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2008 AMC 10A Problem 10

2008 AMC 10A Problem 10AMC 10 Step by Step

Name: Date: September 13, 2026

Quadrilaterals & Polygon Areas · difficulty 2 of 5 · about 2 min · position P1-10

Each of the sides of a square $S_1$ with area $16$ is bisected, and a smaller square $S_2$ is constructed using the bisection points as vertices. The same process is carried out on $S_2$ to construct an even smaller square $S_3$ . What is the area of $S_3$ ?

A)
$\frac{1}{2}$
B)
$1$
C)
$2$
D)
$3$
E)
$4$

Answer

Key insight

Joining the midpoints of a square cuts off four corner triangles that together make half the area, so each step halves the area: 16, 8, 4.

Solution

$S_1$ has area $16$ , so its side is $4$ . Connecting the four midpoints cuts off four right isosceles triangles at the corners, each with legs $2$ and area $\frac12 \cdot 2 \cdot 2 = 2$ . The four corners remove $8$ , so $S_2$ has area $16 - 8 = 8$ .

The same construction always removes exactly half the area: the four corner triangles have legs $\frac{s}{2}$ and total area $4 \cdot \frac12\left(\frac{s}{2}\right)^2 = \frac{s^2}{2}$ . Hence
$[S_3] = \frac{1}{2}[S_2] = \frac{1}{2}\cdot 8 = 4.$

The answer is $\boxed{\textbf{(E)}\ 4}$ .

Why this works

The midpoint square is a tilted square whose side is $\frac{s}{\sqrt2}$ , the hypotenuse of a corner triangle, so its area is $\frac{s^2}{2}$ . Recognizing "midpoints of a square give half the area" as a reusable fact lets you iterate without recomputing: after $k$ steps the area is $16/2^k$ .

The trap

Halving the side length instead of the area, so the area drops by a factor of 4 each step and the answer comes out 1.

Common mistakes

Halving the side length instead of the area, so the area drops by a factor of 4 each step and the answer comes out 1.
Stopping after one step and answering $8$ , or applying the halving three times and answering $2$ .

Techniques

Cut the figure into known shapes (triangles, rectangles, sectors)