Difference between revisions of "2011 AMC 12A Problems/Problem 22"
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== Solution == | == Solution == | ||
| − | First, notice that there must be four rays emanating from <math>X</math> that intersect the four corners of the square region. Depending on the location of <math>X</math>, the number of rays distributed among these four triangular sectors will vary. We start by finding the corner-most point that is <math>100</math>-ray partitional. 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 <math>96</math> rays are divided among the other two triangular sectors, each sector with <math>48</math> rays, thus dividing these two sectors into <math>49</math> triangles of equal areas. Let the distance from this corner point be <math>a</math> and the side of the square be <math>s</math>. From this, we get the equation <math>\frac{a\times s}{2}=\frac{(1-a)\times s}{2}\times\frac1{49}</math>. | + | First, notice that there must be four rays emanating from <math>X</math> that intersect the four corners of the square region. Depending on the location of <math>X</math>, the number of rays distributed among these four triangular sectors will vary. We start by finding the corner-most point that is <math>100</math>-ray partitional. 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 <math>96</math> rays are divided among the other two triangular sectors, each sector with <math>48</math> rays, thus dividing these two sectors into <math>49</math> triangles of equal areas. Let the distance from this corner point to the closest side be <math>a</math> and the side of the square be <math>s</math>. From this, we get the equation <math>\frac{a\times s}{2}=\frac{(1-a)\times s}{2}\times\frac1{49}</math>. |
== See also == | == See also == | ||
{{AMC12 box|year=2011|num-b=21|num-a=23|ab=A}} | {{AMC12 box|year=2011|num-b=21|num-a=23|ab=A}} | ||
Revision as of 23:54, 12 February 2011
Problem
Let 
be a square region and 
an integer. A point 
in the interior or is called n-ray partitional if there are 
rays emanating from that divide into triangles of equal area. How many points are 
-ray partitional but not 
-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. 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 
.
See also
| 2011 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 21 |
Followed by Problem 23 |
| 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 12 Problems and Solutions | |
