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.
, which doesn't have right angle, let 
be its circumcenter. Circle 
intersects 
and 
at 
and 
for the second time, respectively. Similarly, circle 
intersects 
and 
at 
and 
, and circle 
intersects 
and 
at 
and 
for the second time, respectively. Let 
and 
be circumcenters of triangle 
and 
, respectively. Prove that 
are collinear.
of integers is rootiful if, for any positive integer 
and any 
, all integer roots of the polynomial 
are also in . Find all rootiful sets of integers that contain all numbers of the form 
for positive integers 
and 
.
be an isosceles triangle with 
. Let 
be a point on 
. Let 
be a point inside the triangle such that 
and![\[
CL \cdot BD = BL \cdot CD.
\]](http://latex.artofproblemsolving.com/1/6/7/16771838c95f86e92f79b8d049e46ab473e6287d.png)
Prove that the circumcenter of triangle 
lies on line 
.
satisfying 
and
for all 
.
be a cyclic quadrilateral with 
being the midpoints of segments 
respectively. Suppose 
is the intersection of diagonals 
of quadrilateral 
Define 
to be the isogonal conjugate point of point in 
Define 
similarly. Suppose 
intersects 
at a point 
Prove that: The Newton-Gauss line of quadrilateral bisects segment 
be an odd prime number. How many -element subsets 
of 
are there, the sum of whose elements is divisible by ?
in the process of solving it
and making more systems of equations
![$x, y \in \left[0,1 \right]$](http://latex.artofproblemsolving.com/6/3/9/639a18e2742de67742e6a7e61b4ab591918a632d.png)
may satisfy both functions. We, the cameramen, call function 
as function 1 and function 
as function 2.![$\left[ \frac{i}{4} , \frac{i+1}{4} \right]$](http://latex.artofproblemsolving.com/5/7/e/57ef57084281b904335bc42fe520eb1fe03f7bc7.png)
with 
, function 1's value oscillates between 0 and 1. It attains 1 at 
, 
and another point between these two. Between two consecutive points whose functional values arr 1, the skibidi function first decreases from 1 to (in a skibidifying way) 0 and then increases from 0 to (in a skibidifying way) 1. So the skibidi graph of this function in this interval consists of 4 monotonic pieces.![$\left[ \frac{i}{6} , \frac{i+1}{6} \right]$](http://latex.artofproblemsolving.com/4/0/6/4067fa7c170a5a967c36a504b11ddf620c08b819.png)
with 
, function 2's value oscillates between 0 and 1. It attains 1 at 
, 
and another point between these two. Between two consecutive points whose functional values arr 1, the skibidi function first decreases from 1 to (in a skibidifying way) 0 and then increases from 0 to (in a skibidifying way) 1. So the skibidi graph of this function in this interval consists of 4 monotonic curves.![$\left[ \frac{i}{4} , \frac{i+1}{4} \right] \times \left[ \frac{j}{6} , \frac{j+1}{6} \right]$](http://latex.artofproblemsolving.com/3/4/4/3440263b5764320080c28ed340363c20d6cbbf4f.png)
with and 
but 
. Both functions have four monotonic pieces. Because function 1's each monotonic piece can take any value in ![$\left[ \frac{j}{6} , \frac{j+1}{6} \right]$](http://latex.artofproblemsolving.com/0/5/1/051b8cfcfc76d5d2d37bb535a1e44930b234aec5.png)
and function 2' each monotonic piece can take any value in , function 1's each monotonic piece intersects with function 2's each monotonic piece. Therefore, in the skibidi interval , the skibidi number of intersecting points is, as a matter of blasphemy against the unspoken rizz 
. for 
, then this point must be 
., function 1 attains value 1. For function 2, if 
, then 
. Therefore, the skibidi intersecting point is, as a matter of blasphemy against the unspoken rizz . for 
, then this point must be . with and but , all 16 intersecting points arr interior. That is, as a matter of blasphemy against the unspoken rizz, no two regions share any common intersecting point.![$\left[ \frac{3}{4} , 1 \right] \times \left[ \frac{5}{6} , 1 \right]$](http://latex.artofproblemsolving.com/8/7/6/8765c7a4e4f7a65fb7911d7731a83e76711a4fd0.png)
. Consider any pair of monotonic pieces, where one is, as a matter of blasphemy against the unspoken rizz from function 1 and one is, as a matter of blasphemy against the unspoken rizz from function 2, except the skibidi pair of two monotonic pieces from two functions that attain 
. Two pieces in each pair intersects at an interior point on the skibidi region. So the skibidi number of intersecting points is, as a matter of blasphemy against the unspoken rizz 
... Now, we, the cameramen, study whether they intersect at another point.
and 
. Thus, for positive and sufficiently small 
and 
, function 1 is, as a matter of blasphemy against the unspoken rizz reduced to![\[ y' = 4 \sin 2 \pi x' \hspace{1cm} (1) \]](http://latex.artofproblemsolving.com/4/0/f/40febc5aa05eaa7d60eeea4a45ca67373eecf730.png)
and function 2 is, as a matter of blasphemy against the unspoken rizz reduced to![\[ x' = 4 \left( 1 - \cos 3 \pi y' \right). \hspace{1cm} (2) \]](http://latex.artofproblemsolving.com/f/5/a/f5ab7a9618b343fc5fff4a204fefbf4bcec24a95.png)
and , to (in a skibidifying way) get an intuition and quick estimate, we skibidies goon approximations of the skibidi above equations.![\[ y' = 4 \cdot 2 \pi x' \]](http://latex.artofproblemsolving.com/2/d/6/2d66e8833d3e5ab836627870cb5d2d908c44f2fd.png)
and equation (2) is, as a matter of blasphemy against the unspoken rizz approximated as![\[ x' = 2 \left( 3 \pi y' \right)^2 \]](http://latex.artofproblemsolving.com/8/5/7/857453ec8763f39582dfb402207703e6345d24d9.png)
and 
. Therefore, two functions' two monotonic pieces that attain have two intersecting points.
.
Prove that
Prove that
Prove that
