AoPS Academy
, we know that lines 
and 
intersect at 
, lines 
and 
intersect at 
, and lines 
and 
intersect at 
. Suppose that the circumcircle of 
intersects line 
at 
and 
, and the circumcircle of 
intersects line at 
and 
, where 
and 
are collinear in that order. Prove that if lines 
and 
intersect at 
, then 
.
and 
are written on a blackboard, where 
and 
are relatively prime positive integers. At any point, Evan may pick two of the numbers 
and 
written on the board and write either their arithmetic mean 
or their harmonic mean 
on the board as well. Find all pairs 
such that Evan can write 1 on the board in finitely many steps.
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.
.
