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Median Maximum (Posted on 2008-02-03) Difficulty: 3 of 5
Two medians of a triangle have lengths of 12 and 15. How long is the third median when the area of the triangle is a maximum? (Try to solve this without calculus.)

See The Solution Submitted by Brian Smith    
Rating: 3.3333 (3 votes)

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Solution Using Trigonometry Comment 3 of 3 |
Let the length of medians from A,B and C be x,y and z
respectively.
Let mid point of BC be D.
Consider triangle ABD
cosB=(c²+(a/2)²-x²)/ac
but cosB=(a²+c²-b²)/2ac
=> 2(c²+a²/4-x²)=a²+c²-b²
=> c²+b²-a²/2=2x² ----(1)
Similarly
a²+c²-b²/2=2y² -------(2)
a²+b²-c²/2=2z² -----(3)
Subtract (3) from (2)
3/2(c²-b²)=2(y²-z²) -------- (4)
2(1)+(2) =>
3c²+3/2b²=4x²+2y² ----- (5)
(5)+(4) =>
c²=(4/9)*(2x²+2y²-z²) -- (6)
b²=(4/9)*(2x²+2z²-y²) -- (7)
a²=(4/9)*(2z²+2y²-x²) -- (8)
Area = √s(s-a)(s-b)(s-c)
=(1/4)*√{(a+b+c)(a+b-c)(c+a-b)(c-(a-b))}
=(1/4)*√{(2ab+(a²+b²-c²))(2ab-(a²+b²-c²))}
=(1/4)*√(4a²b²-(a²+b²-c²)²)
Sub (6),(7),(8) in above eq
Area=(1/3)*√(4(2z²+2y²-x²)(2x²+2z²-y²)-(5z²-x²-y²)²)
Sub x=12,y=15
Area=(1/3)*√(4(2z²+306)(2z²+63)-(5z²-369)²)
=(1/3)*√(4(4z^4+738z²+19268)-25z^4+3690z²-136161)
=(1/3)*√(-9z^4+6642z²-59049)
=√(738z²-z^4-6561)
=√129600-(z²-369)²
So, Area will be Maximum when z²=369 => z=3√41
and Maximum Area = √129600 = 360.

Edited on February 4, 2008, 1:45 am
  Posted by Praneeth on 2008-02-04 00:36:40

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