We continue discussing the problem on the 2006 sided regular polygon described earlier as
A diagonal of a regular 2006-gon is called odd if its endpoints divide the boundary into
two parts, each composed of an odd number of sides. Sides are also regarded as odd diagonals.
Suppose the 2006-gon has been dissected into triangles by 2003 non-intersecting diagonals.
Find the maximum possible number of isosceles triangles with two odd sides.
We used the lemma earlier that when a dissecting diagonal leaves a shorter path of the boundary of the 2006-gon that consists of n -segments, then the number of iso-odd triangles with vertices on this shorter path does not exceed n/2.
We said that if there is an iso-odd triangle PQS with all vertices on this shorter path and with PQ the diagonal being the longest, then S is the summit of the triangle PQS and the vertices divide the shorter path into four segments of AP, PS, SQ and QB.
Since PQ is part of an iso-odd triangle and is the longest diagonal, an iso-odd triangle cannot have vertices on both the first and the last segment. This means that every iso-odd triangle that fall on the shorter path has all its vertices on one of the four segments.
Given each of these four segments and applying the lemma we started with, we get the sum of the triangles in all these four regions to be less than n/2. Furthermore the two inner segments will have odd number of sides which makes the inequality of the number of triangles to be strictly less than n/2 and leaves one out of two in excess for both these inner edges. Because of this excess, PQS is also covered by this estimate n/2 and the lemma holds.
We can argue the same when the iso-odd triangle with the longest diagonal is drawn the other way with its third vertex not falling on the shorter path. In this case, the triangle is acute or right triangle not the obtuse we had seen earlier. And we have three regions XZ, YZ and XY where XZ and YZ are shorter than XY. For each of the three regions, we apply the lemma and infer that there are no more than n/2 iso-odd triangles in all unless XYZ is one of them. Thus the total number of triangles in the dissection is not greater than n/2.
We can also reason it for arbitrary triangles. Let us say ABC is an iso-odd triangle with AB and BC odd sides. This implies there are odd number of sides between A and B and B and C. We call these sides as belonging to iso odd triangle ABC.
At least one side in each of these triangles do not belong to any other iso-odd triangle because an iso-odd triangle with two sides of equal length and therefore there are even number of sides belonging to it. Eliminating all sides that belong to iso-odd triangles, we get one that isn't. Let us assign these two sides one in each group to the triangle ABC. To each iso-odd triangle, we have thus assigned a pair of sides with no two triangles sharing an assigned side. Therefore there 1003 such iso-odd triangles that can appear.
A diagonal of a regular 2006-gon is called odd if its endpoints divide the boundary into
two parts, each composed of an odd number of sides. Sides are also regarded as odd diagonals.
Suppose the 2006-gon has been dissected into triangles by 2003 non-intersecting diagonals.
Find the maximum possible number of isosceles triangles with two odd sides.
We used the lemma earlier that when a dissecting diagonal leaves a shorter path of the boundary of the 2006-gon that consists of n -segments, then the number of iso-odd triangles with vertices on this shorter path does not exceed n/2.
We said that if there is an iso-odd triangle PQS with all vertices on this shorter path and with PQ the diagonal being the longest, then S is the summit of the triangle PQS and the vertices divide the shorter path into four segments of AP, PS, SQ and QB.
Since PQ is part of an iso-odd triangle and is the longest diagonal, an iso-odd triangle cannot have vertices on both the first and the last segment. This means that every iso-odd triangle that fall on the shorter path has all its vertices on one of the four segments.
Given each of these four segments and applying the lemma we started with, we get the sum of the triangles in all these four regions to be less than n/2. Furthermore the two inner segments will have odd number of sides which makes the inequality of the number of triangles to be strictly less than n/2 and leaves one out of two in excess for both these inner edges. Because of this excess, PQS is also covered by this estimate n/2 and the lemma holds.
We can argue the same when the iso-odd triangle with the longest diagonal is drawn the other way with its third vertex not falling on the shorter path. In this case, the triangle is acute or right triangle not the obtuse we had seen earlier. And we have three regions XZ, YZ and XY where XZ and YZ are shorter than XY. For each of the three regions, we apply the lemma and infer that there are no more than n/2 iso-odd triangles in all unless XYZ is one of them. Thus the total number of triangles in the dissection is not greater than n/2.
We can also reason it for arbitrary triangles. Let us say ABC is an iso-odd triangle with AB and BC odd sides. This implies there are odd number of sides between A and B and B and C. We call these sides as belonging to iso odd triangle ABC.
At least one side in each of these triangles do not belong to any other iso-odd triangle because an iso-odd triangle with two sides of equal length and therefore there are even number of sides belonging to it. Eliminating all sides that belong to iso-odd triangles, we get one that isn't. Let us assign these two sides one in each group to the triangle ABC. To each iso-odd triangle, we have thus assigned a pair of sides with no two triangles sharing an assigned side. Therefore there 1003 such iso-odd triangles that can appear.
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