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.
that satisfy the equation 
for all real numbers 
and 
.
be an interior point of triangle 
. Lines 
and 
intersect sides 
and 
at points 
and 
, respectively. Points 
and 
are on the plane such that 
and 
are parallelograms. Suppose lines 
and 
intersect at a point 
inside triangle . Prove that points 
, 
, , and are concyclic if and only if 
.
and 
of circle 
meet at point 
. The line through parallel to meets again at , and 
meets again at . Let 
and let 
be the reflection of over 
. Show that 
passes through the midpoint of .
(in base 
), such that 
is divisible by 
.
be an acute triangle inscribed in a circle 
. Let 
be the orthogonal projection of onto so that 
is an altitude. Let 
and 
be the orthogonal projections of onto and , respectively. Point 
is such that quadrilateral 
is convex and has the same area as triangle . Is it possible that strictly lies in the interior of circle ? Justify your answer.
grid, each cell contains a card placed face down, with a real number written on the front. Let the number on the card in the 
-th row and 
-th column be 
, where 
, 
. For any 
, 
, if 
, then 
.
such that there exists a positive real constant 
satisfying the following: for any positive integers 
and any grid of numbers adhering to the above property, the player can guarantee that by flipping no more than 
cards, they can identify the smallest number in the grid.
of positive integers such that 
for all positive integers 
such that 
. Prove that the sequence is either bounded or linear.
A sequence of positive integers is 
if there exists a constant such that 
for all 
. A sequence is 
if 
for all 
with reminder 
or 
modulo 
prove that the number triples 
is odd.
be non-negative real numbers. Setting![\[x = \frac{3(a^2 + b^2 + c^2 + d^2)}{2(ab + bc + ca + da + db + dc)}.\]](http://latex.artofproblemsolving.com/3/c/d/3cd85503ab5de8b5af352ec1c30fcf2cb79f09bd.png)
Prove that![\[\frac{a}{b + c + d} + \frac{b}{c + d+a} + \frac{c}{d+a + b } + \frac{d}{a + b + c} \geqslant \frac43 + \frac15\left(x - \frac1x\right).\]](http://latex.artofproblemsolving.com/4/f/d/4fd1f976f9d50bb35869533cdce8461810064992.png)
(2) Let 
be non-negative real numbers. Prove that![\[\sum_{i=1}^{n} \frac{a_i}{\displaystyle \sum_{j=1}^{n} a_j - a_i} \geqslant \frac{n}{n - 1} + \frac{1}{2n-3}\left[\frac{\displaystyle (n - 1)\sum_{i=1}^{n} a_i^2}{\displaystyle \left(\sum_{i=1}^{n} a_i \right)^2 - \sum_{i=1}^{n} a_i^2} - \frac{\displaystyle \left(\sum_{i=1}^{n} a_i \right)^2 - \sum_{i=1}^{n} a_i^2}{\displaystyle (n - 1)\sum_{i=1}^{n} a_i^2}\right].\]](http://latex.artofproblemsolving.com/9/7/5/975e5faa294dc6e4debc130b1edf31bb29e27016.png)
with 
. The circumcircle of has radius 
. There is a point in the interior of the line segment such that 
and the length of 
is . The perpendicular bisector of 
intersects at the points and . is the incentre of triangle 
.
be a polynomial with integer coefficients such that 
and 
for some integer 
. Prove that there exists no integer 
such that 
.
. Let 
such that 
for 
. This yields:
an irrational number. Thus, no such 
exists under these conditions.
an irrational number. Thus, no such exists under these conditions.
just because 
In fact, this clearly wrong when 
and 
isn't irrational when 
, yet such 
and exist as alexheinis pointed out. The rest of your solution is neat, its just this one step that makes an assumption. In fact, I believe the statement is false for all 
except constant and quadratic polynomials. Here is a proof: has degree 
, then take
where 
has degree 
with integer coefficients. Then 
and 
(This type of construction is common when you have one polynomial behaving like a smaller degree polynomial, such as 
, for a set of .) For the quadratic case, take 
. Then and 
give, respectively,
and
The first equation tells us that 
from which we get
and
We can do similar analysis for the second equation:
and
and produce a nonzero, integral value for 
(nonzero since we are looking for quadratics), and hence no such quadratics exist.
is an integer for integers and 
so 
or 
so 
so 
works.
where 
Prove that
Let 
Prove that
Let 
Prove that
and 
.