AMC 10 · 2003 · #23

Grade 8 geometry-2d

Archetype Compound Area by Decomposition / Difference └ Sub-archetype Polygon Triangulation

isosceles-right-trianglearea-differenceratio-proportion identify-subproblems ↑ Prerequisites: area-triangles

📏 Medium solution 💡 2 insights 📊 Diagram

Problem

A regular octagon $ABCDEFGH$ has an area of one square unit. What is the area of the rectangle $ABEF$ ?

Pick an answer.

(A)

$1-\frac{\sqrt2}{2}$

(B)

$\frac{\sqrt2}{4}$

(C)

$\sqrt2-1$

(D)

$\frac{1}2$

(E)

$\frac{1+\sqrt2}{4}$

AMC 10 2003 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

Try it yourself first — the explanation is most useful after you’ve attempted it.

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Toolkit + CCSS Solution

Understand

Restated: A regular octagon $ABCDEFGH$ has area $1$. The four vertices $A$, $B$, $E$, $F$ form a rectangle. Find the area of that rectangle.

Givens: $ABCDEFGH$ is a regular octagon (all sides equal, all angles equal); The whole octagon has area $1$ square unit; $A$ and $B$ are the endpoints of one side; $E$ and $F$ are the endpoints of the opposite side; $ABEF$ is a rectangle, so $AB$ and $EF$ are opposite sides of it; Answer choices: (A) $1-\frac{\sqrt2}{2}$, (B) $\frac{\sqrt2}{4}$, (C) $\sqrt2-1$, (D) $\frac{1}{2}$, (E) $\frac{1+\sqrt2}{4}$

Unknowns: The area of rectangle $ABEF$, given the octagon's area is $1$

Understand

Restated: A regular octagon $ABCDEFGH$ has area $1$. The four vertices $A$, $B$, $E$, $F$ form a rectangle. Find the area of that rectangle.

Plan

Primary tool: #7 Identify Subproblems

Secondary: #1 Draw a Diagram, #4 Introduce a Variable

Comparing a rectangle to a whole octagon looks hard until you cut both into pieces you already know. Tool #1 (Draw a Diagram) reframes the octagon as a big square with its four corners sliced off — corners that are $45$-$45$-$90$ triangles. Tool #4 (Introduce a Variable) names the octagon's side length $s$ so every length becomes a formula. Then tool #7 (Identify Subproblems) splits the job into three easy area calculations: the octagon (square minus four corners), the rectangle (width times height), and their ratio. Since only the ratio matters, the messy $\sqrt2$ terms cancel and the answer falls out cleanly.

Execute — Answer: D

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

See the octagon as a cut-corner square

Slide the octagon so sides $AB$ and $EF$ are horizontal, at the top and bottom.
Draw the smallest square that fits snugly around it.
The square touches the four horizontal-and-vertical sides of the octagon; the four slanted sides of the octagon are exactly the cut-off corners of that square.
Each cut corner is a right triangle whose two short legs are equal, because a regular octagon's corners are sliced at $45^\circ$.

$$\text{octagon} = \text{(big square)} - \text{(four equal corner triangles)}$$

💡 A regular octagon is just a square with its four corners trimmed off evenly.

#4 Introduce a Variable 8.G.B.7 Step 2

Name the side and find the corner legs

Let the octagon's side length be $s$.
Each cut corner is a $45$-$45$-$90$ right triangle, and the slanted octagon side of length $s$ is its hypotenuse.
In a $45$-$45$-$90$ triangle the hypotenuse is $\sqrt2$ times a leg, so each leg equals $\dfrac{s}{\sqrt2}$.

$$\text{leg} = \dfrac{s}{\sqrt2}$$

💡 The corner's slanted edge is the hypotenuse, so its legs shrink by a factor of $\sqrt2$.

#7 Identify Subproblems 7.G.B.6 Step 3

Compute the octagon's area

The big square's side is one octagon side plus two corner legs: $s + 2\cdot\dfrac{s}{\sqrt2} = s + s\sqrt2 = s(1+\sqrt2)$.
Its area is $s^2(1+\sqrt2)^2 = s^2(3+2\sqrt2)$.
Each of the four corner triangles has area $\dfrac{1}{2}\left(\dfrac{s}{\sqrt2}\right)^2 = \dfrac{s^2}{4}$, so the four together are $s^2$.
Subtract to get the octagon's area.

$$\text{octagon} = s^2(3+2\sqrt2) - s^2 = s^2(2+2\sqrt2) = 2s^2(1+\sqrt2)$$

💡 Whole square minus the four trimmed corners leaves the octagon.

#7 Identify Subproblems 6.G.A.1 Step 4

Compute the rectangle's area

Rectangle $ABEF$ has width equal to the top side $AB$, which is $s$.
Its height is the straight-across distance between the top side $AB$ and the bottom side $EF$ — the full height of the octagon, which is the same as the big square's side, $s(1+\sqrt2)$.
Multiply width by height.

$$\text{rectangle} = s \cdot s(1+\sqrt2) = s^2(1+\sqrt2)$$

💡 The rectangle is one side wide and reaches the full height of the octagon.

#7 Identify Subproblems 6.RP.A.3 Step 5

Compare the two areas

Divide the octagon's area by the rectangle's area.
The $s^2(1+\sqrt2)$ factor is common to both, so it cancels and the ratio is exactly $2$.
The octagon is twice the rectangle, which means the rectangle is half the octagon.
Since the octagon's area is $1$, the rectangle's area is $\dfrac{1}{2}$, choice (D).

$$\dfrac{\text{octagon}}{\text{rectangle}} = \dfrac{2s^2(1+\sqrt2)}{s^2(1+\sqrt2)} = 2 \;\Rightarrow\; \text{rectangle} = \tfrac{1}{2}\cdot 1 = \tfrac{1}{2} \Rightarrow \textbf{(D)}$$

💡 The identical $\sqrt2$ factors cancel, leaving a clean $2$-to-$1$ ratio.

[1] #1 6.G.A.1 Slide the octagon so sides $AB$ and $EF$ are horizontal, at the top and bottom.

[2] #4 8.G.B.7 Let the octagon's side length be $s$. Each cut corner is a $45$-$45$-$90$ right

[3] #7 7.G.B.6 The big square's side is one octagon side plus two corner legs: $s + 2\cdot\dfra

[4] #7 6.G.A.1 Rectangle $ABEF$ has width equal to the top side $AB$, which is $s$. Its height

[5] #7 6.RP.A.3 Divide the octagon's area by the rectangle's area. The $s^2(1+\sqrt2)$ factor is

Review

Reasonableness: Half is a believable answer: the rectangle $ABEF$ is the tall middle strip of the octagon, and the two leftover pieces on its left and right sides are mirror images of each other, so together they should roughly balance the middle strip — a split near half. A quick numerical check confirms it: with the octagon drawn on a unit circle its area is $2\sqrt2 \approx 2.83$ and the rectangle is $\sqrt2 \approx 1.41$, whose ratio is exactly $2$. So the rectangle is half of $1$, namely $\tfrac12$.

Alternative: Draw all eight segments from the center to the vertices, cutting the octagon into $8$ equal triangles, each of area $\tfrac18$. Connecting the center to $A$, $B$, $E$, $F$ splits rectangle $ABEF$ into four triangles: $OAB$ and $OEF$ are two of those eight slices, and the other two, $OFA$ and $OBE$, each turn out to have the same area as one slice as well. That makes the rectangle $4\times\tfrac18 = \tfrac12$ of the octagon — again $\tfrac12$.

CCSS standards used (min grade 8)

6.G.A.1 Find areas of polygons by composing into rectangles or decomposing into triangles (Seeing the octagon as a square with four corner triangles removed, and taking the rectangle as width times height.)
8.G.B.7 Apply the Pythagorean Theorem to find unknown side lengths in right triangles (Getting each $45$-$45$-$90$ corner leg as $\frac{s}{\sqrt2}$ from the octagon side (the hypotenuse).)
7.G.B.6 Solve problems involving area of two-dimensional objects composed of triangles and polygons (Computing the octagon's area as the big square minus the four corner triangles.)
6.RP.A.3 Use ratio and rate reasoning to solve real-world and mathematical problems (Dividing octagon area by rectangle area to find the $2$-to-$1$ ratio and the rectangle's share of the total.)

⭐ Turn the octagon into a square with trimmed corners, write every length in terms of one side $s$, and the $\sqrt2$ pieces cancel to show the rectangle is exactly half.

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

See the octagon as a cut-corner square

Name the side and find the corner legs

Compute the octagon's area

Compute the rectangle's area

Compare the two areas

Review

CCSS standards used (min grade 8)

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