Difference between revisions of "2024 AIME I Problems/Problem 12"
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+ | |||
+ | ==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 | ||
+ | <cmath> | ||
+ | \[ | ||
+ | y' = 4 \sin 2 \pi x' \hspace{1cm} (1) | ||
+ | \] | ||
+ | </cmath> | ||
+ | and Function 2 is reduced to | ||
+ | <cmath> | ||
+ | \[ | ||
+ | x' = 4 \left( 1 - \cos 3 \pi y' \right) . \hspace{1cm} (2) | ||
+ | \] | ||
+ | </cmath> | ||
+ | |||
+ | 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 | ||
+ | <cmath> | ||
+ | \[ | ||
+ | y' = 4 \cdot 2 \pi x' | ||
+ | \] | ||
+ | </cmath> | ||
+ | and Equation (2) is approximated as | ||
+ | <cmath> | ||
+ | \[ | ||
+ | x' = 2 \left( 3 \pi y' \right)^2 | ||
+ | \] | ||
+ | </cmath> | ||
+ | |||
+ | 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== | ||
+ | https://youtu.be/PA3aLjPinhI?t=7722 | ||
+ | |||
+ | ~Math Gold Medalist | ||
+ | |||
+ | |||
+ | ==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) | ||
+ | |||
+ | ==Drawing of two solutions near <math>\left( 1, 1 \right)</math>== | ||
+ | |||
+ | https://artofproblemsolving.com/wiki/index.php/File:2024_AIME_I_Problem_12,_two_solutions_near_(1,1).png | ||
+ | |||
+ | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
+ | |||
==See also== | ==See also== |
Latest revision as of 19:41, 18 February 2024
Contents
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
and Function 2 is reduced to
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
and Equation (2) is approximated as
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
https://youtu.be/PA3aLjPinhI?t=7722
~Math Gold Medalist
Video Solution (Rigorous analysis of why there are two solutions near (1,1))
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Drawing of two solutions near ![$\left( 1, 1 \right)$](//latex.artofproblemsolving.com/0/d/6/0d6d5b4cb4c0dd407303e33b79d902ebeb3a5140.png)
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.