- Let me redraw your diagram
- M |\ | \ | \ | \ | \ | \ | \

Q |-------\ N

| 3 |\ | 3| \ |_ _ _

|\_ _ _ _ _Q' P

I take your diagram to mean that MP = 15, and NQ = 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 = 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

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