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TMUA 2016 · Paper 1 · Question 20 of 20

TMUA 2016 Paper 1 Question 20

Coordinate geometry — Shortest path over a solid · unfolding the net. Try it first; the answer and a full worked solution are below.

TMUA 2016 · Paper 1Coordinate geometryShortest path over a solid · unfolding the net5 optionshard

The original question includes a diagram: A square-based pyramid with base PQRS and apex O drawn in perspective, with the hidden edges from P shown dashed; all edges are 20 m.

The diagram shows a square-based pyramid with base PQRS and vertex O. All the edges of the pyramid are of length 20 metres.

Find the shortest distance, in metres, along the outer surface of the pyramid from P to the midpoint of OR.

  1. A10√5 − 2√3
  2. B10√3
  3. C10√5
  4. D10√7
  5. E10√5 + 2√3
Show the answer and worked solution
answer · D
  1. A10√5 − 2√3
  2. B10√3
  3. C10√5
  4. D10√7
  5. E10√5 + 2√3
A shortest surface path becomes a straight line once you unfold the faces it crosses into a plane. P and R are opposite corners of the square, so the route runs across two triangular faces, either OPQ then OQR or OSP then OSR; the two are mirror images, so take the first. Every face is equilateral, so unfolding about OQ places P and R at 20 from O with angle POR = 60∘ + 60∘ = 120∘ in the flattened picture. With M the midpoint of OR, OM = 10, and the cosine rule gives PM2 = 202 + 102 − 2(20)(10)cos 120∘ = 500 + 200 = 700, so PM = 10√7. Any route using the base is longer, since crossing the base alone already costs 20√2.