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

Latest revision as of 10:37, 31 July 2025

Problem

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

Video solution by grogg007

https://youtu.be/PNBxBvvjbcU?t=401

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

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! What we are doing is considering the quadratic with respect to $1$ , where $12x^2$ is the coefficient of $1^2$ , $-xy$ is the coefficient of $1$ , and $-6y^2$ is the constant term.

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

$\frac{xy \pm \sqrt{x^2 y^2 + (12\cdot6\cdot4\cdot x^2 \cdot y^2)}}{24x^2}$

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

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

Rather than putting this equation as zero, the numerators and denominators must be equal [EDIT: We set each value equal to one because the quadratic is taken with respect to the root $1$ ]. 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 from $[-75,75]$ , 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 -clarified by golden_star_123

Video Solution 1 by SpreadTheMathLove

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

Video Solution by yjtest

https://www.youtube.com/watch?v=P7LEfKK1Vew

@@ Line 1: / Line 1: @@
 ==Problem==
 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>.
+==Video solution by [[User:grogg007|grogg007]]==
+https://youtu.be/PNBxBvvjbcU?t=401
 ==Solution 1==
 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,
@@ Line 21: / Line 25: @@
 ~[[User:Mathkiddus|mathkiddus]]
+==Solution 2==
+First, notice that <math>(0,0)</math> 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>.
+Now we can count the edge case of <math>(0,0)</math>. The answer is therefore <math>50+66+1 = \boxed{117}</math>.
+~lprado
+==Solution 3==
+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! What we are doing is considering the quadratic with respect to <math>1</math>, where <math>12x^2</math> is the coefficient of <math>1^2</math>, <math>-xy</math> is the coefficient of <math>1</math>, and <math>-6y^2</math> is the constant term.
+You get: <cmath>\frac{-b \pm \sqrt{b^2-4ac}}{2a}</cmath>
+<cmath>
+\frac{xy \pm \sqrt{x^2 y^2 + (12\cdot6\cdot4\cdot x^2 \cdot y^2)}}{24x^2}
+</cmath>
+<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 [EDIT: We set each value equal to one because the quadratic is taken with respect to the root <math>1</math>]. These two equations simplify to:
+<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 from <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.
+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>.
+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.
+Therefore, the answer is <math>51+67-1 = \boxed{117}.</math>
+-U-King3.14Root
+-LaTeX corrected by Andrew2019
+-clarified by golden_star_123
+==Video Solution 1 by SpreadTheMathLove==
+https://www.youtube.com/watch?v=J-0BapU4Yuk
+==Video Solution by yjtest==
+https://www.youtube.com/watch?v=P7LEfKK1Vew
+==See also==
+{{AIME box|year=2025|num-b=3|num-a=5|n=I}}
+{{MAA Notice}}

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

Page

Toolbox

Search