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

m (→‎Solution 2 (Pure Elementary Algebra))
m (→‎Solution 1)
 
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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>
−
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>.
+
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>.  
  
−
''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.
 +
 
 +
-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.
+
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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