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Difference between revisions of "2023 AMC 12B Problems/Problem 6"
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{{duplicate|[[2023 AMC 10B Problems/Problem 12|2023 AMC 10B #12]] and [[2023 AMC 12B Problems/Problem 6|2023 AMC 12B #6]]}}
 +
 +
==Problem==
 +
When the roots of the polynomial
 +
 +
<cmath>P(x)  = (x-1)^1 (x-2)^2 (x-3)^3 \cdot \cdot \cdot \cdot (x-10)^{10}</cmath>
 +
 +
are removed from the number line, what remains is the union of <math>11</math> disjoint open intervals. On how many of these intervals is <math>P(x)</math> positive?
 +
 +
<math>\textbf{(A)}~3\qquad\textbf{(B)}~7\qquad\textbf{(C)}~6\qquad\textbf{(D)}~4\qquad\textbf{(E)}~5</math>
 +
 
==Solution 1==
 
==Solution 1==
−
Looking at the polynomial, we can use Vieta's formulas. Let's say the roots are r, s, and t. Then
+
The expressions to the power of even powers are always positive, so we don't need to care about those. We only need to care about <math>(x-1)^1(x-3)^3(x-5)^5(x-7)^7(x-9)^9</math>. We need 0, 2, or 4 of the expressions to be negative. The 9 through 10 interval and 10 plus interval make all of the expressions positive. The 5 through 6 and 6 through 7 intervals make two of the expressions negative. The 1 through 2 and 2 through 3 intervals make four of the expressions negative. There are <math>\boxed{\textbf{(C) 6}}</math> intervals.
−
<cmath>r+s+t = -a</cmath>
+
 
−
<cmath>rs+rt+st = b</cmath>
+
~Aopsthedude
−
<cmath>rst = 6</cmath>
+
 
−
We know that r, s, and t are integers, so we can list the possibilities for distinct solutions using the third formula. We find that <math>(r,s,t)</math> can be <math>(1,2,3), (-1,-2,3), (-1,2,-3), (1,2,-3),</math> and <math>(1,-1,-6)</math>. Remember that r, s, and t have to be different. That's 5 solutions. We can also use those values to find the values of a and b to check that none of them are distinct, and we'll find that they are <math>(6,11), (0,-7), (-2,-5), (-4,1),</math> and <math>(-6,-1)</math>. This means that there are <math>\boxed{5}</math>
+
==Solution 2==
 +
The roots of the factorized polynomial are intervals from numbers 1 to 10. We take each interval as being defined as the number behind it. To make the function positive, we need to have an even number of negative expressions. Real numbers raised to even powers are always positive, so we only focus on <math>x-1</math>, <math>x-3</math>, <math>x-5</math>, <math>x-7</math>, and <math>x-9</math>. The intervals 1 and 2 leave 4 negative expressions, so they are counted. The same goes for intervals 5, 6, 9, and 10. Intervals 3 and 4 leave 3 negative expressions and intervals 7 and 8 leave 1 negative expression. The solution is the number of intervals which is <math>\boxed{\textbf{(C) 6}}</math>.
 +
 
 +
~darrenn.cp
 +
~DarkPheonix
 +
 
 +
==Solution 3==
 +
We can use the turning point behavior at the roots of a polynomial graph to find out the amount of intervals that are positive.
 +
 
 +
First, we evaluate any value on the interval <math>(-\infty, 1)</math>. Since the degree of <math>P(x)</math> is <math>1+2+...+9+10</math> = <math>\frac{10\times11}{2}</math> = <math>55</math>, and every term in <math>P(x)</math> is negative, multiplying <math>55</math> negatives gives a negative value. So <math>(-\infty, 0)</math> is a negative interval.
 +
 
 +
We know that the roots of <math>P(x)</math> are at <math>1,2,...,10</math>. When the degree of the term of each root is odd, the graph of <math>P(x)</math> will pass through the graph and change signs, and vice versa. So at <math>x=1</math>, the graph will change signs; at <math>x=2</math>, the graph will not, and so on.
 +
 
 +
This tells us that the interval <math>(1,2)</math> is positive, <math>(2,3)</math> is also positive, <math>(3,4)</math> is negative, <math>(4,5)</math> is also negative, and so on, with the pattern being <math>+,+,-,-,+,+,-,-,...</math> .
 +
 
 +
The positive intervals are therefore <math>(1,2)</math>, <math>(2,3)</math>, <math>(5,6)</math>, <math>(6,7)</math>, <math>(9,10)</math>, and <math>(10,\infty)</math>, for a total of <math>\boxed{\textbf{(C) 6}}</math>.
 +
 
 +
~nm1728 ~ESAOPS (minor edits)
 +
 
 +
==Solution 4==
 +
 
 +
Denote by <math>I_k</math> the interval <math>\left( k - 1 , k \right)</math> for <math>k \in \left\{ 2, 3, \cdots , 10 \right\}</math> and <math>I_1</math> the interval <math>\left( - \infty, 1 \right)</math>.
 +
 
 +
Therefore, the number of intervals that <math>P(x)</math> is positive is
 +
<cmath>
 +
\begin{align*}
 +
1 + \sum_{i=1}^{10} \Bbb I \left\{
 +
\sum_{j=i}^{10} j \mbox{ is even}
 +
\right\}
 +
& = 1 + \sum_{i=1}^{10} \Bbb I \left\{
 +
\frac{\left( i + 10 \right) \left( 11 - i \right)}{2} \mbox{ is even}
 +
\right\} \\
 +
& = 1 + \sum_{i=1}^{10} \Bbb I \left\{
 +
\frac{- i^2 + i + 110}{2} \mbox{ is even}
 +
\right\} \\
 +
& = 1 + \sum_{i=1}^{10} \Bbb I \left\{
 +
\frac{i^2 - i}{2} \mbox{ is odd}
 +
\right\} \\
 +
& = \boxed{\textbf{(C) 6}} .
 +
\end{align*}
 +
</cmath>
 +
 
 +
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 +
 
 +
==Solution 5: Graphing==
 +
Recall two key facts about the roots of a polynomial:
 +
- If a root has an odd multiplicity (e.g. the root appears an odd number of times), then the graph will cross the x-axis.
 +
- If a root has an even multiplicity (e.g. the root appears an even number of times), then the graph will bounce off the x-axis.
 +
 
 +
Sketching the graph and noting the multiplicity of the roots, we see that there are <math>C)6</math> positive intervals.
 +
 
 +
 
 +
==Video Solution 1 by OmegaLearn==
 +
https://youtu.be/taNU5dQ5-sA
 +
 
 +
~OmegaLearn
 +
 
 +
==Video Solution==
 +
 
 +
https://youtu.be/j8z8rup7KHc
 +
 
 +
 
 +
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 +
==Video Solution by Interstigation==
 +
 
 +
https://youtu.be/2C5MVT_LID8
 +
 
 +
~Interstigation
 +
 
 +
==See Also==
 +
{{AMC10 box|year=2023|ab=B|num-b=11|num-a=13}}
 +
{{AMC12 box|year=2023|ab=B|num-b=5|num-a=7}}
 +
{{MAA Notice}}

Latest revision as of 10:24, 27 October 2024

The following problem is from both the 2023 AMC 10B #12 and 2023 AMC 12B #6, so both problems redirect to this page.

Problem

When the roots of the polynomial

\[P(x) = (x-1)^1 (x-2)^2 (x-3)^3 \cdot \cdot \cdot \cdot (x-10)^{10}\]

are removed from the number line, what remains is the union of $11$ disjoint open intervals. On how many of these intervals is $P(x)$ positive?

$\textbf{(A)}~3\qquad\textbf{(B)}~7\qquad\textbf{(C)}~6\qquad\textbf{(D)}~4\qquad\textbf{(E)}~5$

Solution 1

The expressions to the power of even powers are always positive, so we don't need to care about those. We only need to care about $(x-1)^1(x-3)^3(x-5)^5(x-7)^7(x-9)^9$. We need 0, 2, or 4 of the expressions to be negative. The 9 through 10 interval and 10 plus interval make all of the expressions positive. The 5 through 6 and 6 through 7 intervals make two of the expressions negative. The 1 through 2 and 2 through 3 intervals make four of the expressions negative. There are $\boxed{\textbf{(C) 6}}$ intervals.

~Aopsthedude

Solution 2

The roots of the factorized polynomial are intervals from numbers 1 to 10. We take each interval as being defined as the number behind it. To make the function positive, we need to have an even number of negative expressions. Real numbers raised to even powers are always positive, so we only focus on $x-1$, $x-3$, $x-5$, $x-7$, and $x-9$. The intervals 1 and 2 leave 4 negative expressions, so they are counted. The same goes for intervals 5, 6, 9, and 10. Intervals 3 and 4 leave 3 negative expressions and intervals 7 and 8 leave 1 negative expression. The solution is the number of intervals which is $\boxed{\textbf{(C) 6}}$.

~darrenn.cp ~DarkPheonix

Solution 3

We can use the turning point behavior at the roots of a polynomial graph to find out the amount of intervals that are positive.

First, we evaluate any value on the interval $(-\infty, 1)$. Since the degree of $P(x)$ is $1+2+...+9+10$ = $\frac{10\times11}{2}$ = $55$, and every term in $P(x)$ is negative, multiplying $55$ negatives gives a negative value. So $(-\infty, 0)$ is a negative interval.

We know that the roots of $P(x)$ are at $1,2,...,10$. When the degree of the term of each root is odd, the graph of $P(x)$ will pass through the graph and change signs, and vice versa. So at $x=1$, the graph will change signs; at $x=2$, the graph will not, and so on.

This tells us that the interval $(1,2)$ is positive, $(2,3)$ is also positive, $(3,4)$ is negative, $(4,5)$ is also negative, and so on, with the pattern being $+,+,-,-,+,+,-,-,...$ .

The positive intervals are therefore $(1,2)$, $(2,3)$, $(5,6)$, $(6,7)$, $(9,10)$, and $(10,\infty)$, for a total of $\boxed{\textbf{(C) 6}}$.

~nm1728 ~ESAOPS (minor edits)

Solution 4

Denote by $I_k$ the interval $\left( k - 1 , k \right)$ for $k \in \left\{ 2, 3, \cdots , 10 \right\}$ and $I_1$ the interval $\left( - \infty, 1 \right)$.

Therefore, the number of intervals that $P(x)$ is positive is \begin{align*} 1 + \sum_{i=1}^{10} \Bbb I \left\{ \sum_{j=i}^{10} j \mbox{ is even} \right\} & = 1 + \sum_{i=1}^{10} \Bbb I \left\{ \frac{\left( i + 10 \right) \left( 11 - i \right)}{2} \mbox{ is even} \right\} \\ & = 1 + \sum_{i=1}^{10} \Bbb I \left\{ \frac{- i^2 + i + 110}{2} \mbox{ is even} \right\} \\ & = 1 + \sum_{i=1}^{10} \Bbb I \left\{ \frac{i^2 - i}{2} \mbox{ is odd} \right\} \\ & = \boxed{\textbf{(C) 6}} . \end{align*}

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Solution 5: Graphing

Recall two key facts about the roots of a polynomial: - If a root has an odd multiplicity (e.g. the root appears an odd number of times), then the graph will cross the x-axis. - If a root has an even multiplicity (e.g. the root appears an even number of times), then the graph will bounce off the x-axis.

Sketching the graph and noting the multiplicity of the roots, we see that there are $C)6$ positive intervals.


Video Solution 1 by OmegaLearn

https://youtu.be/taNU5dQ5-sA

~OmegaLearn

Video Solution

https://youtu.be/j8z8rup7KHc


~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Video Solution by Interstigation

https://youtu.be/2C5MVT_LID8

~Interstigation

See Also

2023 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
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
2023 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
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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