Difference between revisions of "2011 AMC 10A Problems/Problem 25"
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Revision as of 09:42, 8 May 2011
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 | ||
