Express a Function in Terms of Another Function

Date: 09/18/97 at 13:29:20
From: Migdalia Anes
Subject: Expressing a function in terms of another function

The altitude perpendicular to the hypotenuse of a right triangle is 
12 cm. Express the length of the hypotenuse in terms of the perimeter.

I tried lots of things, but they're wrong. I really don't know how to 
get it! 

Date: 09/24/97 at 17:08:08
From: Doctor Rob
Subject: Re: Expressing a function in terms of another function

Draw a diagram of a right triangle with hypotenuse h, perimeter p, 
and one leg x, other leg p-h-x:

               .
              /|
             / |
            /  |
           /   |
          /    |
       h /     | p-h-x
        /      |
       /       |
      /`-.12   |
     /    `-.  |
    /________`-|
         x

Applying the Pythagorean Theorem to the small triangles gives the two
parts of the hypotenuse as Sqrt[x^2-144] and Sqrt[(p-h-x)^2-144].  
Their sum must be h. The Pythagorean Theorem on the large triangle 
gives another equation.  Together we have:

   x^2 + (p-h-x)^2 = h^2,
   h = Sqrt[x^2-144] + Sqrt[(p-h-x)^2-144].

To remove the radicals from the second equation, rewrite it as

   h - Sqrt[x^2-144] = Sqrt[(p-h-x)^2-144],

and square both sides:

   h^2 - 2*h*Sqrt[x^2-144] + x^2 - 144 = (p-h-x)^2 - 144.

Add 144 to both sides, and replace (p-h-x)^2 by h^2 - x^2 from the 
first equation.

   h^2 - 2*h*Sqrt[x^2-144] + x^2 = h^2 - x^2.

Isolating the radical on the right side, you get

   x^2 = h*Sqrt[x^2-144].

Now square both sides again, and the radicals disappear:

   x^4 = h^2*(x^2-144).

Combine this with the first equation, and you get the system

   x^4 - h^2*x^2 + 144*h^2 = 0,
   x^2 + (p-h-x)^2 - h^2 = 0.

You want to eliminate x from these equations, to get one equation 
involving h and p alone.  Expand the second equation:

   x^2 + p^2 + h^2 + x^2 - 2*h*p - 2*p*x + 2*h*x - h^2 = 0,
   2*x^2 + (2*h-2*p)*x + (p^2-2*h*p) = 0.

Find the quotient Q of the quartic polynomial divided by the quadratic 
one as polynomials in x. Subtract Q times the quadratic one from the 
quartic one. That will leave, fortuitously, a quadratic equation in h.  
Factor it, and set each factor linear in h equal to zero. These can be 
solved for h to give the desired expressions. 

Of course you have to check your answer, since squaring twice could 
have introduced spurious solutions! Actually, you can use the known 
fact that h < p/2 to throw out one of the solutions.

Once you find h in terms of p, there will be two values of x for each 
value of p, because we didn't specify if x was the longer leg or the 
shorter one. Thus x will be the root of a quadratic equation, namely 
the last one given above, with coefficients functions of p.

-Doctor Rob,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/