2025 AIME I Problems/Problem 12

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 $39\sqrt{2},13\sqrt{6},26\sqrt{6}$ . 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

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

Page

Toolbox

Search

2025 AIME I Problems/Problem 12

Problem

Solution

See also