Suppose ABC is an equilateral triangle and P is a point inside the triangle, such that PA = 3 cms., PB = 4 cms., and PC = 5 cms.
Then find the length of the side of the equilateral triangle.
(In reply to Solution
Here is the general case, based on tom's solution:
Let a=|PA|, b=|PB|, c=|PC|, and let s=|AB|=|BC|=|CA|.
Rotate ABC 60 degrees about A. This takes P to P', B to B'=C, and C to C'.
Now, AP=AP', and PAP' = 60 degrees, so triangle PAP' is equilateral with side length a. In particular, |PP'|=a. |P'C|=|P'B'|=|PB|=b, and |PC|=c. So triangle PP'C has sides of lengths a,b,c. Let K be the area of this triangle.
By the law of cosines,
s^2 = a^2 + b^2 - 2ab cos(AP'C).
Now, the angle AP'C = AP'P + PP'C = 60 + PP'C, so
cos(AP'C) = cos(60 + PP'C) = 1/2 cos(PP'C) + sqrt3/2 sin(PP'C).
But by the law of cosines again,
cos(PP'C) = [a^2 + b^2 - c^2]/(2ab),
and by the area formula,
sin(PP'C) = 2K/(ab).
Plugging in these values gives
s^2 = a^2 + b^2 - 2ab[ 1/2 [a^2 + b^2 - c^2]/(2ab) + sqrt3/2 [2K/(ab)] ]
= [a^2 + b^2 + c^2 + K*4sqrt 3]/2,
where again, K is the area of a triangle with side length a,b,c.
In our case a=3,b=4,c=5, and K=6, and we get s^2 = 25+12sqrt3.
Posted by Eigenray
on 2008-03-15 18:24:00