AMC 10 · 2022 · #10

Grade 8 geometry-2d
pythagorean-theoremarea-rectanglescoordinate-geometry convert-to-algebraidentify-subproblems ↑ Prerequisites: pythagorean-theorem
📏 Medium solution 💡 2 insights 📊 Diagram
Problem
An index card is a rectangle with diagonal 8 cm. From two opposite corners, 1 cm × 1 cm squares are removed. The two innermost vertices of those notches (one from each square) are 4√(2) cm apart. Find the area of the original rectangle.

Pick an answer.

(A)
14
(B)
$10\sqrt{2}$
(C)
16
(D)
$12\sqrt{2}$
(E)
18

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

How to solve
Strategy Draw a Diagram

Tool #1 (Draw a Diagram) — place the rectangle in coordinates with one cut corner at (0,0) and the other at (l,w). Mark the inner notch-vertices at (1,1) and (l-1, w-1). The picture makes both length facts directly visible: the diagonal of the rectangle is the segment from (0,0) to (l,w), and the segment between the inner notches is from (1,1) to (l-1, w-1) — same direction, just shortened by √(2) at each end. Tool #7 (Identify Subproblems) — turn the two length facts into two equations using the Pythagorean theorem (or distance formula). Tool #13 (Convert to Algebra) — combine the two equations using the identity (l+w)² = l² + w² + 2lw to get lw directly, without ever solving for l and w individually.

1STEP 1

Put the card on a grid; the two cut corners sit at (0,0) and (l, w), so their inner notch-vertices are (1,1) and (l-1, w-1).

P₁ = (1, 1), P₂ = (l-1, w-1)
2STEP 2

The diagonal is the hypotenuse of the right triangle with legs l and w, so Pythagoras with diagonal 8 gives l²+w²=64.

l² + w² = 8² = 64
3STEP 3

Between the inner vertices the changes are l-2 and w-2; length 4√(2) gives (l-2)²+(w-2)²=32 by the distance formula.

(l-2)² + (w-2)² = (4√(2))² = 32
4STEP 4

Expand the second equation and substitute l²+w²=64; the squares cancel and it collapses to l+w=10.

64 - 4(l+w) + 8 = 32 → 4(l+w) = 40 → l + w = 10
5STEP 5

Feed l+w=10 and l²+w²=64 into the identity (l+w)²=l²+w²+2lw to read the product lw directly — the area.

10² = 64 + 2lw → 2lw = 36 → lw = 18 → (E)
Answer
18
Plug back: with l+w=10 and lw=18, solve the quadratic t² - 10t + 18 = 0 to get t = 5 ± √(7), so {l, w} = {5+√(7), 5-√(7)}. Check the diagonal: l² + w² = (5+√(7))² + (5-√(7))² = (25+10√(7)+7) + (25-10√(7)+7) = 64 ✓. Check the inner distance: (l-2)² + (w-2)² = (3+√(7))² + (3-√(7))² = (9+6√(7)+7)+(9-6√(7)+7) = 32 ✓. Area = 18 matches choice (E). The other choices fail: e.g. 14 comes from forgetting the +8 when expanding; 16 comes from a sign error.
💡Key takeaway

This AMC 10 problem only needs Grade 8 Pythagorean and algebra you already know — place the rectangle on a grid, write l²+w²=64 and (l-2)²+(w-2)²=32, get l+w=10, and use (l+w)² = l²+w²+2lw to read off area = lw = 18.