Difference between revisions of "2016 AMC 12A Problems/Problem 23"
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WLOG assume <math>a</math> is the largest. Scale the triangle to <math>1,{b}/{a},{c}/{a}</math> or <math>1,x,y</math> with <math>x,y \in [0,1]</math> and equally likely to be any pair of numbers within the interval. Then, <math>x+y>1</math>, meaning the solution is <math>\boxed{\textbf{(C)}\;1/2}</math>, as shown in the graph below. | WLOG assume <math>a</math> is the largest. Scale the triangle to <math>1,{b}/{a},{c}/{a}</math> or <math>1,x,y</math> with <math>x,y \in [0,1]</math> and equally likely to be any pair of numbers within the interval. Then, <math>x+y>1</math>, meaning the solution is <math>\boxed{\textbf{(C)}\;1/2}</math>, as shown in the graph below. | ||
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===Solution 2: Conditional Probability=== | ===Solution 2: Conditional Probability=== |
Revision as of 17:37, 27 March 2020
Contents
Problem
Three numbers in the interval are chosen independently and at random. What is the probability that the chosen numbers are the side lengths of a triangle with positive area?
Solution
Solution 1: Super WLOG
WLOG assume is the largest. Scale the triangle to
or
with
and equally likely to be any pair of numbers within the interval. Then,
, meaning the solution is
, as shown in the graph below.
Solution 2: Conditional Probability
WLOG, let the largest of the three numbers drawn be . Then the other two numbers are drawn uniformly and independently from the interval
. The probability that their sum is greater than
is
Solution 3: Calculus
When , consider two cases:
1) , then
2), then
is the same. Thus the answer is
.
Solution 4: Geometry
The probability of this occurring is the volume of the corresponding region within a cube, where each point
corresponds to a choice of values for each of
and
. The region where, WLOG, side
is too long,
, is a pyramid with a base of area
and height
, so its volume is
. Accounting for the corresponding cases in
and
multiplies our answer by
, so we have excluded a total volume of
from the space of possible probabilities. Subtracting this from
leaves us with a final answer of
.
Solution 5: More Calculus
The probability of this occurring is the volume of the corresponding region within a cube, where each point
corresponds to a choice of values for each of
and
. We take a horizontal cross section of the cube, essentially picking a value for z. The area where the triangle inequality will not hold is when
, which has area
or when
or
, which have an area of
Integrating this expression from 0 to 1 in the form
Solution 6: Geometry in 2-D
WLOG assume that is the largest number and hence the largest side. Then
. We can set up a square that is
by
in the
plane. We are wanting all the points within this square that satisfy
. This happens to be a line dividing the square into 2 equal regions. Thus the answer is
.
[][] diagram for this problem goes here (z by z square)
Solution 7: More WLOG, Complementary Probability
The triangle inequality simplifies to considering only one case: . Consider the complement (the same statement, except with a less than or equal to). Assume (WLOG)
is the largest, so on average
(now equal to becomes a degenerate case with probability
, so we no longer need to consider it). We now want
, so imagine choosing
at once rather than independently. But we know that
is between
and
. The complement is thus:
. But keep in mind that we choose each
and
randomly and independently, so if there are
ways to choose
together, there are
ways to choose them separately, and therefore the complement actually doubles to match each case (a good example of this is to restrict b and c to integers such that if
, then we only count this once, but in reality: we have two cases
, and
; similar reasoning also generalizes to non-integral values). The complement is then actually
. Therefore, our desired probability is given by
Solution 8: 3D geometry
We can draw a 3D space where each coordinate is in the range [0,1]. Drawing the lines and
We have a 3D space that consists of two tetrahedrons. One is a regular tetrahedron with side length
and the other has 3 sides of length
and 3 sides of length
The volume of this region is
. Hence our solution is
Solution 9: Cheap Solution
Pretend that the values of ,
, and
are integers ranging from
. Test out the probability of the first few values of
(for example,
). Since real numbers contain infinite increments, the answer is the limit as
approaches infinity of
, which is easily hypothesized as
, or
-solution by fidgetboss_4000 get rect
See Also
2016 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 22 |
Followed by Problem 24 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.