A point in an equilateral triangle

[BMAD]

A point P is inside an equilateral triangle of side length d such that the distances to the vertices are given by a, b, c. Find the formula relating a,b,c,d.

 

Try with the case a,b,c = 3,4,5

[BMAD]

*bump*

[bonanova]

We can start thinking about the problem by looking for extrema for the sum s of of a, b and c.

 

By inspection, s = 2d at a vertex. That is its maximum.

At the centroid, s = 3^1/2d = ~ 1.733 d. That is its minimum.

Now I'm distracted to think what the curves of constant s look like.

 

But I think the next step is to write the law of cosines a bunch of times.

More later. 

[bonanova]

Let a b c = 3 4 5 as suggested, and solve for d.

 

[1] Assume a value for d.

Initial value of 6 seemed reasonable.

 

[2] Calculate the area A of the triangle two ways.

1. Use Heron's formula to get areas A₁ A₂ A₃ of the three triangles defined by the vertices and the interior point.
   A_sum = A₁ + A₂ + A₃.
    
2. Use the area formula for equilateral triangles: A_calc = (sqrt(3)/4) d²

[3] Compare A_sum to A_calc and adjust d until they are equal.

 

This gives d =~ 6.766432567.

 

I suppose the two area calculations could be equated to get a formula.

[gavinksong]

Let a b c = 3 4 5 as suggested, and solve for d.

 

[1] Assume a value for d.

Initial value of 6 seemed reasonable.

 

[2] Calculate the area A of the triangle two ways.

1. Use Heron's formula to get areas A₁ A₂ A₃ of the three triangles defined by the vertices and the interior point.

   A_sum = A₁ + A₂ + A₃.

    

2. Use the area formula for equilateral triangles: A_calc = (sqrt(3)/4) d²

[3] Compare A_sum to A_calc and adjust d until they are equal.

 

This gives d =~ 6.766432567.

 

I suppose the two area calculations could be equated to get a formula.

 

You also touched on this, but you could also use the law of cosines:

a² + b² - 2ab cos(x₁) = b² + c² - 2bc cos(x₂) = a² + b² - 2ab cos(x₃)

x₁ + x₂ + x₃ = 2π

 

Edit -- decided to use subscripts and the actual symbol for π (which is alt + p on macs btw)

Edited February 28, 2015 by gavinksong

[bonanova]

The same result is found using 3(a⁴ + b⁴ + c⁴ + d⁴) = (a² + b² + c² + d²)².

And I suppose this result could be derived a la my last post.

I'm not sufficiently motivated.   

[bonanova]

Let a b c = 3 4 5 as suggested, and solve for d.

 

[1] Assume a value for d.

Initial value of 6 seemed reasonable.

 

[2] Calculate the area A of the triangle two ways.

1. Use Heron's formula to get areas A₁ A₂ A₃ of the three triangles defined by the vertices and the interior point.

   A_sum = A₁ + A₂ + A₃.

    

2. Use the area formula for equilateral triangles: A_calc = (sqrt(3)/4) d²

[3] Compare A_sum to A_calc and adjust d until they are equal.

 

This gives d =~ 6.766432567.

 

I suppose the two area calculations could be equated to get a formula.

 

You also touched on this, but you could also use the law of cosines:

a² + b² - 2ab cos(x₁) = b² + c² - 2bc cos(x₂) = a² + b² - 2ab cos(x₃)

x₁ + x₂ + x₃ = 2π

 

Edit -- decided to use subscripts and the actual symbol for π (which is alt + p on macs btw)

 

I played with LOC for a bit, but OP asked for the relationship among a b c d.

I found I could not eliminate all the angles, only one of them.

 

The same result is found using 3(a⁴ + b⁴ + c⁴ + d⁴) = (a² + b² + c² + d²)².

And I suppose this result could be derived a la my last post.

I'm not sufficiently motivated.   

 

This is the relationship among a b c d.

I didn't derive it, or use it, opting instead in the specific case of a b c = 3 4 5 for iterative solution.