MHB Im assuming a factorization problem?

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The discussion revolves around a problem involving finding the shortest path, with participants noting that the shortest path identified is 13. There is uncertainty about how to justify this path as the shortest, with suggestions that a "fuzzy" justification may suffice. Participants also discuss the relevance of obstacles in determining the path length, mentioning that without certain obstacles, the path could potentially be as short as 9. The conversation includes a reminder for users to show effort in their questions and to attach images directly to posts for clarity. The topic remains focused on understanding the shortest path and its justification.
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Hi Ilikebugs and welcome to MHB! :D

It is preferable that you attach images to your post instead of linking to an image site, which may be unreliable. I've attached the image you linked to.

Also, we ask that users show some effort when posting questions so we may best know how to help. Why do you think the problem may involve factoring?
 
Well I have to justify why the shortest path is the shortest one. I can't find any way to justify why. The shortest path I've found is 13.
 
Ilikebugs said:
Well I have to justify why the shortest path is the shortest one. I can't find any way to justify why. The shortest path I've found is 13.

Hi Ilikebugs! That's a nice problem! ;)

The shortest I've found is 13 as well and I have no "hard" justification.
Perhaps only a "fuzzy" justification is needed?
Such as that it's the shortest path the goes between the King's Lake and the King's Forest.
And we might mention that without those 2 obstacles, we could get to 9.
Btw, what makes you think it's a factorization problem?
 

Thread 'Trigonometry problem of interest'

Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
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