1985 AMC 8 第 24 题

先试着解答 1985 AMC 8 第 24 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 1985 AMC 8 解答,或核对答案。

所有题目均经美国数学协会(MAA)官方合法授权使用。

24.

在一个魔术三角形中,六个整数 1010 至 1515 各被放入一个圆内,使三角形每条边上三个数的和 SS 都相同。SS 的最大可能值是
In a magic triangle, each of the six whole numbers 1010 through 1515 is placed in one of the circles so that the sum SS of the three numbers on each side of the triangle is the same. The largest possible value for SS is
3636
3737
3838
3939
4040
答案:D
知识点:幻方最优化
小提示:
把三条边的和相加时,每个顶点数被计算两次,每个中点数被计算一次
Adding the three side sums counts each corner number twice and each midpoint number once
大提示:
设 CC 为三个顶点数之和,则 3S=75+C3S = 75 + C;要使它最大,应把三个最大的数放在顶点上
Let CC be the sum of the corner numbers. Then 3S=75+C;3S = 75 + C; put the three largest numbers at the corners
解答:
设 CC 为三个顶点数之和。把三条边的和相加可得 3S=75+C3S = 75 + C,因为六个数之和为 7575,且每个顶点数都被多计算了一次。要使 SS 最大,应把三个最大的数 1313、1414、1515 放在顶点,此时 C=42C = 42。
于是 3S=75+42=1173S = 75 + 42 = 117,所以 S=39S = 39。这个值可以达到:在顶点 1313 与 1414 之间放 1212,在 1414 与 1515 之间放 1010,在 1515 与 1313 之间放 1111。
所以正确答案是 D。
Let CC be the sum of the corner numbers. Adding the three side sums gives 3S=75+C,3S = 75 + C, because the six numbers sum to 7575 and every corner is counted one extra time. To maximize S,S, place the three largest numbers 13,13, 14,14, 1515 at the corners, giving C=42.C = 42.
Then 3S=75+42=117,3S = 75 + 42 = 117, so S=39.S = 39. This is achievable: put 1212 between corners 1313 and 14,14, put 1010 between 1414 and 15,15, and put 1111 between 1515 and 13.13.
Thus, the correct answer is D .
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