Difference between revisions of "2016 AMC 12A Problems/Problem 23"
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===Solution 1: Super WLOG=== | ===Solution 1: Super WLOG=== | ||
− | WLOG assume <math>a</math> is the largest. Scale the triangle to <math>1,{b}/{a},{c}/{a}</math> Then the solution is <math>\boxed{\textbf{(C)}\;1/2}</math> (Insert graph with square of side length 1 and the line <math> | + | WLOG assume <math>a</math> is the largest. Scale the triangle to <math>1,{b}/{a},{c}/{a}</math> Then the solution is <math>\boxed{\textbf{(C)}\;1/2}</math> (Insert graph with square of side length 1 and the line <math>x+y=1</math> that cuts it in half) |
===Solution 2: Conditional Probability=== | ===Solution 2: Conditional Probability=== |
Revision as of 16:34, 21 March 2016
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 Then the solution is (Insert graph with square of side length 1 and the line that cuts it in half)
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)
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.