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.
cards labeled 
lined up in a random order on his desk. He calls a pair 
of these cards swapped if 
and the card labeled 
is to the left of the card labeled 
. For instance, in the sequence of cards 
, there are three swapped pairs of cards, 
, 
, and 
.
card to its left, then it now has cards to its right. He then picks up the card labeled 
and reinserts it in the same manner, and so on until he has picked up and put back each of the cards 
exactly once in that order. (For example, the process starting at would be 
.)
+1 w
be a sequence of real numbers, and let 
be a fixed positive integer less than . We say an index 
with 
is good if there exists some 
with 
such that 
, where the indices are taken modulo . Let 
be the set of all good indices. Prove that 
.
such that for any positive real numbers 
and 
,
is tangent at 
to the circle with center 
; point 
is interior to the circle; and 
intersects the circle at 
. If 
, 
, and 
, then the radius of the circle is
, with 
.Initially, he folds the sheet along a straight line 
, where 
is a point on the side 
, so that vertex is located on side 
, as shown in the figure. Then folds the sheet again along a straight line 
, where 
is a point on side , so that vertex 
lies on the line ; and finally folds the sheet along the line 
. Matías observed that the vertices and were located on the same point of segment after making the folds. Calculate the measure of the angle 
.
be relatively prime positive integers. Calculate 
are different at 
. If 
are also different at , we call the ordered -tuple 
lucky. For which positive integers , one can find a lucky -tuple?
, 
be the orthocenter of it and 
be any point on the side 
. The points 
are on the segments 
, respectively, such that the points 
and 
are cyclic. The segments 
and 
intersect at 
is a point on 
such that 
is tangent to the circumcircle of triangle 
at 
and 
intersect at 
. Prove that the points 
and lie on the same line.
. Prove that
denote a triangular grid of side length consisting of 
pegs. Charles the Otter wishes to place some rubber bands along the pegs of such that every edge of the grid is covered by exactly one rubber band (and no rubber band traverses an edge twice). He considers two placements to be different if the sets of edges covered by the rubber bands are different or if any rubber band traverses its edges in a different order. The ordering of which bands are over and under does not matter.
different ways to cover 
using exactly two rubber bands; the full list is shown below, with one rubber band in orange and the other in blue.
denote the total number of ways to cover 
with any number of rubber bands. Compute the remainder when is divided by 
.
be a triangle, and let 
be its altitudes. Let 
be its orthocenter, and let 
and 
be the circumcenters of triangles 
and 
. Let 
be the second intersection of the circumcircles of triangles 
and 
. Prove that the lines 
, , and -median of 
are concurrent.
to is 
thus, 
equivalent to 
,
,
,
.
.
?
![[asy]
size(10cm);
draw((5,7)--(11,7)--(3,11)--cycle);
label("$A(5,7)$", (5,7),S);
label("$B(11,7)$", (11,7),S);
label("$C(3,y)$", (3,11),W);
[/asy]](http://latex.artofproblemsolving.com/d/5/e/d5ed8caf884979a843424f5f8958763603b9776c.png)
,
.
![[asy]
size(10cm);
draw((5,7)--(11,7)--(3,11)--cycle);
draw((3,11)--(3,7)--(5,7),red);
draw((3,7.5)--(3.5,7.5)--(3.5,7));
label("$A(5,7)$", (5,7),S);
label("$B(11,7)$", (11,7),S);
label("$C(3,y)$", (3,11),W);
label("$D(3,7)$", (3,7),SW);
[/asy]](http://latex.artofproblemsolving.com/6/c/3/6c3d5e8cefbc0744689ea5ca762f90f5209701e9.png)
as the point at which 
.![$[\triangle ABC] = [\triangle BCD] - [\triangle ACD]$](http://latex.artofproblemsolving.com/f/3/0/f308cc469812756561fc3e1e8503d82e6b1c9e6a.png)
,
are 
and 
.
are 
Now,![\begin{align*}
[\triangle ABC] = 12 &= \frac{1}{2}\cdot (y-7)\cdot 8 - \frac{1}{2}\cdot (y-7)\cdot 2 \\
&= 3(y-7).
\end{align*}](http://latex.artofproblemsolving.com/4/a/3/4a37bae26ebbd9a4b33446698be5d3b5e83e9ab4.png)
Dividing by 
gives 
so 
because 
so the diff is in the y intercept so it can be 4 units more or less, clearly by the graph and the equation 
it needs to be 
