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Difference between revisions of "2025 AIME I Problems/Problem 4"

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==Problem==
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Find the number of ordered pairs <math>(x,y)</math>, where both <math>x</math> and <math>y</math> are integers between <math>-100</math> and <math>100</math> inclusive, such that <math>12x^2-xy-6y^2=0</math>.
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==Solution 1==
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We begin by factoring, <math>12x^2-xy-6y^2=(3x+2y)(4x-3y)=0.</math> Since the RHS is <math>0</math> we have two options,
 +
 +
<math>\underline{\text{Case 1:}}\text{ } 3x+2y = 0</math>
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 +
In this case we have, <math>y=\frac{-3x}{2}.</math> Using the bounding on <math>y</math> we have,
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<cmath>-100\le\frac{-3x}{2}\le 100.</cmath>
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<cmath>\frac{200}{3}\ge x \ge \frac{-200}{3}.</cmath>
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In addition in order for <math>y</math> to be integer <math>2 | x,</math> so we substitute <math>x=2k.</math> <cmath>\frac{200}{3}\ge 2k \ge \frac{-200}{3}.</cmath> <cmath>\frac{100}{3}\ge k \ge \frac{-100}{3}.</cmath> From this we have solutions starting from <math>-33</math> to <math>33</math> which is <math>67</math> solutions.
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<math>\underline{\text{Case 2: }}\text{ } 4x-3y = 0</math>
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 +
On the other hand, we have, <math>y=\frac{4x}{3}.</math> From bounds we have,
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<cmath>-100\le\frac{4x}{3}\le 100.</cmath>
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<cmath>-75 \le x \le 75.</cmath>
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In this case, for <math>y</math> to be integer <math>3 | x,</math> so we substitute <math>x=3t.</math> <cmath>-75 \le 3t \le 75.</cmath> <cmath>-25 \le t \le 25.</cmath> This gives us <math>51</math> solutions.
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Finally we overcount one case which is the intersection of the <math>2</math> lines or the point <math>(0,0).</math> Therefore our answer is <math>67+51-1=\boxed{117}</math>
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~[[User:Mathkiddus|mathkiddus]]
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==Solution 2==
 +
First, notice that (0,0) is a solution.
 +
 
 +
Divide the equation by <math>y^2</math>, getting <math>12(\frac{x}{y})^2-\frac{x}{y}-6 = 0</math>. (We can ignore the <math>y=0</math> case for now.) Let <math>a = \frac{x}{y}</math>. We now have <math>12a^2-a-6=0</math>. Factoring, we get <math>(4a-3)(3a+2) = 0</math>. Therefore, the graph is satisfied when <math>4a=3</math> or <math>3a=-2</math>. Substituting <math>\frac{x}{y} = a</math> back into the equations, we get <math>4x=3y</math> or <math>3x=-2y</math>.
 +
 
 +
Remember that both <math>x</math> and <math>y</math> are bounded by <math>-100</math> and <math>100</math>, inclusive. For <math>4x=3y</math>, the solutions are <math>(-75,-100), (-72,-96), (-69, -92), \dots, (72,96), (75,100)</math>. Remember to not count the <math>x=y=0</math> case for now. There are <math>25</math> positive solutions and <math>25</math> negative solutions for a total of <math>50</math>.
 +
 
 +
For <math>3x-2y</math>, we do something similar. The solutions are <math>(-66,99), (-64,96), \dots, (64, -96), (66, -99)</math>. There are <math>33</math> solutions when <math>x</math> is positive and <math>33</math> solutions when <math>x</math> is negative, for a total of <math>66</math>.
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 +
Now we can count the edge case of <math>(0,0)</math>. The answer is therefore <math>50+66+1 = \boxed{117}</math>.
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 +
~lprado
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==Solution 3==
 +
 
 +
Please help with LaTex Formatting:
 +
 
 +
You can use the quadratic formula for this equation: <math>12x^2 - xy - 6y^2 = 0</math>;
 +
Although this solution may seem to be misleading, it works!
 +
 
 +
You get: <cmath>\frac{-b \pm \sqrt{b^2-4ac}}{2a}</cmath>
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<cmath>= \frac{xy \pm \sqrt{x^2y^2+(12\cdot6\cdot4\cdotx^2\cdoty^2)}}{24x^2}</cmath>
 +
 
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<cmath>= \frac{xy \pm\sqrt{289x^2 y^2}}{24x^2}</cmath>
 +
 
 +
<cmath>= \frac{18xy}{24x^2}\text{, and }\frac{-16xy}{24x^2}</cmath>
 +
 
 +
Rather than putting this equation as zero, the numerators and denominators must be equal. These two equations simplify to:
 +
 
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<cmath>3y = 4x;</cmath> <cmath>-2y = 3x;</cmath>
 +
 
 +
As <math>x</math> and <math>y</math> are between <math>-100</math> and <math>100</math>, for the first equation, <math>x</math> can be between <math>(-75,75)</math>, but <math>x</math> must be a multiple of <math>3</math>, so there are:
 +
 
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<math>((75+75)/3) + 1 = 51</math> solutions for this case.
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For <cmath>-2y = 3x:</cmath>
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<math>x</math> can be between <math>(-66, 66)</math>, but <math>x</math> has to be a multiple of <math>2</math>.
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Therefore, there are <math>(66+66)/2 + 1 = 67</math> solutions for this case.
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However, the one overlap would be <math>x = 0</math>, because y would be <math>0</math> in both solutions.
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Therefore, the answer is <math>51+67-1 = \boxed{117}.</math>
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-U-King3.14Root
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-LaTeX corrected by Andrew2019, though idk if this is what you wanted to say
 +
 
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==Video Solution 1 by SpreadTheMathLove==
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https://www.youtube.com/watch?v=J-0BapU4Yuk
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 +
==See also==
 +
{{AIME box|year=2025|num-b=3|num-a=5|n=I}}
 +
 
 +
{{MAA Notice}}

Latest revision as of 12:59, 16 February 2025

Problem

Find the number of ordered pairs $(x,y)$, where both $x$ and $y$ are integers between $-100$ and $100$ inclusive, such that $12x^2-xy-6y^2=0$.

Solution 1

We begin by factoring, $12x^2-xy-6y^2=(3x+2y)(4x-3y)=0.$ Since the RHS is $0$ we have two options,

$\underline{\text{Case 1:}}\text{ } 3x+2y = 0$

In this case we have, $y=\frac{-3x}{2}.$ Using the bounding on $y$ we have, \[-100\le\frac{-3x}{2}\le 100.\] \[\frac{200}{3}\ge x \ge \frac{-200}{3}.\] In addition in order for $y$ to be integer $2 | x,$ so we substitute $x=2k.$ \[\frac{200}{3}\ge 2k \ge \frac{-200}{3}.\] \[\frac{100}{3}\ge k \ge \frac{-100}{3}.\] From this we have solutions starting from $-33$ to $33$ which is $67$ solutions.

$\underline{\text{Case 2: }}\text{ } 4x-3y = 0$

On the other hand, we have, $y=\frac{4x}{3}.$ From bounds we have, \[-100\le\frac{4x}{3}\le 100.\] \[-75 \le x \le 75.\] In this case, for $y$ to be integer $3 | x,$ so we substitute $x=3t.$ \[-75 \le 3t \le 75.\] \[-25 \le t \le 25.\] This gives us $51$ solutions.

Finally we overcount one case which is the intersection of the $2$ lines or the point $(0,0).$ Therefore our answer is $67+51-1=\boxed{117}$

~mathkiddus

Solution 2

First, notice that (0,0) is a solution.

Divide the equation by $y^2$, getting $12(\frac{x}{y})^2-\frac{x}{y}-6 = 0$. (We can ignore the $y=0$ case for now.) Let $a = \frac{x}{y}$. We now have $12a^2-a-6=0$. Factoring, we get $(4a-3)(3a+2) = 0$. Therefore, the graph is satisfied when $4a=3$ or $3a=-2$. Substituting $\frac{x}{y} = a$ back into the equations, we get $4x=3y$ or $3x=-2y$.

Remember that both $x$ and $y$ are bounded by $-100$ and $100$, inclusive. For $4x=3y$, the solutions are $(-75,-100), (-72,-96), (-69, -92), \dots, (72,96), (75,100)$. Remember to not count the $x=y=0$ case for now. There are $25$ positive solutions and $25$ negative solutions for a total of $50$.

For $3x-2y$, we do something similar. The solutions are $(-66,99), (-64,96), \dots, (64, -96), (66, -99)$. There are $33$ solutions when $x$ is positive and $33$ solutions when $x$ is negative, for a total of $66$.

Now we can count the edge case of $(0,0)$. The answer is therefore $50+66+1 = \boxed{117}$.

~lprado

Solution 3

Please help with LaTex Formatting:

You can use the quadratic formula for this equation: $12x^2 - xy - 6y^2 = 0$; Although this solution may seem to be misleading, it works!

You get: \[\frac{-b \pm \sqrt{b^2-4ac}}{2a}\] 

\[= \frac{xy \pm \sqrt{x^2y^2+(12\cdot6\cdot4\cdotx^2\cdoty^2)}}{24x^2}\] (Error compiling LaTeX. Unknown error_msg)

\[= \frac{xy \pm\sqrt{289x^2 y^2}}{24x^2}\]

\[= \frac{18xy}{24x^2}\text{, and }\frac{-16xy}{24x^2}\]

Rather than putting this equation as zero, the numerators and denominators must be equal. These two equations simplify to:

\[3y = 4x;\] \[-2y = 3x;\]

As $x$ and $y$ are between $-100$ and $100$, for the first equation, $x$ can be between $(-75,75)$, but $x$ must be a multiple of $3$, so there are:

$((75+75)/3) + 1 = 51$ solutions for this case.

For \[-2y = 3x:\]

$x$ can be between $(-66, 66)$, but $x$ has to be a multiple of $2$.

Therefore, there are $(66+66)/2 + 1 = 67$ solutions for this case.

However, the one overlap would be $x = 0$, because y would be $0$ in both solutions.

Therefore, the answer is $51+67-1 = \boxed{117}.$

-U-King3.14Root -LaTeX corrected by Andrew2019, though idk if this is what you wanted to say

Video Solution 1 by SpreadTheMathLove

https://www.youtube.com/watch?v=J-0BapU4Yuk

See also

2025 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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