Difference between revisions of "2012 AMC 10A Problems/Problem 25"
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Without loss of generality, assume that <math>n\geq x \geq y \geq z \geq 0</math>. Then the set of points <math>(x,y,z)</math> is a tetrahedron, or a triangular pyramid. The point <math>(x,y,z)</math> distributes uniformly in this region. If this is not easy to understand, read Solution II. | Without loss of generality, assume that <math>n\geq x \geq y \geq z \geq 0</math>. Then the set of points <math>(x,y,z)</math> is a tetrahedron, or a triangular pyramid. The point <math>(x,y,z)</math> distributes uniformly in this region. If this is not easy to understand, read Solution II. | ||
− | The altitude of the tetrahedron is <math>n</math> and the base is an isosceles right triangle with a leg length <math>n</math>. The volume is <math>V_1\dfrac{n^3}{6}</math>. As shown in the first figure in red | + | The altitude of the tetrahedron is <math>n</math> and the base is an isosceles right triangle with a leg length <math>n</math>. The volume is <math>V_1=\dfrac{n^3}{6}</math>. As shown in the first figure in red. |
<asy> | <asy> | ||
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Since <math>n\geq x \geq y \geq z \geq 0</math>, we have <math>x-y\geq1</math>, <math>y-z\geq1</math>, <math>z-x\geq1</math>. | Since <math>n\geq x \geq y \geq z \geq 0</math>, we have <math>x-y\geq1</math>, <math>y-z\geq1</math>, <math>z-x\geq1</math>. | ||
− | The region of points <math>(x,y,z)</math> satisfying the condition is show in the second Figure in black | + | The region of points <math>(x,y,z)</math> satisfying the condition is show in the second Figure in black. It is a tetrahedron, too. |
<asy> | <asy> | ||
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So the probability is <math>p=\dfrac{V_2}{V_1}=\dfrac{(n-2)^3}{n^3}</math>. | So the probability is <math>p=\dfrac{V_2}{V_1}=\dfrac{(n-2)^3}{n^3}</math>. | ||
− | Substitude <math>n</math> by the values in the choices, we will find that when <math>n=10</math>, <math>p=\frac{512}{1000}>\frac{1}{2}</math>, when <math>n=9</math>, <math>p= | + | Substitude <math>n</math> by the values in the choices, we will find that when <math>n=10</math>, <math>p=\frac{512}{1000}>\frac{1}{2}</math>, when <math>n=9</math>, <math>p=\frac{343}{729}<\frac{1}{2}</math>. So <math>n\geq 10</math>, the answer is <math>(\text{D})</math>. |
Revision as of 08:25, 14 March 2012
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
Solution I:
Since are all reals lacated 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 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 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
, the answer is
.
Solution II:
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 |