UC知道数学比赛的问题
来源：百度知道 编辑：UC知道 时间：2024/06/24 13:08:43
最好能够在今天以内拿到答案（我明天就要比赛了）
如果可以在今天拿到答案的话
我会追加分数的
题目（超难的）
Q1) The sum of three numbers is 4, the sum of their squares is 10 and the sum of their cubes is 22.
What is the sum of their fourth powers?
Q2) In a regular polygon there are 2 diagonals such that the angle between them is 50 degrees.
What is the smallest number of sides of the polygon for which this is possible?
如果可以在今天拿到答案的话
我会追加分数的
题目（超难的）
Q1) The sum of three numbers is 4, the sum of their squares is 10 and the sum of their cubes is 22.
What is the sum of their fourth powers?
Q2) In a regular polygon there are 2 diagonals such that the angle between them is 50 degrees.
What is the smallest number of sides of the polygon for which this is possible?
1.
a+b+c=4
a^2+b^2+c^2=10
a^3+b^3+c^3=22
要求
a^4+b^4+c^4
注意到
ab+bc+ac=((a+b+c)^2-a^2-b^2-c^2)/2=3
(a+b+c)(a^2+b^2+c^2)
=a^3+b^3+c^3+ab(a+b)+bc(b+c)+ac(a+c)
=a^3+b^3+c^3+(ab+bc+ac)(a+b+c)-3abc
=22+4(ab+bc+ac)-3abc
=22+2(16-10)-3abc
=34-3abc=40
得abc=-2
那么
(a+b+c)(a^3+b^3+c^3)
=a^4+b^4+c^4+ab(a^2+b^2)+bc(b^2+c^2)+ac(a^2+c^2)
=a^4+b^4+c^4+(ab+bc+ac)(a^2+b^2+c^2)-abc(a+b+c)
=a^4+b^4+c^4+11(ab+bc+ac)+2*4
=a^4+b^4+c^4+33+8
=88
所以a^4+b^4+c^4=47
2.
正多边形两条对角线夹角50度
问最少是几边形?
正n多边形任两条对角线的夹角都是180/(n-2)的倍数
那么180/(n-2)*m=50,m,n都是整数
即
180/(n-2)=50/m
也就是18m-5(n-2)=0
因18与5互质，所以n-2最小18，n最小20
多边形最少是20边形
很多数学高手英语都很烂
你确定参加的是数学比赛吗
好几年不用英语了，有的词语忘记了什么意思了