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HMMT 二月 2004 · 几何 · 第 2 题

HMMT February 2004 — Geometry — Problem 2

专题
Contest Math / 竞赛数学
难度
L3
来源
HMMT

题目详情

英文原题

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

英文解析

  1. A parallelogram has 3 of its vertices at (1, 2), (3,8), and (4, 1). Compute the sum ofthe possible x -coordinates for the 4 th vertex.
    Solution: 8
    There are 3 possible locations for the 4 th vertex. Let ( a, b ) be its coordinates. If it isopposite to vertex (1 , 2), then since the midpoints of the diagonals of a parallelograma +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 fourthvertex are (0 , 9) and (2 , − 5), and all of these choices do give parallelograms. So theanswer is 6 + 0 + 2 = 8.