AIME 2004 I · 第 9 题

Solution

We let $AB=3, AC=4, DE=6, DG=7$ for the purpose of labeling. Clearly, the dividing segment in $DEFG$ must go through one of its vertices, without loss of generality $D$. The other endpoint ($D'$) of the segment can either lie on $\overline{EF}$ or $\overline{FG}$. $V_2$ is a trapezoid with a right angle then, from which it follows that $V_1$ contains one of the right angles of $\triangle ABC$, and so $U_1$ is similar to $ABC$. Thus $U_1$, and hence $U_2$, are $3-4-5\,\triangle$s.

Suppose we find the ratio $r$ of the smaller base to the larger base for $V_2$, which consequently is the same ratio for $V_1$. By similar triangles, it follows that $U_1 \sim \triangle ABC$ by the same ratio $r$, and since the ratio of the areas of two similar figures is equal to the square of the ratio of their corresponding lengths, it follows that $[U_1] = r^2 \cdot [ABC] = 6r^2$.

• If $D'$ lies on $\overline{EF}$, then $ED' = \frac34 \cdot 6 = \frac{9}{2},\, 8$ from maintaining the $3-4-5$ triangular ratio; the latter ($\frac43 \cdot 6 = 8$) can be discarded as extraneous since we need $ED'. Therefore,$D'F = \frac{5}{2}$, and the ratio$r = \frac{D'F}{DG} = \frac{5}{14}$. The area of$[U_1] = 6\left(\frac{5}{14}\right)^2$ in this case.

• If $D'$ lies on $\overline{FG}$, then $GD' = \frac{21}{4},\, \frac{28}{3}$; the latter can be discarded as extraneous since we need $GD'. Therefore,$D'F = \frac{3}{4}$, and the ratio$r = \frac{D'F}{DE} = \frac{1}{8}$. The area of$[U_1] = 6\left(\frac{1}{8}\right)^2$ in this case.

Of the two cases, the second is smaller; the answer is $\frac{3}{32}$, and $m+n = \boxed{035}$.