2000 AMC 10 Problem 16

Place the lower-left lattice point at the origin. Reading the grid: $A = (0, 3)$, $B = (6, 0)$, $C = (4, 2)$, $D = (2, 0)$.

Line $AB$ drops $3$ over a run of $6$, so its slope is $-\tfrac12$ and its equation is $y = 3 - \tfrac12 x$. Line $CD$ rises $2$ over a run of $2$, slope $1$, passing through $(2, 0)$: $y = x - 2$.

At $E$ both hold:
$$x - 2 = 3 - \tfrac12 x \quad\Longrightarrow\quad \tfrac32 x = 5 \quad\Longrightarrow\quad x = \tfrac{10}{3}.$$

Now avoid the distance formula: $E$ lies on segment $AB$, and along that segment the horizontal coordinate runs from $0$ at $A$ to $6$ at $B$. So $E$ sits $\frac{10/3}{6} = \frac59$ of the way from $A$ to $B$, and
$$AE = \tfrac59 \cdot AB = \tfrac59 \sqrt{6^2 + 3^2} = \tfrac59 \cdot 3\sqrt5 = \tfrac{5\sqrt5}{3}.$$

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

A lattice diagram is a coordinate system in disguise, so line equations and their intersection are the most reliable route. The finishing trick, that a point's position along a segment is proportional to its horizontal displacement, saves the messy square root of $(10/3)^2 + (5/3)^2$. Tagging note: the problem is stated on a lattice, so coordinate geometry is the primary topic even though similar triangles also solve it.

Extend $DC$ upward until it meets the top row at $F = (5, 3)$; $AF$ and $DB$ are both horizontal, hence parallel. Triangles $AEF$ and $BED$ are then similar, giving $\frac{AE}{EB} = \frac{AF}{DB} = \frac54$, so $AE = \frac59 AB = \frac59 \cdot 3\sqrt5 = \frac{5\sqrt5}{3}$.

• Misreading the grid (it is 7 points wide and 4 tall, so A = (0,3) and B = (6,0)), or reporting the whole length AB instead of AE.
• Solving $x - 2 = 3 - \tfrac12 x$ carelessly (for example getting $x = 2$ or $x = \tfrac{5}{3}$), which shifts the answer to another choice.
• Using $EB$'s fraction $\tfrac49$ instead of $AE$'s fraction $\tfrac59$, which gives choice (A).