7. 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 unitial and final positions.
Answer: $\frac{2}{3} \sqrt{2 \sqrt{2}+5}$ feet
Solution: The distance is the hypothenus 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
$$
\begin{aligned}
\sqrt{\left.\left(\frac{\sqrt{2}}{3}+\frac{4}{3}\right)^{2}+\frac{\sqrt{2}}{3}\right)^{2}} & =\sqrt{\frac{8}{9} \sqrt{2}+\frac{20}{9}} \\
& =\frac{2}{3} \sqrt{2 \sqrt{2}+5}
\end{aligned}
$$