Difference between revisions of "2019 AMC 12A Problems/Problem 17"
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<math>s_3 - 5s_2 + 8s_1 = 39</math> | <math>s_3 - 5s_2 + 8s_1 = 39</math> | ||
− | <math>39</math> can be written as <math>13s_0</math> | + | <math>39</math> can be written as <math>13s_0</math>, giving |
<math>s_3 = 5s_2 - 8s_1 + 13s_0</math> | <math>s_3 = 5s_2 - 8s_1 + 13s_0</math> | ||
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Therefore, <math>a = 5, b = -8</math>, and <math>c = 13</math> by matching coefficients. | Therefore, <math>a = 5, b = -8</math>, and <math>c = 13</math> by matching coefficients. | ||
− | <math>5 - 8 + 13 = \boxed{\textbf{(D)}10}</math>. | + | <math>5 - 8 + 13 = \boxed{\textbf{(D) } 10}</math>. |
==Solution 3== | ==Solution 3== | ||
− | Let <math>p, q</math>, and <math>r</math> be the roots of the polynomial. By Vieta's, we have | + | Let <math>p, q</math>, and <math>r</math> be the roots of the polynomial. By Vieta's Formulae, we have |
<math>p+q+r = 5</math> | <math>p+q+r = 5</math> | ||
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<math>pqr=13</math>. | <math>pqr=13</math>. | ||
− | |||
We know <math>s_k = p^k + q^k + r^k</math>. Consider <math>(p+q+r)(s_k) =5s_k</math>. | We know <math>s_k = p^k + q^k + r^k</math>. Consider <math>(p+q+r)(s_k) =5s_k</math>. | ||
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<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 <math>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)</math> | |
− | <math>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)</math> | ||
<math> = s_{k+1} + p^{k-1} (pq + pr + qr) + q^{k-1} (pq + pr + qr) + r^{k-1} (pq + pr + qr)</math> | <math> = s_{k+1} + p^{k-1} (pq + pr + qr) + q^{k-1} (pq + pr + qr) + r^{k-1} (pq + pr + qr)</math> | ||
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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>. | ||
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− | |||
==Video Solution== | ==Video Solution== | ||
For those who want a video solution: https://www.youtube.com/watch?v=tAS_DbKmtzI | For those who want a video solution: https://www.youtube.com/watch?v=tAS_DbKmtzI | ||
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==See Also== | ==See Also== |
Revision as of 21:58, 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.