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Difference between revisions of "2024 AIME I Problems/Problem 12"
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==Solution 2==
 
==Solution 2==
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We will denote <math>h(x)=4g(f(x))</math> for simplicity. Denote <math>p(x)</math> as the first equation and <math>q(y)</math> as the graph of the second. We notice that the graph of <math>f(x)</math> oscillates between <math>y=0</math> and <math>y=1</math>, and the graph of <math>g(x)</math> oscillates between <math>x=0</math> and <math>x=1</math>. The intersections are thus all in the square <math>(0,0)</math>, <math>(0,1)</math>, <math>(1,1)</math>, and <math>(1,0)</math>.
+
We will denote <math>h(x)=4g(f(x))</math> for simplicity. Denote <math>p(x)</math> as the first equation and <math>q(y)</math> as the graph of the second. We notice that both <math>f(x)</math> and <math>g(x)</math> oscillate between 0 and 1. The intersections are thus all in the square <math>(0,0)</math>, <math>(0,1)</math>, <math>(1,1)</math>, and <math>(1,0)</math>. Every <math>p(x)</math> wave going up and down crosses every <math>q(y)</math> wave. Now, we need to find the number of times each wave touches 0 and 1.
  
  
−
Every <math>p(x)</math> wave going up and down crosses every <math>q(y)</math> wave. Now, we need to find the number of times each wave touches 0 and 1. We notice that <math>h(x)=0</math> occurs at <math>x=-\frac{3}{4}, -\frac{1}{4}, \frac{1}{4}, \frac{3}{4}</math>, and <math>h(x)=1</math> occurs at <math>x=-1, -\frac{1}{2}, 0,\frac{1}{2},1</math>. A sinusoid passes through each point twice during each period, but it only passes through the extrema once. <math>p(x)</math> has 1 period between 0 and 1, giving 8 solutions for <math>p(x)=0</math> and 9 solutions for <math>p(x)=1</math>, or 16 up and down waves. <math>q(y)</math> has 1.5 periods, giving 12 solutions for <math>q(y)=0</math> and 13 solutions for <math>q(y)=1</math>, or 24 up and down waves. This amounts to <math>16\cdot24=384</math> intersections.
+
We notice that <math>h(x)=0</math> occurs at <math>x=-\frac{3}{4}, -\frac{1}{4}, \frac{1}{4}, \frac{3}{4}</math>, and <math>h(x)=1</math> occurs at <math>x=-1, -\frac{1}{2}, 0,\frac{1}{2},1</math>. A sinusoid passes through each point twice during each period, but it only passes through the extrema once. <math>p(x)</math> has 1 period between 0 and 1, giving 8 solutions for <math>p(x)=0</math> and 9 solutions for <math>p(x)=1</math>, or 16 up and down waves. <math>q(y)</math> has 1.5 periods, giving 12 solutions for <math>q(y)=0</math> and 13 solutions for <math>q(y)=1</math>, or 24 up and down waves. This amounts to <math>16\cdot24=384</math> intersections.
  
  

Revision as of 22:26, 3 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 \[y=4 g(f(\sin (2 \pi x))) \quad\text{ and }\quad x=4 g(f(\cos (3 \pi y))).\]

Graph

https://www.desmos.com/calculator/wml09giaun

Solution 1

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 $\boxed{385}$

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 $384$.

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 $h(x)=4g(f(x))$ for simplicity. Denote $p(x)$ as the first equation and $q(y)$ as the graph of the second. We notice that both $f(x)$ and $g(x)$ oscillate between 0 and 1. The intersections are thus all in the square $(0,0)$, $(0,1)$, $(1,1)$, and $(1,0)$. Every $p(x)$ wave going up and down crosses every $q(y)$ wave. Now, we need to find the number of times each wave touches 0 and 1.


We notice that $h(x)=0$ occurs at $x=-\frac{3}{4}, -\frac{1}{4}, \frac{1}{4}, \frac{3}{4}$, and $h(x)=1$ occurs at $x=-1, -\frac{1}{2}, 0,\frac{1}{2},1$. A sinusoid passes through each point twice during each period, but it only passes through the extrema once. $p(x)$ has 1 period between 0 and 1, giving 8 solutions for $p(x)=0$ and 9 solutions for $p(x)=1$, or 16 up and down waves. $q(y)$ has 1.5 periods, giving 12 solutions for $q(y)=0$ and 13 solutions for $q(y)=1$, or 24 up and down waves. This amounts to $16\cdot24=384$ intersections.


However, we have to be very careful when counting around $(1, 1)$. At this point, $q(y)$ has an infinite downwards slope and $p(x)$ is slanted, giving us an extra intersection; thus, we need to add 1 to our answer to get $\boxed{385}$.

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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