By Hopf H.

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**Example text**

2 Let I0 be the center of the incircle of triangle ABC, and let I1 , I2 , and I3 be the centers of the excircles. Let r0 , r1 , r2 , and r3 be the radii of these circles, and let R be the radius of the circumcircle. 3 I0 = (ar0 , br0 , cr0 ), I1 = (−ar1 , br1 , cr1 ), I2 = (ar2 , −br2 , cr2 ), I3 = (ar3 , br3 , −cr3 ) . 5) an elegant formula for the area of the triangle that has the centers of the excircles as vertices. Analogously, we ﬁnd [I0 I2 I3 ] = 2R(s − a), [I0 I3 I1 ] = 2R(s − b), and [I0 I1 I2 ] = 2R(s − c) .

Because of its importance, there exist coordinate systems that are associated to the triangle that is being considered. In this chapter, we discuss two of these systems. If P is a point in the plane of triangle ABC, we will denote the distances from P to respectively BC, CA, and AB by x ¯, y¯, and z¯, where x ¯ is considered to be positive if P lies on the same side of BC as A, and the analogue holds A Q P B C R Fig. 1. for y¯ and z¯. 1, x ¯, y¯, and z¯ are positive for P ; for Q, x ¯ and y¯ are positive, and z¯ is negative; and for R, y¯ is positive, and x ¯ and z¯ negative.

AB = 13 b 3, AC = 13 c 3, and ∠B AC = 60◦ + A . In triangle B AC , we ﬁnd: √ (B C )2 = 16 (a2 + b2 + c2 + 4 3 [ABC]) , where [ABC] denotes the area of triangle ABC. It follows from the symmetry of the expression that A B C is an equilateral triangle. We can easily convince ourselves that the theorem does not hold, for example, for the centers of the squares based on the sides. The points A , B , and C have more interesting properties; in particular, AA , BB , and CC O. 1007/978-0-387-78131-0 11, c Springer Science+Business Media, LLC 2008 48 11 Inequalities in a Triangle are concurrent.