Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.1 National Olympiad Prove it Philippines

Problem:

An ant situated at point $A$ decides to walk 1 foot east, then $\frac{1}{2}$ foot northeast, then $\frac{1}{4}$ foot east, then $\frac{1}{8}$ foot northeast, then $\frac{1}{16}$ foot east and so on (that is, the ant travels alternately between east and northeast, and the distance travelled is decreased by half every time the ant changes its direction). The ant eventually reaches a certain point $B$ . Determine the distance between the ant's initial and final positions.

Solution

Solution:

The distance is the hypotenuse of a right triangle. The length of its base is
$\begin{aligned} 1+\frac{1}{2 \sqrt{2}}+\frac{1}{4}+\frac{1}{8 \sqrt{2}}+\frac{1}{16}+\frac{1}{32 \sqrt{2}}+\cdots & =\left(1+\frac{1}{4}+\frac{1}{16}+\cdots\right)+\frac{1}{2 \sqrt{2}}\left(1+\frac{1}{4}+\frac{1}{16}+\cdots\right) \\ & =\frac{1}{1-\frac{1}{4}}\left(1+\frac{1}{2 \sqrt{2}}\right) \\ & =\frac{\sqrt{2}}{3}+\frac{4}{3} \end{aligned}$
Its height is
$\begin{aligned} \frac{1}{2 \sqrt{2}}+\frac{1}{8 \sqrt{2}}+\frac{1}{32 \sqrt{2}}+\cdots & =\frac{1}{2 \sqrt{2}}\left(1+\frac{1}{4}+\frac{1}{16}+\cdots\right) \\ & =\frac{1}{1-\frac{1}{4}}\left(\frac{1}{2 \sqrt{2}}\right) \\ & =\frac{\sqrt{2}}{3} \end{aligned}$
The distance is
$\begin{aligned} \sqrt{\left(\frac{\sqrt{2}}{3}+\frac{4}{3}\right)^2+\left(\frac{\sqrt{2}}{3}\right)^2} & =\sqrt{\frac{8}{9} \sqrt{2}+\frac{20}{9}} \\ & =\frac{2}{3} \sqrt{2 \sqrt{2}+5} \end{aligned}$

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.