实数 ${a_1} , {a_2} , \cdots,{a_{2013}}$ 满足 ${a_1} + {a_2} + \cdots + {a_{2013}} = 0$，且$$\left| {{a_1} - 2{a_2}} \right| = \left| {{a_2} - 2{a_3}} \right| = \cdots = \left| {{a_{2012}} - 2{a_{2013}}} \right|= \left| {{a_{2013}} - 2{a_1}} \right|.$$求证：${a_1} = {a_2} = \cdots = {a_{2013}} = 0$．
【难度】
【出处】
2013年北京大学等三校联考自主招生保送生测试
【标注】
【答案】
略
【解析】
令 ${b_1} = {a_1} - 2{a_2}$，${b_2} = {a_2} - 2{a_3}$，$\cdots$，${b_{2013}} = {a_{2013}} - 2{a_1}$，则$$\left| {{b_1}} \right| = \left| {{b_2}} \right| = \cdots = \left| {{b_{2013}}} \right|, {b_1} + {b_2} + \cdots + {b_{2013}} = 0.$$设 $\left| {{b_1}} \right| = \left| {{b_2}} \right| = \cdots = \left| {{b_{2013}}} \right| = m $，则 $ {b_1} , {b_2} , \cdots ,{b_{2013}} $ 或者为 $ m $ 或者为 $ - m $，
设其中有 $ x $ 个 $ m $，$ (2013 - x) $ 个 $ - m$，则$$ {b_1} + {b_2} + \cdots + {b_{2013}} = mx + \left({ - m} \right)\left({2013 - x} \right)= m\left({2x - 2013} \right).$$由于 $2x - 2013 \ne 0$，因此 $m = 0$．
于是$$ {b_1} = {b_2} = \cdots = {b_{2013}} = 0,$$进而易得$$ {a_1} = {a_2} = \cdots = {a_{2013}} = 0.$$
设其中有 $ x $ 个 $ m $，$ (2013 - x) $ 个 $ - m$，则$$ {b_1} + {b_2} + \cdots + {b_{2013}} = mx + \left({ - m} \right)\left({2013 - x} \right)= m\left({2x - 2013} \right).$$由于 $2x - 2013 \ne 0$，因此 $m = 0$．
于是$$ {b_1} = {b_2} = \cdots = {b_{2013}} = 0,$$进而易得$$ {a_1} = {a_2} = \cdots = {a_{2013}} = 0.$$
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