Difference between revisions of "2016 AMC 12A Problems/Problem 23"
(→Solution 7: More WLOG) |
(→Solution 7: More WLOG, Complementary Probability) |
||
Line 44: | Line 44: | ||
=== Solution 7: More WLOG, Complementary Probability === | === Solution 7: More WLOG, Complementary Probability === | ||
− | Consider the complement. Assume (WLOG) <math>a</math> is the largest, so on average <math>a=1/2</math>. We now want <math>b+c<1/2</math>, so imagine choosing <math>b+c</math> at once rather than independently. But we know that <math>b+c</math> is between <math>0</math> and <math>2</math>. The complement is thus: <math>(1/2-0)/2=1/4</math>. 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 <math>b+c</math> 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 <math>b+c=3</math>, then we only count this once, but in reality: we have two cases <math>1+2</math>, and <math>2+1</math>; similar reasoning also generalizes to non-integral values). The complement is then actually <math>2(1/4)=1/2</math>. Therefore, our desired probability is given by <math>1-\text{complement}=1/2, C</math> | + | The triangle inequality simplifies to considering only one case: <math>\text{the smallest side+ the second smallest side} > \text{the largest side}</math>. Consider the complement (the same statement, except with a less than or equal to). Assume (WLOG) <math>a</math> is the largest, so on average <math>a=1/2</math> (now equal to becomes a degenerate case with probability <math>0</math>, so we no longer need to consider it). We now want <math>b+c<1/2</math>, so imagine choosing <math>b+c</math> at once rather than independently. But we know that <math>b+c</math> is between <math>0</math> and <math>2</math>. The complement is thus: <math>(1/2-0)/2=1/4</math>. 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 <math>b+c</math> 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 <math>b+c=3</math>, then we only count this once, but in reality: we have two cases <math>1+2</math>, and <math>2+1</math>; similar reasoning also generalizes to non-integral values). The complement is then actually <math>2(1/4)=1/2</math>. Therefore, our desired probability is given by <math>1-\text{complement}=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 15:27, 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, 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
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.