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

Daniel finds a rectangular index card and measures its diagonal to be $8$ centimeters.
Daniel then cuts out equal squares of side $1$ cm at two opposite corners of the index card and measures the distance between the two closest vertices of these squares to be $4\sqrt{2}$ centimeters, as shown below. What is the area of the original index card?

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

Pick an answer.

(A)

(B)

$10\sqrt{2}$

(C)

(D)

$12\sqrt{2}$

(E)

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Toolkit + CCSS Solution

Understand

Restated: 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 $4\sqrt{2}$ cm apart. Find the area of the original rectangle.

Givens: Original rectangle has length $l$, width $w$, and diagonal $8$ cm; $1 \times 1$ squares removed at two opposite corners; Distance between the two inner notch-vertices is $4\sqrt{2}$ cm; Answer choices: (A) $14$, (B) $10\sqrt{2}$, (C) $16$, (D) $12\sqrt{2}$, (E) $18$

Unknowns: The area $lw$ of the original rectangle

Understand

Plan

Primary tool: #1 Draw a Diagram

Secondary: #7 Identify Subproblems, #13 Convert to Algebra

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 $\sqrt{2}$ 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 $(l+w)^2 = l^2 + w^2 + 2lw$ to get $lw$ directly, without ever solving for $l$ and $w$ individually.

Execute — Answer: E

#1 Draw a Diagram 6.G.A.3 Step 1

Place the rectangle with corners $(0,0), (l, 0), (l, w), (0, w)$.
The cut squares are at the opposite corners $(0,0)$ and $(l, w)$.
Their inner notch-vertices are $(1,1)$ and $(l-1, w-1)$.

$$P_1 = (1, 1),\;P_2 = (l-1, w-1)$$

💡 Sticking the rectangle on a coordinate grid turns each labeled length into a Pythagorean computation.

#7 Identify Subproblems 8.G.B.7 Step 2

The diagonal of the rectangle is the hypotenuse of the right triangle with legs $l$ and $w$.
Apply the Pythagorean theorem with diagonal $= 8$.

$$l^2 + w^2 = 8^2 = 64$$

💡 Length-width-diagonal forms a right triangle — the classic Pythagorean setup.

#7 Identify Subproblems 8.G.B.8 Step 3

The segment between the two inner notch-vertices has horizontal change $(l-1)-1 = l-2$ and vertical change $(w-1)-1 = w-2$.
Its length is $4\sqrt{2}$, so by the distance formula:

$$(l-2)^2 + (w-2)^2 = (4\sqrt{2})^2 = 32$$

💡 Distance between two points in the plane is also a Pythagorean computation on coordinate differences.

#13 Convert to Algebra 8.EE.C.8 Step 4

Expand the second equation: $(l-2)^2 + (w-2)^2 = l^2 - 4l + 4 + w^2 - 4w + 4 = (l^2+w^2) - 4(l+w) + 8 = 32$.
Substitute the first equation $l^2+w^2 = 64$.

$$64 - 4(l+w) + 8 = 32 \;\Rightarrow\; 4(l+w) = 40 \;\Rightarrow\; l + w = 10$$

💡 Expanding and substituting collapses the two equations into a single linear fact about $l+w$.

#13 Convert to Algebra 7.EE.A.1 Step 5

Use the identity $(l+w)^2 = l^2 + 2lw + w^2$ to extract $lw$ directly, without finding $l$ and $w$ separately.
Plug in $l+w = 10$ and $l^2+w^2 = 64$.

$$10^2 = 64 + 2lw \;\Rightarrow\; 2lw = 36 \;\Rightarrow\; lw = 18 \;\Rightarrow\; \textbf{(E)}$$

💡 The square of a sum identity lets us trade $(l+w)^2$ and $l^2+w^2$ directly for the product $lw$ — the very thing we want (the area).

[1] #1 6.G.A.3 Place the rectangle with corners $(0,0), (l, 0), (l, w), (0, w)$. The cut square

[2] #7 8.G.B.7 The diagonal of the rectangle is the hypotenuse of the right triangle with legs

[3] #7 8.G.B.8 The segment between the two inner notch-vertices has horizontal change $(l-1)-1

[4] #13 8.EE.C.8 Expand the second equation: $(l-2)^2 + (w-2)^2 = l^2 - 4l + 4 + w^2 - 4w + 4 = (

[5] #13 7.EE.A.1 Use the identity $(l+w)^2 = l^2 + 2lw + w^2$ to extract $lw$ directly, without f

Review

Reasonableness: Plug back: with $l+w=10$ and $lw=18$, solve the quadratic $t^2 - 10t + 18 = 0$ to get $t = 5 \pm \sqrt{7}$, so $\{l, w\} = \{5+\sqrt{7},\;5-\sqrt{7}\}$. Check the diagonal: $l^2 + w^2 = (5+\sqrt{7})^2 + (5-\sqrt{7})^2 = (25+10\sqrt{7}+7) + (25-10\sqrt{7}+7) = 64$ ✓. Check the inner distance: $(l-2)^2 + (w-2)^2 = (3+\sqrt{7})^2 + (3-\sqrt{7})^2 = (9+6\sqrt{7}+7)+(9-6\sqrt{7}+7) = 32$ ✓. 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.

Alternative: Tool #1 + Tool #8 (Analyze the Units) — observe that the diagonal of the inner segment lies along the same diagonal line of the rectangle. Its length is $l\sqrt{2} \cdot (\cos\theta)$-style ... actually a cleaner shortcut is: the inner segment $P_1 P_2$ is the original diagonal shrunk by $\sqrt{2}$ at each end (because each corner is pushed in by $(1,1)$, a vector of length $\sqrt{2}$ along the diagonal direction only when $l = w$). When $l \ne w$ the components of the corner-shift along and perpendicular to the diagonal differ, so this shortcut needs the algebra to make precise — the substitution-based path above is the cleanest.

CCSS standards used (min grade 8)

6.G.A.3 Draw polygons in the coordinate plane given coordinates for the vertices (Placing the rectangle and notch-vertices on a coordinate grid: $(0,0), (l,0), (l,w), (0,w), (1,1), (l-1, w-1)$.)
8.G.B.7 Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Relating the rectangle's diagonal $8$ to its sides via $l^2 + w^2 = 64$.)
8.G.B.8 Apply the Pythagorean theorem to find distance between two points in a coordinate system (Distance from $(1,1)$ to $(l-1, w-1)$: $(l-2)^2 + (w-2)^2 = 32$.)
8.EE.C.8 Analyze and solve pairs of simultaneous linear equations (Subtracting $l^2+w^2=64$ from the expanded second equation to isolate $l+w = 10$.)
7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions (Using $(l+w)^2 = l^2 + 2lw + w^2$ to extract $lw = 18$ from $l+w$ and $l^2+w^2$.)

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