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.
is called lucky if it has at least two distinct prime divisors and can be written in the form:![\[
n = p_1^{\alpha_1} + \cdots + p_k^{\alpha_k}
\]](http://latex.artofproblemsolving.com/7/4/4/744a5ccaeb9476ebd7d999c395762cb6e99a7a71.png)
where 
are distinct prime numbers that divide . (Note: it is possible that has other prime divisors not among .) Prove that for every prime number 
, there exists a lucky number such that 
.
be a sequence of positive real numbers such that for every 
, we have:![\[
a_n = \max_{1 \leq i \leq 2025} a_{n-i} - \min_{1 \leq i \leq 2025} a_{n-i}
\]](http://latex.artofproblemsolving.com/e/3/0/e30e9e533eb9a41cf847484e676ac4522edda665.png)
Prove that there exists a natural number 
such that for all 
, the following holds:![\[
a_n < \frac{1}{1404}
\]](http://latex.artofproblemsolving.com/0/1/f/01f1a81eed2230332d51a15501ef2a6ad3a7ec82.png)
, 
, 
, 
, where and 
are odd primes and![\[2^ap^b=(p+2)^c-1.\]](http://latex.artofproblemsolving.com/5/8/5/58518a8ebed14836b72af743ff100eae2efba5e2.png)
, such that for any positive integer 
and non-negative reals 
, the following inequality always holds:
Here 
and 
denote the median and arithmetic mean of 
, respectively.
be an isosceles triangle with 
, and 
an arbitrary point on side 
. The internal angle bisector of 
meets the circumcircle of 
again at 
, and the internal angle bisector of 
meets it again at 
. Show that lines 
, 
and 
are concurrent.
be non-negative real numbers. Prove that
be a triangle such that 
. Points 
lie on segments 
, respectively, such that lines 
and 
are the angle bisectors of triangle 
. Find the measure of angle 
.
satisfy 
prove in equality.
has circumcircle 
and circumcenter 
. A circle 
with center 
intersects the segment at points 
and 
, such that 
, , , and 
are all different and lie on line in this order. Let 
and 
be the points of intersection of and , such that , , , , and lie on in this order. Let 
be the second point of intersection of the circumcircle of triangle 
and the segment 
. Let 
be the second point of intersection of the circumcircle of triangle 
and the segment 
.
and 
are different and intersect at the point 
. Prove that lies on the line 
.
be a point inside a triangle 
such that![\[ \angle APB - \angle ACB = \angle APC - \angle ABC.
\]](http://latex.artofproblemsolving.com/a/6/e/a6e3dc27d0457682fa22af4f918f58d9cd8403bc.png)
, 
be the incenters of triangles 
, 
, respectively. Show that the lines 
, 
, 
meet at a point.
satisfying
Prove
is a triangle and 
are midpoints of 
. Line 
intersects 
in and circumcircle of in . 
are on 
such that 
. 
is on circumcircle of that 
is diameter. The point of intersection of 
and is 
. 
are on 
that 
. Prove that 
and 
are perpendicular.
is a triangle and are midpoints of . Line intersects in and circumcircle of in . are on such that . is on circumcircle of that is diameter. The point of intersection of and is . are on that . Prove that and are perpendicular.
. First of all I think that there's a mistake , 
must be the midpoints of 
respectively...
is a parallelogram and because 
so 
is cyclic and 
and because 
is cyclic too, so the quadrilaterals and are similar. Now assume that 
and 
and 
. and are similar so we have 
, so 
is cyclic too so because 
is a diameter of the circumcircle of we have 
, and because is cyclic we have : 
and that means 
or 
is perpendicular to 
.
is a parallelogram and because so is cyclic and
and because is cyclic too, so the quadrilaterals and are similar.
and 
, .... is there a problem? 
, i.e. exist at least two quadrilaterals with the same angles and which yet aren't similarly.
( sorry for having much confidence 
)
be a fixed convex quadrilateral inscribed in the circle 
with the diameter ![$[AC]$](http://latex.artofproblemsolving.com/0/9/3/0936990e6625d65357ca51006c08c9fe3e04ba0c.png)
, i.e. 
be two mobile points so that 
, the line 
does not separate the points , and the line 
separates the points , . and 
, where 
, have the same values of the its (corresponding to the writing) angles, i.e. and 
, 
and yet they never aren't similarly, i.e. 
is cyclic ![\[\angle T'AS'=180-\angle CAB=180-\angle TPS=\angle TES.\]](http://latex.artofproblemsolving.com/c/b/f/cbf1e582eb64292ee447f175ddeaf16403c0da12.png)
Also, since 
and 
, we have ![\[\angle EST=\angle EPT=\angle PAQ=\angle AE'T'=\angle AS'T'.\]](http://latex.artofproblemsolving.com/f/b/4/fb4b3401a7206dc3df0171b1ca586b0a53c9f20a.png)
Hence the triangles 
and 
are similar (with the same orientation). to . However the corresponding sides 
and 
, as well as 
and 
are perpendicular, so the third pair of sides, and must also be perpendicular.
:
and , .... is there a problem? 
right angles , Do you think they are all Similar? I dont think so.
Prove that
Where 
,such that 
P
