Partial Polynomial

Algebra Level 3

If $x,y,z,p,q,r$ are real numbers with $p, q, r$ all distinct such that $\frac{1}{x+p}+\frac{1}{y+p}+\frac{1}{z+p}=\frac{1}{p}$ $\frac{1}{x+q}+\frac{1}{y+q}+\frac{1}{z+q}=\frac{1}{q}$ $\frac{1}{x+r}+\frac{1}{y+r}+\frac{1}{z+r}=\frac{1}{r}$ Find the numerical value of $\dfrac{1}{p}+\dfrac{1}{q}+\dfrac{1}{r}$ .

The answer is 0.

1 solution

Mark Hennings
Dec 2, 2020

Let us assume that $p,q,r$ are distinct (otherwise this result is not true - try $x=y=z=2$ , $p=q=r=1$ ).

If $f(X) = (X+x)(X+y)(X+z) = X^3 + aX^2 + bX + c$ , the equations tell us that $f(p) = pf'(p)$ , $f(q) = qf'(q)$ and $f(r) = rf'(r)$ , so that $p,q,r$ are the zeros of $Xf'(X) - f(X) \; = \; 2X^3 + aX^2 - c$ Thus $pq+pr+qr=0$ , while $pqr = \tfrac12c$ , and hence $p^{-1} + q^{-1} + r^{-1} = \boxed{0}$ .

Partial Polynomial

The answer is 0.

1 solution

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