2007 AMC 12A 第 19 题

The area of $\triangle ABC$ fixes the altitude from $A$ to $BC,$ so $A$ lies on $y=18$ or $y=-18$

Likewise $A$ lies on one of two lines parallel to $DE$; the four intersection points average to the center of a parallelogram

The altitude $h$ from $A$ in $\triangle ABC$ satisfies $2007=\tfrac12\cdot 223\cdot h,$ so $h=18.$ Thus $A$ lies on $y=18$ or $y=-18.$

Line $DE$ has equation $x-y-300=0.$ The condition on $\triangle ADE$ similarly places $A$ on one of two lines parallel to $DE.$

The four possible positions of $A$ are the vertices of a parallelogram whose center is the intersection of $y=0$ with line $DE,$ namely $(300,0).$ Hence the sum of the four $x$-coordinates is $4\cdot 300=1200.$

Thus, the correct answer is E.