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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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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>.
 
  
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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
 
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<cmath>a+b+c=0</cmath>
 
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<cmath>8a+4b+2c=4</cmath>
 
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<cmath>27a+9b+3c=16</cmath>
 
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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>.
 
 
==Solution 2==
 
==Solution 2==
 
So 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>. 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.  
 
So 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>. 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.  

Revision as of 08:05, 5 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

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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