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#1 h Jun 8, 2005, 3:09 PM • 1 Y

Y by Adventure10

A convex polygon is given, no two of whose sides are parallel. For each side we consider the angle the side subtends at the vertex farthest from the side. Show that the sum of these angles equals 180 degrees.

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Remike

#2 h Jun 9, 2005, 5:19 AM • 1 Y

Y by Adventure10

I suppose that at least the number of vertices must be odd. Otherwise what do you mean by "side subtends at the vertex"?

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#3 h Jun 9, 2005, 12:27 PM • 2 Y

Y by Adventure10, Mango247

Well, for each side there must be a vertex farthest from that side, then you draw the angle using that side and that vertex, so as far as I understand, the number of sides can be even...

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Remike

#4 h Jun 9, 2005, 2:42 PM • 2 Y

Y by Adventure10, Mango247

warut_suk wrote:

Well, for each side there must be a vertex farthest from that side, then you draw the angle using that side and that vertex, so as far as I understand, the number of sides can be even...

But I don't understand how you can draw an angle using a side and a vertex. You can only draw an angle using two sides!

And one more question. There is only the following condition in the statement: "no two sides are parallel". But this is not sufficient to the fact that for each side there is only one vertex farthest from that side. For example consider regular pentagon. Perhaps no two sides and diagonals may not be parallel.

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#5 h Jun 10, 2005, 5:40 AM • 2 Y

Y by Adventure10, Mango247

Why? You join each endpoint of the side to the vertex. That's the angle you want.
And say the regular pentagon is ABCDE then D is farthest from AB, E is to BC, etc.

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#6 h Aug 22, 2005, 1:29 PM • 2 Y

Y by Adventure10, Adventure10

I'll just sketch some steps, since the details might become a bit dry and boring to deal with.

First of all, let $\mathcal P=A_1A_2,\ldots A_n$ be the polygon. The indices are modulo $n$ , i.e. $A_{n+1}=A_1$ . For each side draw two segments, connecting the two endpoints of the side to the vertex which is farthest away from the side. We will call these simply segments.

Try to show that every two segments must intersect.

Show that if $A_i$ is the vertex farthest away from $A_uA_{u-1}$ , while $A_j$ is the vertex farthest away from $A_uA_{u+1}$ , then all the sides on the path $A_iA_j$ which does not contain $A_u$ have $A_u$ as the vertex which is farthest away.

Use these results to conclude that if $T_1,T_2,\ldots,T_k$ are the vertices (listed in a counter-clockwise direction; again, $T_{k+1}=T_1$ ) of $\mathcal P$ which are endpoints of at least two segments each, then $k$ is odd, and if for each $T_i$ we draw the two segments which form the largest angle with vertex $T_i$ , then these two segments have $T_{i+\frac{k-1}2},T_{i+\frac{k+1}2}$ as second endpoints, and the sum of angles that we are looking for is $\sum_{i=1}^k\angle T_{i+\frac{k-1}2}T_iT_{i+\frac{k+1}2}$ . The conclusion is easy to draw from here.

I'm sure a picture will make everything clear.

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