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Equilateral Triangle (Posted on 2003-12-18) Difficulty: 3 of 5
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.

No Solution Yet Submitted by Ravi Raja    
Rating: 4.0000 (5 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution analytical solution (no computer used) | Comment 14 of 19 |
s=sqrt(25+12sqrt(3))

Proof:
s=side of triangle
a=angle PCB

in PCB and PAC - theorem of cosinus:
16=25+s²-10s(cos(a))
9=25+s²-10s(cos(pi/3-a))

s²-10s(cos(a))+9=0
s²-5s(cos(a))-5s√3(sin(a))=0

eliminate a
cos(a)=(s²+9)/(10s)
sin(a)=sqrt(1-cos(a)*cos(a))

the result is an equation (bi-square):
s^4 - 50s^2 + 193 = 0
the unique solution satisfying the side of a triangle is:

s=sqrt(25+12√3)

  Posted by luminita on 2004-01-08 08:44:39
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