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.
and 
is closest to which integer?
be a quadrilateral inscribed in a circle 
. The lines 
and 
meet at 
, the lines 
and 
meet at 
, and the diagonals 
and 
meet at 
. Let 
be the midpoint of the segment 
, and let 
be the common point of the segment 
and the circle . Prove that the circumcircle of the triangle 
and are tangent to one another.
be an acute triangle with orthocenter 
, and let 
be a point on the side , lying strictly between 
and 
. The points and 
are the feet of the altitudes from and , respectively. Denote by 
is the circumcircle of 
, and let 
be the point on such that 
is a diameter of . Analogously, denote by 
the circumcircle of triangle 
, and let 
be the point such that 
is a diameter of . Prove that 
and are collinear.
be an acute triangle. Define 
as its incenter. Let 
and 
be the incircle's tangent points to 
and 
, respectively. Let 
be the midpoint of 
. Let 
be the intersection point of a perpendicular line passing through to 
. Line 
intersects the circumcircle of 
at 
. The circumcircle of 
intersects line 
at 
. Prove that quadrilateral 
is cyclic., the diagonal bisects neither the angle nor the angle 
. The point lies inside and satisfies ![\[\angle PBC=\angle DBA\quad\text{and}\quad \angle PDC=\angle BDA.\]](http://latex.artofproblemsolving.com/c/2/0/c20761f3eadd054958f40259f3d1c05f26279783.png)
Prove that is a cyclic quadrilateral if and only if 
.
be a cyclic quadrilateral in which the lines and meet at a point . Let be the point of the line 
, different from , such that 
. We construct the parallelograms 
and 
. Prove that the points 
lie on the same circle.
point 
was chosen such that 
and 
. Prove that if perpendicular from to passes from the intersection of diagonals of , then 
.
, 
. The internal angle bisector of 
and the external angle bisector of intersect the ray at points 
and , respectively. Given that 
, prove that 
.
and 
intersect at two points 
and 
. Let 
be the line tangent to these circles at 
and 
, respectively, so that lies closer to than . Let 
be the line parallel to and passing through the point , with 
on and 
on . Lines 
and 
meet at 
; lines 
and meet at 
; lines 
and meet at 
. Show that 
.
be an acute triangle. Let be the altitude on , and let be any interior point on . Lines 
, when extended, intersect 
at 
respectively. Prove that 
.
such that 
(This is to eliminate any cases where a number pops up such that 
which would case a huge headache). Consider how many 
s divide into 
. There are 
of them. Assume 
. Since 
is odd, 
. Then the first digit of base 
has value 
. The remainder of the number has value 
. ![\[\left\lfloor \dfrac{a_k n^{k-1}}{(2n)^{k-2}} \right \rfloor\]](http://latex.artofproblemsolving.com/2/5/c/25cc6b3b5c2d567481c42a3dee5a917b5a4d92d5.png)
is the 2nd digit from the left. This equals 
. Since 
, we have that ![\[\left\lfloor \dfrac{n}{2^{n-2}} \right \rfloor \le \left\lfloor \dfrac{a_{k-1}n}{2^{n-2}} \right\rfloor < 2n.\]](http://latex.artofproblemsolving.com/c/6/4/c64006c8c048e7aa924eab6447e1e5598a19b719.png)
If we let 
, we notice that 
is odd, so 
. Similar logic can be applied for the 3rd digit, and the 4th, etc. Then (?) the first digit equals 
, the second digit is greater than or equal to 
, etc. Since all these values can be arbitrarily large if is arbitrarily large, then no matter the value of 
, we can choose a such that all digits are greater than or equal to values greater than .
.
