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

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<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>
 
As this is a polynomial identity, and it is true for infinitely many <math>s</math>, it must be true for all <math>s</math> (since a polynomial with infinitely many roots must in fact be the constant polynomial <math>0</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>
 
As this is a polynomial identity, and it is true for infinitely many <math>s</math>, it must be true for all <math>s</math> (since a polynomial with infinitely many roots must in fact be the constant polynomial <math>0</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>
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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>.
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We can express <math>p^2 + q^2 + r^2 = (p+q+r)^2 - 2(pq + qr + pr)</math>, and by Vieta's Formulas, we know that this expression is equal to <math>484</math>. Vieta's also gives <math>pq + qr + pr = 80</math> (which we also used to find <math>p^2+q^2+r^2</math>), so the answer is <math>484 -240 = \boxed{\textbf{(B) } 244}</math>.  
  
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''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. Extensive time is also spent covering this process (partial fraction decomposition) in the Art of Problem Solving's Intermediate Algebra course.
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''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.
  
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-Edit from countmath1
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-very small latex edit from countmath1 :)
 +
 
 +
Minor rephrasing for correctness and clarity ~ Technodoggo
  
 
==Solution 2 (Pure Elementary Algebra)==
 
==Solution 2 (Pure Elementary Algebra)==
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Solution 1 uses a trick from Calculus that seemingly contradicts the restriction <math>s\not\in\{p,q,r\}</math>. I am going to provide a solution with pure elementary algebra.
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Solution 1 uses a trick from Calculus that seemingly contradicts the restriction <math>s\not\in\{p,q,r\}</math>. Here is a solution with pure elementary algebra.
 
<cmath>A(s-q)(s-r) + B(s-p)(s-r) + C(s-p)(s-q)=1</cmath>
 
<cmath>A(s-q)(s-r) + B(s-p)(s-r) + C(s-p)(s-q)=1</cmath>
 
<cmath>s^2(A+B+C)-s(Aq+Ar+Bp+Br+Cp+Cq)+(Aqr+Bpr+Cpq-1)=0</cmath>
 
<cmath>s^2(A+B+C)-s(Aq+Ar+Bp+Br+Cp+Cq)+(Aqr+Bpr+Cpq-1)=0</cmath>
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==See Also==
 
==See Also==
  
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[[Category:Intermediate Algebra Problems]]
 
{{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}}

Latest revision as of 23:12, 23 August 2024

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 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 \[\frac1A + \frac1B + \frac1C = p^2 + q^2 + r^2 - pq - qr - pr\] We can express $p^2 + q^2 + r^2 = (p+q+r)^2 - 2(pq + qr + pr)$, and by Vieta's Formulas, we know that this expression is equal to $484$. Vieta's also gives $pq + qr + pr = 80$ (which we also used to find $p^2+q^2+r^2$), so the answer is $484 -240 = \boxed{\textbf{(B) } 244}$.

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.

-very small latex edit from countmath1 :)

Minor rephrasing for correctness and clarity ~ Technodoggo

Solution 2 (Pure Elementary Algebra)

Solution 1 uses a trick from Calculus that seemingly contradicts the restriction $s\not\in\{p,q,r\}$. Here is a solution with pure elementary algebra. \[A(s-q)(s-r) + B(s-p)(s-r) + C(s-p)(s-q)=1\] \[s^2(A+B+C)-s(Aq+Ar+Bp+Br+Cp+Cq)+(Aqr+Bpr+Cpq-1)=0\] \[\begin{cases} A+B+C=0 & (1)\\ Aq+Ar+Bp+Br+Cp+Cq=0 & (2)\\ Aqr+Bpr+Cpq=1 & (3) \end{cases}\] From $(1)$ we get $A=-(B+C)$, $B=-(A+C)$, $C=-(A+B)$, substituting them in $(2)$, we get $Ap + Bq + Cr=0$ $(4)$

$(4)- (1) \cdot r$, $A(p-r)+B(q-r)=0$ $(5)$

$(3) - (1) \cdot pq$, $Aq(r-p)+Bp(r-q)=1$ $(6)$

$(6) + (5) \cdot p$, $A(r-p)(q-p)=1$

$A = \frac{1}{(r-p)(q-p)}$, by symmetry, $B = \frac{1}{(r-q)(p-q)}$, $C = \frac{1}{(q-r)(p-r)}$

The rest is similar to solution 1, we get $\boxed{\textbf{(B) } 244}$

~isabelchen

Video Solution by Richard Rusczyk

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

Video Solution by TheBeautyofMath

https://youtu.be/zw5CCPcT5IU

~IceMatrix

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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