# Math Help - pentagram geometry

1. ## pentagram geometry

Hi everyone . I have been bogged down by this question for ages . I confirm through 2 different methods what the answer is, and the course book (isbn 0435516043 - section Mics. 5.10 question 11) gives a diffenet answer. It is either a typing error or I am screwing up somewhere. Can someone please try this out to help me confirm. Here is the question.

You have a pentagram. At its centre is a naturally occuring pentagon (five pointed). then of course the five equilateral triangles around that, making up the pentagram. the longest side of these equilateral triangles is x. Also there is a bisect in the pentagon. check out the image:

well, it looks like i can not upload a picture from my computer. So I will have to describe it instead.

Take a circle. A pentagon fits inside the circle, with its edges just kinssing the circumference. therefore, each side of the pentagon has a curve slightly above it, i.e. the perimeter of the circle. The length of these arcs is 1 unit. Therefore I calculated the circumference to be 5 units. The sides of the pentagon have equilateral triangles on them, such that, the pentagon and the triangles form a pentagram. the two equal sides of each equilateral triangle is marked x. Also, it happens that if you take a chord line in the pentagon to form another equilateral triangle, its base side (the side of the triangle that is not one of the two equal sides) is also equal to x. Now if you draw this out, hopefully you will be seeing roughly what I am seeing. This is all the information I was given along with the following question:

The shape below is known as a pentagram. Find x.

I calculated x to be 1.52 (2 d.p.). this answer came about from using two mathematical methods. Now, the book, however gave the answer to be 1.1618. What the hell, I thought .

Please, could someone, or as many as possible go through this question to find out who is wrong here, the book editor or me (most likely me). If any of you do find me to be wrong, please take me through how you work the answer out, to be as what the book says?

Many thanks!

2. Hello, revolution!

We do not have equilateral triangles . . .

You have a regular pentagram.
At its centre is a naturally occuring pentagon.
Then, of course, the five equilateral (?) triangles around that. . . . . no
Code:
              P
*
/ \
/   \
A /  72°\ B
T o - - - o - - - o - - - o Q
\     /     108°\     /
\   /           \   /
\ /             \ /
E o               o C
\             /

The pentagon is $ABCDE.$
The "tips" of the pentagrm are: $P, Q, R, S, T.$

An interior angle of a regular pentagon is 108°:
. . $\angle EAB \:=\:\angle ABC \:=\: 108^o$

Hence: . $\angle PAB \:=\:\angle PBA \:=\:72^o$

Therefore: . $\angle P \:=\:36^o$

Those triangles are isosceles.

It can be shown that the Golden Mean
. . is deeply embedded in this diagram.

For example: . $\frac{PA}{AB} \:=\:\frac{1 + \sqrt{5}}{2} \:=\:\phi$

3. thanks soroban , you almost did it. The book I am working from assumes no knowledge of the golden ratio. So I have to use trigonometry to solve the problem. The diagram you drew was helpful, and I shall exbound upon it a little more to show you what the problem is.

Apparently according to my book line BD = AT = x. x is what i need to find. The only other helpful point the book gives me, is that if you drew an imaginary circle around the pentagon, so that the pentagon itself fitted perfectly into the circle, its circumference would be equal to 5. Therefore I worked out the radius, r as 0.8 (1 d.p.).

Thus we can find line AE, which we will call "b".

b = 2r sin (a/2)
= 0.94, (2 d.p.)

where a/2 = 36' (degrees). a is obtained from 360/5 (the five equal quadrants of the polygon) For example, if the centre of the Polygon was “V“, then angle AVE = 72‘. So let us call the centre of the polygon V. Therefore angle AVE = 72'.

Now that I know b (line AE) I can work out x from using trigonometry on the triangle ATE. If I split triangle ATE into 2 right-angled triangles such that angle ATE (36 degrees) becomes 18 degrees, I should be able to work out x. Here goes;

Because

0.5b/x = sin 18
it implies that
x = b/(2 sin 18)
= 1.52, (2 d.p.)

This of course was different to the answer given in the book, 1.1618 .
So naturally I wanted to confirm x through another method. Here is how I worked x out from using another method.

Consider triangle BVD where V is the centre of the pentagon. Angle BVD is 144'. Therefore angle VBD is 18'. Remembering that line BD = x and line VB
= r, it follows that

x = 2r cos 18
= 1.6 cos 18
= 1.52, (2 d.p.)

Again, same answer I thought. Hmm, I will try using pythagoress theorem and see whether that brings out the same answer.

The half way point on line BD, we shall call point "M", such that M = x/2, because line BD is also equal to x (as previousely mentioned). Therefore triangle MVB (remembering that point V is the centre of the pentagon) forms a right angled triangle. Note also that angle VBM = 18' = 180-90-36.
Now,

q = r sin 18
= 0.25, (2 d.p.)

Therefore applying the theorm of pythagoress,

x = 2(r^2 - q^2)^(1/2)
= 1.52, (2 d.p.)

Again I thought, 1.52. How come the answer at the back of the book is saying x = 1.1618.

How do you work out x? I think it is a printing error, but it seems quite a specific number for a printing error.

Could you please work out x (without reffering to the golden ratio) and show me how you do it?
Your a genius if you can figure this out, I've been brainstorming this for the past 2 days now with no success, what ever method I use.