The Golden Ratio

[jay_shark]

Take any regular pentagon with sides a and diagonal b .

Show that a/b = b/(b+a)

[jay_shark]

Bump

[rsk111]

This is going to be hard without a diagram, but whatever.

Draw two diagonals from any vertex of the pentagon to create an isosceles triangle with base length of a and the other two sides of length b.

We'll call this triangle xyx. Point x will be the vertex opposite the side of length a. The other two vertices will be y and z. Hence segment xy and xz are of length b and segment yz is of lenght a.

angle xyz = angle xzy = 72 degrees (this is trivially provable, so I'm not going to spend the time writing it out)

Therefore angle yxz = 180-72-72=36 degrees.

Now well put a point on the segment xy called w and create a triangle (wyz) such that angle wzy=36 degrees.

angle wyz = angle xyz = 72 (because w and x are on the same line (i.e., these are the exact same angle)

angle ywz = 180 - angle wyz - angle wzy = 180-72-36=72

Now we know that triangles xyz and wyz are similar since they have three angles that are the same.

Since triangle wyz is isosceles and we know segment yz is of length a, then segment wz must also be of length a.

Triangle wxz is also isosceles (once again trivially provable, so won't go through it) segment wx = segment wz, so segment wx also has a length of a.

Therefore segment wy = segment xy - segment wx = b-a.

Since triangle xyz and triangle wyz are similar the ratios of any two sides are equal. Hence,

length of segment xz/length of segment yz=length of segment yz/lenght of segment wy. Substituting the lenghts

b/a=a/(b-a)

Now we do some simple algebra
b(b-a)/a=a
b(b-a)=a2
b2-ab=a2
b2=a2+ab
b2=a(a+b)
b2/a+b=a
b/a+b=a/b

Edit to add that the proof can be done without figuring out the sizes of the angles, but I did this to make it easier to follow.

[jay_shark]

Very nice solution .

I was aware of the similar triangle argument which is neat .

I am ware of two beautiful solutions to this problem . One is the argument you presented , and another proof which makes use of another well known identity in circle geometry .

Does anyone know which one it is ?

[Siegmund]

It's an interesting take on a classic puzzle to present this as a golden ratio calculation... I learned the converse puzzle, the construction of the regular pentagon with straightedge and compass, at a fairly young age, and managed to catch several people by surprise with it who though it couldn't be done.

[jay_shark]

Yes I was very intrigued by this problem when I first encountered it and I remember solving it using similar triangles much like the argument already presented here .
However ,an old friend of mine presented another lovely solution .

Notice that any regular pentagon can be inscribed in a circle . Any 4 vertices on a regular pentagon is cyclic since 72+108 =180 (opposite angles sum to 180)Now we may apply Ptolemy's equality .

That is , for any cyclic quadrilateral with sides a,b,c,d in that order and diagonal e1 and e2, the following holds :
a*c+ b*d=e1*e2

Now apply this to the regular pentagon .
we get a*a + a*b = b*b
a(a+b) = b*b
or (a+b)/b = b/a

[Silent A]

Very nice.

Now, if only Ptolemy's equality could actually described as "well known".

If I've never heard of a certain mathematical identity, it's not well known. Period. /images/graemlins/smile.gif

[Enrique]

The solutions posted are quite pretty, but there is one algebraic way of proving this as well.

If we consider the roots of unity of x^5 - 1. They form a circle.
But calculating them is not so hard because 1 + x + x^2 + x^3 + x^4 = 0 is symmetric, hence if we let z = x + 1/x, then 1 + x + x^2 + x^3 + x^4 = z^2 + z - 1 = 0.

Using this we can calculate the coordinates of the pentagon and then we could get every distance we desire.
A lot of simple algebra that yields the desired equality.