Concurrent lines proof for a regular 18-gon

Question

Let $X_1 X_2 \dotsb X_{18}$ be a regular 18-gon. Show that $X_1 X_{10}$, $X_2 X_{13}$, and $X_3 X_{15}$ are concurrent.

What would be the best way to prove this? I am actually struggling understanding 'concurrent' as I am not sure how they are concurrent in a regular 18-gon

Answer 1

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Let $A,B,C$ be the intersection point of $X_{13}X_1$ with $X_{15}X_3$, $X_{13}X_3$ with $X_{10}X_1$, $X_{13}X_2$ with $X_1X_3$ respectively.

Then, we get $$\begin{align}\frac{AX_1}{X_{13}A}\cdot\frac{CX_3}{X_1C}\cdot\frac{BX_{13}}{X_3B}&=\frac{X_1X_3\sin 40^\circ}{X_{13}X_3\sin 20^\circ}\cdot\frac{X_3X_{13}\sin 10^\circ}{X_1X_{13}\sin 10^\circ}\cdot\frac{X_1X_{13}\sin 30^\circ}{X_1X_3\sin 70^\circ}\\\\&=\frac{\sin 40^\circ\sin 30^\circ}{\sin 20^\circ\sin 70^\circ}\\\\&=\frac{\frac 12\sin 40^\circ}{\sin 20^\circ\sin 70^\circ}\\\\&=\frac{\sin 20^\circ\cos 20^\circ}{\sin 20^\circ \sin 70^\circ}\\\\&=1\end{align}$$ The claim follows from Ceva's theorem.

Answer 2

That's how they can be concurrent. I'm currently working on a proof :)

Answer 3

Lemma 1. If $p$ and $q$ are distinct points on the unit circle such that $pq \ne -1$ then the line through $p$ and $q$ intersects the real axis at $z = \frac{p+q}{1+pq}$.

For arbitrary $p,q$ (not necessarily on the unit circle) let $z = \lambda p + (1-\lambda)q \,\text{,}\,\lambda \in \mathbb{R}$ be a point on the line through $p,q$. Point $z$ lies on the real axis iff $z = \bar z\,$, which is an equation that can be solved for $\lambda\,$, then substituting back gives in the end $z=\frac{p \bar q - \bar p q}{p - \bar p - (q - \bar q)}\,$.

For $p,q$ on the unit circle $\bar p = \frac{1}{p}$ and $\bar q = \frac{1}{q}\,$, and the previous expression simplifies to $z=\frac{p+q}{1+pq}$.

Lemma 2. Let $\omega = \operatorname{cis}(2 \pi / 18)= e^{i\,2 \pi / 18}\,$, then $\;\omega^9 = -1\,$ and $\;\omega^6 - \omega^3+1=0\;$.

The first part follows directly from $\omega^9 = \operatorname{cis}(9 \cdot 2 \pi / 18) = \operatorname{cis}(\pi)=-1\,$.
Rewriting it as $0 = \omega^9+1 = (\omega^3+1)(\omega^6-\omega^3+1)\,$ gives the second part, since $\omega^3 \ne -1\,$.

Proof. Assume WLOG that vertices $X_k$ lie on the unit circle in the complex plane, with $X_1\equiv 1$. Then $X_k \equiv \omega^{k-1}$ for $k=1,2,\cdots,18\,$. Since $X_{10} \equiv \omega^9 = -1$ side $X_1X_{10}$ lies on the real axis, and the problem reduces to proving that the intersections of $X_2X_{13}$ and $X_3X_{15}$ with the real axis coincide. Noting that by Lemma 2 $X_{13}\equiv \omega^{12}=-\omega^3\,$, $X_{15}\equiv \omega^{14}=-\omega^5\,$ and using the formula from Lemma 1 this is equivalent to:

$$ \require{cancel} \frac{\omega-\omega^3}{1-\omega^4} = \frac{\omega^2-\omega^5}{1-\omega^7} \\[5px] \frac{\cancel{\omega}\bcancel{(1-\omega^2)}}{\bcancel{(1-\omega^2)}(1+\omega^2)} = \frac{\omega^\cancel{2}(1-\omega^3)}{1-\omega^7} \\[5px] 1 - \omega^7 = \omega(1+\omega^2-\omega^3-\omega^5) \\[5px] 1-\omega^3+\omega^6-\omega(1-\omega^3+\omega^6) = 0 $$

The latter equality is satisfied because $1-\omega^3+\omega^6=0$ by Lemma 2, which completes the proof.

Answer 4

Let $X_2X_{13}$, $X_5X_{14}$, $X_8X_{15}$ meet at $A$ (by symmetry about $X_5X_{14}$). Then $\triangle X_2X_3X_{17}$ and $\triangle AX_{15}X_5$ have parallel corresponding sides, so they are homothetic about some point $B$, which lies on $X_2X_{13}$, $X_3X_{15}$, and also $X_1X_{10}$ (by symmetry about $X_1X_{10}$).