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Scorpiour's Modding Queue (Modding Team)
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简直为难人 我数学是体育老师教的你造么
简直为难人 我数学是体育老师教的你造么
19803508
I admit that's a good usage of your alias, whatever it actually means.blahpy wrote:
blahpy numbers
123456780
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不算惹 太难，一张纸半管笔美丽
不算惹 太难，一张纸半管笔美丽
333566793
478294496
It doesn't mean anything lolnanda2009 wrote:
I admit that's a good usage of your alias, whatever it actually means.blahpy wrote:
blahpy numbers
I only just remembered this puzzle was here, looks like no-one found the actual solution yet though lol
333566793
~ 066606660 ~
*Fort runs
*Fort runs
Topic Starter
Quiz queue now:
First one is Mathematics:
The one who posts the correct answer first will earn a modding ticket from me
one person per post, no edit allowed!
First one is Mathematics:
The one who posts the correct answer first will earn a modding ticket from me
one person per post, no edit allowed!
k^3/8
Topic Starter
手误手误wmfchris wrote:
k^3/8
xyz = (k-x)(k-y)(k-z)
x = k-x
y = k-y
z = k-z
2x = k
2y = k
2z = k
x = k/2
y = k/2
z = k/2
xyz = (k^3)/8
i don't know about here but you want <= but = is part of <= i guess...
x = k-x
y = k-y
z = k-z
2x = k
2y = k
2z = k
x = k/2
y = k/2
z = k/2
xyz = (k^3)/8
i don't know about here but you want <= but = is part of <= i guess...
Topic Starter
if x=4, y=6 z=5 and k = 10, then?xxdeathx wrote:
xyz = (k-x)(k-y)(k-z)
x = k-x
y = k-y
z = k-z
2x = k
2y = k
2z = k
x = k/2
y = k/2
z = k/2
xyz = (k^3)/8
i don't know about here but you want <= but = is part of <= i guess...
x=kaa/(aa+bc)
y=kbb/(bb+ca)
z=kcc/(cc+ab)
然后我懒了
y=kbb/(bb+ca)
z=kcc/(cc+ab)
然后我懒了
It seems that when x=y=z=1/2k, it is the maximum of xyz, but I can't explain in detail to make a proof orz
设x+a=y+b=z+c=k，则xyz=abc
所以，根据均值定理，
k>=2*sqrt(xa)
k>=2*sqrt(yb)
k>=2*sqrt(zc)
所以k^3>=8*sqrt(xyzabc)
又xyz=abc
所以k^3>=8*(xyz)
所以xyz<=(k^3)/8
得证！
https://osu.ppy.sh/s/178495
所以，根据均值定理，
k>=2*sqrt(xa)
k>=2*sqrt(yb)
k>=2*sqrt(zc)
所以k^3>=8*sqrt(xyzabc)
又xyz=abc
所以k^3>=8*(xyz)
所以xyz<=(k^3)/8
得证！
https://osu.ppy.sh/s/178495
令Z≥X≥Y，所以XYZ的最大值为Z^3且等于(K-Z)的三次方
---> Z^3 = (K-Z)^3
---> Z=K-Z
---> Z=1/2K
所以Z^3=(1/2K)^3 , 这个是最大值
所以XYZ小于等于K^3/8
---> Z^3 = (K-Z)^3
---> Z=K-Z
---> Z=1/2K
所以Z^3=(1/2K)^3 , 这个是最大值
所以XYZ小于等于K^3/8
if x, y, z < k/2, substitute them into xyz to get xyz < k^3 / 8Scorpiour wrote:
if x=4, y=6 z=5 and k = 10, then?xxdeathx wrote:
xyz = (k-x)(k-y)(k-z)
x = k-x
y = k-y
z = k-z
2x = k
2y = k
2z = k
x = k/2
y = k/2
z = k/2
xyz = (k^3)/8
i don't know about here but you want <= but = is part of <= i guess...
if x, y, z > k/2, then k-x < k/2, k-y < k/2, k-z < k/2
then xyz = (k-x)(k-y)(k-z) < k^3 / 8
for effort...please?
As lgdry15 got the correct answer I will perform a standard trick towards the solution.
Let u = (k-x)/x, v = (k-y)/y, w = (k-z)/z (The case for x/y/z = 0 is obvious, so WLOG suppose they are positive.)
Then x = k/(u+1), y = k/(v+1), z = k/(w+1)
So that the equality becomes uvw = 1 and to show that (u+1)(v+1)(w+1) >= 8k^3.
Since u,v,w > 0, the result is obvious by AM-GM.
Let u = (k-x)/x, v = (k-y)/y, w = (k-z)/z (The case for x/y/z = 0 is obvious, so WLOG suppose they are positive.)
Then x = k/(u+1), y = k/(v+1), z = k/(w+1)
So that the equality becomes uvw = 1 and to show that (u+1)(v+1)(w+1) >= 8k^3.
Since u,v,w > 0, the result is obvious by AM-GM.
math is too hard =/
Topic Starter
lgdry15 got a modding ticket~~
and also thanks for other participants~
more quiz are coming
and also thanks for other participants~
more quiz are coming
Topic Starter
yeah i got this.. i didnt watch 2 months worth of anime for nothing..
wait jklols what is this ._.
wait jklols what is this ._.
OMG my mistake orz.
Name : Ore no imouto ga konna ni kawaii wake ga nai
Episode : 15
Season : 2010 / Season 4 ( autumn)
#1 Ore no Imouto ha Konna ni Kawaii Wake ga Nai episode 6 season 1
#2 Hyouka Episode 22 (I think there was only 1 season)
#3 Yosuga no Sora epsideo 9 (only 1 season)
#2 Hyouka Episode 22 (I think there was only 1 season)
#3 Yosuga no Sora epsideo 9 (only 1 season)