2004 AMC 12B Solutions

Solutions 2004 55 th AMC 12 B 39. (E) The rotation takes (−3, 2) into B = (2, 3), and the reflection takes B intoC = (3, 2).A(-3,2)(2,3)BC(3,2)10. (A) The area of the annulus is the difference between the areas of the two circles,which is πb 2 − πc 2 . Because the tangent XZ is perpendicular to the radius OZ,b 2 − c 2 = a 2 , so the area is πa 2 .11. (D) Each score of 100 is 24 points above the mean, so the five scores of 100represent a total of (5)(24) = 120 points above the mean. Those scores must bebalanced by scores totaling 120 points below the mean. Since no student scoredmore than 76 − 60 = 16 points below the mean, the number of other students inthe class must be an integer no less than 120/16. The smallest such integer is8, so the number of students in the class is at least 13. Note that the conditionsof the problem are met if 5 students score 100 and 8 score 61.ORIf there are k students in the class, the sum of their scores is 76k. If the fivescores of 100 are excluded, the sum of the remaining scores is 76k − 500. Sinceeach student scored at least 60, the sum is at least 60(k − 5). Thus76k − 500 ≥ 60(k − 5),so k ≥ 12.5. Since k must be an integer, k ≥ 13.12. (C) Let a k be the k th term of the sequence. For k ≥ 3,a k+1 = a k−2 + a k−1 − a k , so a k+1 − a k−1 = −(a k − a k−2 ).Because the sequence begins2001, 2002, 2003, 2000, 2005, 1998, . . . ,it follows that the odd-numbered terms and the even-numbered terms each formarithmetic progressions with common differences of 2 and −2, respectively. The2004 th term of the original sequence is the 1002 nd term of the sequence 2002,2000, 1998,. . ., and that term is 2002 + 1001(−2) = 0.