[/quote]
,
integers 
,
,
be positive integers such that no one of them is divisible by any other member of the sequence. Then![\[
a_1 \geq 2^k,
\]](http://latex.artofproblemsolving.com/1/f/e/1fecb6d6811bfc4aff7ccefa20412f0a5b860066.png)
where 
is defined by the inequalities![\[
3^k < 2n < 3^{k+1}.
\]](http://latex.artofproblemsolving.com/2/8/f/28f12a76029727a828167f1cb8e118b84d710d62.png)
This estimate for 
is the best possible.
![\[
\max \left( (a_i, a_j) \right) > \frac{38n}{147} - c,
\]](http://latex.artofproblemsolving.com/6/5/d/65df2fa2c65d90d0548cb3c11dd36ea42c4060e6.png)
where 
is a constant independent of 
,
denotes the greatest common divisor of 
and 
.
,
satisfying 
and ![$ (x + 1)[f(x)]^2 - 1$](http://latex.artofproblemsolving.com/9/c/1/9c1e80d035c923dd13711bff624d8a1a3e6a3b15.png)
is an odd function.
be the inscribed circle of a triangle 
.
and 
be the tangency points of 
and 
,
and 
intersect 
and 
,
and 
intersects 
at 
,
intersects 
at 
, and the tangent line of 
at 
.
and 
is a prime and 
,![\[
n! \sum_{pi+j=n} \frac{1}{p^i i! j!} \equiv 0 \pmod{p}.
\]](http://latex.artofproblemsolving.com/1/6/6/166f07aa9e69c0eb35833971acb9b04932fea087.png)
Prove that
Let 
Prove that![$$ (a+b)(a+1)(b+1) \leq \frac{3}{2}+\sqrt[3]{6}+\sqrt[3]{36}$$](http://latex.artofproblemsolving.com/f/5/e/f5e68204507aa9c3e71c28844bbee3a804546b94.png)
,
,
,
be an acute triangle with 
.
points on the segment 
such that 
.
is a point on segment 
intersects 
and 
at 
and 
,
and 
intersect at 
and 
,
is cyclic.
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Points 
lie on a circle 
and point 
lies outside the circle. The given points are such that (i) lines 
and 
are tangent to , (ii) 
are collinear, and (iii) 
. Prove that 
bisects 
.
lie on a circle 
and point 
lies outside the circle. The given points are such that (i) lines 
and 
are tangent to , (ii) 
are collinear, and (iii) 
. Prove that 
bisects 
.
and Z is outside 
Though it's not competition-oriented, I believe it's around the same audience.
be the midpoint of 
is cyclic. Let 
Then, 
so 
colinear.
.
,
,
By Power of a Point and cyclic quadrilaterals, 
So 
are on the same circle, which implies our desired conclusion.
is similar to 
,
is similar to 
.
.
is a median
be the point at infinity along lines 
and 
,
.
is harmonic,![\[-1 = (A,C ;B , D) \stackrel{E}{=} (A, C; M, P_{\infty}) \]](http://latex.artofproblemsolving.com/3/b/a/3bac455ec07839e6b78f06a8fd19884b103e48b3.png)
so 
,
.
cyclic. Now let 
.
and clearly 
where 
is the center of 
,
and thus 
.
is perpendicular to 
is cyclic.
and the tangent criterion, we have ![\[\angle BMP = \angle BMD = \angle DBP = \angle BOP\]](http://latex.artofproblemsolving.com/a/3/6/a363b3afbc14fcf16302abdb30b3f845fadecc28.png)
as desired.
,
and 
let the center of the circle be 
and let 
so we have 
we have 
from which it follows that 
so we are done. 
and 
be complex numbers corresponding to 
and 
respectively. Then, by the formula for the intersection of tangents, we get 
Then, since 
and 
must have the same measures, so it follows that 
Then, since 
and 
lie on the unit circle while 
we have
Therefore, the midpoint of 
It now suffices to show that 
and 
are collinear. By the complex number collinearity condition, this is equivalent to
concluding the proof. 
,
.
are all concyclic.![\[ \angle{BMP} = \angle{MED} = \angle{BDP}, \]](http://latex.artofproblemsolving.com/2/d/d/2dddfc10aac77de9281ee88e4f3cf7a47c46cf03.png)
where the last part comes from the so-called Tangent Lemma. In particular, 
is a cyclic quadrilateral, so 
.
is clearly cyclic 
)
tells us that 
,
is a symmedian. Also, 
is cyclic with 
,
implies 
,
.
,
is cyclic. But since 
has diameter 
,
and hence 
.
;
is isosceles.
is cyclic. We can do so by proving 
:![\[\angle{BXO} = \angle{BED} = \frac{\overarc{BD}}{2} = \frac{\angle{BOD}}{2} = \angle{BOP}\]](http://latex.artofproblemsolving.com/f/a/7/fa7332189b642ee5ad9f46b558a4b8e5aae1b7d4.png)
,
(this can be shown by proving 
,
,
(as 
by the tangent condition). Since 
and then project through 
onto 
is cyclic and by the 
criterion, 
.
is cyclic. This means 
,
and we're done. 
,
