AMC 8 · 2004 · #14

Grade 6 geometry-2d

Archetype Coordinate Plane Figure Computation └ Sub-archetype Lattice or Non-Aligned Polygon Area

coordinate-geometryarea-trianglesspatial-visualization coordinate-geometryidentify-subproblems ↑ Prerequisites: coordinate-geometryarea-triangles

📏 Medium solution 💡 2 insights 📊 Diagram

Problem

On a 10 × 10 geoboard, a quadrilateral is drawn by connecting the pegs (4,0), (0,5), (3,4), (10,10) in order. Find the area enclosed by this quadrilateral.

Pick an answer.

(A)

(B)

$18\frac12$

(C)

$22\frac12$

(D)

(E)

AMC 8 2004 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

The shape is given by coordinates, so Tool #1 (Draw a Diagram) is the first move: plotting the four pegs reveals the quadrilateral is concave at V₃. A concave polygon is messy to handle as one piece, so Tool #7 (Break Into Subproblems) splits it along diagonal V₁V₃ into two ordinary triangles. Each triangle's vertices are integer coordinates, so Tool #13 (Use Algebra) finishes the job via the coordinate area formula. Adding the two triangle areas gives the total.

1STEP 1

Plot the four pegs and connect them in order — the shape dents inward at V₃, so it's concave, not convex.

V₁=(4,0), V₂=(0,5), V₃=(3,4), V₄=(10,10)

2STEP 2

Cut along diagonal V₁V₃ — it stays inside the concave shape — into triangle A = V₁V₂V₃ and triangle B = V₁V₃V₄; total area = A + B.

Area(V₁V₂V₃V₄) = Area(A) + Area(B)

3STEP 3

Plug triangle A's vertices (4,0), (0,5), (3,4) into the coordinate area formula to get $\frac{11}{2}$ .

Area(A) =

\frac{1}{2}

| 4(5-4) + 0(4-0) + 3(0-5) | =

\frac{1}{2}

|4 + 0 - 15| =

\frac{11}{2}

4STEP 4

Do the same for triangle B's vertices (4,0), (3,4), (10,10) to get 17.

Area(B) =

\frac{1}{2}

| 4(4-10) + 3(10-0) + 10(0-4) | =

\frac{1}{2}

|-24 + 30 - 40| =

\frac{34}{2}

= 17

5STEP 5

Add the two triangle areas: $\frac{11}{2}$ + 17 = $\frac{221}{2}$ , which is answer (C).

Total =

\frac{11}{2}

+ 17 =

\frac{11}{2}

\frac{34}{2}

\frac{45}{2}

\frac{221}{2}

→ (C)

Answer

\frac{221}{2}

Cross-check with Pick's theorem for the whole quadrilateral. The boundary lattice points come from the four edges: V₁V₂ has gcd(4,5)=1, V₂V₃ has gcd(3,1)=1, V₃V₄ has gcd(7,6)=1, and V₄V₁ has gcd(6,10)=2, contributing one interior lattice point at (7,5). So B=4+1=5 boundary points. Counting interior pegs inside the concave region (using the diagram) gives I=21. Pick's theorem: Area = I +

\frac{B}{2}

- 1 = 21 + 2.5 - 1 = 22.5. Matches (C). Also, the answer must be a multiple of

\frac{1}{2}

for any geoboard polygon, ruling out anything that isn't a half-integer.

💡Key takeaway

When a coordinate shape looks weird or dented, draw it first and split it along a diagonal. Two clean triangles are always easier than one concave quadrilateral, and the area formula handles each piece on its own.

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