2004 AMC 10B Problem 18

The middle triangle $DBF$ is what is left of $ACE$ after cutting off three corner triangles: $ABF$ (at $A$), $BCD$ (at $C$) and $DEF$ (at $E$). So find those three areas and subtract.

1. The big triangle. Since $12^2 + 16^2 = 144 + 256 = 400 = 20^2$, the right angle is at $C$. Its area is $\frac{1}{2} \times 12 \times 16 = 96$.
2. Put it on a grid so heights are easy to read. Put $C$ at $(0,0)$, $A$ at $(0,12)$ straight up, and $E$ at $(16,0)$ straight right.
- $B$ is on $AC$ with $AB = 3$, so $B$ is $3$ below $A$: $B = (0, 9)$.
- $D$ is on $CE$ with $CD = 4$: $D = (4, 0)$.
- $F$ is on $EA$ with $EF = 5$, which is $\frac{1}{4}$ of $EA = 20$. Going all the way from $E$ to $A$ moves $16$ left and $12$ up, so a quarter of the way moves $4$ left and $3$ up: $F = (12, 3)$.
3. Corner at $C$: triangle $BCD$ has a right angle at $C$, with legs $BC = 9$ and $CD = 4$. Area $= \frac{1}{2} \times 9 \times 4 = 18$.
4. Corner at $A$: triangle $ABF$. Use $AB = 3$ as the base; it lies on the vertical line $AC$. The height is how far $F$ is from that line, which is its $x$-value, $12$. Area $= \frac{1}{2} \times 3 \times 12 = 18$.
5. Corner at $E$: triangle $DEF$. Use $DE = 16 - 4 = 12$ as the base; it lies on the bottom edge. The height is how high $F$ is, $3$. Area $= \frac{1}{2} \times 12 \times 3 = 18$.
6. Subtract: $[DBF] = 96 - 18 - 18 - 18 = 42$. (Here $[DBF]$ means the area of triangle $DBF$.)
7. The ratio is
$$\frac{42}{96} = \frac{7}{16}.$$

The answer is $\boxed{\textbf{(E)}\ \frac{7}{16}}$.

A triangle drawn inside another, with its corners on the sides, is usually easiest to find by subtracting the corner pieces. For each corner piece, pick a base that lies along a side of the big triangle, and the height is just how far the third point is from that side. Here every side is split $1$ to $3$, which is why all three corners come out equal.

Two triangles that share a corner angle have areas in the same ratio as the products of the two sides at that angle. Triangle $ABF$ shares angle $A$ with $ACE$, and uses $\frac{3}{12} = \frac{1}{4}$ of side $AC$ and $\frac{15}{20} = \frac{3}{4}$ of side $AE$. So its area is $\frac{1}{4} \times \frac{3}{4} = \frac{3}{16}$ of $96$, which is $18$. The same fractions appear at $C$ and at $E$, so the middle triangle is $1 - 3 \times \frac{3}{16} = \frac{7}{16}$ of the whole.