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2002 AMC 10A Problem 8

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2002 AMC 10A Problem 8AMC 10 Step by Step
Name:                             Date: September 13, 2026
Quadrilaterals & Polygon Areas · difficulty 2 of 5 · about 2 min · position P1-10

Betsy designed a flag using blue triangles, small white squares, and a red center square, as shown. Let BB be the total area of the blue triangles, WW the total area of the white squares, and PP the area of the red square. Which of the following is correct?

figure

  1. A)

    B=WB = W

  2. B)

    W=PW = P

  3. C)

    B=PB = P

  4. D)

    3B=2P3B = 2P

  5. E)

    2P=W2P = W

Answer

A

Key insight

Cut the flag into a 4-by-4 grid of equal squares; the 12 border squares are each exactly half white and half blue, so B = W.

Solution

Give the flag side length 88. The red square in the middle has side 44, so P=16P = 16, and the flag's total area is 6464.

The white squares are tilted 45∘45^\circ and each has diagonal 22 (a quarter of the flag's side), so each has area 2⋅22=2\frac{2 \cdot 2}{2} = 2. There are 1212 of them (three along each side), giving W=24W = 24.

Everything that is neither red nor white is blue:
B=64−16−24=24.B = 64 - 16 - 24 = 24.

So B=WB = W, and the answer is (A) B=W\boxed{\textbf{(A)}\ B = W}.

Why this works

Rather than measuring irregular blue triangles, measure the regular pieces (the whole flag, the red square, the white squares) and let subtraction find the rest. Equivalently, slice the flag into a 4×44\times4 grid of 2×22\times2 cells: the four central cells are red, and each of the 1212 border cells contains exactly two half-diamonds of white, i.e. half of its area, leaving the other half blue. Looking for a repeating unit cell is the standard move in tiling-style area problems.

The trap

Guessing from the picture that blue covers more than white because the blue pieces are scattered triangles.

Common mistakes

  • Guessing from the picture that blue covers more than white because the blue pieces are scattered triangles.
  • Computing a white square's area as 22=42^2 = 4 from its diagonal instead of d22=2\frac{d^2}{2} = 2, which would wrongly give W=48W = 48 and B=0B = 0.

Techniques

Cut the figure into known shapes (triangles, rectangles, sectors) · Exploit symmetry to reduce work or pair up objects