Difference between revisions of "2011 AMC 10A Problems/Problem 25"
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==Problem 25== | ==Problem 25== | ||
− | Let <math>R</math> be a square region and <math>n\ge4</math> an integer. A point <math>X</math> in the interior of <math>R</math> is called <math>n\text{-}ray</math> partitional if there are <math>n</math> rays emanating from <math>X</math> that divide <math>R</math> into <math>n</math> triangles of equal area. How many points are 100-ray partitional but not 60-ray partitional? | + | <!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>Let <math>R</math> be a square region and <math>n\ge4</math> an integer. A point <math>X</math> in the interior of <math>R</math> is called <math>n\text{-}ray</math> partitional if there are <math>n</math> rays emanating from <math>X</math> that divide <math>R</math> into <math>n</math> triangles of equal area. How many points are 100-ray partitional but not 60-ray partitional?<!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude> |
<math>\text{(A)}\,1500 \qquad\text{(B)}\,1560 \qquad\text{(C)}\,2320 \qquad\text{(D)}\,2480 \qquad\text{(E)}\,2500</math> | <math>\text{(A)}\,1500 \qquad\text{(B)}\,1560 \qquad\text{(C)}\,2320 \qquad\text{(D)}\,2480 \qquad\text{(E)}\,2500</math> |
Revision as of 19:11, 11 November 2015
Contents
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 1
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
.
Solution 2
We may assume that the square has coordinates
. Suppose that
is
-ray partitional.
By definition, there exist
rays from
which divide
into
triangles of equal area, and
of these rays intersect the vertices of
. Let
be the number of triangles that share a side with
, respectively. Then
Let the common area of the triangles be
. Each of the triangles that share a side with
have common height
; since they have the same area, they must have the same base
. Hence
. Similarly,
Substituting gives
and
Since
and
and
are integers,
are rational. Let
be integers such that
and
Thus we have derived a necessary condition for
to be
-ray partitional:
Conversely, if
satisfies the above condition, then it is
-ray partitional since we can define
in terms of
as above.
To count the points that are -ray partitional, it suffices to count the ordered pairs of rationals
in the interior of
such that
divides
and
divides
. This is just
which has
points.
We need to subtract the points that are -ray partitional and
-ray partitional; those are the ordered pairs of rationals
in the interior of
such that
divides
and
divides
. This is just
which has
points.
Hence the answer is .
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 |
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