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 
estimate the value of 
, 
, 
, 
be distinct circles such that , are externally tangent at 
, and , are externally tangent at the same point . Suppose that and ; and ; and ; and meet at 
, 
, 
, 
, respectively, and that all these points are different from . Prove that![\[
\frac{AB\cdot BC}{AD\cdot DC}=\frac{PB^2}{PD^2}.
\]](http://latex.artofproblemsolving.com/0/4/e/04ead9928d0c767a2209f610885e30f6b6e0a4e9.png)
points arranged in the form of a 
grid. What is the maximum number of points that can be chosen among them so that no four of them form the vertices of either an isosceles trapezium or a rectangle whose parallel sides are parallel to the grid lines?
be an acute triangle, and let 
be a variable interior point on the minor arc 
of its circumcircle. Let and 
be the feet of the perpendiculars from to lines 
and 
, respectively. Let 
be the intersection of line 
and the perpendicular from to 
. Let 
be the line through parallel to 
. Prove that as varies along minor arc , the line always passes through a fixed point. (Specifically: prove that there is a point 
, determined by triangle , such that no matter where is on arc , line passes through .)
points 
and lie on the segments 
and , respectively, and are such that 
is a parallelogram. The circle with center the midpoint 
of the segment 
and radius 
and the circle of diameter 
intersect for the second time at the point 
. Prove that the lines 
and 
intersect in a point.
is a triangle, and 
and are points on the segments and respectively, such that 
and 
. Suppose that is the midpoint of 
and 
is the point of intersection between and . Prove that triangle 
is similar to triangle .
of a square table each have length 
, where is a positive integer. For how many ordered 4-tuples 
of nonnegative integers can we cut a piece of length 
from the end of leg 
and still have a stable table?
be a semicircle with diamater . Let be the midpoint of . Let 
be points on the same semiplane with with respect to the line such that 
is a parallelogram. Let 
and 
. Let 
be the incenter of 
. Let 
and 
intersects 
at . Let the intersection point of 
and be and let the perpendicular projection of onto be . Prove that is midpoint of
is situated inside the parallelogram 
such that 
. Prove that 
.
be an acute triangle and 
is orthocenter. Let be the intersection of 
and and be the intersection of 
and . The circumcircle of 
cuts the circumcircle of at 
. Prove that the angle bisectors of 
and 
concur at a point on 
, and 
are its altitudes. A point is chosen on the arc 
of the circumcircle of . If 
, show that the midpoint of lies on 
.
is 
, then every term in the summation will be 
, so the answer is just .
defined as (
denotes greatest integer function, 
denotes the fractional part function)
Suppose 
where 
Then 
and 
Now suppose 
Then we have 
Note that 
and 
by our established inequalities. Thus, 
Thus in all cases of 
our assumption holds, and it is thus true.
Rearranging our claim, we have that 
Thus
All terms in the sum collapse except for the first term. The first term is simply 
and thus we have proven this sum is equal to 
.