AoPS Academy
consecutive even numbers, such that the square of their product divides the sum of their squares.
be a positive integer. Find all real solutions 
to the system:![\[a_1^2 + a_1 - 1 = a_2\]](http://latex.artofproblemsolving.com/a/4/e/a4e26b83c929b82800576bac11e12de0c3aa1cf6.png)
![\[ a_2^2 + a_2 - 1 = a_3\]](http://latex.artofproblemsolving.com/9/9/5/995e8eec93828b82901b0639463f907f5ff95d08.png)
![\[\hspace*{3.3em} \vdots \]](http://latex.artofproblemsolving.com/e/e/1/ee15e1fefe2ad550d3c9b1c51207eaf66fbfa41f.png)
![\[a_{n}^2 + a_n - 1 = a_1\]](http://latex.artofproblemsolving.com/7/c/7/7c727d1f2044c1bf7c80981e1074bb3cde807b8f.png)
the points 
and 
are the feet of the altitudes through 
and 
respectively. The incenters of the triangles 
and 
are 
and 
respectively; the circumcenters of the triangles 
and 
are 
and 
respectively. Prove that 
and 
are parallel.
with 
prime and ![\[a^p+b^p=p!.\]](http://latex.artofproblemsolving.com/f/e/b/feb27894b1f6497559eeb094cfe92256bc26688e.png)
where 
and 
.
![[asy]size(10cm);
import markers;
pair X1=(0,2),X2=(0,-2),Y1=(sqrt(7),1),Y2=(-sqrt(7),1),Z1=(sqrt(7),-1),Z2=(-sqrt(7),-1);
fill(X1--Y1--Z1--cycle^^X1--Y1--Z2--cycle,cyan);draw(X1--Y1--Z1--cycle^^X1--Y1--Z2--cycle,blue);
draw(anglemark(Y1,Z2,X1,10)^^anglemark(Z1,X1,Y1,10),blue);
draw(anglemark(X1,Y1,Z2,10,12)^^anglemark(Y1,Z1,X1,10,12),blue);
D(MP("X_1\;(0,2)",X1,N)--MP("Y_1\;(\sqrt{7},1)",Y1,NE)--MP("Z_1\;(\sqrt{7},-1)",Z1,SE)--MP("X_2\;(0,-2)",X2,S)--MP("Z_2\;(-\sqrt{7},-1)",Z2,SW)--MP("Y_2\;(-\sqrt{7},1)",Y2,NW)--X1);
label("$\triangle X_1Y_1Z_1\sim\triangle Z_2X_1Y_1$",(0,-3));
dot(X1^^X2^^Y1^^Y2^^Z1^^Z2);
[/asy]](http://latex.artofproblemsolving.com/7/a/0/7a0905fae32e34c8667c745d10a5f5e9d6735020.png)
where and .
(though slightly annoying, I'll follow the notation of the diagram from the official solution). Now let 
be the point such that 
is a parallelogram. From this we already know that 
.
such that 
( on the same side of 
with respect to 
). In other words, we consider the spiral similarity with center 
that maps to , and define as the image of 
via this transformation. Now, by definition we get that 
is similar to our original triangle . We shall now prove that 
is similar to the original triangle as well. More precisely, we'll show that 
.
, where in the last step we have used that 
, due to the similarity. So it is enough to prove that 
.
be the lengths of the sides opposite to 
in our original triangle. Now 
, hence the ratio of the similarity is 
. Therefore, 
, and we have 
, as wanted.
are similar, and this holds without any extra requirements on the initial triangle .
is constructed such that 
is a parallelogram. Notice this implies that 
is a parallelogram as well, because it has a pair of opposite sides that are parallel and of equal length.
(respective sides are equal), and also 
(two halves of a parallelogram). So we have 6 similar triangles now.
and 
are congruent (two halves of a parallelogram). So if happens to be similar to the initial triangle , then all 8 triangles 
are similar and we are done.
if and only if 
in the starting triangle. Provided that this is the case, we further need 
, that is 
. Now, 
, and because of the similarity we have that 
, hence 
. So we need 
, and if we set 
we have 
, which is the same to saying that 
and 
for some 
. 
AND 
??
in the plane such that all of the triangles 
?
