Emergency problems #2

Emergency problems #2 - MATH 348: Solved Emergency...

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MATH 348: Solved Emergency Problems, Week 2 Leonid Chindelevitch August 25, 2008 1 Monday, July 14, 2008 1. Problem: ABCD is a cyclic quadrilateral with extended sides AD,BC meeting at Q and extended sides BA,CD meeting at P . Prove that the quadrilateral EFGH determined on ABCD by the bisectors of the angles P and Q is always a rhombus. Solution: Since ABCD is cyclic, the exterior angle DCQ is equal to the interior angle at the opposite vertex, A . Since QE bisects the angle at Q , the angles of Δ AQE are equal respectively to the angles of Δ CQG . Hence CGQ = AEQ . But CGQ = PGE (they are vertically oppo- site), so PEG = PGE and Δ PEG is isoceles. The bisector of P is then the perpendicular bisector of the base EG . Thus H and F , on the perpendicular bisector, are equidistant from E and G . Similarly, E and G are equidistant from H and F , and EFGH is a rhombus. 2. Problem: Suppose two circles Q and R intersect in A and B . A point P on the arc of Q which lies outside R is projected through A and B to 1
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determine chord CD of R . Prove that no matter where P is chosen on its arc, the length of the chord CD is always the same. Solution: Let P and P 0 denote positions of P corresponding to chords CD and C 0 D 0 . We have PAP 0 = PBP 0 , PAP 0 = CAC 0 , PBP 0 = DBD 0 = CAC 0 = DBD 0 . Therefore the arcs CC 0 and DD 0 correspond to equal angles. Adding the arc CD 0 to each one gives C 0 D 0 = CD because the corresponding arcs are equal. 3. Problem: ON is the radius which is perpendicular to chord AB in a circle, center O , meeting AB at M . P is any point on the major arc AB which is not diametrically opposite N . PM and PN determine, respec- tively, Q and R on the circle and AB . Which is longer longer: RN or MQ ? Solution: Let the reflection of PN in the diameter NON 0 be P 0 N . Be- cause AB is perpendicular to ON , this reflection carries R into the inter- section of R 0 of P 0 N and AB . Hence RN = R 0 N . Now PP 0 and AB are 2
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both perpendicular to diameter NON 0 . Hence they are parallel, giving equal corresponding angles NR 0 M and NP 0 P . But, in the same segment we have NP 0 P = NQP . Thus NR 0 M = NQM , so QNMR 0 is cyclic. Since R 0 N subtends a right angle at the circumference of this circle (at M ), it is a diameter. However, the angle subtended at the circumference by the chord QM , namely MNQ , is seen NOT to be a right angle (since NN 0 is a diameter of the given circle, NQN 0 is a right angle, so QNM , in Δ QNN 0 , is not). Consequently the chord QM is less than the diameter R 0 N = RN . 4.
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Emergency problems #2 - MATH 348: Solved Emergency...

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