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.
of positive integers is called central if for every positive integer 
, the arithmetic mean of the first 
terms of the sequence is equal to .
of positive integers such that for every central sequence 
there are infinitely many positive integers with 
.
be prime numbers such that their sum isn't divisible by 
. Find the all 
positive integer quadruples satisfy:
be a natural number. Prove that,![\[ \left\lfloor \frac{n}{1} \right\rfloor+ \left\lfloor \frac{n}{2} \right\rfloor + \cdots + \left\lfloor \frac{n}{n} \right\rfloor + \left\lfloor \sqrt{n} \right\rfloor \]](http://latex.artofproblemsolving.com/5/6/0/560a9568b2c09512cac1d317a011931a439e6a02.png)
and the sets 
If 
has exactly three elements, prove that 
cannot be an integer.
such that
holds for all 
.
such that:
with all 
is real number
be a triangle, and let 
, 
, and 
be collinear points such that 
, 
, and 
. Points 
, 
, and 
are the reflections of , , and over 
, 
, and 
, respectively. Prove that if 
is a nondegenerate triangle, then its circumcenter lies on the circumcircle of .
be three positive real numbers satisfying 
Prove that
There is a very elegant proof :-D Could anyone think of it?
such that is a multiple of 
and has exactly positive factors.
, done.
and 
are given on side of convex quadrilateral 
(with closer than to 
). It is known that 
and 
. Prove that 
.
+ 
..... (1)
- 
....(2)
= 180
=
is cyclic, which implies 
. This also implies that 
. However, we note that 
which implies is cyclic. To finish, we note that 
which implies 
. 
is cyclic because 
.
. We then proceed to angle chase. 
because is cyclic and opposite angles in a cyclic quadrilateral add to 
.
by supplementary angles and 
because the angles in a triangle add up to .
, which shows that 
. Finally, because you just subtract equal angles 
and 
. 
and 
, if we add those equalities to the desired conclusion, we then need to prove that is cyclic., we know that is cyclic, so let 
and 
. Let 
. We then have 
, and by sums of angles in triangle we have 
, so is cyclic as desired. 
, 
is cyclic, wherefore
So is cyclic as well, yielding 
or![\[\angle FAC=\angle BAC-\angle BAE-\angle EAF=\angle CDB-\angle CDF-\angle FDE=\angle EDB,\]](http://latex.artofproblemsolving.com/d/0/1/d01bda978ce395ccda1aa19db70f77153ed72658.png)
as desired. 
is cyclic. Because of this, we can set 
and 
. Now, setting 
and 
, we can angle chase to get that 
and 
. Clearly, 
, so is cyclic. Finally, we have 
, which implies that . We are done. 
be the intersection of 
and 
Since 
is cyclic. As such, 
and 
We also have 
so 
As such, 
As such, 
so 
and is cyclic. As such, 
so 
as desired. 
, we know that 
is cyclic.
, we just need to prove that or that is cyclic
, is cyclic, as desired.
, is a cyclic quadrilateral. It follows that 
However, we also have 
. Thus, 
and 
, so it follows that 
or is cyclic. Then, 
, but we have 
, so 
as desired.
if and only if is cyclic.
and 
is cyclic. This means that 
Therefore, 
which means that 
proving that is cyclic, as desired.
. So, quadrilateral is cyclic.
.
. is cyclic. Hence 
.
is a cyclic quadrilateral. Take point 
on line 
such that 
. Then by angle chasing,
which implies quadrilateral being cyclic. Thus
as desired.
. Since is cyclic, 
. 
and 
also equal to so 
and 
. 
so is cyclic. Hence so 
, we know that is cyclic. This also means that 
, implying 
.
. If we substitute in angles (
from earlier and 
from the given), we find that 
. Thus, is cyclic, as the opposite angles in the quadrilateral are supplementary.
by the cyclic condition, and subtracting off some equal angles yields 
, as desired.
![$[(x-2)^2+(y-4)^2][ (x-1)^2 + (y-2)^2] = 36$](http://latex.artofproblemsolving.com/b/1/0/b108dc2cac5b7491fb83418eb36c7184f6d5d30c.png)
then
.
.
.
,
,
is cyclic, so 
.
.
,
Hence, 
,
Q.E.D.
so, 
,
to intersect 
and 
at 
and 
,
are cyclic.
is cyclic. Now note that 
.
,
.
after simplifying, so quadrilateral 
,
as required. This was a lot easier in hindsight.
.
.![\[ \measuredangle ABC = \measuredangle ABE = \measuredangle BAE + \measuredangle AEB = \measuredangle BAE + \angle AEF \]](http://latex.artofproblemsolving.com/0/e/8/0e8883554573b4877b050977e489a1b68d3e27fd.png)
and that ![\[ \measuredangle ADC = \angle ADF + \angle FDC \]](http://latex.artofproblemsolving.com/b/3/6/b36a40fce2548385508623223ab4710492d843c6.png)
Finally, ![\[ \measuredangle ABC = \measuredangle BAE + \measuredangle AEF = \measuredangle BAE + \measuredangle ADF = \measuredangle FDC + \measuredangle ADF = \measuredangle ADC \]](http://latex.artofproblemsolving.com/6/6/4/664d33dd786dd8481cb5785135a0014d483438fd.png)
![[asy]
pair a, b, c, d, e, f;
a = dir(140); d = dir(220); f = dir(300); e = dir(30); b = 1.3 * e - 0.3 * f; c = intersectionpoints((100*e-99*f)--(100*f-99*e),circumcircle(a,d,b))[1];
draw(a--e--f--d--cycle,orange); draw(d--c--f,blue); draw(e--b--a,blue); draw(circumcircle(a, b, c),blue); draw(circle((0,0),1),orange);
dot("$A$", a, dir(120));
dot("$D$", d, dir(240));
dot("$F$", f, dir(330));
dot("$E$", e, dir(30));
dot("$B$", b, dir(30));
dot("$C$", c, dir(270));
[/asy]](http://latex.artofproblemsolving.com/c/f/a/cfab0fa694eb45c43d633409951a99b4f16775d8.png)
is cyclic. Because 
implies that 
,
,