Let $ABC$ be a [triangle](https://artofproblemsolving.com/wiki/index.php/Triangle) with sides 3, 4, and 5, and $DEFG$ be a 6-by-7 [rectangle](https://artofproblemsolving.com/wiki/index.php/Rectangle). A segment is drawn to divide triangle $ABC$ into a triangle $U_1$ and a trapezoid $V_1$ and another segment is drawn to divide rectangle $DEFG$ into a triangle $U_2$ and a trapezoid $V_2$ such that $U_1$ is similar to $U_2$ and $V_1$ is similar to $V_2.$ The minimum value of the area of $U_1$ can be written in the form $m/n,$ where $m$ and $n$ are [relatively prime](https://artofproblemsolving.com/wiki/index.php/Relatively_prime) positive integers. Find $m+n.$
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](https://artofproblemsolving.com/wiki/index.php/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 $[U_1] = r^2 \cdot [ABC] = 6r^2$.



[asy] defaultpen(linewidth(0.7)); pointpen = black; pair D=(0,0),E=(0,6),F=(7,6),G=(7,0),H=(4.5,6); draw(MP("D",D)--MP("E",E,N)--MP("F",F,N)--MP("G",G)--cycle); draw(D--D(MP("D'",H,N)),dashed); label("\(U_2\)",(1,3)); label("\(V_2\)",(5,3)); MP("7",(D+G)/2,S); MP("6",(D+E)/2,W); MP("9/2",(E+H)/2,N); [/asy]



[asy] defaultpen(linewidth(0.7));  pointpen = black; pair D=(0,0),E=(0,6),F=(7,6),G=(7,0),H=(7,21/4); draw(MP("D",D)--MP("E",E,N)--MP("F",F,N)--MP("G",G)--cycle); draw(D--D(MP("D'",H,NW)),dashed); label("\(V_2\)",(2,3)); label("\(U_2\)",(5,1)); MP("7",(D+G)/2,S); MP("6",(D+E)/2,W); MP("21/4",(G+H)/2,(-1,0)); [/asy]


If $D'$ lies on $\overline{EF}$, then $ED' = \frac{9}{2},\, 8$; the latter can be discarded as extraneous. 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. 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 $m+n = \boxed{035}$.