[...]... , B1, C1 correspond to vertices A, B, C, respectively) and circumscribed about triangle A0 B0 C0 (where A0 lies on BC, B0 on C A, and C0 on AB) Of all such possible triangles, determine the one with maximum area, and construct it 1.2 Employing Algebraic Inequalities A large variety of geometric problems on maxima and minima can be solved by using appropriate algebraic inequalities Conversely, many... Chapter 1 Methods for Finding Geometric Extrema 24 and therefore P ≥ 9+ 9 2 = 27 2 One checks easily that P = 27 2 if and only if x = y = z = showing that the perimeter of X Y Z is minimal precisely when X , Y , and Z are the midpoints of the corresponding edges of the cube ♠ 1 , 2 As we mentioned earlier in this section, when solving geometric problems on maxima and minima by means of algebraic inequalities... N , and P on the sides BC, C A, and AB, respectively, such that the perimeter of M N P is minimal First, we consider a simpler version of this problem Fix an arbitrary point P on AB We are now going to find points M and N on BC and C A, respectively, such that M N P has minimal perimeter (This minimum of course will depend on the choice of P.) Let P be the reflection of the point P in the line BC and. .. an angle X OY and two points A and B interior to it, find points C and D on O X and OY , respectively, such that the length of the broken line AC D B is a minimum 1.1.18 Given an angle X OY and a point A on O X , find points M and N on OY and O X , respectively, such that the sum AM + M N is a minimum 1.1.19 There are given an angle with vertex A and a point P interior to it Show how to construct a line... k1 and k2 Assume that the total length of the path is t and it has a common point M with k2 first and then a common point N with k1 Denote by D the common point of k2 and the altitude through C in ABC and by the line through D parallel to AB Adding the constant h/2 to t and using the triangle inequality, one gets t+ h ≥ AM + M N + N B = AM + M P + P N + N B ≥ AP + P B, 2 where P is the intersection... inequalities can be interpreted geometrically as such problems A typical example is the wellknown arithmetic mean geometric mean inequality, x+y √ ≥ x y (x, y ≥ 0), 2 which is equivalent to the following: Of all rectangles with a given perimeter the square has maximal area In this section we solve several geometric problems on maxima and minima using classical algebraic inequalities As one would expect, in using... triangle is degenerate if and only if P lies on the circumcircle of ABC More exactly: For each point P in the plane the inequality AP + B P ≥ C P holds true The equality occurs if and only if P is on the arc AB of the circumcircle of ABC Solution Let, for instance, C P ≥ AP and C P ≥ B P Consider the 60◦ counterclockwise rotation ϕ about A, and let ϕ carry P to P Then AP = AP and ∠P AP = 60◦ , so AP... unique solution of the problem Notice that since AX 0 and B X 0 make equal angles with , the pair of line segments AX 0 and B X 0 has the same property ♠ The main feature used in the solutions of the above two problems was that among the broken lines connecting two given points A and B the straight line segment AB has minimal length The same elementary observation will be used in the solutions of several... through P with endpoints on the sides of the angle and such that 1 1 + BP CP is a maximum 1.1.20 Given a convex quadrilateral ABC D, draw a line through C, intersecting the extensions of the sides AB and AD at points M and K , such that 1 1 + [BC M] [DC K ] is a minimum 1.1.21 An angle O X Y is given and a point M in its interior Find points A on O X and B on OY such that O A = O B and the sum M A + M B... Inscribe a quadrilateral of minimal perimeter in a given rectangle 1.1.30 Among all quadrilaterals ABC D with AB = 3, C D = 2, and ∠ AM B = 120◦ , where M is the midpoint of C D, find the one of minimal perimeter 1.1.31 Let ABC D E F be a convex hexagon with AB = BC = C D, D E = E F = F A, and ∠BC D = ∠E F A = 60◦ Let G and H be points interior to the hexagon such that the angles AG B and D H E are both 120◦ . old one. Altogether the book contains hundreds of geometric problems on maxima or minima. Despite the great variety of problems considered—from very old and classical ones like the ones mentioned. maximal. Section 2.4 deals with some problems on maxima and minima arising in combinatorial geometry. Chapter 3 collects some geometric problems on maxima and minima that could not be put into any. these classical problems are discussed in Chapter 1, which presents several different methods for solving geometric problems on maxima and minima. One of these concerns applications of geometric transformations,