Difference between revisions of "2009 AMC 12A Problems/Problem 23"
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The answer is <math>r_4 - r_1 = (-r_1) - r_1 = - 150u = 450 + 300\sqrt {2}</math>, and <math>450 + 300 + 2 = 752\ \mathbf{(D)}</math>. | The answer is <math>r_4 - r_1 = (-r_1) - r_1 = - 150u = 450 + 300\sqrt {2}</math>, and <math>450 + 300 + 2 = 752\ \mathbf{(D)}</math>. | ||
+ | |||
+ | == Note == | ||
+ | Actually it is not necessary to solve any quadratic equations, if one utilizes the two facts about the quadratic <math>ax^2+bx+c</math> (<math>a>0</math>) that (i) the difference of the two quadratic roots equals to <math>\sqrt{\Delta}/a</math>, and (ii) that the minimum value of a quadratic equals to <math>-\Delta/4a</math>, where <math>\Delta=b^2-4ac</math>. Here is a possible adjustment to the solution: | ||
+ | |||
+ | Without loss of generality we may "shift" <math>f</math>,<math>g</math>,<math>x_1,x_2,x_3,x_4</math> <math>50</math> units to the left, then the differences of <math>x_i</math> remain the same, <math>x_3</math> and <math>x_2</math> are symmetrical about <math>0</math>, so <math>x_2=-75</math>, <math>x_3=75</math>. The relationship of <math>f</math>,<math>g</math> becomes <math>g(x)=-f(-x)</math>. So we may write: | ||
+ | |||
+ | <cmath>f(x)=a(x-b)^2 + m</cmath> | ||
+ | <cmath>g(x)=-a(x+b)^2 - m</cmath> | ||
+ | |||
+ | Again without loss of generality, we can assume <math>a>0</math> and <math>b>0</math> (Short argument is needed here instead of the lazy "wlog"). Also, the vertex of <math>f</math> is <math>(b,m)</math>, so <math>m=g(b)=-4ab^2-m</math>, or <math>m=-2ab^2 <0</math>. | ||
+ | |||
+ | Since <math>x_2,x_4</math> are roots of <math>f</math>, we have the following relationship of the roots: | ||
+ | <cmath>x_2 + x_4 = 2b</cmath> | ||
+ | <cmath>x_4 - x_2 = \sqrt{\Delta}/a = \sqrt{-4am}/a = 2\sqrt{-m/a} = 2\sqrt{2b^2} = 2\sqrt{2}b = \sqrt{2} (x_2+x_4) </cmath> | ||
+ | |||
+ | So <math>(\sqrt{2}-1)x_4 = 75(1+\sqrt{2})</math>, or <math>x_4 = 75(1+\sqrt{2})^2 = 75(3+2\sqrt{2})</math>. | ||
+ | |||
+ | Therefore <math>x_4-x_1=2x_4=450+300\sqrt{2}</math>. | ||
== See also == | == See also == |
Revision as of 01:57, 4 December 2012
Contents
Problem
Functions and
are quadratic,
, and the graph of
contains the vertex of the graph of
. The four
-intercepts on the two graphs have
-coordinates
,
,
, and
, in increasing order, and
. The value of
is
, where
,
, and
are positive integers, and
is not divisible by the square of any prime. What is
?
Solution
![[asy] import graph; size(250);Label k; k.p=fontsize(6); int ymax = 400, ymin = -400; real rt = 175+150*2^.5; real f(real x){return 1/400*(x-125)*(x+rt);} real g(real x){return -f(100-x);} xaxis(-600,600,Ticks(k, 5),Arrows(6));yaxis(ymin,ymax,Ticks(k, 5),Arrows(6)); draw(graph(f,-450,300),red+linewidth(0.8),Arrows(6));draw(graph(g,-200,550),blue+linewidth(0.8),Arrows(6)); draw((50,ymax)--(50,ymin),linetype("4 4"),Arrows(4));dot((-rt,0));dot((100+rt,0));dot((-25,0));dot((125,0)); [/asy]](http://latex.artofproblemsolving.com/e/a/8/ea8ac43e7941f88d8b25a44e8f402a22b1617cac.png)
The two quadratics are rotations of each other about
. Since we are only dealing with differences of roots, we can translate them to be symmetric about
. Now
and
. Say our translated versions of
and
are
and
, respectively, so that
. Let
be a root of
and
a root of
by symmetry. Note that since they each contain each other's vertex,
,
,
, and
must be roots of alternating polynomials, so
is a root of
and
a root of
![\[p(x) = a(x - 75)(x - x_1) \\ q(x) = - a(x + 75)(x + x_1)\]](http://latex.artofproblemsolving.com/6/6/0/6609a050f7441e7849d1e2eb5629cfa50f3aaa9d.png)
The vertex of is half the sum of its roots, or
. We are told that the vertex of one quadratic lies on the other, so
![\begin{eqnarray*} p\left(\frac {75 + x_1}{2}\right) & = & a\left(\frac {75 - x_1}{2}\right)\left(\frac { - 75 + x_1}{2}\right) \\ & = & - \frac {a}{4}(x_1 - 75)^2 \\ - \frac {a}{4}(x_1 - 75)^2 & = & q\left(\frac {75 + x_1}{2}\right) \\ & = & - a\left(\frac {x_1 + 225}{2}\right)\left(\frac {3x_1 + 75}{2}\right) \\ & = & - \frac {a}{4}(x_1 + 225)(3x_1 + 75) \end{eqnarray*}](http://latex.artofproblemsolving.com/1/b/c/1bca712494ac3ccd8bd454be08ab880255e2d26b.png)
Let and divide through by
, since this is a timed competition and it will drastically simplify computations. We know
and that
, or
![\begin{eqnarray*} 0 & = & (3u + 1)(u + 3) - (u - 1)^2 \\ & = & 3u^2 + 10u + 3 - (u^2 - 2u + 1) \\ & = & 2u^2 + 12u + 2 \\ & = & u^2 + 6u + 1 \end{eqnarray*}](http://latex.artofproblemsolving.com/3/6/9/369c373ca02974656cb1caba542ca177c3397fdc.png)
So . Since
,
.
The answer is , and
.
Note
Actually it is not necessary to solve any quadratic equations, if one utilizes the two facts about the quadratic (
) that (i) the difference of the two quadratic roots equals to
, and (ii) that the minimum value of a quadratic equals to
, where
. Here is a possible adjustment to the solution:
Without loss of generality we may "shift" ,
,
units to the left, then the differences of
remain the same,
and
are symmetrical about
, so
,
. The relationship of
,
becomes
. So we may write:
Again without loss of generality, we can assume and
(Short argument is needed here instead of the lazy "wlog"). Also, the vertex of
is
, so
, or
.
Since are roots of
, we have the following relationship of the roots:
So , or
.
Therefore .
See also
2009 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 22 |
Followed by Problem 24 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |