Difference between revisions of "2016 AMC 12A Problems/Problem 23"
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The probability of this occurring is the volume of the corresponding region within a <math>1 \times 1 \times 1</math> cube, where each point <math>(x,y,z)</math> corresponds to a choice of values for each of <math>x, y,</math> and <math>z</math>. The region where, WLOG, side <math>z</math> is too long, <math>z\geq x+y</math>, is a pyramid with a base of area <math>\frac{1}{2}</math> and height <math>1</math>, so its volume is <math>\frac{\frac{1}{2}\cdot 1}{3}=\frac{1}{6}</math>. Accounting for the corresponding cases in <math>x</math> and <math>y</math> multiplies our answer by <math>3</math>, so we have excluded a total volume of <math>\frac{1}{2}</math> from the space of possible probabilities. Subtracting this from <math>1</math> leaves us with a final answer of <math>\frac{1}{2}</math>. | The probability of this occurring is the volume of the corresponding region within a <math>1 \times 1 \times 1</math> cube, where each point <math>(x,y,z)</math> corresponds to a choice of values for each of <math>x, y,</math> and <math>z</math>. The region where, WLOG, side <math>z</math> is too long, <math>z\geq x+y</math>, is a pyramid with a base of area <math>\frac{1}{2}</math> and height <math>1</math>, so its volume is <math>\frac{\frac{1}{2}\cdot 1}{3}=\frac{1}{6}</math>. Accounting for the corresponding cases in <math>x</math> and <math>y</math> multiplies our answer by <math>3</math>, so we have excluded a total volume of <math>\frac{1}{2}</math> from the space of possible probabilities. Subtracting this from <math>1</math> leaves us with a final answer of <math>\frac{1}{2}</math>. | ||
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+ | == Solution 4: More Calculus == | ||
+ | |||
+ | The probability of this occurring is the volume of the corresponding region within a <math>1 \times 1 \times 1</math> cube, where each point <math>(x,y,z)</math> corresponds to a choice of values for each of <math>x, y,</math> and <math>z</math>. 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 <math>x + y < z</math>, which has area <math>\frac{z^2}{2}</math> or when <math>x+z<y or y+z<x</math>, which have an area of <math>\frac{(1-z)^2}{2}+\frac{(1-z)^2}{2} = (1-z)^2.</math> Integrating this expression from 0 to 1 in the form | ||
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+ | <math>\int_0^1 \frac{z^2}{2} + (1-z)^2 dz = \frac{z^3}{2} - z^2 + z \biggr |_0^1 = \frac{1}{2} -1 + 1 = \frac{1}{2}</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 19:45, 5 February 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: Logic
WLOG the largest number is 1. Then the probability that the other two add up to at least 1 is .
Thus the answer is .
Solution 2: Calculus
When , consider two cases:
1) , then
2), then
is the same. Thus the answer is .
Solution 3: 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 4: 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 , which have an area of Integrating this expression from 0 to 1 in the form
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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