Area of Hexagon - Geometry Challenge

[Mateus Buarque]

Determine the area of the painted hexagon, knowing that the area of triangle ABC is 120cm^2

IMG Link: https://m.imgur.com/a/WtdsW

I tried using Heron´s formula, but just ended up with a bunch of terms and one more variable.

Sidenote: I guess part of it is figuring out that the side lenghts don´t matter, just the actual area. That is because variables x, y and z were just a way to show the sides were divided into three equal parts.

 

[haruspex]

I guess part of it is figuring out that the side lenghts don´t matter,

If you need to prove your answer, yes. But if you just want to get the answer you could cheat and assume that, so make them all equal.

 

[Mateus Buarque]

If you need to prove your answer, yes. But if you just want to get the answer you could cheat and assume that, so make them all equal.

Definetely :). I just said that because the answer is invariant, so they should just cancel out in the end.

 

[mfb]

If you can use non-trivial results, or put together various coordinates, you can write the hexagon as sum of triangles and then follow approaches similar to these here.

 

[haruspex]

Definetely :). I just said that because the answer is invariant, so they should just cancel out in the end.

Not sure if you are saying you are happy to assume it is invariant or feel the need to prove it.

I think proving it is not hard. Consider a linear transformation which skews it parallel to one side. I.e. cut it into very thin strips parallel to one side and drag the strips along in proportion to distance from that side until the vertex strip has the vertex opposite the midpoint of the stationary side. Straight lines are still straight, areas are preserved, and the ratios in which the internal lines divide the triangle's sides are preserved.

Having made it equilateral, I decided to ignore the given 1/3 ratio and consider three lines from each vertex. The middle one bisects the far side, while the other two cut it in the ratio x:1-x. This divides the triangle into 30 regions of 5 different shapes.
(Keeping it as generic x helps because having calculated a region's area you can swap x with 1-x to get areas of two regions for the price of one.)
You can figure out all the areas in a sequence of steps, pretty much just using the cosine rule over and over. I found it useful to find the angle formed at a vertex by one of the triangle's sides and the third ray from it at the vertex. I.e, at a vertex of you five lines, two of the sides of the triangle, a median, and the two intermediate rays. So I mean the angle between a triangle side and the further of the two intermediate rays. It's not hard to show the tan of this angle is (1-x)/(1+x).

No doubt there's a much smarter way, but I did finally get the answer.

 

[.Scott]

I have compute the ratio of the area of triangle ABC to that of the hexagon for:
- A right isosceles triangle;
- An equilateral triangle;
- A 3/4/5 triangle;
- A couple of other random triangles.
In all cases the ratio is the same integer value.

 

[mfb]

I have compute the ratio of the area of triangle ABC to that of the hexagon for:
- A right isosceles triangle;
- An equilateral triangle;
- A 3/4/5 triangle;
- A couple of other random triangles.
In all cases the ratio is the same integer value.

You can transform every triangle into every other triangle while preserving area ratios. It has to be the same for every triangle.

 

[.Scott]

You can transform every triangle into every other triangle while preserving area ratios. It has to be the same for every triangle.

I was in the process of editing my post to include a description of that linear transform.
As a software engineer, you can guess how I did it.
Should I post that code, or would that be too much of a hint?

 

[haruspex]

I was in the process of editing my post to include a description of that linear transform.
As a software engineer, you can guess how I did it.
Should I post that code, or would that be too much of a hint?

Is it simpler than the transformation I described in post #5?
The OP seems to have lost interest; no response in six weeks.

 

[mfb]

Every affine transformation preserves area ratios, and it should be clear how to get from an arbitrary triangle to any other triangle via affine transformations.