Difference between revisions of "2019 AMC 10A Problems/Problem 24"
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− | + | ==Problem== | |
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+ | Let <math>p</math>, <math>q</math>, and <math>r</math> be the distinct roots of the polynomial <math>x^3 - 22x^2 + 80x - 67</math>. It is given that there exist real numbers <math>A</math>, <math>B</math>, and <math>C</math> such that <cmath>\dfrac{1}{s^3 - 22s^2 + 80s - 67} = \dfrac{A}{s-p} + \dfrac{B}{s-q} + \frac{C}{s-r}</cmath>for all <math>s\not\in\{p,q,r\}</math>. What is <math>\tfrac1A+\tfrac1B+\tfrac1C</math>? | ||
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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> | ||
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+ | ==Solution== | ||
+ | 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> | ||
+ | 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>. | ||
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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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+ | ==Video Solution== | ||
+ | Richard Rusczyk: https://www.youtube.com/watch?v=GI5d2ZN8gXY | ||
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+ | ==See Also== | ||
+ | |||
+ | |||
+ | {{AMC10 box|year=2019|ab=A|num-b=23|num-a=25}} | ||
+ | {{MAA Notice}} |
Revision as of 22:06, 17 January 2021
Contents
Problem
Let , , and be the distinct roots of the polynomial . It is given that there exist real numbers , , and such that for all . What is ?
Solution
Multiplying both sides by yields As this is a polynomial identity, and it is true for infinitely many , it must be true for all (since a polynomial with infinitely many roots must in fact be the constant polynomial ). This means we can plug in to find that . Similarly, we can find and . Summing them up, we get that By Vieta's Formulas, we know that and . Thus the answer is .
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 (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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.