Interesting functional equation with geometry
by User21837561, May 9, 2025, 8:14 AM
For an acute triangle
, let
be the circumcentre,
be the orthocentre, and
be the centroid.
Let
satisfy the following condition:

Prove that
is constant.




Let


Prove that

This post has been edited 2 times. Last edited by User21837561, Yesterday at 4:10 PM
Reason: Wrong year for the source
Reason: Wrong year for the source
Equilateral triangle formed by circle and Fermat point
by Mimii08, May 8, 2025, 10:36 PM
Hi! I found this interesting geometry problem and I would really appreciate help with the proof.
Let ABC be an acute triangle, and let T be the Fermat (Torricelli) point of triangle ABC. Let A1, B1, and C1 be the feet of the perpendiculars from T to the sides BC, AC, and AB, respectively. Let ω be the circle passing through points A1, B1, and C1. Let A2, B2, and C2 be the second points where ω intersects the sides BC, AC, and AB, respectively (different from A1, B1, C1).
Prove that triangle A2B2C2 is equilateral.
Let ABC be an acute triangle, and let T be the Fermat (Torricelli) point of triangle ABC. Let A1, B1, and C1 be the feet of the perpendiculars from T to the sides BC, AC, and AB, respectively. Let ω be the circle passing through points A1, B1, and C1. Let A2, B2, and C2 be the second points where ω intersects the sides BC, AC, and AB, respectively (different from A1, B1, C1).
Prove that triangle A2B2C2 is equilateral.
greatest volume
by hzbrl, May 8, 2025, 9:56 AM
A large sphere with radius 7 contains three smaller balls each with radius 3 . The three balls are each externally tangent to the other two balls and internally tangent to the large sphere. There are four right circular cones that can be inscribed in the large sphere in such a way that the bases of the cones are tangent to all three balls. Of these four cones, the one with the greatest volume has volume
. Find
.


Why is the old one deleted?
by EeEeRUT, Apr 16, 2025, 1:33 AM
For a positive integer
, let
be all positive integers smaller than
that are coprime to
. Find all
such that
for all 
Here
is the largest positive integer that divides both
and
. Integers
and
are coprime if
.
Proposed by Paulius Aleknavičius, Lithuania







Here






Proposed by Paulius Aleknavičius, Lithuania
This post has been edited 2 times. Last edited by EeEeRUT, Apr 18, 2025, 12:56 AM
Reason: Authorship
Reason: Authorship
Problem 4 of Finals
by GeorgeRP, Sep 10, 2024, 4:39 PM
The diagonals
,
, and
of a hexagon
inscribed in a circle
intersect at a point
, and the acute angle between any two of them is
. Let
be the radius of the circle tangent to segments
and
and internally tangent to
; the radii
,
,
,
, and
are defined similarly. Prove that
![\[
r_{AB}r_{CD} + r_{CD}r_{EF} + r_{EF}r_{AB} = r_{BC}r_{DE} + r_{DE}r_{FA} + r_{FA}r_{BC}.
\]](//latex.artofproblemsolving.com/0/d/6/0d66d6685bbc97c7553e9ed3ba14ea0dc3328b49.png)
















![\[
r_{AB}r_{CD} + r_{CD}r_{EF} + r_{EF}r_{AB} = r_{BC}r_{DE} + r_{DE}r_{FA} + r_{FA}r_{BC}.
\]](http://latex.artofproblemsolving.com/0/d/6/0d66d6685bbc97c7553e9ed3ba14ea0dc3328b49.png)
(n+1)2^n, (n+3)2^{n+2} not perfect squares for the same n
by parmenides51, Apr 29, 2019, 9:57 AM
Prove that there is not a positive integer
such that numbers
are both perfect squares.


angle chasing with 2 midpoints, equal angles given and wanted
by parmenides51, Dec 11, 2018, 10:02 PM
In the triangle
,
and
are the midpoints of sides
and
respectively. Point
lies inside the triangle. Let
. Prove that
.








This post has been edited 1 time. Last edited by parmenides51, Mar 26, 2024, 2:37 PM
IMO 2010 Problem 3
by canada, Jul 7, 2010, 4:41 PM
Find all functions
such that
is a perfect square for all 
Proposed by Gabriel Carroll, USA

![\[\left(g(m)+n\right)\left(g(n)+m\right)\]](http://latex.artofproblemsolving.com/5/a/c/5ac9fce315e8a524a3a01b345cfe76bc76056d68.png)

Proposed by Gabriel Carroll, USA
This post has been edited 1 time. Last edited by djmathman, Jun 22, 2015, 12:52 AM
Reason: formatting
Reason: formatting
Game
by Pascual2005, Jun 7, 2005, 5:48 PM












Proposed by A. Slinko & S. Marshall, New Zealand
The oldest, shortest words — "yes" and "no" — are those which require the most thought.
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