How Many Equilateral Triangles are Needed for Perfect Coverage?

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The discussion centers around determining the minimum number of identical equilateral triangles needed for perfect coverage when one triangle is randomly selected and others can be slid to cover it. The initial guess for the minimum number is four, based on the symmetry properties of equilateral triangles. Participants question the definition of "randomly" placing the triangles, suggesting a uniform distribution but acknowledging the complexities due to infinite placement possibilities. The conversation highlights the informal nature of the problem, noting it is more of a puzzle than a formal mathematical question. Overall, the challenge lies in balancing geometric symmetry with the nuances of probability in triangle placement.
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Homework Statement



We are using a drawing program in computer and we place x number of identical equilateral triangles(of same length of edges) randomly. So whenever we choose a triangle on the screen randomly(each has an equal number of possibility of being selected), we can slide the other triangles(without rotating them) in any way to cover the chosen one. In order to able to do this for each chosen triangle and for each different placements of the other triangles, what is the minimum number of triangles,that is x, placed on the screen?



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The Attempt at a Solution



This problem is neither homework nor coursework; it is a challenging - i think -math puzzle i saw on the internet. I didn't solve it formally but i guess that the minimum number of triangles is 4 using the symmetry(both relative and intrinsic) of the equilateral triangles.
 
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First you will have to tell exactly what you mean by "place x number of identical equilateral triangles(of same length of edges) randomly". While we can assume that choosing one of the triangles "randomly" means using the uniform distribution, since there are only a finite number of triangles, there are an infinite number of ways to place triangles on the screen. What probability distribution are you using for "randomly"?
 
I have thought of it as uniform distribution, too, without taking into account the technical problem that might occur with the infinite number of possibilities. I didn't take the probability and random variables at the university so i don't know what random actually means here but i doubt the puzzlewriter uses it in a technical manner, since it is not a formal math question but a puzzle.
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
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