2019 AMC 10 A

Complete problem set with solutions and individual problem pages

Problem 24 Hard

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 \frac{1}{s^3-22s^2+80s-67}=\frac{A}{s-p}+\frac{B}{s-q}+\frac{C}{s-r} for all s\notin\left\{ p,q,r\right\}. What is \frac{1}{A}+\frac{1}{B}+\frac{1}{C}? (2019 AMC 10A Problem, Question#24)

  • A.

    243

  • B.

    244

  • C.

    245

  • D.

    246

  • E.

    247

Answer:B

Multiplying both sides by (s-p)(s-q)(s-r) yields

1=A(s-q)(s-r)+B(s-p)(s-r)+C(s-p)(s-q) As this is a polynomial identity, and it is true for infinitely many s, it must be true for all s (since a polynomial with infinitely many roots must in fact be the constant polynomial 0). This means we can plug in s=p to find that \frac{1}{A}=(p-q)(p-r). Similarly, we can find \frac{1}{B}=(q-p)(q-r) and \frac{1}{C}=(r-p)(r-q).

Summing them up, we get that \frac{1}{A}+\frac{1}{B}+\frac{1}{C}=p^2+q^2+r^2-pq-qr-pr By Vieta's Formulas, we know that p^2+q^2+r^2=(p+q+r)^2-2(pq+qr+pr)=324 and qp+qr+pr=80.

Thus the answer is 324-80=\boxed{(\text{B})244} .

Note: This process of substituting in the 'forbidden' values in the original identity is a standard technique for partial fraction decomposition, as taught in calculus classes.

Multiplying by (s-p) on both sides, we find that

\frac{s-p}{s^3-22s^2+80s-67}=A+\frac{B(s-p)}{s-q}+\frac{C(s-p)}{s-r} As s\rightarrow p, notice that the B and C terms on the right will cancel out and we will be left with only A. Hence, A=\underset{s\rightarrow p}{\text{lim}}\frac{s-p}{s^3-22s^2+80s-67} ,which by L'Hospital's rule becomes \underset{s\rightarrow p}{\text{lim}}\frac{1}{3s^2-44s+80}=\frac{1}{3p^2-44p+80}. We can reason similarly to find B and C. Adding up the reciprocals and using Vieta's Formulas, we have that \frac{1}{A}+\frac{1}{B}+\frac{1}{C}=3(p^2+q^2+r^2)-44(p+q+r)+240=3(22^2-2(80))-44(22)+240=\boxed{(\text{B})244}.

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