2026 AMC 8 Problem 13: Solution

Work through 2026 AMC 8 Problem 13 on coordinate geometry with a written solution, hints, and a visual explanation.

GeometryLevel 3 · AMC-style

The figure below shows a tiling of 1 × 1 unit squares. Each row of unit squares is shifted horizontally by half a unit relative to the row above it. A shaded square is drawn on top of the tiling. Each vertex of the shaded square is a vertex of one of the unit squares. In square units, what is the area of the shaded square?

A square drawn over staggered rows of unit squares. From its left vertex to its top vertex, the displacement is three units right and one unit up; to its bottom vertex it is one unit right and three units down.
Diagram 1
  • A. 10
  • B. 21/2
  • C. 32/3
  • D. 11
  • E. 34/3
Hints
  1. Assign coordinates to the grid, keeping in mind that alternate rows are shifted by half a unit.
  2. Find the horizontal and vertical distance between two adjacent vertices of the shaded square.
  3. For a square, area equals side length squared — you can use the squared distance directly without ever taking a square root.
Read the step-by-step solution

Answer: A · 10

  1. Use coordinate geometry

    Place one vertex at A = (0, 0). The adjacent vertex B is at (1, 3) on the shifted grid.

  2. Compute the area

    Side length = √(1² + 3²) = √10. Area = (√10)² = 10.

    √(1 + 9) = √10, area = 10

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