AMC 10 · 2022 · #5

Grade 8 geometry-2d

area-rectanglessimilar-trianglespythagorean-theorem convert-to-algebraidentify-subproblems ↑ Prerequisites: pythagorean-theorem

📏 Medium solution 💡 2 insights

Problem

Inside a unit square ABCD, place points P, Q, R, S on the four sides so that APQCRS is an equilateral convex hexagon — every side the same length s. Find s.

Pick an answer.

(A)

$\frac{\sqrt{2}}{3}$

(B)

$\frac{1}{2}$

(C)

$2 - \sqrt{2}$

(D)

$1 - \frac{\sqrt{2}}{4}$

(E)

$\frac{2}{3}$

AMC 10 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

Without a picture, six points and six equal sides are a tangle. Tool #1 (Draw a Diagram) places A, P, Q, C, R, S on the square and immediately reveals the structure: two right triangles cut off at corners B and D, each isosceles with legs 1-s. Tool #7 (Identify Subproblems) names the real question — find s given that one of these triangles has hypotenuse s. Tool #13 (Convert to Algebra) writes that Pythagorean condition as s² = 2(1-s)² and isolates s. A pure guess-and-check (#6) on the choices would also work but the algebra is short and clean here.

1STEP 1

Draw the square; P on AB gives PB = 1 - s, and QC = s makes BQ = 1 - s, so corner B is cut off by a small right triangle.

AP = QC = s → PB = BQ = 1 - s

2STEP 2

Equal legs 1 - s meeting at right-angled B make △ BPQ an isosceles right triangle with hypotenuse PQ = s; corner D mirrors it.

△ BPQ: legs = 1-s, hypotenuse = PQ = s

3STEP 3

Pythagoras on the isosceles right triangle turns geometry into algebra: s² = 2(1 - s)².

PQ² = BP² + BQ² → s² = (1-s)² + (1-s)² = 2(1-s)²

4STEP 4

Both sides positive, so take the root: s = √(2)(1 - s); gather s terms to get s = $\frac{√(2)}{1 + √(2)}$ .

s = √(2) (1-s) = √(2) - √(2) s → s(1 + √(2)) = √(2) → s =

\frac{√(2)}{1 + √(2)}

5STEP 5

Multiply by the conjugate √(2) - 1: the denominator collapses to 1, leaving s = 2 - √(2) → (C).

s =

\frac{√(2)}{1+√(2)}

\frac{√(2)-1}{√(2)-1}

\frac{√(2)(√(2)-1)}{2-1}

= 2 - √(2) → (C)

Answer

2 - √(2)

Numerical sanity. s = 2 - √(2) ≈ 2 - 1.414 = 0.586, which sits between 0 and 1 — required since P lies strictly inside AB. Cross-check the corner triangle: legs 1 - s ≈ 0.414, so hypotenuse √(2)(1-s) ≈ 1.414 · 0.414 ≈ 0.586, matching s. The competing root 2 + √(2) ≈ 3.414 is rejected because it exceeds 1.

💡Key takeaway

This AMC 10 problem only needs Grade 8 Pythagorean theorem you already know — the cut-off corner triangle has equal legs 1-s and hypotenuse s, which gives s = 2 - √(2).