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

(Solution 2 (limits))
(Video Solution)
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<math>\textbf{(A) }243\qquad\textbf{(B) }244\qquad\textbf{(C) }245\qquad\textbf{(D) }246\qquad\textbf{(E) } 247</math>
 
<math>\textbf{(A) }243\qquad\textbf{(B) }244\qquad\textbf{(C) }245\qquad\textbf{(D) }246\qquad\textbf{(E) } 247</math>
  
==Solution 1==
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==Solution==
 
Multiplying both sides by <math>(s-p)(s-q)(s-r)</math> yields
 
Multiplying both sides by <math>(s-p)(s-q)(s-r)</math> yields
 
<cmath>1 = A(s-q)(s-r) + B(s-p)(s-r) + C(s-p)(s-q)</cmath>
 
<cmath>1 = A(s-q)(s-r) + B(s-p)(s-r) + C(s-p)(s-q)</cmath>
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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>. Thus the answer is <math>324 -80 = \boxed{\textbf{(B) } 244}</math>.
 
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>. Thus the answer is <math>324 -80 = \boxed{\textbf{(B) } 244}</math>.
  
''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.
+
''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)==
+
==Video Solution==
Multiplying by <math>(s-p)</math> on both sides, we find that
+
Richard Rusczyk: https://www.youtube.com/watch?v=GI5d2ZN8gXY
<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>. Hence, <math>A = \lim_{s \to p} \frac{s-p}{s^3 - 22s^2 + 80s - 67}</math>, which by L'Hôpital's rule becomes <math>\lim_{s \to p}\frac{1}{3s^2 - 44s + 80} = \frac{1}{3p^2 - 44p + 80}</math>. We can reason similarly to find <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==
 +
  
 
{{AMC10 box|year=2019|ab=A|num-b=23|num-a=25}}
 
{{AMC10 box|year=2019|ab=A|num-b=23|num-a=25}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 22:06, 17 January 2021

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 $(s-p)(s-q)(s-r)$ 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.

Video Solution

Richard Rusczyk: https://www.youtube.com/watch?v=GI5d2ZN8gXY

See Also

2019 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 23
Followed by
Problem 25
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All AMC 10 Problems and Solutions

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