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

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==Solution 1==
 
==Solution 1==
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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 </math>(-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>
+
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>. 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+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>(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>
 
<cmath>Q(x+1) = Q(x)</cmath>
−
From here, we know that </math>Q(x) = C<math> for a constant </math>C<math>, 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}$.
+
From here, we know that <math>Q(x) = C</math> for a constant <math>C</math>, 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>.
  
 
==Solution 2==
 
==Solution 2==

Revision as of 11:48, 6 March 2016

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 $(0-1)P(0+1) = (0+2)P(0)\Rightarrow P(0) = -\frac{1}{2}P(1) = 0$. 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 $C$, 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

So 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 $P(x)$. 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 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

As above, we find that $P(2)=4$. 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 \[P(n+1)=\frac{(n+2)!/3!}{(n-1)!}P(2).\] Therefore, as $P(2)=4$, 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 $\boxed{109}$.

See also

2016 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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