2011 AMC 10A Problems/Problem 25
Problem 25
Let be a square region and
an integer. A point
in the interior of
is called
partitional if there are
rays emanating from
that divide
into
triangles of equal area. How many points are 100-ray partitional but not 60-ray partitional?
Solution
First, notice that there must be four rays emanating from that intersect the four corners of the square region. Depending on the location of
, the number of rays distributed among these four triangular sectors will vary. We start by finding the corner-most point that is
-ray partitional (let this point be the bottom-left-most point). We first draw the four rays that intersect the vertices. At this point, the triangular sectors with bases as the sides of the square that the point is closest to both do not have rays dividing their areas. Therefore, their heights are equivalent since their areas are equal. The remaining
rays are divided among the other two triangular sectors, each sector with
rays, thus dividing these two sectors into
triangles of equal areas. Let the distance from this corner point to the closest side be
and the side of the square be
. From this, we get the equation
. Solve for
to get
. Therefore, point
is
of the side length away from the two sides it is closest to. By moving
to the right, we also move one ray from the right sector to the left sector, which determines another
-ray partitional point. We can continue moving
right and up to derive the set of points that are
-ray partitional. In the end, we get a square grid of points each
apart from one another. Since this grid ranges from a distance of
from one side to
from the same side, we have a
grid, a total of
-ray partitional points. To find the overlap from the
-ray partitional, we must find the distance from the corner-most
-ray partitional point to the sides closest to it. Since the
-ray partitional points form a
grid, each point
apart from each other, we can deduce that the
-ray partitional points form a
grid, each point
apart from each other. To find the overlap points, we must find the common divisors of
and
which are
and
. Therefore, the overlapping points will form grids with points
,
,
, and
away from each other respectively. Since the grid with points
away from each other includes the other points, we can disregard the other grids. The total overlapping set of points is a
grid, which has
points. Subtract
from
to get
.
See Also
2011 AMC 10a (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last Question | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |