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考題24-25 2015 AMC 10A

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AMC10數學競賽是美國高中數學競賽中的一項,是針對高中一年級及初中三年級學生的數學測試,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。不管是對高校申請還是今后在數學領域的發展都極其有利!那么接下來跟隨小編來看一下AMC10數學競賽真題以及官方解答吧:

Problem 24For some positive integers?, there is a quadrilateral??with positive integer side lengths, perimeter?, right angles at??and?,?, and?. How many different values of??are possible?

SolutionLet??and??be positive integers. Drop a perpendicular from??to??to show that, using the Pythagorean Theorem, thatSimplifying yields?, so?. Thus,??is one more than a perfect square.

The perimeter??must be less than 2015. Simple calculations demonstrate that?is valid, but??is not. On the lower side,??does not work (because?), but??does work. Hence, there are 31 valid??(all??such that??for?), and so our answer is

Problem 25Let??be a square of side length?. Two points are chosen independently at random on the sides of?. The probability that the straight-line distance between the points is at least??is?, where?,?, and??are positive integers with?. What is??

Solution 1Divide the boundary of the square into halves, thereby forming??segments. Without loss of generality, let the first point??be in the bottom-left segment. Then, it is easy to see that any point in the??segments not bordering the bottom-left segment will be distance at least??apart from?. Now, consider choosing the second point on the bottom-right segment. The probability for it to be distance at least??apart from??is??because of linearity of the given probability. (Alternatively, one can set up a coordinate system and use geometric probability.)

If the second point??is on the left-bottom segment, then if??is distance??away from the left-bottom vertex, then??must be up to??away from the left-middle point. Thus, using an averaging argument we find that the probability in this case is

(Alternatively, one can equate the problem to finding all valid??with??such that?, i.e.??is outside the unit circle with radius?)

Thus, averaging the probabilities gives

Our answer is?.

Solution 2Let one point be chosen on a fixed side. Then the probability that the second point is chosen on the same side is?, on an adjacent side is?, and on the opposite side is?. We discuss these three cases.

Case 1: Two points are on the same side. Let the first point be??and the second point be??in the?-axis with?. Consider??a point on the unit square??on the Cartesian plane. The region??has the area of?. Therefore, the probability that??is?.

Case 2: Two points are on two adjacent sides. Let the two sides be??on the x-axis and??on the y-axis and let one point be?and the other point be?. Then??and the distance between the two points is?. As in Case 1,??is a point on the unit square?. The area of the region??is??and the area of its complementary set inside the square (i.e.??) is?. Therefore, the probability that the distance between?and??is at least??is?.

Case 3: Two points are on two opposite sides. In this case, the probability that the distance between the two points is at least??is obviously?.

Thus the probability that the probability that the distance between the two points is at least??is given byTherefore?,?, and?. Thus,??and the answer is

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