Difference between revisions of "2016 AMC 12A Problems/Problem 23"
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[][] diagram for this problem goes here (z by z square) | [][] diagram for this problem goes here (z by z square) | ||
+ | === Solution 7: More WLOG === | ||
+ | Consider the complement. Assume <math>a</math> is the largest, so on average <math>a=1/2</math>. We now want <math>b+c<1/2</math>, but we know that b+c is between 0 and 2. The complement is thus: (1/2-0)/2=1/4. But keep in mind that we choose each <math>b</math> and <math>c</math> randomly and independently, so if there are <math>k</math> ways to choose b+c together, there are <math>2k</math> 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 b+c=3, then we only count this once, but in reality: we have two cases 1+2, and 2+1; similar reasoning also generalizes to non-integral values. The complement is then actually 2(1/4)=1/2. Therefore, our desired probability is given by 1-complement=<math>1/2, C</math> | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2016|ab=A|num-b=22|num-a=24}} | {{AMC12 box|year=2016|ab=A|num-b=22|num-a=24}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 15:19, 28 July 2017
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 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)
Solution 7: More WLOG
Consider the complement. Assume is the largest, so on average . We now want , but we know that b+c is between 0 and 2. The complement is thus: (1/2-0)/2=1/4. But keep in mind that we choose each and randomly and independently, so if there are ways to choose b+c 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 b+c=3, then we only count this once, but in reality: we have two cases 1+2, and 2+1; similar reasoning also generalizes to non-integral values. The complement is then actually 2(1/4)=1/2. Therefore, our desired probability is given by 1-complement=
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 |
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