Difference between revisions of "2024 AIME I Problems/Problem 12"
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~Xyco | ~Xyco | ||
+ | |||
+ | ==Solution 3 (Rigorous analysis of why there are two solutions near (1,1))== | ||
+ | |||
+ | We can easily see that only <math>x, y \in \left[0,1 \right]</math> may satisfy both functions. | ||
+ | We call function <math>y = 4g \left( f \left( \sin \left( 2 \pi x \right) \right) \right)</math> as Function 1 and function <math>x = 4g \left( f \left( \cos \left( 3 \pi y \right) \right) \right)</math> as Function 2. | ||
+ | |||
+ | For Function 1, in each interval <math>\left[ \frac{i}{4} , \frac{i+1}{4} \right]</math> with <math>i \in \left\{ 0, 1, \cdots, 3 \right\}</math>, Function 1's value oscillates between 0 and 1. It attains 1 at <math>x = \frac{i}{4}</math>, <math>\frac{i+1}{4}</math> 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 <math>\left[ \frac{i}{6} , \frac{i+1}{6} \right]</math> with <math>i \in \left\{ 0, 1, \cdots, 5 \right\}</math>, Function 2's value oscillates between 0 and 1. It attains 1 at <math>y = \frac{i}{6}</math>, <math>\frac{i+1}{6}</math> 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 <math>\left[ \frac{i}{4} , \frac{i+1}{4} \right] \times \left[ \frac{j}{6} , \frac{j+1}{6} \right]</math> with <math>i \in \left\{ 0, 1, \cdots, 3 \right\}</math> and <math>j \in \left\{0, 1, \cdots , 5 \right\}</math> but <math>\left( i, j \right) \neq \left( 3, 5 \right)</math>. | ||
+ | Both functions have four monotonic pieces. | ||
+ | Because Function 1's each monotonic piece can take any value in <math>\left[ \frac{j}{6} , \frac{j+1}{6} \right]</math> and Function 2' each monotonic piece can take any value in <math>\left[ \frac{i}{4} , \frac{i+1}{4} \right]</math>, Function 1's each monotonic piece intersects with Function 2's each monotonic piece. | ||
+ | Therefore, in the interval <math>\left[ \frac{i}{4} , \frac{i+1}{4} \right] \times \left[ \frac{j}{6} , \frac{j+1}{6} \right]</math>, the number of intersecting points is <math>4 \cdot 4 = 16</math>. | ||
+ | |||
+ | Next, we prove that if an intersecting point is on a line <math>x = \frac{i}{4}</math> for <math>i \in \left\{ 0, 1, \cdots, 4 \right\}</math>, then this point must be <math>\left( 1, 1 \right)</math>. | ||
+ | |||
+ | For <math>x = \frac{i}{4}</math>, Function 1 attains value 1. | ||
+ | For Function 2, if <math>y = 1</math>, then <math>x = 1</math>. | ||
+ | Therefore, the intersecting point is <math>\left( 1, 1 \right)</math>. | ||
+ | |||
+ | Similarly, we can prove that if an intersecting point is on a line <math>y = \frac{i}{6}</math> for <math>i \in \left\{ 0, 1, \cdots, 6 \right\}</math>, then this point must be <math>\left( 1, 1 \right)</math>. | ||
+ | |||
+ | Therefore, in each region <math>\left[ \frac{i}{4} , \frac{i+1}{4} \right] \times \left[ \frac{j}{6} , \frac{j+1}{6} \right]</math> with <math>i \in \left\{ 0, 1, \cdots, 3 \right\}</math> and <math>j \in \left\{0, 1, \cdots , 5 \right\}</math> but <math>\left( i, j \right) \neq \left( 3, 5 \right)</math>, all 16 intersecting points are interior. | ||
+ | That is, no two regions share any common intersecting point. | ||
+ | |||
+ | Next, we study region <math>\left[ \frac{3}{4} , 1 \right] \times \left[ \frac{5}{6} , 1 \right]</math>. | ||
+ | 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 <math>\left( 1 , 1 \right)</math>. | ||
+ | Two pieces in each pair intersects at an interior point on the region. | ||
+ | So the number of intersecting points is <math>4 \cdot 4 - 1 = 15</math>. | ||
+ | |||
+ | Finally, we compute the number intersection points of two functions' monotonic pieces that both attain <math>\left( 1, 1 \right)</math>. | ||
+ | |||
+ | One trivial intersection point is <math>\left( 1, 1 \right)</math>. | ||
+ | Now, we study whether they intersect at another point. | ||
+ | |||
+ | Define <math>x = 1 - x'</math> and <math>y = 1 - y'</math>. | ||
+ | Thus, for positive and sufficiently small <math>x'</math> and <math>y'</math>, 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 <math>x'</math> and <math>y'</math>, 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 <math>x' = \frac{1}{8^2 \cdot 18 \pi^4}</math> and <math>y' = \frac{1}{8 \cdot 18 \pi^3}</math>. | ||
+ | Therefore, two functions' two monotonic pieces that attain <math>\left( 1, 1 \right)</math> have two intersecting points. | ||
+ | |||
+ | Putting all analysis above, the total number of intersecting points is <math>16 \cdot 4 \cdot 6 + 1 = \boxed{\textbf{(385) }}</math>. | ||
+ | |||
+ | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
+ | |||
+ | ==Video Solution (Rigorous analysis of why there are two solutions near (1,1))== | ||
+ | |||
+ | https://youtu.be/XjXwWfFzSrM | ||
+ | |||
+ | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
==See also== | ==See also== |
Revision as of 00:24, 4 February 2024
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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