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Transcript
PUZZLE
# 016
SEPTEMBER 19, 2016
PUZZLE
# 016
THE
FIVE-POINTED
STAR
You draw a five-pointed star, like a pentagram, with
a single pen stroke of five lines. Each of the five lines
have a length of exactly 1 unit (Fig. 1).
The Question: What is the distance between two
adjacent points of the pentagram?
x
Fig 1: The five-pointed star
METHOD
Triangle A
A
Triangle B
x
C
x
D
Triangle C
x
Triangle D
x
PROCESS
How We Can Look At The Problem:
The goal is to find the distance between
two adjacent points of the star (x).
x
B
1
1 unit
We can approach the problem by
dividing up the star, and pentagon that
is formed by connecting the points, into
more manageable geometry.
What We Know:
x
Triangle A
We know that the length of the lines
making up the star are 1 unit long.
0.5 units
By dividing the pentagram into smaller
geometry, we can use a right angle
triangle that has the line connecting
two adjacent points of the star (x) as
the hypotenuse (Fig. 2).
Fig 2: Triangle A with base 0.5 units, and hypotenuse x
What We Can Find:
360º/5
= 72º
72º
2
In order to solve for x, we need this
height, or the angles of Triangle A
(Fig. 3).
Fig 3: The angle of a pentagon at the center
180º total in
triangle
54º
Triangle B
3
54º
72º
The base of Triangle A is 0.5 units, but
we are missing the height.
= (180º - 72º)/2
= 108º/2
= 54º
108º
From the center, the angle between
two lines connected to the points of the
star is 72º.
Solving Triangle B:
Knowing the angle at the center, we
can solve Triangle B. Here we find that
the angle between the edges of the
pentagon is 108º
This means the top angle of Triangle A,
is 54º (Fig. 4).
Fig 4: Angles at edge of the pentagon
SOLUTION
Triangle A
x = 0.618
54º
units
Sin(54º) = 0.5/x
x = 0.5/[Sin(54º)]
x = 0.618 units
36º
0.5 units
Using Trigonometry:
Now that we have the angles of
Triangle A, we can use sine to solve for x
(Fig. 5).
The distance between two adjacent
points of this five-pointed star is 0.618
units.
Fig 5: Use sine to solve for x: x = 0.618 units