Bounds on Area Involving Lattice Size

The lattice size of a lattice polygon $P$ was introduced and studied by Schicho, and by Castryck and Cools in relation to the problem of bounding the total degree and the bi-degree of the defining equation of an algebraic curve. In this paper we establish sharp lower bounds on the area of plane convex bodies $P\subset\mathbb{R}^2$ that involve the lattice size of $P$. In particular, we improve bounds established by Arnold, and B\'ar\'any and Pach. We also provide a classification of minimal lattice polygons $P\subset\mathbb{R}^2$ of fixed lattice size $\operatorname{ls_\square}(P)$.


Introduction
This paper is devoted to providing sharp lower bounds on the area of plane convex bodies P , which involve the lattice size of P .This invariant was formally introduced by Schicho, and Kastryck and Cools in [8,15], although it had appeared implicitly earlier in the work of Arnold [3], Bárány and Pach [5], Brown and Kaspzyck [6], and Lagarias and Ziegler [13].The lattice size was further studied in [1,2,10], and [11].
We next reproduce the definition of the lattice size from [8] applying it now to a plane convex body P .
Definition 1.1.The lattice size ls X (P ) of a convex body P ⊂ R 2 with respect to a set X ⊂ R 2 is the smallest real non-negative l such that φ(P ) is contained in the l-dilate lX of X for some transformation φ, which a combination of multiplication by a unimodular matrix and a translation by an integer vector.
It was shown in [10,11] that in dimension 2 a so-called reduced basis computes both ls □ (P ) and ls ∆ (P ), and that in dimension 3 it computes ls □ (P ), but not necessarily ls ∆ (P ).One can then use the generalized basis reduction algorithm, described and analyzed in [10,12,14] to find the lattice size of a lattice polygon, which, as explained in [10,11], outperforms the "onion skins" algorithm of [8,15].See Definition 2.4 and Theorem 2.7 for the definition of a reduced basis and for the precise formulation of the described results from [10] and [11].
One of the questions that we address in this paper is the following: What is the smallest possible nonzero area A(P ) of a lattice polygon P of fixed lattice size ls □ (P ) or ls ∆ (P )?In Theorem 4.1 we prove a sharp bound A(P ) ≥ 1  2 ls ∆ (P ) and describe the lattice polygons on which this bound is attained.Since ls ∆ (P ) ≥ ls □ (P ) it follows that A(P ) ≥ 1  2 ls □ (P ), and we show in Corollary 4.3 that this bound is sharp.
In the last section of the paper we classify inclusion-minimal lattice polygons P with fixed lattice size ls □ (P ), see Theorem 5.4.This classification provides an alternative proof of Corollary 4.3.Note that a classification of inclusion-minimal lattice polygons P with fixed lattice width w(P ) was provided in [7].
In both of the above bounds it is crucial that P is a lattice polygon, since a plane convex body P of fixed lattice size ls ∆ (P ) or ls □ (P ) may have an arbitrarily small area.Hence, in the case of plane convex bodies, it makes sense to look for lower bounds on the area that involve one of the lattice sizes, ls ∆ (P ) or ls □ (P ), together with the lattice width w(P ) of P .In the case of the lattice size with respect to the unit square, such a bound was essentially proved by Fejes-Tóth and Makai in [9], where they showed that for a plane convex body P one has A(P ) ≥ 3  8 w(P ) 2 , and that this bound is attained at conv{(0, 0), w, w 2 , w 2 , w }, where w = w(P ).By a simple rescaling argument we establish in Theorem 3.1 a sharp bound A(P ) ≥ 3  8 w(P ) ls □ (P ).In our main result, Theorem 3.3, we establish a version of the Fejes-Tóth-Makai result where we bound the area in terms of ls ∆ (P ) and w(P ).For a plane convex body P we show that A(P ) ≥ 1  4 w(P ) ls ∆ (P ) and describe convex bodies on which this bound is attained.The idea of inscribing a lattice polygon inside a small multiple of the unit square had appeared in [3,5,6,13], before the lattice size was introduced formally.Both [3] and [5] are devoted to estimating the order of the number of lattice polygons of given area, up to the lattice equivalence, with [5] improving the result of [3].It is shown in one of the steps of the argument in [3] that for any lattice convex polygon P ⊂ R 2 of nonzero area A(P ) there exists its lattice-equivalent copy inside a square of size 36A(P ).In terms of the lattice size this means that ls □ (P ) ≤ 36A(P ).Hence our result in Corollary 4.3 improves Arnold's bound from [3] to a sharp one replacing a constant of 36 with 2.
A similar result is proved in [5, Lemma 3]: For a convex lattice polygon P with nonzero area A(P ) there exists a lattice-equivalent copy of P inside a rectangle [0, w] × [0, h] with wh < 4A(P ).In Theorem 3.1 we improve the bound of Bárány and Pach to a sharp bound A(P ) ≥ 3  8 w(P ) ls □ (P ).Note that our result from Theorem 4.1 can also be reformulated in the spirit of [3,5]: For any lattice polygon P of nonzero area A(P ) there exists a latticeequivalent copy of P contained in 2A(P )∆.

Definitions
Recall that a plane convex body P ⊂ R 2 is a compact convex subset of R 2 with non-empty interior.Given (a, b) ∈ R 2 , the width of P in the direction (a, b) is w (a,b) (P ) = max Consider the Minkowski sum of P with −P , its reflection in the origin, and let K := (P + (−P )) * be the polar dual of the sum.Then K is origin-symmetric and convex and it defines a norm on R 2 by For details see, for example, [4].We then have This, in particular, implies that u → w u (P ) is a convex function on R 2 .Recall that a vector u = (a, b) ∈ Z 2 is called primitive if gcd(a, b) = 1.The lattice width of P , denoted by w(P ), is the minimum of w u (P ) over all primitive directions u.
A lattice polygon is a convex polygon all of whose vertices have integer coordinates.An integer square matrix A is called unimodular if det A = ±1.Two convex bodies in R 2 are called lattice-equivalent if one of them is the image of the other under a map which is a composition of multiplication by a unimodular matrix and a translation by an integer vector.
Definition 2.1.The lattice size ls ∆ (P ) of a convex body P ⊂ R 2 with respect to the standard simplex is the smallest l ≥ 0 such that the l-dilate l∆ contains a lattice-equivalent copy of P .
Definition 2.4.A basis (u 1 , u 2 ) of the integer lattice Z 2 ⊂ R 2 is called reduced with respect to a convex body P ⊂ R 2 if w u 1 (P ) ≤ w u 2 (P ) and w u 1 ±u 2 (P ) ≥ w u 2 (P ).
A fast algorithm for finding a reduced basis with respect to a convex body P ⊂ R 2 was given in [12].It was shown in [10] and [11] that if the standard basis is reduced then one can easily find ls □ (P ) and ls ∆ (P ), as we summarize in Theorem 2.7 below.Definition 2.5.Let P ⊂ R 2 be a plane convex body.We define nls ∆ (P ) to be the smallest l ≥ 0 such that φ(P ) ⊂ l∆ where φ is the composition of a multiplication by a matrix of the form ±1 0 0 ±1 and a translation by an integer vector.Equivalently, if we let l 1 (P ) := max then nls ∆ (P ) is the smallest of the four l i (P ).
Proof.Let (a, b) ∈ Z 2 be a primitive direction.Without loss of generality we can assume that |a| ≥ |b|.We then have and so we get w (a,b) (P ) ≥ (|a| − |b|)h.Hence for |a| > |b| we conclude that w (a,b) (P ) ≥ h.Hence directions (±1, ±1) are the only primitive directions with respect to which the width of P could be less than h.Since no two out of these four directions can be used as rows to form a unimodular matrix, the conclusion follows.□ Definition 2.10.Let P ⊂ R 2 be a lattice polygon and let p ∈ P be one of its vertices.We say that a lattice polygon Q is obtained from P by dropping p if Q is the convex hull of all the lattice points of P , except p.

3.
Lower bounds on the area of a plane convex body in terms of its width and lattice size.
Consider a convex body P ⊂ R 2 .Let w = w(P ) be its width and A(P ) be its area.It was shown in [9] that A(P ) ≥ 3 8 w 2 and that this bound is attained only on the convex bodies that are lattice-equivalent to conv{(0, 0), w, w 2 , w 2 , w }.We formulate a result which is a straight-forward corollary of this bound.Theorem 3.1.Let P ⊂ R 2 be a convex body with h = ls □ (P ) and w = w(P ).Then for the area A(P ) of P we have A(P ) ≥ 3 8 wh.This bound is sharp and attained only on the convex bodies P that are lattice-equivalent to conv{(0, 0), w, w 2 , w 2 , w }.
Remark 3.2.It was shown in Lemma 3 of [5] that for any lattice convex polygon P ⊂ R 2 there exist numbers w, h ≥ 0 with wh ≤ 4A(P ), such that [0, w] × [0, h] contains a latticeequivalent copy of P .Note that Theorem 3.1 strengthens this result replacing the constant of 4 with 8  3 , and also extends the result to plane convex bodies.Proof of Theorem 3.1.Let the standard basis be reduced with respect to P .Then by Theorem 2.7 w (1,0) (P ) = w and w (0,1) (P ) = h.Let P ′ be the image of P under the map (x, y) → (x, w h y).We next check that the standard basis is also reduced with respect to P ′ .Denote by (x 1 , y 1 ) and (x 2 , y 2 ) points in P that, correspondingly, minimize and maximize x + y over P .Then we have w (1,1 Similarly, first reflecting P in the line y = w/2, we conclude that w (1,−1) (P ′ ) ≥ w, so we have checked that the standard basis remains reduced as we pass from P to P ′ .We conclude that w(P ′ ) = w, and hence, as shown in [9], we have A(P ′ ) ≥ 3 8 w 2 and this bound is attained at P ′ which are lattice-equivalent to conv{(0, 0), w, w 2 , w 2 , w }.This implies that A(P ) = h w A(P ′ ) ≥ 3 8 wh and this bound can be attained only if P is latticeequivalent to conv{(0, 0), w, h 2 , w 2 , h }.Since the standard basis is reduced with respect to P , if w ≥ h/2, we have w h, which implies w = 0. We conclude that the bound is attained exactly at P which are lattice-equivalent to conv (0, 0), w, w 2 , w 2 , w .□ Let P be a convex body with l = ls ∆ (P ), h = ls □ (P ), and w = w(P ).We can assume that P ⊂ [0, h] 2 and hence P ⊂ [0, h] 2 ⊂ 2h∆, so we conclude that ls ∆ (P ) ≤ 2 ls □ (P ).This implies that A(P ) ≥ 3  8 wh ≥ 3 16 wl.We now improve this bound to a sharp bound A(P ) ≥ wl 4 .
Theorem 3.3.Let P ⊂ R 2 be a convex body with l = ls ∆ (P ) and w = w(P ).Then for the area A(P ) of P we have A(P ) ≥ wl 4 .This bound is attained only at P which are lattice-equivalent to conv (0, 0), w, w 2 , w 2 , w .Proof.Let the standard basis be reduced with respect to P .Then by Theorem 2.7 we have w = w (1,0) (P ) and h := ls □ (P ) = w (0,1) (P ), so we can assume P ⊂ Π := [0, w] × [0, h].We can also assume that P ⊂ l∆, where l = ls ∆ (P ) and P touches all three sides of l∆.
Suppose first that h = w and hence P ⊂ Π = [0, w] 2 .Pick points p 1 , p 2 , p 3 , and p 4 in P , one on each side of Π, and points q 1 , q 2 , q 3 , and q 4 that maximize and minimize over P the linear functions x + y and x − y, as depicted in the first diagram of Figure 3.Note that some of these eight points may coincide.Let Q = conv{p 1 , p 2 , p 3 , p 4 , q 1 , q 2 , q 3 , q 4 }.
If we move q 4 within Π along the support line x + y = l , the area of the triangle with the vertices p 3 , q 4 , p 4 will be the smallest when q 4 is on y = w or x = w, depending on the slope of the line connecting p 3 and p 4 .Therefore, we can move q 4 to the boundary of Π not increasing the area, and preserving the width and both lattice sizes.Similarly, we move q 1 , q 2 , and q 3 along the corresponding support lines to the boundary of Π.If we end up with the case when there is one of the q i on each side of Π, as in the second diagram of Figure 3, we pass to R = conv{q 1 , q 2 , q 3 , q 4 }, depicted in the third diagram.
Note that passing from P to Q to R we did not increase the area and did not change the minima and the maxima in the directions (1, 0), (0, 1), (1, ±1).Hence the standard basis remains reduced and there is no change in l 1 , l 2 , l 3 and l 4 .Hence R has the same l, w, and h as P , and ls Let q 1 = (0, b), q 2 = (a, 0), q 3 = (w, c), and q 4 = (l − w, w).Note that since the standard basis is reduced we have w where we used 2w ≥ l which holds true since (w, w) is on x + y = l or outside of l∆.In We conclude that in this case the inequality is strict.Each of the q i 's may slide along the support lines in one of the two directions.Due to the symmetry in the line x = y, we can assume that q 4 slides toward y = w, as in Figure 3. Since for each of q 1 , q 2 , q 3 we have have two choices, there are eight cases total, one of which we just covered and will refer to as Case 0. The remaining seven cases are depicted in Figure 4, where we also introduce the notation for the coordinates for some of the q i 's.In each of these cases, we move the q i 's to the boundary of Π not increasing the area, after which we drop (see Definition 2.10) the p i 's on the sides of Π where we now have a q i .We next cover each of these seven cases.
In Case 1, let q 3 = (c, 0) and q 4 = (l − w, w), as depicted in the first diagram in Figure 4. Then we have l 3 (R) = w + c ≥ l, and hence the slope of the line connecting (l − w, w) to (c, 0) is negative.This implies that we can slide p 3 to (w, w − c) and then, unless c = w, drop (c, 0), which reduces this case to Case 0.
In Case 2, l 2 (R) = 2w − a ≥ l implies that the slope of the line connecting (0, a) to (l − w, w) is at least 1, so we can slide (0, b) to (w − b, w) not increasing the area.This reduces Case 2 to Case 6.
We have , provided that l > 3 2 w.Suppose next that l ≤ 3 2 w.We have l 4 (S) = w + b ≥ l and w (1,1) (P ) = l − a ≥ w and hence a ≤ l − w ≤ b.We conclude that T = conv{(0, l − w), (w, 0), (l − w, w)} is contained in S and we get where we used l 3 (S) = 2w ≥ l, so 2w − l ≥ 0. The inequality turns into equality if and only if l = 3 2 w and P is lattice-equivalent to T , that is, to conv (w, 0) , w 2 , w , 0, w
In Case 7, if c ≤ a, we can move p 2 to (w − c, 0) and we are in Case 6.If c > a we move p 2 to (a, 0) and end up in Case 4, which completes the argument.
It remains to explain how the general case reduces to the case when h = w.Let P ′ be the image of P under the map (x, y) → (x, w h y).Then, as we have shown in the proof of Theorem 3.1, the standard basis is also reduced with respect to P ′ .Hence w(P ′ ) = ls □ (P ′ ) = w and ls ∆ (P ′ ) is the smallest of l 1 (P ′ ), l 2 (P ′ ), l 3 (P ′ ), and l 4 (P ′ ).Pick (x 1 , y 1 ) ∈ P that satisfies x 1 + y 1 = l.Then (x 1 , w h y 1 ) ∈ P ′ and we have Similarly, for i = 2, 3, 4 we get l i (P ′ ) ≥ w h l i (P ) ≥ w h l 1 (P ) = w h l and hence we can conclude that ls ∆ (P ′ ) ≥ w h l.
From the above argument we know that A(P ′ ) ≥ ls ∆ (P ′ ) w(P ′ )

4
. Together with ls ∆ (P ′ ) ≥ w h l and A(P ′ ) = w h A(P ) this implies A(P ) ≥ wl 4 .The inequality for P ′ turns into equality if and only if P ′ = conv (0, 0), w, w 2 , w 2 , w .For such P ′ we have , h .
We get A(P ) = 3 8 wh, ls ∆ (P ) = l 1 (P ) = w 2 + h and w(P ) = w.It follows that A(P ) ≥ w(P ) ls ∆ (P ) 4 turns into equality only if 3 8 wh = w 4 w 2 + h , which is equivalent to h = w.□ 4. Proving A(P ) ≥ 1 2 ls ∆ (P ) for lattice polygons P .If P ⊂ R 2 is a lattice polygon with nonzero area then its width is at least 1 and Theorem 3.3 implies that A(P ) ≥ 1  4 ls ∆ (P ).In this section we will improve this bound to a sharp bound A(P ) ≥ 1 2 ls ∆ (P ).We will also observe that this implies that A(P ) ≥ 1 2 ls □ (P ), which is again a sharp bound.Theorem 4.1.Let P ⊂ R 2 be a convex lattice polygon of nonzero area A(P ) and lattice size l = ls ∆ (P ).Then A(P ) ≥ 1 2 ls ∆ (P ).This bound is sharp and is attained exactly at lattice polygons P that are lattice-equivalent to one of the following: } for l = 3.
Remark 4.2.It follows from this theorem that for any lattice polygon P of nonzero area A(P ) there exists a lattice-equivalent copy of P contained in 2A(P )∆.
Proof of Theorem 4.1.Let l = l 1 (P ) = ls ∆ (P ) so that P ⊂ l∆.For l = 1 and 2 the conclusion is clear, so we will assume that l ≥ 3. We first consider the case where P contains one of the vertices of l∆.Note that φ : x y → −1 −1 1 0 x y + l 0 maps l∆ to itself rotating its vertices in the counterclockwise direction.Using map φ together with the reflection in the line y = x we can assume that P contains the origin and point (c, l − c) with l/2 ≤ c ≤ l.Denote I = [(0, 0), (c, l − c)].Let m = max (x,y)∈P x be attained at p ∈ P .Since we have max (x,y)∈P (y − x) ≥ l − m.Let this maximum be attained at q ∈ P .

Figure 7. P contains the origin
Let Q = conv{(0, 0), (c, l − c), p, q}.Since the slope of I is at most 1 we can move p and q to (m, l − m) and (0, l − m) correspondingly not increasing the area of Q, as illustrated This inequality is in fact strict since a = l − 1 would imply b = c = l and since we assumed that a ≥ 2. If b = 1 then 2A(T ) = l − 1 and l 2 (T ) = l − 1, which implies T ⊊ P .Even if (0, 0) ∈ P we would still have l 4 (P ) < l and hence P has to contain a lattice point (x, y) ∈ l∆ such that x ≥ c + 1 or y ≥ l − c + 1.In the first of these two cases the area of conv{T ∪ (x, y)} would be minimal if (x, y) = (c + 1, l − c − 1), and in the second the area would be minimal if (x, y) = (c − 1, l − c + 1).See Figure 9 for an illustration.In both cases we get 2A(P ) ≥ 2(l − 1) > l.
If c = 1 we have l 2 (T ) = l − 1 and hence P contains a point with the x-coordinate equal to at least 2, which implies 2A(P ) ≥ 2(l − 1) > l. □ Corollary 4.3.Let P ⊂ R 2 be a convex lattice polygon with nonzero area A(P ) and ls □ (P ) = h.Then A(P ) ≥ 1 2 ls □ (P ) and this inequality turns into equality if and only if P is lattice-equivalent to conv{(0, 0), (h, 0), (0, 1)}.Remark 4.4.It was shown in [3] that any convex lattice polygon P ⊂ R 2 has a latticeequivalent copy inside a square of size 36A(P ).The corollary strengthens this result replacing the constant of 36 with 2.
Definition 5.1.We say that that a lattice polygon P with ls □ (P ) = h is minimal if there is no lattice polygon P ′ properly contained in P such that ls □ (P ′ ) = h.
In this section we will classify all the minimal lattice polygons P of fixed lattice size ls □ (P ) = h.This classification will provide an alternative proof for Corollary 4.  Proof.The first claim follows from Proposition 2.9.To prove the second claim, suppose first that a + b ≥ h.We need to show that for any lattice polygon P properly contained in T we have ls □ (P ) < h.We have Note that if we drop (a, h) from T to get to P (see Definition 2.10) then w (1,−1) (P ) < h.Also, w (0,1) (P ) < h and hence Also, w (1,1) (P ) ≥ h + a > h and hence the standard basis is reduced and by Theorem 2.7 we have ls □ (P ) = h, so T is not minimal.□ Furthermore, if Q satisfies one of these inequalities it cannot satisfy the other.Also, if Q satisfies one of them, it is lattice-equivalent to a quadrilateral that satisfies the other.Proof.Since P touches all four sides of [0, h] 2 , by Proposition 2.9, we have ls □ (P ) = h.Now assume that one of the two inequalities is satisfied.Then If we drop one of the vertices from Q to get P , we would also have w (1,0) (P ) < h or w (0,1) (P ) < h and hence we would be able to use one of 1 0 1 ±1 , 0 1 1 ±1 and a lattice translation to fit P into a smaller square, so Q is minimal.If neither of the two inequalities holds, we have w (1,1) (Q) ≥ h and w (1,−1) (Q) ≥ h.If both of these inequalities are strict, then after we drop a vertex to get from Q to P we would have w (1,1) (P ) ≥ h and w (1,−1) (P ) ≥ h, so the standard basis would still be reduced.We would also have either w (1,0) (P ) = h or w (0,1) (P ) = h, so we can conclude that ls □ (P ) = h and hence Q is not minimal.
A similar argument works if we have w We would need to drop a vertex that does not change the width in the direction (1, 1) in the first case and in the direction of (1, −1) in the second.
If we have w (1,1) (Q) = h and w (1,−1) (Q) = h then max{a, c} + max{b, d} = h and min{a, b} + min{c, d} = h.Since we have min{a, b} ≤ max{b, d} and min{c, d} ≤ max{a, c} we can conclude min{a, b} = max{b, d} and min{c, d} = max{a, c}.Hence we get a = b = c = d = h/2, and then Q contains a horizontal segment of lattice length h, so Q is not minimal.
Note that Q cannot satisfy both inequalities since then we would get h < min{a, b} + min{c, d} ≤ max{a, c} + max{b, d} < h.
If Q satisfies the second inequality then the transformation (x, y) Proof.We can assume that P ⊂ [0, h] 2 .Suppose first that P has lattice points on all four sides of [0, h] 2 .One way this can happen is when P contains a segment connecting two opposite vertices of the square.Then, since by Example 2.8 we have ls □ (I) = h and P is minimal, we conclude that P is lattice-equivalent to I. Next, P could be a lattice triangle, one of whose vertices is a vertex of [0, h] Suppose next that the standard basis is reduced and hence P ⊂ [0, h] 2 .If P touches all four sides of [0, h] 2 , we are done by the above.If P touches only three sides, we can assume, switching the basis vectors, that w (1,0) (P ) = h, w (0,1) (P ) < h, and also w (1,±1) (P ) ≥ h.
If w (1,1) (P ) = h or w (1,−1) (P ) = h we can use one of 1 0 1 ±1 to reduce to the case of P touching all four sides of h□.Hence we can assume that w (1,±1) (P ) > h.If P contains the entire segment I = [(0, 0), (h, 0)], by the minimality of P and by Example 2.8 we have P = I.Otherwise, we can assume that (h, 0) ̸ ∈ P .Let (a, 0) with a < h be the rightmost point of P in [(0, 0), (h, 0)] and assume that a > 0. We drop (a, 0) to get from P to P ′ , see the first diagram in Figure 12.Since 0 < a < h we have (a, 1) ∈ P and hence the width in the directions (1, ±1) could drop by at most 1.Hence we have w (1,0) (P ′ ) = h, w (1,±1) (P ′ ) ≥ h and we can conclude that ls □ (P ′ ) = h, so P is not minimal.Finally, let a = 0. Suppose that the highest point of P is on the line y = c and let (b, c) be the leftmost point of P on this line.We then drop (b, c) to get from P to P ′ .If c > 1 and b < h then (b, c − 1) ∈ P and, as above, we conclude that P is not minimal.If c = 1 or b = h, with the exception of the case (b, c) = (0, 1), we have P ⊂ conv{(0, 0), (h, 0), (h, h)}, but then w (1,−1) (P ) ≤ w (1,−1) conv{(0, 0), (h, 0), (h, h)} = h.
If (b, c) = (0, 1) then (h, 1) ∈ P and P is not minimal.□ This classification leads to an alternative argument for Corollary 4.3.If P is of nonzero area and contains a lattice segment of lattice length h, then A(P ) ≥ h/2 and in this case this inequality turns into equality if and only if P is lattice-equivalent to conv{(0, 0), (h, 0), (0, 1)}.It remains to show that the strict form of this inequality holds for triangles T and quadrilaterals Q from Theorem 5.4.For triangle T , we get A(T ) = (h 2 − ab)/2 > h/2 since ab < h(h − 1) as a, b ≤ h − 1.

Figure 3 .
Figure 3. Case 0 reduction order for the inequality to be attained we would need w = l/2 and also b = c or l = w + a.Let w = l/2 and b = c.Then we have l 4 (R) = w + b ≥ l and c + w ≤ l since (c, w) ∈ l∆.Hence b = c = l/2, but this contradicts l 3 (R) = 2w − c ≥ l.If w = l/2 and l = w + a, then a = l/2 and this contradicts l 4 (R) = 2w − a ≥ l.