试题呈现设a>n>c且a+b+c=1,a~2+b~2+c~2=1,求a+b的取值范围.文[1]中采用构造方程的方法,将问题转化为根的分布问题,去除技巧,解法自然,不失为好方法.但观察式子中的变量a,b,c,如果将其中的a,b看作变量,c看作常量的话,将式子变形为a+b=1-c,a~2+b~2=1-c~2,考虑到方程有解,直接将问题转化为给定范围内解决直线与圆相交问题.另解:由a>b>c,得1=a+b+c>3c,故c<1/3①;若存在a,b满足a+b+c=1,a~2+b~2+c~2=1,则圆心(O,O)到直线的距离d=|c-1|/2~(1/2)< The test questions assume that a> n> c and a + b + c = 1, a ~ 2 + b ~ 2 + c ~ 2 = 1 and find the range of a + b. , The problem is transformed into the distribution of roots, removal techniques, natural solution, after all, a good way to look at the variables in the formula a, b, c, if one of the a, b as a variable, c as a constant, The equation is transformed into a + b = 1-c, a ~ 2 + b ~ 2 = 1-c ~ 2, taking into account the solution of the equation, the problem is directly converted into a given range to solve the problem of intersecting lines and circles. Solution: if a> b> c, 1 = a + b + c> 3c, so c <1 / 3①; if a, b satisfy a + b + c = 1, a ~ 2 + b ~ 2 + c ~ 2 = 1, the distance from the center of the circle (O, O) to the straight line d = | c-1 | / 2 ~ (1/2) <