A polynomial $p(x)$ with real coefficients is said to be almeriense if it is of the form: $$ p(x) = x^3+ax^2+bx+a $$ And its three roots are positive real numbers in arithmetic progression. Find all almeriense polynomials such that $p\left(\frac{7}{4}\right) = 0$
Problem
Source: Spain Mathematical Olympiad 2020 P1
Tags: algebra, polynomial, arithmetic sequence, Spain
RagvaloD
14.07.2020 19:35
If roots are $ m-d,m,m+d$ then $3m=-a, 3m^2-d^2=b, m(m^2-d^2)=-a \to m^2-d^2=3 \to d^2=m^2-3,b=2m^2+3$ and $7^3+a7^2*4+b*7*4^2+a*4^3=0$ or $ 343-588m+(2m^2+3)*112-192m=0 \to m=\frac{7}{4},d=\frac{1}{4}$ or $m=\frac{97}{56},d=\frac{1}{56}$ $a=\frac{-21}{4},b=\frac{73}{8}$ or $a=\frac{-291}{56},b=\frac{14113}{1568}$
Tintarn
14.07.2020 19:37
So let the roots be $\{c,c+cd,c+2cd\}$ with $c,d>0$ where one of them has to be $\frac{7}{4}$ and by Vieta
\[c+(c+cd)+(c+2cd)=c(c+cd)(c+2cd).\]Hence
\[3d+3=c^2(d+1)(2d+1)\]and hence
\[3=c^2(2d+1).\]In particular, $c^2<3$ and hence $c<\frac{7}{4}$. So we have two cases:
1. $c(d+1)=\frac{7}{4}$. Then $48(d+1)^2=49(2d+1)$ and hence $d=\frac{1}{6}, c=\frac{3}{2}$ which works.
In this case, the roots are $\frac{3}{2}, \frac{7}{4}$ and $2$ and so
\[p(x)=x^3-\frac{21}{4}x^2+\frac{73}{8}x-\frac{21}{4}.\]2. $c(2d+1)=\frac{7}{4}$. Then $c=\frac{12}{7}$ and $d=\frac{1}{96}$ which works.
In this case, the roots are $\frac{96}{56}, \frac{97}{56}$ and $\frac{98}{56}=\frac{7}{4}$ and so
\[p(x)=x^3-\frac{291}{56}x^2+\frac{14113}{1568}x-\frac{291}{56}.\]