Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2009 AMC 12A Problems/Problem 23"

m (→‎Solution: added space)
m (still trying to add space)
Line 16: Line 16:
 
The two quadratics are <math>180^{\circ}</math> rotations of each other about <math>(50,0)</math>. Since we are only dealing with differences of roots, we can translate them to be symmetric about <math>(0,0)</math>. Now <math>x_3 = - x_2 = 75</math> and <math>x_4 = - x_1</math>. Say our translated versions of <math>f</math> and <math>g</math> are <math>p</math> and <math>q</math>, respectively, so that <math>p(x) = - q( - x)</math>. Let <math>x_3 = 75</math> be a root of <math>p</math> and <math>x_2 = - 75</math> a root of <math>q</math> by symmetry. Note that since they each contain each other's vertex, <math>x_1</math>, <math>x_2</math>, <math>x_3</math>, and <math>x_4</math> must be roots of alternating polynomials, so <math>x_1</math> is a root of <math>p</math> and <math>x_4</math> a root of <math>q</math>
 
The two quadratics are <math>180^{\circ}</math> rotations of each other about <math>(50,0)</math>. Since we are only dealing with differences of roots, we can translate them to be symmetric about <math>(0,0)</math>. Now <math>x_3 = - x_2 = 75</math> and <math>x_4 = - x_1</math>. Say our translated versions of <math>f</math> and <math>g</math> are <math>p</math> and <math>q</math>, respectively, so that <math>p(x) = - q( - x)</math>. Let <math>x_3 = 75</math> be a root of <math>p</math> and <math>x_2 = - 75</math> a root of <math>q</math> by symmetry. Note that since they each contain each other's vertex, <math>x_1</math>, <math>x_2</math>, <math>x_3</math>, and <math>x_4</math> must be roots of alternating polynomials, so <math>x_1</math> is a root of <math>p</math> and <math>x_4</math> a root of <math>q</math>
  
−
<center><cmath>p(x) = a(x - 75)(x - x_1) \\
+
<center><cmath>p(x) = a(x - 75)(x - x_1) </center>
−
q(x) = - a(x + 75)(x + x_1)</cmath></center>
+
<center></cmath>q(x) = - a(x + 75)(x + x_1)</center>
  
 
The vertex of <math>p(x)</math> is half the sum of its roots, or <math>\frac {75 + x_1}{2}</math>. We are told that the vertex of one quadratic lies on the other, so
 
The vertex of <math>p(x)</math> is half the sum of its roots, or <math>\frac {75 + x_1}{2}</math>. We are told that the vertex of one quadratic lies on the other, so

Revision as of 11:21, 20 July 2017

Problem

Functions $f$ and $g$ are quadratic, $g(x) = - f(100 - x)$, and the graph of $g$ contains the vertex of the graph of $f$. The four $x$-intercepts on the two graphs have $x$-coordinates $x_1$, $x_2$, $x_3$, and $x_4$, in increasing order, and $x_3 - x_2 = 150$. The value of $x_4 - x_1$ is $m + n\sqrt p$, where $m$, $n$, and $p$ are positive integers, and $p$ is not divisible by the square of any prime. What is $m + n + p$?

$\textbf{(A)}\ 602\qquad \textbf{(B)}\ 652\qquad \textbf{(C)}\ 702\qquad \textbf{(D)}\ 752 \qquad \textbf{(E)}\ 802$

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]

The two quadratics are $180^{\circ}$ rotations of each other about $(50,0)$. Since we are only dealing with differences of roots, we can translate them to be symmetric about $(0,0)$. Now $x_3 = - x_2 = 75$ and $x_4 = - x_1$. Say our translated versions of $f$ and $g$ are $p$ and $q$, respectively, so that $p(x) = - q( - x)$. Let $x_3 = 75$ be a root of $p$ and $x_2 = - 75$ a root of $q$ by symmetry. Note that since they each contain each other's vertex, $x_1$, $x_2$, $x_3$, and $x_4$ must be roots of alternating polynomials, so $x_1$ is a root of $p$ and $x_4$ a root of $q$

\[p(x) = a(x - 75)(x - x_1) </center> <center>\]q(x) = - a(x + 75)(x + x_1)

The vertex of $p(x)$ is half the sum of its roots, or $\frac {75 + x_1}{2}$. 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*}

Let $x_1 = 75u$ and divide through by $75^2$, since this is a timed competition and it will drastically simplify computations. We know $u < - 1$ and that $(u - 1)^2 = (3u + 1)(u + 3)$, 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*}

So $u = \frac { - 6\pm\sqrt {32}}{2} = - 3\pm2\sqrt2$. Since $u < - 1$, $u = - 3 - 2\sqrt2$.

The answer is $r_4 - r_1 = (-r_1) - r_1 = - 150u = 450 + 300\sqrt {2}$, and $450 + 300 + 2 = 752\ \mathbf{(D)}$.

Note

Actually it is not necessary to solve any quadratic equations, if one utilizes the two facts about the quadratic $ax^2+bx+c$ ($a>0$) that (i) the difference of the two quadratic roots equals to $\sqrt{\Delta}/a$, and (ii) that the minimum value of a quadratic equals to $-\Delta/4a$, where $\Delta=b^2-4ac$. Here is a possible adjustment to the solution:

Without loss of generality we may "shift" $f$,$g$,$x_1,x_2,x_3,x_4$ $50$ units to the left, then the differences of $x_i$ remain the same, $x_3$ and $x_2$ are symmetrical about $0$, so $x_2=-75$, $x_3=75$. The relationship of $f$,$g$ becomes $g(x)=-f(-x)$. So we may write:

\[f(x)=a(x-b)^2 + m\] \[g(x)=-a(x+b)^2 - m\]

Again without loss of generality, we can assume $a>0$ and $b>0$ (Short argument is needed here instead of the lazy "wlog"). Also, the vertex of $f$ is $(b,m)$, so $m=g(b)=-4ab^2-m$, or $m=-2ab^2 <0$.

Since $x_2,x_4$ are roots of $f$, we have the following relationship of the roots: \[x_2 + x_4 = 2b\] \[x_4 - x_2 = \sqrt{\Delta}/a = \sqrt{-4am}/a = 2\sqrt{-m/a} = 2\sqrt{2b^2} = 2\sqrt{2}b = \sqrt{2} (x_2+x_4)\]

So $(\sqrt{2}-1)x_4 = 75(1+\sqrt{2})$, or $x_4 = 75(1+\sqrt{2})^2 = 75(3+2\sqrt{2})$.

Therefore $x_4-x_1=2x_4=450+300\sqrt{2}$.

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_12A_Problems/Problem_23&oldid=86474"