2011 AMC 12A Problem 22

Attempt Problem 22 of the 2011 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2011 AMC 12A solutions, or check the answer key.

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22.

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?

15001500

15601560

23202320

24802480

25002500

Answer: C
Concepts:arealattice pointcomplementary counting
Difficulty rating: 2460
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 1/n.1/n. The triangles whose bases partition the bottom side together have area y/2,y/2, so their number is ny/2.ny/2. Similarly, the numbers along the top, left, and right sides are n(1y)/2,n(1-y)/2, nx/2,nx/2, and n(1x)/2.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 (i/50,j/50)(i/50,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 i/50=c/30i/50=c/30 and j/50=d/30j/50=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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