2018-11-29 重点归纳

**AMC10数学竞赛**是美国高中数学竞赛中的一项，是针对高中一年级及初中三年级学生的数学测试，该竞赛开始于2000年，分A赛和B赛，于每年的2月初和2月中举行，学生可任选参加一项即可。不管是对高校申请还是今后在数学领域的发展都极其有利！那么接下来跟随小编来看一下**AMC10数学竞赛真题**以及**官方解答**吧：

All the numbers are written in a array of squares, one number in each square, in such a way that if two numbers are consecutive then they occupy squares that share an edge. The numbers in the four corners add up to . What is the number in the center?

Quick testing shows thatis a valid solution. , and the numbers follow the given condition. The center number is found to be .

First let the numbers bewith the numbers around the outsides and in the middle. We see that the sum of the four corner numbers is . If we switch and , then the corner numbers will add up to and the consecutive numbers will still be touching each other. The answer is .

Consecutive numbers share an edge. That means that it is possible to walk from to by single steps north, south, east, or west. Consequently, the squares in the diagram with different shades have different parity:

But there are only four even numbers in the set, so the five darker squares must contain the odd numbers, which sum to Therefore if the sum of the numbers in the corners is , the number in the center must be , which is answer .

The sum of an infinite geometric series is a positive number , and the second term in the series is . What is the smallest possible value of

The sum of an infinite geometric series is of the form:where is the first term and is the ratio whose absolute value is less than 1.

We know that the second term is the first term multiplied by the ratio. In other words:

Thus, the sum is the following:

Since we want the minimum value of this expression, we want the maximum value for the denominator, . The maximum x-value of a quadratic with negative is .

Plugging into the quadratic yields:

Therefore, the minimum sum of our infinite geometric sequence is . (Solution by akaashp11)

After observation we realize that in order to minimize our sum with being the reciprocal of r. The common ratio has to be in the form of with being an integer as anything more than divided by would give a larger sum than a ratio in the form of .

The first term has to be , so then in order to minimize the sum, we have minimize .

The smallest possible value for such that it is an integer that's greater than is . So our first term is and our common ratio is . Thus the sum is or . Solution 2 by No_One

We can see that if is the first term, and is the common ratio between each of the terms, then we can getAlso, we know that the second term can be expressed as notice if we multiply by , we would getThis quadratic has real solutions if the discriminant is greater than or equal to zero, orThis yields that or . However, since we know that has to be positive, we can safely conclude that the minimum possible value of is .

以上就是小编对**AMC10数学竞赛真题**以及解析的介绍，希望对你有所帮助，如果想了解更多关于AMC数学竞赛报考点、**南京AMC数学竞赛培训**、美国数学竞赛AMC有用吗以及AMC学习资料等信息请持续关注**AMC数学竞赛网**。

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