One punch, four layers: a folding demonstration that produced four holes
Ms. Brennan demonstrates a folding item. She takes a 16 cm by 16 cm square of paper, folds the top edge down onto the bottom edge, creases, then folds the left edge over onto the right edge and creases again. The result is an 8 cm by 8 cm packet with four layers; the corner where the two creases meet is the centre of the original sheet. She punches one hole 2 cm in from each of those two creased edges. She opens the paper fully and shows four holes in a small 2-by-2 cluster around the middle of the sheet. The child must choose which of four drawings shows the unfolded sheet. Drawing 1: one hole. Drawing 2: two holes on a diagonal. Drawing 3: four holes in a 2-by-2 cluster near the middle. Drawing 4: four holes, one near each corner of the sheet. The child chooses Drawing 1 and says, 'you only made one hole.'
1. How many holes are in the unfolded sheet and where are they? Show the reasoning a child would need.
Read the worked answer
Count layers, then reflect.
- Layers: the first fold doubles the paper to two layers and the second doubles it again to four, so one punch passes through four layers and produces four holes. This is the step the child skipped - she counted punches in the world (one) instead of holes in the paper (four).
- Position: each crease is a mirror line. The punch sits 2 cm from the horizontal midline crease and 2 cm from the vertical midline crease, so unfolding reflects it 2 cm to the other side of each line. The four holes form a small 2-by-2 cluster straddling the centre of the original sheet - Drawing 3.
Two checks worth modelling aloud: the holes must be mirror-symmetric about both crease lines, so any option whose holes are not symmetric about both midlines is out; and no hole can lie near an outer edge, because the punch was 2 cm from the creases and therefore 6 cm from the nearest outer edge. Drawing 4 is the answer to a punch in a different location, not to a different number of folds.
2. The child says 'you only made one hole.' What exactly is the misconception, and what is the shortest demonstration that fixes it?
Read the worked answer
She is counting the punch, not the paper. Her model of the event is 'one tool, one hole', which is true of a single sheet and false of a four-layer packet; she has not connected the folding to the number of layers.
Verbal explanation rarely shifts this, so the fix is to unfold one crease at a time with the paper in her hands. Punch, then open only the second fold: two holes appear, visibly mirror images about the crease. Open the last fold: four. At each stage ask her to predict the next count before she opens it, so the doubling becomes something she has stated rather than something she has watched. If she predicts 'two more' on the second opening she is adding rather than doubling, which needs another round.
The 30-second substitute is a stack of four separate sheets: punch through, count four, then point out that the folded packet is the same four sheets.
3. Suppose the punch had gone through the packet's outer corner, or exactly on a crease. How do the answers change, and what does that tell you the item is really measuring?
Read the worked answer
Three cases, one principle: the number of holes equals the number of distinct positions the punch occupies once all the layer-mapping is undone, and coincidences collapse.
- 2 cm from the two creased edges (the scenario): four separate holes in a central cluster.
- 2 cm in from the packet's outer corner: still four holes, because the punch now sits near a corner of the original sheet, so the holes land one near each corner - Drawing 4 would be correct. Same count, different position.
- Exactly on one crease: the hole straddles the fold, so unfolding joins two half-holes into one full hole. Two holes, not four.
- Exactly at the intersection of the two creases: four quarter-holes become a single hole at the centre - one hole.
That last case matters for grading. The child gave the answer 'one hole', which is right only for a punch on the crease intersection. The item is therefore measuring position tracking as much as counting, and a correct count with wrong placement - or the reverse - is a partly solved item, not a wrong one.