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Difference between revisions of "2025 AIME I Problems/Problem 12"
Line 1: Line 1:
 
==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>
 +
==Solution 1==
 +
Rewriting we have <math>z=75-x-y.</math>
  
−
==Solution==
+
From the inequality <math>x-yz<y-zx</math> we can rewrite to get,
 +
<cmath>x-y(75-x-y)< y-x(75-x-y).</cmath> <cmath>76x-76y+y^2-x^2<0.</cmath> <cmath>(x+y+76)(x-y)<0.</cmath>
 +
 
 +
Similarly from the inequality <math>y-zx<z-xy</math> we rewrite to get,
 +
<cmath>y-x(75-x-y)<(75-x-y)-xy.</cmath>
 +
<cmath>2y + 2xy + x^2 - 74x - 75 < 0.</cmath>
 +
<cmath>(x + 1)(2y + x - 75) < 0.</cmath>
 +
Our next step is a visual which can be seen by roughly graphing the two inequalities. The first inequality is similar to a bow tie(you have to graph to see it lol) with bounds of <math>x-y=0</math> and <math>76+x+y=0.</math> The second one is a bow tie with edges of <math>x+1=0</math> and <math>2y+x-75=0.</math>Here is the region of our solutions.
 +
<asy>
 +
import graph;
 +
size(400, 400);
 +
real xmin = -250, xmax = 250;
 +
real ymin = -150, ymax = 150;
 +
draw((xmin,0)--(xmax,0), black+0.8bp, Arrow);
 +
draw((0,ymin)--(0,ymax), black+0.8bp, Arrow);
 +
pair A = (-38, -38),
 +
    B = (-10, -10),
 +
    C = (-227, 151),
 +
    D = (-10, 85/2),
 +
    E = (25, 25);
 +
path L1 = (xmin, xmin)--(xmax, xmax);
 +
path L2 = (xmin, -76-xmin)--(xmax, -76-xmax);
 +
path L3 = ((-10, ymin)--(-10, ymax));
 +
path L4 = (xmin, (75-xmin)/2)--(xmax, (75-xmax)/2);
 +
fill(B--D--E--cycle, lightgreen);
 +
draw(L1, blue);
 +
draw(L2, blue);
 +
draw(L3, red);
 +
draw(L4, red);
 +
dot(A, black); label("$(-38,-38)$", A, SW);
 +
dot(B, black); label("$(-1,-1)$", B, SE);
 +
dot(C, black); label("$(-227,151)$", C, NW);
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dot(D, black); label("$(-1,38)$", D, NE);
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dot(E, black); label("$(25,25)$", E, SE);
 +
</asy>
 +
It is simple to find the intersections of those which are <math>(-1,38),(25,25)</math> and <math>(-1,-1).</math> The sidelengths of this triangle are <math>39\sqrt{2},26\sqrt{6},13\sqrt{6}</math> which is a <math>30-60-90</math> gives us an area of <cmath>\frac{1}{2}(39\sqrt{2})(13\sqrt{6})=507\sqrt{3}\implies\boxed{510}.</cmath>
 +
~mathkiddus ~plang2008
 +
==Solution 2==
 
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>.
 
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>.
  

Revision as of 21:48, 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 1

Rewriting we have $z=75-x-y.$

From the inequality $x-yz<y-zx$ we can rewrite to get, \[x-y(75-x-y)< y-x(75-x-y).\] \[76x-76y+y^2-x^2<0.\] \[(x+y+76)(x-y)<0.\]

Similarly from the inequality $y-zx<z-xy$ we rewrite to get, \[y-x(75-x-y)<(75-x-y)-xy.\] \[2y + 2xy + x^2 - 74x - 75 < 0.\] \[(x + 1)(2y + x - 75) < 0.\] Our next step is a visual which can be seen by roughly graphing the two inequalities. The first inequality is similar to a bow tie(you have to graph to see it lol) with bounds of $x-y=0$ and $76+x+y=0.$ The second one is a bow tie with edges of $x+1=0$ and $2y+x-75=0.$Here is the region of our solutions. [asy] import graph; size(400, 400); real xmin = -250, xmax = 250; real ymin = -150, ymax = 150; draw((xmin,0)--(xmax,0), black+0.8bp, Arrow); draw((0,ymin)--(0,ymax), black+0.8bp, Arrow); pair A = (-38, -38), B = (-10, -10), C = (-227, 151), D = (-10, 85/2), E = (25, 25); path L1 = (xmin, xmin)--(xmax, xmax); path L2 = (xmin, -76-xmin)--(xmax, -76-xmax); path L3 = ((-10, ymin)--(-10, ymax)); path L4 = (xmin, (75-xmin)/2)--(xmax, (75-xmax)/2); fill(B--D--E--cycle, lightgreen); draw(L1, blue); draw(L2, blue); draw(L3, red); draw(L4, red); dot(A, black); label("$(-38,-38)$", A, SW); dot(B, black); label("$(-1,-1)$", B, SE); dot(C, black); label("$(-227,151)$", C, NW); dot(D, black); label("$(-1,38)$", D, NE); dot(E, black); label("$(25,25)$", E, SE); [/asy] It is simple to find the intersections of those which are $(-1,38),(25,25)$ and $(-1,-1).$ The sidelengths of this triangle are $39\sqrt{2},26\sqrt{6},13\sqrt{6}$ which is a $30-60-90$ gives us an area of \[\frac{1}{2}(39\sqrt{2})(13\sqrt{6})=507\sqrt{3}\implies\boxed{510}.\] ~mathkiddus ~plang2008

Solution 2

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
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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