Four cylindrical drums, each of radius 1 m, are placed such that each one touches other two drums. A sewer pipe is to be placed through the gap of these drums. What could be the maximum possible radius for this pipe?

Correct — B, (√2 − 1) m. Four drums of radius 1 m, each touching two others, sit as a two-by-two block: their axes are at the corners of a square whose side is 2 m, because two touching circles of radius 1 have their centres 2 m apart. The gap the pipe must pass through is at the centre of that square. The distance from the centre of a square of side 2 to any corner is half the diagonal, which is (2√2)/2 = √2. Subtract the 1 m radius of a drum and what is left between the drum surface and the middle is √2 − 1, about 0.41 m. That is the largest circle that fits in the gap, so the pipe's maximum radius is √2 − 1 metres.

• (a)(√2 + 1) m — Adds the drum radius instead of subtracting it, giving about 2.41 m — larger than the drums themselves. A pipe of that radius could not sit in a gap between four one-metre drums.
• (c)(π√2 − 1) m — Introduces pi, which appears in circumference and area but not in a distance between centres. The whole calculation is Pythagoras on a square, so no pi can enter.
• (d)(π√2 + 1) m — Carries the same misplaced pi and adds the radius as well, giving over 5 m. A sanity check on size is enough to discard it.

Circle-packing questions reduce to distances between centres. Two circles that touch externally have their centres separated by the sum of their radii, so a block of four equal touching circles gives a square of centres whose side is twice the radius. The largest circle that fits in the middle gap has radius equal to the half-diagonal minus one radius. For unit circles the half-diagonal is √2, so the gap circle has radius √2 − 1.

Two checks would have saved anyone here. First, dimensions: every quantity in the problem is a length, and pi is a pure number that arrives with curvature of a boundary, not with the distance between two points, so an answer of the form π√2 − 1 has no route into a Pythagoras calculation. Second, magnitude: √2 − 1 is about 0.41 m, comfortably smaller than the 1 m drums, whereas π√2 − 1 is about 3.4 m and π√2 + 1 about 5.4 m, both absurd for a gap between one-metre drums.

• Two circles touching externally have their centres apart by the sum of their radii.
• Four equal touching circles of radius r have centres at the corners of a square of side 2r.
• The centre of a square of side s is at s/√2 from each corner, so for s = 2 the distance is √2.
• The largest circle in the central gap has radius √2 − 1, roughly 0.41 m for unit drums.
• Pi enters lengths only through arcs and circumferences, never through a straight distance between centres.

• Adding the drum radius instead of subtracting it.
• Letting pi into a problem that contains no arc length or area.
• Taking the full diagonal rather than half of it as the centre-to-corner distance.

A packing item where two of the four options can be eliminated on inspection because they contain a quantity the geometry cannot generate.

No directly related past PYQ was found.

• practice — not a real PYQ

  Three circles of radius 1 unit touch one another externally. What is the distance between the centres of any two of them?

  • (a)1 unit
  • (b)2 units
  • (c)√3 units
  • (d)2√3 units

  Answer(b) 2 units — the centres are apart by the sum of the radii, and they form an equilateral triangle of side 2.

• practice — not a real PYQ

  Four circles of radius r are placed so that each touches two others, with their centres at the corners of a square. The radius of the largest circle that fits in the central gap is

  • (a)r(√2 − 1)
  • (b)r(√2 + 1)
  • (c)r/√2
  • (d)r√2

  Answer(a) r(√2 − 1) — the half-diagonal is r√2 and one radius r is subtracted.