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Equilateral Triangle: Area Formula and Proof


Date: 06/16/98 at 16:41:48
From: Andy
Subject: Equilateral Triangle Area

Is there a formula and proof to find the area of an equilateral 
triangle given a point on its interior and the lengths of the 
segments from the point to the three vertices? I have tried 
inscribing it in a circle and using Heron's formula.  

Thank you.


Date: 06/25/98 at 14:44:44
From: Doctor Floor
Subject: Re: Equilateral Triangle Area

Hi Andy,

Thank you for sending your question in to Dr. Math. 

This is not really an easy one. I will give an example of how to 
compute the side of the equilateral triangle when the three distances 
to the three vertices are 6, 8 and 10. I will also give the general 
formula to use for this.

First I'll draw a picture and introduce some names:

                C
               / \
              /   \            AB = AC = BC = a
             /     \           Angle names: t1 = APB t2 = BPC t3 = CPA
            /  10   \          
           /         \
          /    P      \
         /  6      8   \
        A---------------B

t1+t2+t3 = 2*pi. This yields cos(t1+t2+t3) = 1.
But we can also use a lot of trig to rewrite cos(t1+t2+t3). The 
trig formulas to use will be stated below, and will be given numbers 
[1], [2], etc. This gives:

1 = cos(t1+t2+t3)  
= cos(t1)cos(t2+t3) - sin(t1)sin(t2+t3) (use [1]) 
= cos^2(t1) - sin(t1)sin(t2)cos(t3) - sin(t1)cos(t2)sin(t3) ([2],[3])
= cos^2(t1) - sin(t1)sin(t2)cos(t3) + cos(t1)cos(t2)cos(t3)
            - cos(t1)cos(t2)cos(t3) - sin(t1)cos(t2)sin(t3)
= cos^2(t1) + cos(t3)(-sin(t1)sin(t2) + cos(t1)cos(t2))
            - cos(t1)cos(t2)cos(t3) - sin(t1)cos(t2)sin(t3)
= cos^2(t1) + cos(t3)cos(t1+t2)
            - cos(t1)cos(t2)cos(t3) - sin(t1)cos(t2)sin(t3) ([1])
= cos^2(t1) + cos^2(t3) - sin(t1)cos(t2)sin(t3) 
    - cos(t1)cos(t2)cos(t3) ([3])
= cos^2(t1) + cos^2(t3) - sin(t1)cos(t2)sin(t3) 
    + cos(t1)cos(t2)cos(t3) - 2cos(t1)cos(t2)cos(t3)
= cos^2(t1) + cos^2(t3) + cos(t2)(-sin(t1)sin(t3)+cos(t1)cos(t3))
    - 2cos(t1)cos(t2)cos(t3)
= cos^2(t1) + cos^2(t2) + cos^2(t3) - 2cos(t1)cos(t2)cos(t3) ([1],[3])


Formulas: [1]: cos(A+B)=cos(A)cos(B)-sin(A)sin(B)
          [2]: sin(A+B)=sin(A)cos(B)-cos(A)sin(B)
          [3]: when A+B+C=2*pi, then cos(A)=cos(2*pi - A)=cos(B+C)

So we derived:

  2cos(t1)cos(t2)cos(t3)-cos^2(t1)-cos^2(t2)-cos^2(t3)+1 = 0 (*)

Now let's take a look at triangle ABP

           P
          /t1\
        6/     \ 8
        /        \
       A-----------B
             a

We can apply the cosine-rule here to get: 

 a^2 = 6^2 + 8^2 - 2*6*8*cos(t1) or
   cos(t1) = (100 - a^2)/96

And in the same way we find:

   cos(t2) = (164 - a^2)/160
   cos(t3) = (136 - a^2)/120

These three we can substitute into (*) to get, after simplifying:

  - a^6 + 200a^4 -3088a^2 = 0   

So, since a cannot be zero (it must be a triangle-side!), we can 
factor out -a^2, to get:

  a^4 - 200a^2 + 3088 = 0

This is quadratic in a^2, so you find:

  a^2 = 100 + 48 SQRT(3) and a^2 = 100 - 48 SQRT(3).

So you find two possible values for a, one for P inside ABC and one 
for P outside ABC.

If you use d1, d2, d3 instead of 6, 8 and 10, you get the following 
formula:

  a^4 - (d1^2 + d2^2 + d3^2)a^2 
      + (d1^4 + d2^4 + d3^4 - d1^2*d2^2 - d1^2*d3^2 - d2^2*d3^2) = 0

And from here you can compute the roots.

Of course, when you've calculated a, the area of the triangle is not a 
problem any more.

I hope I have helped you enough with the problem. If you have some 
question on this, or another math question, send it again to Dr. Math!

Best regards,

- Doctor Floor, The Math Forum
  http://mathforum.org/dr.math/   


Date: 05/22/2000 at 21:17:19
From: Joe Mercer
Subject: Equilateral Triangle Area

I submit the following much, much, much easier solution to the 
equilateral triangle problem with a point inside that is a, b and c 
units from the vertices (the problem is to find the length of a side):  

Let ABC be the equilateral triangle and P a point inside that is a, b 
and c units from the respective vertices. Assume a < b < c. Rotate 
triangle APB about A through 60 degrees so AB coincides with AC, and 
label the new position of P as P' (note that AP' = AP = a and P'C = 
PB = b). Connect PP'. Then triangle APP' is equilateral, so angle 
AP'P is 60 degrees and PP' = a. Also P'C = b, and so triangle PP'C 
is completely determined. Thus angle PP'C can be found directly with 
the Law of Cosines (in the case that a, b and c are Pythagorean, it 
is 90 degrees). So angle AP'C = 60 + angle PP'C, and from another 
application of the Law of Cosines one can immediately find the side 
AC.


Date: 09/02/2000 at 14:14:54
From: Doctor Floor
Subject: Re: Equilateral Triangle Area

Hi,

My original reply to the question was made unnecessarily complex 
by the long derivation of the trig formula 
2cos(t1)cos(t2)cos(t3)-cos^2(t1)-cos^2(t2)-cos^2(t3)+1 = 0. 
In fact the derivation of this formula can be done rather simply.

We note that t1+t2+t3 = 2*pi, so we find

   cos(t1) = cos(2*pi-t1) 
           = cos(t2+t3)
           = cos(t2)cos(t3) - sin(t2)sin(t3).

Now we derive

   cos(t1) - cos(t2) cos(t3) = -sin(t2)sin(t3)
   (cos(t1) - cos(t2) cos(t3))^2 = sin^2(t2) sin^2(t3)
   cos^2(t1) - 2cos(t1)cos(t2)cos(t3) + cos^2(t2)cos^2(t3) =
   (1-cos^2(t2))(1-cos^2(t3))

which is easily rewritten to 
2cos(t1)cos(t2)cos(t3)-cos^2(t1)-cos^2(t2)-cos^2(t3)+1 = 0.

Best regards,
- Doctor Floor, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Geometry
High School Triangles and Other Polygons
High School Trigonometry

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