Difference between revisions of "2019 AMC 10A Problems/Problem 24"
(→Solution 2 (limits)) |
(→Video Solution) |
||
(12 intermediate revisions by 10 users not shown) | |||
Line 5: | Line 5: | ||
<math>\textbf{(A) }243\qquad\textbf{(B) }244\qquad\textbf{(C) }245\qquad\textbf{(D) }246\qquad\textbf{(E) } 247</math> | <math>\textbf{(A) }243\qquad\textbf{(B) }244\qquad\textbf{(C) }245\qquad\textbf{(D) }246\qquad\textbf{(E) } 247</math> | ||
− | ==Solution | + | ==Solution== |
Multiplying both sides by <math>(s-p)(s-q)(s-r)</math> yields | 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> | <cmath>1 = A(s-q)(s-r) + B(s-p)(s-r) + C(s-p)(s-q)</cmath> | ||
Line 11: | Line 11: | ||
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>. | 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>. | ||
− | ''Note'': | + | ''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 | + | ==Video Solution== |
− | + | Richard Rusczyk: https://www.youtube.com/watch?v=GI5d2ZN8gXY | |
− | |||
− | |||
− | |||
==See Also== | ==See Also== | ||
+ | |||
{{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}} |
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.