Sunday Times Teaser 3230 – Polytunnel
by Howard Williams
Published Sunday August 18 2024 (link)
I have extended my market garden by the addition of a polytunnel. The polyethylene cover is supported at intervals by identical sets of three vertical pillars supporting metal arcs, of negligible thickness, resting on the level ground on both sides. In each set, the highest pillar is in the middle of the cross-section, at the highest point of the tunnel. The other two pillars are of equal height, being three quarters of the height of the central pillar. These shorter pillars are positioned on each side, 2/11 of the ground-level width of the tunnel from each edge.
Each metal arc is part of a circle (less than a semicircle), and the highest point is exactly 3m above the ground.
What is the ground-level width of the polytunnel in cm?
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Too easy as a programming challenge so here is a manual solution
The diagram shows the arrangement of the polytunnel supports and gives the lengths of the various dimensions in terms of the variables \(h\) (polytunnel height), \(w\) (polytunnel half-width) \(r\) (polytunnel radius). We can express the square of the polytunnel radius in two ways \[r^2=(r-h)^2+w^2\] \[r^2=(r-h/4)^2 +(7w/11)^2\] These can be simplified to show that: \[2rh=h^2+w^2\] \[968rh=121h^2+784w^2\] We can now equate \(rh\) from these two equations and simplify to show that\[w=11h/10\] Hence knowing \(h\) we can now calculate the full width of the polytunnel.