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2012 AMC 10A Problems/Problem 25

Revision as of 14:30, 15 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)--(1,1,1)--(1,1,0), 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), 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((1, 0.9, 0.9)--(1, 1, 1), dashed+red); draw((1, 0.9, 0)--(1, 1, 0), dashed+red); draw((0,0,0)--(1,1,0)--(1,1,1)--(0, 0, 0)--(0.1, 0, 0), dashed+red); draw((0.1, 0, 0)--(1, 0.9, 0)--(1, 0, 0)--(0.1, 0, 0)); draw((1, 0, 0)--(1, 0.9, 0.9)--(1, 0.9, 0)); draw((0.1, 0, 0)--(1, 0.9, 0.9)); //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 $n\geqslant 10$.

So the answer is D.

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 $V_0=n^3$.

Cubic.png

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 cuting 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.

2.png

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

3.png

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

4B.png

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

5.png

After cutting off the slice corresponding to $|z-x|< 1$, we have 6 congruent triangular prisms.

6B.png

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.

7.png

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 D.


If there is no choice for selection, we can also find the minimum value of the integer $n$ if we do not substitude $n$ 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.

See Also

2012 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
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All AMC 10 Problems and Solutions
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