As shown in the figure, p' is the moving point on BD, and p is the midpoint of BD:

p ' p = a *(ì2/2)* tanM

P ' a = a *(ì2/2)/cosM = P ' c

P ' b = P b-P ' P = a *(ì2/2)( 1-tanM)

So p 'b+p 'a+p 'c = a * [√ 2 * secm+(√ 2/2) (1-tanm)] = f (m).

Take the derivative of F(M):

f'(m)=a*[(√2*secm*tanm)-(√2/2)(secm)^2]，

If F'(M)=0, there are:

a*(√2*secm*tanm)=a*(√2/2)*(secm)^2

At the same time, divided by a*secM≠0 on the left and right sides of the equation, we get:

√2*tanM=(√2/2)*secM

SinM= 1/2 is obtained.

So m = 30.

Once again, it is proved to be a minimum point, and the second derivative of F(M) is as follows:

f''(m)=√2*a[(secm)^3+(tanm)^2*secm]-√2*a(secm)^2*tanm

Replace m = 30 with f'' (30) = √ 2 * a [8/(3 * √ 3)+2/(3 * √ 3)]-√ 2 * a (4/3 * √ 3/3).

=√2 * a[ 10/(3 *√3)-4/(3 *√3)]=(6√6)* a/9 & gt； 0

The first derivative is equal to zero, and the second derivative is greater than zero, then this point is the minimum point.

When m = 30,

f(M)= a *[(√2 * 2/√3)+(√2/2)( 1-√3/3)]= a *[2√6/3+√2/2-√6/6]

=a/6*[3√2-3√6], because the title gives F(M)=√2+√6. So a=2.

Fermat point is the point of 120.

After calculation

AP 'c = BP 'c = AP 'b = 120 is fermat point, so I wrote it at the beginning.

This is why bees always divide their hives into hexagons when building them, saving material A.

Thanks to the landlord's special triangle, otherwise I can't solve this problem at all.

Please pay more attention to the concept and nature of fermat point! There is obviously something wrong with the solution upstairs, and the answer is wrong. For ordinary people, we can use SIN 15 = 1/4 (√ 6-√ 2).

Cos 15 = 1/4 (√ 6+√ 2) guesses that this problem must be related to the angle of 15, which is indeed the case, BAP' = 15 = BCP', and we can try to solve the problem by omitting the above proof!