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A square from similar rectangles Sergei Dorichenko, Mikhail Skopenkov, Olesya Malinovskaya Problem. When is it possible to tile a square by rectangles similar to a given one? Leading questions — I have a thought! — said Boa and opened his eyes. — A thought. And I think it. — Which thought? — asked Monkey. — It takes time to explain... — Wow! — jumped Monkey. — What a good thought! May I also think it a bit? G. Oster, “Boa’s grandmother” 1. Form a square from rectangles m × n, where m and n are integers. 2. A designer was offered to make square window frames. In Figures A,B it is shown how the panes should be adjacent to each other and how they should be oriented (with the short side or with the long one to the top). Can all panes in each frame be similar rectangles? 3. Is it possible to dissect a square into 3 similar, but not equal rectangles? 4. Is it possible to dissect a square into 5 squares? 5. All shelves in Figure C, and all scraps, from which consists the piecework in Figure D, are squares. Are the board and the piecework also squares? 6. Is it possible to tile the plane by pairwise distinct squares with integer sides? A B C D √ 7. Is it possible to dissect a square into rectangles of side ratio 2 + 2? The same question √ √ √ is for 2 − 2, for 3 + 2 2, and for 3 − 2 2. √ √ 8. Is the number 1 + 2 the square of a number of the form a + b 2, where a and b are rational? A tiling by rectangles is drawn on a rectangular sheet of paper. One may cut the sheet of paper along any straight line segment into two rectangles, then repeat the same operation separately for each of the obtained parts, and so forth. If it is possible to represent the initial tiling in this way, then the tiling is called simple. For instance, the tilings shown in Figure A,B are simple, but those in Figure С,D are difficult. It is suggested that you solve the following 4 problems for simple tilings first and for arbitrary tilings afterwards. 9. Which rectangles can be (simply) tiled by rectangles with one of the sides equal to 1? 10. Which rectangles may be (simply) dissected into squares? √ 11. Is it possible to√dissect a square (simply) into rectangles of side ratio 2? The same √ √ 3 3 question is for 1 + 2, for 1 + 2, and for 2. √ 12. All numbers, which can be represented in the form x = a + b 2 with rational a, b, are called good. For which good x a square may be (simply) tiled by rectangles of side ratio x? Summer Conference of the Tournament of Towns 2014 A square from similar rectangles Sergei Dorichenko, Mikhail Skopenkov, Olesya Malinovskaya The Freiling–Laczkovich–Rinne–Szekeres Theorem (1994). For r > 0 the following 3 conditions are equivalent: (1) a square can be tiled by rectangles of side ratio r; (2) for some positive rational numbers c1 , . . . , cn we have 1 c1 r + = 1; 1 c2 r + 1 c3 r + · · · + cn r (3) the number r is a root of a nonzero polynomial with integer coefficients such that all the complex roots of the polynomial have positive real parts. From dissections to systems of linear equations. You road I enter upon and look around, I believe you are not all that is here, I believe that much unseen is also here. Walt Whitman “Song of the Open Road”. 13. Prove the implication 2) =⇒ 1) in the Freiling–Laczkovich–Rinne–Szekeres Theorem. 14. Prove that a good number√x (for the definition, see Problem 12) can be uniquely √ represented in the form x = a + b 2, where a and b are rational. The number x̄ = a − b 2 is called conjugate to x. Prove that if x and y are good then x + y, x − y, x · y, x/y are also good. How are conjugates to these numbers expressed through x̄ and ȳ? 15. A rectangle 1 × x is dissected into squares x1 × x1 , x2 × x2 , . . . , xn × xn . Prove that numbers x, x1 , . . . , xn satisfy a system of linear equations with integer coefficients. Does the following equality of areas follow from your system of equations: x = x21 + · · · + x2n ? Try to find a system of equations implying this equality. 16. Suppose that a system of linear equations in variables x1 , . . . , xn has a unique solution. Prove that x1 , . . . , xn can be expressed in the coefficients of the equations with the help of operations of addition, subtraction, multiplication and division. In particular, show that if all the coefficients are rational, then x1 , . . . , xn are rational. 17. Alice tiled a rectangle A with 7 squares A1 , . . . , A7 and a rectangle B with 7 squares B1 , . . . , B7 . It is known that the sides of all the squares are horizontal or vertical. The horizontal sides of rectangles A and B are equal. It turned out that for every pair of indices i, j: • if Ai is adjacent on the left to Aj by a part of the vertical edge, then Bi is adjacent on the left to Bj by a part of the vertical edge; • if Ai is adjacent from above to Aj by a part of the horizontal edge, then Bi is adjacent from above to Bj by a part of the horizontal edge. Prove that for every i = 1, . . . , 7 the sides of the squares Ai and Bi are equal. 18. The Dehn theorem. A rectangle can be tiled by squares (not necessarily equal) if and only if the ratio of two orthogonal sides of the rectangle is rational. Summer Conference of the Tournament of Towns 2014 From dissections to the roots of polynomials Here is the test of wisdom, Wisdom is not finally tested in schools. . . Walt Whitman "Song of the Open Road". 19. A rectangle is dissected into several rectangles of side ratio r. Prove that the sides of the initial rectangle are as P (r) : Q(r), where P (x) and Q(x) are polynomials with integer coefficients. 20. We can choose these polynomials so that P (−x)/Q(−x) ≡ −P (x)/Q(x). P (z) > 0. 21. If x > 0, then P (x)/Q(x) > 0. If Re z > 0, then Re Q(z) 22. A square is dissected into several rectangles of side ratio r. Prove that r is a root of a nonzero polynomial with integer coefficients. 23. Prove the implication 1) =⇒ 3) in the Freiling–Laczkovich–Rinne–Szekeres Theorem. 24. The Euclid Algorithm. Given a rectangle of side ratio r. Let us cut off it a square by one straight-line cut. Then repeat the procedure with the remained rectangle, and so on. Show that at some step we obtain a square if and only if r is rational. 25. The Euler-Lagrange Theorem. Prove that in this process two similar rectangles appear if and only if r is irrational but is a root of a quadratic trinomial with integer coefficients. 26. The Foster-Cauer Reactance Theorem. Let R(z) = P (z)/Q(z) be a ratio of polynomials with integer coefficients such that the degree of P is greater than the degree of Q and R(−z) = −R(z) for all complex z. Then the following 5 conditions are equivalent: (1) (2) (3) (4) if Re z > 0, then Re R(z) > 0; if R(z) = 1, then Re z > 0; if R(z) = 0, then Re z = 0 and R0 (z) > 0; for some integer n ≥ 0 and real d1 > 0, a1 > b1 > a2 > · · · > bn ≥ 0 R(z) = d1 z n Y z 2 + a2 k=1 z2 + k , 2 bk (5) for some integer m ≥ 1 and rational d1 , . . . , dm > 0 R(z) = d1 z + 1 d2 z + · · · + 1 . dm z 27. Prove the implication 3) =⇒ 2) in the Freiling–Laczkovich–Rinne–Szekeres Theorem. 28. If a square can be dissected into rectangles similar to a given one, then it can be dissected simply (see definition in page 1). 29. Given a polynomial with integer coefficients and its root r, invent an algorithm of dissecting a square into rectangles of side ratio r (if such a dissection exists). 30. The Yu–Chen Problem. Consider the following operation on a set of real numbers. ab bc ca Take any three elements a, b, c and add the numbers 1/a, a + b, a+b+c , a+b+c , a+b+c to the set. Then the process is repeated. Prove that if a rectangle of side ratio k can be tiled by rectangles of side ratio r then k can be obtained from r by these operations. 3 A square from similar rectangles Sergei Dorichenko, Mikhail Skopenkov, Olesya Malinovskaya Question. When a cube can be dissected into parallelepipeds similar to a given one? What is next? Faust Where will we go, then? Mephistopheles Where you please. The little world, and then the great, we’ll see. With what profit and delight, This term, you’ll be a parasite! Goethe. “Faust” Nobody knows the answer to the Question about the Cube raised above. For the following Theorems, we do not know simple elementary proofs. We are interested even in partial progress. 31. The Prasolov-Skopenkov Theorem (2011) For r > 0 the following three conditions are equivalent: 1) A rectangle of side ratio r can be dissected into rectangles of side ratio r in such a way that not all of the dissection rectangles are homothetic to the initial rectangle. 2) For some positive rational numbers ci the following equality is held 1 1 = . c1 r + r 1 c2 r + 1 c3 r + · · · + cn r 3) The number r2 is a root of a nonzero polynomial with integer coefficients, such that its number of negative roots is 1 less than its degree. 32. The Laczkovich–Szegedy Theorem (1999). A square can be dissected into triangles similar to a given one if and only if the given triangle has either angles (π/8, π/4, 5π/8), or (π/4, π/3, 5π/12), or (π/12, π/4, 2π/3), or is right-angled with the ratio of cathetus equal to a root of some polynomial with integer coefficients such that all its real roots are positive. 33. The Su–Ding Theorem (2005). Suppose that all the vertices of a polygon (not necessarily convex) have rational coordinates, and all its sides are parallel to the coordinate axes. Then it can be dissected into rectangles of side ratio r if and only if a square can be dissected into rectangles of side ratio r. 34. The Kenyon Theorem (1998) Let A1 A2 A3 A4 A5 A6 be an L-shape hexagon, and its vertices are enumerated starting from the vertex of a non-convex angle (see Fig. E). Then the hexagon A1 A2 A3 A4 A5 A6 can be dissected into squares if and only if the system of linear ( equations A3 A4 · x + A1 A2 · y = A2 A3 , A3 A2 A1 A5 A6 · z − A1 A2 · y A6 = A6 A1 ; has a solution x, y, z with nonnegative rational x, y, z. 35. The Kenyon Theorem (1998) A trapezoid is dissected into trapezoids. The bases of all trapezoids of the dissections are parallel, and the ratio of the midline and the altitude is rational for every trapezoid of the dissection. Then for the initial trapezoid the ratio of the midline and the altitude is also rational. A4 A5 E Summer Conference of the Tournament of Towns 2014 A square from similar rectangles Sergei Dorichenko, Mikhail Skopenkov, Olesya Malinovskaya Where to find solutions 1. Where to find solutions Solution of most problems are available in the text written by the Taiwan participants of the conference Hung–Hsun Yu and Brian Chen. An elementary proof of the main Freiling– Laczkovich–Rinne–Szekeres theorem is given in [7]. Acknowledgements Many problems are taken from the K. Kokhas collection and the papers [9, 2]. The authors are grateful to K. Kuyumzhiyan for help with the translation and to G. Chelnokov and A. Zaslavsky for help with solution checking. References [1] R. L. Brooks, C. A. B. Smith, A. H. Stone, and W. T. Tutte, The dissection of rectangles into squares, Duke Math. J. 7 (1940), 312–340. [2] S. Dorichenko, M. Skopenkov, A square from similar rectangles, in preparation (in Russian). С. Дориченко, М. Скопенков, Квадрат из подобных прямоугольников, препринт, http://arxiv.org/abs/1305.2598. [3] M. Gardner, Squaring the square. In: The 2nd Scientific American Book of Mathematical Puzzles and Diversions, University of Chicago Press, 1987, 256 p. М. Гарднер, Квадрирование квадрата. В: Математические головоломки и развлечения, М.:Мир, 1999. [4] C. Freiling and D. Rinne, Tiling a square with similar rectangles, Math. Res. Lett. 1 (1994), 547–558. [5] M. Laczkovich and G. Szekeres, Tiling of the square with similar rectangles, Discr. Comp. Geom. 13 (1995), 569–572. [6] Jiří Matoušek, Thirty-three Miniatures: Mathematical and Algorithmic Applications of Linear Algebra, Amer. Math. Soc., 2010, 182 p. [7] M. Prasolov and M. Skopenkov, Tilings by rectangles and alternating current, J. Combin. Theory A 118:3 (2011), 920–937, http://arxiv.org/abs/1002.1356. [8] M. Prasolov, M. Skopenkov and B. Frenkin, Invariants of polygons, 19-th summer conference International mathematical Tournament of towns, Belarus, Minsk, 2007. М. Прасолов, М. Скопенков и Б. Френкин, Инварианты многоугольников, 19-я летняя конференция международного математического Турнира городов, Беларусь, Минск, 2007. http://www.turgor.ru/lktg/2007/1/index.php Summer Conference of the Tournament of Towns 2014 [9] M. Skopenkov, M. Prasolov, S. Dorichenko, Dissections of a metal rectangle, Kvant 3 (2011), 10-16 (in Russian). М. Скопенков, М. Прасолов, С. Дориченко, Разрезания металлического прямоугольниа, Квант 3 (2011), 10-16.http://arxiv.org/abs/ 1011.3180 [10] M. Skopenkov, V. Smykalov, A. Ustinov, Random walks and electric networks, Mat. Prosv. 3rd ser. 16 (2012), 25-47 (in Russian). М. Скопенков, В. Смыкалов, А. Устинов, Случайные блуждания и электрические цепи, Мат. Просв., 3-я серия 16 (2012), 2547. http://www.mccme.ru/free-books/matprosh.html [11] I.M. Yaglom, How to Dissect a Square?, Math. Bibl., Nauka, Moscow, 1968, 112 p. (in Russian). И.М. Яглом, Как разрезать квадрат? Мат. Библ., М.: Наука, 1968, 112 стр. http://ilib.mirror1.mccme.ru/djvu/yaglom/square.htm [12] S. Wagon, Fourteen proofs of a result about tiling a rectangle, Amer. Math. Monthly 94:7 (1987). 6