1. Distinct Distances in the Plane
Assaf Lev-Ran
Date: 12/1/15

2. Reminder
Let 𝑃 be a set of 𝑛 points. We want to bound the size of the set: 𝑑 𝑝 1 , 𝑝 𝑝 1 , 𝑝 2 ∈𝑃}
Our aim is to prove that it's Ω(𝑛/log⁡𝑛).
This bound is almost tight, since the upper bound is 𝑂(𝑛/ log 𝑛 ), Using the 𝑛 × 𝑛 grid.

3. The Plan
First we'll show a reduction to the problem of incidences between points and lines in 3D (Elekes – Sharir framework)
After that we'll show a bound for this problem by Guth and Katz.
Let's begin!

4. The Reduction
Let 𝑃 be a set of 𝑠 points. We'll denote the number of distinct distances by 𝑥.
We'll denote the distances in 𝛿 1 ,…, 𝛿 𝑥 .
We'll also denote 𝐸 𝑖 =|{(𝑝,𝑞)|𝑑(𝑝,𝑞)= 𝛿 𝑖 }|

5. Let us define 𝑄={ 𝑎,𝑏,𝑐,𝑑 ∈ 𝑃 4 | (𝑎,𝑏)≠(𝑐,𝑑) and |𝑎𝑏|=|𝑐𝑑|>0}.
Now since 𝑖 𝐸 𝑖 =2 𝑠 2 = 𝑠 2 −𝑠 we get by Cauchy Schwarz that: 𝑄 = 𝑖=1 𝑥 2 𝐸 𝑖 2 ≥ 𝑖=1 𝑥 𝐸 𝑖 −1 2 ≥
1 𝑥 𝑖=1 𝑥 ( 𝐸 𝑖 −1) 2 = 𝑠 2 −𝑠−𝑥 2 𝑥
So if we'll prove that |𝑄|=𝑂( 𝑠 3 log 𝑠 ), we'll get 𝑥 =Ω(𝑠/log⁡𝑠). 𝑐 𝑑

6. A proper rigid motion is a transformation of the plane that preserves distances and orientation.
Every proper rigid motion is a pure translation or a pure rotation (around some point).

7. For every (𝑎,𝑏,𝑐,𝑑) such that |𝑎𝑏|=|𝑐𝑑|>0, there exists a unique rotation 𝜏 such that τ 𝑎 =𝑐 and 𝜏 𝑏 =𝑑.
Therefore, every element in 𝑄 corresponds to a unique rotation 𝜏.

8. Since every pair (𝑎,𝑐) has a unique translation 𝜏 such that 𝜏(𝑎)=𝑐, the first three elements in every quadruple (𝑎,𝑏,𝑐,𝑑) determine the fourth.
Hence, there are at most 𝑠 3 quarduples that correspond to a translation.
What about rotations? c a d b

9. If τ is a rotation, its multiplicity is |𝜏 𝑃 ∩ 𝑃| (the number of points of 𝑃 that 𝜏 takes to another point in 𝑃)
We'll define 𝑆 𝑘 ( 𝑆 ≥𝑘 ) the number of rotations of multiplicity 𝑘 (≥𝑘)

10. If 𝜏 has multiplicity 𝑘 then it contributes 𝑘 2 quarduples to 𝑄.
So we have: 𝑄 = 𝑘=2 𝑘 2 𝑆 𝑘 = 𝑘=2 𝑘 2 ( 𝑆 ≥𝑘 − 𝑆 ≥𝑘+1 ) = 𝑆 ≥2 + 𝑘≥3 𝑘−1 𝑆 ≥𝑘
If we'll show 𝑆 ≥𝑘 = 𝑂( 𝑠 3 / 𝑘 2 ) we'll get 𝑄 =𝑂 𝑠 𝑘≥3 1 𝑘 =𝑂( 𝑠 3 log 𝑠 ) as we wanted.

11. A rotation can be determined by three elements: its center 𝑐=(𝑥,𝑦) and its angle 𝜃.

12. We’ll map every rotation 𝜏 to a point in ℝ 3 : (𝑥,𝑦, cot 𝜃 2 )
Now, let 𝑎,𝑏 be points in ℝ 2 . The locus of all rotations that map 𝑎 to 𝑏 is: 𝑎 𝑥 + 𝑏 𝑥 2 , 𝑎 𝑦 + 𝑏 𝑦 2 ,0 +𝑡 𝑏 𝑦 − 𝑎 𝑦 2 , 𝑎 𝑥 − 𝑏 𝑥 2 ,1 for 𝑡∈ℝ It’s a line! a
𝑏−𝑎 cot 𝜃 2
𝑎+𝑏 2
𝜃 2
𝑏−𝑎 2 b

13. So, to show that 𝑆 ≥𝑘 is 𝑂( 𝑠 3 / 𝑘 2 ) we need to bound the number of points in ℝ 3 that coincide with at least 𝑘 lines.
We have an incidence problem, and we'll solve it with algebraic methods.
Onwards we go!

14. Incidences in ℝ 3
Theorem (Guth & Katz, 2010): Let 𝑃 be a set of 𝑚 points and 𝐿 a set of 𝑛 lines in ℝ 3 , such that (i) each point of 𝑃 is incident in at least three lines in 𝐿, and (ii) no plane contains more than 𝑞 lines of 𝐿. then: 𝐼 𝑃,𝐿 =𝑂( 𝑚 1/2 𝑛 3/4 + 𝑚 2/3 𝑛 1/3 𝑞 1/3 +𝑚+𝑛) this bound is tight in the worst case.

15. How does it help us?
We can deduce that if 𝐿 is a set of 𝑛 lines in ℝ 3 , then the number 𝑚 ≥𝑘 (i.e. the number of points that coincide with at least 𝑘 lines) for 𝑘≥3 satisfy: 𝑚 ≥𝑘 =𝑂( 𝑛 3/2 𝑘 2 + 𝑛𝑞 𝑘 3 + 𝑛 𝑘 )

16. In our problem, every plane contains at most O( 𝑛 ) lines, because every two lines must intersect and therefore be the locus of four distinct points.
A set of pairwise disjoints pairs contains 𝑂(𝑠) elements. Hence, there are at most 𝑂 𝑠 =𝑂( 𝑛 ) lines on one plane and the assumption is applied.
Since 𝑛=𝑂( 𝑠 2 ) we get that the number of points that incident in at least 𝑘 lines is 𝑂( 𝑠 3 / 𝑘 2 ) as we want.
𝑙 𝑎,𝑏
𝑙 𝑎,𝑐
𝑎→𝑏,𝑐

17. What about 𝑘=2?
Using methods from the 19th century, it can be shown that 𝑚 ≥2 is 𝑂 𝑛 3/2 =𝑂( 𝑠 3 ).
We won’t be showing this today...

18. Proof of the main theorem
We want to use the polynomial partition technique. To do that, we need to define a parameter 𝑟.
So let 𝑟 be: 𝑟= 𝑐 𝑚 3/2 𝑛 3/4 for 𝑎 ′ 𝑛 1/2 ≤𝑚≤𝑎 𝑛 3/2 𝑐 𝑛 3/2 for 𝑚>𝑎 𝑛 3/ for suitable constants 𝑐, 𝑎 and 𝑎′ whose values will be determined later. We’ll also demand 𝑟≤𝑚.

19. In a previous lecture, it was mentioned that there exists a trivariate polynomial 𝑓 of degree 𝐷 =𝑂( 𝑟 1/3 ) such that every connected component in ℝ 3 \ 𝑍(𝑓) contains at most 𝑂(𝑚/𝑟) points of 𝑃 (where 𝑍(𝑓) is the zero set of 𝑓).
Moreover, the number of components (we'll call them “cells”) is 𝑂( 𝐷 3 )=𝑂(𝑟).

20. Let 𝑃 0 be the number of points from 𝑃 in 𝑍(𝑓).
Let 𝐿 0 be the number of lines from 𝐿 that are fully contained in 𝑍(𝑓).
Let 𝑃′=𝑃\ 𝑃 0 and 𝐿′=𝐿\ 𝐿 0 .
Then: 𝐼(𝑃,𝐿) = 𝐼( 𝑃 0 , 𝐿 0 ) + 𝐼( 𝑃 0 ,𝐿′) + 𝐼(𝑃′,𝐿′).
Let's estimate each term of this sum separately.

21. Estimating 𝐼( 𝑃 0 ,𝐿′)
Every line in 𝐿′ contains at most 𝐷 points from 𝑃 0 (otherwise it was in 𝑍(𝑓) because the polynomial is of degree 𝐷, and therefore intersects 𝑍(𝑓) in at most 𝐷 points).
Hence, 𝐼 𝑃 0 , 𝐿 ′ ≤|𝐿′|𝐷=𝑂(𝑛 𝑟 1/3 ).

22. Estimating 𝐼(𝑃′,𝐿′)
Let us denote the cells of ℝ 3 \ 𝑍(𝑓) in 𝐾 1 ,…, 𝐾 𝑡 where 𝑡=𝑂(𝑟).
Let 𝑃 𝑖 be the number of points from 𝑃 in 𝐾 𝑖 , and 𝐿 𝑖 be the number of lines from 𝐿 that cross 𝐾 𝑖 . Also, let 𝑛 𝑖 =| 𝐿 𝑖 | and 𝑚 𝑖 =| 𝑃 𝑖 |=𝑂(𝑚/𝑟).
Every line crosses at most 𝐷+1 cells (why?) so 𝑖 𝑛 𝑖 ≤𝑛(1+𝐷)=𝑂(𝑛 𝑟 1/3 ).
Clearly, 𝐼(𝑃′,𝐿′)= 𝑖 𝐼( 𝑃 𝑖 . 𝐿 𝑖 )
Let's look at the cases of 𝑟.

23. Case (i): 𝑚≤𝑎 𝑛 3/2
By projecting the points to a plane and using the bound of incidences on plane, we have 𝐼( 𝑃 𝑖 , 𝐿 𝑖 )=𝑂( 𝑚 𝑖 2 + 𝑛 𝑖 ).
Therefore, 𝐼 𝑃 ′ , 𝐿 ′ =𝑂 𝑖=1 𝑡 𝑚 𝑖 2 + 𝑛 𝑖 =𝑂 𝑟 𝑚 𝑟 2 +𝑛 𝑟 1/3 =𝑂( 𝑚 2 𝑟 +𝑛 𝑟 1/3 )
Since in this case, 𝑟= 𝑐 𝑚 3/2 𝑛 3/4 , we get (splitting to cases 𝑚> 𝑐 ′ 𝑛 1/2 and 𝑚< 𝑐 ′ 𝑛 1/2 ): 𝐼(𝑃′,𝐿′)=𝑂( 𝑚 1/2 𝑛 3/4 +𝑛)

24. Case (ii): 𝑚>𝑎 𝑛 3/2
There are 𝑂(𝑟) cells, and every line intersects 𝑂( 𝑟 1/3 ) cells.
Therefore, there are 𝑛 𝑟 1/3 intersections between the lines and cells, so every cell contains 𝑛 𝑟 2/3 lines in the average case.
If a cell contains 𝑘𝑛 𝑟 2/3 lines, we'll split it to 𝑘 cells with 𝑛 𝑟 2/3 lines each. Every cell has in the worst case 𝑚/𝑟 points, and there are still 𝑂(𝑟) cells.

25. 𝑛=𝑂( 𝑚 2/3 ), and, similarly to what we saw before, 𝐼( 𝑃 𝑖 , 𝐿 𝑖 )=𝑂( 𝑚 𝑖 + 𝑛 𝑖 2 )
We still have 𝑂(𝑟) cells, and each contains 𝑚/𝑟 points and 𝑛 𝑟 2/3 lines. So we have (using 𝑟=𝑐 𝑛 3/2 ): 𝐼(𝑃′,𝐿′)=𝑂( 𝑚 𝑟 𝑛 𝑟 2/ )𝑂(𝑟)=𝑂(𝑚+ 𝑛 2 𝑟 1/3 ) = 𝑂(𝑚+ 𝑛 3/2 ) = 𝑂(𝑚).

26. Let’s Take a Break.

27. Estimating 𝐼( 𝑃 0 , 𝐿 0 )
Reminder: we want to bound the number of points on 𝑍(𝑓) and the number of lines that are fully contained in 𝑍(𝑓).
First, we assume 𝑓 has no linear factor. If it does, we'll divide it by linear factors and handle them later.
To simplify the notation, we'll call the new polynomial 𝑓.

28. Now we'll apply a pruning process on the lines and points from 𝑃 0 and 𝐿 0 .
At each step we'll remove a point of 𝑃 0 that is incident to at most two (surviving) lines of 𝐿 0 .
If such a point doesn't exist, we'll remove a line of 𝐿 0 that is incident to fewer than 4𝐷 (surviving) points.

29. The removal of lines caused a loss of at most 4| 𝐿 0 |𝐷 incidences.
The removal of points caused a loss of at most 2 incidences for each step, which will be charged to an incidence either with a line of 𝐿′ or with a previously removed line of 𝐿 0 . Overall this accumulates to at most 8 𝐿 0 𝐷+2 𝐿 ′ 𝐷≤8𝑛𝐷.

30. In summary, we removed at most 12𝑛𝐷 incidences.
Now, every point in 𝑃 0 is incident to at least 3 lines in 𝐿 0 , and every line in 𝐿 0 is incident to at least 4𝐷 points in 𝑃 0 .
To simplify the notation, we'll call the remaining points and lines 𝑃 0 and 𝐿 0 , respectively.

31. Now, we classify 𝑃 0 to two types:
If 𝑝 is incident to three non-coplanar lines of 𝐿 0 , then 𝑝 is a critical point of 𝑍(𝑓).
If 𝑝 is not critical, and all the lines of 𝐿 0 that 𝑝 is incident to are in the same plane (this plane is tangent to 𝑍(𝑓) at 𝑝), then 𝑝 is a flat point of 𝑍(𝑓).
There exist three polynomials 𝜋(𝑓) of degree at most 3𝐷−4, which vanish at every flat point of 𝑍(𝑓). (we saw a sketch of the proof two weeks ago)

32. Each line in 𝐿 0 is incident either to 𝐷 critical points or 3𝐷 flat points.
Each line of the first kind is a critical line because each of its partial derivative is a 𝐷−1 degree polynomial that vanishes in 𝐷 points and therefore vanishes on the whole line.
Each line of the second kind is a flat line because every 𝜋(𝑓) is a 3𝐷−4 degree polynomial that vanishes at 3𝐷 points.

33. We’ll use a lemma that we saw two weeks ago:
Let 𝑓 and 𝑔 be trivariate polynomials of degrees 𝐷 𝑓 and 𝐷 𝑔 , respectively. If 𝑓 and 𝑔 do not have common factor then the number lines contained in the common zero set 𝑍 𝑓 ∩𝑍(𝑔) is at most 𝐷 𝑓 𝐷 𝑔 .

34. As a result, 𝑍(𝑓) can contain at most 𝐷(𝐷−1) critical lines and 𝐷(3𝐷−4) flat lines.
Thus, the number of (surviving) lines in 𝐿 0 is at most 𝐷 𝐷−1 +𝐷 3𝐷−4 < 𝑛 0 :=4 𝐷 2 = 𝑂( 𝑟 2/3 ).
WLOG assume | 𝐿 0 |= 𝑛 0 . We may also assume that 𝑛 0 ≤𝛽𝑛, for sufficiently small 𝛽.

35. We claim that the new set of points and lines satisfies the assumption of the theorem.
Indeed, each point in 𝑃 0 is incident to at least 3 lines, and no plane contains more than 𝐷 lines.
This happens because 𝑓, restricted to a plane ℎ, becomes a bivariate polynomial of degree at most 𝐷. But 𝑓 has no linear factors, and therefore doesn't vanish on ℎ and can vanish for at most 𝐷−1 lines of ℎ.
We’ll denote the maximum possible value of 𝐼(𝑃,𝐿) with parameter 𝑞 by 𝐼(𝑃,𝐿,𝑞).
We got at most 𝐼(𝑚,𝛽𝑛, 𝛽𝑛 1/2 )+𝑂(𝑛𝐷) incidences.

36. Linear Factors
If (the original) 𝑓 had linear factors, it might vanish on at most 𝐷 planes. We'll denote them in ℎ 1 ,…, ℎ 𝑠 (𝑠≤𝐷)
We'll process them iteratively.

37. Handling ℎ 𝑖
Let 𝑃 0𝑖 be the points of 𝑃 0 in ℎ 𝑖 \ 𝑗<𝑖 ℎ 𝑗 , and 𝐿 0𝑖 be the lines of 𝐿 0 that are contained (except for a finite number) in ℎ 𝑖 \ 𝑗<𝑖 ℎ 𝑗 .
We will consider only the points of 𝑃 0𝑖 that are incident to at least two lines of 𝐿 0𝑖 .
Let 𝑚 𝑖 =| 𝐿 0𝑖 | and 𝑛 𝑖 =| 𝑃 0𝑖 |. (by assumption, 𝑛 𝑖 <𝑞)
Using the Szemerédi – Trotter theorem we get: 𝑂 𝑚 𝑖 2/3 𝑛 𝑖 2/3 + 𝑚 𝑖 + 𝑛 𝑖 = 𝑂( 𝑚 𝑖 2/3 𝑛 𝑖 1/3 𝑞 1/3 + 𝑚 𝑖 + 𝑛 𝑖 ) incidences on ℎ 𝑖 .

38. We got 𝑂( 𝑚 𝑖 2/3 𝑛 𝑖 1/3 𝑞 1/3 + 𝑚 𝑖 + 𝑛 𝑖 ) for each of the planes.
Sum them up we get, by Hölder inequality: 𝑂 ∑ 𝑚 𝑖 2/3 ∑ 𝑛 𝑖 1/3 𝑞 1/3 +𝑚+𝑛 =𝑂( 𝑚 2/3 𝑛 1/3 𝑞 1/3 +𝑚+𝑛) incidences.

39. The Other Cases Cases we haven't handled yet:
incidence between points and lines contained in another plane not yet processed.
incidences between points that are incident only with one (relevant) line
Any of these incidences occur at crossings of a line and a plane (that doesn't contain the line).
There are only 𝑂(𝑛𝐷) such crossings.

40. Putting Everything Together
Consider all that we have seen, and substitute 𝐷 with its value according to the range of 𝑚. We’ll get: 𝐼 𝑚,𝑛,𝑞 ≤ 𝑂 𝑛 for 𝑚< 𝑎 ′ 𝑛 1/ 𝐼 𝑚,𝛽𝑛, 𝛽𝑛 1/2 +𝐵 𝑚 1/2 𝑛 3/4 + 𝑚 2/3 𝑛 1/3 𝑞 1/3 +𝑛 for 𝑎 ′ 𝑛 1/2 ≤𝑚≤𝑎 𝑛 3/2 𝐼 𝑚 3 ,𝑛,𝑞 +𝐼 𝑚,𝛽𝑛, 𝛽𝑛 1/2 +𝐵 𝑚 2/3 𝑛 1/3 𝑞 1/3 + 𝑛 3/2 for 𝑚>𝑎 𝑛 3/ For an appropriate constant 𝐵.

41. The solution is 𝐼 𝑚,𝑛,𝑞 ≤𝐴( 𝑚 1/2 𝑛 3/4 + 𝑚 2/3 𝑛 1/3 𝑞 1/3 +𝑛) for a sufficiently large constant 𝐴. we'll prove it by Induction on 𝑚,𝑛.
The base cases are easily established for large enough 𝐴.
For the induction step, we'll split the proof for the different cases of 𝑚.

42. Induction Step for 𝑎′ 𝑛 1/2 ≤𝑚≤𝑎 𝑛 3/2
By the induction hypothesis we have: 𝐼 𝑚,𝑛,𝑞 ≤𝐴 𝑚 1/2 𝛽𝑛 3/4 + 𝑚 2/3 𝛽𝑛 1/3 𝛽𝑛 1/6 +𝛽𝑛 𝑚 1/2 𝛽𝑛 3/4 + 𝑚 2/3 𝛽𝑛 1/3 𝛽𝑛 1/6 +𝛽𝑛 +𝐵 𝑚 1/2 𝑛 3/4 + 𝑚 2/3 𝑛 1/3 𝑞 1/3 +𝑛 𝑚 1/2 𝑛 3/4 + 𝑚 2/3 𝑛 1/3 𝑞 1/3 +𝑛 =𝐴 𝛽 3/4 +𝐵 𝑚 1/2 𝑛 3/4 +𝐴 𝛽 1/2 𝑚 2/3 𝑛 1/2 +𝐵 𝑚 2/3 𝑛 1/3 𝑞 1/3 +(𝐴𝛽 +𝐵)𝑛 .

43. for 𝑚≤𝑎 𝑛 3/2 we have 𝑚 2/3 𝑛 1/2 ≤ 𝑎 1/6 𝑚 1/2 𝑛 3/4 , so this bound is at most: 𝐴 𝛽 3/4 +𝐵+𝐴 𝑎 1/6 𝛽 1/2 𝑚 1/2 𝑛 3/4 +𝐵 𝑚 2/3 𝑛 1/3 𝑞 1/3 + 𝐴𝛽+𝐵 𝑛 we choose small enough 𝛽 to ensure that 𝛽< 1 2 and 𝛽 3/4 + 𝑎 1/6 + 𝑏 1/2 < 1 2 , and 𝐴>2𝐵. This yields our bound.

44. Induction Step for 𝑚<𝑎 𝑛 3/2
By the induction hypothesis we have:

45. We got that the bound is at most: 𝐴 /2 + 𝛽 3/4 + 𝑎 1/6 𝛽 1/2 𝑚 1/2 𝑛 3/4 + 𝐴 3 2/3 +𝐵 𝑚 2/3 𝑛 1/3 𝑞 1/3 + 𝐴 1+𝛽 +𝐵 𝑛 3/2
Since 𝑚>𝑎 𝑛 3/2 we get 𝑛 3/2 < 𝑎 −1/2 𝑚 1/2 𝑛 3/4 , so if 𝑎 is sufficiently large, 𝛽 is sufficiently small and 𝐴 is sufficiently large, we get our bound.
This completes the induction step and thus the whole proof.

46. D. Q. E.

47. If Time Remains: the Lower Bound
What about the lower bound?
First, we present a construction that gives us 𝐼(𝑃,𝐿)=Ω( 𝑚 1/2 𝑛 3/4 ).

48. The construction
Taking 𝑙 and 𝑘, we construct an integer grid of 𝑘×2𝑘𝑙×2𝑘𝑙.
𝑃 will be the set of grid points, so 𝑚=4 𝑘 3 𝑙 2 .
𝐿 will be all the lines in the form: 𝑦 = 𝑎𝑥+𝑏, 𝑧=𝑐𝑥+𝑑, for 𝑎,𝑐 ∈ 1,…,𝑙 , and 𝑏,𝑑 ∈ {1,…,𝑘𝑙}.
𝑛= 𝑘 2 𝑙 4
Each line in 𝐿 has exactly 𝑘 incidences with the points in 𝑃.
Therefore, 𝐼 𝑃,𝐿 = 𝑘 3 𝑙 4 = Θ(𝑚 1/2 𝑛 3/4 ).

49. It's not Enough!
We need to ensure that the other assumption of the theorem hold.
To ensure that every point is incident to at least three lines, we'll add axis parallel lines of the grid to 𝐿.
We increased 𝐿 by 4 𝑘 2 𝑙+4 𝑘 2 𝑙 2 <8 𝑘 2 𝑙 2 , and this is negligible in comparison to 𝑘 2 𝑙 4 .
In addition, it can be shown that any plane contains at most 𝑂( 𝐿 1/2 )=𝑂(𝑘 𝑙 2 ) points.

50. The Next Term We want to construct the bound Ω( 𝑚 2/3 𝑛 1/3 𝑞 1/3 ).
For it to be the largest term, we need:
𝑞> 𝑛 1/2 , otherwise 𝑚 1/2 𝑛 3/4 is bigger.
𝑞> 𝑛 2 / 𝑚 2 , otherwise 𝑂(𝑛) is bigger.
𝑚=Ω( 𝑛 1/2 ), otherwise 𝑂(𝑛) is bigger.
𝑚=𝑂(𝑛𝑞), because it can be shown from the proof of the upper bound that any construction that satisfies the assumptions has 𝑚=𝑂(𝑛𝑞).

51. The Construction
Take 𝑛/𝑞 planes in general position, and place 𝑞 lines and 𝑚𝑞/𝑛 points on each of them with Θ( 𝑚𝑞 𝑛 2/3 𝑞 2/3 ) incidences.

52. Take 𝑛/𝑞 planes in general position, and place 𝑞 lines and 𝑚𝑞/𝑛 points on each of them with Θ( 𝑚𝑞 𝑛 2/3 𝑞 2/3 ) incidences.
From the assumptions below we get 𝑚𝑞/𝑛=Ω(1) and 𝑐 𝑞 1/2 <𝑚𝑞/𝑛< 𝑐 ′ 𝑞 2 , so we can use this construction.
In this construction, indeed every point is incident to at least three lines.
So we got 𝑛/𝑞 Θ 𝑚𝑞/𝑛 2/3 𝑞 2/3 =Θ( 𝑚 2/3 𝑛 1/3 𝑞 1/3 ) incidences.

53. Thank You!