Difference between revisions of "2019 AMC 12A Problems/Problem 17"
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==Solution 1== | ==Solution 1== | ||
− | Applying | + | Applying [https://artofproblemsolving.com/wiki/index.php/Newton's_Sums Newton's Sums], we have<cmath>s_{k+1}+(-5)s_k+(8)s_{k-1}+(-13)s_{k-2}=0,</cmath>so<cmath>s_{k+1}=5s_k-8s_{k-1}+13s_{k-2},</cmath>we get the answer as <math>5+(-8)+13=10</math>. |
==Solution 2== | ==Solution 2== | ||
Line 24: | Line 24: | ||
<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> | ||
Line 32: | Line 32: | ||
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== | ||
+ | |||
+ | 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>pq+qr+rp = 8</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>. | ||
+ | |||
+ | <math>5s_k = [p^{k+1} + q^{k+1} + r^{k+1}] + p^k q + p^k r + pq^k + q^k r + pr^k + qr^k</math> | ||
+ | |||
+ | Using <math>pqr = 13</math> and <math>s_{k-2} = p^{k-2} + q^{k-2} + r^{k-2}</math>, we see | ||
+ | <math>13s_{k-2} = p^{k-1}qr + pq^{k-1}r + pqr^{k-1}</math>. | ||
+ | |||
+ | 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 | ||
+ | <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>. | ||
+ | |||
+ | -gregwwl | ||
+ | |||
+ | ==Solution 4== | ||
+ | |||
+ | Let <math>r,s,t</math> be the roots of <math>x^3-5x^2+8x-13</math>. Then: | ||
+ | |||
+ | <math>r^3=5r^2-8r+13</math> \\ | ||
+ | <math>s^3=5s^2-8s+13</math> \\ | ||
+ | <math>t^3=5t^2-8t+13</math> | ||
+ | |||
+ | If we multiply both sides of the equation by <math>r^k</math>, where <math>k</math> is a positive integer, then that won't change the coefficients, but just the degree of the new polynomial and the other term's exponents. We can try multiplying to find <math>r^4+s^4+t^4</math>, but that is just to check. So then with the above information about <math>r^3,s^3,t^3</math>, we see that: | ||
+ | |||
+ | <math>r^k=5r^{k-1}-8r^{k-2}-13r^{k-3}</math>, | ||
+ | <math>s^k=5s^{k-1}-8s^{k-2}-13s^{k-3}</math>, | ||
+ | <math>t^k=5t^{k-1}-8t^{k-2}-13t^{k-3}</math> | ||
+ | |||
+ | <math>s_k=r^k+s^k+t^k</math> | ||
+ | |||
+ | Then: <math>s_k=5s_{k-1}-8s_{k-2}+13s_{k-3}</math> | ||
+ | |||
+ | This means that <math>s_{k+1}=5s_{k}-8s_{k-1}+13s_{k-2}</math>, as expected. So we have <math>a=5, b=-8, c=13</math>. So our answer is <math>5-8+13=\boxed{\textbf{(D) } 10}</math> | ||
+ | |||
+ | -IzhanAli | ||
+ | |||
+ | ==Video Solution== | ||
+ | For those who want a video solution: https://www.youtube.com/watch?v=tAS_DbKmtzI | ||
==See Also== | ==See Also== |
Latest revision as of 22:18, 12 March 2023
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, we haveso
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, .
-gregwwl
Solution 4
Let be the roots of
. Then:
\\
\\
If we multiply both sides of the equation by , where
is a positive integer, then that won't change the coefficients, but just the degree of the new polynomial and the other term's exponents. We can try multiplying to find
, but that is just to check. So then with the above information about
, we see that:
,
,
Then:
This means that , as expected. So we have
. So our answer is
-IzhanAli
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 |
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