AMC 10 · 2009 · #14

Grade 8 geometry-2d
area-rectanglesratio-proportionsystems-of-equations convert-to-algebraidentify-subproblems ↑ Prerequisites: area-rectangles
📏 Medium solution 💡 2 insights 📊 Diagram
Problem
Four identical rectangles are set in a pinwheel around a small square, and together they fill a larger square. The big square's area is 4 times the small square's area. Find the ratio of a rectangle's longer side to its shorter side.

Pick an answer.

(A)
3
(B)
$\sqrt {10}$
(C)
$2 + \sqrt2$
(D)
$2\sqrt3$
(E)
4

AMC 10 2009 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Introduce a Variable

The figure hides two squares whose sides are built from the rectangle's two side lengths. Name those lengths, read the two square sides off the picture, and the area fact becomes a single equation to solve.

1STEP 1

Read the two squares off the figure

One edge of the big square is a full long side plus a short side; the leftover gap in the middle is a long side minus a short side.

outer side = ℓ + s, inner side = ℓ - s
2STEP 2

Turn the area fact into an equation

Call the shorter side s and the longer side ℓ. Area is side squared, so (ℓ + s)² must be 4 times (ℓ - s)².

(ℓ + s)² = 4(ℓ - s)²
3STEP 3

Take the square root of both sides

Both sides are squares of positive lengths and 4 = 2², so the positive square root clears the exponents.

ℓ + s = 2(ℓ - s)
4STEP 4

Solve for the side ratio

Expanding gives ℓ + s = 2ℓ - 2s, so ℓ = 3s and the longer side is triple the shorter — choice (A).

ℓ + s = 2ℓ - 2s → 3s = ℓ → ℓ/s = 3
Answer
3
Test with s = 1, ℓ = 3. Outer side = 1 + 3 = 4 and inner side = 3 - 1 = 2, giving areas 16 and 4. Indeed 16 = 4 × 4, so the area condition holds and the ratio 3 is confirmed. It also matches the sample rectangle in the figure.
💡Key takeaway

If one square has 4 times the area of another, its side is twice as long, so turn the area clue into a side clue first.

  • Read the two squares off the figure
  • Turn the area fact into an equation
  • Take the square root of both sides
  • Solve for the side ratio