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moneybird
Polygon inscribed in circle!
A regular polygon, with 49 sides, is inscribed in a circle. How many triangles with vertices that are vertices of the polygon have the centre of the circle in their interior?
I have not gotten the solution yet
For an 'n' sided polygon, if we label each vertex of the polygon from 1 to n and consider the triangle from the center of the circle to the first two vertices, we get this diagram: |dw:1324760887506:dw| Now consider some vertex labelled 'i'. We need to pick the position of this vertex such that it lies between the projections of vertex 1 and 2 through the center of the triangle - this will ensure that a triangle drawn between vertices 1, 2 and 'i' will contain the center of the circle. This leads us to this inequality:\[ \begin{align} \pi+\frac{2\pi}{n}&>&\frac{2\pi}{n}*(i-1)&>\pi\\ \therefore n+2&>&2(i-1)&>n\\ \therefore \frac{n+2}{2}&>&i-1&>\frac{n}{2}\\ \therefore \frac{n+4}{2}&>&i&>\frac{n+2}{2} \end{align}\]so, for a polygon with 49 sides we get:\[26.5>i>25.5\]and since 'i' has to be an integer, there is only only one solution which is \(i=26\). So for the edge between vertices 1 and 2, we can construct just one triangle to vertex 26 which contains the center of the circle. Since this polygon has 49 edges, we can therefore form 49 such triangles. So I think the answer is 49.
i think i have got the solution to this, it comes out to be a series, for a polygon with n sides, if n is odd then ans is n-1/2 for even n ans is n-4/2 i got that by anlysing the cases of hexagon, septagon and octagon, and seeing the pattern