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2012 AMC 10A Problems/Problem 25

Revision as of 08:28, 14 March 2012 by Lww (talk | contribs) (→‎Solution)

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 $n\geqslant x \geqslant y \geqslant z \geqslant 0$. 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.

[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 $|x-y|\geqslant1$, $|y-z|\geqslant1$, $|z-x|\geqslant1$.

Since $n\geqslant x \geqslant y \geqslant z \geqslant 0$, we have $x-y\geqslant1$, $y-z\geqslant1$, $z-x\geqslant1$.

The region of points $(x,y,z)$ satisfying the condition is show in the second Figure in black. 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 $n\geqslant 10$. 

The answer is D.

Alternatively, $\&#036; Solution II: 

Because$ (Error compiling LaTeX. Unknown error_msg)x$,$y$, and$z$are chosen independently and at random from the interval$[0,n]$, which means that$x$,$y$, and$z$distributes uniformly and independently in the interval$[0,n]$. So the point$(x, y, z)$distributes uniformly in the cubic$0\geqslant x, y, z \geqslant n$, as shown in the figure below. The volume of this cubic is$V_0=n^3$.

   As we want to find the probablity of the incident$ (Error compiling LaTeX. Unknown error_msg)A=\{ |x-y|\geqslant 1, |y-z|\geqslant1, |z-x|\geqslant 1 \}$, we should find the volume of the region of points such that$|x-y|\geqslant 1$,$|y-z|\geqslant 1$,$|z-x|\geqslant 1$and$0\geqslant x, y, z \geqslant n$.
   
Now we will find the region$ (Error compiling LaTeX. Unknown error_msg)\left{ (x,y,z)\ | \ 0\geqslant x, y, z \geqslant n, |x-y|\geqslant 1, |y-z|\geqslant 1, |z-x|\geqslant 1 \right} $.
   
   It is the result of the cubic by substracting by 3 slices corresponding to$ (Error compiling LaTeX. Unknown error_msg)|x-y|< 1$,$|y-z|< 1$, and$|z-x|< 1$, respectively.
   
   After cutting off a slice corresponding to$ (Error compiling LaTeX. Unknown error_msg)|x-y|< 1$, we get two triangular prisms, as shown in the figure.
   
   In order to observe it clearly, we rotate the object by the$ (Error compiling LaTeX. Unknown error_msg)z$axis, as shown.
   
   We can draw the slice corresponding to$ (Error compiling LaTeX. Unknown error_msg)|y-z|< 1$on the object.
   
   After cutting off the slice corresponding to$ (Error compiling LaTeX. Unknown error_msg)|y-z|< 1$, we have 4 pieces left.
   
   After cutting off the slice corresponding to$ (Error compiling LaTeX. Unknown error_msg)|z-x|< 1$, we have 6 congruent triangular prisms. 
   
   Here we draw all the pictures in colors in order to explain the solution clearly. That does not mean that the students should do it in this way in the examination. They can draw a figure with lines only, in this way.
   
   Every triangular prism has an altitude$ (Error compiling LaTeX. Unknown error_msg)n-2$and a base of isoceless right triangle with leg length$n-2$, so the volume is$(n-2)^3/6$.     Then the volume of the region$\left{ (x,y,z)\ | \ 0\geqslant x, y, z \geqslant n, |x-y|\geqslant 1, |y-z|\geqslant 1, |z-x|\geqslant 1 \right} $is$V_A=6*(n-2)^3/6$=$(n-2)^3$.
   
   So the probability of the incident$ (Error compiling LaTeX. Unknown error_msg)A$is$P(A)=$\dfrac{V_A}{V_0}$=$\dfrac{(n-2)^3}{n^3}$.
   
   Then we can get the answer the same as Solution I.
   
   If there is no choice for selection, we can find the minimum value of the integer $n$ without a calculator in this way if we do not substitude $n$ by the possible values one by one.
   
   Let $P(A)>1/2$, i.e.,  then $\dfrac{(n-2)^3}{n^3}>dfrac{1}{2}$,  so $\dfrac{n-2}{n}>dfrac{1}{\sqrt[^3\!]{2}}$, or $1-\dfrac{2}{n}>dfrac{1}{\sqrt[^3\!]{2}}$, hence $n>dfrac{2\sqrt[^3\!]{2}}{\sqrt[^3\!]{2}-1}$.

See Also

2012 AMC 10A (Problems • Answer Key • Resources)
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