HMMT 二月 2004 · 几何 · 第 2 题

HMMT February 2004 — Geometry — Problem 2

A parallelogram has 3 of its vertices at (1, 2), (3,8), and (4, 1). Compute the sum of the possible x -coordinates for the 4th vertex.

解析

A parallelogram has 3 of its vertices at (1, 2), (3,8), and (4, 1). Compute the sum of the possible x -coordinates for the 4th vertex. Solution: 8 There are 3 possible locations for the 4th vertex. Let ( a, b ) be its coordinates. If it is opposite to vertex (1 , 2), then since the midpoints of the diagonals of a parallelogram a +1 b +2 3+4 8+1 coincide, we get ( , ) = ( , ). Thus ( a, b ) = (6 , 7). By similar reasoning for 2 2 2 2 the other possible choices of opposite vertex, the other possible positions for the fourth vertex are (0 , 9) and (2 , − 5), and all of these choices do give parallelograms. So the answer is 6 + 0 + 2 = 8.