A hard question on solids for P6, with a step-by-step solution, diagrams and the answer. Common in the Hua Xia Cup's P5 and P6 first rounds and finals: painted small cubes, water levels and the surface area of stacks.
The question
Some identical small cubes are stacked into a solid.
Seen from the front, the right and the top, it looks as shown (every square of the top view has at least one cube).
1, 2, 3… are the columns from left to right;
A, B, C are the rows from front to back.
What is the difference between the most and the fewest cubes the solid can have?
Show the hints (3)
💡Hint 1: Write in each square of the top view how many cubes stand there. The front view shows each column's tallest; the right view each row's tallest.
💡Hint 2: One tall square can serve a column and a row at once. How few tall squares can you get away with?
💡Hint 3: Most: each square as tall as the smaller of its column's front height and its row's side height. Fewest: a column and a row of the same height share one tall square; every other square gets just 1.
Worked solution
Have a go first, then read the solution step by step.
1.In each square of the top view, write the most it can hold: no taller than its column seen from the front (orange), nor its row seen from the right (blue) — so the smaller of the two.
2.Most = (4 + 4 + 3) + (4 + 4 + 3) = 22. The front and right views both come out as given.
3.The top view has 3 × 2 = 6 squares; each needs at least 1 cube.
4.Each column's height seen from the front, and each row's seen from the right, must be reached by some square. A column and a row of the same height can share one square (green); every other square holds just 1.
5.Fewest = (4 + 1 + 3) + (1 + 4 + 1) = 14.
6.The difference is 22 − 14 = 8.
⚠️Common mistake: not letting a column and a row of the same height share a square, so the fewest comes out too many.
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