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

Revision as of 06:07, 31 October 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 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 (Pure Elementary Algebra)

Solution 1 uses a trick from Calculus contradicting the restriction $s\not\in\{p,q,r\}$ . I am going to provide 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

@@ Line 13: / Line 13: @@
 ''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 (Pure Elementary Math)==
+==Solution 2 (Pure Elementary Algebra)==
-Solution 1 uses a trick from Calculus that 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 contradicting the restriction <math>s\not\in\{p,q,r\}</math>. I am going to provide 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>s^2(A+B+C)-s(Aq+Ar+Bp+Br+Cp+Cq)+(Aqr+Bpr+Cpq-1)=0</cmath>
@@ Line 30: / Line 30: @@
 <math>(6) + (5) \cdot p</math>, <math>A(r-p)(q-p)=1</math>
-<math>A = \frac{1}{(r-p)(q-p)}</math>, by symmetry, <math>B = \frac{1}{(r-q)(p-q)}</math>, <math>B = \frac{1}{(q-r)(p-r)}</math>
+<math>A = \frac{1}{(r-p)(q-p)}</math>, by symmetry, <math>B = \frac{1}{(r-q)(p-q)}</math>, <math>C = \frac{1}{(q-r)(p-r)}</math>
 The rest is similar to solution 1, we get <math>\boxed{\textbf{(B) } 244}</math>

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