AMC 10 · 2011 · #25

Grade 7 geometry-2dcounting

Archetype Lattice / Grid Pattern from Small Cases └ Sub-archetype Lattice Constraint Analysis

coordinate-geometryprinciple-of-inclusion-exclusion pattern-recognitioncomplementary-counting ↑ Prerequisites: coordinate-geometry

📏 Long solution 💡 4 insights

Problem

Take a unit square and a point X strictly inside it. For a whole number n at least 4, call X n-ray partitional if you can shoot n rays out of X that split the square into n triangles all having the same area. Count the points that are 100-ray partitional but are not 60-ray partitional.

Pick an answer.

(A)

1500

(B)

1560

(C)

2320

(D)

2480

(E)

2500

AMC 10 2011 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Introduce a Variable

Naming the point X = (x, y) turns a vague geometry question into an equation about areas. A diagram of the four rays hitting the corners shows why each side of the square must carry a whole number of equal triangles, which pins x and y to a grid of fractions. Once the grid is known, counting is a systematic list, spotting the shared 60-grid is a pattern, and the final 'not 60-ray' twist is a count-the-complement subtraction.

1STEP 1

Four rays must hit the corners

With X at (x, y), four rays must aim at the corners, cutting the square into side triangles of area y/2, (1-y)/2, x/2, and (1-x)/2.

bottom=y/2, top=(1-y)/2, left=x/2, right=(1-x)/2

2STEP 2

Each side holds whole tiles

Pieces on one side share a height, so equal areas force equal bases and every tile is 1/n — giving y = 2a/n and x = 2c/n for whole a, c.

y/2=a/n → y=2a/n, x=2c/n

3STEP 3

Count the 100-ray points

At n = 100, x = c/50 and y = a/50 with a, c each from 1 to 49 (interior), a grid of 49 by 49 = 2401 points.

x=c/50, y=a/50, 1 ≤ a,c ≤ 49 → 49²=2401

4STEP 4

Find the shared 60-ray points

A 60-ray point needs x = c'/30, so c/50 = c'/30 makes c a multiple of 5; the nine such c per axis share 9 by 9 = 81 points.

c/50=c'/30→ 3c=5c'→ 5 ∣ c, c∈{5,…,45} (9) → 9²=81

5STEP 5

Subtract the overlap

The ask excludes the 60-ray points, so subtract them from the full set: 2401 - 81 = 2320, choice (C).

2401-81=2320

Answer

2320

The result 2320 sits just below the full 2401 count, which is right because only a thin 9 by 9 layer of points also lands on the coarser 60-ray grid. The construction never let a or c reach 0 or 50, so every counted point is genuinely interior, matching the problem's requirement. The value also matches choice (C) and neatly avoids the trap answers 2500 = 50² and 2401 that come from forgetting the interior cap or the overlap.

💡Key takeaway

A point works for n rays exactly when both its coordinates are fractions over n/2, so count the whole 100-grid (49 by 49 = 2401) and take away the spots that also fit the 60-grid (9 by 9 = 81) to get 2320.

Four rays must hit the corners
Each side holds whole tiles
Count the 100-ray points
Find the shared 60-ray points
Subtract the overlap

Four rays must hit the corners

Each side holds whole tiles

Count the 100-ray points

Find the shared 60-ray points

Subtract the overlap

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