Two Folds, One Punch: Counting Holes in an 8 cm Sheet
A square sheet of paper 8 cm on a side is labeled with corner A at the top-left, B at the top-right, C at the bottom-right, and D at the bottom-left. Use an origin at A, with x increasing to the right along AB and y increasing downward along AD, so the vertices are A(0,0), B(8,0), C(8,8), and D(0,8). The sheet is folded once along the vertical line x = 4, bringing the right half onto the left half, and the crease is kept sharp. It is then folded along the horizontal line y = 4, bringing the bottom half up onto the top half. The resulting packet lies entirely over the region 0 <= x <= 4, 0 <= y <= 4. A single punch is then driven straight through the packet at the point (2, 2), and the sheet is unfolded completely. No tearing, stretching, or additional folding occurs.
1. How many holes does the unfolded sheet show, and at what coordinates?
Read the worked answer
Four holes. Work backward through the folds one at a time.
Undo fold 2 (horizontal crease at y = 4). The punch at (2, 2) was in the top half; the bottom half was folded up on top of it, so the same punch passed through the layer that unfolds to the mirror image across y = 4: y = 8 - 2 = 6. Holes so far: (2, 2) and (2, 6).
Undo fold 1 (vertical crease at x = 4). Each of those holes was duplicated into the right half by reflection across x = 4: x = 8 - 2 = 6. Holes: (6, 2) and (6, 6).
Final set: (2, 2), (6, 2), (2, 6), (6, 6).
Check by layer count: two folds in half give 4 layers everywhere in the packet, and a punch through a 4-layer region produces 4 holes. The four holes are the corners of a 4 x 4 square centered in the sheet - a symmetry check worth doing, since two perpendicular half-folds must produce a pattern with both mirror symmetries.
2. Now suppose the punch is placed at exactly (4, 2), on the vertical crease. How many holes appear, and where? Justify the count from the geometry of the fold rather than from a layer-counting shortcut.
Read the worked answer
Two holes: (4, 2) and (4, 6).
At x = 4 the paper is not doubled by a mirror image - the two halves are joined at the crease and share that line. A punch centered on the crease straddles both halves as one continuous opening, so unfolding the vertical fold does not create a second hole; it merely opens the single hole symmetrically about x = 4.
The horizontal fold still behaves normally: y = 2 reflects to y = 8 - 2 = 6, giving a second hole.
So the layer-count shortcut (4 layers, therefore 4 holes) fails here, and that is the trap. Layer counting gives the number of thicknesses penetrated, not the number of distinct holes: where two layers are joined along an edge, their two 'holes' merge into one. The safe rule is to reflect the punch point across each crease and then delete duplicates - here the x-reflection of (4, 2) is (4, 2) itself.
3. The sheet is folded identically again (right onto left about x = 4, then bottom up onto top about y = 4), but this time the punch is aimed at (5, 1). How many holes result? Explain why the arithmetic must not be applied blindly here.
Read the worked answer
Zero holes.
The answer is not obtained by reflecting (5, 1) and counting; it comes from asking whether any paper exists at that point in the folded state.
After the first fold, the entire right half (x > 4) has been carried over to the left. The packet occupies only 0 <= x <= 4. The point (5, 1) lies in the region that is now empty space. The punch passes through air and damages nothing; unfolding therefore reveals no holes at all.
This is the mirror image of a mistake students make in the opposite direction - assuming every punch must produce at least as many holes as there are layers. Two checks keep the reasoning honest:
- Locate the punch in packet coordinates first (is it inside 0 <= x <= 4, 0 <= y <= 4?), and only then reflect.
- Count layers at that specific point, not layers in the packet on average. A fold does not double thickness uniformly if the flap is smaller than the base, and a point outside the flap's footprint is single-layered even when the packet is nominally 'folded.'
If the punch were instead at (3.5, 1) - inside the packet but still within the flap's footprint - the answer would be four holes: (3.5, 1), (4.5, 1), (3.5, 7), (4.5, 7).