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"

(→‎Solution)
m (→‎Solution 1 (BASH, DO NOT ATTEMPT IF INSUFFICIENT TIME))
Line 5: Line 5:
 
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>
 
===Note===
 
===Note===
−
While this solution might seem unreliable (it probably is), the only parts where counting the intersection might be tricky is near the origin.
+
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>.
  
 
==See also==
 
==See also==

Revision as of 19:41, 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 $\boxed{384}$

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=214305"