AoPS Academy
The altitudes 
, intersect
. The tangents at 
and 
of intersect at 
. Let 
be the midpoint of 
. 
intersects 
, 
intersects 
at 
lie on the same circle.
intersects the circle at the second point 
The line intersects at the second point 
. Prove that 
are collinear.
complex numbers with 
such that
Find the value of 
.
denote the set of natural numbers and let 
denote 
ordered pairs in 
. Prove that there exist 
distinct elements in the set 
for 
whose product is a perfect cube.
Where 
are positive integers and 
is a non-square positive integer. If 
denotes the n-th set of solution to the equation with 
Then, prove that 
have real roots and real coefficients with 
. What is the maximum value of 
?
, all the roots added will give 
. I also learned that if every single root, multiplies every single root, it will give 
. I also learned that if all the roots are multiplied, it will give 
. Is this right? And is there any purpose for these equations?
such that:
and the set
and we are supposed to count the number of sets (not pairs!) 
, such that 
, 
and such that the Euclidean distance between and 
is an integer. Ok?
that will qualify for this count will be such that and have either the same 
-coordinate or the same 
-coordinate. Let's call these trivial pairs and count them first. For every point 
there will be exactly ten other points in 
with the same -coordinate and ten more with the same -coordinate, and as there is no overlap, there are 20 such points in total for every pair . As there are 
points in , this produces 
pairs of points we need to count as trivial pairs. However, as sets, there are just 
of them. is an element in such that 
share neither - nor -coordinate, then there are exactly two points 
such that 
is a non-degenerate rectangle with each side parallel to one of the coordinate axes, and qualifies for our count if and only if the two sidelengths of that rectangle and the length of its diagonal form a Pythagerean triple, and, due to the geometric dimensions of , the sidelengths, i.e. the first two elements of the Pythagorean triple, must be smaller or equal to 10. But there are just two such Pythagorean triples, namely 
and 
. with sidelengths either 3 and 4 or 6 and 8 such that 
and such that the sides of these rectangles are parallel to the coordinate axes. Every such rectangle contains two pairs of points, and 
that qualify. We do the counting by keeping track of the bottom-left point of the possible rectangles.-axis, the bottom-left corner can vary from 
to 
, which produces 
such rectangles, and if the side with length 3 is parallel to the -axis, we get another 56 rectangles. That's 112 rectangles here.-axis, the bottom-left corner can vary from to 
, which produces 
such rectangles, and in the other orientation we get another 15 rectangles, so 30 rectangles here.
and 
are complex numbers such that
and 
are the roots of the cubic equation 
then compute 
and 
Show that
Let 
Show that
,
.
.
(I won't be proving this here). This new information allows us to utilize the mean inequality chain as follows:
Similarly, applying AM-HM yields 
.
.
