Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2024 AIME I Problems/Problem 12"

m (→‎Solution 1 (BASH, DO NOT ATTEMPT IF INSUFFICIENT TIME))
m (→‎Solution 1 (BASH, DO NOT ATTEMPT IF INSUFFICIENT TIME))
Line 4: Line 4:
 
==Solution 1 (BASH, DO NOT ATTEMPT IF INSUFFICIENT TIME)==
 
==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>
 
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>
 +
(and yes, I did use this on the real AIME and it worked)
 
===Note===
 
===Note===
 
While this solution might seem unreliable (it probably is), the only parts where counting the intersection might be tricky is near <math>(1,1)</math>. Make sure to count them as two points and not one, or you'll get <math>383</math>.
 
While this solution might seem unreliable (it probably is), the only parts where counting the intersection might be tricky is near <math>(1,1)</math>. Make sure to count them as two points and not one, or you'll get <math>383</math>.

Revision as of 19:42, 2 February 2024

Problem

Define $f(x)=|| x|-\tfrac{1}{2}|$ and $g(x)=|| x|-\tfrac{1}{4}|$. Find the number of intersections of the graphs of \[y=4 g(f(\sin (2 \pi x))) \quad\text{ and }\quad x=4 g(f(\cos (3 \pi y))).\]

Solution 1 (BASH, DO NOT ATTEMPT IF INSUFFICIENT TIME)

If we graph $4g(f(x))$, we see it forms a sawtooth graph that oscillates between $0$ and $1$ (for values of $x$ between $-1$ and $1$, which is true because the arguments are between $-1$ and $1$). Thus by precariously drawing the graph of the two functions in the square bounded by $(0,0)$, $(0,1)$, $(1,1)$, and $(1,0)$, and hand-counting each of the intersections, we get $\boxed{384}$ (and yes, I did use this on the real AIME and it worked)

Note

While this solution might seem unreliable (it probably is), the only parts where counting the intersection might be tricky is near $(1,1)$. Make sure to count them as two points and not one, or you'll get $383$.

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2024_AIME_I_Problems/Problem_12&oldid=214307"