We can use the identity: sin(A-B) = sinAcosB - cosAsinB

sin(A-B+C) = sinAcos(B-C) + cosAsin(B-C)

Given: sin(A-B+C) = 1/2

We know that A is acute, so sinA > 0. Therefore, we can divide both sides by sinA:

sin(A-B+C)/sinA = sinAcos(B-C)/sinA + cosAsin(B-C)/sinA

Using the fact that sinAcos(B-C) + cosAsin(B-C) = sin(A-B), we can simplify:

sin(A-B)/sinA = cos(B-C) + cotA sin(B-C)

We can use the Law of Sines to write sinA = a/2R, where a is the length of side BC and R is the radius of the circumcircle of triangle ABC. Similarly, sin(B-C) = (b-c)/2R, where b and c are the lengths of sides AC and AB, respectively. We can also use the fact that cotA = (b^2 + c^2 - a^2)/(4K), where K is the area of triangle ABC.

Substituting these into our equation, we get:

(b-c)/a = cos(B-C) + (b^2 + c^2 - a^2)/(4Ka)

Multiplying both sides by 4Ka, we get:

4K(b-c) = a^2cos(B-C) + b^2 + c^2 - a^2

Using the Law of Cosines, we can write a^2 = b^2 + c^2 - 2bc cosA. Substituting this into our equation, we get:

4K(b-c) = 2bc cosA cos(B-C) + b^2 + c^2 - b^2 - c^2 + 2bc cosA

Simplifying, we get:

4K(b-c) = 2bc cosA (cos(B-C) + 1)

K = (1/2)bc sinA, so we can substitute this in:

2(b-c)sinA = 2bc cosA (cos(B-C) + 1)

Dividing by 2sinA, we get:

(b-c)/a = cos(B-C) + 1/sinA

Since sin(A-B+C) = 1/2, we know that A-B+C = 30 degrees (or pi/6 radians). Therefore, we can use the fact that sin(pi/6) = 1/2 to get sinA = 2sin(B-C).

Substituting this in, we get:

(b-c)/a = cos(B-C) + 1/(2sin(B-C))

Multiplying both sides by 2sin(B-C), we get:

2(b-c) = 2a cos(B-C) + sin(B-C)

Using the Law of Cosines again, we can write a^2 = b^2 + c^2 - 2bc cosA = b^2 + c^2 - 2bc cos(B+C). Substituting this in, we get:

2(b-c) = 2(b^2 + c^2 - 2bc cos(B+C)) cos(B-C) + sin(B-C)

Simplifying,