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

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== Problem ==
 
== Problem ==
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[[Function]]s <math>f</math> and <math>g</math> are [[quadratic]], <math>g(x) = - f(100 - x)</math>, and the graph of <math>g</math> contains the [[vertex]] of the graph of <math>f</math>. The four <math>x</math>-intercepts on the two graphs have <math>x</math>-coordinates <math>x_1</math>, <math>x_2</math>, <math>x_3</math>, and <math>x_4</math>, in increasing order, and <math>x_3 - x_2 = 150</math>. The value of <math>x_4 - x_1</math> is <math>m + n\sqrt p</math>, where <math>m</math>, <math>n</math>, and <math>p</math> are positive integers, and <math>p</math> is not divisible by the square of any prime. What is <math>m + n + p</math>?
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[[Function]]s <math>f</math> and <math>g</math> are [[quadratic]], <math>g(x) = - f(100 - x)</math>, and the graph of <math>g</math> contains the [[vertex]] of the graph of <math>f</math>. The four <math>x</math>-intercepts on the two graphs have <math>x</math>-coordinates <math>x_1</math>, <math>x_2</math>, <math>x_3</math>, and <math>x_4</math>, in increasing order, and <math>x_3 - x_2 = 150</math>. Then <math>x_4 - x_1 = m + n\sqrt p</math>, where <math>m</math>, <math>n</math>, and <math>p</math> are positive integers, and <math>p</math> is not divisible by the square of any prime. What is <math>m + n + p</math>?
  
 
<math>\textbf{(A)}\ 602\qquad \textbf{(B)}\ 652\qquad \textbf{(C)}\ 702\qquad \textbf{(D)}\ 752 \qquad \textbf{(E)}\ 802</math>
 
<math>\textbf{(A)}\ 602\qquad \textbf{(B)}\ 652\qquad \textbf{(C)}\ 702\qquad \textbf{(D)}\ 752 \qquad \textbf{(E)}\ 802</math>
  
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== Solution ==
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== Solution (Alcumus)==
 
<center><asy>
 
<center><asy>
 
import graph; size(250);Label k; k.p=fontsize(6); int ymax = 400, ymin = -400; real rt = 175+150*2^.5;
 
import graph; size(250);Label k; k.p=fontsize(6); int ymax = 400, ymin = -400; real rt = 175+150*2^.5;
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& = & - \frac {a}{4}(x_1 + 225)(3x_1 + 75) \end{eqnarray*}</cmath></center>
 
& = & - \frac {a}{4}(x_1 + 225)(3x_1 + 75) \end{eqnarray*}</cmath></center>
  
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Let <math>x_1 = 75u</math> and divide through by <math>75^2</math>, since this is a timed competition and it will drastically simplify computations. We know <math>u < - 1</math> and that <math>(u - 1)^2 = (3u + 1)(u + 3)</math>, or
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Let <math>x_1 = 75u</math> and divide through by <math>75^2</math>, since it will drastically simplify computations. We know <math>u < - 1</math> and that <math>(u - 1)^2 = (3u + 1)(u + 3)</math>, or
  
 
<center><cmath>\begin{eqnarray*} 0 & = & (3u + 1)(u + 3) - (u - 1)^2 \\
 
<center><cmath>\begin{eqnarray*} 0 & = & (3u + 1)(u + 3) - (u - 1)^2 \\
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So <math>u = \frac { - 6\pm\sqrt {32}}{2} = - 3\pm2\sqrt2</math>. Since <math>u < - 1</math>, <math>u = - 3 - 2\sqrt2</math>.  
 
So <math>u = \frac { - 6\pm\sqrt {32}}{2} = - 3\pm2\sqrt2</math>. Since <math>u < - 1</math>, <math>u = - 3 - 2\sqrt2</math>.  
  
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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>.
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The answer is <math>x_4 - x_1 = (-x_1) - x_1 = - 150u = 450 + 300\sqrt {2}</math>, and <math>450 + 300 + 2 = 752\ \mathbf{(D)}</math>.
  
 
== Note ==
 
== Note ==

Latest revision as of 01:04, 1 April 2024

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 $x_3 - x_2 = 150$. Then $x_4 - x_1 = m + n\sqrt p$, 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 (Alcumus)

[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 $q$

\[p(x) = a(x - 75)(x - x_1)\]
\[q(x) = - a(x + 75)(x + x_1)\]

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 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 $x_4 - x_1 = (-x_1) - x_1 = - 150u = 450 + 300\sqrt {2}$, 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

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