Difference between revisions of "2019 AMC 10A Problems/Problem 24"

Revision as of 09:42, 9 October 2020

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 $\frac1A + \frac1B + \frac1C = 3(p^2 + q^2 + r^2) - 44(p + q + r) + 240 = 3(22^2 - 2(80)) - 44(22) + 240 = \boxed{\textbf{(B) } 244}$

Solution 3

First multiply both sides by $s^3 - 22s^2 + 80s - 67$ so $\sum A(s-q)(s-r)=1$ . Now $s$ can be arbitrarily close to $p,q,r$ , not affecting anything, so we can set $s$ equal to $p,q,r$ to get $\sum \frac{1}{A} = \sum (p-q)(p-r) = \sum p^2 - \sum qr = (p+q+r)^2 - 3\sum qr = 484-240=\boxed{244}$ by Vieta's.

~Lcz

@@ Line 20: / Line 20: @@
 <cmath>\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}</cmath>
-==Solution 3==
+===Solution 3===
 First multiply both sides by <math>s^3 - 22s^2 + 80s - 67</math> so <math>\sum A(s-q)(s-r)=1</math>. Now <math>s</math> can be arbitrarily close to <math>p,q,r</math>, not affecting anything, so we can set <math>s</math> equal to <math>p,q,r</math> to get <cmath>\sum \frac{1}{A} = \sum (p-q)(p-r) = \sum p^2 - \sum qr = (p+q+r)^2 - 3\sum qr = 484-240=\boxed{244}</cmath> by Vieta's.

2019 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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