Consider n points in the plane, not all on a line. We want to find two triangles determined by the points with area ratio as close as possible to 1. Ophir and Pinchasi (2014) showed that in any set of n points in the plane with no three on a line, there are two triples \(\{a,b,c\}\) and \(\{a',b',c'\}\) of points such that the triangles \(\triangle abc\) and \(\triangle a'b'c'\) have almost the same area in the precise sense that

$$\begin{aligned} \left| \frac{\triangle abc}{\triangle a'b'c'}-1\right| < \frac{60\log ^{1/3} n}{n^{2/3}}. \end{aligned}$$

We present the following two improvements of this result.

FormalPara Theorem 1

Given a set S of n non-collinear points in the plane, there exist distinct points \(a,b,c,d\in S\) such that c and d are both not on the line through a and b, and

$$\begin{aligned} \frac{1}{r} \le \frac{\triangle abd}{\triangle abc}\le r \end{aligned}$$

where \(r=3^{3/(n-3)} = 1+\frac{3\ln 3}{n}+O(1/n^2)\).

FormalPara Theorem 2

Given a set S of n non-collinear points in the plane, there exist non-collinear triples of points \(\{a,b,c\}\) and \(\{a',b',c'\}\) from S such that

$$\begin{aligned} \frac{1}{r} \le \frac{\triangle abc}{\triangle a'b'c'}\le r \end{aligned}$$

where \(r = 1 + O(\frac{\log n}{n^2})\).

The proof of Theorem 1 is a simple pigeon-hole argument, that can be generalised as follows to higher dimensions.

FormalPara Corollary 3

Given a set S of n points that span d-dimensional Euclidean space, there exist \(d+2\) distinct points \(a_1,a_2,\dots ,a_d,b,c\in S\) such that \(a_1,\dots ,a_d\) span a hyperplane not containing b and c, and the ratio between the volume of the simplices with vertex sets \(\{a_1,\dots ,a_d,b\}\) and \(\{a_1,\dots ,a_d,c\}\) lies in [1/r, r], where \(r=1+O_d(\frac{\log n}{n^2})\).

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.

FormalPara Proposition 4

There exists a set of n points \(p_1,\dots ,p_n\) in the plane such that whenever \(\frac{1}{14}\le \frac{\triangle p_i p_j p_k}{\triangle p_{i'} p_{j'} p_{k'}}\le 14\), 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 \(\mathbb {R}\), there exist two distinct pairs \(\{a,b\}\) and \(\{a',b'\}\) of points from S such that

$$\begin{aligned} \left| \frac{|a-b|}{|a'-b'|}-1\right| \le \frac{9\log n}{n^2}. \end{aligned}$$

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 \(\mathbb {R}\), 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.

FormalPara 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 \(\{(s,n^{5s})|s\in S\}\) is \( > rsim 1+1/n^2\).

We say that a set \(\{a_1,a_2,\dots ,a_n\}\) of n integers is a Sidon set if the sums \(a_i+a_j\), \(i\le j\), are all different. Ophir and Pinchasi noted that the example of Erdős and Turán (1941) of a Sidon set of n integers from \(\{1,2,\dots ,n^2+O(n)\}\) is also an example of n points in \(\mathbb {R}\) 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 \(\mathbb {R}\) with a slightly better lower bound of \(4/n^2\). This construction additionally has a ratio of \(\Theta (n)\) between the minimum and maximum distance in the set, where a Sidon set has ratio \(\Theta (n^2)\).

FormalPara Proposition 6

There exists a set of n points on the real line such that for any two distinct pairs \(\{a,b\}\) and \(\{c,d\}\) from the set with \(|a-b|\ge |c-d|\), we have

$$\begin{aligned} 1+\frac{4}{n^2}+O(1/n^3)\le \left| \frac{a-b}{c-d}\right| \le O(n). \end{aligned}$$

1 Proofs

Proof of Theorem 1

Choose \(a,b,c\in S\) such that \(\triangle abc\) has maximum area among all triples of points from S. Without loss of generality we may apply an affine transformation so that \(\triangle abc\) becomes an equilateral triangle of side length 1, as in Figure 1.

Fig. 1
figure 1

Apply an affine transformation so that \(\triangle abc\) is equilateral with side length 1

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Fig. 2
figure 2

Pigeon-hole principle inside a trapezium

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Let \(\triangle def\) be the triangle with sides parallel to the sides of \(\triangle abc\) and such that a, b, c are midpoints of the edges of \(\triangle def\). Then all n points are inside \(\triangle def\). Let p be the centroid of \(\triangle abc\) (and \(\triangle def\)). Consider the three lines through p parallel to the three sides of \(\triangle abc\). At least n/3 points must lie on the side of one of these lines that is opposite to the parallel side of \(\triangle abc\). Without loss of generality the trapezium degh contains at least n/3 of the points. Let \(k=\lceil n/3\rceil -1\). Subdivide the trapezium using k parallel lines of height \(\frac{1}{2\sqrt{3}}r^i\), \(i=0,1,\dots ,k-1\), above the line bc, where r is chosen such that \(\frac{1}{2\sqrt{3}}r^k=\sqrt{3}/2\) (Figure 2). Since \(k<n/3\), there are two points in at least one of the regions, say \(q_1\) and \(q_2\). Then \(\frac{1}{r}\le \triangle bcq_1/\triangle bcq_2\le r\), and since \(k\ge n/3 - 1\),

$$\begin{aligned} r = 3^{1/k}\le 3^{3/(n-3)} = 1+ 3\ln 3/n + O(1/n^2). \end{aligned}$$

\(\square \)

Proof of Proposition 4

Let \(p_i=(2^{2^i},2^{2^{i+1}})\), \(i=1,\dots ,n\). If \(i<j<k\), then the area of \(\triangle p_i p_j p_k\) is

$$\begin{aligned} \triangle p_i p_j p_k = \frac{1}{2}\begin{vmatrix}1&1&1 \\ 2^{2^i}&2^{2^j}&2^{2^k} \\ 2^{2^{i+1}}&2^{2^{j+1}}&2^{2^{k+1}} \end{vmatrix} = \frac{1}{2}(2^{2^k}-2^{2^j})(2^{2^k}-2^{2^i})(2^{2^j}-2^{2^i}). \end{aligned}$$

Thus \(\triangle p_ip_jp_k \le \frac{1}{2} 2^{2^{k+1}+2^j}\) and

$$\begin{aligned} \triangle p_ip_jp_k&\ge \frac{1}{2}(2^{2^k}-2^{2^2})(2^{2^k}-2^{2^1})(2^{2^j}-2^{2^1})\\&= \frac{1}{2} 2^{2^{k+1}+2^j}(1-2^{4-2^k})(1-2^{2-2^k})(1-2^{2-2^j}) \end{aligned}$$

We now consider two distinct triples \(\{i<j<k\}\) and \(\{i'<j'<k'\}\), where without loss of generality, \(k\ge k'\). If \(k>k'\) then

$$\begin{aligned} \frac{\triangle p_{i} p_{j} p_{k}}{\triangle p_{i'} p_{j'} p_{k'}}&\ge 2^{2^{k+1}-2^k + 2^j - 2^{j'}}(1-2^{4-2^4})(1-2^{2-2^4})(1-2^{2-2^2})\\&\ge 2^{2^k + 2^2 - 2^{k-2}}(1-2^{4-2^4})(1-2^{2-2^4})(1-2^{2-2^2})\\&\ge 2^{2^4 + 2^2 - 2^{4-2}}(1-2^{4-2^4})(1-2^{2-2^4})(1-2^{2-2^2}) > 2^{15}. \end{aligned}$$

If \(k=k'\) then without loss of generality, \(j\ge j'\). If \(j>j'\), then

$$\begin{aligned} \frac{\triangle p_{i} p_{j} p_{k}}{\triangle p_{i'} p_{j'} p_{k'}}&\ge 2^{2^j - 2^{j-1}}(1-2^{4-2^3})(1-2^{2-2^3})(1-2^{2-2^3})\\&\ge 2^4(1-2^{4-2^3})(1-2^{2-2^3})(1-2^{2-2^3}) > 14. \end{aligned}$$

Therefore, if the ratio is at most 14, then \(j=j'\) and \(k=k'\). \(\square \)

Proof of Theorem 2

Without loss of generality, the maximum-area triangle \(\triangle abc\) 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 \(4/\sqrt{3}\). Thus S is contained in \(\triangle def\) of Figure 1, so any two points are at distance \(\le 8/\sqrt{3}\).

Assume that for any two distinct triangles \(\triangle xyz\) and \(\triangle x'y'z'\),

$$\begin{aligned} \max \left\{ \frac{\triangle xyz}{\triangle x'y'z'}, \frac{\triangle x'y'z'}{\triangle xyz}\right\} \ge 1+ \frac{6\log n}{n^2}. \end{aligned}$$

We next show that the distance between any two points \(p,p'\in S\) is \( > rsim 4\log n/n^2\). Since the perpendicular distance from p to some edge of \(\triangle abc\), say ab, is \(\ge 2/3\), we obtain

$$\begin{aligned} \frac{\triangle abp'}{\triangle abp} \le 1 + \frac{pp'}{2/3}.\end{aligned}$$

Similarly, since \(p'\) is at perpendicular distance \(\ge 2/3-pp'\) from ab, we obtain

$$\begin{aligned} \frac{\triangle abp}{\triangle abp'} \le 1 + \frac{pp'}{2/3-pp'}. \end{aligned}$$

It follows that

$$\begin{aligned} 1+ \frac{6\log n}{n^2} \le 1 + \frac{pp'}{2/3-pp'}, \end{aligned}$$

from which \(pp' > rsim 4\log n/n^2\) follows.

Among all \(\left( {\begin{array}{c}n\\ 3\end{array}}\right) \) triples of points, the \(\left( {\begin{array}{c}n\\ 3\end{array}}\right) -n^2\) smallest areas are all \(\le (1+\frac{6\log n}{n^2})^{-n^2}\sim n^{-6}\). Suppose that each pair of points belongs to more than 6 triangles of area \( > rsim n^{-6}\). Then there are at least \(7\left( {\begin{array}{c}n\\ 2\end{array}}\right) /3 > n^2\) triangles of area \( > rsim n^{-6}\), a contradiction.

Therefore, some pair of points \(\{p,q\}\) belongs to at most 6 triangles of area \( > rsim n^{-6}\), hence to at least \(n-8\) triangles \(\triangle pqp_i\), \(i=1,\dots ,n-8\), each of area \(\lesssim n^{-6}\). Since the distance between p and q is \( > rsim 4\log n/n^2\), the perpendicular distance of any \(p_i\) to the line \(\ell \) through p and q is \(\lesssim 1/(2n^4\log n)\).

We now choose coordinates so that \(\ell \) becomes the x-axis. Then each \(p_i=(x_i,\varepsilon _i)\), where \(|\varepsilon _i|\lesssim 1/(2n^4\log n)\). Since \(\triangle abc\) has width 2, one of its three vertices, say \(a=(x,h)\), is at distance \(|h|\ge 1\) from \(\ell \). By the result of Ophir and Pinchasi applied to \(x_1,x_2,\dots ,x_{n-8}\), there are two pairs \(\{i,j\}\) and \(\{s,t\}\) such that

$$\begin{aligned} \frac{|x_i-x_j|}{|x_s-x_t|} = 1 + O(\log n/n^2).\end{aligned}$$

We next show that the ratio between the areas of \(\triangle ap_ip_j\) and \(\triangle ap_sp_t\) is asymptotically the same as \(|x_i-x_j|/|x_s-x_t|\).

We claim that \(\triangle ap_ip_j = |h(x_i-x_j)|(1+o(\log n/n^2))\). Indeed,

$$\begin{aligned} \pm 2\triangle ap_ip_j&= \begin{vmatrix} 1&1&1\\ x&x_i&x_j\\ h&\varepsilon _i&\varepsilon _j\end{vmatrix} = \begin{vmatrix} 1&1&1\\ 0&x_i-x&x_j-x\\ h&\varepsilon _i&\varepsilon _j\end{vmatrix}\\&= h(x_j-x_i)\left( 1 - \frac{\varepsilon _i(x_j-x)}{h(x_j-x_i)} + \frac{\varepsilon _j(x_i-x)}{h(x_j-x_i)}\right) . \end{aligned}$$

Since \(|x_j-x|\le p_jp\le 8/\sqrt{3}\), \(|h|\ge 1\), and

$$\begin{aligned} |x_j-x_i|\ge p_ip_j-|\varepsilon _i|-|\varepsilon _j| > rsim \frac{4\log n}{n^2} - \frac{1}{n^4\log n} > rsim \frac{4\log n}{n^2}, \end{aligned}$$

we obtain

$$\begin{aligned} \left| \frac{\varepsilon _i(x_j-x)}{h(x_j-x_i)}\right| = O\left( \frac{1}{n^2\log ^2 n}\right) . \end{aligned}$$

Similarly,

$$\begin{aligned} \left| \frac{\varepsilon _j(x_i-x)}{h(x_j-x_i)}\right| = O\left( \frac{1}{n^2\log ^2 n}\right) , \end{aligned}$$

and it follows that

$$\begin{aligned} 2\triangle ap_ip_j = |h(x_i-x_j)|(1+O(1/(n^2\log ^2 n))). \end{aligned}$$

Similarly,

$$\begin{aligned} 2\triangle ap_sp_t = |h(x_s-x_t)|(1+O(1/(n^2\log ^2 n))), \end{aligned}$$

and we conclude that

$$\begin{aligned} \frac{\triangle ap_ip_j}{\triangle ap_sp_t} = \frac{|x_i-x_j|}{|x_s-x_t|}(1+O(1/(n^2\log n))) = 1 + O(\log n/n^2). \end{aligned}$$

\(\square \)

Proof of Proposition 5

Let S be a Sidon set of n elements from \(\{1,2,\dots ,N\}\) where \(N=n^2+O(n)\). Write \(p_s=(s,n^{5s})\) and \(q_s=(s,0)\) for each \(s\in S\). Consider three \(s,t,u\in S\) with \(s<t<u\).

Then

$$\begin{aligned} 2\triangle p_sp_tp_u = \begin{vmatrix} 1&1&1\\ s&t&u \\ n^{5s}&n^{5t}&n^{5u} \end{vmatrix} = n^{5u}(t-s) + n^{5t}(s-u) + n^{5s}(u-t) \end{aligned}$$

and

$$\begin{aligned} 2\triangle q_sq_tp_u = \begin{vmatrix} 1&\quad 1&\quad 1\\ s&\quad t&\quad u \\ 0&\quad 0&\quad n^{5u} \end{vmatrix} = n^{5u}(t-s). \end{aligned}$$

Since the ratio between these two areas is close to 1, we can replace \(\triangle p_sp_tp_u\) with \(\triangle q_sq_tp_u\) in our calculations. Specifically,

$$\begin{aligned} \left| \frac{\triangle p_sp_tp_u}{\triangle q_sq_tp_u}-1\right|&= \left| \frac{n^{5t}(s-u)+n^{5s}(u-t)}{n^{5u}(t-s)}\right| \\&\le n^{5(t-u)}\left| \frac{s-u}{t-s}\right| + n^{5(s-u)}\left| \frac{u-t}{t-s}\right| \\&< n^{-5} N + n^{-10} N \sim \frac{1}{n^3}. \end{aligned}$$

Consider distinct triples \(\{a<b<c\}\) and \(\{d<e<f\}\) of elements from S, where we assume without loss of generality that \(c\ge f\). Then

$$\begin{aligned} \frac{\triangle p_ap_bp_c}{\triangle p_dp_ep_f} > rsim \left( 1-\frac{1}{n^3}\right) ^2\frac{\triangle q_aq_bp_c}{\triangle q_dq_ep_f} = \left( 1-\frac{1}{n^3}\right) ^2\frac{n^{5c}(b-a)}{n^{5f}(e-d)}. \end{aligned}$$

If \(c=f\) then \(\{a,b\}\ne \{d,e\}\), and we assume without loss of generality that \(b-a > e-d\). We obtain

$$\begin{aligned} \frac{\triangle p_ap_bp_c}{\triangle p_dp_ep_f} \ge \left( 1-\frac{1}{n^3}\right) ^2\frac{b-a}{e-d}\ge \left( 1-\frac{1}{n^3}\right) ^2\frac{N}{N-1} > rsim 1 + \frac{1}{n^2}. \end{aligned}$$

On the other hand, if \(c>f\) then the ratio is even larger:

$$\begin{aligned} \frac{\triangle p_ap_bp_c}{\triangle p_dp_ep_f} \ge \left( 1-\frac{1}{n^3}\right) ^2n^5\frac{1}{N} > rsim n^3. \end{aligned}$$

\(\square \)

Proof of Proposition 6

Fix \(\varepsilon >0\), and let \(p_i=(1+\varepsilon )^i-1\), \(i=0,1,\dots ,n-1\). Then for any integer a, b with \(0\le a<b\le n-1\),

$$\begin{aligned} p_b-p_a = (1+\varepsilon )^b-(1+\varepsilon )^a. \end{aligned}$$

Take any \(a,b,c,d\in \{0,\dots ,n-1\}\) with \(a<b\), \(c<d\), \(\{a,b\}\ne \{c,d\}\) and without loss of generality, \(b-a\le d-c\). If \(b-a=d-c\) then without loss of generality, \(a<c\), and

$$\begin{aligned} \frac{p_d-p_c}{p_b-p_a} = \frac{(1+\varepsilon )^d-(1+\varepsilon )^c}{(1+\varepsilon )^b-(1+\varepsilon )^a} = (1+\varepsilon )^{c-a}\frac{(1+\varepsilon )^{d-c}-1}{(1+\varepsilon )^{b-a}-1} = (1+\varepsilon )^{c-a}\ge 1+\varepsilon . \end{aligned}$$

If \(b-a<d-c\), then, setting \(b-a=k\in \{1,2,\dots ,n-2\}\),

$$\begin{aligned} \frac{p_d-p_c}{p_b-p_a}&\ge \frac{p_d-p_{d-b+a-1}}{p_b-p_a} = (1+\varepsilon )^{(d-b+a-1)+(n-1-b)} \frac{p_{k+1}-p_0}{p_{n-1}-p_{n-1-k}}\\&\ge \frac{(1+\varepsilon )^{k+1}-1}{(1+\varepsilon )^{n-1}-(1+\varepsilon )^{n-1-k}}. \end{aligned}$$

This last expression will be \(\ge 1+\varepsilon \) if and only if \((1+\varepsilon )^{k+1}+(1+\varepsilon )^{n-k}\ge (1+\varepsilon )^n+1\). If we use the Binomial Theorem to expand this up to second order, we obtain

$$\begin{aligned}&1+(k+1)\varepsilon +\left( {\begin{array}{c}k+1\\ 2\end{array}}\right) \varepsilon ^2+O(k^3\varepsilon ^3)\\&\qquad +1+\varepsilon (n-k)+\left( {\begin{array}{c}n-k\\ 2\end{array}}\right) \varepsilon ^2+O((n-k)^3\varepsilon ^3)\\&\quad \ge 1+n\varepsilon +\left( {\begin{array}{c}n\\ 2\end{array}}\right) \varepsilon ^2+O(n^3\varepsilon ^3)+1, \end{aligned}$$

which is equivalent to \(\varepsilon \ge \left( \left( {\begin{array}{c}n\\ 2\end{array}}\right) -\left( {\begin{array}{c}k+1\\ 2\end{array}}\right) -\left( {\begin{array}{c}n-k\\ 2\end{array}}\right) \right) \varepsilon ^2+O(n^3\varepsilon ^3)\), that is, we need the inequality \(1\ge k(n-k-1)\varepsilon +O(n^3\varepsilon ^2)\) to hold for all \(k=1,2,\dots ,n-2\). Since \(k(n-k-1)\le \left( \frac{n-1}{2}\right) ^2\), we obtain that we need \(\varepsilon \le \frac{4}{n^2} +O(\frac{1}{n^3}) +O(n\varepsilon ^2)\). Thus we can take \(\varepsilon = \frac{4}{n^2}+O(\frac{1}{n^3})\).

This shows that we obtain a minimum ratio of \(1+\frac{4}{n^2}+O(\frac{1}{n^3})\). \(\square \)

Instead of using points where the successive distances \(p_{i+1}-p_i\) form a geometric progression, as in the above proof, we can also use an arithmetic progression. If we take the n points \(p_0=0\), \(p_i=\sum _{j=0}^{i-1} (1+j\varepsilon )\), \(i=1,2,3,\dots ,n-1\), then a calculation shows that we obtain the same optimal asymptotics of \(\varepsilon = \frac{4}{n^2}+O(\frac{1}{n^3})\).

2 Final remarks

We did not touch on the problem of Ophir and Pinchasi on whether there exist in any set of n elements of \(\mathbb {R}\) two pairs with ratio better than \(1+O(\log n/n^2)\), but we did find the sets of points for which the smallest ratio \(>1\) is a maximum when \(n\le 4\). Thus consider a set \(S\subset \mathbb {R}\) of n points that maximizes

$$\begin{aligned} \min \left\{ \frac{|a-b|}{|c-d|} : a,b,c,d\in S, |a-b|\ge |c-d|>0\right\} \end{aligned}$$

among all sets of n points in \(\mathbb {R}\).

If \(n=3\), it is easy to see that there is a unique extremal set up to similarity, namely \(S=\{a<b<c\}\) such that \(\frac{c-b}{b-a}\) equals the golden ratio \((1+\sqrt{5})/2\).

For \(n=4\) the problem is already non-trivial, as there are 6 different distances. Using a case analysis, we can show that up to similarity there are two extremal sets. One of them is the above geometric progression construction \(\{0,1,1+r,1+r+r^2\}\), where r is the unique real root of the cubic polynomial \(r^3-r-1\). The other configuration is \(\{0,1,r,r^2\}\). The number

$$\begin{aligned}r = \root 3 \of {\frac{9+\sqrt{69}}{18}} + \root 3 \of {\frac{9-\sqrt{69}}{18}} = 1.3247179572\dots \end{aligned}$$

is known as the plastic number of van der Laan (1960), which is closely related to the golden ratio (Aarts et al. 2001; Rush 2012; Stewart 1996).