Difference between revisions of "2010 AIME I Problems/Problem 6"
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=== Solution 4 === | === Solution 4 === | ||
− | Let <math>Q(x) = P(x) - (x^2-2x+2)</math>, then <math>0\le Q(x) \le (x-1)^2</math>. Therefore, <math>0\le Q(x+1) \le x^2 \Rightarrow Q(x) = Ax^2</math> for some real value A. | + | 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(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> | <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>. | ||
+ | |||
+ | === 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 == | == See Also == |
Latest revision as of 20:39, 20 November 2023
Contents
Problem
Let be a quadratic polynomial with real coefficients satisfying
for all real numbers
, and suppose
. Find
.
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]](http://latex.artofproblemsolving.com/8/6/4/8642816d5b50f02498e44e455af5a0ae225ce976.png)
Let ,
. Completing the square, we have
, and
, so it follows that
for all
(by the Trivial Inequality).
Also, , so
, and
obtains its minimum at the point
. Then
must be of the form
for some constant
; substituting
yields
. Finally,
.
Solution 2
It can be seen that the function must be in the form
for some real
and
. This is because the derivative of
is
, and a global minimum occurs only at
(in addition, because of this derivative, the vertex of any quadratic polynomial occurs at
). Substituting
and
we obtain two equations:
![$P(11) = 99a + c = 181$](http://latex.artofproblemsolving.com/1/2/8/1280d75a5502cedf7bcf254923bdcf8b62164e8c.png)
![$P(1) = -a + c = 1$](http://latex.artofproblemsolving.com/0/3/9/0391d44108cc4deef2a67c6c1ab842904914dd26.png)
Solving, we get and
, so
. Therefore,
.
Solution 3
Let ; note that
. Setting
, we find that equality holds when
and therefore when
; this is true iff
, so
.
Let ; clearly
, so we can write
, where
is some linear function. Plug
into the given inequality:
, and thus
For all ; note that the inequality signs are flipped if
, and that the division is invalid for
. However,
,
and thus by the sandwich theorem ; by the definition of a continuous function,
. Also,
, so
; plugging in and solving,
. Thus
, and so
.
Solution 4
Let , then
(note this is derived from the given inequality chain). Therefore,
for some real value A.
.
Solution 5
Let . Plugging in
to the expressions on both sides of the inequality, we see that
. We see from the problem statement that
. Since we know the vertex of
lies at
, by symmetry we get
as well. Since we now have three equations, we can solve this trivial system and get our answer of
.
Solution 6
Similar to Solution 5, let . Note that
is a vertex of the polynomial. Additionally, this means that
(since
is the minimum
point). Thus, we have
. Therefore
. Moreover,
. And so our polynomial is
. Plug in
to get
.
Solution 7
Very similar to Solution 6, start by noticing that is between two polynomials, we try to set them equal to find a point where the two polynomials intersect, meaning that
would also have to intersect that point (it must be between the two graphs). Setting
, we find that
. Note that both of these graphs have the same vertex (at
), and so
must also have the same vertex
. Setting
(this is where we have a vertex at
), we plug in
and find that
. Evaluating
when
(our intended goal), we find that
.
See Also
2010 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.