This is an additional question for the senior high school entrance examination in Zhejiang Province in 2009. I think this is quite difficult ... If P is a point on the plane of △ABC, and ∠ APB = ∠ BPC = ∠ CPA = 120, then point P is called the fermat point of △ABC. (1) If point p is an acute angle.

(2) As shown in Figure 5, make an equilateral △ACB' outside the acute angle △ABC to connect BB'.

Proof: BB' passes fermat point P, BB'=PA+PB+PC.

Take B as the vertex and rotate BPC 60 degrees outside BC to get BDE. According to fermat point's definition and rotation, there are:

1) ∠APB= 120 degrees

2) ∠BDE=∠BPC= 120 degrees

3) A, P, D and E four-point * * * line

4) Delta ·BPD is an equilateral triangle.

5) < CBE = 60 degrees

Because ∠ABC=60 degrees, so

6) Abe = ABC+CBE = 120 degrees

According to 4) and 6):

7) ambulatory blood pressure+ambulatory blood pressure = 60 degrees

because