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Difference between revisions of "2025 AIME I Problems/Problem 12"
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==Problem==
 
==Problem==
 
The set of points in <math>3</math>-dimensional coordinate space that lie in the plane <math>x+y+z=75</math> whose coordinates satisfy the inequalities <cmath>x-yz<y-zx<z-xy</cmath>forms three disjoint convex regions. Exactly one of those regions has finite area. The area of this finite region can be expressed in the form <math>a\sqrt{b},</math> where <math>a</math> and <math>b</math> are positive integers and <math>b</math> is not divisible by the square of any prime. Find <math>a+b.</math>
 
The set of points in <math>3</math>-dimensional coordinate space that lie in the plane <math>x+y+z=75</math> whose coordinates satisfy the inequalities <cmath>x-yz<y-zx<z-xy</cmath>forms three disjoint convex regions. Exactly one of those regions has finite area. The area of this finite region can be expressed in the form <math>a\sqrt{b},</math> where <math>a</math> and <math>b</math> are positive integers and <math>b</math> is not divisible by the square of any prime. Find <math>a+b.</math>
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==Solution==
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Consider <math>x-yz<y-zx<z-xy</math>. From <math>x-yz<y-zx</math>, we find <math>z(y-x)>x-y</math>. Thus, if <math>x>y</math>, then <math>z<-1</math>, and if <math>x<y</math>, then <math>z>-1</math>. Similarly, taking another pair of the inequalities yields <math>y>-1</math> when <math>z>x</math> and <math>y<-1</math> when <math>x>z</math>. Finally, taking the third pair yields <math>x>-1</math> if <math>z>y</math> and <math>x<-1</math> if <math>z<y</math>.
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Consider the first two resultant pairs of inequalities. Taking them pairwise (one from the first set and one from the second set) results in four cases:
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1. Combining <math>z<-1</math> if <math>x>y</math> and <math>y>-1</math> if <math>z>x</math> yields <math>-1>z>x>y>-1</math>, a contradiction.
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2. Combining <math>z<-1</math> if <math>x>y</math> and <math>y<-1</math> if <math>z<x</math> yields <math>x,-1>y,z</math>.
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3. Combining <math>z>-1</math> if <math>x<y</math> and <math>y>-1</math> if <math>z>x</math> yields <math>y,z>-1,x</math>.
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4. Combining <math>z>-1</math> if <math>x<y</math> and <math>y<-1</math> if <math>z<x</math> yields <math>-1>y>x>z>-1</math>, a contradiction.
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Now we have only two satisfactory inequalities. We now consider the third pair of inequalities (<math>x>-1</math> if <math>z>y</math> and <math>x<-1</math> if <math>z<y</math>). Taking the two sets pairwise:
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1. Combining <math>x,-1>y,z</math> and <math>x>-1</math> if <math>z>y</math> yields <math>x>-1>z>y</math>. Consider some valid <math>x,y,z</math> that satisfy <math>x+y+z=75</math> and <math>x>-1>z>y</math>. We can infinitely increase <math>x</math> while decreasing <math>y</math> by the same amount, leading to another valid triple, so this case is infinite and we do not consider this case (for instance, if <math>x=100,y=-13,z=-12</math>, then <math>x=100+a,y=-13-a,z=-12</math> is a valid triple for all nonnegative <math>a</math>).
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2. Combining <math>y,z>-1,x</math> and <math>x>-1</math> if <math>z>y</math> yields <math>z>y>x>-1</math>. This case is finite due to the lower bound.
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3. Combining <math>x,-1>y,z</math> and <math>x<-1</math> if <math>z<y</math> yields <math>-1>x>y>z</math>. There are no possible solutions since <math>x,y,z</math> are negative from this inequality, but at least one must be positive to satisfy <math>x+y+z=75</math>, a contradiction.
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4. Combining <math>y,z>-1,x</math> and <math>x<-1</math> if <math>z<y</math> yields <math>y>z>-1>x</math>. By the same argument as in Case 1, this is an infinite case.
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Thus we are tasked with finding the area of the figure formed by all triples <math>x,y,z</math> satisfying <math>x+y+z=75</math> and <math>z>y>x>-1</math>. We consider edge cases, so we maximize each variable by the largest amount possible to find three triples <math>(77,-1,-1),(38,38,-1),(25,25,25)</math>. We assume that these are the only edge cases (so the figure forms a triangle), and we can use the [[Distance Formula]]. We find that the three side lengths of our triangle are <math>39\sqrt{2},13\sqrt[6},26\sqrt{6}</math>. These side lengths just so happen to form a <math>30-60-90</math> triangle with legs <math>13\sqrt{6}</math> and <math>39\sqrt{2}</math>, so the area of the triangle is
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<cmath>\frac{1}{2}\cdot13\sqrt{6}\cdot39\sqrt{2}=507\sqrt{3}</cmath>
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Thus the answer is <math>507+3=\boxed{510}</math>. ~eevee9406
  
 
==See also==
 
==See also==

Revision as of 21:45, 13 February 2025

Problem

The set of points in $3$-dimensional coordinate space that lie in the plane $x+y+z=75$ whose coordinates satisfy the inequalities \[x-yz<y-zx<z-xy\]forms three disjoint convex regions. Exactly one of those regions has finite area. The area of this finite region can be expressed in the form $a\sqrt{b},$ where $a$ and $b$ are positive integers and $b$ is not divisible by the square of any prime. Find $a+b.$

Solution

Consider $x-yz<y-zx<z-xy$. From $x-yz<y-zx$, we find $z(y-x)>x-y$. Thus, if $x>y$, then $z<-1$, and if $x<y$, then $z>-1$. Similarly, taking another pair of the inequalities yields $y>-1$ when $z>x$ and $y<-1$ when $x>z$. Finally, taking the third pair yields $x>-1$ if $z>y$ and $x<-1$ if $z<y$.


Consider the first two resultant pairs of inequalities. Taking them pairwise (one from the first set and one from the second set) results in four cases:


1. Combining $z<-1$ if $x>y$ and $y>-1$ if $z>x$ yields $-1>z>x>y>-1$, a contradiction.

2. Combining $z<-1$ if $x>y$ and $y<-1$ if $z<x$ yields $x,-1>y,z$.

3. Combining $z>-1$ if $x<y$ and $y>-1$ if $z>x$ yields $y,z>-1,x$.

4. Combining $z>-1$ if $x<y$ and $y<-1$ if $z<x$ yields $-1>y>x>z>-1$, a contradiction.


Now we have only two satisfactory inequalities. We now consider the third pair of inequalities ($x>-1$ if $z>y$ and $x<-1$ if $z<y$). Taking the two sets pairwise:


1. Combining $x,-1>y,z$ and $x>-1$ if $z>y$ yields $x>-1>z>y$. Consider some valid $x,y,z$ that satisfy $x+y+z=75$ and $x>-1>z>y$. We can infinitely increase $x$ while decreasing $y$ by the same amount, leading to another valid triple, so this case is infinite and we do not consider this case (for instance, if $x=100,y=-13,z=-12$, then $x=100+a,y=-13-a,z=-12$ is a valid triple for all nonnegative $a$).

2. Combining $y,z>-1,x$ and $x>-1$ if $z>y$ yields $z>y>x>-1$. This case is finite due to the lower bound.

3. Combining $x,-1>y,z$ and $x<-1$ if $z<y$ yields $-1>x>y>z$. There are no possible solutions since $x,y,z$ are negative from this inequality, but at least one must be positive to satisfy $x+y+z=75$, a contradiction.

4. Combining $y,z>-1,x$ and $x<-1$ if $z<y$ yields $y>z>-1>x$. By the same argument as in Case 1, this is an infinite case.

Thus we are tasked with finding the area of the figure formed by all triples $x,y,z$ satisfying $x+y+z=75$ and $z>y>x>-1$. We consider edge cases, so we maximize each variable by the largest amount possible to find three triples $(77,-1,-1),(38,38,-1),(25,25,25)$. We assume that these are the only edge cases (so the figure forms a triangle), and we can use the Distance Formula. We find that the three side lengths of our triangle are $39\sqrt{2},13\sqrt[6},26\sqrt{6}$ (Error compiling LaTeX. Unknown error_msg). These side lengths just so happen to form a $30-60-90$ triangle with legs $13\sqrt{6}$ and $39\sqrt{2}$, so the area of the triangle is

\[\frac{1}{2}\cdot13\sqrt{6}\cdot39\sqrt{2}=507\sqrt{3}\]

Thus the answer is $507+3=\boxed{510}$. ~eevee9406

See also

2025 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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