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Difference between revisions of "2019 AMC 12A Problems/Problem 17"

(I offered another solution using factoring and Vieta's)
(→‎Solution 5)
 
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==Solution 1==
 
==Solution 1==
−
Applying Newton's Sums (see [https://artofproblemsolving.com/wiki/index.php/Newton%27s_Sums this link]), we get the answer as <math>10</math>.
+
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==
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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 <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
  
−
<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>
+
==Solution 5 ==
 +
Let the roots be <math>r</math>, <math>s</math>, and <math>t</math>. We know <math>r^2+s^2+t^2 = (r+s+t)(r+s+t) - 2(rs+st+tr)</math>.Continuing, we have:
  
−
<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} + q^{k-1} + r^{k-1})(pq + pr + qr)</math>
+
<math>r^3+s^3+t^3 = (r^2+s^2+t^2)(r+s+t) - (rs+st+tr)(r+s+t)+3rst</math>
  
−
<math> 5s_k + 13s_{k-2} = s_{k+1} + 8s_{k-1}</math>
 
  
−
Rearrange to get
+
<math>r^4+s^4+t^4 = (r^3+s^3+t^3)(r+s+t) - (rs+st+tr)(r^2+s^2+t^2)-(rst)(r+s+t)</math>
−
<math>s_{k+1} = 5s_k - 8s_{k-1} + 13s_{k-2}</math>
+
 
 +
 
 +
<math>r^5+s^5+t^5 = (r^4+s^4+t^4)(r+s+t) - (rs+st+tr)(r^3+s^3+t^3) - (rst)(r^2+s^2+t^2)</math>
 +
 
 +
 
 +
Clearly, the answer is <math>5-8+13 = \boxed{\textbf{(D)} 10}</math>
  
−
So, <math>a+ b + c = 5 -8 + 13 = \boxed{\textbf{(D)}10}</math>.
+
-skibbysiggy
  
 
==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
−
 
  
 
==See Also==
 
==See Also==

Latest revision as of 12:52, 3 October 2024

Problem

Let $s_k$ denote the sum of the $\textit{k}$th powers of the roots of the polynomial $x^3-5x^2+8x-13$. In particular, $s_0=3$, $s_1=5$, and $s_2=9$. Let $a$, $b$, and $c$ be real numbers such that $s_{k+1} = a \, s_k + b \, s_{k-1} + c \, s_{k-2}$ for $k = 2$, $3$, $....$ What is $a+b+c$?

$\textbf{(A)} \; -6 \qquad \textbf{(B)} \; 0 \qquad \textbf{(C)} \; 6 \qquad \textbf{(D)} \; 10 \qquad \textbf{(E)} \; 26$

Solution 1

Applying Newton's Sums, we have\[s_{k+1}+(-5)s_k+(8)s_{k-1}+(-13)s_{k-2}=0,\]so\[s_{k+1}=5s_k-8s_{k-1}+13s_{k-2},\]we get the answer as $5+(-8)+13=10$.

Solution 2

Let $p, q$, and $r$ be the roots of the polynomial. Then,

$p^3 - 5p^2 + 8p - 13 = 0$

$q^3 - 5q^2 + 8q - 13 = 0$

$r^3 - 5r^2 + 8r - 13 = 0$

Adding these three equations, we get

$(p^3 + q^3 + r^3) - 5(p^2 + q^2 + r^2) + 8(p + q + r) - 39 = 0$

$s_3 - 5s_2 + 8s_1 = 39$

$39$ can be written as $13s_0$, giving

$s_3 = 5s_2 - 8s_1 + 13s_0$

We are given that $s_{k+1} = a \, s_k + b \, s_{k-1} + c \, s_{k-2}$ is satisfied for $k = 2$, $3$, $....$, meaning it must be satisfied when $k = 2$, giving us $s_3 = a \, s_2 + b \, s_1 + c \, s_0$.

Therefore, $a = 5, b = -8$, and $c = 13$ by matching coefficients.

$5 - 8 + 13 = \boxed{\textbf{(D) } 10}$.

Solution 3

Let $p, q$, and $r$ be the roots of the polynomial. By Vieta's Formulae, we have

$p+q+r = 5$

$pq+qr+rp = 8$

$pqr=13$.

We know $s_k = p^k + q^k + r^k$. Consider $(p+q+r)(s_k) =5s_k$.

$5s_k = [p^{k+1} + q^{k+1} + r^{k+1}] + p^k q + p^k r + pq^k + q^k r + pr^k + qr^k$

Using $pqr = 13$ and $s_{k-2} = p^{k-2} + q^{k-2} + r^{k-2}$, we see $13s_{k-2} = p^{k-1}qr + pq^{k-1}r + pqr^{k-1}$.

We have \[\begin{split} 5s_k + 13s_{k-2} &= s_{k+1} + (p^k q + p^k r + p^{k-1}qr) + (pq^k + pq^{k-1}r + q^k r) + (pqr^{k-1} + pr^k + qr^k) \\ &= s_{k+1} + p^{k-1} (pq + pr + qr) + q^{k-1} (pq + pr + qr) + r^{k-1} (pq + pr + qr) \\ &= s_{k+1} + (p^{k-1} + q^{k-1} + r^{k-1})(pq + pr + qr) \\ &= 5s_k + 13s_{k-2} = s_{k+1} + 8s_{k-1}\end{split}\]

Rearrange to get $s_{k+1} = 5s_k - 8s_{k-1} + 13s_{k-2}$

So, $a+ b + c = 5 -8 + 13 = \boxed{\textbf{(D) } 10}$.

-gregwwl

Solution 4

Let $r,s,t$ be the roots of $x^3-5x^2+8x-13$. Then:

$r^3=5r^2-8r+13$ \\ $s^3=5s^2-8s+13$ \\ $t^3=5t^2-8t+13$

If we multiply both sides of the equation by $r^k$, where $k$ 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 $r^4+s^4+t^4$, but that is just to check. So then with the above information about $r^3,s^3,t^3$, we see that:

$r^k=5r^{k-1}-8r^{k-2}-13r^{k-3}$, $s^k=5s^{k-1}-8s^{k-2}-13s^{k-3}$, $t^k=5t^{k-1}-8t^{k-2}-13t^{k-3}$

$s_k=r^k+s^k+t^k$

Then: $s_k=5s_{k-1}-8s_{k-2}+13s_{k-3}$

This means that $s_{k+1}=5s_{k}-8s_{k-1}+13s_{k-2}$, as expected. So we have $a=5, b=-8, c=13$. So our answer is $5-8+13=\boxed{\textbf{(D) } 10}$

-IzhanAli

Solution 5

Let the roots be $r$, $s$, and $t$. We know $r^2+s^2+t^2 = (r+s+t)(r+s+t) - 2(rs+st+tr)$.Continuing, we have:


$r^3+s^3+t^3 = (r^2+s^2+t^2)(r+s+t) - (rs+st+tr)(r+s+t)+3rst$


$r^4+s^4+t^4 = (r^3+s^3+t^3)(r+s+t) - (rs+st+tr)(r^2+s^2+t^2)-(rst)(r+s+t)$


$r^5+s^5+t^5 = (r^4+s^4+t^4)(r+s+t) - (rs+st+tr)(r^3+s^3+t^3) - (rst)(r^2+s^2+t^2)$


Clearly, the answer is $5-8+13 = \boxed{\textbf{(D)} 10}$

-skibbysiggy

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
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All AMC 12 Problems and Solutions

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