Pells equation
by Entrepreneur, May 15, 2025, 3:56 PM
A Pells Equation is defined as follows
Where
are positive integers and
is a non-square positive integer. If
denotes the n-th set of solution to the equation with
Then, prove that 








Inequalities
by sqing, May 14, 2025, 3:46 AM
Inequalities
by sqing, May 13, 2025, 11:31 AM
Incircle concurrency
by niwobin, May 11, 2025, 4:28 PM
Triangle ABC with incenter I, incircle is tangent to BC, AC, and AB at D, E and F respectively.
DT is a diameter for the incircle, and AT meets the incircle again at point H.
Let DH and EF intersect at point J. Prove: AJ//BC.
DT is a diameter for the incircle, and AT meets the incircle again at point H.
Let DH and EF intersect at point J. Prove: AJ//BC.
Weird locus problem
by Sedro, May 11, 2025, 3:12 AM
Points
and
are in the coordinate plane such that
. Let
denote the locus of all points
in the coordinate plane satisfying
, and let
be the midpoint of
. Points
and
are on
such that
and
. The value of
can be expressed in the form
, where
and
are relatively prime positive integers. Find
.


















Inequalities
by sqing, May 9, 2025, 8:50 AM
Let
Prove that







This post has been edited 1 time. Last edited by sqing, May 9, 2025, 8:58 AM
trigonometric functions
by VivaanKam, Apr 29, 2025, 8:29 PM
Hi could someone explain the basic trigonometric functions to me like sin, cos, tan etc.
Thank you!
Thank you!
IOQM P23 2024
by SomeonecoolLovesMaths, Sep 8, 2024, 10:20 AM
Consider the fourteen numbers,
. The smallest natural numebr
such that they leave distinct remainders when divided by
is:



This post has been edited 1 time. Last edited by SomeonecoolLovesMaths, Sep 26, 2024, 12:05 PM
L
The centroid of ABC lies on ME [2023 Abel, Problem 1b]
by Amir Hossein, Mar 12, 2024, 4:49 PM
In the triangle
, points
and
lie on the side
, with
. Also,
is the midpoint of
. Show that the centroid of
lies on
.









Problem 63: Concurrent lines
by henderson, Mar 26, 2017, 12:23 PM























Let
be a circle of center
and
be four arbitrary points on this circle. Denote by
the point of intersection of the lines
and
and by
the line passing through
that is perpendicular to
If
and
then the point
is the midpoint of the segment 
The converse of the butterfly theorem is also true. More precisely,

to the degenerated quadrilateral 





































![\[ \frac{E_2D}{\sin \angle E_2SD}=\frac{E_2S}{\sin \angle E_2DS} \ \text{and} \ \frac{F_2D}{\sin \angle F_2SD}=\frac{F_2S}{\sin \angle F_2DS} \]](http://latex.artofproblemsolving.com/3/2/7/327c10c92a2c88cf341c3b999920a63a8a6e3649.png)

![\[ \frac{E_2D}{F_2D}=\frac{\sin \angle F_2DS}{\sin \angle E_2DS}=\frac{\sin \angle FAO}{\sin \angle EAO} .\]](http://latex.artofproblemsolving.com/9/4/0/940ffb84973aa1be6fe971938e7b90ae9498699f.png)
![\[ \frac{EA}{FA}=\frac{\sin \angle FAO}{\sin \angle EAO} \]](http://latex.artofproblemsolving.com/f/f/8/ff808600bd2b1fea9bec32189f45f1f70a43edb8.png)

Thus, we obtain
![\[\frac{E_2D}{F_2D}=\frac{EA}{FA}.\]](http://latex.artofproblemsolving.com/4/f/4/4f45bb351b05224e333950b7e43dde651dd0c10c.png)
![\[\angle E_2DF_2=\angle E_2DS+\angle F_2DS= \angle EAO+\angle FAO=\angle EAF,\]](http://latex.artofproblemsolving.com/1/c/4/1c45c34c179990c6860886604c5d495e806d160e.png)
![\[\triangle DE_2F_2 \sim \triangle AEF ,\]](http://latex.artofproblemsolving.com/8/3/f/83f3ac70ae2a94534f0f2eb8e0cf6b1b4e39ffef.png)


Let









Therefore, the lines









This post has been edited 4 times. Last edited by henderson, Mar 30, 2017, 4:49 PM
"Do not worry too much about your difficulties in mathematics, I can assure you that mine are still greater." - Albert Einstein
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