be a triangle such that 
and 
is obtuse. Point 
is the circumcenter of triangle 
is the reflection of 
in 
.
be an arbitrary point on line 
is in between 
.
cuts the circumcircle of 
.
and 
cut 
respectively. Prove that 
is tangent to the circumcircle of triangle 
.
and circumcenter 
.
be the orthocenter and 
be the incenter of triangle 
be the midpoint of the arc 
,
be a point on the arc 
of 
.
be the circumcenter of triangle 
.
and 
meet on 
non-negative real numbers, then we have:
| 
with 
are positive integers and 
is prime 
)
such that 
and:
of positive integers such that 
is divisible by 
.
be a set consisting of non-negative integers such that:
,
,
and 
.
such that![\[
f\left(\frac{x^2 - f(x)}{f(x) - 1}\right) = x
\]](http://latex.artofproblemsolving.com/8/0/0/8009a4709ffd6708023d67db2b544e605d1fbe68.png)
for all real numbers 
satisfying 
.
and 
such that![\[ {a^2+b\over b^2-a}\quad\mbox{and}\quad{b^2+a\over a^2-b} \]](http://latex.artofproblemsolving.com/e/3/b/e3bc96af9f635e1c10e21dc91894c8afbf113a6d.png)
lie on segments 
,
and 
intersect at point 
and 
intersect at a second point 
.
.
such that 
.
and 
are similar.
,
with integer coefficients and for any prime 
,
of at most 
integers, such that![\[\bigcup_{s \in S} \{s,\; Q(s),\; Q(Q(s)),\; Q(Q(Q(s))),\; \dots\}\]](http://latex.artofproblemsolving.com/f/9/1/f919f2391121c2b0551d29d4f09ea6cde631a8ba.png)
contains a complete residue system modulo 
Y by Rounak_iitr
Dear Mathlinkers,
1. ABC a triangle
2. D, E two point on the segment BC so that BD = DE= EC
3. M, N the midpoint of ED, AE
4. H the orthocenter of the acutangle triangle ADE
5. 1, 2 the circumcircle of the triangle DHM, EHN
6. P, Q the second point of intersection of 1 and BM, 2 and CN
7. U, V the second points of intersection of 2 and MN, PQ.
Prove : UV is parallel to PM.
Sincerely
Jean-Louis
1. ABC a triangle
2. D, E two point on the segment BC so that BD = DE= EC
3. M, N the midpoint of ED, AE
4. H the orthocenter of the acutangle triangle ADE
5. 1, 2 the circumcircle of the triangle DHM, EHN
6. P, Q the second point of intersection of 1 and BM, 2 and CN
7. U, V the second points of intersection of 2 and MN, PQ.
Prove : UV is parallel to PM.
Sincerely
Jean-Louis
is cyclic, and we also have 
cyclic, so 
.
