2010 AMC 10A Problem 7

Put the start at the origin with north as the positive $y$-direction. After the first mile she is at $(0, 1)$.

A one-mile run northeast moves $\dfrac{\sqrt2}{2}$ east and $\dfrac{\sqrt2}{2}$ north (the legs of a $45$-$45$-$90$ triangle with hypotenuse $1$). A one-mile run southeast moves $\dfrac{\sqrt2}{2}$ east and $\dfrac{\sqrt2}{2}$ south. The north and south parts cancel, and the east parts add to $\sqrt2$. So after three miles she is at $(\sqrt2, 1)$.

The straight-line distance back to the origin is
$$\sqrt{(\sqrt2)^2 + 1^2} = \sqrt{2 + 1} = \sqrt3.$$

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

Diagonal moves at $45^\circ$ split into equal north and east parts of length $\frac{\sqrt2}{2}$. Tracking the two components separately turns a zig-zag path into a single right triangle whose hypotenuse is the answer. Alternatively, notice the NE and SE legs are perpendicular, so they form a right isosceles triangle whose hypotenuse ($\sqrt2$, due east) replaces both legs.

• Treating the two diagonal legs as making net eastward progress of 2 miles, giving sqrt(5), or assuming the return leg is just 1 mile.
• Forgetting that the southeast leg cancels the northward part of the northeast leg, leaving her higher than $1$ mile north.