SEO - How would you do this geometry optimization problem?

How would you do this geometry optimization problem?

A square of maximum area is inscribed in a regular pentagon with area 43. An equilateral triangle of maximum area is then inscribed in the square. Finally, a circle of maximum area is inscribed in the triangle. What is the area of the circle?
Scythian, I'm not exactly sure about your "sharing a vertex" reasoning. Can you please expand? When I tried drawing this out, there were no shared vertices.
Both answers are good. I'm sending it to a vote. Thanks for answering.

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For reasons of symmetry, the pentagon, square, and equilateral triangle can all share the same vertex, and embedded with a vertical axis symmetry. Using nothing more than the Law of Sines, given the length of the side x of the pentagon, the side of the triangle is just

y = x(Sin(90)/Sin(75))(Sin(108)/Sin(63))

and the area of the circle inside would be (1/12)πy². However, if the area of the pentagram is 43, x then must be

x = (2√43) / √√(25+10√5)

Putting all of this together, the numerical answer for the area is about 7.9909. The exact value is more of a hassle, the derivation of which I'll leave to the reader's imagination:

A = (43π/3√10)(√3 - 2)(√5 - 3)(√5 + 1)(√(5 - 2√5))(2√2 - √(5 - √5))

Addendum: For those who wonder, the triangle formed between the pentagon and the square has angles 9, 108, and 63 degrees. Furthermore, trignometric functions of angles which are multiples of 3 always have an algebraic value.

Addendum 2: As (Ω) Dr D suggests, the square could be fitted inside the pentagon in such a way that the base of it is parallel to the base of the pentagon. Let me check to see if it can be bigger that way.

Okay, for a pentagon of side 1, the maximum square that can be placed inside with one side parallel to one of the pentagon has a side length of 1.0605. The maximum square that can be placed inside with one vertex coinciding with one of the pentagon's has a side length of 1.0674. I'll keep looking at this problem to see where I could have gone wrong.

Yes, a pentagon of side length 4.99931 has an area of 43, as (Ω) Dr D says.

Addendum 3: whitesox08, finding the maximum size equilateral inside of a square is trivial, it's not hard to show that they must share a vertex. Finding the maximum size square inside of a pentagon is harder, since 1) they can share a vertex, or 2) they can have sides that are parallel. In both cases, there has to be a shared line of symmetry, otherwise the sides of the box inside the pentagon wouldn't be equal.

Addendum 4: (Ω) Dr D, please look at that expression x*(3/2 * sec36 - 1) again. This is the only place where we differ. It does not seem right, even for a square with a side parallel to one of the pentagon's. I think the area of the circle in this case should be 7.8872, as compared to 7.9909.
I'm getting a different answer. I"ll check it over just in case. But I'm finding that if the square shares the same vertex as the pentagon, it won't fit optimally.

First up, the area of the pentagon. Let the side be x.

Dividing the area into 3 triangles,

A = 2 * 1/2 * x^2 * sin108 + 1/2 * x^2 * (2cos36)^2 sin36

= 43

I'm getting x = 4.999 =~ 5

For the maximum square, I'm setting two of the sides parallel to one of the sides of the pentagon. All corners of the square touch the sides of the pentagon.

y = side of square

Applying trigs,

y = x*(3/2 * sec36 - 1)

For the maximum triangle, a common vertex is shared.

z = side of triangle.

z = y * sec15

For the circle, it's simply the incircle

r = z / (2√3)

I'm getting an area close to 49π/16 = 9.616

As I said, I have to check this over.

*EDIT*

Scythian: Yes I did make a mistake in simplifying. It should be:

y = x / (tan18 + 1/2 * sec36)

for a final area of 7.887145