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

(→‎Solution 3)
(→‎Solution 3)
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Please help with LaTex Formatting:
 
Please help with LaTex Formatting:
  
−
You can use the quadratic formula for this equation: 12x^2 - xy - 6y^2 = 0;
+
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!
 
Although this solution may seem to be misleading, it works!
  
−
You get: \frac{-b +- \sqrt b^2-4ac}{2a}
+
You get: <cmath>\frac{-b +- \sqrt b^2-4ac}{2a}</cmath>
  
−
= \frac{<math>xy +- \sqrt(x^2y^2+(12*6*4*x^2*y^2)}{24x^2}
+
<math></math>= \frac{<math>xy +- \sqrt(x^2y^2+(12*6*4*x^2*y^2)}{24x^2}<cmath>
  
−
= \frac{xy +- \sqrt289x^2 y^2}{24x^2}
+
</cmath>= \frac{xy +- \sqrt289x^2 y^2}{24x^2}<cmath>
  
−
= \frac{18xy/24x^2</math>, and <math>-16xy}{24x^2}
+
</cmath>= \frac{18xy/24x^2</math>, and <math>-16xy}{24x^2}<cmath>
  
 
Rather than putting this equation as zero, the numerators and denominators must be equal. These two equations simplify to:
 
Rather than putting this equation as zero, the numerators and denominators must be equal. These two equations simplify to:
  
−
</math>3y = 4x<math>; </math>-2y = 3x<math>;  
+
</cmath>3y = 4x<cmath>; </cmath>-2y = 3x</math><math>;  
  
−
As x and y are between </math>-100<math> and </math>100<math>, for the first equation, x can be between </math>(-75,75)<math>, but x must be a multiple of 3, so there are:
+
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:
  
 
</math>((75+75)/3) + 1 = 51<math> solutions for this case.
 
</math>((75+75)/3) + 1 = 51<math> solutions for this case.
  
−
For -2y = 3x:
+
For <cmath>-2y = 3x</cmath>:
  
−
x can be between (-66, 66), but x has to be a multiple of 2.
+
</math>x<math> can be between </math>(-66, 66)<math>, but </math>x<math> has to be a multiple of </math>2<math>.
  
−
Therefore, there are </math>(66+66)/2 + 1 = 67<math> solutions for this case
+
Therefore, there are </math>(66+66)/2 + 1 = 67<math> solutions for this case.
  
−
However, the one overlap would be x = 0, because y would be 0 in both solutions.
+
However, the one overlap would be </math>x = 0<math>, because y would be </math>0<math> in both solutions.
  
−
Therefore, the answer is </math>51+67-1 = \boxed{117}$.
+
Therefore, the answer is </math>51+67-1 = \boxed{117}.$
  
 
-U-King3.14Root
 
-U-King3.14Root

Revision as of 10:25, 15 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: \[\frac{-b +- \sqrt b^2-4ac}{2a}\]

$$ (Error compiling LaTeX. Unknown error_msg)= \frac{$xy +- \sqrt(x^2y^2+(12*6*4*x^2*y^2)}{24x^2}<cmath>

</cmath>= \frac{xy +- \sqrt289x^2 y^2}{24x^2}<cmath>

</cmath>= \frac{18xy/24x^2$ (Error compiling LaTeX. Unknown error_msg), and $-16xy}{24x^2}<cmath>

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

</cmath>3y = 4x<cmath>; </cmath>-2y = 3x$ (Error compiling LaTeX. Unknown error_msg)$;

As$ (Error compiling LaTeX. Unknown error_msg)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 <cmath>-2y = 3x</cmath>:$ (Error compiling LaTeX. Unknown error_msg)x$can be between$(-66, 66)$, but$x$has to be a multiple of$2$.

Therefore, there are$ (Error compiling LaTeX. Unknown error_msg)(66+66)/2 + 1 = 67$solutions for this case.

However, the one overlap would be$ (Error compiling LaTeX. Unknown error_msg)x = 0$, because y would be$0$in both solutions.

Therefore, the answer is$ (Error compiling LaTeX. Unknown error_msg)51+67-1 = \boxed{117}.$

-U-King3.14Root

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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