Problem 27 · 2025 Math Kangaroo
Stretch
Geometry & Measurement
areaarea-decomposition
In the diagram we see a regular hexagon ABCDEF. The point P lies on BC in such a way that the area of the triangle PEF is 64 and the area of the triangle PDE is 42. What is the area of the triangle APF?
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Answer: B — 54
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Hint 1 of 2
For a point on the hexagon's boundary, the triangle to the far parallel side has a fixed area.
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Hint 2 of 2
Triangle PEF (base EF, height = width between opposite sides) equals one-third of the hexagon's area; use the constant-sum property for the other group.
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Approach: opposite-side area is one-third of the hexagon
- PEF uses base EF and the full width to the opposite side, so [PEF] = (1/3)·[hexagon] = 64, giving [hexagon] = 192.
- P lies on BC, so [PBC] = 0 and [PDE] + [PFA] = ½[hexagon] = 96.
- [APF] = 96 − 42 = 54.
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