The following fact will come in handy later in the solution.
-
Below is a picture of the – loop (in blue) and the
– loop (in red) on
.
There are 17 intersection points. Remember that the two loops
intersect at , which is the
same point as the other three vertices of .
There are 17 small squares. Of these , 9 are whole squares in the middle of
, 4 are split into two
pieces over the horizontal edges of , and 4 are split over the
vertical edges of .
Here is a picture of a – loop (in blue) and a – loop (in red) on .
There are 13 intersection points and 13 small squares.
Here is a picture of a – loop (in blue) and a – loop (in red) on .
There are 25 intersection points and 25 small squares.
Here is a picture of a – loop (in blue) and a – loop (in red) on .
There are 2 intersection points and 2 small squares. Here the squares
are harder to see, since both of them pass through the edges of , passing over to the other
side.
It is not a coincidence that the number of small squares is equal to
the number of intersection points in all four cases.
Consider the line of slope through the origin. By
Fact 1, is the lattice point
in the first quadrant that is closest to the origin. Therefore, there
are no lattice points on
other than its endpoints. To illustrate the idea in this solution, the
diagram below shows along
with every unit lattice square through which it passes.
The – loop
continues to wrap around
until it reaches a corner again. Because of how behaves, we can think of the
– loop as the result
of overlaying all of the unit lattice squares in the diagram above on
top of each other. Indeed, if the blue line reaches the top of the
square, it jumps to the bottom with the same -coordinate and continues with the same
slope. This means that after it jumps, what the line does in is the same as what it does as
it continues into the adjacent unit lattice square.
This is true in general. So the number of line segments in the – loop is equal to the number
of unit lattice squares that intersects. By "intersects" we
mean that the line passes through some part of the interior of
the square and not just a vertex.
Therefore, to count the line segments in the – loop, we can count the unit
lattice squares through which passes. Line segment begins in and passes into a new unit lattice
square exactly when it crosses either a vertical line with equation
for some integer or a horizontal line with equation
for some integer . Since passes through no lattice points
other than its endpoints, it will never cross such a vertical and a
horizontal line at the same time. Since starts at and ends at , it must cross such vertical lines and such horizontal lines. Adding one to
account for the unit square from which it originates, this means there
are line
segments in the – loop.
For the – loop, this
gives line segments. By
counting, you can verify that there are indeed unit lattice squares in the diagram
above. If you carefully draw the – loop, you will see that
there are segments.
Since and are positive, is positive. Suppose is a – lattice line that
intersects . Then we must have
, otherwise would be entirely below the graph of
(remember, its slope is
positive). Additionally, we must have that , otherwise would be completely above the graph of
.
The inequality is
equivalent to ,
or . Combining
with , we get , which we will
write as .
So far, we have not used the fact that contains at least one lattice point.
Suppose passes through some
lattice point . That is, and are integers and . Then which can be rearranged
to get . This
implies that there is some integer for which . Substituting into , we get
which is equivalent to . There are exactly integers that satisfy these inequalities, so
there are – lattice lines that
intersect through .
Before moving on, we make the observation that two of the – lattice lines that
intersect only pass through the
vertices and . This means there are actually
– lattice lines that pass
through the interior of .
Since and , the loops are neither
horizontal nor vertical. Therefore each of the small squares has a
unique left-most vertex, which is the vertex with the smallest -coordinate. Each of the small squares
has exactly one left-most vertex, and every point of intersection is the
left-most vertex of exactly one square. This means that the number of
intersection points is equal to the number of squares.
Note that this lends some credibility to counting the four vertices
as one intersection point. See, for example, the and loops from the problem
statement. In that example, the square labelled by is split into two pieces, and its
leftmost vertex is represented by all four vertices of . It might be useful to think about this
for a moment before moving on.
Since and are positive, is negative. Suppose is a – lattice line that
intersects . Since the slope
of is negative, we must have
. Otherwise, would be entirely above the graph
of .
For similar reasoning, we also require that , which means . This can be
rearranged to . Combining with gives . By
essentially the same argument that was used in the solution to part (c),
must take the form for some integer . Therefore, we have and there are exactly integers with this property. These lines are all
different and all intersect , so the answer is .
Before moving on, we note that the lines of slope that pass through and are – lattice lines that
intersect . Therefore, there
are points of
intersection of – lattice lines with , excluding the endpoints. We will
refer to this set of
points several times in the solution to part (f), so we will denote it
by for convenience.
Let be the set of
intersection points of the – loop with the – loop, other than the point
represented by the four corners of . We will show that every point
in corresponds to exactly one
point in , thereby showing that
and have the same number of elements. Since
there are points in , if we include the point in represented by the corners,
this will show that the loops have exactly intersection points in . By part (d), this will show
that the loops divide
into small squares.
The rest of the solution is devoted to showing that and have the same number of elements. We
will use the following fact several times. Its proof is not
included.
Fact 2: If an – lattice line is translated units to the left and units to the right for some integers
and , then the line obtained is also an
– lattice line.
We first observe that every line segment in the – loop is part of some – lattice line. To see why
this is true, consider the solution to part (c). It was discussed there
that the line segments of the – loop can be obtained by
overlaying on the unit lattice
squares through which
passes. For each such unit lattice square, there are integers and so that the square can be translated
units to the left and units down in order to land on . Therefore, if we translate units to the left and units down, it will land exactly on the
segment of the – loop
in question. is part of a
– lattice line, so by
Fact 2 above, we have shown that every line segment in the – loop is part of some – lattice line.
By part (b), there are
segments in the – loop.
By the remark at the end of the solution to part (c), there are – lattice lines that pass
through the interior of . Each of
the segments in the – loop is in the interior of
the square. We conclude that the segments of the – loop are exactly the parts
of – lattice lines that
pass through the interior of .
By very similar reasoning, we can also conclude that the – loop is made precisely of
the segments of – lattice lines that pass
through the interior of .
Consider a point in and suppose it is inside the unit
lattice square with vertices ,
, , and . Let be the – lattice line that
intersects at . The point is in . If we translate and to the left by units and down by units, the resulting lines will
intersect at . By Fact 2,
translating in this way
results in a – lattice
line and translating
results in a – lattice
line. By the previous argument, these will be segments of the and loops, respectively.
Therefore, is a point of
intersection of the two loops. This gives a natural way to translate
points of to points in . In symbols, we send to .
Suppose that and are two different points in . If these points are in different unit
lattice squares, then when they are translated to , they will end up on different line
segments of the – loop,
so they cannot possibly be translated to the same point in . If they are in the same unit lattice
square, then they will be translated to the left and down by exactly the
same amount, so they will end up in different places in . Either way, we can conclude that
different points in are sent to
different points in by the rule
described in the previous paragraph.
Suppose is a point in
. This point is on some line
segment of the – loop,
and we have argued earlier that these segments come from translating
unit lattice squares through which passes (and taking the parts of
along for the ride).
Suppose the segment on which
lies comes from the unit lattice square with vertices , , , and and call this square . The point is in and lies on .
We showed earlier that every line segment on the – loop is part of some – lattice line. Let be the – lattice line on which
lies. If we translate to the right by units and up by units, the result will be a new – lattice line by Fact 2.
Moreover, this line will pass through , which also lies on . Therefore, is a point in that is inside the square . If we translate to the left by and down by , it will be sent to by construction.
We have shown that each of the points in corresponds to exactly one point in
. As explained earlier, if we
consider the four corners of as one intersection point,
then we get a total of
intersection points of the two loops.
As mentioned in the problem, we are taking for granted that the
squares all have the same size. This means that each of them has an area
of exactly .
Referring back to the Euclid problem from the beginning, we were dealing
with the situation where and
, so the area of the squares is
.