AoPS Academy
![\[
\begin{cases}
b^2 + 1 = ac, \\
c^2 + 1 = bd,
\end{cases}
\qquad (1)
\]](http://latex.artofproblemsolving.com/f/0/b/f0b02d1f56bfc12a2f72618b6a12822b1ad5fe0b.png)
where 
are positive integers.
is a solution of (1), then![\[
a = 3b - c, \quad d = 3c - b.
\]](http://latex.artofproblemsolving.com/9/b/d/9bd0e3d39592062f1cc5620b5d13729843d0c805.png)
, an 
-omino is a figure without holes formed by unit squares such that each unit square shares an edge with at least one other square. We say two -ominoes are identical if they have the same shape when rotated or if they are the reflection of one of the other. such that we can partitionate an 
board with pairwise distinct -ominoes.
students stand at distinct vertices of a regular -gon. Let 
and 
be a pair of diametrically opposite points on the regular -gon (i.e., and are separated by 
vertices in either direction). Initially, the handkerchief is at point . Every second, the handkerchief is passed with equal probability to one of the two adjacent positions. The expected number of seconds it takes for the handkerchief to reach for the first time is ______.
. Then the minimum possible surface area of the cylinder is ______.
which intersect at points 
. Let 
be an arbitrary point on 
. Suppose that the lines 
meet 
again at points 
respectively. Prove that the circumcircles of all triangles 
have the same radius.
which intersect at points . Let be an arbitrary point on . Suppose that the lines meet again at points respectively. Prove that the circumcircles of all triangles have the same radius.
is of a fixed length and thus by the Law of Sines, we are done.
subtends is actually the sum of the measures of the arc 
in both circles, from angle chasing, which is obviously constant.
is an isosceles trapezium and hence 
that is length of is constant.
is fixed, 
is also fixed.
.
is fixed the circumradius of 
is also fixed. 
is constant regardless of where is on .
and 
be four points. We are only required to prove 
, which is obvious because of cyclic quadrilaterals. is constant.
where 
is the circumradius of 
. We are required to prove that is constant. We already know that 
is constant.
so 
is constant. This implies is constant, so is constant, implying the result.
and 
on .
is a constant chord, we must have that 
.
and 
respectively for the points and respectively,
and 
.
and 
have the same circumcricle, (and the same circumradius). This implies that 
and 
have the same circumradius too.
