# Elementary proof of a triangular grid lemma

I am looking for an elementary proof of the following lemma, which concerns what Green and Tao call "triangular grids" (see arXiv:1208.4714).

Let $a_1$, $a_2$, $a_3$, $a_4$, $b_1$, $b_2$ be six arbitrary lines in the plane (in general position, for whatever the appropriate sense of "general position" is).

Let $c_1$ be the line through the intersection points $a_3b_1$ and $a_2b_2$.

Let $c_2$ be the line through $a_4b_1$ and $a_3b_2$.

Let $b_3$ be the line through $a_1c_1$ and $a_2c_2$.

Let $c_3$ be the line through $a_4b_2$ and $a_3b_3$.

Let $b_4$ be the line through $a_1c_2$ and $a_2c_3$.

Let $c_4$ be the line through $a_4b_3$ and $a_3b_4$.

Let $b_5$ be the line through $a_1c_3$ and $a_2c_4$.

Let $c_5$ be the line through $a_4b_4$ and $a_3b_5$.

Then the intersection points $b_1c_3$, $b_2c_4$, $b_3c_5$ are collinear.

As a consequence, this triangular grid can be indefinitely continued.

See figure below.

This can be proven using cubic curves, specifically the Cayley-Bacharach theorem: In the dual setting, all the points dual to the given lines lie on a common cubic curve. See the above-mentioned paper of Green and Tao.

But as I said, I was wondering whether an elementary proof exists.

I see there is previous literature on these triangular grids (also called with other names; see reference  in the above paper, and references cited there). I quickly glanced at the literature, and I didn't find what I'm looking for. We can prove this with a cross ratio chase, but first we need an easy lemma.

Lemma: If points $$A,B,C,D$$ are on one line and $$E,F,G,H$$ are on another line, then $$(A,B;C,D) = (E,F;G,H)$$ if and only if the three points $$X = AF\cap BE, Y = BG\cap CF, Z = CH\cap DG$$ lie on a line.

Proof: Let $$P = AG\cap CE, Q = CG\cap XY$$. By Pappus's Theorem, $$P$$ is on line $$XY$$. Projecting through $$G$$, we have $$(A,B;C,D) = (P,Y;Q,DG\cap XY)$$, and projecting through $$C$$, we have $$(E,F;G,H) = (P,Y;Q,CH\cap XY)$$. Thus $$(A,B;C,D) = (E,F;G,H)$$ if and only if $$CH, DG$$, and $$XY$$ meet at a point.

Now we need to name some points. Let $$A = a_2\cap b_1, B = a_3\cap b_1\cap c_1, C = a_4\cap b_1\cap c_2, D = b_1\cap c_3, E = b_1\cap c_4$$, let $$F = a_1\cap b_3\cap c_1, G = a_2\cap b_3\cap c_2, H = a_3\cap b_3\cap c_3, I = a_4\cap b_3\cap c_4, J = b_3\cap c_5$$, and let $$K = b_5\cap c_2, L = a_1\cap b_5\cap c_3, M = a_2\cap b_5\cap c_4, N = a_3\cap b_5\cap c_5, O = a_4\cap b_5$$. Finally, let $$X = AN \cap DK$$.

By two visually obvious applications of the Lemma we have $$(A,B;C,D) = (F,G;H,I) = (K,L;M,N)$$. Thus $$(B,D;A,C) = (L,N;K,M)$$, so by a slightly less obvious application of the Lemma $$X$$ must be on the line $$GH = b_3$$. Applying the Lemma again, since $$X$$ is on the line $$GI = b_3$$ we have $$(E,C;A,D) = (O,M;K,N)$$.

Putting $$(A,B;C,D) = (K,L;M,N)$$ and $$(E,C;A,D) = (O,M;K,N)$$ together, we see that $$(B,C;D,E) = (L,M;N,O)$$ (consider the projective transformation from line $$b_1$$ to line $$b_5$$ taking $$A,C,D$$ to $$K,M,N$$: the first equality says it takes $$B$$ to $$L$$, and the second says it takes $$E$$ to $$O$$). Applying the Lemma in another visually obvious way, we have $$(L,M;N,O) = (G,H;I,J)$$. Thus $$(B,C;D,E) = (G,H;I,J)$$, so by the Lemma the intersection $$DJ\cap EI$$ is on the line connecting $$BH\cap CG, CI\cap DH$$, which is $$b_2$$. Since $$D = b_1\cap c_3$$, $$J = b_3\cap c_5$$, $$EI = c_4$$, this means that $$b_1\cap c_3, b_2\cap c_4, b_3\cap c_5$$ are collinear.

• What does $(A,B;C,D)$ mean? Jun 3, 2015 at 17:29
• @GabrielNivasch The cross ratio of the four points, I would guess. Jun 3, 2015 at 17:31
• @GabrielNivasch en.wikipedia.org/wiki/Cross-ratio
– zeb
Jun 3, 2015 at 18:50
• Perhaps you can also help me with this "power of a point"-like lemma for a parabola: mathoverflow.net/q/203076/27742 Is there a nice geometric proof (using Dandelin spheres perhaps, or I don't know what)? Jun 3, 2015 at 19:32