<pre>
Let me redraw your diagram:

     M
     ||
     || 
     ||  
     ||   
     ||    
     ||     
     ||      
   Q ||------- N
     ||   3   ||
     ||      3|| 
     ||'' '' '' ''||'' '' '' '' '' ''
             Q'  P

I take your diagram to mean that MP <code> 15, and NQ </code> NQ' = 3.  And we
want to find MQ+3.  Now fold rotate triangle MQN around N, so that we
get a new right triangle:

                   N
                   _
                ''''/||
             ''''/   || 
          ''''/      ||  
       ''''/         ||   
   M  /'' '' '' '' '' '' ||'' ''   P
                   Q

Define a <code> NP, b </code> MQ, c <code> QP, d </code> MP.  Then MN = 15-a, and we want to
find b+3.

Now, b is to 3 as 3 is to c, so d <code> b+c </code> b+(9/b).  Triangle MNP is a
right triangle, so

    a^2 + (15-a)^2 = d^2

Now, consider the area S of triangle MNP.  We can consider segment MP
to be the base, in which case we get

    S = 3d/2

Or we can consider segment NP to be the base, in which case we get

    S = a(15-a)/2

So we have

    a(15-a) = 3d

From this we can write

    15^2 = [[[a + (15-a)]]]^2
         = a^2 + 2a(15-a) + (15-a)^2
         = a^2 + (15-a)^2 + 2a(15-a)
         = d^2 + 6d

or

    d^2 + 6d - 225 = 0

which we can solve easily to get d = sqrt(234)-3.  (The other root
is negative and hence cannot be a length.)  Recall that

    d = b+(9/b)

We can rewrite this as

    b^2 - db + 9 = 0

which we solve to get b = (d + sqrt(d^2-36))/2.  The desired length
is then just b+3.

byron elbows
brian@isi.edu
</pre>
