Competition · AMC preparation · step 4 of 4

AMC 8 · 2015 · #19

Grade 6 geometry-2d
Archetype Coordinate Plane Figure Computation └ Sub-archetype Lattice or Non-Aligned Polygon Area
coordinate-geometryarea-trianglesarea-rectanglesfraction-arithmetic coordinate-geometryarea-differenceidentify-subproblems ↑ Prerequisites: coordinate-geometryarea-triangles
📏 Medium solution 💡 3 insights 📊 Diagram
Problem
A triangle has vertices A=(1,3), B=(5,1), and C=(4,4) on a 6 × 5 grid. What fraction of the grid does the triangle cover?

Pick an answer.

(A)
$\frac{1}{6}$
(B)
$\frac{1}{5}$
(C)
$\frac{1}{4}$
(D)
$\frac{1}{3}$
(E)
$\frac{1}{2}$

AMC 8 2015 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Draw a Diagram

The triangle is tilted, so its base and height are not obvious from the picture — Tool #1 (Draw a Diagram) earns its keep by letting us sketch the smallest axis-aligned rectangle that hugs the triangle. Inside that bounding rectangle, the triangle plus three right triangles in the corners fill the whole rectangle, so Tool #16 (Count the Complement) gives us a clean subtraction move: instead of finding the tilted triangle directly, find the three easy right triangles and subtract. Tool #7 (Identify Subproblems) splits the final question into two clean parts — first compute the triangle's area, then compute the requested fraction by dividing by the grid area.

1STEP 1

Find the whole grid area

The whole 6 × 5 grid is one rectangle, so its area — the fraction's denominator — is 30.

grid area = 6 × 5 = 30
2STEP 2

Box the triangle in a rectangle

Box the triangle in the smallest axis-aligned rectangle: width 4, height 3, so its area is 12.

bounding rectangle area = (5-1) × (4-1) = 4 × 3 = 12
3STEP 3

Name the three corner triangles

The box splits into △ ABC plus three corner right triangles — find those easy ones instead.

T₁: A(1,3), C(4,4), (1,4) T₂: C(4,4), B(5,1), (5,4) T₃: A(1,3), B(5,1), (1,1)
4STEP 4

Find each corner triangle's area

Each corner triangle is ½·leg·leg; their areas 1.5, 1.5, and 4 add to 7.

T₁ = 1/2(3)(1) = 1.5 T₂ = 1/2(1)(3) = 1.5 T₃ = 1/2(4)(2) = 4 → T₁+T₂+T₃ = 7
5STEP 5

Subtract the corners

Subtract the corners from the box: 12 − 7 gives △ ABC an area of 5.

Area(△ ABC) = 12 - 7 = 5
6STEP 6

Form the fraction and reduce

Divide triangle area by grid area: 530\frac{5}{30} reduces to 16\frac{1}{6} → choice (A).

(Area(△ ABC))/(grid area) = 5/30 = 1/6 → (A)
Answer
1/6
The triangle is clearly smaller than half the grid and larger than nothing, so the answer should be a small but nonzero fraction. 16\frac{1}{6} of 30 is 5, which matches our computed triangle area. The bounding rectangle (12) is itself 1230\frac{12}{30} = 25\frac{2}{5} of the grid, and the triangle takes up a bit less than half of that rectangle (512\frac{5}{12}), so 25\frac{2}{5} · 512\frac{5}{12} = 16\frac{1}{6} — consistent.
💡Key takeaway

This AMC 8 problem only needs Grade 6 coordinate geometry — box the tilted triangle, subtract the easy corners, then simplify the fraction!

  • Find the whole grid area
  • Box the triangle in a rectangle
  • Name the three corner triangles
  • Find each corner triangle's area
  • Subtract the corners
  • Form the fraction and reduce

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