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Latest revision as of 16:00, 20 January 2021
Contents
Problem
Real numbers , , and are chosen independently and at random from the interval for some positive integer . The probability that no two of , , and are within 1 unit of each other is greater than . What is the smallest possible value of ?
Solution 1
Since are all reals located in , the number of choices for each one is continuous so we use geometric probability.
WLOG(Without loss of generality), assume that . Then the set of points is a tetrahedron, or a triangular pyramid. The point distributes uniformly in this region. If this is not easy to understand, read Solution II.
The altitude of the tetrahedron is and the base is an isosceles right triangle with a leg length . The volume is , as shown in the first figure in red.
Now we will find the region with points satisfying , , .
Since , we have , .
The region of points satisfying the condition is shown in the second figure in black. It is a tetrahedron, too.
The volume of this region is .
So the probability is .
Substituting with the values in the choices, we find that when , , when , . So .
So the answer is .
Solution 2
Because , , and are chosen independently and at random from the interval , which means that , , and distributes uniformly and independently in the interval . So the point distributes uniformly in the cubic , as shown in the figure below. The volume of this cubic is .
As we want to find the probablity of the incident , we should find the volume of the region of points such that , , and .
Now we will find the region .
The region can be generated by cutting off 3 slices corresponding to , , and , respectively, from the cubic.
After cutting off a slice corresponding to , we get two triangular prisms, as shown in the figure.
In order to observe the object clearly, we rotate the object by the axis, as shown.
We can draw the slice corresponding to on the object.
After cutting off the slice corresponding to , we have 4 pieces left.
After cutting off the slice corresponding to , 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 and a base of isoceless right triangle with leg length , so the volume is . Then the volume of the region is =.
So the probability of the incident is =.
Then we can get the answer the same way as Solution I.
The answer is .
If there is no choice for selection, we can also find the minimum value of the integer if we do not substitute by the possible values one by one.
Let , i.e., , so , or , hence .
Now we will estimate the value of without a calculator.
Since =, so = = =.
Now we would get the approximation of and .
In order to avoid compicated computation, we get the approximation with one decimal digit only.
Estimation of .
Since , so .
The mean of 1 and 1.5 with one decimal digit is about 1.3 .
As , so .
The mean of 1 and 1.3 with one decimal digit is about 1.2.
As , so .
Estimation of .
As , so , then .
As , so .
The mean of 1.5 and 1.69 with one decimal digit is about 1.6.
As , so .
Then , i.e.,
,
As , So the minimal value of integer is 10.
Appendix
This solution is motivated by the suggestive formula .
The problem generalizes easily to -dimensional real space . In the general -dimensional case, we are asked to find the probability that a randomly chosen -tuple satisfies for all . To avoid repetition, let us say that is spaced-out if for all .
Let be the -dimensional hyper-cube of side length : Then has volume . Let be the set of spaced-out -tuples . The desired probability is Vol.
The set of -tuples such that there exist distinct indices such that has volume , so we may restrict our attention to -tuples such that for all .
Further, the condition that is spaced-out is "invariant upon permuting the indices"; in other words, if is a permutation of the set of indices , then is spaced-out if and only if is spaced-out. Therefore, we may consider the set of spaced-out -tuples which additionally satisfy . Let us denote this set by . This condition is equivalent to Let us choose new variables for . This change of variables is just a translation of each by the vector ; 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 which satisfy . As above, we can disregard any such that for some . Given any such that for all , there exists exactly one permutation of the indices such that . Since there are permutations of , the desired volume is equal to times the volume of the -dimensional hyper-cube of side length , which is . Hence has volume as well and has volume . Hence the desired probability is .
Solution 3
If , , and are separated by at least one, then by subtracting the minimum space between the three variables (which is ), , , and can be chosen randomly in the interval . The probability is hence is the minimum value in the answer choices that satisfies this inequality. ~ edit by dolphin7
Solution 4
Imagine Points , , as the "starting points" of three "blocks" of real numbers that have length . We are just trying to find the probability that those three "blocks" do not overlap. To do this we can set each unit of into equal little increments, and take the limit of the probability as approaches . (This is because there are indefinitely many real numbers in a given interval.) We see that the total number of arrangements for the three blocks of , without considering the rule that no two blocks shall overlap is . We see that the number of ways to arrange the three blocks such that no three of them are overlapping is simply . Taking the limit as approaches infinity, we obtain our closed form to be Dividing leading coefficients, we obtain . Solving the inequality , we get the least value of as .
-fidgetboss_4000
Video Solution by Richard Rusczyk
https://www.youtube.com/watch?v=HOtyh9qycrI
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.