2019 AMC 10A Problem 24

Question

Let

p,

q,

and

r

be the distinct roots of the polynomial

x^3 - 22x^2 + 80x - 67.

There exist real numbers

A,

B,

and

C

such that

\begin{gathered} \frac{1}{s^3-22s^2+80s-67}\\ =\frac{A}{s-p}+\frac{B}{s-q}\\ \quad+\frac{C}{s-r} \end{gathered}

for all real numbers

s

with

s\notin\{p,q,r\}.

What is

\dfrac1A+\dfrac1B+\dfrac1C?

243

244

245

246

247

Multiply by

(s-p)(s-q)(s-r)

Substitute each root, then use Vieta’s formulas

Multiplying the identity by

(s-p)(s-q)(s-r)

gives

\begin{aligned} 1={}&A(s-q)(s-r)\\ &+B(s-p)(s-r)\\ &+C(s-p)(s-q). \end{aligned}

Setting

s=p,q,r

, in turn, yields

\begin{aligned} \frac1A&=(p-q)(p-r),\\ \frac1B&=(q-p)(q-r),\\ \frac1C&=(r-p)(r-q). \end{aligned}

Adding and expanding gives

\begin{aligned} \frac1A+\frac1B+\frac1C &=p^2+q^2+r^2\\ &\quad-pq-pr-qr. \end{aligned}

By Vieta’s formulas,

p+q+r=22

and

pq+pr+qr=80

, so

\begin{aligned} p^2+q^2+r^2&=22^2-2(80)\\ &=324. \end{aligned}

Therefore the requested value is

324-80=244.

Thus, B is the correct answer.

Problem 24 in other years