Difference between revisions of "2014 AMC 12A Problems/Problem 24"
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\textbf{(D) }302\qquad | \textbf{(D) }302\qquad | ||
\textbf{(E) }303\qquad</math> | \textbf{(E) }303\qquad</math> | ||
+ | |||
==Solution 1== | ==Solution 1== | ||
− | 1. Draw the graph of <math>f_0(x)</math> by dividing the domain into three parts. | + | 1. Draw the graph of <math>f_0(x)</math> 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. | ||
+ | |||
+ | <b>Note:</b> <math>f_n(x) = |f_{n-1}(x)| - 10</math> 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 <math>f_{100}</math>. Notice that the summits start <math>100</math> away from <math>0</math> and get <math>1</math> closer each iteration, so they reach <math>0</math> exactly at <math>f_{100}</math>. | ||
+ | <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> | ||
− | + | <math>f_{100}(x)</math> reaches <math>0</math> at <math>x = -300</math>, then zigzags between <math>0</math> and <math>-1</math>, hitting <math>0</math> at every even <math>x</math>, before leaving <math>0</math> at <math>x = 300</math>. | |
− | + | This means that <math>f_{100}(x) = 0</math> at all even <math>x</math> where <math>-300 \le x \le 300</math>. This is a <math>601</math>-integer odd-size range with even numbers at the endpoints, so just over half of the integers are even, or <math>\frac{601+1}{2} = \boxed{\textbf{(C) }301}</math>. | |
− | (Revised by Flamedragon & Jason,C) | + | (Revised by Flamedragon & Jason,C & [[User:emerald_block|emerald_block]]) |
==Solution 2== | ==Solution 2== | ||
− | First, notice that the | + | First, notice that the recursive rule moves the current value <math>1</math> closer to <math>0</math>. Upon reaching <math>0</math>, it alternates between <math>-1</math> and <math>0</math>. This means that <math>f_{100}(x) = 0</math> exactly when <math>|f_0(x)| \le 100</math> (to reach <math>0</math> in time) and <math>f_0(x)</math> is even (so <math>f_{100}(x) \ne -1</math>). |
+ | |||
+ | Casework each part of <math>f_0(x)</math> (where the expressions in the absolute values do not change sign): | ||
+ | <cmath>x \le -100 \implies f_0(x) = x-(x-100)+(x+100) = x+200</cmath> | ||
+ | so even <math>-300 \le x \le -100</math> work. | ||
+ | <cmath>-100 \le x \le 100 \implies f_0(x) = x-(x-100)-(x+100) = -x</cmath> | ||
+ | so even <math>-100 \le x \le 100</math> work. | ||
+ | <cmath>100 \le x \implies f_0(x) = x+(x-100)-(x+100) = x-200</cmath> | ||
+ | so even <math>100 \le x \le 300</math> work. | ||
+ | |||
+ | Putting these together, all even <math>x</math> where <math>-300 \le x \le 300</math> work. So, the answer is <math>2\cdot150+1 = \boxed{\textbf{(C)}\ 301}</math>. | ||
+ | ~revised by [[User:emerald_block|emerald_block]] | ||
+ | |||
+ | == Video Solution by Richard Rusczyk == | ||
+ | |||
+ | https://artofproblemsolving.com/videos/amc/2014amc12a/383 | ||
+ | |||
+ | ~ dolphin7 | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2014|ab=A|num-b=23|num-a=25}} | {{AMC12 box|year=2014|ab=A|num-b=23|num-a=25}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 16:02, 29 August 2023
Problem
Let , and for
, let
. For how many values of
is
?
Solution 1
1. Draw the graph of by dividing the domain into three parts.
2. Apply the recursive rule a few times to find the pattern.
Note: is used to enlarge the difference, but the reasoning is the same.
3. Extrapolate to . Notice that the summits start
away from
and get
closer each iteration, so they reach
exactly at
.
reaches
at
, then zigzags between
and
, hitting
at every even
, before leaving
at
.
This means that at all even
where
. This is a
-integer odd-size range with even numbers at the endpoints, so just over half of the integers are even, or
.
(Revised by Flamedragon & Jason,C & emerald_block)
Solution 2
First, notice that the recursive rule moves the current value closer to
. Upon reaching
, it alternates between
and
. This means that
exactly when
(to reach
in time) and
is even (so
).
Casework each part of (where the expressions in the absolute values do not change sign):
so even
work.
so even
work.
so even
work.
Putting these together, all even where
work. So, the answer is
.
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.