AB=BC=AC=2，∠A=∠B=∠C=60

Pass p to PE⊥BC in E, pass E to EF⊥AC in F, and pass F to PQ⊥AB in Q.

Then ∠ bpe = ∠ cef = ∠ afq = 30.

Let BP = X.

Then BE=x/2,

So EC=2-x/2

So CF=(2-x/2)/2= 1-x/4.

So af = 2-(1-x/4) =1+x/4.

So AQ = (1+x/4)/2 =1/2+x/8 = x/8+1/2.

That is y=x/8+ 1/2, (0 < x ≤ 2).

2. if point p coincides with point q.

Then BP+AQ=2.

That is, x+y=2.

So x+x/8+ 1/2=2.

The solution is x=4/3.

That is, when the length of BP is equal to 4/3, point P coincides with point Q.

3. When the line segments PE and FQ intersect,

Because PEF = EFQ = 60.

So the triangle surrounded by line segments PE, EF and FQ is still an equilateral triangle.

Its side length is equal to EF length.

Derived from Pythagorean theorem

EF=√3CF=√3( 1-x/4)

So the perimeter of the triangle surrounded by line segments PE, EF and FQ is

C=3EF=3√3( 1-x/4)

When the line segments PE and FQ intersect,

Blood pressure +AQ≥2

That is, x+y≥2

x+x/8+ 1/2≥2

x≥4/3

So when the line segment PE and FQ intersect, (4/3≤x≤2)

Because in C=3√3( 1-x/4), c decreases with the increase of x.

so 3√3( 1-2/4)≤C≤3√3[ 1-(4/3)/4]

That is, 3√3/2≤C≤2√3

So when the line segments PE and FQ intersect,

The range of the triangle perimeter c surrounded by line segments PE, EF and FQ is 3√3/2≤C≤2√3.

The third question is great! Thank you for providing such a good question.