HMMT 二月 2008 · GEN1 赛 · 第 1 题

HMMT February 2008 — GEN1 Round — Problem 1

[ 2 ] Let ABCD be a unit square (that is, the labels A, B, C, D appear in that order around the square). Let X be a point outside of the square such that the distance from X to AC is equal to the distance √ 2 2 from X to BD , and also that AX = . Determine the value of CX . 2

解析

[ 2 ] Let ABCD be a unit square (that is, the labels A, B, C, D appear in that order around the square). Let X be a point outside of the square such that the distance from X to AC is equal to the distance √ 2 2 from X to BD , and also that AX = . Determine the value of CX . 2 5 Answer: 2 A D M N X B C Since X is equidistant from AC and BD , it must lie on either the perpendicular bisector of AB or the perpendicular bisector of AD . It turns that the two cases yield the same answer, so we will just assume the first case. Let M be the midpoint of AB and N the midpoint of CD . Then, XM is perpendicular √ 1 3 1 10 to AB , so XM = and thus XN = , N C = . By the Pythagorean Theorem we find XC = 2 2 2 2 and the answer follows.