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2011 AMC 12A Problem 22

Problem 22 of 25HarderGeometryCounting & Probability

Let RR be a square region and n4n \ge 4 an integer. A point XX in the interior of RR is called nn-ray partitional if there are nn rays emanating from XX that divide RR into nn triangles of equal area. How many points are 100100-ray partitional but not 6060-ray partitional?

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Solution

Scale the square to [0,1]2[0,1]^2 and write X=(x,y).X=(x,y). The rays must include those through the four vertices. Every small triangle has area 1n.\frac{1}{n}. The triangles whose bases partition the bottom side together have area y2,\frac{y}{2}, so their number is ny2.\frac{ny}{2}. Similarly, the numbers along the top, left, and right sides are n(1y)2,\frac{n(1-y)}{2}, nx2,\frac{nx}{2}, and n(1x)2.\frac{n(1-x)}{2}. These four numbers must be positive integers. Hence nn is even and X=(2in,2jn),1i,jn21. \begin{gathered} X=\left(\dfrac{2i}{n},\dfrac{2j}{n}\right), \\ 1\le i,j\le\dfrac n2-1. \end{gathered} Conversely, partitioning each side into the indicated number of equal segments and joining the division points to XX produces nn equal-area triangles. Thus these are exactly the partitional points. For n=100,n=100, the points are (i50,j50)(\frac{i}{50},\frac{j}{50}) with 1i,j49,1\le i,j\le49, giving 492=2401.49^2=2401. Such a point is also 6060-ray partitional exactly when i50=c30\frac{i}{50}=\frac{c}{30} and j50=d30\frac{j}{50}=\frac{d}{30} for integers c,d.c,d. Thus ii and jj must both be multiples of 5.5. There are 99 choices for each, so the overlap has 92=819^2=81 points. So the count is 240181=2320.2401 - 81 = 2320. Thus, the correct answer is C.

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Concepts: area · lattice point · complementary counting

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