AMC 10A, 2019, Problem 24, Question and solution

[Deepan Dutta]

Let \(p\), \(q\), and \(r\) be the distinct roots of the polynomial \(x^3 - 22x^2 + 80x - 67\). It is given that there exist real numbers \(A\), \(B\), and \(C\) such that\[\dfrac{1}{s^3 - 22s^2 + 80s - 67} = \dfrac{A}{s-p} + \dfrac{B}{s-q} + \frac{C}{s-r}\]for all \(s\not\in\{p,q,r\}\). What is \(\tfrac1A+\tfrac1B+\tfrac1C\)?

[Deepan Dutta]

\(s^3 - 22s^2 + 80s - 67\) should be factorised as \((s-p)(s-q)(s-r)\), as \(p,q\) and \(r\) are the roots of the polynomial \(x^3 - 22x^2 + 80x - 67\).
So, \(1=A(s-q)(s-r)+B(s-r)(s-p)+C(s-p)(s-q)\)

Now for \(s=p\), we would get, \(\frac{1}{A}=(p-q)(p-r)\)
Similarly, for \(s=q\), we get, \(\frac{1}{B}=(q-p)(q-r)\)
And, for \(s=r\), we get, \(\frac{1}{C}=(r-p)(r-q)\)
\(\therefore\) Adding them we can get, \(\frac{1}{A}+\frac{1}{B}+\frac{1}{C}=p^2+q^2+r^2-pq-qr-rp\)
Now form the given equation we can obtain the sum of the roots,
\(p+q+r=22\)
Also the sum of the roots taken two at a time \(pq+pr+rq=80\)
So, \(p^2+q^2+r^2=(p+q+r)^2-2(pq+pr+rq)\)
\(\Rightarrow p^2+q^2+r^2=22^2-160\)
\(\Rightarrow p^2+q^2+r^2=484-160=324\)
So, \(\frac{1}{A}+\frac{1}{B}+\frac{1}{C}=324-80=\boxed{244}\)

• This reply was modified 1 year, 3 months ago by Deepan Dutta.

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