AMC 10 · 2011 · #25
Grade 7 geometry-2dLet R be a unit 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?
Pick an answer.
AMC 10 2011 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: 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.
Givens: R is a unit square (side length 1).; X is a point in the interior of R.; n rays from X must cut R into n triangles of equal area.; We compare the case n = 100 with the case n = 60.
Unknowns: The number of interior points that are 100-ray partitional but not 60-ray partitional.
Understand
Restated: 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.
Givens: R is a unit square (side length 1).; X is a point in the interior of R.; n rays from X must cut R into n triangles of equal area.; We compare the case n = 100 with the case n = 60.
Plan
Primary tool: #4 Introduce a Variable
Secondary: #1 Draw a Diagram, #2 Make a Systematic List, #5 Look for a Pattern, #16 Change Focus / Count the Complement
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.
Execute — Answer: C
6.G.A.1 Step 1 Four rays must hit the corners
- Place the square with corners (0,0), (1,0), (1,1), (0,1) and let X sit at (x, y) inside it.
- In any tiling by triangles with vertex X, four of the rays have to go straight to the four corners, because one triangle cannot wrap around a corner.
- These four rays cut the square into four triangles, one resting on each side.
- The triangle on the bottom side has base 1 and height y, so its area is y/2.
- The others are (1 - y)/2 on top, x/2 on the left, and (1 - x)/2 on the right.
💡 A corner is a sharp turn, so a ray must aim straight at it; the four corner rays frame the whole picture.
6.EE.B.6 Step 2 Each side holds whole tiles
- Inside one of the four triangles the remaining rays fan out onto that side, splitting it into smaller triangles that all share the same height (the distance from X to that side).
- Equal area then forces equal bases, so every small triangle has area 1/n.
- That means each of the four side-areas must be a whole-number count of these 1/n tiles.
- Writing the bottom count as a, we need y/2 = a/n, so y = 2a/n; the identical reasoning on the left gives x = 2c/n for a whole number c.
💡 Equal pieces along a side mean that side's area is a whole number of the 1/n tiles.
7.EE.B.4 Step 3 Count the 100-ray points
- Set n = 100.
- Then y = 2a/100 = a/50 and x = c/50.
- Because X is strictly inside, the top count 50 - a and the right count 50 - c must each be at least 1, so a and c each range from 1 to 49.
- Every pair (a, c) gives one distinct interior point, and different pairs give different points.
- That is a 49 by 49 grid of points, so there are 49^2 = 2401 points that are 100-ray partitional.
💡 The interior rule caps each coordinate strictly between the sides, leaving 49 choices per axis.
6.NS.B.4 Step 4 Find the shared 60-ray points
- A point is 60-ray partitional by the very same rule with n = 60: its coordinates must be x = c'/30 and y = a'/30 for whole numbers a', c' from 1 to 29.
- So a 100-ray point x = c/50 is also a 60-ray point only when c/50 equals some c'/30.
- Cross-multiplying gives 30c = 50c', that is 3c = 5c', which forces c to be a multiple of 5.
- The multiples of 5 from 1 to 49 are 5, 10, ..., 45 -- nine of them -- and each gives a valid c' = 3c/5 in range.
- The same holds for y.
- So the shared points form a 9 by 9 grid: 9^2 = 81 points are both 100- and 60-ray partitional.
💡 The two grids line up only where the 1/50 spacing and the 1/30 spacing agree, which is every 1/10 step.
7.EE.B.3 Step 5 Subtract the overlap
- The question asks for points that are 100-ray partitional but NOT 60-ray partitional, so remove the shared points from the full set.
- That leaves 2401 - 81 = 2320 points.
- So the answer is (C).
💡 Count everything that qualifies, then take away the part you must exclude.
6.G.A.1 Place the square with corners (0,0), (1,0), (1,1), (0,1) and let X sit at (x, y) 6.EE.B.6 Inside one of the four triangles the remaining rays fan out onto that side, spli 7.EE.B.4 Set n = 100. Then y = 2a/100 = a/50 and x = c/50. Because X is strictly inside, 6.NS.B.4 A point is 60-ray partitional by the very same rule with n = 60: its coordinates 7.EE.B.3 The question asks for points that are 100-ray partitional but NOT 60-ray partiti Review
Reasonableness: 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 = 50^2 and 2401 that come from forgetting the interior cap or the overlap.
Alternative: Overlay both grids directly. The 100-ray points lie on lines spaced 1/50 apart; the 60-ray points lie on lines spaced 1/30 apart. Two such lines coincide only on the common 1/10 grid (since 1/10 is the coarsest spacing that is a whole multiple of both 1/50 and 1/30). The interior 1/10 grid is 9 by 9 = 81 points, the same overlap, so again 2401 - 81 = 2320.
CCSS standards used (min grade 7)
6.G.A.1Find area of triangles, special quadrilaterals, and polygons by composing (Computing the four corner-triangle areas y/2, (1-y)/2, x/2, (1-x)/2 from base and height.)6.EE.B.6Use variables to represent numbers and write expressions to solve problems (Naming X = (x, y) and the sector counts a, c to turn equal-area into the equations y = 2a/n and x = 2c/n.)7.EE.B.4Use variables to represent quantities and construct simple equations and inequalities (Applying the interior constraint 1 <= a, c <= 49 to count the 49 by 49 grid of 100-ray points.)6.NS.B.4Find greatest common factor and least common multiple of two numbers (Using divisibility (3c = 5c' forces 5 | c) to find the 9 shared coordinates on each axis.)7.EE.B.3Solve multi-step real-world problems posed with positive and negative rational numbers (Subtracting the 81 shared points from the 2401 total to reach the final count 2320.)
⭐ 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.
⭐ 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.
More like this
Same archetype — closest grade level first.