[求助]两道关于多边形对角线的数学题

xiliuhuang

32. How many diagonals does a polygon with 21 sides have, if one of its vertices does not connect to any diagonal?

 

a)      21

 

b)      170

 

c)      340

 

d)     357

 

e)      420

 

33. How many diagonals does a polygon with 18 sides have if three of its vertices do not send any diagonal?

a)      90

 

b)      126

 

c)      210

 

d)     264

 

e)      306

EXPLAINATION

32. The best answer is B.

We have 20 vertices linking to 17 others each: that is 17*20=340. We divide that by 2 since every diagonal connects two vertices. 340/2=170. The vertex that does not connect to any diagonal is just not counted.

 

33. The best answer is A.

 

We have 15 Vertices that send diagonals to 12 each (not to itself and not to the two adjacent vertices). 15*12=180. Divide it by 2 since any diagonal links 2 vertices = 90. The three vertices that do not send a diagonal also do not receive any since the same diagonal is sent and received. Thus they are not counted.

 我的问题是：如果用另一种方法做32题，先忽略 not connect的那一个点，考虑21个点连出来的对角线应该是=（18*21）/2=189 再减去本从 not connect点发出的18条，剩下171条。比答案的170多一条。但33题中，用两种方法做结果的一样的都是90.

     请问怎么解释32题中两种方法做的结果不一样？

ninihao

32.

A polygon of 21 sides has 21 vertices.
Number of straight lines among 21 points = 21C2 = 210
Of those lines, number of diagonals = 210 - 21 = 189 (21 is the number of sides of the polygon)
Therefore, number of diagonals through all but one vertex = 189 - 18 = 171

None of the choices. Closest is B (170)

xiliuhuang

谢谢

关于你说的Closest is B (170)

可以随便就close了么？