Parallel Packing Squares into a Triangle

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Introduction
Let P be a polygon and let S i be a square for i = 1, 2, . . . One side of P is called the base of P . A collection S 1 , S 2 , . . . is said to be packed into P if their union is contained in P and if these squares have pairwise disjoint interiors. A packing is called parallel if a side of each packed square is parallel to the base of P .
The goal is to pack the squares into P with high density. The area of a polygon R is denoted by |R|. Let (P ) be the greatest number such that any collection of squares of the total area not greater than (P )·|P | can be parallel packed into P .
Assume that S is a square and that T h is a triangle with the interior angles at the base of the measure not greater than 90 • , with the base length 1 and the height h. There are many results concerning packing squares or rectangles. Moon and Moser [11] showed that (S) = 1/2; in this case the most effective usual packing is the parallel packing. In many publications in which squares are placed into a rectangle, when authors write "packing" they mean "parallel packing" (see for example [1,2,5,[8][9][10]12]). Let us add that for packing squares or rectangles of the side lengths not greater than 1 into a large square the most effective packings are not parallel (see [3,7]).
Without loss of generality one can assume that α ≥ β (see Fig. 1).  Figure 2. Two congruent squares in

Upper
Proof. If the area of a square equals It is easy to verify that h ≤ 1/2.

Corollary 3. For any
Note that the worst case appears when one square is packed: one square of the area greater than h |T h | cannot be packed into By Corollary 2 we obtain (T h ) ≤ h . Now the worst case appears when two squares are packed: two congruent squares of the total area greater than h |T h | cannot be packed either along the base of T h or vertically. However, it is possible to pack two congruent squares of the total area equal to h |T h | horizontally (see Fig. 2).
By Corollary 2 we get (T h ) ≤ h . In that case the worst case appears when two squares are packed: two congruent squares of the total area greater than h |T h | cannot be packed either along the base of T h or vertically, but it is possible to pack two congruent squares of the total area equal to h |T h | vertically (see Fig. 3).
In Sect. 5 we will show that (T h ) ≥ h .   Let S 1 and S 2 be the squares of the side lengths a 1 , a 2 , respectively and let

Lemmas
First we show that if S 1 and S 2 cannot be packed along the base of T h , then a 2 1 + a 2 2 > 2h 2 (2h+1) 2 . If S 2 cannot be parallel packed into T h "horizontally" as on Fig. 4, left, then a 1 cot α + a 1 + a 2 + a 2 cot β > 1. Since it follows that a 1 + a 2 > 2h 2h+1 , i.e., a 2 > 2h 2h+1 − a 1 . Thus In that case we show that S 1 and S 2 can be packed into T h "vertically" as on Fig. 4 1 . The function f (a 1 ) = a 1 + p − a 2 1 has a local maximum at a 1 = p/2. This means that a 1 + a 2 ≤ p/2 + p/2 = √ 2p. Since a 1 + a 2 = √ 2p for a 1 = a 2 , it follows that if two congruent squares of the total area p can be parallel packed into T h , then any two squares of the total area p can be packed into T h . By Corollary 2 two congruent squares of the total area greater than 2h 2 /(2h + 1) 2 and not greater than 2h 2 /(h + 2) 2 can be parallel packed into T h "vertically".
Let sT h be the image of T h after a homothety with the ratio s. Clearly, sT h is a triangle of the base length s and the height sh. Since affine transformations preserve relative areas, by Claim 1 and 3 we get the following result.

Corollary 4.
Any square of the area not greater than h · |sT h | can be parallel packed into sT h . Moreover, two squares of the sum of the areas not greater than h · |sT h | can be parallel packed into sT h .
The following two inequalities will be used in the proof of Lemma 3.
Proof. Assume that h > 0 and that x ≤ 1. We will show that .
Observe that Proof. We proceed as in the proof of Lemma 1. Observe that α β m 2 = 1, k = 3 and n k = 1 Figure 6. Division of L 1

Packing Squares into Trapezoids
In the main packing method squares will be packed in trapezoid-shape layers.
In this section an algorithm for packing squares into trapezoids is described. 2 a 1 , then L 1 is partitioned into smaller trapezoids. Let m 2 be an integer such that 2 −m2−1 a 1 < a 2 ≤ 2 −m2 a 1 . Then L 1 is divided into 2 m2 trapezoids L + 1,1 , . . . , L + 1,2 m 2 of the height h 2 = 2 −m2 a 1 (see Fig. 6). The square S 2 is packed into L + 1,1 as far to the left as possible, i.e., in the left-bottom corner. New trapezoids are defined as follows: L 2,1 is the part of L + 1,1 lying to the right of S 2 and L 2,i = L + 1,i for i = 2, 3, . . . , 2 m2 (it is possible that L 2,1 is a triangle). -Assume that squares S 1 , . . . , S k−1 were packed and that trapezoids L k−1,i for i = 1, 2, . . . , 2 m k−1 (of the height h k−1 ) were defined. Next square S k is placed in the following way. If a k > 1 2 h k−1 , then let j be the smallest integer such that b(L k−1,j ) ≥ a k (1 + cot β) (we find the lowest lying trapezoid into which S k can be packed). Then S k is packed into L k−1,j in the left-bottom corner. New trapezoids are defined: L k,j is the part of L k−1,j lying to the right of S k and L k,i = L k−1,i for i = j (it is possible that some trapezoids are triangles). For example, squares S 2 , S 3 and S 4 are packed on Fig. 8 (left) by this rule. Moreover, squares S 4 , . . . , S 10 are packed by this rule on Fig.  9, left. If a k ≤ 1 2 h k−1 , then each trapezoid L k−1,i is partitioned into smaller trapezoids. Let n k be an integer such that 2 Fig. 6, right). Let j be the smallest integer such that b(L + k−1,j ) ≥ a k (1 + cot β) (we find the lowest lying trapezoid into which S k can be packed). The square S k is packed into L + k−1,j in the left-bottom corner. New trapezoids are defined: L k,j is the part of L + k−1,j lying to the right of S k and L k,i = L + k−1,i for i = j and i ∈ {1, 2, . . . , 2 m k }, where m k = m k−1 n k (it is possible that some trapezoids are triangles). For example, squares S 3 and S 9 are packed on Fig. 9 (right) by this rule.  Fig. 5, left). Moreover assume that a 1 (1 + cot α) + a 2 (1 + cot β) ≤ b(L). If S z is the first square that cannot be packed into L by the L-method, then |S 1 |+. . .+|S z−1 | > h · |L|. Proof. Since any homothety preserves relative areas, without loss of generality one can assume that b(L) = 1.
Consider four cases. Fig. 7) Let x = a z /a 1 . Since a z cannot be packed into L, it follows that a 1 cot α+ a 1 + a 2 + a z + a z cot β > 1. Recall that cot α + cot β = 1/h and cot α ≤ 1 2h . By 1 < a 1 cot α + a 1 + a 2 + a z + a z cot β = a 1 cot α + a 1 we get Denote by (L) the ratio of the sum of the areas of squares packed into L to the area of L. Only two squares are packed in L, therefore Since the function f (a 1 ) = 2ha . − a z−1 , with the height a 1 and the base angles with the same measures as the base angles of L (see Fig. 8, where z = 5). Moreover, let sL − be the image of L − after a homothety with the ratio s such that the larger base of sL − is equal to 1. By Case 1, the density of packing squares sS 1 and sS 2 into sL − is greater than h . This implies that the density of packing squares S 1 and S 2 into L − is greater than h . The density of packing squares S 3 , . . . , S z−1 into the rectangle R L = (a 3 +. . .+a z−1 )×a 1 is greater than 1/2 ≥ h . Clearly, |L − |+|R L | = |L|. As a consequence, the density of packing squares S 1 , S 2 , . . . , S z−1 into L is greater than h .
Let k be the smallest integer such that a k < a 1 /2 (see Fig. 9, left). Obviously, k ≤ z. Let L k−1,1 be the part of L 1 lying to the right of S k−1 , i.e., either the trapezoid of the height h r = h(L k−1,1 ), of the base length b r = b(L k−1,i ) and with the base angles measuring 90 • and β or the triangle and L + k−1,2 with disjoint interiors, where L k−1,1 = L + k−1,1 ∪ L + k−1,2 (it is possible that trapezoids are triangles).
Assume that S v is the first square that cannot be packed into L + k−1,1 . Clearly a v ≤ h r /2 and a z ≤ h r /2 (see Fig. 10, right).
Observe that (see Fig. 10, right) Let  a 1 h r ). Obviously, p 1 is equal to the area of the gray rectangle and ya 1 is equal to the base length of the gray rectangle on Fig. 10, left. Since and a z ≤ a v ≤ h r /2, we get p 2 > p 1 . Hence, to show that First assume that k = 2. We argue as in Case 1. By conditions a 1 cot α + a 1 + ya 1 + 1 2 a 1 + 1 2 a 1 cot β = 1, cot α + cot β = 1/h and cot α ≤ 1/(2h), we get Consequently,  Figure 11. Subcase 3C Now we estimate the packing density: By Lemma 2, a 2 1 + p 1 > |L| · 2h · h /(2h) = h · |L|. If k > 2, then we argue as in Case 2.
This assumption implies that a square of the side length greater than h r /2 cannot be packed in L k−1,1 . By Case 1 and Case 2, The total area of squares packed in two trapezoids contained in L k−1,1 (see Fig. 11, right) is not smaller than In this subcase, similarly as in Subcase 3a, it suffices to check that 1 is equal to the area of the gray rectangle on Fig. 11, left). Observe that Fig. 9, right). We proceed in the same way as in Case 3. The sum of the areas of squares packed into two trapezoids L + l,i , L + l,i+1 contained in the trapezoid L + l,i ∪L + l,i+1 is greater than the area of the corresponding "gray" rectangle of the height h(L + l,i ) and the base length b(L + l,i ) − h(L + l,i ) · (1 + cot β) contained in L + l,i ∪ L + l,i+1 .

Main Result
Theorem 1. Any collection of squares of the total area not greater than h ·|T h | can be parallel packed into T h .
Proof. Let S 1 , S 2 , . . . be a collection of squares of the total area not greater than h · |T h |. There is no loss of generality in assuming that a 1 ≥ a 2 ≥ . . ., where a i denotes the side length of S i . Squares will be packed into trapezoid-shape layers in such a way that the packing density in each layer, i.e., the ratio of the sum of packed squares to the area of the layer, is greater than h .
If a 1 (1 + cot α) + a 2 (1 + cot β) ≤ 1 (see Fig. 13), then let L 1 be the trapezoid contained in T h with the base length 1, the height a 1 and the base angles measuring α and β. This trapezoid is called the basic layer. Squares S 1 , S 2 , . . . are packed into L 1 by the L-method. Denote by S n1 the first square that cannot be packed into L 1 (n 1 = 3 on Fig. 13). By Lemma 3, the total area of squares packed into L 1 is not smaller than h · |L 1 |.
If a 1 (1 + cot α) + a 2 (1 + cot β) > 1 and at the same time a 1 + a 2 ≤ 2h 2+h (this case is possible for h > 1), then we pack S 1 and S 2 "vertically". The first condition implies that 1 < (a 1 + a 2 )(2h + 1)/(2h) (see the proof of Claim 3). Let λ = a 1 + a 2 . The so-called double layer L 1 of the height λ is created for packing only two squares, see Fig. 12. In that case n 1 = 3 (S n1 is the first square that is not packed into L 1 ) and λ > 2h/(2h + 1). It is easy to verify that a 2 1 +a 2 2 ≥ (λ/2) 2 +(λ/2) 2 = λ 2 /2. Recall that h ≤ 1/2 while the packing density in the double layer is not smaller than 1/2: Since the sum of the areas of the squares packed into L 1 is greater than h · |L 1 |, it follows that the total area of the remaining squares is smaller than h · |T 1 |, where T 1 = T h \ L 1 .
For packing squares S n1 , S n1+1 , . . . the new layer L 2 of the base length 1 − a 1 cot α − a 1 cot β, with the base angles measuring α and β and with the  Figure 12. a 1 + a 2 > 2h/(2h + 1) Figure 13. Packing method height either a n1 or a n1 +a n1+1 is created directly above L 1 (see Fig. 13, where L 2 is a double layer). If a n1 (1 + cot α) + a n1+1 (1 + cot β) ≤ b(L 2 ), then we place S n1 , S n1+1 , . . . into the basic layer L 2 of the height a n1 by the L-method (denote by S n2 the first square that cannot be packed into L 2 ). Otherwise, we pack S n1 and S n1+1 into the double layer of the height a n1 + a n1+1 and we take n 2 = n 1 + 2. Assume that the layers L 1 , . . . , L p were created and that S np is the first square that cannot be packed into L p . The total area of squares packed into each L i (i = 1, . . . , p) is greater than p ·|L i |. This implies that a 2 np+1 +a 2 np+2 + . . . is smaller than h times the area of T p = T h \ p i=1 L i . By Corollary 4, S np can be packed into T p . Moreover by Claim 3, S np and S np+1 can be packed into T p provided z ≥ n p + 1. There is an empty space in T h to create either the basic or the double layer to pack S np and S np+1 . This means that all the squares can be packed into T h .