Maths olympiad question · P5–P6★★★★ Expert
Counting shapes: olympiad question 9
An expert question on counting shapes for P5–P6, with a step-by-step solution, diagrams and the answer. Every year of the Hua Xia Cup, P1 to P6, has shape-counting questions. The hard part is counting each shape once and missing none.
The question
A sheet of squared paper has 3 rows of 4 squares, and three squares have a dot, as shown: ① in row 1, square 3, ② in row 1, square 4, ③ in row 2, square 4 (rows from the top, squares from the left).
How many rectangles in the picture contain no dot at all? (Squares count as rectangles.)
Show the hints (3)
- Hint 1: Count all rectangles, then take away those with a dot. A rectangle may hold one, two or three dots — how do you avoid taking any away twice?
- Hint 2: A rectangle holding several cells must cover the whole block they span: its left side left of all of them, its right side right of all, and the same up and down. Multiply the choices for the four sides.
- Hint 3: Holding at least one = the three single counts (46) − the three pair counts (23) + the all-three count (6).
Worked solution
Have a go first, then read the solution step by step.
- 1.All rectangles: two of 5 upright lines, 10 ways; two of 4 across lines, 6: 60.
- 2.Holding a given dot (left × right × top × bottom choices): ① 3 × 2 × 1 × 3 = 18; ② 4 × 1 × 1 × 3 = 12; ③ 4 × 1 × 2 × 2 = 16, 46 in all.
- 3.Holding a given two (covering the block they span): ①② 3 × 1 × 1 × 3 = 9; ①③ 3 × 1 × 1 × 2 = 6; ②③ 4 × 1 × 1 × 2 = 8, 23 in all.
- 4.Holding all three: 3 × 1 × 1 × 2 = 6.
- 5.Holding at least one: 46 − 23 + 6 = 29 (two-dot ones were taken off twice, so add them back once; three-dot ones went off three times and back on three times, so take them off once more).
- 6.Holding none: 60 − 29 = 31.
- ⚠️Common mistake: taking off the single-dot counts only, forgetting rectangles with two dots went twice; or, after adding the pairs back, forgetting the all-three ones need one more step.
Answer: 31 rectangles