[/quote]
,
be an isosceles trapezoid with 
.
.
and 
the circumcircles of the triangles 
and 
,
be the crossing point of the tangent to 
with the tangent to 
is tangent to 
is a positive integer. Call all positive divisors of 
and 
beautiful divisors and difference of any 
we have 
.
.
,
,
,
,
meet the sides 
,
,
at points 
,
,
,
be the incenter of triangle 
,
be the foot of the perpendicular from 
.
is cyclic, then the center of its circumcircle lies on segment 
.
.
intersects 
,
at 
passes from circumcenter of 
.
be an infinite sequence of real numbers in the interval 
.![\[ \frac{a_1}{1},\frac{a_2}{2},\frac{a_3}{3},\frac{a_4}{4},\ldots.\]](http://latex.artofproblemsolving.com/3/c/e/3cef10f63e9c2dde90b29c742ece8fd31e0a8e94.png)
with rational coordinates lie on a plane. It turned out that the distance between every pair of points is an integer. Prove that there exist points 
with integer coordinates such that 
for every pair 
and 
,
and 
.
and internally tangent to 
.
and 
intersect at a point lying on 
be inscribed in the circle 
.
intersects 
at points 
be the reflection of 
lies on the circle 
and 
coincide.
on the board. For each procces, the stutent deletes the number x and replaces it with either 
or 
or 
.
on the board?
satisfying
Prove
:![\[
1 \cdot 3 \cdot 5 \cdot \ldots \cdot (2n + 1) \mid (4n + 3)(4n + 5) \cdot \ldots \cdot (8n + 3).
\]](http://latex.artofproblemsolving.com/3/1/a/31a0eac3c7ca298c8fa221083dc94e127e705784.png)
and 
.
?
and 
.
.
.
Y by
Let 
with 
being the minimum length, and 
as the intersection of the tangents of 
at points 
and 
. Internal angle bisectors of 
intersects circle with radius 
at 
and 
respectively. Define 
as its incenter.
Suppose
are the circumcenters of 
, and 
intersects at 
. Prove that
![\[ \frac{YM}{XN} = \frac{CH - AH}{BH - AH} \]](//latex.artofproblemsolving.com/0/0/d/00dbe699194c4e6a810bc487b579545b6c09ad80.png)
where 
is the orthocenter of .
Proposed by Jonathan Christian, Indonesia
with 
being the minimum length, and 
as the intersection of the tangents of 
at points 
and 
. Internal angle bisectors of 
intersects circle with radius 
at 
and 
respectively. Define 
as its incenter.Suppose
are the circumcenters of 
, and 
intersects at 
. Prove that![\[ \frac{YM}{XN} = \frac{CH - AH}{BH - AH} \]](http://latex.artofproblemsolving.com/0/0/d/00dbe699194c4e6a810bc487b579545b6c09ad80.png)
where 
is the orthocenter of .Proposed by Jonathan Christian, Indonesia
This post has been edited 4 times. Last edited by IndoMathXdZ, Sep 4, 2020, 9:30 AM
Reason: @Thanks below Yeah, there is indeed a typo and kinda my fault for not rechecking the statement.
Reason: @Thanks below Yeah, there is indeed a typo and kinda my fault for not rechecking the statement.
should be 
