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.
An isosceles triangle is called odd if it has two odd sides. A triangle in the dissection above which is odd and isosceles is called iso-odd.
We make an observation as follows:
Let AB be one of the dissecting diagonals and let Lambda be one of the shorter part of the boundary of the 2006-gon with endpoints A, B. Suppose that lambda consists of n segments. Then the number of iso-odd triangles with vertices on Lambda does not exceed n/2.
We can prove this by induction . For the initial case of n = 2, there can only be one such triangle. So this holds. Let us assume this observation holds for all lambda of length < n. Then we consider the case when Lambda == n. Let PQ be the longest diagonal which is a side of an iso-odd triangle PQS with all vertices on Lamda. If there is no such triangle, then we are not increasing the number of iso-odd triangles and hence the observation holds. If on the other hand, there is such a triangle, then every triangle whose vertices lie on lambda is right angled or obtuse with the maximum possible when it form the triangle PQS with the summit at S. Then we may assume that the five points APSQB lie on lambda and partition lambda into four pieces of AP, PS, SQ and QB each where possibly the outer two segments reduce to a point. Since PQ is the longest diagonal and PQS is an iso-odd triangles, it cannot have vertices on both AP and QB the outer segments. Therefore every iso -odd triangle lying within lambda has its vertices on just one of four pieces.Each of these pieces we can apply the same induction and we can add the four inequalities,we get the required number to not exceed n/2.
The intuition behind this approach is that there are only three vertices we need to pick to for such triangles and given that they can lie arbitrarily along the segments we categorize them into forms that the triangle can take and where those vertices may fall. Once we have a lemma for the number of such triangles, we can easily study it in each of the categories
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.
An isosceles triangle is called odd if it has two odd sides. A triangle in the dissection above which is odd and isosceles is called iso-odd.
We make an observation as follows:
Let AB be one of the dissecting diagonals and let Lambda be one of the shorter part of the boundary of the 2006-gon with endpoints A, B. Suppose that lambda consists of n segments. Then the number of iso-odd triangles with vertices on Lambda does not exceed n/2.
We can prove this by induction . For the initial case of n = 2, there can only be one such triangle. So this holds. Let us assume this observation holds for all lambda of length < n. Then we consider the case when Lambda == n. Let PQ be the longest diagonal which is a side of an iso-odd triangle PQS with all vertices on Lamda. If there is no such triangle, then we are not increasing the number of iso-odd triangles and hence the observation holds. If on the other hand, there is such a triangle, then every triangle whose vertices lie on lambda is right angled or obtuse with the maximum possible when it form the triangle PQS with the summit at S. Then we may assume that the five points APSQB lie on lambda and partition lambda into four pieces of AP, PS, SQ and QB each where possibly the outer two segments reduce to a point. Since PQ is the longest diagonal and PQS is an iso-odd triangles, it cannot have vertices on both AP and QB the outer segments. Therefore every iso -odd triangle lying within lambda has its vertices on just one of four pieces.Each of these pieces we can apply the same induction and we can add the four inequalities,we get the required number to not exceed n/2.
The intuition behind this approach is that there are only three vertices we need to pick to for such triangles and given that they can lie arbitrarily along the segments we categorize them into forms that the triangle can take and where those vertices may fall. Once we have a lemma for the number of such triangles, we can easily study it in each of the categories
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