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Difference between revisions of "2025 AIME I Problems/Problem 12"

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Thus the answer is <math>507+3=\boxed{510}</math>. ~eevee9406
 
Thus the answer is <math>507+3=\boxed{510}</math>. ~eevee9406
 +
 +
==Solution 2==
 +
Decomposing the inequality chain:
 +
<cmath>x-yz<y-zx \quad \text{and} \quad y-zx<z-xy</cmath>
 +
which is equivalent to
 +
<cmath>(x-y)(z+1)<0 \quad \text{and} \quad (y-z)(x+1)<0</cmath>
 +
Substituting <math>z</math> with <math>z=75-x-y</math> and simplifying yields
 +
<cmath>(x-y)(x+y-76)>0 \quad \text{and} \quad (x+2y-75)(x+1)<0</cmath>
 +
See that the solution to the first inequality is
 +
<cmath>x-y>0, \, x+y-76>0 \quad \text{(I)} \quad \text{or} \quad x-y<0, \, x+y-76<0 \quad \text{(II)} </cmath>
 +
Applying a similar method results in the solution to the second:
 +
<cmath>x+2y-75>0, \, x+1<0 \quad \text{(III)} \quad \text{or} \quad x+2y-75<0, \, x+1>0 \quad \text{(IV)}</cmath>
 +
Trying each grouping (i.e. let <math>\text{(I)}</math> and <math>\text{(III)}</math>, <math>\text{(I)}</math> and <math>\text{(IV)}</math>, <math>\text{(II)}</math> and <math>\text{(III)}</math>, or <math>\text{(II)}</math> and <math>\text{(IV)}</math> be satisfied at the same time) and graphing shows that when <math>\text{(II)}</math> and <math>\text{(IV)}</math> are both satisfied, a triangle whose vertices are <math>(-1,38)</math>, <math>(-1,-1)</math>, and <math>(25,25)</math> is formed. Further calculations show that the area of the triangle is <math>507</math>. However, this is not the final answer. We have projected the original shape to the <math>xy</math>-plane by substituting <math>z</math>. We know that for a surface defined by the equation <math>z=f(x,y)</math>, the area element <math>dS</math> for this surface is given by
 +
<cmath>dS=\sqrt{1+(f_x)^2+(f_y)^2}dxdy</cmath>
 +
where <math>f_x</math> and <math>f_y</math> are the partial derivatives of the function <math>f(x,y)</math> with respect to <math>x</math> and <math>y</math>. For the plane <math>x+y+z=75</math> where <math>f(x,y)=75-x-y</math>, computation gives
 +
<cmath>f_x=-1, f_y=-1</cmath>
 +
Substituting these into the original equation to get
 +
<cmath>dS=\sqrt{3}dxdy</cmath>
 +
This implies that to find the area of the original shape, we have to multiply the area of its projection on the <math>xy</math>-plane by <math>\sqrt{3}</math>. Therefore, the area of the original shape is <math>507\sqrt{3}</math>, with final answer <math>\boxed{510}</math>.
 +
 +
~[[User:Bloggish|Bloggish]]
  
 
==See also==
 
==See also==

Revision as of 20:59, 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 $39\sqrt{2},13\sqrt{6},26\sqrt{6}$. 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

Solution 2

Decomposing the inequality chain: \[x-yz<y-zx \quad \text{and} \quad y-zx<z-xy\] which is equivalent to \[(x-y)(z+1)<0 \quad \text{and} \quad (y-z)(x+1)<0\] Substituting $z$ with $z=75-x-y$ and simplifying yields \[(x-y)(x+y-76)>0 \quad \text{and} \quad (x+2y-75)(x+1)<0\] See that the solution to the first inequality is \[x-y>0, \, x+y-76>0 \quad \text{(I)} \quad \text{or} \quad x-y<0, \, x+y-76<0 \quad \text{(II)}\] Applying a similar method results in the solution to the second: \[x+2y-75>0, \, x+1<0 \quad \text{(III)} \quad \text{or} \quad x+2y-75<0, \, x+1>0 \quad \text{(IV)}\] Trying each grouping (i.e. let $\text{(I)}$ and $\text{(III)}$, $\text{(I)}$ and $\text{(IV)}$, $\text{(II)}$ and $\text{(III)}$, or $\text{(II)}$ and $\text{(IV)}$ be satisfied at the same time) and graphing shows that when $\text{(II)}$ and $\text{(IV)}$ are both satisfied, a triangle whose vertices are $(-1,38)$, $(-1,-1)$, and $(25,25)$ is formed. Further calculations show that the area of the triangle is $507$. However, this is not the final answer. We have projected the original shape to the $xy$-plane by substituting $z$. We know that for a surface defined by the equation $z=f(x,y)$, the area element $dS$ for this surface is given by \[dS=\sqrt{1+(f_x)^2+(f_y)^2}dxdy\] where $f_x$ and $f_y$ are the partial derivatives of the function $f(x,y)$ with respect to $x$ and $y$. For the plane $x+y+z=75$ where $f(x,y)=75-x-y$, computation gives \[f_x=-1, f_y=-1\] Substituting these into the original equation to get \[dS=\sqrt{3}dxdy\] This implies that to find the area of the original shape, we have to multiply the area of its projection on the $xy$-plane by $\sqrt{3}$. Therefore, the area of the original shape is $507\sqrt{3}$, with final answer $\boxed{510}$.

~Bloggish

See also

2025 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
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All AIME Problems and Solutions

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