2012 AMC 10A Problems/Problem 25
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
?
Solutions
Solution 1
Since are all reals located in
, the number of choices for each one is infinite.
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 III.
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 show in the second Figure in black. It is a tetrahedron, too.
The volume of this region is .
So the probability is .
Substitude by the values in the choices, we will find that when
,
, when
,
. So
.
So the answer is D.
Solution II
This solution is motivated by the suggestive formula .
We generalize to -dimensional real space
. Suppose we are asked to find the probability that a randomly chosen
-tuple
satisfies
for all
. The total set of
-tuples in
has volume
. Let
be the set of triples
which satisfy
for all
. The desired probability is
. The set of triples
such that there exist distinct indices
such that
has volume
, so we may restrict our attention to the triples such that
for all
.
Further, the condition that for all
is invariant upon permuting the indices. Therefore we may consider the set of
-tuples
which satisfy
for all
and
; let us denote this set by
. This condition is equivalent to
Let us choose new variables
for
; then the set of
-tuples
which satisfy
has volume
. Hence
has volume
as well and
has volume
. Hence the desired probability is
.
Solution III
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 cuting 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 D.
If there is no choice for selection, we can also find the minimum value of the integer if we do not substitude
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.
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 |