2000 AMC 10 Problem 16

1. Set up coordinates. Put the bottom-left dot at $(0, 0)$, with each step between dots equal to $1$. Reading the picture: $A = (0, 3)$, $B = (6, 0)$, $C = (4, 2)$, $D = (2, 0)$.
2. Line $AB$. From $A$ to $B$ you go $6$ right and $3$ down, so for every $1$ step right the line drops $\tfrac12$. Starting at height $3$: $y = 3 - \tfrac12 x$.
3. Line $CD$. From $D$ to $C$ you go $2$ right and $2$ up, so the line rises $1$ for every $1$ right. It passes through $(2, 0)$: $y = x - 2$.
4. Where they cross. At $E$ both equations give the same height:
$$x - 2 = 3 - \tfrac12 x.$$
Add $\tfrac12 x$ and $2$ to both sides: $\tfrac32 x = 5$, so $x = \tfrac{10}{3}$.
5. How far along $AB$ is $E$? Going from $A$ to $B$, the $x$-coordinate goes from $0$ to $6$ at a steady rate. $E$ has $x = \tfrac{10}{3}$, and $\tfrac{10}{3} \div 6 = \tfrac{5}{9}$. So $E$ is $\tfrac59$ of the way from $A$ to $B$.
6. Length of $AB$. By the Pythagorean theorem, $AB = \sqrt{6^2 + 3^2} = \sqrt{45} = 3\sqrt5$.
7. Length of $AE$.
$$AE = \tfrac59 \cdot 3\sqrt5 = \tfrac{5\sqrt5}{3}.$$

The answer is $\boxed{\textbf{(B)}\ \frac{5\sqrt{5}}{3}}$.

A grid of dots is really a coordinate plane, so writing the two lines as equations and solving them together is the most reliable way to find where they cross. The last trick, using how far along the $x$-values $E$ is, avoids working with the messy coordinates of $E$ in the distance formula.

Use similar triangles. Extend line $DC$ up to the top row of dots; it reaches $F = (5, 3)$. The top row $AF$ and the bottom row $DB$ are parallel, so triangles $AEF$ and $BED$ have the same angles and are similar. Their matching sides are $AF = 5$ and $BD = 4$, so $AE : EB = 5 : 4$. That makes $AE = \tfrac59 AB = \tfrac59 \cdot 3\sqrt5 = \tfrac{5\sqrt5}{3}$.