已知 $f\left( x \right)$ 为一元二次函数，且 $a,f\left( a \right),f\left( {f\left( a \right)} \right),f\left( {f\left( {f\left( a \right)} \right)} \right)$ 为正项等比数列，求证：$f\left( a \right) = a$．
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【出处】
2012年北京大学保送生试题
【标注】
【答案】
略
【解析】
由 $a,f\left( a \right),f\left( {f\left( a \right)} \right),f\left( {f\left( {f\left( a \right)} \right)} \right)$ 成等比数列得$$\dfrac{{f\left( a \right)}}{a} = \dfrac{{f\left( {f\left( a \right)} \right)}}{{f\left( a \right)}} = \dfrac{{f\left( {f\left( {f\left( a \right)} \right)} \right)}}{{f\left( {f\left( a \right)} \right)}},$$下面用反证法，若 $f(a)\ne a$，则有$$ \dfrac{{f\left( {f\left( a \right)} \right) - f\left( a \right)}}{{f\left( a \right) - a}} = \dfrac{{f\left( {f\left( {f\left( a \right)} \right)} \right) - f\left( {f\left( a \right)} \right)}}{{f\left( {f\left( a \right)} \right) - f\left( a \right)}}.$$所以，三点 $A\left( {a,f\left( a \right)} \right)$，$B\left( {f\left( a \right),f\left( {f\left( a \right)} \right)} \right)$，$C\left( {f\left( {f\left( a \right)} \right),f\left( {f\left( {f\left( a \right)} \right)} \right)} \right)$ 满足$${k_{AB}} = {k_{BC}},$$所以 $A,B,C$ 三点共线，与 $A,B,C$ 三点在抛物线上矛盾，所以 $ f\left( a \right) = a$．
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