A $4 \times 4$ square piece of paper is cut into two identical pieces along its diagonal. The resulting triangular pieces of paper are each cut into two identical pieces.

![](https://cdn.mathpix.com/cropped/2024_04_20_ac36362783317e0251fdg-076.jpg?height=212&width=916&top_left_y=1195&top_left_x=564)

Each of the four resulting pieces is cut into two identical pieces. Each of the eight new resulting pieces is finally cut into two identical pieces. The length of the longest edge of one of these final sixteen pieces of paper is
(A) 1
(B) 2
(C) $\frac{1}{2}$
(D) $\frac{1}{\sqrt{2}}$
(E) $2 \sqrt{2}$
We label the stages in this process as Stage 0 (a square), Stage 1 (2 triangles), Stage 2 (4 triangles), Stage 3 (8 triangles), and Stage 4 (16 triangles).

We want to determine the length of the longest edge of one of the 16 triangles in Stage 4. At Stage 1, we have two right-angled isosceles triangles with legs of length 4.

Consider a general right-angled isosceles triangle $A B C$ with legs $A B$ and $B C$ of length $a$.

Since this is a $45^{\circ}-45^{\circ}-90^{\circ}$ triangle, its hypotenuse $A C$ has length $\sqrt{2}$ a.

We split the triangle into two equal pieces by bisecting the right-angle at $B$ :

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-151.jpg?height=288&width=499&top_left_y=1767&top_left_x=867)

Since $\triangle A B C$ is isosceles, then this bisecting line is both an altitude and a median. In other words, it is perpendicular to $A C$ at $M$ and $M$ is the midpoint of $A C$.

Therefore, the two triangular pieces $\triangle A M B$ and $\triangle C M B$ are identical $45^{\circ}-45^{\circ}-90^{\circ}$ triangles. The longest edges of these triangles $(A B$ and $C B$ ) are the legs of the original triangle, and so have length $a$.

Since the longest edge of the original triangle was $\sqrt{2} a$, then the longest edge has been reduced by a factor of $\sqrt{2}$.

Since we have shown that this is the case for an arbitrary isosceles right-angled triangle, we can then apply this property to our problem.

In Stage 1, the longest edge has length $4 \sqrt{2}$.

Since the longest edge in Stage 1 has length $4 \sqrt{2}$, then the longest edge in Stage 2 has length $\frac{4 \sqrt{2}}{\sqrt{2}}=4$.

Since the longest edge in Stage 2 has length 4, then the longest edge in Stage 3 has length $\frac{4}{\sqrt{2}}=\frac{2 \sqrt{2} \sqrt{2}}{\sqrt{2}}=2 \sqrt{2}$.

Since the longest edge in Stage 3 has length $2 \sqrt{2}$, then the longest edge in Stage 4 has length $\frac{2 \sqrt{2}}{\sqrt{2}}=2$.

ANSWER: (B)