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 a positive integer. Call all positive divisors of which are different from 
and beautiful divisors.We call a special number when it has at least 
beautiful divisors and difference of any beautiful divisors divides as well. Find all special numbers.
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 
, of equal radius intersect at different points 
and 
. Consider a circle 
externally tangent to at 
and internally tangent to at point 
. Prove that lines 
and 
intersect at a point lying on .
be inscribed in the circle 
. A line through point 
intersects 
and 
at points 
and 
, respectively. Let 
be the reflection of across the midpoint of , and 
be the reflection of across the midpoint of . Prove that:
of triangle across line 
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 
. Is this possible to make the number 
on the board?
and 
.
?
lies inside triangle such that 
and 
. Point is the midpoint of segment 
. Point lies on segment with 
. Prove that 
.
the number of positive integer 
satisfying 
is less than or equal to 
where 
is a constant independent of 
be different real numbers. Prove that![\[\sum_{1 \leqslant i \leqslant n} \prod_{j \neq i} \frac{1-x_{i} x_{j}}{x_{i}-x_{j}}=\left\{\begin{array}{ll}
0, & \text { if } n \text { is even; } \\
1, & \text { if } n \text { is odd. }
\end{array}\right.\]](http://latex.artofproblemsolving.com/6/5/2/6525fc6b451ff5a3c83ff2a0541b063a7cf065c3.png)
satisfying
for all real numbers 
and 
.
.
. On the other hand, since the point F is the point where the circle k, which is the K-excircle of triangle DCK, touches the side CD of this triangle, we have 
. Hence, CE' = DF. Therefore, DF = CE holds if and only if CE = CE'. But since the points E and E' both lie on the segment CD, we have CE = CE' if and only if E = E'. Hence, DF = CE holds if and only if E = E'.
, since O is the center of the circle k. On the other hand, 
, as the point F is the point where the circle k touches the side CD. But since AB || CD, we thus get OH || OF. The lines OH and OF must therefore coincide (in fact, they are parallel and they have a common point - namely, the point O). This yields that the point O lies on the segment HF. Also, since the points H and F lie on the circle k, whose center is O, we have OH = OF. Now, since AB || CD, by Thales, 
; since OH = OF, we thus get HM = EF. Combined with HM || EF (this is equivalent to AB || CD), this yields that the quadrilateral HEFM is a parallelogram; thus, HE || MF, so the point E is the point of intersection of the parallel to the line MF through the point H with the line CD.., then DF = CE.
. Since HM = EF, we thus have 
, and 
. Now, since AB || CD, Thales yields 
, 
(since the points K, H and E are collinear) and 
. Together with , this yields 
, and thus . Thus, Assertion 1 is proven.. Then, . But since AB || CD, Thales yields and 
. Hence, 
, so that 
, and thus the point N is the midpoint of the segment KM. But on the other hand, the point N is the midpoint of the segment E'F. Thus, the diagonals KM and E'F of the quadrilateral KE'MF bisect each other (they have the same midpoint N); hence, this quadrilateral KE'MF is a parallelogram, and it follows that KE' || MF. Hence, the line KE' is the parallel to the line MF through the point H. Consequently, the point E', lying on the line KE' and on the line CD, must be the point of intersection of the parallel to the line MF through the point H with the line CD. But this is exactly the way we characterized the point E in the above. Hence, E = E', and thus, according to the above, DF = CE. Hence, Assertion 2 is established, and the problem is solved.
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 has midpoint 
.
meets 
iff 
.
