2024 AIME I Problems/Problem 12
Contents
[hide]Problem
Define and . Find the number of intersections of the graphs of
Graph
https://www.desmos.com/calculator/wml09giaun
Solution 1
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 . Make sure to count them as two points and not one, or you'll get .
Note 1
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).
Solution 2
We will denote for simplicity. Denote as the first equation and as the graph of the second. We notice that both and oscillate between 0 and 1. The intersections are thus all in the square , , , and . Every wave going up and down crosses every wave. Now, we need to find the number of times each wave touches 0 and 1.
We notice that occurs at , and occurs at . A sinusoid passes through each point twice during each period, but it only passes through the extrema once. has 1 period between 0 and 1, giving 8 solutions for and 9 solutions for , or 16 up and down waves. has 1.5 periods, giving 12 solutions for and 13 solutions for , or 24 up and down waves. This amounts to intersections.
However, we have to be very careful when counting around . At this point, has an infinite downwards slope and is slanted, giving us an extra intersection; thus, we need to add 1 to our answer to get .
~Xyco
Solution 3 (Rigorous analysis of why there are two solutions near (1,1))
We can easily see that only may satisfy both functions. We call function as Function 1 and function as Function 2.
For Function 1, in each interval with , Function 1's value oscillates between 0 and 1. It attains 1 at , and another point between these two. Between two consecutive points whose functional values are 1, the function first decreases from 1 to 0 and then increases from 0 to 1. So the graph of this function in this interval consists of 4 monotonic pieces.
For Function 2, in each interval with , Function 2's value oscillates between 0 and 1. It attains 1 at , and another point between these two. Between two consecutive points whose functional values are 1, the function first decreases from 1 to 0 and then increases from 0 to 1. So the graph of this function in this interval consists of 4 monotonic curves.
Consider any region with and but . Both functions have four monotonic pieces. Because Function 1's each monotonic piece can take any value in and Function 2' each monotonic piece can take any value in , Function 1's each monotonic piece intersects with Function 2's each monotonic piece. Therefore, in the interval , the number of intersecting points is .
Next, we prove that if an intersecting point is on a line for , then this point must be .
For , Function 1 attains value 1. For Function 2, if , then . Therefore, the intersecting point is .
Similarly, we can prove that if an intersecting point is on a line for , then this point must be .
Therefore, in each region with and but , all 16 intersecting points are interior. That is, no two regions share any common intersecting point.
Next, we study region . Consider any pair of monotonic pieces, where one is from Function 1 and one is from Function 2, except the pair of two monotonic pieces from two functions that attain . Two pieces in each pair intersects at an interior point on the region. So the number of intersecting points is .
Finally, we compute the number intersection points of two functions' monotonic pieces that both attain .
One trivial intersection point is . Now, we study whether they intersect at another point.
Define and . Thus, for positive and sufficiently small and , Function 1 is reduced to \[ y' = 4 \sin 2 \pi x' \hspace{1cm} (1) \] and Function 2 is reduced to \[ x' = 4 \left( 1 - \cos 3 \pi y' \right) . \hspace{1cm} (2) \]
Now, we study whether there is a non-zero solution. Because we consider sufficiently small and , to get an intuition and quick estimate, we do approximations of the above equations.
Equation (1) is approximated as \[ y' = 4 \cdot 2 \pi x' \] and Equation (2) is approximated as \[ x' = 2 \left( 3 \pi y' \right)^2 \]
To solve these equations, we get and . Therefore, two functions' two monotonic pieces that attain have two intersecting points.
Putting all analysis above, the total number of intersecting points is .
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution (Rigorous analysis of why there are two solutions near (1,1))
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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 |
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