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 21. What is the smallest possible value of n?
Answer Choices:
A. 7
B. 8
C. 9
D. 10
E. 11
Solution:
It may be assumed that x≤y≤z. Because there are six possible ways of permuting the triple (x,y,z), it follows that the set of all triples (x,y,z) with 0≤x≤y≤z≤n is a region whose volume is 61 of the volume of the cube [0,n]3, that is 61n3. Let S be the set of triples meeting the required condition. For every (x,y,z)∈S consider the translation (x,y,z)↦(x′,y′,z′)=(x,y−1,z− 2). Note that y′=y−1>x=x′ and z′=z−2>y−1=y′. Thus the image of S under this translation is equal to {(x′,y′,z′):0≤x′<y′<z′≤n−2}. Again by symmetry of the possible permutations of the triples (x′,y′,z′), the volume of this set is 61(n−2)3. Because 9373=729343<21 and 10383=1000512>21, the smallest possible value of n is (D)10.
The problems on this page are the property of the MAA's American Mathematics Competitions