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Finding the Base of Parts of a Triangle

Date: 05/22/2000 at 12:40:23
From: Sarah Pawlewski
Subject: Geometry/Algebra

Not even sure if this is solvable analytically. If it's not solvable, 
then can you say why not?

Can you derive an expression for L1 in terms of L2 and L3 such that 

     (area A1)/(Total Area of triangle)  =  10%
     (area A2)/(Total Area of triangle)  =  10%

                     / |  *
                    /  |    *
                   /   |      *
                  /    |        *
                 /     |          *
                /      |            *
               /       |              *
              /        |                *
             /|        |                  *
            / |        |                    *
           /  |        |                   |  *
          /   |        |                   |    *
         /    |        |                   |      *
        /     |        |                   |        *
       /   A1 |        |                   |   A2     *
      /       |        |                   |            *

      <--L1--> <--L2--> <-------L3--------> <----L4------->


Date: 05/22/2000 at 16:47:41
From: Doctor Rob
Subject: Re: Geometry/Algebra

Thanks for writing to Ask Dr. Math, Sarah.

Such an expression can be found. Let the height of the triangle with 
area A1 be H1, the height of the triangle with area A2 by H4, and the 
total height of the triangle be H. Then you have the following 

       H1/L1 = H/(L1+L2)
       H4/L4 = H/(L3+L4)
     L1*H1/2 = (L1+L2+L3+L4)*H/20
     L4*H4/2 = (L1+L2+L3+L4)*H/20

Now you can eliminate H1 and H4 from the last two equations by solving 
for them using the first two equations and substituting. Then you can 
cancel H from both sides of the new last two equations. That gives:

     L1 + L2 + L3 + L4 = 10*L1^2/(L1+L2)
     L1 + L2 + L3 + L4 = 10*L4^2/(L3+L4)

Solve the first of these equations for L4 and substitute that into the 
other one. That will give you an equation in L1, L2, and L3 alone. 
Simplify and put everything on one side over a common denominator. 
Then the numerator must equal zero. Unfortunately, it is a fourth 
degree polynomial in L1 that does not factor into lower degree ones. 
Its coefficients are polynomials in L2 and L3:

   72*L1^4 + (-34*L2-18*L3)*L1^3 + (-13*L2^2-14*L2*L3+L3^2)*L1^2
                       + L2*(L2+L3)*(2*L2+L3)*L1 + L2^2*(L2+L3)^2 = 0

Irreducible quartic equations can be solved in terms of square and 
cube roots of functions of their coefficients, but in this case, where 
the coefficients are polynomials, the expressions involved are very 
complicated, and I hesitate to proceed further. If you have numerical 
values of L2 and L3, this equation might reduce to one that is 
factorable, and then the solution is easier. Alternatively, one could 
solve the equation numerically to obtain approximate values for L1.

So you see that L1 can be expressed in terms of L2 and L3, but not 

- Doctor Rob, The Math Forum
Associated Topics:
College Triangles and Other Polygons
High School Triangles and Other Polygons

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