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Difference between revisions of "2019 AMC 10A Problems/Problem 24"
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==Solution==
 
==Solution==
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Multiplying both sides by <math>(s-p)(s-q)(s-r)</math> on both sides yields
 +
<cmath>1 = A(s-q)(s-r) + B(s-p)(s-r) + C(s-p)(s-q)</cmath>
 +
As this is a polynomial and is true for infinitely many <math>s</math>, it must be true for all <math>s</math>.  This means we can plug in <math>s = p</math> to find that <math>\frac1A = (p-q)(p-r)</math>.  Similarly, we can find <math>\frac1B = (q-p)(q-r)</math> and <math>\frac1C = (r-p)(r-q)</math>.  Summing them up, we get that <cmath>\frac1A + \frac1B + \frac1C = p^2 + q^2 + r^2 - pq - qr - pr</cmath>
 +
By Vieta's formulas, we know that <math>p^2 + q^2 + r^2 = (p+q+r)^2 - 2(pq + qr + pr) = 324</math> and <math>pq + qr + pr = 80</math>.  So the answer is <math>324 -80 = \boxed{\textbf{(B) } 244}</math>
 +
 +
(adapted from djmathman's solution)
 +
==Solution 2 (similar)==
 +
Multiplying by <math>(s-p)</math> on both sides we find that
 +
<cmath>\frac{s-p}{s^3 - 22s^2 + 80s - 67} = A + \frac{B(s-p)}{s-q} + \frac{C(s-p)}{s-r}</cmath>
 +
As <math>s\to p</math>, notice that the <math>B</math> and <math>C</math> terms on the right will cancel out and we will be left with only <math>A</math>.  So, <math>A = \lim_{s \to p} \frac{s-p}{s^3 - 22s^2 + 80s - 67}</math>, which by L'Hospital's rule is equal to <math>\lim_{s \to p}\frac{1}{3s^2 - 44s + 80} = \frac{1}{3p^2 - 44p + 80}</math>.  We can do similarly for <math>B</math> and <math>C</math>.  Adding up the reciprocals and using Vieta's formulas, we have that
 +
<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>
  
 
==See Also==
 
==See Also==

Revision as of 18:16, 9 February 2019

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

Multiplying both sides by $(s-p)(s-q)(s-r)$ on both sides yields \[1 = A(s-q)(s-r) + B(s-p)(s-r) + C(s-p)(s-q)\] As this is a polynomial and is true for infinitely many $s$, it must be true for all $s$. 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 $pq + qr + pr = 80$. So the answer is $324 -80 = \boxed{\textbf{(B) } 244}$

(adapted from djmathman's solution)

Solution 2 (similar)

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 $A$. So, $A = \lim_{s \to p} \frac{s-p}{s^3 - 22s^2 + 80s - 67}$, which by L'Hospital's rule is equal to $\lim_{s \to p}\frac{1}{3s^2 - 44s + 80} = \frac{1}{3p^2 - 44p + 80}$. We can do similarly for $B$ and $C$. Adding up the reciprocals and using Vieta's formulas, we have that \[\frac1A + \frac1B + \frac1C = 3(p^2 + q^2 + r^2) - 44(p + q + r) + 240 = 3(22^2 - 2(80)) - 44(22) + 240 = \boxed{\textbf{(B) } 244}\]

See Also

2019 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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