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.
such that 
is a divisor of 
.
, such that 
and 
.
such that 
is prime, 
with 
and ?
. He says to Cecilia: "If you pick any subset of 10 elements from , it will be disjoint with at least one subset from my list." What is the smallest possible number of subsets Miguel's list can have if what he tells Cecilia is true?
be a triangle and 
points in the same plane such that 
is the orthocenter of triangle 
, is the incenter of triangle 
, and is the centroid of triangle 
. If line 
is parallel to 
, prove that 
is the midpoint of segment .
for which the following statement is true: If positive integers 
and 
are such that 
is a multiple of 
and 
is a multiple of , then each of the numbers and is a multiple of ., define
and
meaning 
is the least common multiple of the numbers 
. Determine whether or not there exists a constant 
such that
for every positive integer .
and 
be positive integers with 
. There are tokens placed initially in the leftmost cell of a horizontal row consisting of cells. A move consists of shifting any token 
, 
, 
, 
, 
, or 
positions to the right. Andrés and Beto alternate turns, starting with Andrés. The winner is the player who moves a token into the rightmost cell. Determine who has a winning strategy in terms of and .
be an acute triangle with circumcenter 
. Draw altitude 
, with 
on side 
. The parallel line to 
passing through intersects line 
at point . Prove that point and the midpoints of sides 
and are collinear.
is mundane if there exist positive integers and less than 
such that 
is a positive integer. Find, with proof, all prime numbers that are not mundane.
be a fixed positive integer. Find the greatest real constant 
that has the following property: Any 
real numbers, not necessarily distinct, that lie in the interval ![$[100,101]$](http://latex.artofproblemsolving.com/8/e/3/8e3fe24e3b899bbf2128eebe08756e1ff963aaa0.png)
, can be partitioned into two groups with sums 
and 
such that
.
, 
and 
.
.
-tromino is a figure made up of three squares, obtained by removing one square from a 
board.
board consisting of 
unit segments. A configuration of several -trominoes is optimal if the -trominoes do not overlap, each one covers exactly three squares of the board, and moreover, no unit segment of the board belongs to two -trominoes. Below is an optimal configuration of 5 -trominoes: for which there exists an optimal configuration of L-trominoes on the board.
points 
(
is between 
and ) and 
are chosen. The points 
and 
are defined as the intersection point of 
with the segments 
and 
, respectively. If 
, 
, 
and 
, prove that 
.
points ( is between and ) and are chosen. The points and are defined as the intersection point of with the segments and , respectively. If , , and , prove that .
by adding 
to both sides it is equivalent to prove 
on the segment such that 
, then it is sufficient to prove that 
, since 
and 
is isosceles we get that 
is parallel to 
so 
Also we have that
So we have that
Claim: 
such that 
.
is isosceles and 
we find that 
thus it remains to prove that 
.
where 
and 
.
, so we need to prove that 
, which is equivalent to 
.
and so we only need to prove that
We prove this by some simple appliance of the Law of Sines!
and 
we have
And since 
is cyclic from the assumption of 
we find that
.
and points 
we have:
where we used also the fact that 
.
implies that quadrilateral 
is cyclic. On the other hand from PoP then we have:
where we used the fact that and 
.
and points 
we have:
Summing together before established results gives us:
as desired.
is cyclic
,
such that 
.
belongs to the segment 
and 
.
.
Hence 
So finally 
.