AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
with 
, the foot of altitudes from 
to the sides 
are 
, respectively. 
is the orthocenter. 
is the midpoint of segment 
. Lines 
and 
intersect at 
. Let the tangents drawn to circumcircle 
from 
and 
intersect at 
. Prove that 
are colinear
and 
, define 
as the greatest exponent 
such that 
divides 
. Prove that there are infinitely many such that 
and infinitely many 
such that 
.
prime and a positive integer. Determine all pairs 
satisfying the equation: 
be positive real numbers with 
. Prove that
and 
satisfy 
, then 
is an even number.
, not all equal to 
, the following inequality holds
for which the equality holds.
,
triangle with 
. We take , 
point on AC, AB respectively such that 
, 
. 
, 
lines intersect at point . If 
, find 
be positive reals such that 
. Prove that ![\[(a-b)^2 + a + b \geq 2\]](http://latex.artofproblemsolving.com/c/f/a/cfa01a5b54a72afe6746f547390e65dbbb82ea64.png)
and 
we define 
as the set of all products of an element of and an element of . For example, if 
and 
then 
We call a set 
of positive integers good if there do not exist sets 
of positive integers, each with at least two elements and such that the sets 
and are the same. Prove that the set of perfect powers greater than or equal to 
is good.
, , no two elements are equal, but any two or three of these sets may have common elements. A perfect power is an integer of the form 
, where 
and 
are integers.)
be a scalene triangle with point 
and 
on the plane such that 
is an equilateral . Let 
intersect 
and 
at and 
respectively and 
intersect 
and 
at and 
respectively .
is obtuse with 
to avoid configurational confusion. Now, note that by the Law of Sines on 
,![\[\frac{AW}{AB} = \frac{\sin \angle ABW}{\sin \angle AWB} = \frac{\sin (120-B)}{ \sin (60-C)}\]](http://latex.artofproblemsolving.com/2/0/3/203dccd87d72f76d2e5ff42fb09a421fe0b393a3.png)
Similarly,![\[\frac{AZ}{AC} = \frac{\sin (60+C)}{\sin (B-60)}\]](http://latex.artofproblemsolving.com/0/5/b/05bb3d5943ae753afa5df4a97be3bf93f3acdd6c.png)
![\[\frac{AY}{AB} = \frac{\sin (B-60)}{\sin (C+60)}\]](http://latex.artofproblemsolving.com/f/2/5/f252c986da7fbabd35ae53a6a58a9ff90f4044a8.png)
and![\[\frac{AX}{AC} = \frac{\sin (60-C)}{\sin (120-B)}\]](http://latex.artofproblemsolving.com/f/8/5/f8518212751d1933a335126f99ba94848d87921d.png)
Now it is easy to see that,![\[\frac{AW}{AZ} = \frac{\sin (120-B)}{ \sin (60-C)} \cdot \frac{\sin (B-60)}{\sin (60+C)} \cdot \frac{AB}{AC}\]](http://latex.artofproblemsolving.com/0/8/8/0885b3a57024fc6384b8cbe633a3d741e60c8e21.png)
and![\[\frac{AY}{AX} = \frac{\sin (B-60)}{\sin (C+60)} \cdot \frac{\sin (120-B)}{\sin (60-C)} \cdot \frac{AB}{AC}\]](http://latex.artofproblemsolving.com/3/3/2/332cdf6eda4a15c3d88391eb296aeb4319cc0e3b.png)
from which we may conclude that![\[\frac{AW}{AZ} = \frac{AY}{AX}\]](http://latex.artofproblemsolving.com/1/d/3/1d3cc00b856005181a33e51b8e54745e27f9834c.png)
implying that 
as desired.
. As 
and 
are equilateral, 
.
