An Island and 2 Planks (level 2)

Geometry Level 4

Suppose you have a perfectly circular Island in the center of a perfectly circular sea, as shown. The radius of the sea is $R$ , and the radius of the island is one half of that. You have 2 non-bouyant planks of equal length, $L$ , which are short of the distance $\frac{R}{2}$ . You cannot glue/ nail or join them, and there is nothing for you to pin them down with. You could position them on top of each other. What is the shortest length of the 2 planks such that you can cross from the shore, to the island over the sea?

\frac{1}{2}R

\left( \frac{\sqrt{17}}{5} - \frac{2}{5} \right)R

\left( \frac{\sqrt{17}}{5} - \frac{1}{3} \right)R

\left( \frac{\sqrt{19}}{4} - \frac{2}{3} \right)R

\left( \frac{\sqrt{19}}{5} - \frac{2}{5} \right)R

1 solution

James Long
Jul 6, 2019

If you arrange the planks as shown in the sketch above (lengths are not correct), you can use pythagoras' theorem to express the relationship of the length of the plank to the radius of the sea, $id \;est$ ,

$R^2 = \left( \frac{L}{2} \right)^2 + \left( \frac{R}{2} + L \right) ^2$

With some re-arrangement, we can see that L is in a quadratic:

$0 = \frac{5}{4}L^2 + RL - \frac{3}{4}R^2$

Now if we take the positive solution from the quadratic formulae:

$L = \frac{-R + \sqrt{R^2 - 4\times\frac{5}{4}\times\frac{-3}{4}\times R^2}}{2\times\frac{5}{4}}$

This gives us the solution:

$L = \left( \frac{\sqrt{19}}{5} - \frac{2}{5}\right)R$

level 3 coming not so soon. But it's coming!

An Island and 2 Planks (level 2)

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