AMC 8 · 2015 · #25

Grade 6 geometry-2d

Archetype Compound Area by Decomposition / Difference

area-rectanglesarea-trianglesspatial-visualizationreflection-symmetry identify-subproblemsarea-difference ↑ Prerequisites: area-rectanglesarea-triangles

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

One-inch squares are cut from the corners of this $5$ inch square. What is the area in square inches of the largest square that can be fitted into the remaining space?

$\textbf{(A) }9\qquad\textbf{(B) }12+4\sqrt{2}\qquad\textbf{(C) }15\qquad\textbf{(D) }9+4\sqrt{5}\qquad\textbf{(E) }21$

Pick an answer.

(A)

(B)

$12+4\sqrt{2}$

(C)

(D)

$9+4\sqrt{5}$

(E)

View mode:

Toolkit + CCSS Solution

Understand

Restated: A $5 \times 5$ square has a $1 \times 1$ square cut out of each of its four corners, leaving a plus-shaped region. We want the area of the largest square that fits inside that plus shape.

Givens: Outer square is $5 \times 5$ (side $5$ inches); Four $1 \times 1$ corner squares are removed; The inscribed square may be tilted; its vertices may touch the inner edges of the cut-out region; Answer choices: (A) $9$, (B) $12 + 4\sqrt{2}$, (C) $15$, (D) $9 + 4\sqrt{5}$, (E) $21$

Unknowns: The area, in square inches, of the largest square that fits in the remaining plus-shaped region

Understand

Plan

Primary tool: #1 Draw a Diagram

Secondary: #7 Identify Subproblems

Tool #1 (Draw a Diagram) is the move that turns this from a scary AMC 8 #25 into a one-line area calculation: when you sketch the plus shape and the biggest tilted square inside it, you can see that the inscribed square's four corners must sit at the four inner corners of the cut-out notches — the only way to push outward as far as possible without hitting a removed unit square. Once the picture is right, Tool #7 (Identify Subproblems) splits the inscribed square into an upright inner $3 \times 3$ square (the part you can color in by eye) plus $4$ congruent right triangles that fill the sides. Adding two simple areas gives the answer; no algebra and no Pythagoras needed.

Execute — Answer: C

#1 Draw a Diagram 3.MD.C.7 Step 1

Sketch the plus shape.
Put the $5 \times 5$ square on grid paper and erase the four unit squares at the corners.
The leftover shape has a small upright square in the middle whose corners are at the four points $(1,1), (4,1), (4,4), (1,4)$ — these are exactly the inner corners of the four notches.

$$\text{Inner upright square has side } 5 - 1 - 1 = 3.$$

💡 Drawing the shape on grid paper makes the side length $3$ jump out: $5$ across, minus $1$ unit notch on each end.

#1 Draw a Diagram 5.G.A.2 Step 2

Now draw the biggest tilted square inside the plus.
Its four vertices must sit at the four points on the outer edges that are farthest from the center while still inside the plus — those are the inner corners of the notches, the midpoints of the long edges of the plus.
Joining $(1,5), (5,4), (4,0), (0,1)$ gives the inscribed square.

$$\text{Vertices of the inscribed square: } (1,5),\; (5,4),\; (4,0),\; (0,1).$$

💡 Plotting the vertices on the grid is exactly the Grade 5 coordinate-plane move.

#7 Identify Subproblems 6.G.A.1 Step 3

Identify the subproblems.
Inside the tilted square sits the upright $3 \times 3$ inner square from Step 1.
The rest of the tilted square is $4$ congruent right triangles, one along each side, each with base $3$ (a side of the inner square) and height $1$ (the depth of a notch).

$$\text{Tilted square area} = (\text{inner } 3\times 3 \text{ square}) + 4 \times (\text{notch triangle}).$$

💡 Decomposing a tilted polygon into an upright rectangle plus right triangles is the Grade 6 area-by-decomposition strategy.

#7 Identify Subproblems 3.MD.C.7 Step 4

Compute the inner square's area.

$$3 \times 3 = 9.$$

💡 Side-times-side for a square is the Grade 3 area formula.

#7 Identify Subproblems 6.G.A.1 Step 5

Compute the four triangle areas.
Each right triangle has legs $3$ and $1$, so its area is $\tfrac{1}{2} \cdot 3 \cdot 1 = \tfrac{3}{2}$.
There are $4$ of them.

$$4 \times \tfrac{1}{2} \cdot 3 \cdot 1 = 4 \times \tfrac{3}{2} = 6.$$

💡 Area of a right triangle with legs $b$ and $h$ is $\tfrac{1}{2}bh$ — Grade 6.

#7 Identify Subproblems 3.MD.C.7 Step 6

Add the pieces.

$$\text{Largest inscribed square area} = 9 + 6 = 15 \;\Rightarrow\; \textbf{(C)}.$$

💡 Total area = sum of non-overlapping pieces — Grade 3 area additivity.

[1] #1 3.MD.C.7 Sketch the plus shape. Put the $5 \times 5$ square on grid paper and erase the f

[2] #1 5.G.A.2 Now draw the biggest tilted square inside the plus. Its four vertices must sit a

[3] #7 6.G.A.1 Identify the subproblems. Inside the tilted square sits the upright $3 \times 3$

[4] #7 3.MD.C.7 Compute the inner square's area.

[5] #7 6.G.A.1 Compute the four triangle areas. Each right triangle has legs $3$ and $1$, so it

[6] #7 3.MD.C.7 Add the pieces.

Review

Reasonableness: The original $5 \times 5$ square has area $25$. The plus-shaped region has area $25 - 4 = 21$, which equals answer (E). The largest square inside the plus must have area less than $21$ but clearly more than $9$ (the upright inner square is already $9$ and we still have room to tilt), so the answer should sit strictly between $9$ and $21$. The four candidate values in that range are $12 + 4\sqrt{2} \approx 17.66$, $15$, and $9 + 4\sqrt{5} \approx 17.94$. Our decomposition gives exactly $15$, the smallest of the three, which matches the fact that any tilted square still has to dodge the four unit-square corners.

Alternative: Tool #16 (Count the Complement): start from the $5 \times 5$ outer square (area $25$) and subtract everything that is NOT inside the inscribed square. Outside the inscribed square but inside the outer square are the $4$ corner unit squares (total area $4$) PLUS $4$ leftover triangles that sit in the outer square but outside the tilted square. Each leftover triangle has legs $1$ and $3$, so area $\tfrac{3}{2}$, total $6$. The inscribed square's area is $25 - 4 - 6 = 15$, confirming (C).

CCSS standards used (min grade 6)

3.MD.C.7 Relate area to multiplication and addition; find areas by tiling and by adding non-overlapping parts (Computing the inner upright square's area as $3 \times 3 = 9$ and combining the non-overlapping pieces ($9 + 6$) into the final $15$.)
5.G.A.2 Represent real-world and mathematical problems by graphing points in the first quadrant of the coordinate plane (Placing the plus shape and the inscribed square on a coordinate grid (vertices at $(1,5), (5,4), (4,0), (0,1)$) so the side length of the inner square and the triangle legs can be read off directly.)
6.G.A.1 Find the area of right triangles, other triangles, and polygons by composing into rectangles or decomposing into triangles (Decomposing the tilted inscribed square into an upright $3 \times 3$ square plus $4$ right triangles with legs $3$ and $1$, and using $\tfrac{1}{2}bh$ for each triangle.)

⭐ Draw the picture first — the tilted square breaks into one $3 \times 3$ square plus four little right triangles, and Grade 6 area-by-pieces does the rest.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 6)

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