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Difference between revisions of "2012 AMC 10A Problems/Problem 25"

(→‎Solution)
(→‎Solution)
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==Solution==
 
==Solution==
−
Without loss of generality, assume that <math>0 \le x \le y \le z \le n</math>.
+
'''Solution I:'''
  
−
Then, the possible choices for <math>x</math>, <math>y</math>, and <math>z</math> are represented by the expression <math>\frac{n^3}{3!}=\frac{n^3}{6}</math>.
+
Since <math>x,y,z</math> are all reals lacated in <math>[0, n]</math>, the number of choices for each one is infinite.
  
−
There are two restrictions: <math>x+1 \le y</math> and <math>y+1 \le z</math>.
+
Without loss of generality, assume that <math>n\geq x \geq y \geq z \geq 0</math>. Then the set of points <math>(x,y,z)</math> is a tetrahedron, or a triangular pyramid. The point <math>(x,y,z)</math> distributes uniformly in this region. If this is not easy to understand, read Solution II.
  
−
Now, let <math>y'=y+1</math>. Then, <math>x+2 \le y'</math> and <math>y' \le z</math>.
+
The altitude of the tetrahedron is <math>n</math> and the base is an isosceles right triangle with a leg length <math>n</math>. The volume is <math>V_1\dfrac{n^3}{6}</math>. As shown in the first figure in red color.
  
−
Then, let <math>x'=x+2</math>. Combining the two inequalities gives us <math>x' \le y' \le z</math>.
+
<asy>
 +
import three;
 +
unitsize(10cm);
 +
size(150);
 +
currentprojection=orthographic(1/2,-1,2/3);
 +
// three - currentprojection, orthographic
 +
draw((1,1,0)--(0,1,0)--(0,0,0),green);
 +
draw((0,0,0)--(0,0,1),green);
 +
draw((0,1,0)--(0,1,1),green);
 +
draw((1,1,0)--(1,1,1),green);
 +
//draw((1,0,0)--(1,0,1),green);
 +
draw((0,0,1)--(1,0,1)--(1,1,1)--(0,1,1)--cycle,green);
  
−
Since <math>0 \le x</math>, then <math>2 \le x'</math>. Thus, <math>2 \le x' \le y' \le z \le n</math>.
+
draw((0,0,0)--(1,0,0)--(1,1,0)--(0,0,0)--(1,0,1)--(1,0,0), red);
 +
draw((1,1,0)--(1,0,1), red);
 +
</asy>
  
−
There are <math>n-2</math> ways to choose each number; the successful choices are represented by <math>\frac{(n-2)^3}{6}</math>.
 
  
−
The probability then, is <math>\text{P}=\frac{\text{successful}}{\text{possible}}=\frac{\frac{(n-2)^3}{6}}{\frac{n^3}{6}}</math> which must be greater than <math>\frac{1}{2}</math>.
+
Now we will find the region with points satisfying  <math>|x-y|\geq1</math>, <math>|y-z|\geq1</math>, <math>|z-x|\geq1</math>.
  
−
Plug in values. Trying <math>(\text{C})</math> gives us <math>\frac{343}{729}</math>, which is less than <math>\frac{1}{2}</math>. Try the next integer, <math>10</math>, which gives us <math>\frac{512}{1000}</math> which is greater than <math>\frac{1}{2}</math>.
+
Since <math>n\geq x \geq y \geq z \geq 0</math>, we have <math>x-y\geq1</math>, <math>y-z\geq1</math>, <math>z-x\geq1</math>.
  
−
Thus, our answer is <math>(\text{D})</math>.
+
The region of points <math>(x,y,z)</math> satisfying the condition is show in the second Figure in black color. It is a tetrahedron, too.
 +
 
 +
<asy>
 +
import three;
 +
unitsize(10cm);
 +
size(150);
 +
currentprojection=orthographic(1/2, -1, 2/3);
 +
// three - currentprojection,  orthographic
 +
draw((1, 1, 0)--(0, 1, 0)--(0, 0, 0), green);
 +
draw((0, 0, 0)--(0, 0, 1), green);
 +
draw((0, 1, 0)--(0, 1, 1), green);
 +
draw((1, 1, 0)--(1, 1, 1), green);
 +
//draw((1, 0, 0)--(1, 0, 1), green);
 +
draw((0, 0, 1)--(1, 0, 1)--(1, 1, 1)--(0, 1, 1)--cycle, green);
 +
 
 +
draw((0, 0, 0)--(1, 0, 1)--(1, 1, 0)--(0, 0, 0),  dashed+red);
 +
draw((0, 0, 0)--(0.1, 0, 0),  dashed+red);
 +
draw((1, 0, 0.9)--(1, 0, 1),  dashed+red);
 +
draw((1, 0.9, 0)--(1, 1, 0),  dashed+red);
 +
 
 +
 
 +
draw((0.1, 0, 0)--(1, 0, 0.9)--(1, 0.9, 0)--(0.1, 0, 0));
 +
draw((1, 0, 0)--(0.1, 0, 0));
 +
draw((1, 0, 0.9)--(1, 0, 0));
 +
draw((1, 0.9, 0)--(1, 0, 0));
 +
</asy>
 +
 
 +
The volume of this region is <math>V_2=\dfrac{(n-2)^3}{6}</math>.
 +
 
 +
So the probability is <math>p=\dfrac{V_2}{V_1}=\dfrac{(n-2)^3}{n^3}</math>.
 +
 
 +
Substitude <math>n</math> by the values in the choices, we will find that when <math>n=10</math>, <math>p=\frac{512}{1000}>\frac{1}{2}</math>, when <math>n=9</math>, <math>p=</math>\frac{343}{729}<\frac{1}{2}<math>. So </math>n\geq 10<math>, the answer is </math>(\text{D})$.
 +
 
 +
 
 +
'''Solution II''':
  
 
== See Also ==
 
== See Also ==
  
 
{{AMC10 box|year=2012|ab=A|num-b=24|after=Last Problem}}
 
{{AMC10 box|year=2012|ab=A|num-b=24|after=Last Problem}}

Revision as of 08:22, 14 March 2012

Problem

Real numbers $x$, $y$, and $z$ are chosen independently and at random from the interval $[0,n]$ for some positive integer $n$. The probability that no two of $x$, $y$, and $z$ are within 1 unit of each other is greater than $\frac {1}{2}$. What is the smallest possible value of $n$?

$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11$

Solution

Solution I:

Since $x,y,z$ are all reals lacated in $[0, n]$, the number of choices for each one is infinite.

Without loss of generality, assume that $n\geq x \geq y \geq z \geq 0$. Then the set of points $(x,y,z)$ is a tetrahedron, or a triangular pyramid. The point $(x,y,z)$ distributes uniformly in this region. If this is not easy to understand, read Solution II.

The altitude of the tetrahedron is $n$ and the base is an isosceles right triangle with a leg length $n$. The volume is $V_1\dfrac{n^3}{6}$. As shown in the first figure in red color.

[asy] import three; unitsize(10cm); size(150); currentprojection=orthographic(1/2,-1,2/3); // three - currentprojection, orthographic draw((1,1,0)--(0,1,0)--(0,0,0),green); draw((0,0,0)--(0,0,1),green); draw((0,1,0)--(0,1,1),green); draw((1,1,0)--(1,1,1),green); //draw((1,0,0)--(1,0,1),green); draw((0,0,1)--(1,0,1)--(1,1,1)--(0,1,1)--cycle,green); draw((0,0,0)--(1,0,0)--(1,1,0)--(0,0,0)--(1,0,1)--(1,0,0), red); draw((1,1,0)--(1,0,1), red); [/asy]


Now we will find the region with points satisfying $|x-y|\geq1$, $|y-z|\geq1$, $|z-x|\geq1$.

Since $n\geq x \geq y \geq z \geq 0$, we have $x-y\geq1$, $y-z\geq1$, $z-x\geq1$.

The region of points $(x,y,z)$ satisfying the condition is show in the second Figure in black color. It is a tetrahedron, too.

[asy] import three; unitsize(10cm); size(150); currentprojection=orthographic(1/2, -1, 2/3); // three - currentprojection, orthographic draw((1, 1, 0)--(0, 1, 0)--(0, 0, 0), green); draw((0, 0, 0)--(0, 0, 1), green); draw((0, 1, 0)--(0, 1, 1), green); draw((1, 1, 0)--(1, 1, 1), green); //draw((1, 0, 0)--(1, 0, 1), green); draw((0, 0, 1)--(1, 0, 1)--(1, 1, 1)--(0, 1, 1)--cycle, green); draw((0, 0, 0)--(1, 0, 1)--(1, 1, 0)--(0, 0, 0), dashed+red); draw((0, 0, 0)--(0.1, 0, 0), dashed+red); draw((1, 0, 0.9)--(1, 0, 1), dashed+red); draw((1, 0.9, 0)--(1, 1, 0), dashed+red); draw((0.1, 0, 0)--(1, 0, 0.9)--(1, 0.9, 0)--(0.1, 0, 0)); draw((1, 0, 0)--(0.1, 0, 0)); draw((1, 0, 0.9)--(1, 0, 0)); draw((1, 0.9, 0)--(1, 0, 0)); [/asy]

The volume of this region is $V_2=\dfrac{(n-2)^3}{6}$.

So the probability is $p=\dfrac{V_2}{V_1}=\dfrac{(n-2)^3}{n^3}$.

Substitude $n$ by the values in the choices, we will find that when $n=10$, $p=\frac{512}{1000}>\frac{1}{2}$, when $n=9$, $p=$\frac{343}{729}<\frac{1}{2}$. So$n\geq 10$, the answer is$(\text{D})$. 


Solution II:

See Also

2012 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Last Problem
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions
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