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Difference between revisions of "2016 AIME I Problems/Problem 11"

(Created page with "We substitute <math>x=2</math> into <math>(x-1)P(x+1)=(x+2)P(x)</math> to get <math>P(3)=4P(2)</math>. Since we also have that <math>\left(P(2)\right)^2 = P(3)</math>, we have...")
 
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We substitute <math>x=2</math> into <math>(x-1)P(x+1)=(x+2)P(x)</math> to get <math>P(3)=4P(2)</math>. Since we also have that <math>\left(P(2)\right)^2 = P(3)</math>, we have that <math>P(2)=4</math> and <math>P(3)=16</math>. We can also substitute <math>x=1</math>, <math>x=0</math>, and <math>x=3</math> into <math>(x-1)P(x+1)=(x+2)P(x)</math> to get that <math>0=P(1)</math>, <math>-1P(1)=2P(0)</math>, and <math>2P(4)=5P(3)</math>. This leads us to the conclusion that <math>P(0)=P(1)=0</math> and <math>P(4)=40</math>.
+
==Problem==
 +
Let <math>P(x)</math> be a nonzero polynomial such that <math>(x-1)P(x+1)=(x+2)P(x)</math> for every real <math>x</math>, and <math>\left(P(2)\right)^2 = P(3)</math>. Then <math>P(\tfrac72)=\tfrac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>.
  
−
We next use finite differences to find that <math>P</math> is a cubic polynomial. Thus, <math>P</math> must be of the form of <math>ax^3+bx^2+cx+d</math>. It follows that <math>d=0</math>; we now have a system of <math>3</math> equations to solve. We plug in <math>x=1</math>, <math>x=2</math>, and <math>x=3</math> to get
+
==Solution 1==
 +
Plug in <math>x=1</math> to get <math>(1-1)P(1+1) = 0 = (1+2)P(1) \Rightarrow P(1) = 0</math>. Plug in <math>x=0</math> to get <math>(0-1)P(0+1) = (0+2)P(0)\Rightarrow P(0) = -\frac{1}{2}P(1) = 0</math>. Plug in <math>x=-1</math> to get <math>(-1-1)P(-1+1) = (-1+2)P(-1)\Rightarrow (-2)P(0)=P(-1)\Rightarrow P(-1) = 0</math>. So <math>P(x) = x(x-1)(x+1)Q(x)</math> for some polynomial <math>Q(x)</math>. Using the initial equation, once again, <cmath>(x-1)P(x+1) = (x+2)P(x)</cmath>
 +
<cmath>(x-1)((x+1)(x+1-1)(x+1+1)Q(x+1)) = (x+2)((x)(x-1)(x+1)Q(x))</cmath>
 +
<cmath>(x-1)(x+1)(x)(x+2)Q(x+1) = (x+2)(x)(x-1)(x+1)Q(x)</cmath>
 +
<cmath>Q(x+1) = Q(x)</cmath>
 +
From here, we know that <math>Q(x) = C</math> for a constant <math>C</math> (<math>Q(x)</math> cannot be periodic since it is a polynomial), so <math>P(x) = Cx(x-1)(x+1)</math>. We know that <math>\left(P(2)\right)^2 = P(3)</math>. Plugging those into our definition of <math>P(x)</math>: <math>(C \cdot 2 \cdot (2-1) \cdot (2+1))^2 = C \cdot 3 \cdot (3-1) \cdot (3+1) \Rightarrow (6C)^2 = 24C \Rightarrow 36C^2 - 24C = 0 \Rightarrow C = 0</math> or <math>\frac{2}{3}</math>. So we know that <math>P(x) = \frac{2}{3}x(x-1)(x+1)</math>. So <math>P(\frac{7}{2}) = \frac{2}{3} \cdot \frac{7}{2} \cdot (\frac{7}{2} - 1) \cdot (\frac{7}{2} + 1) = \frac{105}{4}</math>. Thus, the answer is <math>105 + 4 = \boxed{109}</math>.
  
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<cmath>a+b+c=0</cmath>
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==Solution 2==
−
<cmath>8a+4b+2c=4</cmath>
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From the equation we see that <math>x-1</math> divides <math>P(x)</math> and <math>(x+2)</math> divides <math>P(x+1)</math> so we can conclude that <math>x-1</math> and <math>x+1</math> divide <math>P(x)</math> (if we shift the function right by 1, we get <math>(x-2)P(x) = (x+1)P(x-1)</math>, and from here we can see that <math>x+1</math> divides <math>P(x)</math>). This means that <math>1</math> and <math>-1</math> are roots of <math>P(x)</math>. Plug in <math>x = 0</math> and we see that <math>P(0) = 0</math> so <math>0</math> is also a root.
−
<cmath>27a+9b+3c=16</cmath>
 
  
−
We solve this system to get that <math>a=\frac{3}{2}</math>, <math>b=0</math>, and <math>c=-\frac{3}{2}</math>. Thus, <math>P(x)=\frac{3}{2}x^3-\frac{3}{2}x</math>. Plugging in <math>x=\frac{7}{2}</math>, we see that <math>P\left(\frac{7}{2}\right)=\frac{105}{4}</math>. Thus, <math>m=105</math>, <math>n=4</math>, and our answer is <math>m+n=\boxed{109}</math>.
+
Suppose we had another root that is not one of those <math>3</math>. Notice that the equation above indicates that if <math>r</math> is a root then <math>r+1</math> and <math>r-1</math> is also a root. Then we'd get an infinite amount of roots! So that is bad. So we cannot have any other roots besides those three.
 +
 
 +
That means <math>P(x) = cx(x-1)(x+1)</math>. We can use <math>P(2)^2 = P(3)</math> to get <math>c = \frac{2}{3}</math>. Plugging in <math>\frac{7}{2}</math> is now trivial and we see that it is <math>\frac{105}{4}</math> so our answer is <math>\boxed{109}</math>
 +
 
 +
==Solution 3==
 +
Although this may not be the most mathematically rigorous answer, we see that <math>\frac{P(x+1)}{P(x)}=\frac{x+2}{x-1}</math>. Using a bit of logic, we can make a guess that <math>P(x+1)</math> has a factor of <math>x+2</math>, telling us <math>P(x)</math> has a factor of <math>x+1</math>. Similarly, we guess that <math>P(x)</math> has a factor of <math>x-1</math>, which means <math>P(x+1)</math> has a factor of <math>x</math>. Now, since <math>P(x)</math> and <math>P(x+1)</math> have so many factors that are off by one, we may surmise that when you plug <math>x+1</math> into <math>P(x)</math>, the factors "shift over," i.e. <math>P(x)=(A)(A+1)(A+2)...(A+n)</math>, which goes to <math>P(x+1)=(A+1)(A+2)(A+3)...(A+n+1)</math>. This is useful because these, when divided, result in <math>\frac{P(x+1)}{P(x)}=\frac{A+n+1}{A}</math>. If <math>\frac{A+n+1}{A}=\frac{x+2}{x-1}</math>, then we get <math>A=x-1</math> and <math>A+n+1=x+2</math>, <math>n=2</math>. This gives us <math>P(x)=(x-1)x(x+1)</math> and <math>P(x+1)=x(x+1)(x+2)</math>, and at this point we realize that there has to be some constant <math>a</math> multiplied in front of the factors, which won't affect our fraction <math>\frac{P(x+1)}{P(x)}=\frac{x+2}{x-1}</math> but will give us the correct values of <math>P(2)</math> and <math>P(3)</math>. Thus <math>P(x)=a(x-1)x(x+1)</math>, and we utilize <math>P(2)^2=P(3)</math> to find <math>a=\frac{2}{3}</math>. Evaluating <math>P \left ( \frac{7}{2} \right )</math> is then easy, and we see it equals <math>\frac{105}{4}</math>, so the answer is <math>\boxed{109}</math>
 +
 
 +
==Solution 4==
 +
Substituting <math>x=2</math> into the given equation, we find that <math>P(3)=4P(2)=P(2)^2</math>. Therefore, either <math>P(2)=0</math> or <math>P(2)=4</math>. Now for integers <math>n\ge 2</math>, we know that
 +
<cmath>P(n+1)=\frac{n+2}{n-1}P(n).</cmath>
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Applying this repeatedly, we find that
 +
<cmath>P(n+1)=\frac{(n+2)!/3!}{(n-1)!}P(2).</cmath>
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If <math>P(2)=0</math>, this shows that <math>P(x)</math> has infinitely many roots, meaning that <math>P(x)</math> is identically equal to zero. But this contradicts the problem statement. Therefore, <math>P(2)=4</math>, and we find <math>P(n+1)=\frac{2}{3}(n+2)(n+1)n</math> for all positive integers <math>n\ge2</math>. This cubic polynomial matches the values <math>P(n+1)</math> for infinitely many numbers, hence the two polynomials are identically equal. In particular, <math>P\left(\frac72\right)=\frac23\cdot\frac92\cdot\frac72\cdot\frac52=\frac{105}{4}</math>, and the answer is <math>\boxed{109}</math>.
 +
 
 +
==Solution 5==
 +
We can find zeroes of the polynomial by making the first given equation <math>0 = 0</math>. Plugging in <math>x = 1</math> and <math>x = -2</math> gives us the zeroes <math>1</math> and <math>-1</math>, respectively. Now we can plug in these zeros to get more zeroes. <math>x = -1</math> gives us the zero <math>0</math> (no pun intended). <math>x = 1</math> makes the equation <math>0 \cdot P(2) = 0</math>, which means <math>P(2)</math> is not necessarily <math>0</math>. If <math>P(2) = 0</math>, then plugging in <math>2</math> to the equation yields <math>P(3) = 0</math>, plugging in <math>3</math> to the equation yields <math>P(4) = 0</math>, and so on, a contradiction of "nonzero polynomial". So <math>2</math> is not a zero. Note that plugging in <math>x = 0</math> to the equation does not yield any additional zeros. Thus, the only zeroes of <math>P(x)</math> are <math>-1, 0,</math> and <math>1</math>, so <math>P(x) = a(x + 1)x(x - 1)</math> for some nonzero constant <math>a</math>. We can plug in <math>2</math> and <math>3</math> into the polynomial and use the second given equation to find an equation for <math>a</math>. <math>P(2) = 6a</math> and <math>P(3) = 24a</math>, so:
 +
<cmath>(6a)^2  = 24a \implies 36a^2 = 24a \implies a = \frac23</cmath>
 +
Plugging in <math>\frac72</math> into the polynomial <math>\frac23(x + 1)x(x - 1)</math> yields <math>\frac{105}{4}</math>. <math>105 + 4 = \boxed{109}</math>.
 +
 
 +
== Solution 6 ==
 +
 
 +
Plug in <math>x=2</math> yields <math>P(3)=4P(2)</math>. Since also <math>(P(2))^2=P(3)</math>, we have <math>P(2)=4</math> and <math>P(3)=16</math>. Plug in <math>x=3</math> yields <math>2P(4)=5P(3)</math> so <math>P(4)=40</math>.
 +
 
 +
Repeat the action gives <math>P(2)=4</math>, <math>P(3)=16</math>, <math>P(4)=40</math>, <math>P(5)=80</math>, and <math>P(6)=140</math>.
 +
 
 +
Since <math>P(x)</math> is a polynomial, the <math>k</math>th difference is constant, where <math>k=\deg(P(x))</math>. Thus we can list out the 0th, 1st, 2nd, 3rd, ... differences until we obtain a constant.
 +
 
 +
<cmath>4,16,40,80,140</cmath>
 +
<cmath>12,24,40,60</cmath>
 +
<cmath>12,16,20</cmath>
 +
<cmath>4,4,4</cmath>
 +
 
 +
Since the 3rd difference of <math>P(x)</math> is constant, we can conclude that <math>\deg(P(x))=3</math>.
 +
 
 +
Let <math>P(x)=a_3x^3+a_2x^2+a_1x+a_0</math>. Plug in the values for <math>x</math> and solve the system of 4 equations gives <math>(a_3,a_2,a_1,a_0)=(\frac{2}{3},0,-\frac{2}{3},0)</math>
 +
 
 +
Thus <math>P(x)=\frac{2}{3}x^3-\frac{2}{3}x</math> and <math>P(\frac{7}{2})=\frac{105}{4}\Longrightarrow m+n=\boxed{109}</math>
 +
 
 +
~ Nafer
 +
 
 +
-Note that the coefficient of <math>x^3</math> is <math>\frac{2}{3}</math> because <math>\frac{2}{3}\cdot\deg(x^3)!=\frac{2}{3}\cdot3!=4</math>, the 3rd difference of P(x).
 +
~inaccessibles
 +
 
 +
== See also ==
 +
{{AIME box|year=2016|n=I|num-b=10|num-a=12}}
 +
{{MAA Notice}}

Latest revision as of 15:35, 8 September 2024

Problem

Let $P(x)$ be a nonzero polynomial such that $(x-1)P(x+1)=(x+2)P(x)$ for every real $x$, and $\left(P(2)\right)^2 = P(3)$. Then $P(\tfrac72)=\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$.

Solution 1

Plug in $x=1$ to get $(1-1)P(1+1) = 0 = (1+2)P(1) \Rightarrow P(1) = 0$. Plug in $x=0$ to get $(0-1)P(0+1) = (0+2)P(0)\Rightarrow P(0) = -\frac{1}{2}P(1) = 0$. Plug in $x=-1$ to get $(-1-1)P(-1+1) = (-1+2)P(-1)\Rightarrow (-2)P(0)=P(-1)\Rightarrow P(-1) = 0$. So $P(x) = x(x-1)(x+1)Q(x)$ for some polynomial $Q(x)$. Using the initial equation, once again, \[(x-1)P(x+1) = (x+2)P(x)\] \[(x-1)((x+1)(x+1-1)(x+1+1)Q(x+1)) = (x+2)((x)(x-1)(x+1)Q(x))\] \[(x-1)(x+1)(x)(x+2)Q(x+1) = (x+2)(x)(x-1)(x+1)Q(x)\] \[Q(x+1) = Q(x)\] From here, we know that $Q(x) = C$ for a constant $C$ ($Q(x)$ cannot be periodic since it is a polynomial), so $P(x) = Cx(x-1)(x+1)$. We know that $\left(P(2)\right)^2 = P(3)$. Plugging those into our definition of $P(x)$: $(C \cdot 2 \cdot (2-1) \cdot (2+1))^2 = C \cdot 3 \cdot (3-1) \cdot (3+1) \Rightarrow (6C)^2 = 24C \Rightarrow 36C^2 - 24C = 0 \Rightarrow C = 0$ or $\frac{2}{3}$. So we know that $P(x) = \frac{2}{3}x(x-1)(x+1)$. So $P(\frac{7}{2}) = \frac{2}{3} \cdot \frac{7}{2} \cdot (\frac{7}{2} - 1) \cdot (\frac{7}{2} + 1) = \frac{105}{4}$. Thus, the answer is $105 + 4 = \boxed{109}$.

Solution 2

From the equation we see that $x-1$ divides $P(x)$ and $(x+2)$ divides $P(x+1)$ so we can conclude that $x-1$ and $x+1$ divide $P(x)$ (if we shift the function right by 1, we get $(x-2)P(x) = (x+1)P(x-1)$, and from here we can see that $x+1$ divides $P(x)$). This means that $1$ and $-1$ are roots of $P(x)$. Plug in $x = 0$ and we see that $P(0) = 0$ so $0$ is also a root.

Suppose we had another root that is not one of those $3$. Notice that the equation above indicates that if $r$ is a root then $r+1$ and $r-1$ is also a root. Then we'd get an infinite amount of roots! So that is bad. So we cannot have any other roots besides those three.

That means $P(x) = cx(x-1)(x+1)$. We can use $P(2)^2 = P(3)$ to get $c = \frac{2}{3}$. Plugging in $\frac{7}{2}$ is now trivial and we see that it is $\frac{105}{4}$ so our answer is $\boxed{109}$

Solution 3

Although this may not be the most mathematically rigorous answer, we see that $\frac{P(x+1)}{P(x)}=\frac{x+2}{x-1}$. Using a bit of logic, we can make a guess that $P(x+1)$ has a factor of $x+2$, telling us $P(x)$ has a factor of $x+1$. Similarly, we guess that $P(x)$ has a factor of $x-1$, which means $P(x+1)$ has a factor of $x$. Now, since $P(x)$ and $P(x+1)$ have so many factors that are off by one, we may surmise that when you plug $x+1$ into $P(x)$, the factors "shift over," i.e. $P(x)=(A)(A+1)(A+2)...(A+n)$, which goes to $P(x+1)=(A+1)(A+2)(A+3)...(A+n+1)$. This is useful because these, when divided, result in $\frac{P(x+1)}{P(x)}=\frac{A+n+1}{A}$. If $\frac{A+n+1}{A}=\frac{x+2}{x-1}$, then we get $A=x-1$ and $A+n+1=x+2$, $n=2$. This gives us $P(x)=(x-1)x(x+1)$ and $P(x+1)=x(x+1)(x+2)$, and at this point we realize that there has to be some constant $a$ multiplied in front of the factors, which won't affect our fraction $\frac{P(x+1)}{P(x)}=\frac{x+2}{x-1}$ but will give us the correct values of $P(2)$ and $P(3)$. Thus $P(x)=a(x-1)x(x+1)$, and we utilize $P(2)^2=P(3)$ to find $a=\frac{2}{3}$. Evaluating $P \left ( \frac{7}{2} \right )$ is then easy, and we see it equals $\frac{105}{4}$, so the answer is $\boxed{109}$

Solution 4

Substituting $x=2$ into the given equation, we find that $P(3)=4P(2)=P(2)^2$. Therefore, either $P(2)=0$ or $P(2)=4$. Now for integers $n\ge 2$, we know that \[P(n+1)=\frac{n+2}{n-1}P(n).\] Applying this repeatedly, we find that \[P(n+1)=\frac{(n+2)!/3!}{(n-1)!}P(2).\] If $P(2)=0$, this shows that $P(x)$ has infinitely many roots, meaning that $P(x)$ is identically equal to zero. But this contradicts the problem statement. Therefore, $P(2)=4$, and we find $P(n+1)=\frac{2}{3}(n+2)(n+1)n$ for all positive integers $n\ge2$. This cubic polynomial matches the values $P(n+1)$ for infinitely many numbers, hence the two polynomials are identically equal. In particular, $P\left(\frac72\right)=\frac23\cdot\frac92\cdot\frac72\cdot\frac52=\frac{105}{4}$, and the answer is $\boxed{109}$.

Solution 5

We can find zeroes of the polynomial by making the first given equation $0 = 0$. Plugging in $x = 1$ and $x = -2$ gives us the zeroes $1$ and $-1$, respectively. Now we can plug in these zeros to get more zeroes. $x = -1$ gives us the zero $0$ (no pun intended). $x = 1$ makes the equation $0 \cdot P(2) = 0$, which means $P(2)$ is not necessarily $0$. If $P(2) = 0$, then plugging in $2$ to the equation yields $P(3) = 0$, plugging in $3$ to the equation yields $P(4) = 0$, and so on, a contradiction of "nonzero polynomial". So $2$ is not a zero. Note that plugging in $x = 0$ to the equation does not yield any additional zeros. Thus, the only zeroes of $P(x)$ are $-1, 0,$ and $1$, so $P(x) = a(x + 1)x(x - 1)$ for some nonzero constant $a$. We can plug in $2$ and $3$ into the polynomial and use the second given equation to find an equation for $a$. $P(2) = 6a$ and $P(3) = 24a$, so: \[(6a)^2 = 24a \implies 36a^2 = 24a \implies a = \frac23\] Plugging in $\frac72$ into the polynomial $\frac23(x + 1)x(x - 1)$ yields $\frac{105}{4}$. $105 + 4 = \boxed{109}$.

Solution 6

Plug in $x=2$ yields $P(3)=4P(2)$. Since also $(P(2))^2=P(3)$, we have $P(2)=4$ and $P(3)=16$. Plug in $x=3$ yields $2P(4)=5P(3)$ so $P(4)=40$.

Repeat the action gives $P(2)=4$, $P(3)=16$, $P(4)=40$, $P(5)=80$, and $P(6)=140$.

Since $P(x)$ is a polynomial, the $k$th difference is constant, where $k=\deg(P(x))$. Thus we can list out the 0th, 1st, 2nd, 3rd, ... differences until we obtain a constant.

\[4,16,40,80,140\] \[12,24,40,60\] \[12,16,20\] \[4,4,4\]

Since the 3rd difference of $P(x)$ is constant, we can conclude that $\deg(P(x))=3$.

Let $P(x)=a_3x^3+a_2x^2+a_1x+a_0$. Plug in the values for $x$ and solve the system of 4 equations gives $(a_3,a_2,a_1,a_0)=(\frac{2}{3},0,-\frac{2}{3},0)$

Thus $P(x)=\frac{2}{3}x^3-\frac{2}{3}x$ and $P(\frac{7}{2})=\frac{105}{4}\Longrightarrow m+n=\boxed{109}$

~ Nafer

-Note that the coefficient of $x^3$ is $\frac{2}{3}$ because $\frac{2}{3}\cdot\deg(x^3)!=\frac{2}{3}\cdot3!=4$, the 3rd difference of P(x). ~inaccessibles

See also

2016 AIME I (Problems • Answer Key • Resources)
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