Difference between revisions of "2010 AIME I Problems/Problem 6"
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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>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: | ||
− | <math>f(11) = 99a + c = 181</math> | + | <math>f(11) = 99a + c = 181</math>, and <math>f(1) = -a + c = 1</math>. |
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>. | 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>. |
Revision as of 10:48, 18 March 2010
Contents
[hide]Problem
Let be a quadratic polynomial with real coefficients satisfying
for all real numbers
, and suppose
. Find
.
Solution
![[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
. Substituting
and
we obtain two equations:
, and
.
Solving, we get and
, so
. Therefore,
.
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 |