- Let me redraw your diagram
- get a new right triangle
Define a = NP, b = MQ, c = QP, d = MP. Then MN = 15-a, and we want to find b+3.
Now, b is to 3 as 3 is to c, so d = b+c = b+(9/b). Triangle MNP is a right triangle, so
Now, consider the area S of triangle MNP. We can consider segment MP to be the base, in which case we get
Or we can consider segment NP to be the base, in which case we get
So we have
From this we can write
which we can solve easily to get d = sqrt(234)-3. (The other root is negative and hence cannot be a length.) Recall that
We can rewrite this as
which we solve to get b = (d + sqrt(d^2-36))/2. The desired length is then just b+3.