Jules and the Crooked Shadow Screen

At a glance

1. Move the far-right panel in the bottom row into the gap in the top row directly above the second panel. The result is a 2 ×4 rectangle.
2. The finished rectangle has area 8 square panels and perimeter 12 panel-edges. The perimeter decreases by 4 panel-edges.
3. The moved panel had 1 full-side neighbor before the move and 3 full-side neighbors after it. The panel count stays 8, so the area stays 8. It gains 2 shared sides; each new shared side hides 2 boundary edges, so the perimeter drops by 2 ×2 =4.

1. Move one panel

The original screen is

■ □ ■ ■ □
■ ■ ■ ■ ■

Move the lower-right panel to the first gap:

■ ■ ■ ■
■ ■ ■ ■

2. Find the new measures

For the finished 2 ×4 rectangle,

Equivalently, count 4 edges across the top, 4 across the bottom, and 2 on each side:

The original perimeter is given as 16, so

3. Explain the area and perimeter change

Area counts panels. Jules removes no panel and adds no panel; the same 8 panels are merely rearranged. Therefore, the area remains 8 square panels.

Perimeter counts exposed panel-edges. At the original lower-right location, the moved panel shares only its left side, so it contributes 3 exposed edges. Removing it exposes the right edge of its former neighbor, for a net perimeter decrease of 2.

At the new location, the panel touches panels on its left, right, and below. It contributes only 1 exposed edge, while covering 3 edges that had been exposed. Placing it there causes a further net perimeter decrease of 2. Altogether, the perimeter decreases by 4.

The compact shared-side argument reaches the same result: the moved panel goes from 1 neighbor to 3 neighbors, gaining 2 shared sides. Each shared side replaces 2 exposed edges in the total boundary count, so the perimeter changes by −2 ×2 = −4.

Check

Trace the finished rectangle’s outside boundary: 4 top edges + 4 bottom edges + 2 left edges + 2 right edges = 12. The drawing still contains exactly 8 shaded unit squares.

Accepted responses

• For part 1, accept any unambiguous diagram or arrow that produces the 2 ×4 rectangle with exactly one moved panel.
• For part 2, accept perimeter 12 from formula, skip-counting, or direct boundary tracing, with a stated decrease of 4.
• For part 3, the explanation must distinguish preserved panel count from changed exposed boundary. A correct remove-then-place boundary argument is equivalent to the 1-neighbor-to-3-neighbor argument.