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.
be a cyclic quadrilateral. Let 
be the midpoint of the arc 
of its circumcircle which does not contain 
or 
. Let the lines 
and 
meet at 
and the lines 
and 
meet at 
. Prove that the lines 
and are parallel.
be an acute triangle with circumcenter 
and orthocenter 
. Let 
denote the circumcircle of triangle , and 
the midpoint of 
. The tangents to at 
and , and the line through perpendicular to line 
, determine a triangle whose circumcircle we denote by 
. Define 
and 
similarly., and are concurrent on line .
such that for all positive integers 
and 
,![\[f(m+nf(m))=f(n)^m+2024! \cdot m.\]](http://latex.artofproblemsolving.com/8/b/9/8b90a9326c0fa448a5a574809c184f43513f768f.png)
Jaedon Whyte
, and an odd prime 
not dividing any of them, and an integer 
such that if one considers the infinite sequence 
and looks at the highest power of that divides each of them, these powers are not all zero, and are all at most . Prove that there exists some 
(which may depend on 
) such that whenever divides an element of this sequence, the maximum power of that divides that element is exactly 
.
, let 
be the least prime that does not divide 
, and define 
to be the product of all primes less than . In particular, 
For having 
, define 
. Consider the sequence 
defined by 
and ![\[ x_{n+1} = \frac{x_n p(x_n)}{q(x_n)} \]](http://latex.artofproblemsolving.com/8/2/5/825be3bbef6a29de69aead03392ca7215093f320.png)
for 
. Find all such that 
.
, 
, 
be an infinite sequence of positive integers. Suppose that there is an integer 
such that, for each 
, the number
is an integer. Prove that there is a positive integer such that 
for all 
. be a right triangle with 
. Point lies on the line 
such that is between and . Let 
be the midpoint of 
and let be the seconf intersection point of the circumcircle of 
and the circumcircle of 
. Prove that as varies, the line 
passes through a fixed point.
and 
intersect each other at 
and 
. The common tangent to two circles nearer to touch and at 
and 
respectively. Let 
and 
be the reflection of and respectively with respect to . The circumcircle of the triangle 
intersect circles and respectively at points 
and 
(both distinct from ). Show that the line 
is the second tangent to and .
to 
face-down on a table. On each move, the Detective can point to two cards and inquire if the numbers on them are consecutive. The Magician replies truthfully. After a finite number of moves, the Detective points to two cards. She wins if the numbers on these two cards are consecutive, and loses otherwise.
questions.
= 2∠AMP.
is the midpoint of the arc 
, containing 
, in the circumcircle of 
, and let 
be the -excircle's center. Assume that the external angle bisector of 
intersects at 
. Prove that 
is perpendicular to 
, where 
is the incenter of .
with 
and the circle 
of radius centered at 
Let be the midpoint of 
The line 
intersects a second time at 
Let be a point on such that 
Let be the intersection of 
and Prove that 
cities. For every two cities there exists exactly one direct one-way road between them. We say that a path from 
to 
is a sequence of roads such that one can move from to along this sequence without returning to an already visited city. A collection of paths is called diverse if no road belongs to two or more paths in the collection.
and be two distinct cities in Anisotropy. Let 
denote the maximal number of paths in a diverse collection of paths from to . Similarly, let 
denote the maximal number of paths in a diverse collection of paths from to . Prove that the equality 
holds if and only if the number of roads going out from is the same as the number of roads going out from .
with 
.
, the above expression is equivalent to:
Now by using the substitution 
we can homogenize the above inequality. Thus we have to show that,
which is trivial by Schur's and hence, we are done. 
and so on. Multiply the product of denominators on both sides of the inequality and expand the 
and simplify a bit. The resulting inequality is just a case of Schur's.
We wish to show that 
or 
Expanding gives Schur's inequality, which is true.
hours
, there exist positive real numbers 
such that:![\[
a = \frac{x}{y}, \quad b = \frac{y}{z}, \quad c = \frac{z}{x}.
\]](http://latex.artofproblemsolving.com/4/6/4/464af3e2af4fb20cab9fae3d17ad0ce7dfadd50c.png)
Thus, the given inequality transforms into:![\[
(x + y - z)(-x + y + z)(x - y + z) \leq xyz.
\]](http://latex.artofproblemsolving.com/9/b/9/9b9cb9788281fec4f7c4b6dd1a5c81cf2701ff12.png)
Expanding this, we obtain:![\[
x^3 + y^3 + z^3 + 3xyz \geq x^2(y + z) + y^2(x + z) + z^2(x + y).
\]](http://latex.artofproblemsolving.com/b/7/3/b73dc791b148ac9993bad4f8ea7f4e5250c6ba15.png)
which is trivial by Schur's inequality
, 
. Then the inequality reduces to![\[\frac{(x+y-z)(-x+y+z)(x-y+z)}{xyz} \le 1 \iff \sum_{\text{sym}} x^2y \le x^3+y^3+z^3+3xyz.\]](http://latex.artofproblemsolving.com/0/f/a/0fa9bb71f48dc8bdd60444915d3dcc99ef8f186a.png)
But this is Schurs with 
.
Hence, we have 
By expanding, we have that 
This follows from Schur's.
be positive real numbers so that 
.![\[ \left( a - 1 + \frac 1b \right) \left( b - 1 + \frac 1c \right) \left( c - 1 + \frac 1a \right) \leq 1.
\]](http://latex.artofproblemsolving.com/7/6/e/76e794a62090c8c4b5f3a70ad4d036669418ca69.png)
and 
such that 
,
and 
.![\[ \left( \frac{x-y+z}y \right) \left( \frac{y-z+x}z \right) \left( \frac{z-x+y}x \right) \leq 1. \]](http://latex.artofproblemsolving.com/d/a/6/da6f9bad78d4d92814b774a070ec5a199546e61a.png)
,
and 
.![\[ 8pqr \le (p+q)(q+r)(r+p) \]](http://latex.artofproblemsolving.com/a/0/e/a0e912dc9dde61f92dd1e0d75fcdd321ecf6e623.png)
which is obvious from the AM-GM since at most one of 
can be negaitve..
so we have to prove this
then 
and 
an others then it ia sides of triangle so take
so from it follows. (we will look next step)
. Since 
are positive so
![\[8pqr \le (p+q)(q+r)(r+p)\]](http://latex.artofproblemsolving.com/e/3/9/e39bf0011b2a9ce9a212ac340670d747d32e60fb.png)
which is obvious from the AM-GM since at most one of 
are positive by definition.
and multiply each factor
for example? If not, you can't appy AM-GM. There is a solution in that sense, but not as trivial as the one posted there.
is almost essential to the problem. This basically simplifies out to 
.
,
,
and 
WLOG then we have 
which fails since x,y,z all must be positive. But if 
then the left hand side must be non-positive, and hence the negatives case is considered. We move on by assuming the Left Hand Side is positive.
,
,
.
are side lengths of a triangle. Write 
through ravi substitutions (yay). This becomes 
and therefore we take AM-GM on individual factors in the RHS which finishes the problem. 
for positive reals ![\[
(x - y + z)(y - z + x)(z - x + y) \le xyz.
\]](http://latex.artofproblemsolving.com/7/6/4/76461f23eb9524e2107325ece81d4ba2d8e26750.png)
Now, notice that if 
,
and 
for positive reals 
.![\[
8def \le (d+e)(e+f)(d+f).
\]](http://latex.artofproblemsolving.com/a/4/e/a4e0db3a479faef7aca7659bca20159cce238b7f.png)
Finally, by AM-GM, we have that 
,
and 
.
.
which is Schur's inequality for 
such that
Substituting into the original problem statement, we have to prove that
or equivalently, that
where we derive the second inequality by multiplying the first by 
.
,
,
.
(and all other cyclic variations) and we must prove that
which immediately follows from performing AM-GM three times over the pairs of any two of 
.