Enjoy with this one
by Halykov, Apr 30, 2025, 2:00 PM
Let
be a scalene triangle with circumcircle
and circumcircle
. Denote
and
as the midpoints of
and
, respectively. The altitude from
to
intersects
again at
. The line
meets
a second time at
, and
intersects
at
. The perpendicular bisector of
intersects
at
and
at
. Let the circumcircle of
intersect
again at
, and let the line
intersect
at
.
If
is the reflection of
over
, prove that the intersection of
and
lies on the circumcircle of
.




























If






Conjecture: Intersection of diagonals of cyclic quadrilateral lies on a fixed li
by kieusuong, Apr 30, 2025, 12:28 PM
Let (O) be a fixed circle and let P be a point outside the circle.
From P, draw a line that intersects the circle at points A and B.
Now, from P, draw another line that intersects the circle at points N and M so that quadrilateral ANMB is cyclic (i.e., lies on the circle).
Let AM and BN intersect at point G. Let AN and BM intersect at point T.
Let PJ be the tangent from P to circle (O), and let J be the point of tangency.
Claim (Conjecture):
As the quadrilateral ANMB varies (still inscribed in the circle), the points T, G, and J always lie on a straight line.
Moreover, this line TJ is perpendicular to the fixed chord AB.
I believe this might be a new result and would appreciate any insights or proof ideas.
Attached is a diagram for reference.
From P, draw a line that intersects the circle at points A and B.
Now, from P, draw another line that intersects the circle at points N and M so that quadrilateral ANMB is cyclic (i.e., lies on the circle).
Let AM and BN intersect at point G. Let AN and BM intersect at point T.
Let PJ be the tangent from P to circle (O), and let J be the point of tangency.
Claim (Conjecture):
As the quadrilateral ANMB varies (still inscribed in the circle), the points T, G, and J always lie on a straight line.
Moreover, this line TJ is perpendicular to the fixed chord AB.
I believe this might be a new result and would appreciate any insights or proof ideas.
Attached is a diagram for reference.
1 line solution to Inequality
by ItzsleepyXD, Apr 30, 2025, 9:27 AM
Let
be positive real integer such that
Prove that
such that
and 





a nice problem of nt from PUMaC
by Namisgood, Apr 29, 2025, 9:47 AM
Problem is attached
This post has been edited 1 time. Last edited by Namisgood, Yesterday at 9:47 AM
Reason: Typo
Reason: Typo
Solve $\sin(17x)+\sin(13x)=\sin(7x)$
by Speed2001, Apr 29, 2025, 1:06 AM
How to solve the equation:
Approach: I'm trying to factor
to get
.
Any hint would be appreciated.



Any hint would be appreciated.
Difficult combinatorics problem about distinct sums under shifts
by CBMaster, Apr 27, 2025, 5:22 AM
Problem. Let
be the nonnegative integers in
where
. Define
. Prove that
.
Bonus problem (Open). Can we prove a tighter result than the one above? That is, is there a function
such that
where
, and the statement is still true when
?
Or, is there a function
such that
where
, and the statement is still true when
?.





Bonus problem (Open). Can we prove a tighter result than the one above? That is, is there a function




Or, is there a function




This post has been edited 10 times. Last edited by CBMaster, Apr 27, 2025, 5:43 AM
Reason: .
Reason: .
Inequality with condition a+b+c = ab+bc+ca (and special equality case)
by DoThinh2001, May 2, 2019, 11:41 AM
Let
be real numbers such that
and 
Prove that
and determine the equality cases.
(Edit: Proposed by sir Leonard Giugiuc, Romania)



Prove that

(Edit: Proposed by sir Leonard Giugiuc, Romania)
This post has been edited 1 time. Last edited by DoThinh2001, May 3, 2019, 10:09 AM
Diophantine equation !
by ComplexPhi, Feb 4, 2015, 2:05 PM
Determine all triples
satisfying :
![\[n^{2p}=m^2+n^2+p+1\]](//latex.artofproblemsolving.com/7/4/e/74ee592cb8b3f3a3996ea34f40b78fde9d3672fc.png)
where
and
are integers and
is a prime number.

![\[n^{2p}=m^2+n^2+p+1\]](http://latex.artofproblemsolving.com/7/4/e/74ee592cb8b3f3a3996ea34f40b78fde9d3672fc.png)
where



7^a - 3^b divides a^4 + b^2 (from IMO Shortlist 2007)
by Dida Drogbier, Apr 21, 2008, 2:53 PM
Find all pairs of natural numbers
such that
divides
.
Author: Stephan Wagner, Austria



Author: Stephan Wagner, Austria
The oldest, shortest words — "yes" and "no" — are those which require the most thought.
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