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 there exist integers 
with ![\[x^3_1+x^3_2+\,\ldots\,+x^3_t=2002^{2002}\,?\]](http://latex.artofproblemsolving.com/e/f/a/efaa98e77ac0273284ee3446e4335d8cc4f681cb.png)
and 
of the equation ![\[(a^2+b)(a+b^2) = (a-b)^3.\]](http://latex.artofproblemsolving.com/b/5/7/b57940d59694d6892784e6390487010d5e036604.png)
be a triangle with circumcenter 
, angle bisector 
with 
and altitude 
with 
. The lines 
and 
meet at 
. The circumcircle of 
meets 
at 
and 
meets at 
. The circumcircles of triangles 
and meet at 
. Show that 
is a symmedian in 
that satisfies the following condition.
, the number of positive integers 
that 
is less than 
, points 
are chosen such that when going around the triangle, the points occur in the order 
. It is given that![\[
AD_1 = AD_2 = BE_1 = BE_2 = CF_1 = CF_2.
\]](http://latex.artofproblemsolving.com/4/c/f/4cf9e7bc009736836bcd10c1d8ba46bd3a91cd7b.png)
Prove that the perimeters of the triangles formed by the lines 
and 
are equal.
is chosen. Prove that there exists a prime number that divides infinitely many terms of the sequence 
, where![\[
b_k = a^{k^k} \cdot 2^{2^k - k} + 1.
\]](http://latex.artofproblemsolving.com/1/7/5/1751a60482d729a36c71b77ac9c978e724f40da0.png)
lie on the same circle. If 
and 
, what is the area of octagon ?
, , and 
or 11. However, I didn't know how to find the radius. Can anyone give me a hint?
be a cyclic quadrilateral, and let 
, 
, 
, and be the midpoints of 
, , 
, and 
respectively. Let 
, 
, 
and 
be the orthocenters of triangles 
, 
, 
and 
, respectively. Prove that the quadrilaterals and 
have the same area.
with real coefficients is the average of two monic polynomials of degree with real roots.
of positive integers, for which![$$(a, b) + 3[a, b] = a^3 - b^3$$](http://latex.artofproblemsolving.com/b/8/3/b83f52a0152dc39b6d966880a1d6ca824a0f3786.png)
denotes the greatest common divisor of , and ![$[a, b]$](http://latex.artofproblemsolving.com/d/a/2/da2e551d2ca2155b8d8f4935d2e9757722c9bab6.png)
denotes the least common multiple of .
and 
where each summation runs over 
variables 
and 
Then find the value of 
, 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 
.
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 
