# 2019 AMC 10A Problems/Problem 24

## Problem

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 $\tfrac1A+\tfrac1B+\tfrac1C$? $\textbf{(A) }243\qquad\textbf{(B) }244\qquad\textbf{(C) }245\qquad\textbf{(D) }246\qquad\textbf{(E) } 247$

## Solution

### Solution 1

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 $\frac1A = (p-q)(p-r)$. Similarly, we can find $\frac1B = (q-p)(q-r)$ and $\frac1C = (r-p)(r-q)$. Summing them up, we get that $$\frac1A + \frac1B + \frac1C = p^2 + q^2 + r^2 - pq - qr - pr$$ By Vieta's Formulas, we know that $p^2 + q^2 + r^2 = (p+q+r)^2 - 2(pq + qr + pr) = 324$ and $pq + qr + pr = 80$. Thus the answer is $324 -80 = \boxed{\textbf{(B) } 244}$.

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.

### Solution 2 (limits)

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 $s\to p$, notice that the $B$ and $C$ terms on the right will cancel out and we will be left with only $A$. Hence, $A = \lim_{s \to p} \frac{s-p}{s^3 - 22s^2 + 80s - 67}$, which by L'Hôpital's rule becomes $\lim_{s \to p}\frac{1}{3s^2 - 44s + 80} = \frac{1}{3p^2 - 44p + 80}$. We can reason similarly to find $B$ and $C$. Adding up the reciprocals and using Vieta's Formulas, we have that $$\frac1A + \frac1B + \frac1C = 3(p^2 + q^2 + r^2) - 44(p + q + r) + 240 = 3(22^2 - 2(80)) - 44(22) + 240 = \boxed{\textbf{(B) } 244}$$

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 