[/quote]
,
.
is a polynomial such that 
is divisible by 
,
.
where 
is a positive integer and 
is a prime.(Don't get mad at me,I've used the search function and did not see a correct and complete solution anywhere.)
such that 
divides 
for all positive integers 
and 
with 
.
,
such that 
is a perfect square? (Here 
denotes the integer part of the real number 
?
be an acute triangle with 
,
be the intouch triangle with 
,
,
,
be the intersecttion of the perpendicular from 
to 
with 
,
.
and 
are concylic
the set 
can be partitioned into 
,
,
.
and that 
.
and 
be the midpoints of sides 
,
and 
be the feet of the altitudes from 
and 
in 
,
be the intersection of line 
with the tangent line to 
.
be the intersection point, other than 
.
be the intersection of lines 
and 
.
with centre 
passes through 
respectively. Let 
,
is equidistant from 
is a cyclic quadrilateral inscribed in a circle of radius one unit. If 
,
,
,
,
satisfy 
and 
.
of 
such that 
.
be in an integer. Prove why ?
and 
.
,
can be expressed as 
where 
Y by
*Note: Whenever I have an equation, AoPS gives the error "New users are not allowed to post images in the Community." To circumvent this I've removed all dollar signs from the equations. Sorry for making this unreadable! If anyone has a better solution, please notify me!
Let ABC be a triangle with |AB|\neq|BC| and the circle \omega, passing through the points A and C, intersect sides AB and BC again at points D and E respectively. The tangents to \omega at the points A and E intersect at X. Prove that AC, DE and BX are concurrent.
I stumbled into this result whilst playing around with the nine-point circle. I wasn't able to find this on the internet. If anybody finds it, please reply and give me advice on how to search for such geometry problems! I don't know how to prove this. I can't figure out how to use the tangency.
Let ABC be a triangle with |AB|\neq|BC| and the circle \omega, passing through the points A and C, intersect sides AB and BC again at points D and E respectively. The tangents to \omega at the points A and E intersect at X. Prove that AC, DE and BX are concurrent.
I stumbled into this result whilst playing around with the nine-point circle. I wasn't able to find this on the internet. If anybody finds it, please reply and give me advice on how to search for such geometry problems! I don't know how to prove this. I can't figure out how to use the tangency.
This post has been edited 1 time. Last edited by Gimbrint, Saturday at 10:26 PM
Reason: Added a correction
Reason: Added a correction
and the circle 
and 
are concurrent.
,
.
.
collinear. Now this is a special case of Pascal's theorem (https://en.wikipedia.org/wiki/Pascal%27s_theorem), where the "hexagon" in question is really a quadrilateral, two of whose sides are zero hence why two of the lines in question are tangents.
