triangles, probability, and amazement a connected experience for the classroom jim rahn ...
TRANSCRIPT
TRIANGLES, PROBABILITY, AND AMAZEMENT A CONNECTED EXPERIENCE FOR THE CLASSROOM
JIM RAHN
WWW.JAMESRAHN.COM
Question:If we place six(6) evenly spaced points around the circumference of a circle and then randomly select
three points to form the vertices of a triangle, what is the probability that the triangle formed is a
RIGHT TRIANGLE?
Smith, Richard J., “Equal Arcs, Triangles, and Probability, Mathematics Teacher, Vol. 96, No. 9, December 2003, pp. 618-621.
Make a List of ALL Possible Triangles that can be formed using three of these points. MAKE a List!
Triangle Vertices
ABC
ABD
ABE
ABF
BCD
BCE
BCF
BCA
CDE
CDF
CDA
CDB
TriangleVertices
DEF
DEA
DEB
DEC
EFA
EFB
EFC
EFD
FAB
FAC
FAD
FAERemove all duplicates
TOTAL: 20 Triangles.How many are Right Triangles?
There are three diametersLine Segments AD, BE and CF
What is necessary to be guaranteed a right triangle?
TriangleVertices
ABC
ABD
ABE
ABF
ACD
ACE
ACF
ADE
ADF
AEF
TriangleVertices
BCD
BCE
BCF
BDE
BDF
CDE
CDF
CEF
DEF
EFB
TriangleVertices
ABC
ABD
ABE
ABF
ACD
ACE
ACF
ADE
ADF
AEF
TOTAL: 20 Triangles.How many are Right Triangles?
How many right triangles can be formed with diameter AD?
TriangleVertices
BCD
BCE
BCF
BDE
BDF
CDE
CDF
CEF
DEF
EFB
TriangleVertices
BCD
BCE
BCF
BDE
BDF
CDE
CDF
CEF
DEF
EFB
TriangleVertices
ABC
ABD
ABE
ABF
ACD
ACE
ACF
ADE
ADF
AEF
TOTAL: 20 Triangles.How many are Right Triangles?
How many right triangles can be formed with diameter BE?
TOTAL: 20 Triangles.How many are Right Triangles?
TriangleVertices
ABC
ABD
ABE
ABF
ACD
ACE
ACF
ADE
ADF
AEF
TriangleVertices
BCD
BCE
BCF
BDE
BDF
CDE
CDF
CEF
DEF
EFB
How many right triangles can be formed with diameter CF?
TOTAL: 20 Triangles.
TriangleVertices
DiameterEnd Points?
ABC
ABD AD
ABE BE
ABF
ACD AD
ACE
ACF CF
ADE AD
ADF AD
AEF
TriangleVertices
Diameter End Points?
BCD
BCE BE
BCF CF
BDE BE
BDF
CDE
CDF CF
CEF CF
DEF
EFB BE
Which triangles use diameters AD, BE, or
CF?
There are 12 right triangles
TriangleVertices
Diameter End Points?
BCD
BCE BE
BCF CF
BDE BE
BDF
CDE
CDF CF
CEF CF
DEF
EFB BE
TriangleVertices
DiameterEnd Points?
ABC
ABD AD
ABE BE
ABF
ACD AD
ACE
ACF CF
ADE AD
ADF AD
AEF
TOTAL: 20 Triangles.
Which triangles are equilateral triangles?
There are twelve right triangles
Equilateral
Equilateral
Which triangles are obtuse?
Obtuse
Obtuse
Obtuse
Obtuse
Obtuse
Obtuse
If we place six(6) evenly spaced points around the circumference of a circle and then randomly select
three points to form the vertices of a triangle, what is the probability that the triangle formed is a
RIGHT TRIANGLE?
Smith, Richard J., “Equal Arcs, Triangles, and Probability, Mathematics Teacher, Vol. 96, No. 9, December 2003, pp. 618-621.
12 3
20 5
Using simulation to determine the probability that the vertices of a right triangle is form by randomly selecting three points from six(6) evenly spaced
points around the circumference of a circle.
Place SIX Cubes (two of three different colors) into a bag. Draw out three cubes. If two cubes are of the same color, the triangle is a right triangle! (Repeat 100 times)
Right Triangle
Non Right Triangle
Total
ExperimentalResults
Using simulation to determine the probability that the vertices of a right triangle is form by randomly selecting three points from six(6) evenly spaced
points around the circumference of a circle.
Right Triangle
Non Right Triangle
Total
58 42 100
ExperimentalResults
Compare your results.Gather the results from the class. What does it show?
Using simulation to determine the probability that the vertices of a right triangle is form by randomly generating three numbers from three numbers. Opposite vertices will have the same numbers.
Using your graphing calculator: Type randint(1, 3, 3). This means you will be selecting three numbers from 1,2, and 3. If two digits are the same number, the triangle is a right triangle! (Repeat 100 times)
Right Triangle
Non Right Triangle
Total
ExperimentalResults
=1
=1
=2
=2
=3
=3
Using simulation to determine the probability that the vertices of a right triangle is form by randomly generating three numbers from three numbers. Opposite vertices will have the same numbers.
Right Triangle
Non Right Triangle
Total
58 42 100
ExperimentalResults
=1
=1
=2
=2
=3
=3
Compare your results.Gather the results from the class. What does it show?
If we place three(3) evenly spaced points around the circumference of a circle and then randomly
select three points to form the vertices of a triangle, what is the probability that the triangle
formed is a RIGHT TRIANGLE?
There is only 1 possible triangle and NO Diameters,
Probability three points form a
right triangle is 0
CB
A
If we place four(4) evenly spaced points around the circumference of a circle and then randomly
select three points to form the vertices of a triangle, what is the probability that the triangle
formed is a RIGHT TRIANGLE?
There are 4 possible trianglesBUTThere are TWO Diameters, thus 4 Right Triangles
D
C
B
A
Probability three points form a
right triangle is 4/4 = 1
If we place five(5) evenly spaced points around the circumference of a circle and then randomly
select three points to form the vertices of a triangle, what is the probability that the triangle
formed is a RIGHT TRIANGLE?
There are 10 possible trianglesBUTThere are NO Diameters, thus NO Right Triangles
Probability three points form a
right triangle is 0
Number of equally spaced
points(N)
Total Number of Triangles
(T)
Number of Right Triangles
(R)
Probability a Triangle is Right (R/T)
3 1 0 0/1 = 0
4 4 4 4/4 = 1
5 10 0 0/10 = 0
6 20 12 12/20 = 3/5
Odd > 3 0
8
10
12
:
What patterns do you see in the Total Number of Triangles (T)?
If we place eight(8) evenly spaced points around the circumference of a circle and then randomly select three points to form the vertices of a triangle, what is the probability that the triangle formed is a RIGHT
TRIANGLE?
Number of Total Triangles
How can we determine the total number of triangles?
We will need to determine the total number of triangles that can be formed by using three points.
Method 1 for finding the total number of triangles:
Number of equally spaced points
(N)
Total Number of Triangles(T)
3 1 1
4 4 1+3
5 10 1+3+6
6 20 1+3+6+10
8 1+3+6+10+?
The total number of triangles is the sum of the first n-2 triangular numbers.
The nth triangular number is 2 1
2
n n
8 3
8! 8 7 6 5! 8 7 656
3!5! 3 2 1 5! 3 2C
Method 2: Use combinations because we are choosing 3 points at random from 8 points. As long as we have the same three points selected there is only one triangle that can be formed.
How many right triangle can be found?
What is necessary for the triangle to be a right triangle?
One side of the triangle must be a diameter.
How many diameters can be drawn?
4 Diameters
Probability three points form a right triangle is 24/56 = 3/7
Each diagonal forms 6 right triangle with the remaining vertices
How many right triangles can be formed with each diameter?
B
A
C
DF
G
HHow many right triangles can be formed?
What is the probability of forming a right triangle when three points are selected at random from 8 points equally spaced around a circle?
Number (N) of equally spaced
points
Total Number (T) of Triangles
Number (R) of Right Triangles
Probability a Triangle is Right (R/T)
3 1 0 0/1 = 0
4 4 4 4/4 = 1
5 10 0 0/10 = 0
6 20 12 12/30 = 3/5
Odd > 3 0
8 56 24 24/56 = 3/7
10
12
:
How can we generalize how many right triangles will be formed?
What must be true about the number of points equally spaced around the circle?
If n= number of points is an even number, can we determine the number of right triangles formed?
( 2)( 2)
2 2
n n nn
Number (N) of equally spaced
points
Total Number (T) of Triangles
Number (R) of Right Triangles
Probability a Triangle is Right (R/T)
3 1 0 0/1 = 0
4 4 4 4/4 = 1
5 10 0 0/10 = 0
6 20 12 12/30 = 3/5
Odd > 3 0
8 56 24 24/56 = 3/7
10
12
:
Complete the chart for 10 and 12 points equally spaced around the circle.
10 points equally spaced around a circle
Number of right triangles
Number of total triangles
Probability of forming a right triangle
12 points equally spaced around a circle
Number of right triangles
Number of total triangles
Probability of forming a right triangle
Number (N) of equally spaced
points
Total Number (T) of Triangles
Number (R) of Right Triangles
Probability a Triangle is Right (R/T)
3 1 0 0/1 = 0
4 4 4 4/4 = 1
5 10 0 0/10 = 0
6 20 12 12/30 = 3/5
Odd > 3 0
8 56 24 24/56 = 3/7
10 40 120 40/120=3/9
12 80 220 60/220=3/11
:What patterns do you observe?
What conjecture would you like to make?
If n points are equally spaced on the circumference of a circle and if three points are chosen at random, the probability that the three points will form a right triangle is
3if n iseven
n-10 if n isodd
If we place n evenly spaced points around the circumference of a
circle, where n is an even number greater than 3, and then randomly select three points to form the vertices of a triangle, what is the
probability that the triangle formed is a RIGHT TRIANGLE?
What do you notice about the number of possible triangles?
3
! ( 1) ( 2) ( 3)! ( 1) ( 2)
3!( 3)! 3 2 1 ( 3)! 6n
n n n n n n n nC
n n
Number of
points around
the circle
Number of
Triangles
Using Triangular Numbers
Using Combinations
4 4 1+3 4C3=4
6 20 1+3+6+10 6C3=20
8 56 1+3+6+10+15+21 8C3=56
10 120 1+3+6+10+15+21+28+36 10C3=120
12 220 1+3+6+10+15+21+28+36+45+55
12C3=220
Number of
points around
the circle
Number of
Triangles
Using Triangular Numbers
Using Combinations
4 4 1+3 4C3=4
6 20 1+3+6+10 6C3=20
8 56 1+3+6+10+15+21 8C3=56
10 120 1+3+6+10+15+21+28+36 10C3=120
12 220 1+3+6+10+15+21+28+36+45+55
12C3=220
n
n-2 terms
( 2)( 1)1 3 6
2
n n
3
!
3!( 3)!n
nC
n
3
! ( 1)( 2)( 3)! ( 1)( 2)
3!( 3)! 3!( 3)! 6n
n n n n n n n nC
n n
If we place n evenly spaced points around the circumference of a
circle, where n is an even number greater than 3, and then randomly select three points to form the vertices of a triangle, what is the
probability that the triangle formed is a RIGHT TRIANGLE?
How many right triangles will there be?
n(n-2)# of rightΔ 32P(right triangle being formed)= = =
n(n-1)(n-2)# of totalΔ n-16
( 2)2
nn
There are n/2 diameters. Each forms with (n - 2) right triangles with the remaining vertices. Thus the number of right triangles is:
Number (N) of equally spaced
points
Total Number (T) of Triangles
Number (R) of Right Triangles
Probability a Triangle is Right (R/T)
3 1 0 0/1 = 0
4 4 4 4/4 = 1
5 10 0 0/10 = 0
6 20 12 12/30 = 3/5
Odd > 3 0
8 56 24 24/56 = 3/7
10 120 40 40/120 = 3/9
12 220 60 60/220 = 3/11
:
Even (E) number>3
3/(E – 1)