AoPS Academy
blocks, each weighing at least 
; their total weight is 
. Prove that for every real number 
with 
you can choose a subset of the blocks whose total weight is at least but at most 
.
such that 
.
chessboard. Initially, Azer colors all cells of this chessboard with some colors. Then, Babek takes 
rows and columns and looks at the 
cells in the intersection. Babek wants to have all these cells in a same color, but Azer doesn't. With at least how many colors, Azer can reach his goal?
be non-negative real numbers such that 
Prove that
be a circle, and let a line 
intersect at two points, 
and 
Circles 
and 
are internally tangent to at points 
and 
respectively, and both are tangent to at a common point 
Similarly, circles 
and 
are externally tangent to at and respectively, and are tangent to at points 
and 
respectively. is 
the segment 
has a length of 
and 
find the length of segment 
,
,
, 
, 
is not greater than 
.
and 
be points on a plane such that 
, where is a positive integer. Let 
be the set of all points such that 
, where 
is a real number. The path that traces is continuous, and the value of is minimized. Prove that is rational for all positive integers 
, a point 
is chosen such that the radii of the circumscribed circles of the triangles 
, and 
are equal. Prove that is the intersection point of the heights of the triangle .
, a point is chosen such that the radii of the circumscribed circles of the triangles , and are equal. Prove that is the intersection point of the heights of the triangle .
and 
are angles opposite to equal chords in equal circles, it is 
. Similarly, it is 
and 
. So:
Analogously, we have:
We will argue by contradiction. Suppose, on the contrary, that is not the orthocenter of 
and let 
be the orthocenter of . It is easy to prove that:
Now, consider the circumcircles of 
and 
. Since it is 
(
) and 
(
), both 
and 
are concyclic. This means that the circumcircles of and intersect at three distinct points, namely , and , which is absurd. Thus, the orthocenter of and coincide and the proof is complete.
and 
Prove that
.
Prove that
and 
Prove that
.
.
