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Difference between revisions of "2016 AIME I Problems/Problem 11"
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==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>.
 
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>.
  
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==Solution 1==
 
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>.
 
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 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>.
 
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>.
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==Solution 2==
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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.
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Suppose we had another root that is not 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.
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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>
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== See also ==
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{{AIME box|year=2016|n=I|num-b=10|num-a=12}}
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{{MAA Notice}}

Revision as of 17:19, 4 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

We substitute $x=2$ into $(x-1)P(x+1)=(x+2)P(x)$ to get $P(3)=4P(2)$. Since we also have that $\left(P(2)\right)^2 = P(3)$, we have that $P(2)=4$ and $P(3)=16$. We can also substitute $x=1$, $x=0$, and $x=3$ into $(x-1)P(x+1)=(x+2)P(x)$ to get that $0=P(1)$, $-1P(1)=2P(0)$, and $2P(4)=5P(3)$. This leads us to the conclusion that $P(0)=P(1)=0$ and $P(4)=40$.

We next use finite differences to find that $P$ is a cubic polynomial. Thus, $P$ must be of the form of $ax^3+bx^2+cx+d$. It follows that $d=0$; we now have a system of $3$ equations to solve. We plug in $x=1$, $x=2$, and $x=3$ to get

\[a+b+c=0\] \[8a+4b+2c=4\] \[27a+9b+3c=16\]

We solve this system to get that $a=\frac{3}{2}$, $b=0$, and $c=-\frac{3}{2}$. Thus, $P(x)=\frac{3}{2}x^3-\frac{3}{2}x$. Plugging in $x=\frac{7}{2}$, we see that $P\left(\frac{7}{2}\right)=\frac{105}{4}$. Thus, $m=105$, $n=4$, and our answer is $m+n=\boxed{109}$.

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 $\boxed{109}$

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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