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Question:

Given certain points on a square(including its sides),let these points and the verteces of the squares be the verteces of a certain number of smaller triangles, no vertices of a smaller triangle are on the sides of other smaller triangles. what is the minimum value of the biggest angle ?

I think this is interesting problem,and I think is nice. I guess this answer is $90^{0}$,But I can't prove it.and maybe this is old problem (maybe is Very famous problem)

Thank you

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    $\begingroup$ Your question is unclear. What do you mean by "minimum value of the angles"? You can make an individual angle as small as you want by just choosing a single interior point very close to the midpoint of one side of the square. Even in the diagram you posted above it is obvious that there are many angles less than 90°. $\endgroup$ – mweiss May 16 '14 at 15:42
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    $\begingroup$ Would need the positions of the points maybe? Could you reword the part about the minimum angle? $\endgroup$ – bobbym May 16 '14 at 15:50
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    $\begingroup$ The question is unclear. If you mean the sum of all angles inside the square, then it is dependent only on the number of triangles since the sum of all angles in a triangle is $\pi$ radians. It is constant if you allow the points to move around but the connections to remain fixed. Do you mean the minimum angle as a the minimum from the set of all angles in the diagram? $\endgroup$ – abnry May 16 '14 at 15:58
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    $\begingroup$ What do you mean "no vertices of a smaller triangle are on the sides of other smaller triangles"? $\endgroup$ – abnry May 16 '14 at 15:59
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    $\begingroup$ I think the question is: what is the smallest angle $\alpha$ such that there exists a dissection of the square into triangles satisfying two properties: that none of the triangles has an internal angle greater than $\alpha$; and that no vertex of a triangle touches another triangle except at a vertex. If so, there's an easy lower bound of 67.5 degrees. $\endgroup$ – Peter Taylor May 16 '14 at 16:29
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To make explicit Peter Taylor's lower bound: the corners of the square are $90^\circ$. If you don't split one of them, you have a $90^\circ$ angle. If you do, the largest small angle you can get is $45^\circ$. Then if the other two angles of that triangle are equal, they are $67.5^\circ$ This does not show that there is such a construction. I have found several constructions that result in a smaller square.

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I think the smallest "largest angle" among all the triangles should be

60 degree

for 2 reasons:

  1. 60 degree is the smallest possible value that the largest angle (among all three) in a triangle could be.
  2. You can always draw an equilateral triangle inside the square, and pick wherever other points to complete the graph.
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