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 [6] in the above paper, and references cited there). I quickly glanced at the literature, and I didn't find what I'm looking for.

triangular grid lemma


1 Answer 1


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.

  • $\begingroup$ What does $(A,B;C,D)$ mean? $\endgroup$ Jun 3, 2015 at 17:29
  • $\begingroup$ @GabrielNivasch The cross ratio of the four points, I would guess. $\endgroup$
    – Igor Rivin
    Jun 3, 2015 at 17:31
  • $\begingroup$ @GabrielNivasch en.wikipedia.org/wiki/Cross-ratio $\endgroup$
    – zeb
    Jun 3, 2015 at 18:50
  • $\begingroup$ 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)? $\endgroup$ Jun 3, 2015 at 19:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.