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

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

@@ Line 42: / Line 42: @@
 Although this solution may seem to be misleading, it works!
-You get: <cmath>\frac{-b +- \sqrt b^2-4ac}{2a}</cmath>
+You get: <cmath>\frac{-b \pm \sqrt{b^2-4ac}}{2a}</cmath>
-<math></math>= \frac{<math>xy +- \sqrt(x^2y^2+(12*6*4*x^2*y^2)}{24x^2}<cmath>
+<cmath>= \frac{xy \pm \sqrt{x^2y^2+(12\cdot6\cdot4\cdotx^2\cdoty^2)}}{24x^2}</cmath>
-</cmath>= \frac{xy +- \sqrt289x^2 y^2}{24x^2}<cmath>
+<cmath>= \frac{xy \pm\sqrt{289x^2 y^2}}{24x^2}</cmath>
-</cmath>= \frac{18xy/24x^2</math>, and <math>-16xy}{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:
-</cmath>3y = 4x<cmath>; </cmath>-2y = 3x</math><math>;
+<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:
+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 <cmath>-2y = 3x</cmath>:
+For <cmath>-2y = 3x:</cmath>
-</math>x<math> can be between </math>(-66, 66)<math>, but </math>x<math> has to be a multiple of </math>2<math>.
+<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 </math>x = 0<math>, because y would be </math>0<math> 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}.</math>
 -U-King3.14Root
+-LaTeX corrected by Andrew2019, though idk if this is what you wanted to say
 ==Video Solution 1 by SpreadTheMathLove==

2025 AIME I (Problems • Answer Key • Resources)
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