Competition · AMC preparation · step 4 of 4

AMC 10 · 2023A · #11

Grade 8 geometry-2d

pythagorean-theoremarea-rectanglessimilar-triangles convert-to-algebraidentify-subproblems ↑ Prerequisites: pythagorean-theoremarea-rectangles

📏 Long solution 💡 3 insights 📊 Diagram

Problem

A square of area 3 contains a tilted square of area 2 that touches the outer square once on each side. That tilting cuts off four congruent right triangles from the corners. Find the ratio of the shorter leg to the longer leg of one of those right triangles.

Pick an answer.

(A)

$\frac{1}{5}$

(B)

$\frac{1}{4}$

(C)

$2-\sqrt{3}$

(D)

$\sqrt{3}-\sqrt{2}$

(E)

$\sqrt{2}-1$

AMC 10 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

Start with Tool #1 (Draw a Diagram): a quick labeled sketch shows that along any side of the outer square the two legs of one corner triangle sit end-to-end, so leg + leg equals the outer side √(3). The same picture identifies the hypotenuse with a side of the inner square, √(2), so the Pythagorean theorem gives the second equation. Tool #13 (Convert to Algebra) then turns those two pictorial facts into a clean two-equation system in x (short leg) and y (long leg), which solves to a numerical ratio. Finally Tool #3 (Eliminate Possibilities) double-checks the closed-form answer numerically against the five choices.

1STEP 1

Find the two side lengths

A square's side is the square root of its area, so the outer side is √(3) and the inner side is √(2).

s_out = √(3), s_in = √(2)

2STEP 2

Label the corner triangle

Label the shorter leg x and the longer leg y; along one outer side x + y = √(3), and the hypotenuse is the inner side √(2).

x + y = √(3), hypotenuse = √(2)

3STEP 3

Apply the Pythagorean theorem

By the Pythagorean theorem on the right triangle (legs x, y; hypotenuse √(2)): x² + y² = 2.

x² + y² = (√(2))² = 2

4STEP 4

Combine the two equations

Square x + y = √(3) and subtract x² + y² = 2 to isolate the product: 2xy = 1, so xy = $\frac{1}{2}$ .

(x+y)² = x² + 2xy + y² = 3 → 2xy = 3 - (x²+y²) = 3 - 2 = 1

5STEP 5

Build a quadratic for x and y

With sum √(3) and product $\frac{1}{2}$ , x and y are roots of 2t² - 2√(3) t + 1 = 0, giving t = $\frac{\sqrt{3} ± 1}{2}$ .

2t² - 2√(3) t + 1 = 0 → t = (2√(3) ± √(12 - 8))/4 = (2√(3) ± 2)/4 = (√(3) ± 1)/2

6STEP 6

Form the ratio of roots

The shorter leg is $\frac{\sqrt{3}-1}{2}$ , the longer $\frac{\sqrt{3}+1}{2}$ , so short : long = $\frac{\sqrt{3}-1}{\sqrt{3}+1}$ .

x/y = ((√(3)-1)/2)/((√(3)+1)/2) = (√(3)-1)/(√(3)+1)

7STEP 7

Rationalize the denominator

Multiply numerator and denominator by √(3)-1 to rationalize: $\frac{(\sqrt{3}-1)²}{2}$ = 2 - √(3).

(√(3)-1)/(√(3)+1) · (√(3)-1)/(√(3)-1) = ((√(3)-1)²)/(3-1) = (3 - 2√(3) + 1)/2 = (4 - 2√(3))/2 = 2 - √(3)

8STEP 8

Match against the choices

Numerically 2 - √(3) ≈ 0.268, which matches choice (C); the others give 0.20, 0.25, 0.318, 0.414.

2 - √(3) ≈ 0.268 → (C)

Answer

2-√(3)

Numeric sanity: x + y =

\frac{\sqrt{3}-1}{2}

\frac{\sqrt{3}+1}{2}

= √(3) matches the outer side, and x² + y² =

\frac{(\sqrt{3}-1)² + (\sqrt{3}+1)²}{4}

= (4-2√(3)) +

\frac{4+2\sqrt{3}}{4}

\frac{8}{4}

= 2 matches the squared inner side. Direction check: shorter / longer must be less than 1, and 2 - √(3) ≈ 0.268 ∈ (0,1) — good.

💡Key takeaway

Once you draw the tilted square and label one corner triangle, two facts pop out: the two legs share one outer side (x + y = √(3)), and the hypotenuse is the inner side (x² + y² = 2). From there, Grade 8 algebra of square roots does the rest.

Find the two side lengths
Label the corner triangle
Apply the Pythagorean theorem
Combine the two equations
Build a quadratic for x and y
Form the ratio of roots
Rationalize the denominator
Match against the choices

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