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                                                      考題15-16 2015 AMC 12A

                                                      Harley
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                                                      AMC12是針對高中學生的數學測驗,該競賽開始于2000年,分A賽和B賽,于每年的2月初和2月中舉行,學生可任選參加一項即可。其主要目的在于激發學生對數學的興趣,參予AMC12的學生應該不難發現測驗的問題都很具挑戰性,但測驗的題型都不會超過學生的學習范圍。這項測驗希望每個考生能從競賽中享受數學。那么接下來跟隨小編來看一下AMC12的官方真題以及官方解答吧:

                                                      Problem 15What is the minimum number of digits to the right of the decimal point needed to express the fraction??as a decimal?

                                                      Solution 1We can rewrite the fraction as?. Since the last digit of the numerator is odd, a??is added to the right if the numerator is divided by?, and this will continuously happen because?, itself, is odd. Indeed, this happens twenty-two times since we divide by??twenty-two times, so we will need??more digits. Hence, the answer is

                                                      Solution 2Multiply the numerator and denominator of the fraction by??(which is the same as multiplying by 1) to give?. Now, instead of thinking about this as a fraction, think of it as the division calculation??. The dividend is a huge number, but we know it doesn´t have any digits to the right of the decimal point. Also, the dividend is not a multiple of 10 (it´s not a multiple of 2), so these 26 divisions by 10 will each shift the entire dividend one digit to the right of the decimal point. Thus,??is the minimum number of digits to the right of the decimal point needed.

                                                      Solution 3The denominator is?. Each??adds one digit to the right of the decimal, and each additional??adds another digit. The answer is?.

                                                      Problem 16Tetrahedron??has?,?,?,?,?, and?. What is the volume of the tetrahedron?

                                                      Solution 1Let the midpoint of??be?. We have?, and so by the Pythagorean Theorem??and?. Because the altitude from??of tetrahedron??passes touches plane??on?, it is also an altitude of triangle?. The area??of triangle??is, by Heron´s Formula, given by

                                                      Substituting??and performing huge (but manageable) computations yield?, so?. Thus, if??is the length of the altitude from??of the tetrahedron,?. Our answer is thusand so our answer is

                                                      Solution 2Drop altitudes of triangle??and triangle??down from??and?, respectively. Both will hit the same point; let this point be?. Because both triangle??and triangle??are 3-4-5 triangles,?. Because?, it follows that the??is a right triangle, meaning that?, and it follows that planes??and??are perpendicular to each other. Now, we can treat??as the base of the tetrahedron and??as the height. Thus, the desired volume iswhich is answer

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