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.
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.
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 .
, 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 .
such that 
for any relatively prime natural numbers 
.
is described on a circle 
, and is not a trapezius inscribed in a circle. Let the tangency points of and 
be 
respectively. A circle with a center 
, different from is tangent to the segement and lines 
. A circle with center 
, different from is tangent to segment and lines 
. A circle with center 
, different from is tangent to segment 
and lines . A circle with center 
, different from is tangent to segment 
and lines . Prove that the lines 
are concurrent.
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?
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 . 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 .
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 
.
.
be an isosceles triangle with 
. The point 
is in interior of 
such that 
. Find 
I would especially like to see the solution using tryogonometric Ceva, please 
, please edit the problem
be an isosceles triangle with . The point is in interior of such that . Find I would especially like to see the solution using tryogonometric Ceva, please , please edit the problem
, since they are the same angle between segments 
!
be an isosceles triangle with . The point is in interior of such that 
. Find I would especially like to see the solution using tryogonometric Ceva, please 
and apply sinus law in triangles 
and 
we get that 
and 
so 
,the solution is unique of this trigonometric equation and we can easily observe that 
is a solution because 
or 
which is true so the angle is 
.
be an isosceles triangle with . The point is in interior of such that . Find I would especially like to see the solution using tryogonometric Ceva, please
be an isosceles triangle with . The point is in interior of such that . Find I would especially like to see the solution using tryogonometric Ceva, please 
be an isosceles triangle with . The point is in interior of such that . Find I would especially like to see the solution using tryogonometric Ceva, please and if you want all the problems see 
posts.
and if you want all the problems see posts.
.
.
is equilateral. Then from angle chasing, 
and 
are congruent, so 
and the problem is finished