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.
, no matter how its elements are partitioned into two subsets, at least one of the subsets must contain three numbers 
(
is allowed) such that 
. Find the minimal 
.
and a point 
, let 
be the cevian triangle of with respect to . Let 
be the orthocenter of , and denote the isotomic conjugate of 
with respect to by 
, respectively. Let the centroid of be 
, and denote the isogonal conjugate of with respect to by 
. Prove that
or in brief
be a polynomial with integer coefficients such that 
and 
for some integer 
. Prove that there exists no integer 
such that 
.
be the circumcenter and the orthocenter of an acute triangle 
. Prove that the area of one of the triangles 
, 
and 
is equal to the sum of the areas of the other two.
line segments in the plane such that every two segments cross and no three segments meet at a point. Geoff has to choose an endpoint of each segment and place a frog on it facing the other endpoint. Then he will clap his hands 
times. Every time he claps,each frog will immediately jump forward to the next intersection point on its segment. Frogs never change the direction of their jumps. Geoff wishes to place the frogs in such a way that no two of them will ever occupy the same intersection point at the same time. is odd. is even.
such that a positive integer power of 
equals to a positive integer power of 
.
can be written in the form
where 
for some non-negative 
such that
for every 
.
to the following system:
![\[\frac{\sin^2A+\sin^2B+\sin^2C}{\cos^2A+\cos^2B+\cos^2C} = 2\]](http://latex.artofproblemsolving.com/9/2/a/92a524bbdca92e9dd4091275714bf346bad96cd7.png)
be a parallelogram and points 
be on its exterior. If triangles 
and 
are similar, i.e. 
, prove that triangle 
is similar to these two triangles.
be a triangle and let be a point not lying on any of the three lines 
, 
, or 
. Distinct points 
, 
, and 
lie on lines , 
, and , respectively, such that 
and 
. Show that there exists a point 
on the circumcircle of 
such that 
is similar to 
.
:
Square again and subtract to get 
Thus, 
Roots are 
Negatives are bad because lhs and rhs would have differing signs, and 
fails. Thus, all are bogus except 
. Now observe that for 
, 
, so we must have 
. So, 
, giving 
.
for can be proven by casework.)
:
.
and 
Then 
in terms of 
because the LHS is always positive, so the RHS is also always positive. Now we split it into cases:
This gives 
solving gives 
This gives 
so 
again.
