AMC 8 1985 Problem 24
AMC8 1985 年第 24 题
基本计数原理★★★★☆
In a magic triangle, each of the six whole numbers 10-15 is placed in one of the circles so that the sum, , of the three numbers on each side of the triangle is the same. The largest possible value for is
在一个魔力三角形中，六个整数 10-15 各放入一个圆圈，使得三角形每条边上的三个数之和 相同。 的最大可能值是
- A.36
- B.37
- C.38
- D.39
- E.40
正确答案 · Correct Answer
D
解析 · Solution
解法1
常规解法 · 拆顶点与中点
设三个顶点之和为 、三个中点之和为 ，则 x+y=10+11+12+13+14+15=75。每个顶点在两条边上、每个中点在一条边上，故 3S=2x+y=x+(x+y)=x+75，即
要 最大就让 最大：，故 ，把 13、14、15 放三个顶点即可实现。
巧解 · 试上界
由 ，试 S=40 需 x=45，但顶点和最多 13+14+15=42，不行；试 S=39 需 x=42，恰好成立。故最大为 39。
Solution 1
Elementary · split vertices and midpoints
Let be the sum of the three vertices and the sum of the three midpoints, so x+y=10+11+12+13+14+15=75. Each vertex lies on two sides and each midpoint on one, so 3S=2x+y=x+(x+y)=x+75, giving
Maximizing means maximizing : since , , attained by placing 13, 14, 15 at the vertices.
Slick · test the upper bound
From : trying S=40 needs x=45, impossible (vertex sum is at most 13+14+15=42); S=39 needs x=42, which works. So the largest is 39.
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