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Today I have drawn a regular $18$-gon on a piece of paper. My drawing shows the $18$ vertices of the polygon labeled as $P_1,P_2,\ldots,P_{18}$ in clockwise order, and it also shows all $135$ diagonals.

Now looking at my drawing, it seems to me that the three diagonals $P_1P_{10}$ and $P_2P_{14}$ and $P_4P_{17}$ intersect each other in a common point. I have put a lot of effort into my drawing, and tried to get it as precise as possible, but of course its precision is limited.

Question: Is it indeed true that these three diagonals intersect in a common point?

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  • $\begingroup$ A geogebra drawing says yes. $\endgroup$ – Gabriel Romon Mar 25 '16 at 14:54
  • $\begingroup$ You can add $P_6P_{18}$ as well. $\endgroup$ – Gabriel Romon Mar 25 '16 at 15:10
  • $\begingroup$ @LeGrandDODOM $P_3P_{16}$ also. but that's logical, because that is when you get when you mirror the other two lines in $P_1P_{10}$ $\endgroup$ – Ivo Beckers Mar 25 '16 at 15:20
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To illustrate the puzzle I made the following image:

enter image description here

All angles I entered can be simply calculated using the fact any n-gon has a total of $180 + (n-3) \cdot 180 °$.

Now let's call the intersection of line $b$ and $a$ $X_1$ and the intersection of $b$ and $c$ $X_2$. To prove that $X_1 = X_2$ I'm going to show that $P_1X_1 = P_1X_2$

First, let's look at $X_1$. This one is easy. Because $\triangle P_1P_2X_1$ is isosceles we can immediately conclude that $P_1P_2 = P_1X_1$

$X_2$ is somewhat trickier. But basically it is https://math.stackexchange.com/questions/1713162/given-the-following-quadrilateral-prove-ab-bc but I'll copy the relevant parts: enter image description here

In this image there is a quadrilateral $ABCD$ which is congruent with $P_{18}P_1X_2P_{17}$. We're now going to prove that $AB = BC$ and therefore $P_1X_2 = P_1P_{18} = P_1P_2 = P_1X_1$

Rotate $B$ around $E$ by $60^\circ$, call it $C^\prime$. Then $\triangle C^\prime BC$ and $\triangle BED$ are congruent because $\angle C^\prime BC=\angle BED =20^\circ$, $\angle BC^\prime C=\angle EBD=10^\circ$, and $BE=C^\prime B$. Therefore $BC = DE = AB$

This means that $X_1 = X_2$ so they intersect at the same point

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  • 3
    $\begingroup$ I should note that this is basically what's happening at The World's Hardest Easy Geometry Problem, also known as the famous Langley's adventitious angles. Gerrit Bol classified all multiple intersections of diagonals in regular polyongs in 1936 (by hand!), and later in 1988 Poonen and Rubinstein confirmed it via computers. $\endgroup$ – Fimpellizieri Mar 25 '16 at 19:52

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