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Malfatti's Problems Malfatti's problems 165 Malfatti's problems Emil Kostadinov Mathematics High School “Acad. S. Korolov”, 11 grade Blagoevgrad, Bulgaria Abstract The purpose of the present project is to consider the original Malfatti problem of cutting three circles of maximal combined area from a given triangle as well as various related problems of this type. The project is divided into five sections. In Section 1 some useful formulae for the radii of the Malfatti circles of a triangle are given. In Section 2 a trigono- metric inequality (Zalgaller's inequality) for the angles of an acute triangle is proved. In particular, it is shown that the Malfatti circles never give a solution of the Malfatti problem. In Section 3 we solve the Malfatti problems for equilateral triangles and squares by using an approach different from the one proposed by Zalgaller and Loss. It is based on the so-called dual Malfatti problems. In Section 4 the possibility for solving Malfatti type problems by means of the greedy algorithm is discussed. In the last section of the project we state some open problems that might be objects of future investigations. 1 Introduction In 1803 the Italian mathematician Gianfrancesco Malfatti [3] posed the following prob- lem: Given a right triangular prism of any sort of material, such as marble, how shall three circular cylinders of the same height as the prism and of the greatest possible volume be related to one another in the prism and leave over the least possible amount of material? It is clear that this problem is equivalent to the plane problem of cutting three circles from a given triangle so that the sum of their areas is maximized. As noted in [1], Malfatti, and many others who considered the problem, assumed that the solution would be the three circles that are tangent to each other, while each circle is tangent to two 166 Appendix to Chapter 6 Fig. 1 sides of the triangle (Fig. 1). These circles have became known in the literature as the Malfatti circles, and we refer the reader to [2] and [5] for some historical remarks on the derivation of their radii. In 1929, H. Lob and H. W. Richmond [2] gave a very simple counterexample to the Malfatti conjecture. Namely, they noted that in the case of an equilateral triangle the sum of the areas of the incircle and the two circles tangent to it and inscribed in two angles of the triangle is greater than the sum of the areas of the Malfatti circles. Moreover, M. Goldberg [1] proved in 1967 that the Malfatti circles never give a solution of the Malfatti problem. To the best of the author's knowledge the Malfatti problem was first solved by V. Zal- galler and G. Loss [6] in 1991. They proved that for any triangle ABC with angles \ A \ B \ C the solution of the Malfatti problem is given by the three circles ≤ ≤ k1, k2 and k3, where k1 is the incircle of the triangle, k2 is the circle inscribed in \A and externally tangent to k1, and k3 is either the circle inscribed in \A and externally tangent to k2 (Fig. 2) or the circle inscribed in \B and externally tangent to k1 (Fig. 3), A B depending on whether sin tg . 2 ≥ 4 Fig. 2 Fig. 3 The proof of the above result given by Zalgaller and Loss is very long (more than 25 pages) and at some key points it uses computer computations. In short the proof goes as follows. The first step is to show that it is enough to consider the following 14 configurations of the three circles: Malfatti's problems 167 168 Appendix to Chapter 6 Then configurations 3 – 14 are excluded by using case-by-case arguments showing that the respective three circles have less combined area than those of configurations 1 or 2. The purpose of this project is two-fold. Firstly, we discuss in details one of the main difficulties in the Zalgaller and Loss solution of the Malfatti problem, namely the exclusion of configuration 6. Note that this is equivalent to prove that the Malfatti circles never give a solution of the Malfatti problem. Secondly, we give new solutions of the Malfatti problems for an equilateral triangle or a square by using the so-called dual Malfatti problems. At the end of the project we discuss the possibility of applying the so-called greedy algorithm for solving Malfatti's type problems and state some open related problems. The project is organized as follows. In Section 2 the well-known formulae [2] for the radii of the Malfatti circles of a triangle are derived. In Section 3 a trigonometric inequality for acute-angled triangles, due to Zallgarer [6], is proved. It implies that the Malfatti circles never give a solution of the Malfatti problem. Section 4 is devoted to some problems which in a sense are dual to the Malfatti problems for two circles. More precisely, we find explicit formulae for the side-lengths of the smallest equilateral tri- angle and square containing two non-intersecting circles of given radii. These formulae are used to solve the Malfatti problem for an equilateral triangle or a square. In Section 5 we consider some problems of Malfatti's type and show that they cannot be solved by using the so-called greedy algorithm. In the last section of the project we state some open problems that might be objects of future investigations. 2 Malfatti's circles It is well-known that for any triangle there exist three circles such that each of them is tangent to the other two and to two sides of the triangle. These circles are uniquely determined and are known in the literature as Malfatti's circles. In this section we shall derive some useful formulae for the radii of the Malfatti circles of a triangle. Theorem 2.1 [2]. Let ABC be a triangle with angles α, β, γ and inradius r and let rA, rB and rC be the radii of the Malfatti circles inscribed respectively in the angles A, B and C. Then Malfatti's problems 169 β γ 1 + tg 1 + tg r 4 4 rA = α 2 1 + tg 4 α γ r 1 + tg 1 + tg r = 4 4 (1) B 2 β 1 + tg 4 α β 1 + tg 1 + tg r 4 4 rC = γ : 2 1 + tg 4 Proof. Let OA and OB be the centres of the Malfatti circles inscribed in angles A and B, and let L and M be their tangent points with the side AB (Fig. 4). Then α β AL = r ctg ; MB = r ctg ; A 2 B 2 2 2 LM = (rA + rB) (rA rB) = 2prArB: − − p Fig. 4 α β On the other hand AB = r ctg + ctg and we get the identity 2 2 α β α β r ctg + r ctg + 2pr r = r ctg + ctg : A 2 B 2 A B 2 2 The same reasoning shows that the radii rA; rB and rC of the Malfatti circles are solu- 170 Appendix to Chapter 6 tions of the following system: α β α β r ctg + r ctg + 2pr r = r ctg + ctg A 2 B 2 A B 2 2 β γ β γ r ctg + r ctg + 2pr r = r ctg + ctg (2) B 2 C 2 B C 2 2 γ α γ α r ctg + r ctg + 2pr r = r ctg + ctg : C 2 A 2 C A 2 2 Now we shall check that rA, rB and rC given by formulae (1) satisfy the above system. The uniqueness of the solutions of (2) is proved in [2]. α β γ Set tg = x, tg = y and tg = z. Then 4 4 4 2 α 2 α 1 tg 4 1 x ctg = − α = − 2 2tg 4 2x β 1 tg2 β 1 y2 ctg = − 4 = − 2 β 2y 2tg 4 2 γ 2 γ 1 tg 4 1 z ctg = − γ = − : 2 2tg 4 2z Hence plugging the expressions for rA, rB and rC from (1) in the first identity of (2) we see that we have to prove the identity (1 + y)(1 + z)(1 x2) (1 + x)(1 + z)(1 y2) 1 x2 1 y2 − + − + 1 + z = − + − : 2(1 + x)x 2(1 + y)y 2x 2y Simple algebraic manipulations show that the above identity is equivalent to the identity x + y + z + xy + yz + zx xyz = 1: (3) − α β γ π On the other hand tg + + = tg = 1 and using the formula 4 4 4 4 α β γ α β γ tg + tg ( + ) tg + + = 4 4 4 4 4 4 α β γ 1 tg tg + − 4 4 4 α β γ α β γ tg + tg + tg tg tg tg = 4 4 4 − 4 4 4 α β β γ γ α 1 tg tg tg tg tg tg − 4 4 − 4 4 − 4 4 Malfatti's problems 171 we get α β γ α β β γ α γ α β γ tg + tg + tg + tg tg + tg tg + tg tg tg tg tg = 1: 4 4 4 4 4 4 4 4 4 − 4 4 4 Hence the identity (3) holds true and the Theorem is proved. 3 An inequality for acute triangles In this section we shall prove an inequality of Zalgaller [6] for the angles of an acute triangle which implies that the Malfatti circles never give a solution of the Malfatti prob- lem. In the exposition below we follow [6] and [7].
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