2004 AMC 10B Problem 18

Triangle $ACE$ has a right angle at $C$ (since $12^2 + 16^2 = 20^2$), so its area is $\frac{1}{2} \cdot 12 \cdot 16 = 96$.

The three points cut each side in the ratio $1 : 3$: $AB : BC = 3 : 9$, $CD : DE = 4 : 12$, $EF : FA = 5 : 15$. The middle triangle $DBF$ is what remains after removing the three corner triangles $ABF$, $BCD$, $DEF$.

If two triangles share an angle, their areas are in the ratio of the products of the sides enclosing that angle (area $= \frac{1}{2} xy \sin\theta$). Corner $A$: triangle $ABF$ shares $\angle A$ with $ACE$, with $AB = \frac{1}{4} AC$ and $AF = \frac{3}{4} AE$, so
$$[ABF] = \frac{1}{4} \cdot \frac{3}{4} \cdot 96 = 18.$$
The same fractions $\frac{1}{4}$ and $\frac{3}{4}$ appear at $C$ and at $E$, so $[BCD] = [DEF] = 18$ as well. (At $C$ this is also just $\frac{1}{2} \cdot 9 \cdot 4 = 18$.)

Therefore $[DBF] = 96 - 3 \cdot 18 = 42$, and
$$\frac{[DBF]}{[ACE]} = \frac{42}{96} = \frac{7}{16}.$$

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

An inner triangle with vertices on the sides is almost always found by subtracting the three corner triangles, and each corner triangle is measured by the shared-angle area ratio: scale one enclosing side by $p$ and the other by $q$, and the area scales by $pq$. With the same split ratio on every side, the three corners are equal and the answer is $1 - 3pq$ in general.

Coordinates: $C = (0,0)$, $A = (0,12)$, $E = (16,0)$. Then $B = (0,9)$, $D = (4,0)$, and $F$ is one quarter of the way from $E$ to $A$, so $F = (12, 3)$. Shoelace on $B, D, F$: $\frac{1}{2}\,|0(0 - 3) + 4(3 - 9) + 12(9 - 0)| = \frac{1}{2} \cdot 84 = 42$, giving $\frac{42}{96} = \frac{7}{16}$.

• Computing a corner triangle's area from the two given sides as if they were perpendicular, when only the corner at C is a right angle.
• Using the ratio $\frac{1}{4}$ on both enclosing sides (area fraction $\frac{1}{16}$ per corner) instead of $\frac{1}{4} \cdot \frac{3}{4}$, which yields $\frac{13}{16}$.