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Difference between revisions of "2010 AIME I Problems/Problem 6"

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== Problem ==
 
== Problem ==
 
Let <math>P(x)</math> be a [[quadratic]] polynomial with real coefficients satisfying <math>x^2 - 2x + 2 \le P(x) \le 2x^2 - 4x + 3</math> for all real numbers <math>x</math>, and suppose <math>P(11) = 181</math>. Find <math>P(16)</math>.
 
Let <math>P(x)</math> be a [[quadratic]] polynomial with real coefficients satisfying <math>x^2 - 2x + 2 \le P(x) \le 2x^2 - 4x + 3</math> for all real numbers <math>x</math>, and suppose <math>P(11) = 181</math>. Find <math>P(16)</math>.
  
 
== Solution ==
 
== Solution ==
 +
=== Solution 1 ===
 
<center><asy>
 
<center><asy>
 
import graph; real min = -0.5, max = 2.5; pen dark = linewidth(1);
 
import graph; real min = -0.5, max = 2.5; pen dark = linewidth(1);
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Also, <math>1 = Q(1) \le P(1) \le R(1) = 1</math>, so <math>P(1) = 1</math>, and <math>P</math> obtains its minimum at the point <math>(1,1)</math>. Then <math>P(x)</math> must be of the form <math>c(x-1)^2 + 1</math> for some constant <math>c</math>; substituting <math>P(11) = 181</math> yields <math>c = \frac 95</math>. Finally, <math>P(16) = \frac 95 \cdot (16 - 1)^2 + 1 = \boxed{406}</math>.
 
Also, <math>1 = Q(1) \le P(1) \le R(1) = 1</math>, so <math>P(1) = 1</math>, and <math>P</math> obtains its minimum at the point <math>(1,1)</math>. Then <math>P(x)</math> must be of the form <math>c(x-1)^2 + 1</math> for some constant <math>c</math>; substituting <math>P(11) = 181</math> yields <math>c = \frac 95</math>. Finally, <math>P(16) = \frac 95 \cdot (16 - 1)^2 + 1 = \boxed{406}</math>.
  
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== Solution 2 ==
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=== Solution 2 ===
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It can be seen that the function <math>f(x)</math> must be in the form <math>f(x) = ax^2 - 2ax + c</math> for some real <math>a</math> and <math>c</math>. This is because the [[derivative]] of <math>f(x)</math> is <math>2ax - 2a</math>, and a global minimum occurs only at <math>x = 1</math>. Substituting <math>(1,1)</math> and <math>(11, 181)</math> we obtain two equations:
+
It can be seen that the function <math>P(x)</math> must be in the form <math>P(x) = ax^2 - 2ax + c</math> for some real <math>a</math> and <math>c</math>. This is because the [[derivative]] of <math>P(x)</math> is <math>2ax - 2a</math>, and a global minimum occurs only at <math>x = 1</math> (in addition, because of this derivative, the vertex of any quadratic polynomial occurs at <math>\frac{-b}{2a}</math>). Substituting <math>(1,1)</math> and <math>(11, 181)</math> we obtain two equations:
  
−
<math>f(11) = 99a + c = 181</math>                                     <math>f(1) = -a + c = 1</math>.
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<center><math>P(11) = 99a + c = 181</math>, and <math>P(1) = -a + c = 1</math>.</center>
  
−
Solving, we get <math>a = \frac{9}{5}</math> and <math>c = \frac{14}{5}</math>, so <math>f(x) = \frac{9}{5}x^2 - \frac{18}{5}x + \frac {14}{5}</math>. Therefore, <math>f(16) = \boxed{406}</math>.
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Solving, we get <math>a = \frac{9}{5}</math> and <math>c = \frac{14}{5}</math>, so <math>P(x) = \frac{9}{5}x^2 - \frac{18}{5}x + \frac {14}{5}</math>. Therefore, <math>P(16) = \boxed{406}</math>.
  
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== See also ==
+
=== Solution 3 ===
 +
 
 +
Let <math>y = x^2 - 2x + 2</math>; note that <math>2y - 1 = 2x^2 - 4x + 3</math>. Setting <math>y = 2y - 1</math>, we find that equality holds when <math>y = 1</math> and therefore when <math>x^2 - 2x + 2 = 1</math>; this is true iff <math>x = 1</math>, so <math>P(1) = 1</math>.
 +
 
 +
Let <math>Q(x) = P(x) - x</math>; clearly <math>Q(1) = 0</math>, so we can write <math>Q(x) = (x - 1)Q'(x)</math>, where <math>Q'(x)</math> is some linear function. Plug <math>Q(x)</math> into the given inequality:
 +
 
 +
<math>x^2 - 3x + 2 \le Q(x) \le 2x^2 - 5x + 3</math>
 +
 
 +
<math>(x - 1)(x - 2) \le (x - 1)Q'(x) \le (x - 1)(2x - 3)</math>, and thus
 +
 
 +
<math>x - 2 \le Q'(x) \le 2x - 3</math>
 +
 
 +
For all <math>x > 1</math>; note that the inequality signs are flipped if <math>x < 1</math>, and that the division is invalid for <math>x = 1</math>. However,
 +
 
 +
<math>\lim_{x \to 1}  x - 2 = \lim_{x \to 1}  2x - 3 = -1</math>,
 +
 
 +
and thus by the [[sandwich theorem]] <math>\lim_{x \to 1} Q'(x) = -1</math>; by the definition of a continuous function, <math>Q'(1) = -1</math>. Also, <math>Q(11) = 170</math>, so <math>Q'(11) = 170/(11-1) = 17</math>; plugging in and solving, <math>Q'(x) = (9/5)(x - 1) - 1</math>. Thus <math>Q(16) = 390</math>, and so <math>P(16) = \boxed{406}</math>.
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 +
=== Solution 4 ===
 +
 
 +
Let <math>Q(x) = P(x) - (x^2-2x+2)</math>, then <math>0\le Q(x) \le (x-1)^2</math> (note this is derived from the given inequality chain).  Therefore, <math>0\le Q(x+1) \le x^2 \Rightarrow Q(x+1) = Ax^2</math> for some real value A.
 +
 
 +
<math>Q(11) = 10^2A \Rightarrow P(11)-(11^2-22+2)=100A \Rightarrow 80=100A \Rightarrow A=\frac{4}{5}</math>.
 +
 
 +
<math>Q(16)=15^2A=180 \Rightarrow P(16)-(16^2-32+2) = 180 \Rightarrow P(16)=180+226= \boxed{406}</math>
 +
 
 +
=== Solution 5 ===
 +
 
 +
Let <math>P(x) = ax^2 + bx + c</math>. Plugging in <math>x = 1</math> to the expressions on both sides of the inequality, we see that <math>a + b + c = 1</math>. We see from the problem statement that <math>121a + 11b + c = 181</math>. Since we know the vertex of <math>P(x)</math> lies at <math>x = 1</math>, by symmetry we get <math>81a -9b + c = 181</math> as well. Since we now have three equations, we can solve this trivial system and get our answer of <math>\boxed{406}</math>.
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 +
=== Solution 6 ===
 +
 
 +
Similar to Solution 5, let <math>P(x) = ax^2 + bx + c</math>. Note that <math>(1,1)</math> is a vertex of the polynomial. Additionally, this means that <math>b = -2a</math> (since <math>\frac{-b}{2a}</math> is the minimum <math>x</math> point). Thus, we have <math>P(x) = ax^2 - 2ax + c</math>. Therefore <math>a - 2a + c = 1</math>. Moreover, <math>99a + c = 181</math>. And so our polynomial is <math>\frac{9}{5}x^2 - \frac{18}{5}x + \frac{14}{5}</math>. Plug in <math>x = 16</math> to get <math>\boxed{406}</math>.
 +
 
 +
=== Solution 7 ===
 +
 
 +
Very similar to Solution 6, start by noticing that <math>P(x)</math> is between two polynomials, we try to set them equal to find a point where the two polynomials intersect, meaning that <math>P(x)</math> would also have to intersect that point (it must be between the two graphs). Setting <math>x^2 - 2x + 2 = 2x^2 - 4x + 3</math>, we find that <math>x = 1</math>. Note that both of these graphs have the same vertex (at <math>x = 1</math>), and so <math>P(x)</math> must also have the same vertex <math>(1, 1)</math>. Setting <math>P(x) = ax^2 - 2ax + a + 1</math> (this is where we have a vertex at <math>(1, 1)</math>), we plug in <math>11</math> and find that <math>a = 1.8</math>. Evaluating <math>1.8x^2 - 3.6x + 2.8</math> when <math>x = 16</math> (our intended goal), we find that <math>P(16) = \boxed{406}</math>.
 +
 
 +
== See Also ==
 
{{AIME box|year=2010|num-b=5|num-a=7|n=I}}
 
{{AIME box|year=2010|num-b=5|num-a=7|n=I}}
  
 
[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]
 +
{{MAA Notice}}

Latest revision as of 19:39, 20 November 2023

Problem

Let $P(x)$ be a quadratic polynomial with real coefficients satisfying $x^2 - 2x + 2 \le P(x) \le 2x^2 - 4x + 3$ for all real numbers $x$, and suppose $P(11) = 181$. Find $P(16)$.

Solution

Solution 1

[asy] import graph; real min = -0.5, max = 2.5; pen dark = linewidth(1); real P(real x) { return 8*(x-1)^2/5+1; } real Q(real x) { return (x-1)^2+1; } real R(real x) { return 2*(x-1)^2+1; } draw(graph(P,min,max),dark); draw(graph(Q,min,max),linetype("6 2")+linewidth(0.7)); draw(graph(R,min,max),linetype("6 2")+linewidth(0.7)); dot((1,1)); label("$P(x)$",(max,P(max)),E,fontsize(10)); label("$Q(x)$",(max,Q(max)),E,fontsize(10)); label("$R(x)$",(max,R(max)),E,fontsize(10)); /* axes */ Label f; f.p=fontsize(8); xaxis(-2, 3, Ticks(f, 5, 1)); yaxis(-1, 5, Ticks(f, 6, 1)); [/asy]

Let $Q(x) = x^2 - 2x + 2$, $R(x) = 2x^2 - 4x + 3$. Completing the square, we have $Q(x) = (x-1)^2 + 1$, and $R(x) = 2(x-1)^2 + 1$, so it follows that $P(x) \ge Q(x) \ge 1$ for all $x$ (by the Trivial Inequality).

Also, $1 = Q(1) \le P(1) \le R(1) = 1$, so $P(1) = 1$, and $P$ obtains its minimum at the point $(1,1)$. Then $P(x)$ must be of the form $c(x-1)^2 + 1$ for some constant $c$; substituting $P(11) = 181$ yields $c = \frac 95$. Finally, $P(16) = \frac 95 \cdot (16 - 1)^2 + 1 = \boxed{406}$.

Solution 2

It can be seen that the function $P(x)$ must be in the form $P(x) = ax^2 - 2ax + c$ for some real $a$ and $c$. This is because the derivative of $P(x)$ is $2ax - 2a$, and a global minimum occurs only at $x = 1$ (in addition, because of this derivative, the vertex of any quadratic polynomial occurs at $\frac{-b}{2a}$). Substituting $(1,1)$ and $(11, 181)$ we obtain two equations:

$P(11) = 99a + c = 181$, and $P(1) = -a + c = 1$.

Solving, we get $a = \frac{9}{5}$ and $c = \frac{14}{5}$, so $P(x) = \frac{9}{5}x^2 - \frac{18}{5}x + \frac {14}{5}$. Therefore, $P(16) = \boxed{406}$.

Solution 3

Let $y = x^2 - 2x + 2$; note that $2y - 1 = 2x^2 - 4x + 3$. Setting $y = 2y - 1$, we find that equality holds when $y = 1$ and therefore when $x^2 - 2x + 2 = 1$; this is true iff $x = 1$, so $P(1) = 1$.

Let $Q(x) = P(x) - x$; clearly $Q(1) = 0$, so we can write $Q(x) = (x - 1)Q'(x)$, where $Q'(x)$ is some linear function. Plug $Q(x)$ into the given inequality:

$x^2 - 3x + 2 \le Q(x) \le 2x^2 - 5x + 3$

$(x - 1)(x - 2) \le (x - 1)Q'(x) \le (x - 1)(2x - 3)$, and thus

$x - 2 \le Q'(x) \le 2x - 3$

For all $x > 1$; note that the inequality signs are flipped if $x < 1$, and that the division is invalid for $x = 1$. However,

$\lim_{x \to 1} x - 2 = \lim_{x \to 1} 2x - 3 = -1$,

and thus by the sandwich theorem $\lim_{x \to 1} Q'(x) = -1$; by the definition of a continuous function, $Q'(1) = -1$. Also, $Q(11) = 170$, so $Q'(11) = 170/(11-1) = 17$; plugging in and solving, $Q'(x) = (9/5)(x - 1) - 1$. Thus $Q(16) = 390$, and so $P(16) = \boxed{406}$.

Solution 4

Let $Q(x) = P(x) - (x^2-2x+2)$, then $0\le Q(x) \le (x-1)^2$ (note this is derived from the given inequality chain). Therefore, $0\le Q(x+1) \le x^2 \Rightarrow Q(x+1) = Ax^2$ for some real value A.

$Q(11) = 10^2A \Rightarrow P(11)-(11^2-22+2)=100A \Rightarrow 80=100A \Rightarrow A=\frac{4}{5}$.

$Q(16)=15^2A=180 \Rightarrow P(16)-(16^2-32+2) = 180 \Rightarrow P(16)=180+226= \boxed{406}$

Solution 5

Let $P(x) = ax^2 + bx + c$. Plugging in $x = 1$ to the expressions on both sides of the inequality, we see that $a + b + c = 1$. We see from the problem statement that $121a + 11b + c = 181$. Since we know the vertex of $P(x)$ lies at $x = 1$, by symmetry we get $81a -9b + c = 181$ as well. Since we now have three equations, we can solve this trivial system and get our answer of $\boxed{406}$.

Solution 6

Similar to Solution 5, let $P(x) = ax^2 + bx + c$. Note that $(1,1)$ is a vertex of the polynomial. Additionally, this means that $b = -2a$ (since $\frac{-b}{2a}$ is the minimum $x$ point). Thus, we have $P(x) = ax^2 - 2ax + c$. Therefore $a - 2a + c = 1$. Moreover, $99a + c = 181$. And so our polynomial is $\frac{9}{5}x^2 - \frac{18}{5}x + \frac{14}{5}$. Plug in $x = 16$ to get $\boxed{406}$.

Solution 7

Very similar to Solution 6, start by noticing that $P(x)$ is between two polynomials, we try to set them equal to find a point where the two polynomials intersect, meaning that $P(x)$ would also have to intersect that point (it must be between the two graphs). Setting $x^2 - 2x + 2 = 2x^2 - 4x + 3$, we find that $x = 1$. Note that both of these graphs have the same vertex (at $x = 1$), and so $P(x)$ must also have the same vertex $(1, 1)$. Setting $P(x) = ax^2 - 2ax + a + 1$ (this is where we have a vertex at $(1, 1)$), we plug in $11$ and find that $a = 1.8$. Evaluating $1.8x^2 - 3.6x + 2.8$ when $x = 16$ (our intended goal), we find that $P(16) = \boxed{406}$.

See Also

2010 AIME I (Problems • Answer Key • Resources)
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