AoPS Academy
, the Bank of Cape Town issues coins of denomination 
. Given a finite collection of such coins (of not necessarily different denominations) with total value at most most 
, prove that it is possible to split this collection into 
or fewer groups, such that each group has total value at most 
.
that satisfy the following equation for all real numbers 
:![\[
f\big(f(x - y)\big) = f(x) - f(y).
\]](http://latex.artofproblemsolving.com/c/3/c/c3c423f0ceb2fcb91064c5de218ec88f69c95e6d.png)
such that 
is unbounded and![\[2f(m)f(n)-f(n-m)-1\]](http://latex.artofproblemsolving.com/0/f/7/0f77b5addbf8f5e603eb1b6923cdac795cd132e9.png)
is a perfect square for all integer 
of four distinct positive integers, we denote the sum 
by 
. Let 
denote the number of pairs 
with 
for which 
divides . Find all sets 
of four distinct positive integers which achieve the largest possible value of .
be a parallelogram and let 
and 
be points on the segments 
and 
, respectively, such that 
. Let the lines 
and 
intersect at 
. Show that 
.
.
. Since this takes 
to 
, it means that it's a perspectivity, so the locus of 
is a line. Now look at some particular situations to find this line (for example, consider the cases when is on 
; notice that we don't need the points 
to lie on the segments as long as we choose directions on in a convenient way).
, where we use directed segments. Since 
and 
, we can rewrite this as 
.
. Since BC || DA, we have 
. Thus, we get 
. In other words,
.
, this becomes
.
.. Since this takes to , it means that it's a perspectivity, so the locus of is a line. Now look at some particular situations to find this line (for example, consider the cases when is on ; notice that we don't need the points to lie on the segments as long as we choose directions on in a convenient way). to , it means that it's a perspectivity, so the locus of is a line".
to the points of the line 
, including infinity points). If the intersection point of the two lines is mapped to itself, then there is a theorem stating that the lines formed by the pairs 
pass through a fixed point.
, and the line connecting these two points is mapped to itself. Then the dual of that theorem states that this is a perspectivity, i.e. the intersection points of the pairs 
(
is a line through 
and 
is its image, passing through 
) lies on a line. This intersection point was called . lies actually is.
. as origin points on the lines 
, and consider the positive directions on these lines to be 
respectively. We want to show that 
is a homography between the two lines, so that it will unduce a homography between the pencils of lines through . In order to do this, all we need to do is show that the map is of the form 
, where 
is the coordinate of on the line 
(remember that we have fixed an origin and a direction, so we can regard points on this line as coordinates), and 
is the coordinate of on . From the condition given we get 
, which is precisely a homography.
.From hypothesis we have 
and 
.
taking as the transversal we have 
.From the similar triangles 
and 
we obtain 
.Again from hypothesis we have 
.Plugging in these relations we obtain 
(the last step is from the fact that 
).These together with 
implies that 
.Thus 
and the result follows.
