I have a couple of proofs about a convex 10-gon that I need help with.
"1. Prove there exists a convex 10-gon that has 3 acute, interior angles.
2. Prove there does not exist a convex 10-gon that has 4 acute, interior angles."
Do you understand that the sum of the interior angles is $\displaystyle 8\pi$?
If there were three acute angles the remaining seven angles would sum to more than $\displaystyle \frac{13\pi}{2}$.
Is that possible? Recall that each of those angles would between $\displaystyle 0\;\&\;\pi$
BUT, what would happen if there were four acute angles?
What is the possible sum of the remaining six angles?
Well I do now. I've been sick all weekend but here's what work I got from your post.
For the first one it is possible because $\displaystyle \frac{13\pi}{2}$ is 20.42 and taking 7pi can give you 21.99. (My LaTeX sucks)
I don't know the properties of 10-gons but taking 3/10 of 8 would be 2.4 so where did you get 13/2 instead of 5.6? (I took the pi out to make it simplier but 5.6pi would be what I thought would be the sum you need to go over.
This is the question that has me stuck on how to get what the sum shouldn't be on the second question. (trying to disprove the sums)
If you don’t know the properties of 10-gons, I think it is almost impossible to attempt this problem.
Have a look at this webpage: Polygon -- from Wolfram MathWorld
Actually looking at this problem I think it's easier than I expected. $\displaystyle 1440^\circ$ is the total sum of the interior angles. If 3 angles are acute, then the sum of the remaining angles should be more than $\displaystyle 1170^\circ$. So by taken 1170 divided by 7 I got 167.14 which is below $\displaystyle 180^\circ$ keeping it convex.
For the second proof the remaining angles should be more than $\displaystyle 1080^\circ$ and by taking that divided by 6 I get 180 and going past 180 would make the decagon concave.
I hope this is what you were hinting at in your first post.