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.
. How many complex numbers 
are there such that 
and 
is an integer.
and 
is a subset of 
such that if 
then 
Find the smallest of 
that satisfy the following property: if 
, 
, 
, 
are all the positive divisors of 
with 
, then 
divides 
for every 
.
, in which 
: 
, but the proof is important.
be reals such that 
. Prove that
Let be reals such that 
. Prove that
and 
Prove that
satisfying ![$$(ab+bc+ca)\left[\frac{1}{(a+b)^{2}}+\frac{1}{(b+c)^{2}}+\frac{1}{(c+a)^{2}}\right]\ge \frac{9}{4}+k\cdot\frac{a(a-b)(a-c)+b(b-a)(b-c)+c(c-a)(c-b)}{(a+b+c)^{3}}$$](http://latex.artofproblemsolving.com/f/6/e/f6ed10f7fff1cc94edd8f451e75718a0916a8bfa.png)
holds for all 
be non-negative real numbers such that 
Prove that
be a scalene triangle with 
, and let 
be the foot of the altitude from 
. Let 
be a point in the interior of the segment 
. Let 
be the point on the segment 
such that 
. Similarly, let 
be the point on the segment 
such that 
. The circumcircle of triangle 
intersects segment 
at a second point 
(other than ). Prove that 
.
and be positive integers such that 
. Define 
for 
. Prove that if all the numbers 
are integers, then 
is divisible by an odd prime.
![\[ \frac{\displaystyle\sum_{k=0}^r \binom nk \binom{n-2k}{r-k}}{\displaystyle\sum_{k=r}^n \binom nk \binom{2k}{2r}\left(\frac{3}{4}\right)^{n-k}\left(\frac{1}{2}\right)^{2k-2r}}\]](http://latex.artofproblemsolving.com/9/d/e/9dedd490eaf9c6d22aaf93bbd74d82375adcfa85.png)
\
.
, since scaling up area by some amount will scale perimeter by the square root of that amount, if the perimeters are unequal, they will still be unequal after a scale. Then the perimeter of a triangle that has side length 
is 
. If we can prove that the derivative is positive or negative for the entire interval 
, we will be done.
. We can now just prove 
, both of which are trivial given . Therefore, there exists no noncongruent right triangles with the same area and perimeter.
and 
with 
and 
, then 
and 
.
.
and one angle 
and simplified using those conditions and got this equation
.
, where 
is the hypotenuse. and 
respectively where 
and are the hypotenuses.
----(i)
-----(ii)
----(iii)
, which implies 
and perimeters are same, we get 
. This implies 
. Assuming 
and 
and . Adding and subtracting these two we get 
and 
.
be real numbers such that 
Prove that
,
,
and the other right triangle has legs 
,
.![\[a+b+\sqrt{a^2+b^2}=c+d+\sqrt{c^2+d^2}\]](http://latex.artofproblemsolving.com/8/a/2/8a23f74cf8f85dc0e4ccff1714feec3ef1121d21.png)
and ![\[ab=cd\]](http://latex.artofproblemsolving.com/7/2/6/7260dcc8367798a67fc5b8c4f786038a85a09738.png)
looking for a contradiction. WLOG, 
so we have must have 
because lengths are non-negative and the function 
is strictly increasing over non-negative numbers. This implies 
because 
.
because the square-root function is stricting increasing. Putting this together, 
,
.
.
Using the quadratic formula: 
Therefore, 
or 
