ignore for a second the thing that you have an “inside boundary” where the points have a different weight/cost/reward:

in a standard scenario of a grid of points where all them have the very same weight/cost/reward, because there are no obstacles, the shortest distance between two points is the Manhattan distance, which is x_steps + y_steps, like in this image:

x=11, y=10 → whatever final route that had you move a total of exactly 10 times up and exactly 11 times right, in any combination, will get you from the bottom left corner to your destination point, and all of them will be equal in terms of cost (21 moves, of which 10 up and 11 right)

at this point of the conversation, we can start adding the concept of an area inside of which the costs of movement is halved

in order to get the best outcome, you want to make as many useful movements as possible inside that area, which means in the particular case shown in your picture to get from point Start to point 1, then from 1 to 2 at half the cost, and finally from 2 to destination, like this:

because of the location of the area in this particular example, you will see that the problem is just 3 identical subproblems, again solvable with Manhattan distance

First problem: 6 right + 3 up

Second problem: 2 right + 4 up

Third problem: 3 right + 3 up

any combination of paths you can find using the above rules, will give you the shortest available path in this particular configuration

but what happens if your “half cost area” is in other positions?

for instance, because it’s randomly placed, it could also come in locations like:

in blue it’s visualized a rectangle that is equal to Manhattan distance between Start and Destination: only the points of “half cost area” that lie inside the blue rectangle will contribute to a cheaper path… and the principle is always the very same:

Start - point_1 ; point_1 → point_2 ; point_2 → Destination

it’s important here to distinguish if the associated cost of a movement has to be calculated by distance, like in the above case, or by corners -as initially stated- because the scenarios are much different

if you associate the cost of a given path exclusively to the number of corners, of course the cheapest is always A or B, because both consist of only one corner in total: