Triangles
Find the number of triangles formed by joining the vertices of a 50 - sided polygon:
-
With two sides common with that of polygon
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With only one side with that of polygon.
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No side common with that of polygon.
Enter your answer as the product of answers of the above asked questions.
Try my set : Let's play with polygons .
The answer is 1983750000.
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1 solution
Direct way to solve question 3: Formula of m-sided polygon in a n-sided polygon without common side is:
m n C m − 1 n − m − 1
Therefore, answer to question 3 is 3 5 0 C 2 4 6 = 1 7 ′ 2 5 0
Explanation of the formula:
You pick a first vertice: n ways to do that. But as it could be any of the m you have to place, you divide by m .
You're left with choosing m − 1 non adjacent points on n − 3 places, which is C m − 1 n − 3 + 1 − ( m − 1 ) = C m − 1 n − m − 1 because choosing p non adjacent points on t total points is like choosing ( p points + a right space) on t + 1 − p places, and there is C t t + 1 − p ways to do that.
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There is another way to prove the formula.
Start with a point. There is C m − 1 n − m − 1 polygons having this point (see previous comment for that). If you don't use it, you may take one adjacent point and there is still C m − 1 n − m − 1 polygons. If you don't use it either, you may take one point further: you then have C m − 1 n − 1 − m − 1 = C m − 1 n − m − 2 polygons. If you don't take it, picking the next one gives you C m − 1 n − 1 − m − 3 polygons, and so on.
Formula is C m − 1 n − m − 1 + i = 1 ∑ n − 2 m + 1 C m − 1 n − m − i = C m − 1 n − m − 1 + k = m − 1 ∑ n − m − 1 C m − 1 k .
It can be shown by recursion on n and m that C m − 1 n − m − 1 + k = m − 1 ∑ n − m − 1 C m − 1 k = m n C m − 1 n − m − 1 .
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This is a really interesting way to think about it, and great explanation! Thanks for sharing.
1. There are clearly 50 such triangles, with three vertices next to each other.
2. There are 50 ways to pick the common side. Then, there is 5 0 − 4 = 4 6 non adjacents points to complete the triangle. That is 50·46=2'300 triangles.
3. There is C 3 5 0 = 1 9 ′ 6 0 0 triangles without conditions. Minus triangles of case 1 and 2: 1 9 ′ 6 0 0 − 5 0 − 2 ′ 3 0 0 = 1 7 ′ 2 5 0
Answer is 5 0 ⋅ 2 ′ 3 0 0 ⋅ 1 7 ′ 2 5 0 = 1 ′ 9 8 3 ′ 7 5 0 ′ 0 0 0 .