(1) sets the coincidence after t seconds.
From the meaning of the question, it can be concluded that the m movement distance +AB=N movement distance should be made.
∴2t=t+ 12
The solution is t = 12.
∴ After points M and N move 12 seconds, points M and N coincide (in C).
(2) When an equilateral triangle is formed, AM=AN=MN, that is, △AMN is similar to △ABC.
Let t seconds later form an equilateral triangle.
Before n passes through a:
When △AMN∽△ACB:
By AM/AC=AN/AB
t/ 12 =( 12-2t)/ 12。
The solution is t=4.
N after a:
When m and n are on AC ∠ man = 0, the condition that △AMN is similar to △ABC is not satisfied.
After the two points M and N coincide with C, move again until it stops. Because the distance between m and n is always less than the length of CB, and ∠MAN is always less than 60, the condition that △AMN is similar to △ABC is not satisfied.
To sum up, after moving point M and point N for 4 seconds, an equilateral triangle △AMN can be obtained.
.
(3) existence.
Let t seconds later form an isosceles triangle.
Starting from points m and n moving to the edge of BC,
Then the isosceles triangle with MN as the base can only be △AMN, without △BMN and △CMN.
∫ Take MN as the base
∴AM=AN。
When m moves slower than n, when it can exist:
∠∠B =∠C,AB=AC,AM=AN
∴△ACM≌△ABN
∴CM=BN
From the coincidence of m and n in C after 12s:
1 *(t- 12)= 12-2 *(t- 12)
The solution is t= 16.
When point M and point N move on the edge of BC, an isosceles triangle with MN as the base can be obtained. At this time, the moving time of m and n is 16 seconds.