? Also, what do you find particularly appealing and important about specifically inequalities rather than other branches of mathematics? Thank you :)
36 replies
Advanced topics in Inequalities
va2010 18
N
by sqing
So a while ago, I compiled some tricks on inequalities. You are welcome to post solutions below!
18 replies
A Characterization of Rectangles
G
H
G
H
Source: VN Math Olympiad For High School Students P8 - 2025
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
![\[
AB^2 + CD^2 + AD^2 + BC^2 \geq AC^2 + BD^2,
\]](http://latex.artofproblemsolving.com/a/4/6/a46763d7a72ba44005eade1bffca31e28077efa9.png)
and according to Ptolemy's inequality,![\[
AB \cdot CD + AD \cdot BC \geq AC \cdot BD.
\]](http://latex.artofproblemsolving.com/7/7/c/77c004953365450da97733dd26d294da97b51cdb.png)
Consequently,![\[
(AB+CD)^2 + (AD+BC)^2 \geq (AC+BD)^2.
\]](http://latex.artofproblemsolving.com/8/6/8/8683f36a6457c49d3f6eaff66efe8ffa67105a70.png)
Equality holds if and only if 
is a parallelogram inscribed in a circle, that is, a rectangle.N Quick Reply
G
H
