A simple polygon problem.

[bonanova]

An arbitrary convex polygon has a vector drawn from the midpoint of each of its sides.

The vectors are perpendicular to the sides, pointed outward, and have the same lengths as the sides themselves.

 

What is the sum of these vectors?

[k-man]

The sum is 0. It's pretty easy to see that for any right triangle the sum of such vectors is 0. Any triangle can be subdivided into 2 right triangles and the vectors from the subdivision line will cancel each other out, so every triangle has the sum of such vectors equal 0. Every convex polygon can be subdivided into triangles, so 0 for every convex polygon.

 

Now, a question... Does a polygon even need to be convex? I think this works for any polygon.

[k-man]

0

[bonanova]

Good point. It need not be convex.

I have another solution in mind.

It's not better in any way - a proof it a proof.

 

It wasn't obvious to me that it's obviously true for right triangles.

Wait. Now it's obvious. Nice!

 

Regarding my other solution, my clue would have been:

What if the vectors were drawn inward?

[gavinksong]

I know that k-man already got the answer, but I want to put in my two cents.

Since they're vectors, it doesn't matter whether they're drawn from the midpoint of each side or whatever. All that matters is the length and direction. Length is the same; direction is just perpendicular to the sides. So if we draw the vectors end to end, we just get the shape of the original polygon rotated 90 degrees. Obviously, it's a closed shape, so the sum is zero.

[bonanova]

Yes. So now there are three solutions.

 

I thought at first that it does matter where the vectors are drawn. Consider them as forces and the polygon as a rigid solid. How does the solid move under their influence? Translation and rotation would both have to be considered. Forces can add to zero but still produce torque. But that's not what was asked.

 

Your solution can be visualized as easily as the question (hopefully) was.

[bonanova]

There is a third solution. It involves physics.

 

[spoiler=Clue for the physics solution]What if the vectors pointed inward?

[gavinksong]

There is a third solution. It involves physics.

Clue for the physics solution]What if the vectors pointed inward?

Ok, I'll bite.

I suppose that if the vectors were turned inward, then the net force and the torque would be negated. However, rotating the polygon by 180 degrees should do the same thing because it flips the direction of all of the vectors, but maintains their magnitudes and positions in relation to the centroid. In this case, the net force should also be flipped as well, but the torque shouldn't have changed, so torque = -torque = 0. Can't say anything about the net force though...

[bonanova]

Consider the polygon to be a solid - that it has non-zero thickness - and immerse it in water. The hydrostatic forces on the solid are normal to the surfaces and proportional to their areas and thus to the side lengths of the polygon, as are the vectors. The hydrostatic forces sum to zero, else the solid would move. So the vectors sum to zero also.

 

Or consider inflating it with air. The outward force on the walls is proportional to their area and thus to the side lengths. Again since the object would not move, the vectors sum to zero.