Difference between revisions of "2019 AMC 12A Problems/Problem 17"
Sevenoptimus (talk | contribs) m (Fixed formatting and removed irrelevant attribution) |
Sevenoptimus (talk | contribs) m (Further fixed formatting) |
||
Line 51: | Line 51: | ||
<math>13s_{k-2} = p^{k-1}qr + pq^{k-1}r + pqr^{k-1}</math>. | <math>13s_{k-2} = p^{k-1}qr + pq^{k-1}r + pqr^{k-1}</math>. | ||
− | We have < | + | We have <cmath>\begin{split} 5s_k + 13s_{k-2} &= s_{k+1} + (p^k q + p^k r + p^{k-1}qr) + (pq^k + pq^{k-1}r + q^k r) + (pqr^{k-1} + pr^k + qr^k) \ |
− | + | &= s_{k+1} + p^{k-1} (pq + pr + qr) + q^{k-1} (pq + pr + qr) + r^{k-1} (pq + pr + qr) \ | |
− | + | &= s_{k+1} + (p^{k-1} + q^{k-1} + r^{k-1})(pq + pr + qr) \ | |
− | + | &= 5s_k + 13s_{k-2} = s_{k+1} + 8s_{k-1}\end{split}</cmath> | |
− | |||
− | |||
− | |||
Rearrange to get | Rearrange to get | ||
<math>s_{k+1} = 5s_k - 8s_{k-1} + 13s_{k-2}</math> | <math>s_{k+1} = 5s_k - 8s_{k-1} + 13s_{k-2}</math> | ||
− | So, <math>a+ b + c = 5 -8 + 13 = \boxed{\textbf{(D)}10}</math>. | + | So, <math>a+ b + c = 5 -8 + 13 = \boxed{\textbf{(D) } 10}</math>. |
==Video Solution== | ==Video Solution== |
Revision as of 20:59, 17 February 2019
Problem
Let denote the sum of the th powers of the roots of the polynomial . In particular, , , and . Let , , and be real numbers such that for , , What is ?
Solution 1
Applying Newton's Sums (see this link), we get the answer as .
Solution 2
Let , and be the roots of the polynomial. Then,
Adding these three equations, we get
can be written as , giving
We are given that is satisfied for , , , meaning it must be satisfied when , giving us .
Therefore, , and by matching coefficients.
.
Solution 3
Let , and be the roots of the polynomial. By Vieta's Formulae, we have
.
We know . Consider .
Using and , we see .
We have
Rearrange to get
So, .
Video Solution
For those who want a video solution: https://www.youtube.com/watch?v=tAS_DbKmtzI
See Also
2019 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 16 |
Followed by Problem 18 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.