InterMath | Workshop Support | Write Up Template
Title
Triangles in Hexagons
Problem Statement
Consider all of the possibilities for generating a triangle with three diagonals and/ or sides
of a regular hexagon. In each case, find the probability that a point inside the hexagon is
also inside the triangle. Explain each solution.
Problem setup
Find the probability that a point inside a regular hexagon is also inside an arbitrary
triangle created either by three diagonals and/or sides of a regular hexagon.
Plans to Solve/Investigate the Problem
I plan to construct a regular hexagon, and then look at all of the possible triangles that can
be created. I will then compare the area of each of the triangles to that of the hexagon,
and that should give me the probability that a point inside the hexagon will also be in the
given triangle.
Investigation/Exploration of the Problem
I found a total of three unique triangles that can be created in a hexagon.
Isoceles:
4
_
5
_
m _
_
345 _= _
120 .00
Area
_
U
_
345 _= _
_
14 .01 __
cm _2
T
_
V
_
3
_
6
_
S
_
W
_
m _
_
216 _= _
120 . 00
X
_
_
Isosceles
triangles using
vertices 3 and 5 and 2 and 6 .
_
2
_
Area
_
1
_
216 _= _
_
14.01 __
cm _2
4
_
5
_
m _
_
432_= _
120 . 00
Area
_
U
_
432 _= _
_
14. 01__
cm _2
T
_
V
_
3
_
6
_
S
_
W
_
m_
_
561 _= _
120 .00
Area
_
_
Isosceles
triangles using
vertices 1 and 5 and 2 and 4.
_
561 _= _
_
14.01__
cm _2
X
_
2
_
1
_
4
_
5
_
m _
_
456 _= _
120 .00
Area
_
456 _= _
_
14.01 __
cm _2
U
_
T
_
V
_
3
_
6
_
S
_
_
Isosceles
triangles using
vertices 4 and 6 and 1 and 3.
_
W
_
X
_
m _
_
321 _= _
120 . 00
Area
_
2
_
1
_
321 _= _
_
14.01 __
cm _2
Right:
4
5
m154 = 90.00
Area
3
6
Right triangles using
vertices 1 and 4.
m124 = 90.00
Area
451 = 47.68 cm2
124 = 47.68 cm2
2
1
4
5
m235 = 90.00
Area
235 = 47.68 cm2
3
6
m265 = 90.00
Area
265 = 47.68 cm2
Right triangles using
vertices 2 and 5.
2
1
m346 = 90.00
Area
346 = 47.68 cm2
4
5
3
6
Right triangles using
vertices 3 and 6.
m316 = 90.00
2
1
Area
316 = 47.68 cm2
Equilateral:
Area
135 = 71.52 cm2
m351 = 60.00
4
5
m135 = 60.00
3
6
Equilateral triangle using
vertices 1, 3 and 5
2
1
m513 = 60.00
Area
246 = 71.52 cm2
m246 = 60.00
4
5
m462 = 60.00
3
6
Equilateral triangle using
vertices 2, 3 and 6
m624 = 60.00
2
1
Area 123456 = 84.07 cm2
Area
234 = 14.01 cm2
Area
241 = 28.02 cm2
Area
426 = 42.04 cm2
4
Area
234
Area 123456
Area
241
Area 123456
Area
426
Area 123456
= 0.17
= 0.33
= 0.50
1
6
1
3
1
2
5
= 0.17
U
= 0.33
T
= 0.50
V
3
6
S
W
X
2
1
The probability that a point inside the hexagon will also be in the isosceles triangle is
1
17%, which comes from the fact that the isosceles triangle makes up of the area of the
6
hexagon. The probability that a point inside the hexagon will also be in the right triangle
1
is 33%, which comes from the fact that the right triangle makes up of the area of the
3
hexagon. The probability that a point inside the hexagon will also be in the equilateral
triangle is 50%, which comes from the fact that the equilateral triangle makes up
1
of the
2
area of the hexagon.
Extensions of the Problem
While working this problem I began to wonder how many of each triangle could be found
in a given polygon. I decided to find triangles formed by joining any three vertices and
which have one obtuse angle. From above we can see that the hexagon has 6 possible
obtuse triangles. I then looked at an octagon. An octagon has 24 possible obtuse
triangles.
3
m234 = 135.00
m235 = 112.50
Area
Area
235 = 19.41 cm2
2
234 = 8.04 cm2
4
Area
245 = 19.41 cm2
m245 = 112.50
1
5
Obtuse triangles using
vertices 2 and 5.
6
8
7
m124 = 112.50
Area
3
m134 = 112.50
124 = 19.41 cm2
Area
2
134 = 19.41 cm2
4
m123 = 135.00
Area
123 = 8.04 cm2
1
5
Obtuse triangles
using vertices 1 and 4.
6
8
7
I then decided to look at an 18-gon. Since vertices 1 and 10 is half of the 18-gon, I used
vertices 1 and 9 and found 7 obtuse triangles. Using vertices 1 and 8, I found 6 obtuse
triangles. In all, I found 28 obtuse triangles using vertices 1 and 9 as two of the points on
the triangle. There are a total of 504 obtuse triangles in an 18-gon.
6
5
7
4
8
3
9
2
m123 = 160.00
10
1
11
18
12
17
13
16
15
14
7
5
8
4
m134 = 150.00
9
3
m124 = 150.00
10
2
11
1
12
18
13
17
14
16
15
m145 = 140.00
5
6
4
m135 = 140.00
7
3
8
2
m125 = 140.00
9
1
10
18
11
17
12
16
13
15
14
6
m156 = 130.00
m146 = 130.00
7
5
4
8
9
3
m136 = 130.00
2
10
1
11
m126 = 130.00
12
18
13
17
14
16
15
After doing all of this, I came up with the following equation that can be used to find
obtuse triangles in any regular polygon with an EVEN number of sides:
n
n
n
n n
- 2) + ( - 3) + ( - 4) + … + ( - ), with n being the number of vertices of the
2
2
2
2 2
given even sided regular polygon.
n[(
Author & Contact
Tiffany Graham