Recent content by bob012345

My artwork isn’t as pretty but here it is anyway. Each size triangle is represented only once. There are four shapes out of 22 that are left-right symmetric wrt to the whole figure are multiplied by 4 and the rest by 8.

Perhaps so. I have yet to make a simple regular hexagon work but one can see the answer by inspection!

I think the intent is that it would not count. Only vertices in the original figure count.

I think that’s correct. I was thinking along those lines too as I tried to apply the logical method to this figure; I can get the right number as long as I remove all triplets that have a vertex outside the figure.

Here are my results;

Don’t forget these;

You have caused me to see there are 8 less than I thought. Thanks!

Ok. You can figure the upper bound of total possible triangles and go down from there.

The game is to find the number of triangles in this complicated figure by other than brute force counting and explain your method.

Interesting observation! It appears so if both values are within the range of the functions. Looking at ##(a,s)## being (4,5) vs, (5,4), the area of overlap is the same over the angles but they are different situations. Here are screen shots;

I worked out the overlap of the two squares as a function of angle. Given ##a## is the half edge length of the original square and ##s## is the full edge length of the second square with one corner at the center of the first ; The area function goes as; $$ A_{\text{overlap}} = \begin{cases}...

Agreed. Now the larger square is the 1m square. The overlap depends on the relative orientation until ##s## shrinks to ##\frac{1}{2\sqrt{2}}## then it just becomes ##s^2##. Now it would be interesting to see for ##s## in that range what the overlap vs. angle function is.

The next level is if we let the length ##s## of the larger square vary, what range of values of ##s## will your statement not be true?

Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two...

Well that explains my little mystery from post #18. Here is the curve for the total area of the triangle for c=d normalized to the product cd;