Area of Hexagon - Geometry Challenge

In summary, the conversation discusses different approaches and methods for determining the area of a painted hexagon, based on the known area of triangle ABC. The conversation includes suggestions for using Heron's formula, considering the side lengths, and using linear transformations to transform the hexagon into different shapes. It is concluded that the ratio of the area of triangle ABC to the hexagon is the same for all types of triangles and can be found through affine transformations.
  • #1
Mateus Buarque
6
0
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.
 
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  • #2
Mateus Buarque said:
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.
 
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  • #3
haruspex said:
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.
 
  • #4
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.
 
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  • #5
Mateus Buarque said:
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.
 
  • #6
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.
 
  • #7
.Scott said:
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.
 
  • #8
mfb said:
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?
 
  • #9
.Scott said:
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.
 
  • #10
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.
 

FAQ: Area of Hexagon - Geometry Challenge

What is the formula for finding the area of a hexagon?

The formula for finding the area of a hexagon is: A = (3√3/2) × s², where A is the area and s is the length of the side of the hexagon.

How many sides does a hexagon have?

A hexagon has six sides.

What is the unit of measurement for the area of a hexagon?

The unit of measurement for the area of a hexagon is typically square units, such as square inches or square centimeters.

Can the area of a hexagon be negative?

No, the area of a hexagon cannot be negative. It is a measurement of space and therefore cannot have a negative value.

What is the relationship between the area of a hexagon and its perimeter?

The area of a hexagon is directly proportional to its perimeter. This means that as the perimeter increases, so does the area, and vice versa.

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