Equilateral Triangle: Area Formula and ProofDate: 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/ |
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