AMC8 1996

AMC8 1996 · Q8

AMC8 1996 · Q8. It mainly tests Distance / midpoint, Geometry misc.

Points A and B are 10 units apart. Points B and C are 4 units apart. Points C and D are 3 units apart. If A and D are as close as possible, then the number of units between them is

点A和B相距10个单位。点B和C相距4个单位。点C和D相距3个单位。如果A和D尽可能接近，那么它们之间的单位数是

(A) 0 0

(B) 3 3

(D) 11 11

(E) 17 17

Answer

Correct choice: (B)

正确答案：（B）

Solution

The shortest distance occurs when the four points are collinear with B-C-D towards A, so AD = AB - BC - CD = 10 - 4 - 3 = 3.

最短距离发生在四点共线，B-C-D朝向A时，所以AD = AB - BC - CD = 10 - 4 - 3 = 3。

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