Competition · AMC preparation · step 4 of 4

AMC 8 · 2012 · #25

Grade 6 geometry-2d

Archetype Compound Area by Decomposition / Difference └ Sub-archetype Nested Frame / Concentric Difference

pythagorean-theoremarea-trianglesarea-rectanglesspatial-visualization area-differenceidentify-subproblemsconvert-to-algebra ↑ Prerequisites: pythagorean-theoremarea-triangles

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

An outer square has area 5 and an inner (tilted) square has area 4. Each vertex of the inner square sits on a side of the outer square, splitting that side into two pieces of lengths a and b. Find the product ab.

Pick an answer.

(A)

$\frac{1}5$

(B)

$\frac{2}5$

(C)

$\frac{1}2$

(D)

(E)

AMC 8 2012 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

The figure is a classic "square inside a square" picture. Tool #1 (Draw a Diagram) makes the four corner right triangles visible — each one has legs of length a and b. Tool #7 (Identify Subproblems) then turns the question into a single accounting equation: the four corner triangles fill exactly the gap between the two squares, so their total area equals 5 - 4 = 1. This avoids algebra entirely (no Pythagorean setup, no expanding (a+b)²). For a multiple-choice geometry problem, comparing areas is the simplest path.

1STEP 1

Draw the two squares

The outer square has area 5, so each side is √(5), split into pieces a and b — hence a + b = √(5).

a + b = √(5)

2STEP 2

Look at one corner triangle

Each cut-off corner is a right triangle whose two legs a and b lie on adjacent sides of the outer square.

legs = a and b, right angle at the corner

3STEP 3

Subtract the two square areas

The four triangles fill the region between the squares, so their total area is the area difference: 5 - 4 = 1.

Total triangle area = 5 - 4 = 1

4STEP 4

Write the area using a and b

Each right triangle has area $\frac{1}{2}$ ab, and four congruent ones give 2ab in total.

4 × 1/2ab = 2ab

5STEP 5

Set the two areas equal

Setting the two area expressions equal, 2ab = 1, so ab = $\frac{1}{2}$ — choice (C).

2ab = 1 → ab = 1/2 → (C)

Answer

1/2

Check the magnitude with concrete numbers. We need a + b = √(5) ≈ 2.236 and ab =

\frac{1}{2}

. Solving the quadratic t² - √(5) t +

\frac{1}{2}

= 0 gives t =

\frac{\sqrt{5} ± \sqrt{3}}{2}

, so a ≈ 0.252 and b ≈ 1.984. Both are positive, both fit inside a side of length √(5), and the right triangle with these legs has hypotenuse √(a²+b²) = √((a+b)² - 2ab) = √(5 - 1) = 2 — exactly the inner square's side length. Everything checks.

💡Key takeaway

This AMC 8 problem only needs Grade 6 area reasoning — split the big square into the small square plus four corner triangles, and the rest is a one-step equation.

Draw the two squares
Look at one corner triangle
Subtract the two square areas
Write the area using a and b
Set the two areas equal

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Draw the two squares

Look at one corner triangle

Subtract the two square areas

Write the area using a and b

Set the two areas equal

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