2016 AMC 12A Problems/Problem 23
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 1: Super WLOG
This lemma will be used for the solution: "If a triangle with side lengths a,b,c exists so does a triangle with side lengths ak,bk and ck and vice versa". WLOG . Scale the triangle so we have side lengths of . Then it is easily seen the solution is
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
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)
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