2012 AMC 10A Problems/Problem 25

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 I

Since $x,y,z$ are all reals located 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.

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

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

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

$[asy] import three; unitsize(10cm); size(150); currentprojection=orthographic(1/2, -1, 2/3); // three - currentprojection, orthographic draw((1, 1, 0)--(0, 1, 0)--(0, 0, 0), dashed+green); draw((0, 0, 0)--(0, 0, 1), green); draw((0, 1, 0)--(0, 1, 1), dashed+green); draw((1, 0, 0)--(1, 0, 1), green); draw((0, 0, 1)--(1, 0, 1)--(1, 1, 1)--(0, 1, 1)--cycle, green); draw((1,0,0)--(1,1,0)--(0,0,0)--(1,1,1), dashed+red); draw((0,0,0)--(1,0,0)--(1,1,1), red); draw((1,1,1)--(1,1,0)--(1,0.9,0), red); draw((1, 0.1, 0)--(1, 0.9, 0)--(1, 0.9, 0.8)--cycle); draw((0.2, 0.1, 0)--(1, 0.9, 0.8),dashed); draw((1, 0.1, 0)--(0.2, 0.1, 0)--(1, 0.9, 0),dashed); [/asy]$

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 $n\geqslant 10$ .

So the answer is $\boxed{\textbf{(D)}\ 10}$ .

Solution II

Because $x$ , $y$ , 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 $0\leqslant x, y, z \leqslant n$ , 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 $A=\big\{ |x-y|\geqslant 1, |y-z|\geqslant1, |z-x|\geqslant 1 \big\}$ , 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 $0\leqslant x, y, z \leqslant n$ .

Now we will find the region $\big\{ (x,y,z)\ | \ 0\leqslant x, y, z \leqslant n, |x-y|\geqslant 1, |y-z|\geqslant 1, |z-x|\geqslant 1 \big\}$ .

The region can be generated by cutting off 3 slices corresponding to $|x-y|< 1$ , $|y-z|< 1$ , and $|z-x|< 1$ , respectively, from the cubic.

After cutting off a slice corresponding to $|x-y|< 1$ , we get two triangular prisms, as shown in the figure.

In order to observe the object clearly, we rotate the object by the $z$ axis, as shown.

We can draw the slice corresponding to $|y-z|< 1$ on the object.

After cutting off the slice corresponding to $|y-z|< 1$ , we have 4 pieces left.

After cutting off the slice corresponding to $|z-x|< 1$ , 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 the examination. They can draw a figure with lines only, as shown below.

Every triangular pyramid has an altitude $n-2$ 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 $\big\{ (x,y,z)\ | \ 0\leqslant x, y, z \leqslant n, |x-y|\geqslant 1, |y-z|\geqslant 1, |z-x|\geqslant 1 \big\}$ is $V_A=6\times(n-2)^3/6$ = $(n-2)^3$ .

So the probability of the incident $A$ is $P(A)=\dfrac{V_A}{V_0}$ = $\dfrac{(n-2)^3}{n^3}$ .

Then we can get the answer the same way as Solution I.

The answer is $\boxed{\textbf{(D)}\ 10}$ .

If there is no choice for selection, we can also find the minimum value of the integer $n$ if we do not substitute $n$ by the possible values one by one.

Let $P(A)>1/2$ , i.e., $\dfrac{(n-2)^3}{n^3}>\dfrac{1}{2}$ , so $\dfrac{n-2}{n}>\dfrac{1}{\sqrt[^3\!]{2}}$ , or $1-\dfrac{2}{n}>\dfrac{1}{\sqrt[^3\!]{2}}$ , hence $n>\dfrac{2\sqrt[^3\!]{2}}{\sqrt[^3\!]{2}-1}$ .

Now we will estimate the value of $\dfrac{2\sqrt[^3\!]{2}}{\sqrt[^3\!]{2}-1}$ without a calculator.

Since $a^3-1$ = $(a-1)(a^2+a+1)$ , so $\dfrac{2\sqrt[^3\!]{2}}{\sqrt[^3\!]{2}-1}$ = $\dfrac{2\sqrt[^3\!]{2}\times\left( \sqrt[^3\!]{2}^2+\sqrt[^3\!]{2}+1\right)}{\left( \sqrt[^3\!]{2}-1\right)\left( \sqrt[^3\!]{2}^2+\sqrt[^3\!]{2}+1\right)}$ = $\dfrac{2\times\left( 2+\sqrt[^3\!]{2}^2+\sqrt[^3\!]{2}\right)}{ \sqrt[^3\!]{2}^3-1}$ = $2\times\left( 2+\sqrt[^3\!]{4}+\sqrt[^3\!]{2}\right)$ .

Now we would get the approximation of $\sqrt[^3\!]{4}$ and $\sqrt[^3\!]{2}$ .

In order to avoid compicated computation, we get the approximation with one decimal digit only.

Estimation of $\sqrt[^3\!]{2}$ .

Since $1.5^3=2.25\times1.5>2$ , so $1<\sqrt[^3\!]{2}<1.5$ .

The mean of 1 and 1.5 with one decimal digit is about 1.3 .

As $1.3^3=1.69\times 1.3=2.197>2$ , so $1<\sqrt[^3\!]{2}<1.3$ .

The mean of 1 and 1.3 with one decimal digit is about 1.2.

As $1.2^3=1.44\times 1.2=1.728<2$ , so $1.2<\sqrt[^3\!]{2}<1.3$ .

Estimation of $\sqrt[^3\!]{4}$ .

As $\sqrt[^3\!]{4}=\sqrt[^3\!]{2}^2$ , so $1.2^2<\sqrt[^3\!]{4}<1.3^2$ , then $1.24<\sqrt[^3\!]{4}<1.69$ .

As $1.5^3=2.25\times 1.5=3.375<4$ , so $1.5<\sqrt[^3\!]{4}<1.69$ .

The mean of 1.5 and 1.69 with one decimal digit is about 1.6.

As $1.6^3=(16/10)^3=(2^4/10)^3=2^{12}/10^3=4\times 2^10/10^3=4\times 1.024>4$ , so $1.5<\sqrt[^3\!]{4}<1.6$ .

Then $2\times(2+1.5+1.2)<2\times\left(2+\sqrt[^3\!]{4}+\sqrt[^3\!]{2}\right)<2\times(2+1.6+1.3)$ , i.e., $9.4<2\times\left(2+\sqrt[^3\!]{4}+\sqrt[^3\!]{2}\right)<9.8$ ,

As $n>2\times\left(2+\sqrt[^3\!]{4}+\sqrt[^3\!]{2}\right)$ , So the minimal value of integer $n$ is 10.

Appendix

This solution is motivated by the suggestive formula $\frac{(n-2)^{3}}{n^{3}}$ .

The problem generalizes easily to $k$ -dimensional real space $\mathbb{R}^{k}$ . In the general $k$ -dimensional case, we are asked to find the probability that a randomly chosen $k$ -tuple $(x_{1},\dotsc,x_{k}) \in [0,n]^{k}$ satisfies $|x_{i} - x_{j}| > 1$ for all $i \ne j$ . To avoid repetition, let us say that $(x_{1},\dotsc,x_{k})$ is spaced-out if $|x_{i} - x_{j}| > 1$ for all $i \ne j$ .

Let $C$ be the $k$ -dimensional hyper-cube of side length $n$ : $C = [0,n]^{k} = \big\{(x_{1},\dotsc,x_{k}) \in \mathbb{R}^{k} \;:\; 0 \leqslant x_{i} \leqslant n \text{ for all }i \big\} \;.$ Then $C$ has volume $n^{k}$ . Let $S$ be the set of spaced-out $k$ -tuples $(x_{1},\dotsc,x_{k})$ . The desired probability is Vol $(S)/n^{k}$ .

The set of $k$ -tuples $(x_{1},\dotsc,x_{k})$ such that there exist distinct indices $i, j$ such that $x_{i} = x_{j}$ has volume $0$ , so we may restrict our attention to $k$ -tuples such that $x_{i} \ne x_{j}$ for all $i \ne j$ .

Further, the condition that $(x_{1},\dotsc,x_{k})$ is spaced-out is "invariant upon permuting the indices"; in other words, if $\sigma$ is a permutation of the set of indices $\{1,\dotsc,k\}$ , then $(x_{1},\dotsc,x_{k})$ is spaced-out if and only if $(x_{\sigma(1)},\dotsc,x_{\sigma(k)})$ is spaced-out. Therefore, we may consider the set of spaced-out $k$ -tuples $(x_{1},\dotsc,x_{k})$ which additionally satisfy $x_{1} < \dotsb < x_{k}$ . Let us denote this set by $T$ . This condition is equivalent to $0 \leqslant x_{1} < x_{2} - 1 < \dotsb < x_{i}-(i-1) < \dotsb < x_{k}-(k-1) \leqslant n-(k-1) \;.$ Let us choose new variables $y_{i} = x_{i} - (i-1)$ for $i = 1,\dotsc,k$ . This change of variables is just a translation of each $(x_{1},\dotsc,x_{k})$ by the vector $(0,1,\dots,k-1)$ ; in the above solutions, it corresponds to taking the 6 tetrahedrons and gluing them together to form a cube.

We now compute the volume of the set of $(y_{1},\dotsc,y_{k}) \in [0,n-(k-1)]^{k}$ which satisfy $y_{1} < \dotsb < y_{k}$ . As above, we can disregard any $(y_{1},\dotsc,y_{k})$ such that $y_{i} = y_{j}$ for some $i \ne j$ . Given any $(y_{1},\dotsc,y_{k})$ such that $y_{i} \ne y_{j}$ for all $i \ne j$ , there exists exactly one permutation $\sigma$ of the indices $\{1,\dotsc,k\}$ such that $y_{\sigma(1)} < \dotsb < y_{\sigma(k)}$ . Since there are $k!$ permutations of $\{1,\dotsc,k\}$ , the desired volume is equal to $\frac{1}{k!}$ times the volume of the $k$ -dimensional hyper-cube of side length $n-(k-1)$ , which is $\frac{1}{k!}(n-(k-1))^{k}$ . Hence $T$ has volume $\frac{1}{k!}(n-(k-1))^{k}$ as well and $S$ has volume $(n-(k-1))^{k}$ . Hence the desired probability is $\frac{(n-(k-1))^{k}}{n^{k}}$ .

2012 AMC 10A (Problems • Answer Key • Resources)
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2012 AMC 10A Problems/Problem 25

Contents

Problem

Solution I

Solution II

Appendix

See Also