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

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 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

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

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

T₁ =

\frac{1}{2}

(3)(1) = 1.5 T₂ =

\frac{1}{2}

(1)(3) = 1.5 T₃ =

\frac{1}{2}

(4)(2) = 4 → T₁+T₂+T₃ = 7

5STEP 5

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

Area(△ ABC) = 12 - 7 = 5

6STEP 6

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

\frac{Area(△ ABC)}{grid area}

\frac{5}{30}

\frac{1}{6}

→ (A)

Answer

\frac{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.

\frac{1}{6}

of 30 is 5, which matches our computed triangle area. The bounding rectangle (12) is itself

\frac{12}{30}

\frac{2}{5}

of the grid, and the triangle takes up a bit less than half of that rectangle (

\frac{5}{12}

), so

\frac{2}{5}

\frac{5}{12}

\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!

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