For this problem, our task was to determine

the self-similarity dimension of this shape here.

So first note, it’s the reminder that

this is not exactly the same

as the Sierpinski triangle over here.

So the Sierpinski triangle we remove

a triangle on the middle at every step.

Here in this shape we remove this,

it look like radioactivity symbol to me at every step,

so we remove this three triple triangle.

then from all remain ones

we remove triple triangle and so on.

So end up with this shape.

So what’s its dimension?

Well, we use the familiar formula.

Number of small copies is

magnification factor raise to the dimension D.

So what’s the number of small copies?

So let’s see. 1,2,34,5,6

So I see 6 small copies in here.

And what’s the magnification factor?

Well, that’s 3.

Easiest is to see perhaps along this side,

so here this length

and we have to stretch it 3 times.

Same things to the base,

we would have to stretch it 3 times,

In order for it to be that large,

as the larger shape.

So the magnification factor or the stretch factor is 3.

And then, we need to raise it to the D power.

Alright, so let’s go to work with logs.

Take the log of both sides.

Use the exponent property of logs

to bring the D outside and downstairs,

Then divide both sides by log3

And we’ve done.

log6 over log3

and that’s our dimension.

log6 over log3 there is.

And let’s see, what’s gonna number out of to

an approximate number?

6 log divided by 3 log equals about 1.631

So D approximately 1.631

So that’s the self-similarity dimension of this fractal.