Difference between revisions of "2016 AMC 12A Problems/Problem 23"
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==Solution== | ==Solution== | ||
− | ===Solution 1: Conditional Probability=== | + | ===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, Assume <cmath>a≥b≥c</cmath>. Scale the triangle so we have side lengths of <cmath>1,{b\a},{c\a}</cmath>. Then it is easilt seen the solution is <math>\boxed{\textbf{(C)}\;1/2.}</math> | ||
+ | |||
+ | ===Solution 2: Conditional Probability=== | ||
WLOG, let the largest of the three numbers drawn be <math>a>0</math>. Then the other two numbers are drawn uniformly and independently from the interval <math>[0,a]</math>. The probability that their sum is greater than <math>a</math> is <math>\boxed{\textbf{(C)}\;1/2.}</math> | WLOG, let the largest of the three numbers drawn be <math>a>0</math>. Then the other two numbers are drawn uniformly and independently from the interval <math>[0,a]</math>. The probability that their sum is greater than <math>a</math> is <math>\boxed{\textbf{(C)}\;1/2.}</math> | ||
− | ===Solution | + | ===Solution 3: Calculus=== |
When <math>a>b</math>, consider two cases: | When <math>a>b</math>, consider two cases: | ||
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<math>a<b</math> is the same. Thus the answer is <math>\frac{1}{2}</math>. | <math>a<b</math> is the same. Thus the answer is <math>\frac{1}{2}</math>. | ||
− | ===Solution | + | ===Solution 4: Geometry=== |
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>. | ||
− | === Solution | + | === Solution 5: 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</math> or <math>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 | 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</math> or <math>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> | <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> | ||
− | === Solution | + | === Solution 6: Geometry in 2-D === |
WLOG assume that <math>z</math> is the largest number and hence the largest side. Then <math>x,y \leq z</math>. We can set up a square that is <math>z</math> by <math>z</math> in the <math>xy</math> plane. We are wanting all the points within this square that satisfy <math>x+y > z</math>. This happens to be a line dividing the square into 2 equal regions. Thus the answer is <math>\frac{1}{2}</math>. | WLOG assume that <math>z</math> is the largest number and hence the largest side. Then <math>x,y \leq z</math>. We can set up a square that is <math>z</math> by <math>z</math> in the <math>xy</math> plane. We are wanting all the points within this square that satisfy <math>x+y > z</math>. This happens to be a line dividing the square into 2 equal regions. Thus the answer is <math>\frac{1}{2}</math>. | ||
Revision as of 16:13, 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
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, Assume
\[a≥b≥c\] (Error compiling LaTeX. ! Package inputenc Error: Unicode character ≥ (U+2265))
. Scale the triangle so we have side lengths of
\[1,{b\a},{c\a}\] (Error compiling LaTeX. ! Argument of \a has an extra }.)
. Then it is easilt 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
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 |
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