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.
have real roots and real coefficients with 
. What is the maximum value of 
?
with two altitudes 
and 
.Let 
be the point on the line such that 
.The lines 
and 
intersect at point 
, and 
is the point on segment 
such that 
.Suppose that bisects .Prove that triangle 
is isosceles.
s such that:
such that 
. Prove that
of positive integers with 
satisfying 
is
be a positive integer and let 
and 
be sets of integers satisfying the following conditions:
, 
and is a subset of 
, 
iff 
.
such that,
be a positive integer. Find the number of permutations 
, 
, 
of the
, 
, 
, satisfying
.
positive integer the equation: 
has positive integer solutions.
be an equilateral triangle. Let 
be a point inside of it such that the square root of the distance of to one of the sides is equal to the sum of the square roots of the distances of to the other two sides. Find the geometric place of .
be 
, and 
. What is the sum 
? through 
, shuffle them, and then place them in a stack facedown, turning over the top card next to the stack. They then take turns either drawing the card at the top of the stack into their hand, showing the drawn card to the other player, or drawing the card that is faceup, replacing it with the card on the top of the pile. This is repeated until all cards are drawn, and the player with the largest sum for their cards wins. What is the probability that the player who goes second wins, assuming optimal play?
such that 
is also prime.
vertices of an octagon each red, black, and white, such that no two adjacent sides are the same color?
.
and 
intersect in the points and 
.
.
intersects the first circle in the point 
,
intersects 
in the point 
.
intersects the second circle in the point 
,
intersects in the point 
.
,
.
.
.
(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 
.
.
such that:
From here we must have,
Observe that thus far the RHS of the divisibility statement has been strictly positive, following from the fact that 
for 
.
,
.
.
