An Island and 2 planks (level 1)

Geometry Level 2

Suppose you have a perfectly square Island in a perfectly square sea, as shown. The distance from one edge to the island to the edge of the sea is $D$ . You have 2 non-bouyant planks of equal length, $L$ , which is short of the distance $D$ . You cannot glue/ nail/ join them, but you could position them on top of each other. What is the shortest length of the 2 planks such that you can get over the sea and onto the island from shore?

\frac{2}{3\sqrt{2}} D

\frac{2\sqrt{2}}{3} D

\frac{2\sqrt{3}}{3\sqrt{2}} D

\frac{2}{\sqrt{3}} D

D

1 solution

James Long
Jun 26, 2019

Let's say you arrange the planks as shown in the configuration of the arrows as illustrated below, in the top corner of the sea.

This works as the ends of the planks are all on top of a rigid structure.

We see that the length from the corner to the island to the corner to the sea is expressed using Pythagoras' theorem as a function of $D$ , and is equal to a length and a half of one plank! Which we can express as:

$D^2 + D^2 = (L + \frac{L}{2})^2$

Which is easily re-arranged to make the subject $L$ :

$L = \frac{2\sqrt{2}}{3} D$

level 2 coming soon!

James Long - 1 year, 11 months ago

Nicely explained. Do you happen to know of a proof that this is the best possible? (Perhaps this is getting ahead of things to level 2...)

Chris Lewis - 1 year, 11 months ago

Hello, I'm afraid I do not have a mathematical proof at the moment, I could however, express the length of plank as a function of the angle of the plank off the islands corner, and show by optimisation that the length reaches a minimum at 45 degrees. Would this be something that you are looking for? Kind regards!

James Long - 1 year, 11 months ago

ah, I've just realised that $L$ will in fact be a function of 2 angles, nevertheless, i can still do an optimisation.

James Long - 1 year, 11 months ago

An Island and 2 planks (level 1)

1 solution

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