Induction is too OP
by djmathman, Sep 19, 2013, 3:04 AM
USAMO 1992.5 wrote:
Let 
be a polynomial with complex coefficients which is of degree 
and has distinct zeros. Prove that there exist complex numbers 
such that divides the polynomial
![\[ \left( \cdots \left( (z-a_1)^2 - a_2 \right)^2 \cdots - a_{1991} \right)^2 - a_{1992}. \]](//latex.artofproblemsolving.com/5/8/3/5834ec1483fbf4d5b0924cd014e1481cbcd08fca.png)
be a polynomial with complex coefficients which is of degree 
and has distinct zeros. Prove that there exist complex numbers 
such that divides the polynomial![\[ \left( \cdots \left( (z-a_1)^2 - a_2 \right)^2 \cdots - a_{1991} \right)^2 - a_{1992}. \]](http://latex.artofproblemsolving.com/5/8/3/5834ec1483fbf4d5b0924cd014e1481cbcd08fca.png)
be the 
complex zeroes of 
.
for each of 
.
,
such that the polynomial evaluates to ![\[Q_n(z)=\left( \cdots \left( (z-a_1)^2 - a_2 \right)^2 \cdots - a_{n-1} \right)^2.\]](http://latex.artofproblemsolving.com/9/7/6/976ac82c8bfdce623c4985b2993ed65374717a36.png)
for parameters 
,
does indeed actually exist.
,
such that 
for each 
.
.
is the midpoint of the complex numbers represented by 
and 
.
and 
are diametrically opposite each other with respect to the origin of the Argand plane, which implies that 
.![\[Q_{k+1}(r_1)=Q_{k+1}(r_2)=\cdots=Q_{k+1}(r_{k+1}),\]](http://latex.artofproblemsolving.com/a/a/3/aa3b2991a122f8c98c5ace68c66754dede52fc45.png)
case. By our inductive hypothesis, all values of 
are equal to each other, where 
.
that slides all these values of ![\[ \left(\cdots\left( (z-a_1)^2-a_2\right)^2\cdots-a_{1991}\right)^2-a_{1992}\]](http://latex.artofproblemsolving.com/d/f/a/dfa72dfa66576a0808b3fe8dba9154c162de134a.png)
This post has been edited 1 time. Last edited by djmathman, Sep 19, 2013, 3:22 AM
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