Difference between revisions of "2024 AIME I Problems/Problem 12"
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Define <math>f(x)=|| x|-\tfrac{1}{2}|</math> and <math>g(x)=|| x|-\tfrac{1}{4}|</math>. Find the number of intersections of the graphs of <cmath>y=4 g(f(\sin (2 \pi x))) \quad\text{ and }\quad x=4 g(f(\cos (3 \pi y))).</cmath> | Define <math>f(x)=|| x|-\tfrac{1}{2}|</math> and <math>g(x)=|| x|-\tfrac{1}{4}|</math>. Find the number of intersections of the graphs of <cmath>y=4 g(f(\sin (2 \pi x))) \quad\text{ and }\quad x=4 g(f(\cos (3 \pi y))).</cmath> | ||
| − | ==Solution== | + | ==Solution 1 (BASH, DO NOT ATTEMPT IF INSUFFICIENT TIME)== |
| + | If we graph <math>4g(f(x))</math>, we see it forms a sawtooth graph that oscillates between <math>0</math> and <math>1</math> (for values of <math>x</math> between <math>-1</math> and <math>1</math>, which is true because the arguments are between <math>-1</math> and <math>1</math>). Thus by precariously drawing the graph of the two functions in the square bounded by <math>(0,0)</math>, <math>(0,1)</math>, <math>(1,1)</math>, and <math>(1,0)</math>, and hand-counting each of the intersections, we get <math>\boxed{384}</math> | ||
| + | ===Note=== | ||
| + | While this solution might seem unreliable (it probably is), the only parts where counting the intersection might be tricky is near the origin. | ||
==See also== | ==See also== | ||
Revision as of 19:38, 2 February 2024
Problem
Define 
and 
. Find the number of intersections of the graphs of ![\[y=4 g(f(\sin (2 \pi x))) \quad\text{ and }\quad x=4 g(f(\cos (3 \pi y))).\]](http://latex.artofproblemsolving.com/c/c/9/cc93c9c286edc107f0d7c62bd2c31b959accc4fa.png)
Solution 1 (BASH, DO NOT ATTEMPT IF INSUFFICIENT TIME)
If we graph 
, we see it forms a sawtooth graph that oscillates between 
and 
(for values of 
between 
and , which is true because the arguments are between and ). Thus by precariously drawing the graph of the two functions in the square bounded by 
, 
, 
, and 
, and hand-counting each of the intersections, we get 
Note
While this solution might seem unreliable (it probably is), the only parts where counting the intersection might be tricky is near the origin.
See also
| 2024 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 11 |
Followed by Problem 13 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
