103 Trigonometry Problems

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5. Solutions to Advanced Problems 153 A F A B O M D I E C E C O (F) I D M B Figure 5.8. Second Solution: As shown in Figure 5.8, the incircle touches the sides AB, BC, and CA at D, E, and F , respectively. Let M be the foot of the perpendicular line segments from O to side BC. Then the line OM is the perpendicular bisector of BC, and |BM| =|CM|. From equal tangents, we have |AF |=|AE|, |BD|=|BF|, and |CD|=|CE|. Because c>b, M lies on segment BD. We find that √ 2|OI|=c − b = (|AF |+|FB|) − (|AE|+|EC|) =|FB|−|EC| =|BD|−|DC|. We deduce that |BD|=|BM|+|MD| and |DC| =|CM|−|DM|. Hence √ 2|OI|=2|DM|,or|OI|= √ 2|DM|. Thus lines OI and DM form a 45 ◦ angle, which implies that either OI ⊥ AB or OI ‖ AB. We consider these two cases separately. • First Case: In this case, we assume that OI ⊥ AB. Then OI is the perpendicular bisector of side AB; that is, the incenter lies on the perpendicular bisector of side AB. Thus triangle ABC must be isosceles, with |AC| =|BC|, and so A = B = 45 ◦ and sin A = √ 2 2 . • Second Case: In this case, we assume that OI ‖ AB. Let N be the midpoint of side AB. Then OIFN is a rectangle. Note that ̸ AON = ̸ C, and thus R cos ̸ AON = R cos C =|ON|=|IF|=r. By the solution to Introductory Problem 27, we have cos C = R r = cos A + cos B + cos C − 1,
154 103 Trigonometry Problems implying that cos A = 1 − cos B = 1 − √ 2/2. It follows that √ √ sin A = 1 − cos 2 4 √ 2 − 2 A = . 2 29. [Dorin Andrica] Let n be a positive integer. Find the real numbers a 0 and a k,l , 1 ≤ l
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About the Authors Titu Andreescu re
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Titu Andreescu University of Wiscon
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vi Contents Vectors 41 The Dot Prod
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viii Preface Throughout MOSP, full
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Abbreviations and Notation Abbrevia
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103 Trigonometry Problems
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3 Advanced Problems 1. Two exercise
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3. Advanced Problems 75 9. Find the
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Further Reading 1. Andreescu, T.; F
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Further Reading 213 23. Fomin, D.;
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