Let \( [A_1A_2A_3 \ldots A_{20}] \) be a polygon of 20 sides, in the plane. We represent by \( \measuredangle A_i \) the measure of the internal angle of the polygon in the vertex \( A_i \) for \( i = 1, 2, 3, \ldots , 20 \). Supose that the polygon has the following properties:

Every side has the same length;

\( \measuredangle A_1 = \measuredangle A_3 = \measuredangle A_5 = \cdots = \measuredangle A_{19} = 216^\circ \);

\( \measuredangle A_2 = \measuredangle A_4 = \measuredangle A_6 = \cdots = \measuredangle A_{20} = 108^\circ \)

Show that the lines \( A_1A_{11} \), \(A_2A_{14}\), \( A_3A_{17}\), \(A_5A_{19} \) and \( A_8A_{20} \) have a point in common.

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## Comments

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TopNewestWhat have you tried?

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@Calvin Lin I tried dividing it into two pentagons, but then I wasn't able to do much so I thought about setting up an axis and work with point coordinates and vectors, but it was also a dead-end. Maybe there's some property or something I'm missing. What do you suggest?

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That's a good idea. Are you able to show some set of 3 lines are concurrent?

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@Calvin Lin I found something about Ceva's Theorem online but I didn't know it and don't know how to use it. Is this what you're referring to?

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Nope.

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@Calvin Lin Well, if I divided it into triangles and the lines were angle bissectors, they would intersect at one point but they're not.

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Does it seem likely that there are triangles with those lines as angle bisectors?

If yes, what triangles could we try?

If no, what else can you try?

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@Calvin Lin No, it was a bad idea. I was thinking of setting up a referential. Since I know the angles I can determine the slope and by using Cosine Law a lot I could determine a few lengths and then coordinates. But it seems like too much calculations will be needed and there's no telling if it will actually work. Also, this is a problem similar to Olympiad ones so I don't think I should approach it this way.

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Go back to your idea of dividing it into two (decagons).

Assuming you chose a good set, which 3 lines lie on the same decagon? Why are those lines concurrent?

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@Calvin Lin I can't divide into two pentagons such that 3 lines lie on the same on. Only two do.

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Note that I didn't say pentagons. I said "two (decagons)", where the parenthesis indicates I'm interpreting your statement. There are 10 vertices on a decagon, so we could have 3 lines on them.

Note that we can't split the 20 vertices up into only two pentagons, which is why I thought you meant two decagons.

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@Calvin Lin I'm sorry I'm constantly disturbing you. It's just that I'm given these problems which are just to hard for me, (because I'm in a project in my country in which I get to go to a university and have classes with other high school students like me about Olympiad Math and harder problems, like this one. And I end up not being able to do most of them.) As for the problem, I am not able to split it into two decagons such that three of the lines given are inside one of them. Also, even if I could, I don't know how to see why lines are concurrent unless they're angle bissectors ( which they're not here) or meet at the center of a regular polygon.

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Generally, with such problems, you have to figure out how to solve them, and what tools you have at your disposal. Sometimes, the answer is that you're not equipped yet to deal with them. Given that you've only done a few problems, I do not have enough information about your ability level to draw any conclusion. I advise you to work through the community featured feed, especially at the medium and hard difficulties.

Consider the odd-indexed vertices. Why must those 3 lines be concurrent?

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@Calvin Lin I think that because the line \( [A_1A_{11}] \) divides it into w´two decagons, for some reason, I think \( A_5 \) and \( A_{17} \) are equidistant from that line and so are the other two vertices.

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