Download 20 Triangles.

TRIANGLES, PROBABILITY, AND AMAZEMENT
A CONNECTED EXPERIENCE FOR THE
CLASSROOM
JIM RAHN
WWW.JAMESRAHN.COM
[email protected]
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
Triangle
Vertices
ABC
DEF
ABD
DEA
ABE
DEB
ABF
DEC
BCD
EFA
BCE
EFB
BCF
EFC
BCA
EFD
CDE
FAB
CDF
FAC
CDA
CDB
Remove all duplicates
FAD
FAE
TOTAL: 20 Triangles.
How many are Right Triangles?
Triangle
Vertices
ABC
What is necessary to be
guaranteed a right
triangle?
Triangle
Vertices
BCD
ABD
BCE
ABE
BCF
ABF
BDE
ACD
BDF
ACE
CDE
ACF
CDF
ADE
CEF
ADF
DEF
AEF
EFB
There are three diameters
Line Segments AD, BE and CF
TOTAL: 20 Triangles.
How many are Right Triangles?
Triangle
Vertices
Triangle
Vertices
ABC
BCD
ABD
BCE
ABE
BCF
ABF
BDE
ACD
BDF
ACE
CDE
ACF
CDF
ADE
CEF
ADF
DEF
AEF
EFB
How many right triangles can be
formed with diameter AD?
TOTAL: 20 Triangles.
How many are Right Triangles?
Triangle
Vertices
ABC
ABD
ABE
ABF
ACD
ACE
ACF
ADE
ADF
AEF
Triangle
Vertices
BCD
BCE
BCF
BDE
BDF
CDE
CDF
CEF
DEF
EFB
How many right triangles can be
formed with diameter BE?
TOTAL: 20 Triangles.
How many are Right Triangles?
Triangle
Vertices
Triangle
Vertices
ABC
BCD
ABD
BCE
ABE
BCF
ABF
BDE
ACD
BDF
ACE
CDE
ACF
CDF
ADE
CEF
ADF
DEF
AEF
EFB
How many right triangles can be
formed with diameter CF?
TOTAL: 20 Triangles.
There are 12 right triangles
Triangle Diameter
Vertices End
Points?
Triangle Diameter
Vertices End
Points?
ABC
BCD
ABD
AD
BCE
BE
ABE
BE
BCF
CF
BDE
BE
ABF
ACD
AD
BDF
ACE
CDE
ACF
CF
ADE
AD
ADF
AD
AEF
Which triangles use
diameters AD, BE, or
CF?
CDF
CF
CEF
CF
DEF
EFB
BE
TOTAL: 20 Triangles.
There are twelve right triangles
Triangle Diameter
Vertices End
Points?
Triangle Diameter
Vertices End
Points?
ABC
Obtuse
BCD
Obtuse
ABD
AD
BCE
BE
ABE
BE
BCF
CF
ABF
Obtuse
BDE
BE
ACD
AD
BDF
Equilateral
ACE
Equilateral
CDE
Obtuse
ACF
CF
CDF
CF
ADE
AD
CEF
CF
ADF
AD
DEF
Obtuse
AEF
Obtuse
EFB
BE
Which triangles are
equilateral triangles?
Which triangles are
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?
12 3

20 5
Smith, Richard J., “Equal Arcs, Triangles, and Probability, Mathematics
Teacher, Vol. 96, No. 9, December 2003, pp. 618-621.
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)
Experimental
Results
Right
Triangle
Non Right
Triangle
Total
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.
Compare your results.
Gather the results from
the class. What does it
show?
Experimental
Results
Right
Triangle
58
Non Right
Triangle
42
Total
100
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.
=1
=3
=2
=2
=3
=1
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)
Experimental
Results
Right
Triangle
Non Right
Triangle
Total
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.
=1
=3
=2
=2
Compare your results.
Gather the results from
the class. What does it
show?
=3
=1
Experimental
Results
Right
Triangle
58
Non Right
Triangle
42
Total
100
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,
A
Probability three points form a
right triangle is 0
B
C
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?
A
There are 4 possible triangles
BUT
There are TWO Diameters,
thus 4 Right Triangles
D
B
C
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 triangles
BUT
There are NO Diameters,
thus NO Right Triangles
Probability three points form a
right triangle is 0
Number of
Total Number of
Number of
equally spaced
Triangles
Right Triangles
points
(T)
(R)
(N)
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?
We will need to determine the total
number of triangles that can be
formed by using three points.
Number of
Total Triangles
How can we determine the total number of
triangles?
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  n  2  n  1
2
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.
8 C3 
8! 8  7  6  5! 8  7  6


 56
3!5! 3  2 1 5!
3 2
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
How many right triangles can be
formed with each diameter?
A
Each diagonal forms 6 right triangle with
the remaining vertices
H
B
How 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?
C
G
Probability three points form a right
triangle is 24/56 = 3/7
F
D
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
8
0
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?
n(n  2)
n
(
n

2)

 
2
2
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
8
0
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
3
if n is even
n-1
0
if n is odd
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?
Number
Number
Using Triangular
Using
of
of
Numbers
Combinations
points
Triangles
around
the
n!
n  (n  1)  (n  2)  (n  3)! n  (n  1)  (n  2)
circle


n C3 
3!(n 
3  2 11+3
(n  3)!
64C3=4
4
4 3)!
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
(n  2)(n  1)
1 3  6  
2
n-2 terms
12C3=220
n
n C3 
n C3 
n!
3!(n  3)!
n!
n  (n  1)(n  2)(n  3)! n  (n  1)(n  2)


3!(n  3)!
3!(n  3)!
6
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?
There are n/2 diameters.
Each forms with (n - 2) right triangles with the remaining
vertices. Thus the number of right triangles is: n  (n  2)
2
n(n-2)
# of rightΔ
3
2
P(right triangle being formed)=
=
=
n(n-1)(n-2)
# of totalΔ
n-1
6
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)