Problem:
Let R be a square region and n≥4 an integer. A point X in the interior of R is called n-ray partitional if there are n rays emanating from X that divide R into n triangles of equal area. How many points are 100-ray partitional but not 60-ray partitional?
Answer Choices:
A. 1500
B. 1560
C. 2320
D. 2480
E. 2500
Solution:
Assume without loss of generality that R is bounded by the square with vertices A=(0,0),B=(1,0),C=(1,1), and D=(0,1), and let X=(x,y) be n-ray partitional. Because the n rays partition R into triangles, they must include the rays from X to A,B,C, and D. Let the number of rays intersecting the interiors of AB,BC,CD, and DA be n1,n2,n3, and n4, respectively. Because △ABX∪△CDX has the same area as △BCX∪△DAX, it follows that n1+n3=n2+n4=2n−2, so n is even. Furthermore, the n1+1 triangles with one side on AB have equal area, so each has area 21⋅n1+11⋅y. Similarly, the triangles with sides on BC,CD, and DA have areas 21⋅n2+11⋅(1−x), 21⋅n3+11⋅(1−y), and 21⋅n4+11⋅x, respectively. Setting these expressions equal to each other gives
x=n2+n4+2n4+1=n2(n4+1) and y=n1+n3+2n1+1=n2(n1+1)
Thus an n-ray partitional point must have the form X=(n2a,n2b) with 1≤a<2n and 1≤b<2n. Conversely, if X has this form, R is partitioned into n triangles of equal area by the rays from X that partition AB,BC,CD, and DA into b, 2n−a,2n−b, and a congruent segments, respectively.
Assume X is 100-ray partitional. If X is also 60-ray partitional, then X= (50a,50b)=(30c,30d) for some integers 1≤a,b≤49 and 1≤c,d≤29. Thus 3a=5c and 3b=5d; that is, both a and b are multiples of 5 . Conversely, if a and b are multiples of 5 , then
X=(50a,50b)=⎝⎜⎜⎛3053a,3053b⎠⎟⎟⎞
is 60-ray partitional. Because there are exactly 9 multiples of 5 between 1 and 49, the required number of points X is equal to 492−92=40⋅58=2320.
The problems on this page are the property of the MAA's American Mathematics Competitions