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2014 AMC 12A Problems/Problem 24

Revision as of 19:54, 4 November 2021 by Emerald block (talk | contribs) (→‎Solution 2: revise and elaborate)

Problem

Let $f_0(x)=x+|x-100|-|x+100|$, and for $n\geq 1$, let $f_n(x)=|f_{n-1}(x)|-1$. For how many values of $x$ is $f_{100}(x)=0$?

$\textbf{(A) }299\qquad \textbf{(B) }300\qquad \textbf{(C) }301\qquad \textbf{(D) }302\qquad \textbf{(E) }303\qquad$

Solution 1

1. Draw the graph of $f_0(x)$ by dividing the domain into three parts. [asy] unitsize(0); int w = 250; int h = 125; xaxis(-w,w,Ticks(100.0),Arrows); yaxis(-h,h,Ticks(100.0),Arrows); draw((-100,-h)--(-100,h),dashed); draw((100,-h)--(100,h),dashed); real f0(real x) { return x + abs(x-100) - abs(x+100); } draw(graph(f0,-w,w),Arrows); label("$f_0$",(-w,f0(-w)),W); [/asy]

2. Apply the recursive rule a few times to find the pattern.

Note: $f_n(x) = |f_{n-1}(x)| - 10$ is used to enlarge the difference, but the reasoning is the same. [asy] unitsize(0); int w = 350; int h = 125; xaxis(-w,w,Ticks(100.0),Arrows); yaxis(-h,h,Ticks(100.0),Arrows); draw((-100,-h)--(-100,h),dashed); draw((100,-h)--(100,h),dashed); int s = 10; real f0(real x) { return x + abs(x-100) - abs(x+100); } real f(real x, int k) { real a = abs(f0(x))-s*k; return a >= -s ? a : k%2 == 0 ? abs(x%(2*s)-s)-s : -abs(x%(2*s)-s); } real f1(real x) { return f(x,1); } real f2(real x) { return f(x,2); } real f3(real x) { return f(x,3); } real f4(real x) { return f(x,4); } draw(graph(f0,-w,w,w*2#s),Arrows); draw(graph(f1,-w,w,w*2#s),red,Arrows); draw(graph(f2,-w,w,w*2#s),orange,Arrows); draw(graph(f3,-w,w,w*2#s),lightolive,Arrows); label("$f_0$",(-w,f0(-w)),W); label("$f_1$",(-w,f1(-w)),NW,red); label("$f_2$",(-w,f2(-w)),W,orange); label("$f_3$",(-w,f3(-w)),SW,lightolive); [/asy]

3. Extrapolate to $f_{100}$. Notice that the summits start $100$ away from $0$ and get $1$ closer each iteration, so they reach $0$ exactly at $f_{100}$. [asy] unitsize(0); int w = 350; int h = 125; xaxis(-w,w,Ticks(100.0),Arrows); yaxis(-h,h,Ticks(100.0),Arrows); draw((-100,-h)--(-100,h),dashed); draw((100,-h)--(100,h),dashed); int s = 10; real f0(real x) { return x + abs(x-100) - abs(x+100); } real f(real x, int k) { real a = abs(f0(x))-s*k; return a >= -s ? a : k%2 == 0 ? abs(x%(2*s)-s)-s : -abs(x%(2*s)-s); } real f1(real x) { return f(x,1); } real f2(real x) { return f(x,2); } real f3(real x) { return f(x,3); } real f4(real x) { return f(x,4); } real f98(real x) { return f(x,100#s-2); } real f99(real x) { return f(x,100#s-1); } real f100(real x) { return f(x,100#s); } draw(graph(f0,-w,w,w*2#s),Arrows); draw(graph(f1,-w,w,w*2#s),red,Arrows); draw(graph(f2,-w,w,w*2#s),orange,Arrows); draw(graph(f3,-w,w,w*2#s),lightolive,Arrows); draw(graph(f98,-w,w,w*2#s),heavygreen,Arrows); draw(graph(f99,-w,w,w*2#s),blue,Arrows); draw(graph(f100,-w,w,w*2#s),purple,Arrows); label("$f_0$",(-w,f0(-w)),W); label("$f_1$",(-w,f1(-w)),NW,red); label("$f_2$",(-w,f2(-w)),W,orange); label("$f_3$",(-w,f3(-w)),SW,lightolive); label("$f_{98}$",(-w,f98(-w)),NW,heavygreen); label("$f_{99}$",(-w,f99(-w)),W,blue); label("$f_{100}$",(-w,f100(-w)),SW,purple); [/asy]

$f_{100}(x)$ reaches $0$ at $x = -300$, then zigzags between $0$ and $-1$, hitting $0$ at every even $x$, before leaving $0$ at $x = 300$.

This means that $f_{100}(x) = 0$ at all even $x$ where $-300 \le x \le 300$. This is a $701$-integer odd-size range with even numbers at the endpoints, so just over half of the integers are even, or $\frac{701+1}{2} = \boxed{\textbf{(C) }301}$. (Revised by Flamedragon & Jason,C & emerald_block)

Solution 2

First, notice that the recursive rule moves the current value $1$ closer to $0$. Upon reaching $0$, it alternates between $-1$ and $0$. This means that $f_{100}(x) = 0$ exactly when $|f_0(x)| \le 100$ (to reach $0$ in time) and $f_0(x)$ is even (so $f_{100}(x) \ne -1$).

Casework each part of $f_0(x)$ (where the expressions in the absolute values do not change sign): \[x \le -100 \implies f_0(x) = x-(x-100)+(x+100) = x+200\] so even $-300 \le x \le -100$ work. \[-100 \le x \le 100 \implies f_0(x) = x-(x-100)-(x+100) = -x\] so even $-100 \le x \le 100$ work. \[100 \le x \implies f_0(x) = x+(x-100)-(x+100) = x-200\] so even $100 \le x \le 300$ work.

Putting these together, all even $x$ where $-300 \le x \le 300$ work. So, the answer is $2\cdot150+1 = \boxed{\textbf{(C)}\ 301}$. ~revised by emerald_block

Video Solution by Richard Rusczyk

https://artofproblemsolving.com/videos/amc/2014amc12a/383

~ dolphin7

See Also

2014 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
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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