2003 AMC 10A Problem 22

The point $G$ lies outside the rectangle, beyond side $BC$. Lines from $G$ cross the two parallel sides $BC$ and $AD$, and that always creates two triangles with the same shape. Use them.

1. The pieces of the sides. $BC = AD = 9$. Then $CH = 9 - 6 = 3$ and $EA = 9 - 4 = 5$.
2. Two triangles of the same shape. Look at triangle $GCH$ (small, with base $CH$ on line $BC$) and triangle $GEA$ (large, with base $EA$ on line $AD$). They share the angle at $G$, and because $BC$ is parallel to $AD$, their bases are parallel. So the small triangle is a shrunken copy of the large one, and every length in it is $\frac{CH}{EA} = \frac{3}{5}$ of the matching length in the large one.
3. The heights. The height of the large triangle from $G$ down to line $AD$ is exactly $GF$. The height of the small triangle from $G$ down to line $BC$ is $\frac{3}{5}$ of that, so it is $\frac{3}{5}GF$.
4. The gap between the heights. Both heights are measured straight down from $G$, and line $BC$ sits $AB = 8$ above line $AD$. So the large height is $8$ more than the small one:
$$GF - \frac{3}{5}GF = 8.$$
5. Solve. $\frac{2}{5}GF = 8$, so $GF = 8 \times \frac{5}{2} = 20$.

The answer is $\boxed{\textbf{(B)}\ 20}$.

Two lines through one point that cross a pair of parallel lines always make two triangles of the same shape, and every length (sides and heights alike) scales by the same factor. Here the factor comes from the bases, $3$ and $5$, and the rectangle tells you the two heights differ by $8$. Setting "big height minus small height $= 8$" finishes it.

Use coordinates. Put $D = (0, 0)$, $A = (9, 0)$, $C = (0, 8)$, $B = (9, 8)$. Then $E = (4, 0)$ and $H = (3, 8)$.

• Line $EC$ goes from $(4, 0)$ to $(0, 8)$: it drops $8$ over a run of $4$, so $y = -2x + 8$.
• Line $AH$ goes from $(9, 0)$ to $(3, 8)$: slope $\frac{8}{-6} = -\frac{4}{3}$, so $y = -\frac{4}{3}(x - 9) = -\frac{4}{3}x + 12$.

Setting them equal: $-2x + 8 = -\frac{4}{3}x + 12$ gives $-\frac{2}{3}x = 4$, so $x = -6$ and $y = -2(-6) + 8 = 20$. The point $G = (-6, 20)$ is $20$ above line $AD$ (the $x$-axis), so $GF = 20$.

From the rectangle, $CH = BC - BH = 9 - 6 = 3$ and $EA = AD - DE = 9 - 4 = 5$.

Since $BC \parallel AD$, the lines $GA$ and $GE$ cut the two parallels in similar triangles: $\triangle GCH \sim \triangle GEA$ with ratio $\dfrac{CH}{EA} = \dfrac{3}{5}$.

Extend line $BC$ until it meets line $GF$ at a point $X$. Because $GF$ is perpendicular to $AD$, it is perpendicular to $BC$ as well, so $GX$ and $GF$ are the altitudes from $G$ of the two similar triangles, and $XF$ equals the distance between the parallel sides, $CD = 8$.

Corresponding altitudes are in the similarity ratio:
$$\frac{GX}{GF} = \frac{3}{5}, \qquad GF - GX = 8.$$
So $GF - \tfrac{3}{5} GF = 8$, giving $\tfrac{2}{5} GF = 8$ and $GF = 20$.

The answer is $\boxed{\textbf{(B)}\ 20}$.