求出一條線段的最小值. (難題)

By Pyth thm, AB = 5

Let θ be ∠ABC, then sin θ = 3/5 and cos θ = 4/5

So area of △ABC = 6, hence area of △MBN = 3

Let BM = x, then:

(x/2) (BN) sin θ = 3

x (BN) = 10

BN = 10/x

By cosine law:

MN2 = BM2 + BN2 - 2(BM)(BN) cos θ

= x2 + 100/x2 - 16

So

d(MN2)/dx = 2x - 200/x3

d2(MN2)/dx2 = 2 + 600/x4

When d(MN2)/dx = 0, x = √10

So MN2 is min. when x = √10 which gives MN2 = 4

Finally min. of MN = 2

2013-05-28 15:26:03 補充：

Alternative w/o using diffferentiation:

x^2 + 100/x^2 >= 2√[x^2 (100/x^2)] (AM >= GM)

= 20

Hence MN^2 >= 20 - 16 = 4

MN >= 2

So min. MN = 2