2009 AMC 12A Problem 23

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23.

Functions ff and gg are quadratic, g(x)=f(100x),g(x) = -f(100 - x), and the graph of gg contains the vertex of the graph of f.f. The four xx-intercepts on the two graphs have xx-coordinates x1,x_1, x2,x_2, x3,x_3, and x4,x_4, in increasing order, and x3x2=150.x_3 - x_2 = 150. The value of x4x1x_4 - x_1 is m+np,m + n\sqrt{p}, where m,m, n,n, and pp are positive integers, and pp is not divisible by the square of any prime. What is m+n+p?m + n + p?

602602

652652

702702

752752

802802

Answer: D
Concepts:quadraticparabolasymmetry
Difficulty rating: 2420
Solution:

Because g(x)=f(100x),g(x) = -f(100 - x), the graphs of ff and gg are reflections of each other through the point (50,0),(50, 0), so the four intercepts pair up with x2+x3=x1+x4=100.x_2 + x_3 = x_1 + x_4 = 100.

With x3x2=150,x_3 - x_2 = 150, we get x2=25x_2 = -25 and x3=125.x_3 = 125.

Take x1,x3x_1, x_3 as the roots of f,f, whose vertex has xx-coordinate h=x1+x32,h = \dfrac{x_1 + x_3}{2}, so x1=2h125.x_1 = 2h - 125. The condition that the vertex of ff lies on the graph of gg gives 1=f(h)g(h)=(125h)(h125)(h+25)(3h225), \begin{aligned} 1 &= \frac{f(h)}{g(h)} \\ &= \frac{(125 - h)(h - 125)}{-(h + 25)(3h - 225)}, \end{aligned} which gives h=25±752.h = -25 \pm 75\sqrt{2}. Since x1=2h125x_1 = 2h-125 must be less than x2=25,x_2=-25, we need h<50,h \lt 50, so h=25752.h = -25-75\sqrt{2}.

Then x4=100x1,x_4 = 100 - x_1, so x4x1=3504h=450+3002. \begin{aligned} x_4 - x_1 &= 350 - 4h \\ &= 450 + 300\sqrt{2}. \end{aligned} Hence m+n+p=450+300+2m + n + p = 450 + 300 + 2 =752.= 752.

Thus, the correct answer is D.

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