Problem:
Triangles ABC and ADE have areas 2007 and 7002 , respectively, with B=(0,0),C=(223,0),D=(680,380), and E=(689,389). What is the sum of all possible x-coordinates of A?
Answer Choices:
A. 282
B. 300
C. 600
D. 900
E. 1200 Solution:
Let h be the length of the altitude from A in △ABC. Then
2007=21⋅BC⋅h=21⋅223⋅h
so h=18. Thus A is on one of the lines y=18 or y=−18. Line DE has equation x−y−300=0. Let A have coordinates (a,b). By the formula for the distance from a point to a line, the distance from A to line DE is ∣a−b−300∣/2. The area of △ADE is
7002=21⋅2∣a−b−300∣⋅DE=21⋅2∣a±18−300∣⋅92
Thus a=±18±1556+300, and the sum of the four possible values of a is 4⋅300=1200.
OR
As above, conclude that A is on one of the lines y=±18. By similar reasoning, A is on one of two particular lines l1 and l2 parallel to DE. Therefore there are four possible positions for A, determined by the intersections of the lines y=18 and y=−18 with each of l1 and l2. Let the line y=18 intersect l1 and l2 in points (x1,y1) and (x2,y2), and let the line y=−18 intersect l1 and l2 in points (x3,y3) and (x4,y4). The four points of intersection are the vertices of a parallelogram, and the center of the parallelogram has x-coordinate (1/4)(x1+x2+x3+x4). The center is the intersection of the line y=0 and line DE. Because line DE has equation y=x−300, the center of the parallelogram is (300,0). Thus the sum of all possible x-coordinates of A is 4⋅300=1200.