2010 AMC 10A Problem 7

Track how far east and how far north she is after each part. Then the way back is the long side of one right triangle.

1. First mile, due north. She is $0$ miles east and $1$ mile north of the start.
2. What a diagonal mile does. Running one mile northeast moves her the same amount east as north. Those two amounts are the legs of a right triangle with long side $1$ and two equal short sides. If each short side is $a$, then $a^2 + a^2 = 1$, so $a^2 = \tfrac12$ and $a = \tfrac{\sqrt2}{2}$ (about $0.71$).
3. Second mile, northeast. She moves $\tfrac{\sqrt2}{2}$ east and $\tfrac{\sqrt2}{2}$ north.
4. Third mile, southeast. She moves $\tfrac{\sqrt2}{2}$ east and $\tfrac{\sqrt2}{2}$ south. The north and south parts of these two miles cancel.
5. Where she ends up. East: $\tfrac{\sqrt2}{2} + \tfrac{\sqrt2}{2} = \sqrt2$. North: just the $1$ mile from the start.
6. The way back. It is the long side of a right triangle with legs $\sqrt2$ and $1$:
$$\sqrt{(\sqrt2)^2 + 1^2} = \sqrt{2 + 1} = \sqrt3.$$

The answer is $\boxed{\textbf{(C)}\ \sqrt{3}}$.

Splitting every move into an "east part" and a "north part" turns a zig-zag path into one right triangle, and the Pythagorean theorem finishes it. A mile at $45^\circ$ always splits into two equal parts of $\tfrac{\sqrt2}{2}$ each.

Northeast and southeast are at right angles to each other. So the second and third miles are the two short sides of a right triangle with equal legs $1$ and $1$. Its long side, $\sqrt{1 + 1} = \sqrt2$, points due east. So those two miles together just move her $\sqrt2$ east of the point one mile north of the start, and the way back is $\sqrt{1^2 + (\sqrt2)^2} = \sqrt3$.