Difference between revisions of "2024 AIME I Problems/Problem 12"
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===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>. | ||
− | + | == Note == | |
+ | The answer should be 385 since there are 16 intersections in each of 24 smaller boxes of dimensions 1/6 x 1/4 and then another one at the corner (1,1). | ||
==See also== | ==See also== | ||
{{AIME box|year=2024|n=I|num-b=11|num-a=13}} | {{AIME box|year=2024|n=I|num-b=11|num-a=13}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 13:55, 3 February 2024
Contents
Problem
Define and . Find the number of intersections of the graphs of
Graph
https://www.desmos.com/calculator/wml09giaun
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 (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 . Make sure to count them as two points and not one, or you'll get .
Note
The answer should be 385 since there are 16 intersections in each of 24 smaller boxes of dimensions 1/6 x 1/4 and then another one at the corner (1,1).
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.