Forgot password?
 Register account
View 1785|Reply 2

[组合] 2015清华自主招生题目

[Copy link]

48

Threads

77

Posts

778

Credits

Credits
778

Show all posts

longzaifei Posted 2015-7-25 22:10 |Read mode
一个正十五边形,任取其中三个顶点构成三角形,可以构成多少个钝角三角形?

413

Threads

1431

Posts

110K

Credits

Credits
11100

Show all posts

realnumber Posted 2015-7-26 08:37
我先来试试,顶点依次编号为1,2,3,4,...,15.
1,3连接,这条对角线的一侧有一个顶点,编号2,对应的钝角三角形1个(1,3,2)(说明:(1,3,4)不计算在这里,算在下一条,等等.)
1,4连接,............................................2,3,...................2个(1,4,2),(1,4,3)
........
1,8连接,............................................2,3,4,5,6,7,............6个..
(说明:1,9连接没有了,但9,1连接有的)
以上都可以绕中心$\frac{360\du}{15}x,x=1,2,3,\cdots ,14$,产生钝角三角形而不重复
因此总数为$15\times (1+2+3+4+5+6)=315$个.

48

Threads

77

Posts

778

Credits

Credits
778

Show all posts

 Author| longzaifei Posted 2015-7-27 21:57
回复 2# realnumber


   

Mobile version|Discuz Math Forum

2025-5-31 11:23 GMT+8

Powered by Discuz!

× Quick Reply To Top Edit