Difference between revisions of "2019 AMC 10A Problems/Problem 24"
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<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> | <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> | ||
− | ==Solution 3== | + | ===Solution 3=== |
First multiply both sides by <math>s^3 - 22s^2 + 80s - 67</math> so <math>\sum A(s-q)(s-r)=1</math>. Now <math>s</math> can be arbitrarily close to <math>p,q,r</math>, not affecting anything, so we can set <math>s</math> equal to <math>p,q,r</math> to get <cmath>\sum \frac{1}{A} = \sum (p-q)(p-r) = \sum p^2 - \sum qr = (p+q+r)^2 - 3\sum qr = 484-240=\boxed{244}</cmath> by Vieta's. | First multiply both sides by <math>s^3 - 22s^2 + 80s - 67</math> so <math>\sum A(s-q)(s-r)=1</math>. Now <math>s</math> can be arbitrarily close to <math>p,q,r</math>, not affecting anything, so we can set <math>s</math> equal to <math>p,q,r</math> to get <cmath>\sum \frac{1}{A} = \sum (p-q)(p-r) = \sum p^2 - \sum qr = (p+q+r)^2 - 3\sum qr = 484-240=\boxed{244}</cmath> by Vieta's. |
Revision as of 09:42, 9 October 2020
Contents
[hide]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
Solution 1
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.
Solution 2 (limits)
Multiplying by on both sides, we find that As , notice that the and terms on the right will cancel out and we will be left with only . Hence, , which by L'Hôpital's rule becomes . We can reason similarly to find and . Adding up the reciprocals and using Vieta's Formulas, we have that
Solution 3
First multiply both sides by so . Now can be arbitrarily close to , not affecting anything, so we can set equal to to get by Vieta's.
~Lcz
See Also
Video Solution: https://www.youtube.com/watch?v=GI5d2ZN8gXY&t=53s
2019 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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