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Difference between revisions of "2016 AIME I Problems/Problem 11"
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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.  
 
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.  
  
−
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>
+
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{3}{2}</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>
 +
 
 
== See also ==
 
== See also ==
 
{{AIME box|year=2016|n=I|num-b=10|num-a=12}}
 
{{AIME box|year=2016|n=I|num-b=10|num-a=12}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 17:20, 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 $c = \frac{3}{2}$. 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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