Triangles of nearly equal area

Given any n points in the plane, not all on the same line, there exist two non-collinear triples such that the ratio of the areas of the triangles they determine, differs from 1 by at most O ( log n / n 2 ) . If we furthermore insist that the two triangles have a common edge, then there are two with area ratios differing from 1 by at most O ( 1 / n ) . This improves some results of Ophir and Pinchasi (Discrete Appl. Math. 174 (2014), 122–127). We also give some constructions for these and related problems.


Theorem 2 Given a set S of n non-collinear points in the plane
where r = 1 + O( log n n 2 ). The proof of Theorem 1 is a simple pigeon-hole argument, that can be generalised as follows to higher dimensions.
Corollary 3 Given a set S of n points that span d-dimensional Euclidean space, there exist d + 2 distinct points a 1 , a 2 , . . . , a d , b, c ∈ S such that a 1 , . . . , a d span a hyperplane not containing b and c, and the ratio between the volume of the simplices with vertex sets {a 1 , . . . , a d , b} and {a 1 , . . . , a d , c} lies in [1/r , r ], where r = 1+O d ( log n n 2 ). Theorem 1 is best possible in the sense that we cannot guarantee two triangles with only one vertex in common to have almost the same area.
Proposition 4 There exists a set of n points p 1 , . . . , p n in the plane such that whenever 1 14 ≤ p i p j p k p i p j p k ≤ 14, then {i, j, k} and {i , j , k } have their two largest elements in common.
On the other hand, we do not know if Theorem 2 can be improved. Its proof depends on the following result of Ophir and Pinchasi (2014), for which it is also not known whether it is asymptotically tight.
Given any set S of n elements of R, there exist two distinct pairs {a, b} and {a , b } of points from S such that This result in fact holds for any n-element metric space, where it is best possible up to the constant factor of 9. Ophir and Pinchasi conjecture that for n points in R, there are always two pairs of ratio 1 + c/n 2 . We give the following lower bound for triangle areas, showing that the ratio in Theorem 2 cannot be improved beyond 1 + O(1/n 2 ) either.
Proposition 5 There exists a set S of n real numbers such that the ratio between the area of any two triangles with vertices from the set {(s, n 5s )|s ∈ S} is 1 + 1/n 2 .
We say that a set {a 1 , a 2 , . . . , a n } of n integers is a Sidon set if the sums a i + a j , i ≤ j, are all different. Ophir and Pinchasi noted that the example of Erdős and Turán (1941) of a Sidon set of n integers from {1, 2, . . . , n 2 + O(n)} is also an example of n points in R for which the ratio of the distance between any two distinct pairs differ from 1 by at least 1/n 2 . We next observe that there is a simple construction of n points in R with a slightly better lower bound of 4/n 2 . This construction additionally has a ratio of (n) between the minimum and maximum distance in the set, where a Sidon set has ratio (n 2 ).

Proofs
Proof of Theorem 1 Choose a, b, c ∈ S such that abc has maximum area among all triples of points from S. Without loss of generality we may apply an affine transformation so that abc becomes an equilateral triangle of side length 1, as in Figure 1. Let de f be the triangle with sides parallel to the sides of abc and such that a, b, c are midpoints of the edges of de f . Then all n points are inside de f . Let p be the centroid of abc (and de f ). Consider the three lines through p parallel to the three sides of abc. At least n/3 points must lie on the side of one of these lines that is opposite to the parallel side of abc. Without loss of generality the trapezium degh contains at least n/3 of the points. Let k = n/3 − 1. Subdivide the trapezium using k parallel lines of height 1 Figure 2). Since k < n/3, there are two points in at least one of the regions, say q 1 and q 2 . Then 1 r ≤ bcq 1 / bcq 2 ≤ r , and since k ≥ n/3 − 1, r = 3 1/k ≤ 3 3/(n−3) = 1 + 3 ln 3/n + O(1/n 2 ).

Proof of Proposition 4 Let
Thus p i p j p k ≤ 1 2 2 2 k+1 +2 j and We now consider two distinct triples {i < j < k} and {i < j < k }, where without loss of generality, k ≥ k . If k > k then If k = k then without loss of generality, j ≥ j . If j > j , then Therefore, if the ratio is at most 14, then j = j and k = k .

Proof of Theorem 2
Without loss of generality, the maximum-area triangle abc is equilateral with area 1. Then its height is 2, the distance between its centroid and any side is 2/3, and its side length is 4/ √ 3. Thus S is contained in de f of Figure 1, so any two points are at distance ≤ 8/ √ 3. Assume that for any two distinct triangles x yz and x y z , max x yz x y z , x y z x yz ≥ 1 + 6 log n n 2 .
We next show that the distance between any two points p, p ∈ S is 4 log n/n 2 . Since the perpendicular distance from p to some edge of abc, say ab, is ≥ 2/3, we obtain Similarly, since p is at perpendicular distance ≥ 2/3 − pp from ab, we obtain It follows that from which pp 4 log n/n 2 follows. Among all n 3 triples of points, the n 3 −n 2 smallest areas are all ≤ (1+ 6 log n n 2 ) −n 2 ∼ n −6 . Suppose that each pair of points belongs to more than 6 triangles of area n −6 . Then there are at least 7 n 2 /3 > n 2 triangles of area n −6 , a contradiction. Therefore, some pair of points { p, q} belongs to at most 6 triangles of area n −6 , hence to at least n − 8 triangles pqp i , i = 1, . . . , n − 8, each of area n −6 . Since the distance between p and q is 4 log n/n 2 , the perpendicular distance of any p i to the line through p and q is 1/(2n 4 log n).
We now choose coordinates so that becomes the x-axis. Then each p i = (x i , ε i ), where |ε i | 1/(2n 4 log n). Since abc has width 2, one of its three vertices, say a = (x, h), is at distance |h| ≥ 1 from . By the result of Ophir and Pinchasi applied to x 1 , x 2 , . . . , x n−8 , there are two pairs {i, j} and {s, t} such that We next show that the ratio between the areas of ap i p j and ap s p t is asymptotically the same as |x We claim that ap i p j = |h(x i − x j )|(1 + o(log n/n 2 )). Indeed, Since |x j − x| ≤ p j p ≤ 8/ √ 3, |h| ≥ 1, and we obtain Similarly, and it follows that Similarly, 2 ap s p t = |h(x s − x t )|(1 + O(1/(n 2 log 2 n))), and we conclude that (1 + O(1/(n 2 log n))) = 1 + O(log n/n 2 ).

Proof of Proposition 5
Let S be a Sidon set of n elements from {1, 2, . . . , N } where N = n 2 + O(n). Write p s = (s, n 5s ) and q s = (s, 0) for each s ∈ S. Consider three s, t, u ∈ S with s < t < u. Then 2 p s p t p u = 1 1 1 s t u n 5s n 5t n 5u = n 5u (t − s) + n 5t (s − u) + n 5s (u − t) and 2 q s q t p u = 1 1 1 s t u 0 0 n 5u = n 5u (t − s).
Since the ratio between these two areas is close to 1, we can replace p s p t p u with q s q t p u in our calculations. Specifically, Consider distinct triples {a < b < c} and {d < e < f } of elements from S, where we assume without loss of generality that c ≥ f . Then .
If c = f then {a, b} = {d, e}, and we assume without loss of generality that b − a > e − d. We obtain On the other hand, if c > f then the ratio is even larger: p a p b p c p d p e p f ≥ 1 − 1 n 3 2 n 5 1 N n 3 .