Problem 5 · AMC 8 Stretch
Core
Geometry & Measurement
Fractions, Decimals & Percents
visual-representationlogical-reasoning
Take regular hexagons cut from paper. (a) On one hexagon, fold two OPPOSITE corners in to the center. (b) On another, fold every OTHER corner (3 of them) in to the center. (c) On a third, fold ALL six corners in to the center. For each one, what fraction of the hexagon's area is left showing on top? (For part (b), what fraction is left showing?)
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Answer: (a) 2/3 left (rectangle); (b) 1/2 left (triangle); (c) 1/3 left (hexagon)
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Hint 1 of 4
Split the hexagon into 6 equal triangles, all meeting at the center point. This is the key picture!
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Hint 2 of 4
Folding one corner to the center exactly covers one of those 6 triangles. So folding k corners covers k of the 6 triangles.
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Hint 3 of 4
(a) Fold 2 corners: 2 of 6 triangles get covered. (b) Fold 3 corners. (c) Fold 6 corners.
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Approach: Cut the hexagon into 6 equal triangles and count what folds over
- Cut the regular hexagon into 6 equal triangles that all meet at the center \(O\). Each triangle is \(\tfrac16\) of the hexagon, and folding a corner to the center folds exactly one of these triangles flat.
- (a) Two opposite corners: two triangles fold over, so \(\tfrac26 = \tfrac13\) is covered and \(\tfrac23\) is left showing. The outline becomes a rectangle.
- (b) Three alternate corners: three triangles fold over — half — so \(\tfrac12\) is left showing, and the outline is an equilateral triangle.
- (c) All six corners: every corner flap folds in; the region still showing is \(\tfrac13\) of the original, and the outline is a smaller hexagon.
- Just by counting how many of the 6 triangles fold, you get the area instantly: (a) \(\tfrac23\), (b) \(\tfrac12\), (c) \(\tfrac13\).
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