2016 AMC 12A 第 23 题

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23.

从区间 [0,1][0,1] 中独立随机选取三个数。所选三个数能作为一个面积为正的三角形的边长的概率是多少?

Three numbers in the interval [0,1][0,1] are chosen independently and at random. What is the probability that the chosen numbers are the side lengths of a triangle with positive area?

16\dfrac{1}{6}

13\dfrac{1}{3}

12\dfrac{1}{2}

23\dfrac{2}{3}

56\dfrac{5}{6}

答案:C
知识点:几何概率三角不等式体积
难度评级:2160
解答:

有序三元组 (x,y,z)(x,y,z) 填满体积为 11 的单位立方体。不能构成三角形,恰好意味着某个数大于或等于另外两个数之和。

区域 zx+yz\ge x+y 是一个四面体,顶点为 (0,0,0),(0,0,1),(0,1,1),(1,0,1)(0,0,0),(0,0,1),(0,1,1),(1,0,1),体积为 16\frac16。类似地,区域 xy+zx\ge y+zyx+zy\ge x+z 的体积也各为 16\frac16,而且内部互不相交。因此不能构成三角形的概率为 316=123\cdot\frac16=\frac12,能构成三角形的概率为 112=121-\frac12=\frac12

所以正确答案是 C

The ordered triples (x,y,z)(x,y,z) fill the unit cube of volume 1.1. They fail to form a triangle exactly when one value is at least the sum of the other two.

The region zx+yz\ge x+y is a tetrahedron with vertices (0,0,0),(0,0,1),(0,1,1),(1,0,1)(0,0,0),(0,0,1),(0,1,1),(1,0,1) of volume 16.\frac16. The analogous regions xy+zx\ge y+z and yx+zy\ge x+z also have volume 16\frac16 and have disjoint interiors. So the failure probability is 316=12,3\cdot\frac16=\frac12, and the triangle probability is 112=12.1-\frac12=\frac12.

Thus, the correct answer is C.

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