AMC 10 · 2008 · #10
Grade 6 geometry-2dEach of the sides of a square S1 with area 16 is bisected, and a smaller square S2 is constructed using the bisection points as vertices. The same process is carried out on S2 to construct an even smaller square S3. What is the area of S3?
Pick an answer.
AMC 10 2008 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.
Toolkit + CCSS Solution
Understand
Restated: A square $S_1$ has area $16$. Marking the midpoint of each side and joining those four points makes a smaller square $S_2$. Doing the exact same midpoint construction on $S_2$ makes an even smaller square $S_3$. Find the area of $S_3$.
Givens: $S_1$ is a square with area $16$; $S_2$ has its vertices at the midpoints of the sides of $S_1$; $S_3$ is built from $S_2$ by the identical midpoint construction; Answer choices: (A) $\tfrac{1}{2}$, (B) $1$, (C) $2$, (D) $3$, (E) $4$
Unknowns: The area of the third square $S_3$
Understand
Restated: A square $S_1$ has area $16$. Marking the midpoint of each side and joining those four points makes a smaller square $S_2$. Doing the exact same midpoint construction on $S_2$ makes an even smaller square $S_3$. Find the area of $S_3$.
Givens: $S_1$ is a square with area $16$; $S_2$ has its vertices at the midpoints of the sides of $S_1$; $S_3$ is built from $S_2$ by the identical midpoint construction; Answer choices: (A) $\tfrac{1}{2}$, (B) $1$, (C) $2$, (D) $3$, (E) $4$
Plan
Primary tool: #5 Look for a Pattern
Secondary: #1 Draw a Diagram, #7 Identify Subproblems
The problem repeats one move: join the midpoints, get a smaller square, do it again. Tool #5 (Look for a Pattern) says: figure out what one step does to the area, then that rule applies to every step. To find the rule, Tool #1 (Draw a Diagram) makes the midpoints and the tilted inner square visible, and Tool #7 (Identify Subproblems) splits the big square into the inner square plus four corner triangles so the area is easy to track. Once one step is understood, the pattern hands over $S_3$ with no new work.
Execute — Answer: E
4.MD.A.3 Step 1 Draw the square and mark the midpoints
- Draw $S_1$.
- Its area is $16$, and $4 \times 4 = 16$, so each side has length $4$.
- Bisecting a side splits it into two equal pieces, each of length $2$.
- Mark the midpoint of every side; these four midpoints are the corners of $S_2$.
💡 Area $16$ comes from a side of $4$ because $4\times4=16$, and a bisected side is just two lengths of $2$.
6.G.A.1 Step 2 Cut off the four corner triangles
- The inner square $S_2$ slices a right triangle off each corner of $S_1$.
- At any corner, the two cuts run to the midpoints of the two sides meeting there, so each triangle has two legs of length $2$ with the right angle at the corner.
- Its area is $\tfrac{1}{2}\times 2 \times 2 = 2$.
- There are four such corners, so the triangles remove $4 \times 2 = 8$ in all.
💡 Breaking the big square into an inner square plus four equal corner triangles turns area into simple counting.
6.G.A.1 Step 3 See that one step halves the area
- The area of $S_2$ is what is left after removing the corners: $16 - 8 = 8$.
- That is exactly half of $16$.
- Because the four corner triangles always take away half the square, joining the midpoints of any square leaves a square with half the area.
💡 The removed corners and the inner square end up equal in total, so each is half.
4.OA.C.5 Step 4 Apply the halving pattern again to reach S3
- The step from $S_1$ to $S_2$ used the identical construction that makes $S_3$ from $S_2$, so it halves the area again.
- The areas follow the pattern $16,\ 8,\ 4$: $S_1 = 16$, $S_2 = 8$, and $S_3 = \tfrac{1}{2}\times 8 = 4$.
- So the area of $S_3$ is (E).
💡 Same construction, same effect: halve, then halve again.
4.MD.A.3 Draw $S_1$. Its area is $16$, and $4 \times 4 = 16$, so each side has length $4$ 6.G.A.1 The inner square $S_2$ slices a right triangle off each corner of $S_1$. At any 6.G.A.1 The area of $S_2$ is what is left after removing the corners: $16 - 8 = 8$. That 4.OA.C.5 The step from $S_1$ to $S_2$ used the identical construction that makes $S_3$ fr Review
Reasonableness: Two identical halvings turn $16$ into $16 \times \tfrac{1}{2} \times \tfrac{1}{2} = 4$, matching (E). The answer must be less than $8$ (the area of $S_2$) but still a real square, and $4$ fits; tiny values like $\tfrac{1}{2}$ or $1$ would need far more than two halvings of $16$, so they are too small.
Alternative: Track side lengths with the ratio idea instead. Each inner square has side equal to the diagonal-style segment joining two midpoints, which is $\tfrac{1}{\sqrt{2}}$ of the previous side, so the area ratio is $\left(\tfrac{1}{\sqrt{2}}\right)^2 = \tfrac{1}{2}$ per step. Two steps give $16 \times \tfrac{1}{2} \times \tfrac{1}{2} = 4$, the same choice (E).
CCSS standards used (min grade 6)
4.MD.A.3Apply area and perimeter formulas for rectangles in real-world problems (Reading area $16$ as a side of $4$ (since $4\times4=16$) and splitting each side into two halves of length $2$.)6.G.A.1Find area of triangles, special quadrilaterals, and polygons by composing (Decomposing $S_1$ into the inner square plus four corner right triangles and finding each triangle's area as $\tfrac{1}{2}(2)(2)=2$.)4.OA.C.5Generate a number or shape pattern following a given rule (Recognizing that each identical construction halves the area, giving the pattern $16,\ 8,\ 4$ and the value of $S_3$.)
⭐ Joining the midpoints of a square always leaves exactly half the area, so the areas go $16$, then $8$, then $4$.
⭐ Joining the midpoints of a square always leaves exactly half the area, so the areas go $16$, then $8$, then $4$.
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