- 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