老师想好了两的自然数m，n，满足2≤m≤n≤100，他把m，n的和s告诉了S，积p告诉了P。S，P都诚实可靠，进行了一下对话：
S：我不知道他们的确切值，但我知道你也不知道
P：现在我知道了
S：现在我也知道了
老师证实了他们的表述都是有根据的，结论是正确的，试求m，n

I've seen this problem before, i'll post it next time
btw, you're from shandon? great, my grandpa was from shandon too.
Frank

S：我不知道他们的确切值，但我知道你也不知道
由此推出: S的数不能写成两个素数之和.

于是排除了一大堆. 剩下的用枚举法吧...都列出来, 一个一个试.

也不必列到全部，因为 S 一定是奇数，所以若 S >53 则总可以把 S 写成 S = 53 + (S-53）[否则就找个较大的素数来代替53]。 由于m,n <100 , 所以一定是P=53(S-53) ，从而P就可以知道是什么号码。既然 P 不知，那么S=<53 。之后再用枚举法就可以推论出

(m,n)=(4,13) 是唯一解

我把文章打在这里, 省得以后看不到了再抓瞎.

[quote="[url]http://www.byhh.net/cgi-bin/bbsgcon?board=Job&file=G.1131853512.A&num=1239[/url]"]
[color=blue]一天，鬼谷子随意从2-99中选取了两个数。他把这两个数的和告诉了庞涓， 把这两个数的乘积告诉了孙膑。但孙膑和庞涓彼此不知到对方得到的数。第二天， 庞涓很有自信的对孙膑说：虽然我不知到这两个数是什麽，但我知道你一定也不知 道。随后，孙膑说：我知道了。庞涓说：我也知道了。请问：这两个数是什麽？ [/color]

设欲求的两个数为（X，Y），庞涓知道的和数设为A，孙膑知道的乘积设为B。定义 A的"鬼谷和拆分"为满足m+n=A的整数m、n，且2<= m<=n<=99；定义B的"鬼谷积拆分"为满足p*q=B的整数p、q，且2<=p<=q<=99。

一、 解读"庞涓很有自信的对孙膑说：虽然我不知到这两个数是什麽，但我知道你一定也不知道。"

这说明通过A的所有"鬼谷和拆分"中两个数的乘积都不能知道（X，Y）。

先给出乘积在以下两种情况时就能通过乘积直接知道两个乘数。

结论1、C=D*E,D,E均为素数，这时通过乘积就能知道两个乘数肯定为D,E。

结论2、C=D*E,E为>=53的素数，因为C为2-99之间的两个数的乘积，而E为>=53的素数，所以这两个乘数之一肯定是E，另一个就为D。

下面从分析A的值入手，

（1） A不能为197（99+98），这是2-99之间最大的两个数，孙膑当然能通过B知道这两个数是98、99；

（2） 197>A >=99不能成立，如果A>=99，那么A的一个"鬼谷和拆分"为m+97=A，根据结论2，孙膑就能知道（X，Y）分别为97和B/97；

（3） 99>A >=55不能成立，如果99>A >=55，那么A的一个"鬼谷和拆分"为m+53=A，根据结论2，孙膑就能知道（X，Y）分别为53和B/53；

（4） A不能为<55的偶数，因为任一偶数都能拆成两个素数之和（这是哥德巴赫猜 想的结论，虽然哥德巴赫猜想还没有被证明，但在<55的范围内可以一一试出来），根据结论1，孙膑就能知道（X，Y）就是这两个素数；

（5） A不能为5、7、9、13、19、21、25、31、33、39、43、45、49，因为这些数都能拆成2和另一素数之和，根据结论1，孙膑就能知道（X，Y）就是这两个素数

这样我们只需分别讨论A为11、17、23、27、29、35、37、41、47、51、53这11种情况，也就是说只有A为这11个数之一时，才能"庞涓很有自信的对孙膑说：虽然我不知到这两个数是什麽，但我知道你一定也不知道。"

二、 继续解读"随后，孙膑说：我知道了。"

（1） A=11时，它的"鬼谷和拆分"有（2，9）、（3，8）、（4，7）、（5，6），B只可能为18、24、28、30。

如果B=18，它的"鬼谷积拆分"有（2，9）、（3，6），根据庞涓说的第一句话，（X，Y）不可能是（3，6），孙膑就能知道（X，Y）是（2，9）；

如果B=24，它的"鬼谷积拆分"有（2，12）、（3，8）、（4，6），根据庞涓说的第一句话，（X，Y）不可能是（2，12）和（4，6），孙膑就能知道（X，Y）是（3，8）；

28和30不再讨论。

（2） A=17时，它的"鬼谷和拆分"有（2，15）、（3，14）、（4，13）、（5，12）、（6，11）、（7，10）、（8，9），B只可能为30、42、52、60、66、70、72。

如果B=30，它的"鬼谷积拆分"有（2，15）、（3，10）、（5，6），根据庞涓说的第一句话，（X，Y）不可能是3、10，但是孙膑不能知道（X，Y）是（2，15）还是（5，6）；

如果B=42，它的"鬼谷积拆分"有（2，21）、（3，14）、（6，7），根据庞涓说的第一句话，（X，Y）不可能是（6，7），但是孙膑不能知道（X，Y）是（2，21）还是（3，14）；

如果B=52，它的"鬼谷积拆分"有（2，26）、（4，13），根据庞涓说的第一句话，（X，Y）不可能是（2，26），孙膑就能知道（X，Y）是（4，13）；

如果B=66，它的"鬼谷积拆分"有（2，33）、（3，22）、（6，11），根据庞涓说的第一句话，（X，Y）不可能是（3，22），但是孙膑不能知道（X，Y）是（2，33）还是（6，11）；

如果B=70，它的"鬼谷积拆分"有（2，35）、（5，14）、（7，10），根据庞涓说的第一句话，（X，Y）不可能是（5，14），但是孙膑不能知道（X，Y）是（2，35）还是（7，10）；

如果B=72，它的"鬼谷积拆分"有（2，36）、（3，24）、（4，18）、（6，12）、（8，9），根据庞涓说的第一句话，（X，Y）不可能是（2，36）、（4，18）、（6，12），但是孙膑不能知道（X，Y）是（3，24）还是（8，9）。

只有B=52时才能知道（X，Y）

（3） A=23时，它的"鬼谷和拆分"有（4，19）、（7，16）等，B可能为76、112等。

如果B=76，它的"鬼谷积拆分"有（2，38）、（4，19），根据庞涓说的第一句话，（X，Y）不可能是（2，38），孙膑就能知道（X，Y）是（4，19）；

如果B=112，它的"鬼谷积拆分"有（2，56）、（4，28）、（7，16）、（8，14），根据庞涓说的第一句话，（X，Y）不可能是（2，56）、（4，28）、（8，14），孙膑就能知道（X，Y）是（7，16）；

（4） 在A为27、29、35、37、41、47、51、53时，都至少有两个"鬼谷和拆分"使得孙膑根据B知道（X，Y），这里不再详细讨论，只列出孙膑能确定（X，Y）的A的两个"鬼谷和拆分"。

A=27时，B=50时能确定（X，Y）为（2，25），B=92时能确定（X，Y）为（4，23）。（2，25）、（4，23）是A的"鬼谷和拆分"；

A=29时，B=54时能确定（X，Y）为（2，27），B=168时能确定（X，Y）为（8，21）。（2，27）、（8，21）是A的"鬼谷和拆分"；

A=35时，B=96时能确定（X，Y）为（3，32），B=304时能确定（X，Y）为（16，19）。（3，32）、（16，19）是A的"鬼谷和拆分"；

A=37时，B=232时能确定（X，Y）为（8，29），B=160时能确定（X，Y）为（5，32）。（8，29）、（5，32）是A的"鬼谷和拆分"；

A=41时，B=128时能确定（X，Y）为（4，37），B=288时能确定（X，Y）为（9，32）。（4，37）、（9，32）是A的"鬼谷和拆分"；

A=47时，B=172时能确定（X，Y）为（4，43），B=496时能确定（X，Y）为（16，31）。（4，43）、（16，31）是A的"鬼谷和拆分"；

A=51时，B=188时能确定（X，Y）为（4，47），B=608时能确定（X，Y）为（19，32）。（4，47）、（19，32）是A的"鬼谷和拆分"；

A=53时，B=592时能确定（X，Y）为（16，37），B=672时能确定（X，Y）为（21，32）。（16，37）、（21，32）是A的"鬼谷和拆分"；

三、 再解读"庞涓说：我也知道了。"

通过上面二的分析，只有在A=17时，庞涓才能唯一确定（X，Y）是什么，即（X，Y）=（4，13）

[/quote]

4楼的...
你也在奥数之家混啊...
有空聊聊
MSN:nirvanatear@hotmail.com

请先看这篇引子吧!
   美国斯坦福大学的麦卡锡教授----S先生与P先生谜题。
题目：S先生与P先生谜题
设有两个自然数X、Y，2<=X<=Y<=99,S先生知道这两个数的和S，P先生知道这两个数的积P，他们进行了如下对话：
S：我确信你不知道这两个数是什么，但我也不知道。
P: 一听你说这句话，我就知道这两个数是什么了。
S: 我也是，现在我也知道了。
现在你能通过他们的会话推断出这两个数是什么吗？（当然，S和P先生都是非常聪明的
）
   我想文章的用意是要给出一组"s-p数"使它们满足这一逻辑流,即具有可解性.
依据对话的内容及提设,显然4<=s<=198,而对于p我们不应做过多的讨论_它的可选区间远
比s的来的宽,从谈话中亦可见s更具主动力.
   由于单方面从s或p是得不解,因此s=4,5,197,198可唯一拆分不满足提设.
   而s先生说"我确信你不知道这两个数是什么，但我也不知道".请注意几个关键词由x,
y的和式s即能确定其积式的不可拆分,显然s先生具有实足的把握其值不可拆分为两个素数
的和,因此对于6<=s<=196的偶数显然不合适_当然前提是哥德巴赫猜想成为哥德巴赫定理
,不过对于这么小的值已验证为真的_局部成立,还有不少其间的奇数也是不合适的_如果它
们减去二后还是素数的话.
   最后问题的重心都剩下的那部分s值啦!
   p先生有幸得知这一点后,当然s只得奇偶拆分,而p先生能马上在这有力的启示之下得出
x,y的值,则对于p数的奇偶唯一拆分我们不应有所怀疑的.这样的数具有通式P=2^n*p,p为
素数(当然,除2外).
   由于s先生也立马得出结论_当然,他也不蠢,所以对于s的奇偶拆分的种数也不应多于两个,而这种拆分的原则是s-2^2,s-2^3,s-2^4,...s-2^m当且仅当只有一个差值是素数,其中(s-2^m)*[s-2^(m+1)]<0,全当谜题中的s先生与p先生都各得其所,他们很高兴而我也是高
兴的.
   我算出得几个值是:
s-p,x-y
17-52,4-13
29-208,16-13
41-148,4-37
53-592,16-37
...,...
   用算法求解颇费周折.
   如上所述,便有了s-p开始时的对白了.
   以上只是我的一点个人看法,如有错谬还请各位仁兄多多指正.


Hi guys, does any of you play ping-pong (table tennis)? if you do, then I like to play with you at ambleside youth centre which is located in west vancouver, here is its [url=http://maps.google.com/maps?f=q&hl=en&q=1018+Marine+Drive,+west+vancouver,+bc,canada&sll=37.0625,-95.677068&sspn=39.047881,59.853516&ie=UTF8&ll=49.326996,-123.141568&spn=0.007859,0.014613&z=16&iwloc=addr&om=1]address[/url],

 [url=http://www.westvancouver.ca/article.asp?c=962]this site[/url] says what time you can drop in there, except on fridays it closes at 9pm. for more information reply to this post, then i am going to get back to you very soon. We can have really good time. hope to see you. :)

i play ping-pong but i live in toronto :blush:

uh... i play ping pong but i live in richmond. kinda far from west van.  :(

but i know you'll eventually find ppl to play w/ you in west van cuz ping pong's becoming a really popular sport esp in the past three years.   :)

[quote="samangharib"]i wish you would be here. Anyway, how is your ping-pong? Do you play for a team or ...?
Me myself, I play it for just fun.[/quote]

i used to be on my primary school's ping-pong team and did a really bad job against other's school's ping-pong team. i won the championship in my elementary ping-pong tournment last year :lol:

hey, it doesn't matter if you don't live in west vancouver or you can't get there easily. if you do live in a town around vancouver ( which would be in a greater vancouver )  and you have or know a place which we can play ping-pong for free like this youth centre, let us know, so we can meet each other there on saturday or on sunday or other holidays. I really like to have some games with you, mathlinkers.

[quote="JingyuanMo"]If you come to UBC,then I will play with you...that center is in downtown...don't want to drive there when OIL PRICE is so  high...[/quote]

That's really good. Give its address and if you are free this saturday, we can have a game anytime you say.

[quote="JingyuanMo"]give me your detail information...name,school,etc...maybe sometime after june 22,i have exams and provincials...[/quote]
split my username after "n", that's my first and last name, for school, i don't attend any school right now, to play ping pong you don't need to know anything but maybe my first name, which you have it now, hope to see you after your exams.
Good luck with them.


It is true. If $ \frac{n\plus{}k\minus{}1}{2n}\le a_{k}\le\frac{n\plus{}k}{2n}$, then $ a_{k\plus{}1}\equal{}a_{k}(1\plus{}\frac{a_{k}}{n})$
satisfyed $ \frac{n\plus{}k}{2n}<a_{k\plus{}1}<\frac{n\plus{}k\plus{}1}{2n}$.

Use
$ \frac{1}{a_{k}}\minus{}\frac{1}{a_{k}\plus{}1}\equal{}\frac{1}{n\plus{}a_{k}}$
This relative help us find the solution of the equation. :)


geometry
circumcircle
ratio
cyclic quadrilateral
projective geometry
similar triangles
geometry theorems

I know nothing of harmonic quadrilateral, please post any properties you know about it.
Thanks!

I googled Harmonic Quadrilateral. The first result is the blank page of mathworld, second is mecrazywong's topic. And the third one only tells us: A cyclic quadrilateral for which the product of the opposite sides is the same, is called a harmonic quadrilateral.

yea, I got the same thing, then also some stuff like harmonic ranges, which I couldn't see how it connected. so could someone say properties of this mysterious object, please?

Well, the tangents to the circumcircle of the harmonic quadrilater in the opposite vertice points of it meet on a diagonal of the quadrilateral. A diagonal of a harmonic quadrilateral is a symedian of the triangle, in which the quadrilateral is divided by the other diagonal. And there is another property concerning angles made by joining the midpoint of the diagonal to two neighbouring vertices, but I don't remember it well.

Let $ABCD$ a convex cyclic quadrilateral. I note $S\in AC\cap BD,\ E\in AA\cap CC,\ F\in BB\cap DD,$
$AB=a,\ BC=b,\ CD=c,\ DA=d$. Then the following six properties are equivalently:
$1.\ ac=bd$. (ABCD - [u]harmonic quadrilateral[/u]);
$2.\ E\in BD;$
$3.\ F\in AC;$
$4.$ The point $S$ is the foot of a symmedian in at least one from the triangles $ABD,\ ABC,\ BCD,\ ACD$
from the points $A,\ B,\ C,\ D$ respectiv.
(In this case, the point $S$ is the foot of symmedians of the other three mentioned triangles as well.)
$5.\ (ACSF)$ - [u]harmonic division[/u];
$6.\ (BDSE)$ - [u]harmonic division[/u].
[u]Remark.[/u] $AA$ - the tangent in the point $A\in w$ at the circumcircle $w$ of the quadrilateral $ABCD$.
If we note $M\in AB\cap CD,\ N\in BC\cap AD$, then (from the [u]Pascal's theorem[/u])  $MN\equiv EF$.
See http://www.mathlinks.ro/Forum/viewtopic.php?t=43202

[i]General remark about notation.[/i] If the point $A$ belongs to the circle $w$ then $\boxed{\ AA\ }$ is notation for the tangent in the point $A\in w$ to the circle $w$.

@Virgil Nicula: Thank you very much i got that point.
Now please can you help me with the proof's ?

Davron

[quote="Virgil Nicula"]
$4.$ The point $S$ is the feet of the simedian in at least one from the triangles $ABD,\ ABC,\ BCD,\ ACD$
from the points $A,\ B,\ C,\ D$ respectively.
[/quote]
What 's "simedian" mean, Virgilnicula?
I don't find it in my dictionary :( .

http://en.wikipedia.org/wiki/Symmedian
http://mathworld.wolfram.com/Symmedian.html

See also [b]Darij[/b]'s website : http://de.geocities.com/darij_grinberg/ . :)

For the triangle $ABC$ the ray $[AN$, where $N\in (BC)$, is the $A$ - symmedian iff $\widehat {BAN}\equiv \widehat {CAM}$, where $M$ is the middlepoint of the segment $[BC .$ Exactly and for more knowledge see the links indicated by amiable [i]Mathmanman[/i].

[quote="Ashegh"]dear Davroni , i will promis u, not to prove them by inversion.i will post the simple solutions here as soon as possible.[/quote]

Yes, please post the proofs ...

Here is the proof!
I'll only prove that ac=bd implies the other properties.

2,3) Let the line $EB$ meet the circumference (another time) at $D'$. By similar triangles, we get:
$\frac{AB}{AD'}= \frac{EB}{EA}= \frac{EB}{EC}= \frac{CB}{CD'}$
Then: $\frac{BA}{BC}= \frac{D'A}{D'C}$
We want to show that $D = D'$. This is true because the locus of points such that $\frac{XA}{XC}= \frac{BA}{BC}$ is an Apollonian circle with center on AC, which can meet our circumference at most in two points. One is B, one is D = D'.

4) The fact that AS is a symmedian of ABC is the firse fact that you learn about symmedians... which I'll not prove here.

5,6) We use the points M,N as you defined them. From basical properties of poles and polars, me have M,N,E,F collinear. In the triangle MND, the cevians MA,NC,ED are concurrent in B. From the ruler construction of the harmonic conjugate, we have $(MNEF) =-1$. From B we project the line $EF$ to the line $AC$ and, by invariance of cross-ratios, $(ACSF) =-1$.

hey nice proof with the Apollonian circle, edriv  :)
For the harmonic points you may have used some properties which I don't
know ( but that's because i am not used to it  :D ).
Nevertheless I'll open a new topic later where everybody can post all the interesting properties and useful lemmas about poles 'n polars and harmonic points.

Now here goes my proof which uses Steiner theorem:

First I'll proof that $E \in BD$ implies $ac=bd$:

$E \in BD$ $\Rightarrow$ $DS$= $D$-symmedian
By Steiner theorem we have : $\frac{SA}{SC}= \frac{d^{2}}{c^{2}}=\frac{a^{2}}{b^{2}}$ $\Rightarrow ac=bd$.

Now I'll prove that $ac=bd$ implies $E \in BD$:

Construct $T$ such that it's the intersection of $AC$ and $BE$:

hence $T$ is the feet of the $B$-symmedian of triangle ABC.

So by Steiner theorem we have :
$\frac{TA}{TC}=\frac{a^{2}}{b^{2}}=\frac{d^{2}}{c^{2}}$.
$\Rightarrow$ $T$ is the feet of the $D$- symmedian of triangle ADC. (by the converse of Steiner theorem).
$\Rightarrow$ $T \in DE$
$\Rightarrow AC\cap BD ={T}$ (so $T=S$) and $E \in BD$.
Similarly we have $F \in AC \Leftrightarrow \frac{d}{c}=\frac{a}{b}$ or we can also use La Hire's theorem to prove that $E \in BD \Leftrightarrow F \in AC$.

Now let's finally prove that $(ACSF)=-1$:
We already know $\frac{SA}{SC}=\frac{d^{2}}{c^{2}}$.
Moreover we have by similar triangles: $\frac{FA}{FB}=\frac{a}{b}$ and $\frac{FB}{FC}=\frac{a}{b}$
$\Rightarrow \frac{FA}{FC}=\frac{a^{2}}{b^{2}}$
$\Rightarrow \frac{SA}{SC}=\frac{FA}{FC}$.
The proof for $(BDSE)=-1$ is similar.

here the link of this new topic : http://www.mathlinks.ro/Forum/viewtopic.php?p=768945#768945

I hope we could collect as many properties as possible  :wink:

[quote="Virgil Nicula"]Let $ABCD$ a convex cyclic quadrilateral. I note $S\in AC\cap BD,\ E\in AA\cap CC,\ F\in BB\cap DD,$
$AB=a,\ BC=b,\ CD=c,\ DA=d$. Then the following six properties are equivalently:
$1.\ ac=bd$. (ABCD - [u]harmonic quadrilateral[/u]);
$2.\ E\in BD;$
$3.\ F\in AC;$[/quote]

Lemma 3 of [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=103771#103771]http://www.mathlinks.ro/Forum/viewtopic.php?p=22938 post #5[/url] states that 2. is equivalent to 3., and in the proof of that lemma I actually showed more - I showed that 2. equivalent to 1. and 3. is equivalent to 1.

[quote="Virgil Nicula"]$4.$ The point $S$ is the foot of a symmedian in at least one from the triangles $ABD,\ ABC,\ BCD,\ ACD$
from the points $A,\ B,\ C,\ D$ respectively.
(In this case, the point $S$ is the foot of symmedians of the other three mentioned triangles as well.)
$5.\ (ACSF)$ - [u]harmonic division[/u];
$6.\ (BDSE)$ - [u]harmonic division[/u].
[u]Remark.[/u] $AA$ - the tangent in the point $A\in w$ at the circumcircle $w$ of the quadrilateral $ABCD$.[/quote]

You cannot really state that 5. and 6. are [i]equivalent[/i] to 3. and 2. because 5. and 6. make no sense without assuming 3. and 2., respectively (in order to speak of a harmonic division, the four points must be collinear).

[b]PS.[/b] If you don't recognize this topic - I am merging together lots of stuff in Geometry Theorems and Formulas. For instance, the "Inversion" topic is more or less a compilation of 6 different topics about inversion now.

  Darij


geometry
3D geometry
function
algebra
domain
Support
MATHCOUNTS

Can anybody solve this =) :rotfl:

Scott (a part-time mathematics professor) has decided that he wants to open an ice cream stand and therefore design a large plywood ice cream cone to go on top of his new building. Being a math professor, Scott has decided to model the ice cream scoop part of the design (the top part) with the relationship s(x) = square root of (64 - x2) and the ice cream cone part of the design (the bottom part) with the relationship c(x) = |x| - 8.  x will be measured in feet.

3.	a.	Graph the ice cream scoop on a graphing calculator and submit your best reproduction of the graph with you project.
	b.	Graph the ice cream cone on a graphing calculator and submit your best reproduction of the graph with your project.
	c.	Is the ice cream scoop relationship s(x) = square root of (64 - x2) a function? What is the domain of s(x)?
	d.	Is the ice cream cone relationship c(x) = |x| - 8 a function? What is the domain of c(x)?
	e.	Graph the scoop and the cone together on the same set of axes (make this a very large graph as you will be adding to it).
	f. 	Scott wants to put crossed support beams behind the plywood ice cream cone at the x = -3 and x = 3 lines. In order to do this, he needs to know s(x) and c(x) at those points. What are s(-3), s(3), c(-3) and c(3) (be accurate to at least 4 decimal places)?
	g. 	To build his crossed support beams, Scott needs to know how long to make his beams. What is the distance from the point (-3, s(-3)) across to the point (3, c(3))? What about from (-3, c(-3)) to (3, s(3))? Be accurate to at least 2 decimal places.
	h.	Show what his crossed support beams will look like on your graph from part (e). Label all four points mentioned in part (g).
	i.	What are the x- and y- intercepts of s(x) and c(x)? Include those on your graph from part (e).
	j. 	At what values of x will the scoop be 6 feet high? Be accurate to 2 decimal places.
	k. 	Design a grid pattern of your choice with at least 8 lines for the cone (horizontal, vertical and slanted lines are okay). What are the equations of the grid lines? How long are the grid lines? Sketch the grid pattern on a graph of the cone including all points of intersection with c(x).

I don't know about the sound, but it sure [i]looks[/i] like homework.
And why is this in mathcounts anyway?

Edit:
Right, ok. Moved. About 10 seconds before I posted.

[quote="solafidefarms"]Actually, from my understanding, homework is allowed, so long as you explain where you got stuck...But since we don't use domains or graphing calculators generally in Mathcounts, to Getting Started this goes.[/quote]

do my calc hw, kthx ._.

[quote="solafidefarms"]Actually, from my understanding, homework is allowed, so long as you explain where you got stuck...But since we don't use domains or graphing calculators generally in Mathcounts, to Getting Started this goes.[/quote]

Concept questions are allowed because they make you to learn.

Homework questions are not allowed because you're not learning anything when you simply copy the answer from us.

This topic is now locked. Please do not post homeworks.


integration
function
real analysis
real analysis unsolved

Let $(a_{n})_{n\geq 1}$ be a sequence defined as \[ a_{n}=\int_{0}^{1}\frac{x^{2n}}{1+x} dx, n\geq 1. \]
   Prove that the sequence $(na_{n})_{n\geq 1}$ is convergent and compute its limit.

 [i] Nicolae Musuroaia [/i]

Well, it is strange, but i've got $a_n = ln2 -1 +\frac{1}{2n}$.

A not-quite-rigorous justification of that result:

When $n$ is large, the function is almost entirely concentrated in a very small interval around 1. So we may as well approximate $1 + x = 2$ in the denominator. Then we're integrating $\int_0^1 \frac{n}{2}x^{2n}dx = \frac{n}{4n+2}$.

edit: I see kunny already has this... beat me to it!  :blush:  ;)


calculus
derivative
ratio
calculus computations

Show that the differential equation $ \frac {dy}{dx}=\frac{x^2-y^2}{y}$ can be transformed into a linear differential equation by means of the substitution $ y=\sqrt[]z$ .  Hence find the general solution of the equation.

This is the start of a solution given:
$ y=\sqrt[]z$

$ \frac{dy}{dz}=\frac{dy}{dz}\frac{dz}{dx}=\frac{1}{2}z{}^-{}^1{}^/{}^2\frac{dz}{dx}$

Can someone explain to me how $ \frac{dy}{dz}=\frac{dy}{dz}\frac{dz}{dx}$?

There is a typo in that solution: it should be

$ \frac {dy}{dx} \equal{} \frac {dy}{dz} \frac {dz}{dx}$.

By the way, this is a Bernoulli's differential equation.

That's just the chain rule.  Technically, the following is NOT true and is mathematically gibberish but if it helps you understand it, use it:
$ \frac{dy}{dx}\equal{}\frac{dy}{dz}\cdot\frac{dz}{dx}\equal{}\frac{dy}{\not dz}\frac{\not dz}{dx}\equal{}\frac{dy}{dx}$
If you need to see a proof of the chain rule, you could probably find it on wikipedia or mathworld or someplace else online.

JRav

Is it really gibberish? I always understood that these were just fractions (in the limit, of course), in which case this makes perfect sense.

hello, by this substitution you will get
$ \frac{1}{2\sqrt{z}}\frac{dz}{dx}=\frac{x^2-z}{\sqrt{z}}$.
This we can write in the form $ z^'+2z=2x^2$.
Sonnhard.

[quote="jimi"]Is it really gibberish? I always understood that these were just fractions (in the limit, of course), in which case this makes perfect sense.[/quote]
It is pure nonsense.  That is just Leibnizian notation.  $ \frac{dy}{dx}$ is not a fraction, it's just notation.  Leibniz wrote this because he considered the derivative a ratio of infinitesimals $ dy$ and $ dx$.  You can't actually divide $ dy$'s, $ dx$'s and so on because they don't really mean anything.  A better way to write the chain rule is $ (f\circ g(x))'\equal{}f'(g(x))g'(x)$.


I understood all of these problems, and I think I got the right answers.  Please confirm if my answers are right or not.  No solution neccessary, unless you want to add one.  (These should be pretty easy.)

1) What is the sum of the reciprocals of the roots of the equation $ax^2+bx+c=0$, where a, b, and c are reals?

2) What is the sum of the reciprocals of the roots of the equation $a_nx^n+a_{n-1}x^{n-1}...+a_2x^2+a_1x+a_0$, where $a_i$ are reals?

My Answers:

[hide="Number 1"]$-\frac{b}{c}$ [/hide]
[hide="Number 2"]$-\frac{a_1}{a_0}$[/hide]

OK, cool.  Those were practice problems from Mathematical Ideas by Adeel Khan, and I was just checking to make sure that I got them right.  :D

(this post is meant to explain the underlying concepts that can be used for the problems in the first post of this thread)

Ok, so let's say we have a polynomial $P(x) = a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$.  Vieta's formulas tell us that the sum of [i]all[/i] of the roots (including complex) of this equation equals $-\frac{a_{n-1}}{a_n}$ and the product of the roots equals $(-1)^n\frac{a_0}{a_n}$.  More generally, the sum of products of the roots taken $k$ at a time equals $(-1)^k\frac{a_{n-k+1}}{a_n}$.

Now, an intriguing problem is to find the polynomial whose roots are reciprocals of the roots of $P(x)$.

To figure this out we [i]transform[/i] the polynomial.  We do this by putting a $1/x$ in place of each $x$.  Thus, we have \[ a_n\left(\frac 1x\right)^n+a_{n-1}\left(\frac 1x\right)^{n-1}+\cdots+a_1\left(\frac 1x\right)+a_0. \]

But this isn't a polynomial (polynomials have to have all terms of the form $x^k$ where $k$ is a nonnegative integer).  To make this a polynomial, we multiply through by $x^n$ to get \[ a_n+a_{n-1}x+\cdots+a_1x^{n-1}+a_0x^n. \]

In other words, the polynomial that has roots that are reciprocals of $P(x)$'s is the polynomial formed by reversing the coefficients of $P(x)$!

wow, i knew how to find the sum of the reciprocals the long way, but that's really cool


integration
logarithms
trigonometry
calculus
limit
calculus computations

Perhaps, contour integration might help!

yes, by complex integration I already have solved it. it's interesting to see by real integration.

Well... I hope this solution may be justified, since I interchanged sum and integral where LDCT seems not be able to be applied directly. Even I doubt whether my answer is correct. Numerical calculation argues different values whenever I try! [img]http://myhome.shinbiro.com/~sosQED/images/m_0507.gif[/img]

By suitable integration by substitution, it suffices to evaluate when $ a = 1$.

Put $ I = - \int_{0}^{\infty} \frac {\ln | \sin (\alpha x)|}{1 + x^2} \, dx$. By the identity $ \ln |\sin x| = - \ln 2 - \sum_{n = 1}^{\infty} \frac {1}{n} \cos (2n x),$ we obtain

\begin{eqnarray*}I & = & \int_{0}^{\infty} \frac {1}{1 + x^2} \left( \ln 2 + \sum_{n = 1}^{\infty} \frac {1}{n} \cos (2n \alpha x) \right) \, dx \\
& = & \frac {\pi}{2}\ln 2 + \sum_{n = 1}^{\infty} \frac {1}{n} \int_{0}^{\infty} \frac {\cos (2n \alpha x)}{1 + x^2} \, dx \quad \quad (\longleftarrow \text{Yes, here!}) \\
& = & \frac {\pi}{2}\ln 2 + \frac {\pi}{2} \sum_{n = 1}^{\infty} \frac {e^{ - 2n |\alpha|}}{n} \\
& = & \frac {\pi}{2}\ln 2 - \frac {\pi}{2} \ln(1 - e^{ - 2 |\alpha|}). \end{eqnarray*}

So

$ \int_{0}^{\infty} \frac {\ln \sin^2 (\alpha x)}{1 + x^2} \, dx = \pi \ln \left( \frac {1 - e^{ - 2 |\alpha|}}{2} \right).$

yes , it's true, you can change integral and sum since series uniformal converge.
Great  :)

does the integral even exist ?  :(  i mean, shouldn't it   diverge to $ \minus{}\infty$ ?

what you probably want  is ( for  $ a,\; m \neq 0$  ) ,

$ PV \; \int_0^\infty \frac{\ln\;\sin^2\, mx}{a^2\plus{}x^2} \;dx$

$ \;\equal{}\;\lim_{\varepsilon \to 0^\plus{}}\;\; \sum_{k\equal{}0}^\infty \;\; \int_{\frac{k\pi}{2m}\plus{}\varepsilon } ^ {\frac{(k\plus{}1)\pi}{2m}\minus{}\varepsilon} \quad  \minus{} \;\underbrace{ \frac{\bigg|\; \ln\;\sin^2\, mx\;\bigg| }{a^2\plus{}x^2} }_{\equal{}\;\; \textbf f(x) } \;dx$

even with $ PV$, i am not sure   exactly what the behavior is......   seems to me  the integral should still diverge..  :maybe:

[quote="misan"]does the integral even exist ?  [/quote]
Yes, it does.
Yes there is point's where :
 $ \sin^{2}mx \equal{} 0$
and therefore $ \ln 0 \equal{}\minus{}\infty$ but this $ \minus{}\infty$ is logarifmical.
 for example :
$ \int_{0}^{1} \ln x dx < \infty$


given: f(x)=2x+f(x-1) and f(1)=4, find f(9)

A general sum for f(x):
\[ f(x) \equal{} 2 \plus{} \sum_{i \equal{} 1}^x 2x \equal{} 2 \plus{} 2\sum_{i \equal{} 1}^x i \equal{} 2 \plus{} x(x \plus{} 1)
\]
So f(9)=92.

EDIT: Thanks to SnipedYou

[hide="solution"]

$ f(9) \equal{} 2(9) \plus{} f(8)$
$ f(9) \equal{} 2(9) \plus{} 2(8) \plus{} f(7)$ because $ f(8) \equal{} 2(8) \plus{} f(7)$
$ f(9) \equal{} 2(9 \plus{} 8 \plus{} 7 \plus{} ... \plus{} 2) \plus{} f(1)$
$ f(9) \equal{} 2\left(\frac {9\cdot 10}{2} \minus{} 1\right) \plus{} 4 \equal{} 2(44) \plus{} 4 \equal{} 92$[/hide]

someperson01:

Your mistake comes from simplifying. Correction:

$ f(x) \equal{} 2 \plus{} \sum_{i \equal{} 1}^{x} 2i \equal{} 2 \plus{} 2 \sum_{i \equal{} 1}^{x} i \equal{} 2 \plus{} 2\cdot \frac {x(x \plus{} 1)}{2} \equal{} 2 \plus{} x(x \plus{} 1)$
So $ f(9) \equal{} 92$

Watch your notation to, the bottom number ($ i$ in this case) should be what's being summed.

Wow, I have no idea whatsoever where I got that (x+2) from. Thanks. :(


algorithm function trigonometry calculus integration algebra polynomial

The standard algorithm used in our classes for finding Fourier series of a function is:

1. Checking whether Dirichlet's conditions are fulfilled (we could go without this, though... but since most of them fulfill the conditions, it's helpful to know that the series we are calculating exist :) )
2. finding coefficients using the formulas
3. proving uniform convergence of the series to validate step 2.

Step 3 is in most cases done using Weierstrass M test. But in some cases, the TA omitted the proof ('as trivial'), but I couldn't see the argument for uniform convergence in it:

i.e. $ \sum\frac{\sin(2k\minus{}1)x}{2k\minus{}1}$, $ \sum \frac{(\minus{}1)^n\sin nx}{n}$ etc.

Weierstrass doesn't help, and it seems to me that Abel and Dirichlet don't help either: logical question arose: do the series converge uniformly at all? (browsing ML brought me to http://www.mathlinks.ro/viewtopic.php?t=225693).

But if the series above don't converge uniformly, how can I justify the integration used to find its coefficients?

Those particular series do not converge uniformly. They converge pointwise (and Abel does help, applied appropriately) to functions with jump discontinuities. Since the uniform limit of continuous functions is continuous, that proves that the convergence cannot be uniform.

Interesting, that was the exact argument I gave in an argument today on the convergence of these series (it was more like: "Look, all functions you expanded in FS have jump discontinuities-how can you expect that a continuous function-the series converges to the original, discontinuous function :) "). But still, I don't know how to show that my Fourier expansion is valid without uniform convergence...

The first question to ask is "Valid in what sense?". For instance, it is easy to prove the general theorem that the Fourier series of any square integrable function converges to it in the $ L^2$-sense. If you want pointwise convergence for piecewise smooth function, then [url=http://eom.springer.de/d/d032520.htm]Dini criterion[/url] is your best friend. Of course, you need to learn its proof once but then you do not need to think of the issue in each particular case separately.

When dealing with piecewise continous functions, we would define its (pointwise convergent) Fourierseries as being equal to $ \frac {1}{2}\left(f(x \minus{} ) \plus{} f(x \plus{} )\right)$.

Example (not doing all of the calculations):

Let $ f$ be a periodic function with period $ 2\pi$ such that:
(how can I use the \cases environment here?) (thanks)

$ f(x) \equal{} \begin{cases} \minus{} 1 & x\in( \minus{} \pi,0) \\
1 & x\in(0,\pi) \\
0 & x\in\{\minus{}\pi, 0, \pi\} \end{cases}$

Obviously $ f$ is discontinous but still piecewise continous.

We get as Fourierseries:

g(x):= $ \frac {4}{\pi} \sum_{m \equal{} 0}^{\infty} \frac {\sin(2m \plus{} 1)x}{2m \plus{} 1}$

We get for example that

$ g(0) \equal{} 0 \equal{} \frac {1}{2}\left( \minus{} 1 \plus{} 1\right) \equal{} \frac {1}{2}\left(f(0 \minus{} ) \plus{} f(0 \plus{} )\right)$

Thank you, fedja, Dini's criterion was exactly what I was looking for.

@Tyl:

[code]f(x)=\begin{cases}-1 & x\in(-\pi,0) \\
0 & x\in(0,\pi)\end{cases}[/code]

gives $ f(x)\equal{}\begin{cases}\minus{}1 & x\in(\minus{}\pi,0) \\
0 & x\in(0,\pi)\end{cases}$

[quote="Tyl"]When dealing with piecewise continous functions, we would define its (pointwise convergent) Fourierseries as being equal to $ \frac{1}{2}\left(f(x\minus{}) \plus{} f(x\plus{})\right)$.[/quote]

Ah, but piecewise continuity is not enough for pointwise convergence! You cannot just "define" the sum to be something. The whole point was to show that the sum (understood in the usual sense as the limit of partial sums) equals something. You are right in the sense that it cannot equal anything else but $ \frac 12[f(x\minus{})\plus{}f(x\plus{})]$ if the function is piecewise continuous and the series converges but it may just diverge at $ x$.

Besides the Dini Criterion, I also like the theorem that the Fourier series of a function of bounded variation on $ \mathbb{T}$ converges pointwise everywhere (to the average of the right and left hand limits.) (Being of bounded variation overcomes fedja's objection - but there are piecewise continuous, even continuous, functions that are not of bounded variation.)

Fejer's theorem is also another way to talk about validity. Given an $ L^1$ function, we know we can reconstruct the function from the information contained in its Fourier coefficients. We just have to give up on using the limit of the partial sums and use Cesaro means instead.

[quote="fedja"]
Ah, but piecewise continuity is not enough for pointwise convergence! You cannot just "define" the sum to be something. The whole point was to show that the sum (understood in the usual sense as the limit of partial sums) equals something. You are right in the sense that it cannot equal anything else but $ \frac 12[f(x \minus{} ) \plus{} f(x \plus{} )]$ if the function is piecewise continuous and the series converges but it may just diverge at $ x$.[/quote]

Yes, you are right. Upon rereading what I wrote I noticed that the word "define" is wrong. Sufficient for pointwise convergence would be periodic with piecewise smoothness, would it not?

[quote="fedja"]The first question to ask is "Valid in what sense?". [/quote]

Just to answer this question, for the sake of completeness.

The formulae for $ a_n$ and $ b_n$ coefficients in Fourier expansion are derived by term-by-term integration. To validate the derivation of coefficients, we need the uniform convergence of both the original and integrated series. In most cases, checking uniform continuity of the final Fourier series is sufficient to 'approve' the formulae. And then, when the series do not converge uniformly, the process of deriving coefficients' formulae at first glance lacks rigor-we couldn't have integrated it term-by-term.

Uniform convergence is a straitjacket; we don't need to so confine Fourier analysis. Many very interesting functions have Fourier series that don't converge uniformly. Fedja's question "valid in what sense?" has several different productive answers. I'll give you my favorite.

We can start by assuming that the functions we are talking about are "very nice": $ C^2,$ $ C^{\infty},$ trig polynomials - whatever we feel like using. Since we can interchange sums and integrals freely, we establish the basic formalities and formulas.

Now, the Fourier coefficients are defined by an integral:
\[ \widehat{f}(k)\equal{}\frac1{2\pi}\int_0^{2\pi}f(x)e^{\minus{}ikx}\,dx.\]
This much makes sense whenever $ f$ is integrable - that is, for $ f\in L^1.$

At that point, the most powerful result that confers "validity", at least to me, is Fejer's Theorem:

Define $ S_nf(x)\equal{}\sum_{k\equal{}\minus{}n}^n\widehat{f}(k)e^{ikx}.$

Define $ \sigma_Nf(x)\equal{}\frac1{N\plus{}1}\sum_{n\equal{}0}^{N}S_nf(x)\equal{} \sum_{k\equal{}\minus{}N}^N\left(1\minus{}\frac{|k|}{N\plus{}1}\right)\widehat{f}(k)e^{ikx}.$

The various conclusions of Fejer's theorem include these:

$ \sigma_Nf$ tends to $ f$ in $ L^1$ for all $ f\in L^1.$ That is, $ \lim_{N\to\infty}\|\sigma_Nf\minus{}f\|_1\equal{}0.$

$ \sigma_Nf$ tends to $ f$ uniformly for all $ F\in C(\mathbb{T}).$

$ \sigma_Nf(x)$ tends to $ f(x)$ almost everywhere for all $ f\in L^1.$

Among the consequences of this: if $ \widehat{f}(k)\equal{}0$ for all $ k$ then $ f\equal{}0$ as an element of $ L^1,$ meaning that $ f$ is zero almost everywhere. Equivalently, if $ \widehat{f}(k)\equal{}\widehat{g}(k)$ for all $ k,$ then $ f\equal{}g$ as an element of $ L^1,$ and $ f\equal{}g$ almost everywhere. We also get that trig polynomials are dense in $ L^p$ for $ 1\le p<\infty$ and trig polynomials are dense in $ C$ (the latter being the trigonometric version of the Weierstrass Approximation Theorem.)

One way to use these ideas to test validity: We have $ \sum_{k\equal{}\minus{}\infty}^{\infty}c_ke^{ikx}\equal{}f(x)$ in some sense of limit (possibly pointwise convergence but some other things are possible.) Is that a Fourier representation? Take the $ f$ that we now have and compute the integrals that define $ \widehat{f}(k).$ If $ \widehat{f}(k)\equal{}c_k$ for all $ k,$ then we have a valid representation.


[quote="Farenhajt"]Solve the system of equations:

$\begin{cases}|x-1|+|y-5|=1\\ y=5+|x-1|\end{cases}$[/quote]

[hide]sub the second equation into the first to get $2|x-1]=1$

$x=\frac32,\frac12$
$y=\frac{11}{2}$[/hide]

[quote="Farenhajt"]Solve the system of equations:

$\begin{cases}|x-1|+|y-5|=1\\ y=5+|x-1|\end{cases}$[/quote]

$y+|y-5|=6$, $y<5\implies 5=6$, $y>5 \implies y=\frac{11}{2}$, then
$|x-1|=\frac{1}{2}\implies x=\frac{3}{2}, \frac{1}{2}$

How about graphing? That's useful too. :D

[quote="Chocolate milk"]$y+|y-5|=6$, $y<5\implies 5=6$, $y>5 \implies y=\frac{11}{2}$
  Call me stupid but i don't understand it[/quote]

$y<5$
$y+|y-5|=y-y+5=5\ne6$

$y>5$
$y+|y-5|=2y-5=6 \Rightarrow y=\frac{11}{2}$

Thanks you very much. Now I understand. :)


vector
advanced fields
advanced fields unsolved

Let $ x$ be a vector in $ R^n$.
The problem is to minimize $ max(x_i)$
satisfying $ Ax \leq b$
and $ x_i \geq 0$

How to express this problem as a linear program?
I guess its best to start with
maximize $ z$
s.t $ Ax \leq b$
$ x_i \geq 0$
and
$ z \equal{} x_i \plus{} s_i$ for all i between 1 and n.
but the last constraint of
$ s_1 s_2 ... s_n \equal{} 0$
is not linear....

[quote="kevinatcausa"]How about adding a new variable $ z$ like you suggest, but then just throwing in the additional constraints $ z \geq x_i$, and then minimizing $ z$?[/quote]
You are right  :blush:


How many ordered triples of integers $(a,b,c)$ satisfy \[ |a+b|+c=19 \quad \text {and} \quad ab+|c|=97? \] \[ \text {(A) } 0 \qquad \text {(B) } 4 \qquad \text {(C) } 6 \qquad \text {(D) } 10 \qquad \text {(E) } 12  \]

[hide="Partial Solution"]First let's assume c is non negative. Try $c=0$.

We have $|a+b|=19$ and $ab=97$. Hmm...looks like a quadratic. a and b will be the roots of $x^2+-19x+97$. Now let's take the discriminant, which is $19^2-4 \times 97$ (remember that the coefficient of the x term's sign does not affect the discriminant). Hmm  :roll: ...I think this is negative  :P . Well, that means a and b would be imaginary  :!: .  Notice that if $c=1$ (without bothering to write the quadratic), we get a discriminant of $18^2 - 4 \times 96$. As c gets larger, we have an increase of 4 each time, but it also decreases quadratically, so as c gets larger our discriminant will always be negative.

Now let's try c is negative. First try $c=-1$.

We have $a+b=+-20$ and $ab=96$. Our quadratic is $x^2 +- 20x +96$. So, we have a discriminant of $20^2 - 4 \times 96 = 16$. Oooo we have a solution here. We either have $a=12$ and $b=8$, $a=-12$ and $b=-8$, or the reverse with a and b. So we have solutions (12,8,-1), (-12,-8,-1), (8,12,-1), and (-8,-12,-1).

Okay, what we need to figure out now is when will our discriminant be a square? First let's expess it in terms of c (for negative c). We have $|a+b|=19+|c|$ and $ab=97-|c|$. So, our discriminant is $(19+|c|)^2 +4(|c|-97) = c^2+38|c|+361+4|c| - 27 = c^2 +42|c| - 27$. We must figure out when this is a square.[/hide]

[hide]Say $c\geq 0$, then $c=97-ab$, substitute into first equation,

$|a+b| + (97-ab)=19$, then either

$a+b-ab=-78$, => $(a-1)(b-1)=79$, or
$-a-b-ab=-78$ => $(a+1)(b+1)=79$
Quickly check the solutions to each equation we see that none of them satisfy  $|a+b| \leq 19$ (since$c \geq 0$)

So then $c \leq 0$, then
$|a+b| + (ab-97)=19$, then either
$a+b+ab=116$, => $(a+1)(b+1)=117$
or
$-a-b+ab=116$ => $(a-1)(b-1)=117$

$117=3^2*13$, so there are 6 positive divisors, 12 integer divisors. In the first equation, we need $a+b \geq 0$, note that if (a, b) is a solution, then (-a-2, -b-2) is also a solution, and exactly one of a+b, -(a+b)-4 is positive, hence we have 12/2=6 pair of solutions. Similarly we have 6 pair of solutions in the second equation, hence 12 in total.[/hide]


In any given year, the dates (represented as month/day) 4/4, 6/6, 8/8, 10/10 and 12/12 all fall on the same day of the week.  June 3, 2020 is a Wednesday.  What day of the week is December 15, 2020?

[hide="Solution"]June 6,2020 is a Saturday. December 12,2020 is a Saturday. December 15,2020 is a Tuesday.[/hide]


Hi
im currently a sophomore and ive participated in some math competitions when i was younger but not in the past couple years..

this sophomore year i got 132.5 on AMC and on AIME-- im not sure what my score was but i know i didnt do so well. (i didnt know a lot of the problems+made some silly mistakes. ) Well i think if i didnt make any mistakes i could have gotten about 7 of them right..

This year i wasnt told about the test ahead of time and was busy with some other stuff so i only did some past amc problems for the preparation (i went through 1990-2005 amc problems and no aime--so it makes sense that i messed up on AIME i guess)
now im trying to prepare for the next year AMC/AIME(my goal is to quallify for the USAMO.. hopefully. considering this year was my first math competition in many years and i was not too well prepared for it, i think i can make it next year if i try hard.)

im going to start out this summer with AOPS books.. i heard they are good? but i dont know which one i should start with.  would you recommend that i start with volume 1 or volume 2?
also are they easy to follow? im a little concerned bc i think i'll pretty much have to teach myself.

i am also thinking about taking aime problem series class on this website. would you say that  classes here are helpful (for getting better on AIME quickly!) when you dont have much time for preparation and havent exactly done any preparation?

i really appreciate your time to read this and hopefully some tips!

AMC 10 or AMC 12?

Yes, the AoPS books and courses are [i]very[/i] good.

Regarding AoPS 1 vs. AoPS 2, you should probably get both of them if you don't mind spending a bit of extra money. You may want to do a bit of skipping around AoPS 1, but there'll probably be a fair amount of new material. It also may even be useful to read some ideas you already know, as the AoPS books have an approach that helps you deeply comprehend the topics and apply them to new situations. In particular, I recommend that you look at the [url=http://www.artofproblemsolving.com/Books/AoPS_B_TOC.php?page_id=1]table of contents listed on the site[/url] for each book and decide how familiar you are with the material.

Yeah I would say they're easy to follow? Remember, if you have a question about a problem, don't be afraid to post here.

Regarding courses, sure, AIME Problem Series would be a great idea if you want to improve on AIME. You may also want to try some of the introductory and maybe intermediate courses to help improve your problem solving. Look [url=http://www.artofproblemsolving.com/Classes/AoPS_C_ClassesS.php]here[/url] for tests that determine whether you're ready to take a course and whether you don't need a course.

[quote="eryaman"][quote="h0425"]amc12.

thank you very much. the links help a lot![/quote]Wow, a 132.5 is quite impressive...I would normally expect someone that did so well to do better on AIME, but I guess you never know.  :?[/quote]

Not really.  Some people are better at multiple choice, and some vice-versa.  Especially if one takes the AIME for the first time. it is quite difficult, despite a high AMC 12 score.

[quote="h0425"]
...This year i wasnt told about the test ahead of time and was busy with some other stuff so i only did some past amc problems for the preparation... [/quote]
i feel you pain, this past AMC my school told me 3 days in advance. I had report to do and did not get as much practice time as i would have liked.
I bought both books but they just got here so I couldn't really tell you too much about them.


How would I simplify $S=n+(n/2)+(n/3)+...+(n/x)$ into a formula?  I believe this will take us to a cycle repeated infinite times.

Masoud Zargar

The limit as $x$ approaches $\infty$ is going to make $S=\infty$ (because $\sum^{\infty}_{x=1}\frac{1}{x}=\infty$), so I doubt it's going to expand all the way to infinity.

It won't expand to infinity, but since you have to have an LCD, you will also need a fraction for the numerator, and hence an LCD.  This cycle will continue on forever, so will will alway need an LCD.  This will lead you to no formula, unless there is a way that I am not aware of.

Masoud Zargar


number theory
relatively prime
algebra unsolved
algebra

Prove that if $x,y$ and $z$ are pairwise relatively prime positive integers, and if $\frac{1}{x} + \frac{1}{y} = \frac{1}{z}$, then $x+y, x-z, y-z$ are perfect squares of integers.

Hmmm... rewrite it as $(z(x+y) = xy.$ Since $z$ is coprime with $x$ and $y$ it follows that $z=1$ and the equation reduces to $x+y=xy$, which is $(x-1)(y-1) = 1.$ Thus $x=y=2$ which are not coprime....a contradiction....

So, do you meant $\gcd(x,y,z) = 1$ instead of pairwise relatively prime?

Pierre.

[quote="pbornsztein"]Hmmm... rewrite it as $ (z(x+y) = xy. $ Since $ z $ is coprime with $ x $ and $ y $ it follows that $ z=1 $ and the equation reduces to $ x+y=xy $, which is $ (x-1)(y-1) = 1. $ Thus $ x=y=2 $ which are not coprime....a contradiction....

So, do you meant $ \gcd(x,y,z) = 1 $ instead of pairwise relatively prime?

Pierre.[/quote]

No, what I posted there is what the original problem said.

Pierre, is this same as your solution (I didn't try the problem yet till now)?

Solution:

By given defintion of the problem, we have $(x,y) = 1, (y,z) = 1, (x,z) = 1$. Hence, $\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{x+y}{xy} = \dfrac{1}{z}$.

Cross multiplying this gives result of $z(x+y) = xy$ and dividing by $z$ gives $x+y = \dfrac{xy}{z}$. However, $z$ is coprime to $x$ and $y$ and only way RHS is an integer is if $z = 1$. So, we get

$x+y = xy\\ (x-1)(y-1) = 1$

which is true if $x = 2, y = 2$. But since this condition contradicts the given definition on the problem, there is NO SOLUTION.

Right?  :D

Just wondering, is $(x,y) = (2,2)$ only pair that makes above equation true? I'm not sure on how to find the solution to that equation in positive integer other than the obvious pair (2,2).

[quote="Silverfalcon"]Pierre, is this same as your solution?[/quote]

Yes...

[quote="Silverfalcon"]
Just wondering, is $(x,y) = (2,2)$ only pair that makes above equation true?[/quote]

Clearly not, because the equation is homogeneous so that if $(x,y,z)$ is a solution then $(kx,ky,kz)$ is a solution too for each positive integer $k$.

In fact, the solutions are the triples of the form $(pq(q+c), pc(q+c), pqc)$ where $p,q,c$ are any positive integers. (1)

From the initial remark, we may restrict ourselves to the solutions $(x,y,z)$ such that $\gcd(x,y,z) = 1.$
Let $x=da$ and $z=dc$ where $\gcd(a,c) = 1.$
Then the equation is $c(da+y) = ay.$ Thus $c$ divides $ay$ and is coprime to $a$, so that $c$ divides $y$, and $y=cb.$
The equation is now $da+cb=ab.$ Thus $a$ divides $cb$ and is coprime with $b$ so that $b=ka.$
And the equation is now $d+ck = ka.$ It follows that $k$ divides $d$, but since $\gcd(x,y,z) = 1$ it forces $k=1.$
And the equation reduces to $a=q+c.$ Which leads to the triples (1) (the factor $p$ comes from the initial remark).

Note that in the case $p=1$, each of the number is indeed a perfect square. Therefore, I hink the correct statement is $\gcd(x,y,z)=1$...

Pierre.


If $ \log x\minus{}5 \log 3\equal{}\minus{}2$, then $ x$ equals:

$ \textbf{(A)}\ 1.25 \qquad
\textbf{(B)}\ 0.81 \qquad
\textbf{(C)}\ 2.43 \qquad
\textbf{(D)}\ 0.8 \qquad
\textbf{(E)}\ \text{either 0.8 or 1.25}$

[hide="Solution"]
$ \log x\minus{}5\log 3\equal{}\minus{}2 \implies \log x\equal{}5\log 3\minus{}2 \implies \log x\equal{}\log 243\minus{}\log 100$

Thus, $ \log x\equal{}\log \frac{243}{100}$

$ x\equal{}2.43$

Answer: C [/hide]

[hide="Alternatively..."]$ \log x\minus{}5\log 3\equal{}\log x\minus{}\log 3^5\equal{}\log\frac{x}{3^5}\equal{}\minus{}2\implies x\equal{}3^5/100\equal{}\boxed{\textbf{(C)}\ 2.43}$.[/hide]


[quote="creatorvn"]
- $ \frac {2005}{2004^2 \plus{} 1}$[/quote]
Is it $ \minus{} \frac {2005}{2004^2 \plus{} 1}$ or $ \frac {2005}{2004^2 \plus{} 1}$ ?
My idea is prove that exitst $ n\in N$ so that $ n < LHS < n \plus{} 1$

[url]http://www.mathlinks.ro/viewtopic.php?t=194258[/url]

This has been posted twice? (Homework?)

tdl, your idea is the same one I used


induction
search
combinatorics proposed
combinatorics

There are 20 flowers and 10 colours, each colour is painted on 2 flowers. Put them in 10 baskets in order to satisfy that there are different ways of choices from each basket a flower so that we have enough 10 colours. Let k be the number of different ways of choices that can be done. Prove that $k=2^n$.

What's the reason of having $n$ in the problem? Do you want to solve for a generalization?

I see that it is really joke!

We will use induction on number of flowers $2m$.
If $m=1$ then we have two flowers with the same color, and statement is clear.

Suppose $m>1$. There are two general cases.

Case 1. There are basket which contains exactly one flower. In this case anyway  we must choose this flower. Consider another basket which contains flower this the same color and exclude this flower from consideration. So we have $2(m-1)$ flowers and $m-1$ baskets. Apply induction.

Case 2. Each basket contains exactly 2 flowers. Consider first basket. Suppose it contains flowers of colors $a$ and $b$. Choose arbitrary flower from it, for example, $a$. Then in order to obtain a flower of color $b$ we need to search for second basket with color $b$. Obtained basket contains colors $b$ and $c$. We choose $b$ and search for basket with color $c$. And so on.
There is a moment when new color we looking for is $a$, i.e. we obtained chain of baskets: $(a,b)$, $(b,c)$, ..., $(y,z)$, $(z,a)$.
It is possible that we chose color $b$ from first basket. In this case we will obtain the same chain in reverse order:  $(a,b)$, $(z,a)$, $(y,z)$, ...,$(b,c)$.
Thus obtained chain gives us exactly two ways of choice from its elements. And we have some remaining baskets which accordingly to induction proposition give us $2^n$ ways of proper choice of flowers. So common number of ways to choose $m$ flowers is $2\times 2^n=2^{n+1}$, i.e. it is a power of 2.


AwesomeMath
summer program
AMC
USA(J)MO
USAMO
PROMYS
Ross Mathematics Program

Have you guys received letter from Awesome math summer program? Because the directors are Dr. Titu and Dr.Feng,  the program seems attractive to me. But I have planned to go to summer school(English 11.  I mean if I don't take summer school this year, I will have 3 English credit plus an Esl credit, which is just enough to graduate, but I think it will be nicer for college to see me 4 English credits ) this year, although I haven't signed up yet. What do you think about this? I do not know much about this program, so I do not know if it's worth to give up summer school. I am junior now, and I qualified for this year's USAMO, but not MOP.

What is your impression about this program?

I appreciate any suggestion or opinion.

Definitely go to AwesomeMath if you can.

[quote="Go Beyond"]Well, ESL is one of the English credits, I think. Also, you can make more English credits in senior year. The college will not only see your credits in or before junior.
Those camps are interesting, and also help you improve your English skills.
So try it![/quote]

I don't see how Awesomemath might improve one's english skills, but keep in mind that he's trying to get credits, not necessarily improve his english.

I think it depends on how far you want to go in competitive math. You only have one year left, and you'll only make MOP if you finish in the top 12 in the USAMO next year, so you may decide that it's not worth it. Keep in mind that Awesomemath is more or less only for contest math, as opposed to a program like MathCamp. If you think that having an extra english credit will make a difference in terms of colleges, then go for it.

I have asked my counsellor, and she called the college admission office, and said I can go to the math program.

Now, I just want to ask that do you think awesomemath summer program is a good one? I don't know much about it.

Thank you for replying

Awsome math is...awsome, but you should apply quickly since deadline is coming up (if not past, im not sure).  Also, there's hcissm, promys, and [b]ross[/b], and ross is the best. :D You can always go to:
[url]http://www.ams.org/outreach/mathcamps.html[/url]
to see a list.

[quote="Phelpedo"]
I think it depends on how far you want to go in competitive math. You only have one year left, and you'll only make MOP if you finish in the top 12 in the USAMO next year, so you may decide that it's not worth it. Keep in mind that Awesomemath is more or less only for contest math, as opposed to a program like MathCamp. If you think that having an extra english credit will make a difference in terms of colleges, then go for it.[/quote]

For all practical purposes, $MOSP\neq competitive math$. Also, you have a lot more than one year left - you have your life left. In terms of just competitive math, you have four years of the Putnam (in America), or other undergrad contests in other countries. I don't know the quality of AwesomeMath, but you should probably go if you want to.

No one can be sure because AwesomeMath is new, although the same people that run MOSP also run AwesomeMath, so I would guess that it's a very good program.

Not to mention that many of those people write, grade, review, and/or run the AMC contests  ;)

[quote="jb05"]
For all practical purposes, $MOSP\neq competitive math$.[/quote]

?????... Last time I checked the goal of MOP is to train people for the future IMO team.

[quote="Phelpedo"]
I think it depends on how far you want to go in competitive math. You only have one year left, and you'll only make MOP if you finish in the top 12 in the USAMO next year, so you may decide that it's not worth it. [/quote]

Why is it not worth it?

Thank you for all your suggestions
I've decied to apply

[quote="chess64"][quote="beta"][quote="jb05"]
For all practical purposes, $MOSP\neq competitive math$.[/quote]

?????... Last time I checked the goal of MOP is to train people for the future IMO team.[/quote]

yeah jb05 MOP is very much equal to competitive math[/quote]

Either way, we are talking about Awesomemath, which is purely geared towards preparation for the USAMO. My point was that, for someone entering senior year without a whole lot of chance to make the IMO, it might be better to focus on the college application.

Actually, I would put it this way:

AwesomeMath is very much geared towards problem solving. Just because they do problem solving doesn't mean it's meant specifically for the USAMO. If you've never been exposed to that stuff before, it could really change your perception of math as a whole. For example, I had a very weak understanding of algebra until I did a lot of Mathcounts problems. On the surface, it doesn't seem like one English credit is *that* much of a deal, and colleges would be interested in what you do with your summers, but obviously the only person who can judge is you.

[quote="Phelpedo"]
Either way, we are talking about Awesomemath, which is purely geared towards preparation for the USAMO. My point was that, for someone entering senior year without a whole lot of chance to make the IMO, it might be better to focus on the college application.[/quote]

Very good advice. Very good point. Phelpedo has the wisdom not only in math.

Awesomemath states "this is an initiative in response to numerous requests coming from the parents and the teachers of bright students who have not yet shined at the Olympiad level".

I believe that rising 8 and 9th (maybe 10th) graders will be greatly benefited from the program. Also if you are able to get MOP, you have graduated from this program.

[quote="chess64"][quote="beta"][quote="jb05"]
For all practical purposes, $MOSP\neq competitive math$.[/quote]

?????... Last time I checked the goal of MOP is to train people for the future IMO team.[/quote]

yeah jb05 MOP is very much equal to competitive math[/quote]

Maybe I wasn't clear. I meant that MOSP is not the extent of competitive math--that the view that "there is no point to competitive math after MOSP is over" is completely wrong. As you said, the goal is to train people for the IMO team, but if you fail to go to the IMO, the purpose of your (mathematical) life is not completely over. In case you haven't noticed, there is such a thing as a professional mathematician. ALL they do is problem solving.

The USAMO and Putnam are very similar. If you practice for the USAMO, you will do better on the Putnam and in professional math.

[quote="chess64"][quote="jb05"]In case you haven't noticed, there is such a thing as a professional mathematician. ALL they do is problem solving.[/quote]

Actually I think most mathematicians don't do much problem solving at all... I could be wrong though.[/quote]

They sort of do in that similar skills (to some degree) are required. While an excellent olympiad problem solver may not be good at research, it is very doubtful that a successful professor couldn't solve an easy olympiad problem.

Which do you think is better?
Awesomemath or Promys?

This is a good thread, but it belongs in [url=http://www.artofproblemsolving.com/Forum/index.php?f=136]Other U.S. Contests and Programs[/url], where I would have expected to find it.

There are lots of good summer programs. There has never been a session of Awesome Math before, but it has some great math coaches connected with the program.

There's so many useful ideas being thrown around here, but I'd just like to throw in some of my own:

Someone said that not making IMO or even MOSP isn't the end of the world... a good point but also a practical one considering like .0025% of people taking the AMC make it to IMO, a goal that if put upon yourself only would cause unhappiness.

I think doing something only for the intention of getting into college isn't a good way to plan your schedule... but as reassurance I think a college would be much more impressed if you attended a challenging math program that you were actually interested in than taking an english course, but if you like english it might be a tough decision.

And... it never hurts to apply to awesome math, because if you don't get in I assume you can always go back to the english class as a backup.

I would apply first and then decide if you get in. (Though you better hurry, the application is due on the 15th of May.)

It depends on what you plan to do on college as well.

For instance, if you plan to major on mathematics, having 4 credits of English will be sufficient. However, it might be better if you consider sharpening your skills on English since you're still on ESL. This definitely applies for SAT and maybe, TOEFL although it's easier on English (I think that's what I heard). I used word "better" because although AwesomeMath or other summer programs are good, you already showed your uniqueness by qualifying to USAMO.

But that's just me. Some people choose to do hardcore on math over other things and they do help on admissions.

If you want some information about AweomeMath go to the Math Jam transcript at [url]http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=157?[/url]


[quote="thanhnam2902"][color=darkblue]Let $ a, b, c, d, e > 0$. Prove that: $ \frac {9}{a} \plus{} \frac {16}{b} \plus{} \frac {25}{c} \plus{} \frac {36}{d} \plus{} \frac {47}{e}\ge \frac {2025}{a \plus{} b \plus{} c \plus{} d \plus{} e}$[/color][/quote]

Try $ a\equal{}3$, $ b\equal{}4$, $ c\equal{}5$, $ d\equal{}6$ and $ e\equal{}7$ for a counterexample.

Just use Cauchy in Engel form:

$ \frac {9}{a} \plus{} \frac {16}{b} \plus{} \frac {25}{c} \plus{} \frac {36}{d} \plus{} \frac {49}{e}\equal{}\frac {3^2}{a}\plus{}\frac{4^2}{b}\plus{}...\plus{}\frac{7^2}{e}$
Per Cauchy in Engel form:
$ \frac {3^2}{a}\plus{}\frac{4^2}{b}\plus{}\frac{5^2}{c}\plus{}\frac{6^2}{d}\plus{}\frac{7^2}{e} \geq \frac{(3\plus{}4\plus{}5\plus{}6\plus{}7)^2}{a\plus{}b\plus{}c\plus{}d\plus{}e}\equal{}\frac{625}{a\plus{}b\plus{}c\plus{}d\plus{}e}$

Problem solved.

I saw this right away from #5 on [url]http://www.artofproblemsolving.com/Resources/Papers/MildorfInequalities.pdf[/url].


Prove that choosing any $n+1$ numbers from $1,2,..,2n$ there exist $2$, $a$ and $b$, so that $a|b$.

another classic:
[hide="hint"]
it is quite obvious you need pigeonhole here.  Moreover, $n+1$ numbers implies there should be $n$ holes.  out the the $2n$ numbers, what kind of number has a total of $n$ of them???
[/hide]

pigeon hole!

(is it possible that there is a stronger statement, i.e. taking n numbers from 2n also has the same property for n>1)

yes we can take support from the peageon hole priciple.
there are n different odd number. if we have to select n+1 numbers from them so we have to select atleast one even number hence 1st is proved.
Now for the second we have to think in the other way. there are first numbers in the set which also have numbers which are double from them. if we leave those n numbers and select another n numbers then for selecting one more number we have to select that from the first set(in which we multiply 2 we get the number which is present in the set). hence it is proved.

[hide]The idea is to use the pigeonhole/box principle and parity of the numbers....Engel[/hide]

Masoud Zargar

If you haven't gotten it yet, this should make it very obvious:
[hide]The factorization of every number into $2^{a}b$ with b odd. Or, as boxedexe said, parity.[/hide]


What is the probability that a random arrangement of the letters in the word `SEVEN' will have both E's next to each other?

OK, in all, there are $ 60$ rearrangements ($ \frac{5!}{2}$).  Then, there are four different cases we must consider:

[i]Case 1: EE---[/i]

6 ways.

[i]Case 2: -EE--[/i]

6 ways.

[i]Case 3: --EE-[/i]

6 ways.

[i]Case 4: ---EE[/i]

6 more ways.

Thus, the answer is $ \frac{6\plus{}6\plus{}6\plus{}6}{60}$, or $ \boxed{\frac25}$.

You don't need four cases.

Treat the double-E as a single letter.  Then there are $ 4!\equal{}24$ ways to arrange those, out of $ \frac{5!}{2}\equal{}60$ total, so we have $ \frac{24}{60}\equal{}\boxed{\frac{2}{5}}$.

What are you trying to say @3above, many alcumus problems have been posted on web2.0calc.com, and there's nothing we can do about it. But putting the link on the alcumus forum... Why would you do that, it'll give people the idea to go there for answers and use it to cheat.

sniped :furious:

[quote=roribaki]what is web2.0calc.com?[/quote]

you don't want to know.

It's sort of like an AoPS forum, but a lot of cheating goes on there, and posting of copyrighted AoPS material. (just don't use it)

lmao i've answered questions on there oops
1-2 years ago tho

It's just sad looking at the amount of AoPS material being asked on sites like web2.0calc :wallbash:

[quote=Wukong][quote=nt-red][quote=Wukong]i assume you mean 3above.
but still it means they cheat using 2.0.
exposed.
expect to be banned from alcumus, xin.[/quote]

Wait you can get banned from alcumus  :huh:[/quote]

yes you can
for cheating
spamming wrong problems
etc.[/quote]

really? wow i thought you could only get banned from FTW! and stuff

you can get banned from anything on AoPS

you can get banned from specific forums too
but that's off topic.

never use sites like web2.0 calc to cheat on aops problems please i know im just being a karen saying this when I know nobody here has but sometimes I just can't stress it enough looking at the amount of aops material posted on there :wallbash:

Ok, the post has been reported and issue has been resolved, no need for anymore posts on this thread (unless it's about the actual problem).

[hide=my sol]Place S,V,N like this: _S_V_N_. The spaces stand for the slots in which you may place the EE pair. We do 4c1*3!=24 desirable outcomes. The total number of outcomes is 5!/2!=60. The answer is 24/60=2/5.[/hide]

idk what's been going on earlier in this thread but anyway...
[hide=Solution]To arrange five letters, we have $5\times4\times3\times2\times1=120$ ways.
However, this will count E[i]E[/i]SVN and [i]E[/i]ESVN as two things. Therefore, we divide by $2$ to correct for overcounting, leaving us with $60$ total possible words.

Now we count the successful outcomes, where both 'E's are next to each other. With simple counting, we see that we have $4$ successful outcomes.

However, for each of these $4$ successful outcomes, the S, V and N can be placed in $3\times2\times1=6$ different ways. $4\times6=24$ successes.

$\frac{24}{60}=\boxed{\frac25}$
[/hide]


Which one out of [b] Pyridine[/b] and [b]Pyrrole[/b] is more stable to reactivity??

This is a good question, but the answer depends on the type of reaction we are talking about.


How many ways are there to choose 3 cards from a standard deck of 52 cards, if all three cards must be of different suits?

You can pick any card for the first draw: 52 total.
Now for the second draw, there are: 13 cards x 3 suits=39.
Now for the last draw, there are: 13 cards x 2 suits=26.

However, I believe that order matters (excuse me if I'm wrong) because it seems to me that for example:

$ \clubsuit A, \heartsuit 5, \diamondsuit 3$
Is different from say:
$ \heartsuit 5,\diamondsuit 3, \clubsuit A$

So we would have: $ 52\cdot 39\cdot 26\equal{}52728$

Order doesn't matter, because I put 8788 and got it right.  It never says order matters.  If it was, How many ways can you pick two fruits from an apple, an orange, a banana, and a pear would you think that order mattered?

If the question was "How many ways can you pick two fruits from an apple, an orange, a banana, and a pear, without respect to order?" then the given answer is not to be questioned.  When I worked out the cards problem, I debated about significance of order, then made my best guess and guessed wrong.  It was not clear to me.  Is there an understanding in the discipline of probability that gives a specific meaning to the phrase "how many ways can you pick?"  Does that language always mean that order does not matter?

I'm not sure birdmath, I think you just have to use your common sense and you get a feel for it after doing lots of problems.

I did 4 ways to choose the suit that isn't picked and $ 13^3$ ways of picking a card from that suit and got 8788 as my final answer.  But I'm kind of not sure why when you pick the suits using combinations you don't need to divide by 3! to make sure the order of the cards doesn't matter.  Can someone clarify?

[quote="AIME15"]I have a better solution than the one provided:

There are $ 52$ ways to pick the first card.  Then there are $ 39$ ways to pick the second, because we don't want any of the first suit. Similarly, there are $ 26$ ways to pick the third.  However, we have counted each arrangement $ 3!$ times.  Thus our answer is $ \frac{52 \times 39 \times 26}{3!}\equal{}\boxed{8788}$.[/quote]

I used what you said, but I always confuse between "order does/doesn't matter". So I forgot the /3! part and failed.

I'm very sorry for the 8 year bump, but why can't we do:
[hide=My (incorrect) Solution]$52$ choices for the first card (can be any card);
$39$ for the second card (can be any card from the remaining three sets);
and $26$ for the third card (can be any card from the remaining two sets)
$52*39*26$ $=$ $52$,$728$[/hide]
This answer is incorrect, but I don't see why.


tan(3pi/11) + 4sin(2pi/11)
How can I get the value of this formula?  :help: please~

Denote $ \alpha\equal{}\frac\pi{11}$,

\[ S\equal{}\tan3\alpha\plus{}4\sin2\alpha\equal{}\frac{\sin3\alpha\plus{}2\sin5\alpha\minus{}2\sin\alpha}{\cos3\alpha}.\]

Then $ S>0$ and $ S^2\equal{}11$, because the latter can be easily transformed to

\[ 1\plus{}2(\cos2\alpha\plus{}\cos4\alpha\plus{}\cos6\alpha\plus{}\cos8\alpha\plus{}\cos10\alpha)\equal{}0,\]
which is trivial. Hence the answer is $ \sqrt{11}$.

Thank you.
Sorry not to understand sudden "S²=11". :(
Can you explain it to me more precisely? I'm poor at math  :blush:

$ S^2\equal{}11$ is equivalent to
\[ (\sin3\alpha\plus{}2\sin5\alpha\minus{}2\sin\alpha)^2\equal{}11\cos^23\alpha.\]
Just remove parentheses in the LHS and use formulae
\[ 2\sin\alpha\sin\beta\equal{}\cos(\alpha\minus{}\beta)\minus{}\cos(\alpha\plus{}\beta),\quad\text{etc.}\]


Let $ X_n\equal{}\{1,2,...,n\}$,where $ n \geq 3$.
We define the measure $ m(X)$ of $ X\subset X_n$ as the sum of its elements.(If $ |X|\equal{}0$,then $ m(X)\equal{}0$).
A set $ X \subset X_n$ is said to be even(resp. odd) if $ m(X)$ is even(resp. odd).
(a)Show that the number of even sets equals the number of odd sets.
(b)Show that the sum of the measures of the even sets equals the sum of the measures of the odd sets.
(c)Compute the sum of the measures of the odd sets.

There's almost certainly a bijective proof, but this might actually be cleaner, and was certainly more straightforward.

Let $ g(x) \equal{} \prod_{i \equal{} 1}^n (1 \plus{} x^i)$.  Then $ g(x) \equal{} \sum_{X \subset X_n} x^{m(X)}$.

Then the total number of even sets is the sum of the coefficients of the even part of $ g$, or $ \frac {g(1) \plus{} g( \minus{} 1)}{2} \equal{} \frac {g(1)}{2} \equal{} 2^{n \minus{} 1}$; since there are $ 2^n$ subsets in total, this is half of the sets.

Differentiating, $ g'(x) \equal{} \sum_{X \subset X_n} m(X) x^{m(X) \minus{} 1}$.

The sum of the measures of even sets is the sum of coefficients in the odd part of $ g'$, or $ \frac {g'(1) \minus{} g'( \minus{} 1)}{2}.$ By the product rule, $ g'(x) \equal{} \sum_{i \equal{} 1}^n ix^{i \minus{} 1} \prod_{j \in [n], i \ne j} (1 \plus{} x^j)$. Since $ (1 \plus{} x)(1 \plus{} x^3) \mid g$, we know $ \minus{} 1$ is a double root of $ g$ and is therefore a root of $ g'$. So we have $ \frac {g'(1) \minus{} g'( \minus{} 1)}{2} \equal{} \frac {g'(1)}{2} \equal{} (1 \plus{} 2 \plus{} \dots \plus{} n)2^{n \minus{} 2} \equal{} 2^{n \minus{} 2}\binom{n \plus{} 1}{2}$. Similarly, the sum of the measures of the odd sets is $ \frac {g'(1) \plus{} g'( \minus{} 1)}{2} \equal{} \frac {g'(1)}{2} \equal{} \binom{n \plus{} 1}{2}2^{n \minus{} 2}$; the two quantities are equal.

EDIT: Actually... maybe not.

For each $ S \subset X_n$ Pair $ S$ with $ S \oplus \{1\}$ (using $ \oplus$ for symmetric difference). Then this pairing is a bijection between odd and even sets, giving (a). Restricting to sets containing $ \{1\}$, the map $ T \mapsto T \oplus \{3\}$ is a bijection between odd and even sets, so half of all sets containing $ \{1\}$ are even; therefore the sum of the measures of all even sets equals that in all odd sets. Restricting to sets containing $ \{k\}$, for any $ k \ne 1$, the bijection $ S \mapsto S \oplus \{1\}$ takes even sets to odd sets, so every $ k$ is contained in $ 2^{n\minus{}2}$ even sets (and $ 2^{n\minus{}2}$ odd sets); therefore the sum of the measures of all even (odd) sets is $ (1 \plus{} \dots \plus{} n)2^{n\minus{}2} \equal{} \binom{n\plus{}1}{2}$.

So basically $ g'(\minus{}1)\equal{}0$ implies b)
The total sum of the measures of all subsets of $ X_n$ contains each element $ i$ exactly $ 2^{n\minus{}1}\equal{}|\mathcal{P}(X_n/\{i\})|$-times.
So the total sum is $ 2^{n\minus{}1}\binom{n\plus{}1}{2}$ so by b) the answer to c) is $ 2^{n\minus{}2}\binom{n\plus{}1}{2}$


function
real analysis
LaTeX
advanced fields
advanced fields unsolved

Hi,

since I don't have a lambda, i'll use 'l' instead.

let {f_n} be a series of measurable functions, with Lebesgue measure in X=[0,1]
{f_n} is bounded almost everywhere in X [theres a set A such that l(A)=0 and for
 each x in X\A,  the series {f_n(x)} is bounded]

prove that for every epsilon (denoted 'e' here) e>0, theres a positive number c(e)=c>0
and a measurable set E(e)=E, such that both 1&2 apply:
1)  l(X\E) < e
2) for each x in E, and n>=1,  |f_n(x)| <= c

I don't even know how to start.
Any help is gladly accepted.

    Thanks

Property 2) tells you what the set $E$ is going to be: $E=\{x\in X\colon \sup_{n}|f_{n}(x)|\le c\}$, where $c$ is a constant to be chosen. Letting $c=1,2,3,\dots$, we get an increasing sequence of sets $E_{1},E_{2},E_{3},\dots$. First prove that $\bigcup_{n=1}^{\infty}E_{n}=X\setminus A$. Then deduce the existence of $n$ such that $\lambda(X\setminus E_{n})<\epsilon$. Finally, learn $\LaTeX$.


AoPSwiki
Alcumus
videos
algebra
binomial theorem

The binomial theorem basically tells you the $ x^ny^m$ term in $ (x \plus{} y)^z$. where $ n \plus{} m$ [i]has[/i] to equal $ z$. Basically there is a formula, but its easier just to know that the $ x^ny^m$ term in $ (x \plus{} y)^z$ is $ \frac {z!}{m! \cdot n!}$. So the $ x^6y^{141}$ term in $ (x \plus{} y)^{147}$ is $ \frac {147!}{6!\cdot 141!}$ (note that 6+141=147. If m+n$ \neq$z then you cant apply the formula because there is no such term). Here are more practical question:
Find the $ x^3y^5$ term in $ (x \plus{} y)^8$.[hide="Answer"]Did you get $ \frac {8!}{3!5!}$?[/hide]
Find the $ x^2y^4$ term in $ (x \plus{} y)^7$. [hide="Answer"]Did you get $ \frac {7!}{4!2!}$? If you did thats wrong. Since 4+2 $ \neq$ 7, there is no such term as $ x^2y^4$ in $ (x \plus{} y)^7$. [/hide]

This can be generalized. In general, the $ x^{a_{1}}x^{a_{2}}x^{a_{3}}\dots x^{a_{y}}$ term in the expression $ (x_1 \plus{} x_2 \plus{} x_3 \plus{} \dots \plus{} x_k)^n$, given that $ a_1 \plus{} a_2 \plus{} a_3 \plus{} \dots \plus{} a_y \equal{} n$, is $ \boxed{\frac {n!} {a_1!a_2!a_3!\dots a_y!}}$.

An explanation is found [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=297768]here[/url]

the binomial theorem is also sometimes referred to as newton's sum and it basically provides a fast approach to expand binomials.

The Binomial Theorem is a tool to expand $ (a + b)^n$.

The Binomial Theorem is:
$ \small{(a + b)^n = a^n + \binom n1 a^{n-1} b^1 + \binom n2 a^{n - 2} b^2 + \binom n 3 a^{n - 3} b^3 + \binom n4 a^{n - 4} b^4 + ... + \binom  b^n}$

If this is too complicated for you, try to understand it like this.

The Binomial Theorem is a short cut to expand $ (a + b)^n$.
For example, we all know that

$ (a + b)^2 = a^2 + 2ab + b^2$

and that

$ (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$

Do you see how there is no coefficient behind $ a^n$ and $ b^n$ in any $ (a + b)^n$? This is because it is the circumstance where only the $ a$'s multiply, or only the $ b$'s. There are coefficients behind everything else because there is a multiple of all of the others. The closer the exponent behind $ a$'s and $ b$'s are, the greater the coefficient. Let's try $ (a + b)^4$

$ (a + b)^4 = a^4 + 4a^3b + 6 a^2b^2 + 4 a^3b + b^4$

You see? The coefficients are also combinations. The exponent of $ a + b$ is part of the combinations. The first expression after $ a^n$ of $ (a + b)^n$ will have a coefficient of $ \binom n 1$, or, it will be equal to $ n$. The second after $ a^n$ will have a coefficient of $ \binom n 2$. This goes on until the last expression, which will have a coefficient of 1.

The exponent's of  a and b in each expression will also add up to $ n$ in $ (a + b)^n$.

[quote="Cliu0301"]!?!!!?!?!?!?!?!?!??!?!?!?!?!?!?! AoPSWiki confused me -.-

Can someone please explain to me the Binomial Theorem? I would be very grateful.[/quote]

Why not exploit Alcumus?  Since it's under renovation, you can view all the [url=http://www.artofproblemsolving.com/Alcumus/Videos/VideoList.php]Alcumus Videos[/url].

[quote="TheDeity"]The Binomial Theorem is a tool to expand $ (a + b)^n$.

The Binomial Theorem is:
$ \small{(a + b)^n = a^n + \binom n1 a^{n - 1} b^1 + \binom {n - 1}{2} a^{n - 2} b^2 + \binom {n - 2}{3} a^{n - 3} b^3 + \binom {n - 3}{4} a^{n - 4} b^4 + ... + \binom {n}{n} b^n}$

[/quote]

Umm... The binomial theorem is
\[{ (a + b)^n = \binom{n}{0} a^n + \binom {n}{1} a^{n - 1} b^1 + \binom {n}{2} a^{n - 2} b^2 + \binom {n}{3} a^{n - 3} b^3 + \binom {n}{4} a^{n - 4} b^4 + \dots + \binom {n}{n} b^n}\]

Basically you can think of $ (a+b)^n$ as $ (a+b)$ multiplied and laid out $ n$ times as $ \underbrace{(a+b)(a+b)\dots(a+b)}_\text{n times}.$

Then for each of the terms in the expansion of $ (a+b)^n,$ we choose 0 of them at first, then 1 of those n terms, etc, for the b term in this case, (but we could also have done thing with $ a$ here). And that is where the coefficients come from. So, the coefficient is basically $ \binom{n}{k},$ which basically implies how many of those $ n$ terms as we have written out of $ (a+b)$ you choose the $ b$ terms out of.

And if that totally confused you, check out the Alcumus videos. They are extremely helpful.  :)

[quote="limac"][quote="TheDeity"]The Binomial Theorem is a tool to expand $ (a + b)^n$.

The Binomial Theorem is:
$ \small{(a + b)^n = a^n + \binom n1 a^{n - 1} b^1 + \binom {n}{2} a^{n - 2} b^2 + \binom {n}{3} a^{n } b^3 + \binom {n}{4} a^{n - 4} b^4 + ... + b^n}$

[/quote]

Umm... The binomial theorem is
\[{ (a + b)^n = \binom{n}{0} a^n + \binom {n}{1} a^{n - 1} b^1 + \binom {n}{2} a^{n - 2} b^2 + \binom {n}{3} a^{n - 3} b^3 + \binom {n}{4} a^{n - 4} b^4 + \dots + \binom {n}{n} b^n}\]
[/quote]

That's the exact same thing I said. $ \binom nn$ and $ \binom n0$ are equal to 1.

TheDiety, Hmm...I am pretty sure this is what you had:

$ \small{(a\plus{}b)^{n}\equal{} a^{n}\plus{}\binom n1 a^{n\minus{}1}b^{1}\plus{}\binom{n\minus{}1}{2}a^{n\minus{}2}b^{2}\plus{}\binom{n\minus{}2}{3}a^{n\minus{}3}b^{3}\plus{}\binom{n\minus{}3}{4}a^{n\minus{}4}b^{4}\plus{}...\plus{}\binom{n}{n}b^{n}}$

The n-1 choose 2, n-2 choose 3 as the coefficients was wrong.

But now your's looks correct.

[quote="Cliu0301"]!?!!!?!?!?!?!?!?!??!?!?!?!?!?!?! AoPSWiki confused me -.-

Can someone please explain to me the Binomial Theorem? I would be very grateful.[/quote]

The Binomial Theorem states that the expansion of $ (x\plus{}y)^n$ is $ \binom{n}{0}x^n\plus{}\binom{n}{1}x^{n\minus{}1}y^1\plus{}\binom{n}{2}x^{n\minus{}2}y^2\plus{}\cdots\plus{}\binom{n}{k}x^ky^{n\minus{}k}\plus{}\cdots\plus{}\binom{n}{n\minus{}1}x^1y^{n\minus{}1}\plus{}\binom{n}{n}y^n$ where $ \binom{n}{r}\equal{}\frac{n!}{r!(n\minus{}r)!}$

Actually there is a multinomial theorem, I don't know much about that, but I think the trinomial theorem comes from 'Pascal's Pyramid'


modular arithmetic
inequalities
number theory proposed
number theory

Given a positive integer $ n$, find all integers $ (a_{1},a_{2},\cdots,a_{n})$ satisfying the following conditions:
$ (1): a_{1}\plus{}a_{2}\plus{}\cdots\plus{}a_{n}\ge n^2;$
$ (2): a_{1}^2\plus{}a_{2}^2\plus{}\cdots\plus{}a_{n}^2\le n^3\plus{}1.$

If $ \sum a_i^2\le n^3$ then, we have $ n\ge\sqrt{\frac{\sum a_i^2}{n}}\ge\frac{\sum|a_i|}{n}\ge n$.  Thus, we must have $ a_1\equal{}a_2\equal{}\cdots \equal{}a_n$ which has solution $ a_i\equal{}n$ for all $ n$.

Otherwise, $ \sum a_i^2\equal{} n^3\plus{}1$.  Suppose $ \sum a_i\ge n^2\plus{}1$.  Then, we have $ \sqrt{n^2\plus{}\frac{1}{n}}\equal{}\sqrt{\frac{\sum a_i^2}{n}}\ge\frac{\sum|a_i|}{n}\ge n\plus{}\frac{1}{n}$ which is clearly false.  Thus, $ \sum a_i\equal{}n^2$.  It is well known that $ k\equiv k^2\pmod{2}$ for all integer $ k$.  Thus, we have $ n^3\plus{}1\equal{}\sum a_i^2\equiv\sum a_i\equal{}n^2\pmod{2}$.  That is, we have $ n^3\plus{}1\equiv n^2\pmod{2}$, an impossibility.

Thus, the only solution is $ a_i\equal{}n$ for all $ i$.

Is my solution correct? Can someone help me? This is what I have so far:

Using RMS-AM inequality, we have
\[
\frac{n^2-a_i}{n-1} \leq \frac{(a_1+a_2+\cdots a_n)-a_i}{n-1} \leq \sqrt{\frac{(a_{1}^{2}+a_{2}^{2}+\cdots a_{n}^2)-a_i^2}{n-1}} \leq \sqrt{\frac{(n^3+1-a_{i}^2)}{n-1}}
\]
Expanding and simplifying, we obtain
\[
n\cdot a_n^2-2n^2\cdot a_n+n^3-n+1\leq0
\]
This implies that
\[
n-\sqrt{\frac{n-1}{n}} \leq a_{i} \leq n+\sqrt{\frac{n-1}{n}}
\]
Since $a_i$ is integer, we conclude that $a_i=n$ for all $i$.

[quote="LoveMath4ever"][quote="anything--"]
\[
\frac{(a_1+a_2+\cdots a_n)-a_i}{n-1} \leq \sqrt{\frac{(a_{1}^{2}+a_{2}^{2}+\cdots a_{n}^2)-a_i^2}{n-1}}
\][/quote]

I think it is wrong!  :blush:[/quote]
no, it is true because it is equivalnet to:
$(n-1)(\sum_{j\neq i}{a_j}^2) \ge (\sum_{j\neq i}{a_j})^2$
which is true by cauchy.

Without loss of generality, let $a_n \ge a_{n-1} \ge \cdots \ge a_1$.

Let $\sum_{i = 1}^{n-1}a_i = A_1$ and $\sum_{i = 1}^{n-1}a_i^2 = A_2$. By Cauchy-Schwarz,
$$A_1^2 = \left(\sum_{i = 1}^{n-1}a_i \cdot 1\right)^2 \le (n-1)A_2.$$
By the given inequalities, we find that
$$(n-1)(n^3 + 1 - a_n^2) \ge (n-1)A_2 \ge A_1^2 \ge (n^2 - a_n)^2.$$
Expanding out both sides, we see that
$$n^4 + n - na_n^2 - n^3 - 1 + a_n^2 \ge n^4 - 2n^2a_n + a_n^2.$$
Rearranging both sides, we find that
$$2n^2a_n \ge n^3 + 1 + na_n^2 - n.$$
By the definition of $a_n$ and the given inequalities, we may let $a_n = b + n$ where $b \ge 0$ is an integer. Then, we have that
$$2n^2(b+n) \ge n^3 + 1 + n(b+n)^2 - n = 2n^3 + 2bn^2 + bn - n + 1.$$
Simplifying further, we see that
$$0 \ge bn - n + 1.$$
Thus, we have that $b = 0$. Since $a_n$ is the maximum $a_i$ and the sum of all the $a_i$ is greater than or equal to $n^2$, we must then have that all the $a_i$ are $n$ since we would otherwise have that $\sum_{i = 1}^n a_i < n^2$. Thus, the number of ordered $n$-tuples is $1$ for all $n$.

Just consider the sum of squares $S:=\sum_{i=1}^n (a_i-n)^2$. Then by using (1) and (2) we get
$0 ~\le~ S ~\le~ (\sum_{i=1}^na_i^2) -2n(\sum_{i=1}^na_i) +\sum_{i=1}^nn^2 ~\le (n^3+1)-2n(n^2)+n\cdot n^2 ~=~ 1$.

This means that at least $n-1$ of the squares $(a_i-n)^2$ are $0$, while the last one is either $0$ or $1$.
$\bullet\,$ If all the squares are $0$, this yields the feasible solution $(n,n,n,\ldots,n)$.
$\bullet\,$ If $n-1$ of the squares are $0$ and the last one is $1$, we get that $n-1$ of the $a_i$
$~~~~~$ are equal to $n$, and the last one is either $n-1$ or $n+1$.
$~~~~~$ The case with $n-1$ collides with (1), and the case with $n+1$ collides with (2).

Hence there is only the solution $(n,n,n,\ldots,n)$.

Only the trivial permutation $(n, n, n, \cdots, n)$ works. Obviously if the sum of the squares is equal to $n^3$, we are done by Cauchy. Now, suppose there exist $a_1, a_2, \cdots, a_n$ not equal such that $$a_1^2+a_2^2+\cdots+a_n^2 = n^3+1.$$ The transformation $(a_1, a_2) \to (n, a_1+a_2-n)$ for unequal $a_1, a_2$ decreases the sum of squares by at least $2(n-a)(n-b) > 1$, which contradicts minimality.


Balkan junior olympiad, already solved. Check for it.

Indeed, already posted here two times.

http://www.mathlinks.ro/Forum/viewtopic.php?t=14273
http://www.mathlinks.ro/Forum/viewtopic.php?t=1045 (with solution)

If I don't misunderstand something, it is enough that the numbers a, b, c are > -1 (not necessarily $\geq$ 1).

  Darij


vector
topology
group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $G=PSL_2(\mathbb{K})$ in its natural permutation working. (so Mx = x' with x and x' being 2x1 vectors) ($\mathbb{K}$ is an arbitrary field)
Let $M=\mathbb{K}(1,0)$ and $L=\mathbb{K}(0,1)$. Let $A=G_M$ and $B=G_L$.

(a) Show that $G = \langle A,B \rangle$.

(b) Show that for every $a \in A$ there are unique $b,c \in B$ such that $bac$ 'switches' L and M. ($bacM=L$ and $bacL=M$)

I didn't solve it. :(
Anyone here able to solve it?

I don't think I understand the question :(

I just had a horrible exam today and that is why it took up to now before I got to answering.  Is this a HVM question?

This might help for a

remember that $(G_{x})^g=G_{x^g}$ and remember also that $A$ and $B$ both work transitive on the other points on the projective line.  The two combined gives us that the generated group contains $G_x$ for every x on the line.  This gives us that the generated group will be two transitive. This means that every element in $PSL(2,K)$ can be written as a product of elements in $A$ and $B$, multiplied with an element in $PSL(2,K)$ fixing $(1,0)$ and $(0,1)$.  That last element is of course also in the generated group.  I am beginning to wonder whether you need so much properties for the first theorem.  Just this 'lemma' might suffice : when you take a two transitive permutation representation on a set of at least three elements, two stabilizers generate everything?

Some elaboration (have topology exam really soon  :( )

you can prove my lemma in the last line by this  other lemm :

if $G$ has a transitive subgroup $Y$ then $G_x$ and $Y$ together generate the entire group.

this is true because every element in the group, $g$ can be written as something that sticks in $x$  and something that goes from $x$ to $x^g$ and is in Y (remember Y is transitive)

I think I have finally found a solution... only 7 months late. Seen afterwards, I think it's the same as fredbel6's.

(a) It is well-known that $G$ works sharp $2$-transitive on the set of vector lines $V$. This means that $A$ works sharp transitive on $V\backslash \{M\}$. Now $G_{(L^{m})}=(G_{L})^{m}$ (where $x^{y}$ means $y^{-1}xy$) says that all stabilizers of all vector lines are in the group $\langle A,B \rangle$.

Now we will prove that $\langle A,B \rangle$ works $2$-transitive on $V$. Assume we want to convert $(g,h)$ into $(x,y)$. Take the element of $G_{g}$ converting $(g,h)$ into $(g,y)$, afterwards take the element of $G_{y}$ converting $(g,y)$ into $(x,y)$. Because of the first part of this paragraph those elements are in $\langle A,B \rangle$.

So we have a $2$-transitive working, and as $\langle A,B \rangle\le G$ and $G$ worked sharp $2$-transitive, this means we must have $\langle A,B \rangle=G$.

(b) Now we have that for each $a\in A$ there are unique $X,Y$ such that $(L,X)$ maps to $(Y,L)$ (being $X=L^{(a^{-1})}$ and $Y=L^{a}$). Because $B$ works sharp transitive, there are unique $b$ mapping $(L,M)$ to $(L,X)$ and unique $c$ mapping $(Y,L)$ to $(M,L)$.

Phew... not easy, but nice. :)


geometry
geometric transformation
rotation
geometry proposed

Find the number of the straight lines which intersect some two sides of
a given equilateral triangle and its incircle at points $L, O, V, E$
consequently in such a way that $LO=OV=VE.$

The average of answers by Kyiv TST participants is 9.3928...

[hide="Answer"]12[/hide]
[hide="Hint 1"]Let $ABC$ be the equilateral triangle. Rotation argument gives that it is sufficient to consider the case $L\in AB, E\in BC$ and multiply the answer by 3.[/hide]
[hide="Hint 2"]Denote by $M, N$ the points where the incircle touches $AB, BC.$ Then $ML^2=LO\cdot LV=OE\cdot VE=EN^2.$ Hence $BL=BE$ or $AL=BE.$ Is it clear now how to finish the solution? :? [/hide]


Could you post some problems from  of

RMT (revista matematice Timisoara) $\leq 2004$, and
recreatii matematice  $\leq 2004$

specially problems concerning :

sequence/ serie
functions
integral
matrices
group/ring/ field

Thank you in advance

It is impossible to find RMT and reacreatii matematice in Paris.

Recreatii Matematice website: http://recreatiimatematice.uv.ro
Some of the problems are translated in English. I can send you a basic math dictionary Romanian_English, if you want me to.
As for RMT, I'll post some problems soon.

Let $a,b$ two elements of a group, $ord(a)=m$ and $a^{n}b^{p}=b^{r}a^{n}$.
If $p$ divides $r$, then find $ord(b)$.
Florentin Smarandache, USA.


1) Find all $f: \mathbb R \to \mathbb R$ such that
\[(x^2 \plus{} xy \plus{} y^2)(f(x) \minus{} f(y)) \equal{} f(x^3) \minus{} f(y^3)\]
for all $x,y \in \mathbb R$.

2) Find all $f: \mathbb R \to \mathbb R$ such that
\[ (f(x^2) \plus{} f(xy) \plus{} f(y^2))(x \minus{} y) \equal{} f(x^3) \minus{} f(y^3)\]
for all $x,y \in \mathbb R$.

[quote="mahanmath"]Hi ! :D
Find all $ f: R\rightarrow R$ such that

1) $ (x^2 + xy + y^2)(f(x) - f(y)) = f(x^3) - f(y^3)$[/quote]

Let $ P(x,y)$ be the assertion $ f(x^3) - f(y^3) = (f(x) - f(y))(x^2 + xy + y^2)$

If $ f(x)$ is solution, so is $ f(x) + c$. So we'll look for solutions such that $ f(0) = 0$

Then $ P(x,0)$  $ \implies$ $ f(x^3) = x^2f(x)$ and the equation becomes $ x^2f(x) - y^2f(y) = (f(x) - f(y))(x^2 + xy + y^2)$ $ \iff$ $ y(x + y)f(x) = x(x + y)f(y)$

Setting $ y = 1$, this implies $ (x + 1)f(x) = x(x + 1)f(1)$ and so $ f(x) = xf(1)$ $ \forall x\ne - 1$
Then $ P(2, - 1)$ $ \implies$ $ f( - 1) = - f(1)$ and so $ f(x) = xf(1)$ $ \forall x$

And it's easy to check thas this indeed is a solution.

Hence the solutions $ \boxed{f(x) = ax + b}$

[quote="mahanmath"]Find all $ f: R\rightarrow R$ such that

2) $ f(x^2) + f(xy) + f(y^2)(x - y) = f(x^3) - f(y^3)$[/quote]

Setting $ y=x$, we get $ 2f(x^2)  = 0$ and so $ f(x)=0$ $ \forall x\ge 0$

The equation becomes  $ f(xy) = f(x^3) - f(y^3)$
Using $ y<0$ and $ x=y^2>0$, we get $ f(y^3) = - f(y^3)$ and so $ f(x)=0$ $ \forall x< 0$

Hence the unique solution $ \boxed{f(x)=0}$ $ \forall x$

Sorry . I have made a typo in part 2 and I edit it now . :blush:
I mean $ (f(x^{2}) \plus{} f(xy) \plus{} f(y^{2}))(x \minus{} y) \equal{} f(x^{3}) \minus{} f(y^{3})$

2)
$ y\equal{}0$ gives: $ xf(x^2)\plus{}2xf(0)\equal{}f(x^3)\minus{}f(0)$
Set now: $ x\equal{}1 \Rightarrow f(0)\equal{}0$.

So we have $ xf(x^2)\equal{}f(x^3)$ and therefore $ f(\minus{}x)\equal{}\minus{}f(x)$.

Using this in the first equation gives: $ (x\minus{}y)f(xy)\plus{}xf(y^2)\minus{}yf(x^2)\equal{}0$
Writing the same equation with $ \minus{}y$ gives : $ \minus{}(x\plus{}y)f(xy)\plus{}xf(y^2)\plus{}yf(x^2)\equal{}0$

Adding both lines and $ y\equal{}1$ gives us the solution: $ \boxed{f(x)\equal{}f(1)x}$

[quote="pfanni"]2)
$ y \equal{} 0$ gives: $ xf(x^2) \plus{} 2xf(0) \equal{} f(x^3) \minus{} f(0)$
Set now: $ x \equal{} 1 \Rightarrow f(0) \equal{} 0$.

So we have $ xf(x^2) \equal{} f(x^3)$ and therefore $ f( \minus{} x) \equal{} \minus{} f(x)$.

Using this in the first equation gives: $ (x \minus{} y)f(xy) \plus{} xf(y^2) \minus{} yf(x^2) \equal{} 0$
Writing the same equation with $ \minus{} y$ gives : $ \minus{} (x \plus{} y)f(xy) \plus{} xf(y^2) \plus{} yf(x^2) \equal{} 0$

Adding both lines and $ y \equal{} 1$ gives us the solution: $ \boxed{f(x) \equal{} f(1)x}$[/quote]

Quite nice and ok (I did not succeed up to now to show $ f(0)\equal{}0$)

Just a detail : all your proof lead only to mandatory condition. So $ f(x)$ must be $ xf(1)$ but you need to check back in the original equation to show that this indeed is a solution (sometimes, it may lead to constraints on $ f(1)$ for example).


abstract algebra
superior algebra
superior algebra solved

Let $G$ be the group of invertible upper triangle $2\times 2$ matrices;
$H$ be the group of invertible diagonal matrices;
$K$ be the group of upper triangle matrices with diagonal entries 1.

Does it hold that $G\cong H\times K$?

I realised it soon after I posted it.

It is not isomorphic... Since $H$ and $K$ are abelian but $G$ is not.


Aunt Minnie's age plus the square of Uncle Paul's age is 7308. Uncle Paul's age plus the square of Aunt Minnie's age is 6974. What is the sum of Aunt Minnie's age and Uncle Paul's age?

Thanks

Hint:

[hide]Theres only really one way these problems can be solved. Try playing around with adding/subtracting the equations until you can get something that factors.[/hide]

For Algebra class, best to use TripleM's hint.  For a MATHCOUNTS National Competition, try this hint:

[hide]If Aunt Minnie were a newborn babe, how old would Uncle Paul be?  If Aunt Minnie had reached the ripe old age of 200, how old would Uncle Paul be?[/hide]

Aunt Minnie's age plus the square of Uncle Paul's age is 7308. Uncle Paul's age plus the square of Aunt Minnie's age is 6974. What is the sum of Aunt Minnie's age and Uncle Paul's age?

[hide]

m + p^2 = 7308

p + m^2 = 6974

m = 7308 - p^2

p + (7308 - p^2)^2 = 6974

solve and... p = 85

m = 7308 - 85^2 = 83

85 + 83 = 168

[/hide]

Would you like to tell us how you solved the quartic equation? :)

Sure thing:

p + (7308 - p^2)^2 = 6974

p+ (7308 - p^2)(7308 - p^2) = 6974

p + 7308^2 + (-2*7308)p^2 + p^4 -6974 =0

p^4 + (-2*7308)p^2 + p +53399890 = 0

p/q  (obviously q=1) : all possible rational zeros.  Age will probably be an integer so...

53399890 = 2*5*17*314117

Now think about what the problem is asking (obviously Paul isnt 314117 yrs old!, and ages are + so you can ignore the  :pm: )  Just being logical eliminates a whole bunch of solutions to test.

So some logical ages are: 34 (ie 2*17), 85 (ie 5*17), 17, and maybe 2

Try 34: (could also use synthetic division)

p + (7308 - p^2)^2 = 6974

34 + (7308 - 34^2)^2 = 37847138 which doesnt = 6974

Try 85:

p + (7308 - p^2)^2 = 6974

85 + (7308 - 85^2)^2 = 6974 so p=85

Summary:  I used the rational zeros test, finding the factors of p and q, and finding p/q s that would be logical ages, then tested the solutions and found that 85 worked.

Confession:  I graphed it  when I originally solved the problem (shame on me).

=)

--Julie

Heh. It could be a slight problem to show that 314117 is a prime factor of 53399890 in a competition too :D

Try using one of the above hints: they should lead to a much simpler answer.

this is confusiing....and im already in 6th grade!!

shame on [u]ME[/u]


inequalities
search
real analysis
real analysis unsolved

Let $a_{i}>0, \sum a_{i}<\infty$. Prove that $\sum (a_{1}\dots a_{n})^{\frac1n}<\infty$. I tried to use standart inequality but it doesnt help.

[quote="Ronx"]By A-G we get $(a_{1}\dots a_{n})^{\frac1n}\leq \frac{a_{1}+\dots+a_{n}}{n}$, but it doesnt prove that convergence exists.[/quote]

Well, $a_{n}$ converges, so the Cesaro mean converges (to the same limit actually).

$\sum (a_{1}\dots a_{n})^{\frac{1}{n}}\leq \sum(\frac{a_{1}+\dots+a_{n}}{n}) \leq \sum (a_{1}+\dots+a_{n})<\infty$.
Is it now okey?

Edit: I have corrected it now,but I dont understand what you mean,I dont see a problem with my prof.Can you give a counterexample??

[quote="Ronx"]Well, Mr.Doe, the inequality you wrote is wrong(second part). Phils, we need to prove that the SUM $\sum_{n=1}^{\infty}(a_{1}\dots a_{n})^{\frac1n}\leq\infty$. I'm not sure, may be it is possible to find a counterexample?[/quote]

Yes, I see now. I was being careless.

Search for Carleman's inequality, this must have been discussed on the forum.

Yep. Carleman's inequality is the key of a solution. Its proof was posted by pbornsztein in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=inequality+carleman&t=5016]this topic[/url].


calculus
integration
trigonometry
HMMT
geometric series
calculus computations

Express $ \int_{0}^{2\pi} \frac{1}{a\plus{}b\cos\theta} d\theta$ in terms of $ a$ and $ b$.

[hide="What I have so far"]Seeing this as the sum of an infinite geometric series we have $ \frac{1}{a} \int_{0}^{2\pi} \frac{1}{1\plus{}\frac{b}{a}\cos\theta} d\theta\equal{}\frac{1}{a} \int_{0}^{2\pi} \sum_{n\equal{}0}^{\infty} (\minus{}\frac{b}{a\cos\theta})^{n} d\theta$. I don't know how to proceed any further. Any hints?[/hide]

The cosine in the geometric series goes in the numerator. That is, you should have

$ \int_0^{2\pi}\sum_{n \equal{} 0}^{\infty}\frac {( \minus{} 1)^nb^n}{a^{n \plus{} 1}}\cos^n\theta\,d\theta.$

There are, of course, other ways to do this (which I'm sure will get suggested here quite soon). But the geometric series idea will work.

You do need to know $ \int_0^{2\pi}\cos^n\theta\,d\theta.$

For odd $ n,$ this is zero.

For even $ n:$ $ \int_0^{2\pi}\cos^n\theta\,d\theta \equal{} \frac {n \minus{} 1}{n}\int_0^{2\pi}\cos^{n \minus{} 2}\theta\,d\theta.$

That recursion should give you what you need.

[hide="The answer"]
[hide="One specific claim"]If $ f(x) \equal{} 1 \plus{} \frac12x^2 \plus{} \frac {1\cdot3}{2\cdot 4}x^4 \plus{} \cdots \equal{} \sum_{k \equal{} 0}^{\infty}\frac {1\cdot3\cdot5\cdots(2k \minus{} 1)}{2\cdot4\cdot6\cdots(2k)}x^{2k},$

then $ f(x) \equal{} (1 \minus{} x^2)^{ \minus{} 1/2}.$ One way to find this is seing that $ f$ satisfies the equation $ \frac {f'(x)}x \equal{} (xf(x))',$ and then solving that differential equation.[/hide]

[hide="Using that"]
The integral is $ \frac {2\pi}{a}\left(1 \minus{} \frac {b^2}{a^2}\right)^{ \minus{} 1/2} \equal{} \frac {2\pi}{\sqrt {a^2 \minus{} b^2}}.$[/hide][/hide]

Of course, this comes with the restriction that $ |b|\le |a|.$ Else the integral diverges.

http://www.mathlinks.ro/viewtopic.php?t=311562
http://www.mathlinks.ro/viewtopic.php?t=311906

This is rather a famous problem. Many ways are known. See also HMMT 2009 calculus test. :)

Kent, I think you're missing some $ \theta$s.   :D

[color=green][O.K., fixed now. - K.M.][/color]


linear algebra
matrix
inequalities unsolved
inequalities

Let $a_1,...,a_k, b_1,...,b_k$ be different integers. Prove that
   $\prod_{i,j=1,k}{|a_i-b_j|}\geq (4/9)^k \prod_{i<j}{|a_i-a_j|}\cdot\prod_{i<j}{|b_i-b_j|}$.

Mysterious and beautiful. Is there a more precise source?

The $a,b$ expressions are similar to what one gets in computing the determinant of a matrix with entries $f(a_i - b_j)$.

I do not have more precise source, sorry. Too bad nobody is interested in this beautiful problem. I guess there are more interesting things to be done first. ;)


solucionar en los enteros positivos $ x^2\equal{}y^z\minus{}3$, para $ z\minus{}1$ no divisible por $ 4$.

Caso 1: $ z$ impar. Como $ z\minus{}1$ es par pero no es divisible por 4 se sigue que $ z\minus{}1\equiv{2}(mod.4)$ osea $ z$ es de la forma $ 4k\plus{}3$.
Si $ y$ es par, como $ z\ge 3$ se sigue que $ x^2\equiv{\minus{}3}(mod.8)$ algo contradictorio, luego $ y$ es impar. Claramente $ x$ es par, luego $ 3\equiv{y^z}(mod.4)$, se sigue que $ y\equiv{3}(mod.4)$
La ecuacion es equivalente a $ x^2\plus{}4\equal{}y^z\plus{}1\to x^2\plus{}4\equal{}(y\plus{}1)(y^{z\minus{}1}\minus{}...\minus{}y\plus{}1)$
Ver que $ y^{z\minus{}1}\minus{}...\minus{}y\plus{}1\equiv{1\minus{}3\plus{}1\minus{}3\plus{}...\plus{}1\minus{}3\plus{}1}\equiv{z}\equiv{3}(mod.4)$, luego existe un $ p$ primo de la forma $ 4t\plus{}3$ que divide a $ x^2\plus{}4$.
Claramente $ 1\equal{}MCD(4,p)\equal{}MCD(x^2,p)$, luego $ x^2\equiv{\minus{}4}(mod.p)\to (x\cdot 2^{\minus{}1})^2\equiv{\minus{}1}(mod.p)$, algo contradictorio porque -1 no es residuo cuadrado para los $ p$ de la forma $ 4t\plus{}3$. No hay solución en este caso.

Caso 2: $ z$ par. Sea $ z\equal{}2w$, Trabajando la ecuación: $ 3\equal{}y^{2w}\minus{}x^2\equal{}(y^w\minus{}x)(y^w\plus{}x)$, de donde es facil ver que las soluciones son (x,y,z)=(1,2,2).


Let [b]ABCD[/b] be a convex quadrilateral inscribed in the circle $C$ and the mobile point $M\in C$. I note $X\in AM\cap BD,\ Y\in AB\cap CM$. Prove that $\widehat{ABD}\equiv\widehat{ADC}\Longleftrightarrow $ the line [b]XY[/b] has a fixed direction.

if XY has a fixed direction, then its direction at M = A and M = D is the same. When M = A, XY is the tangent at A; when M = D, XY is line CD, so the angle line AD makes with both of these should be the same; this yields the angle relation.

Given the angle relation, we know CD is parallel to the tangent at A so apply Pascal to AMCDBA to get that XY passes through the point at infinity corresponding to these two lines; i.e. it has a fixed direction.

The your convincing solution is strongly. This problem is a particular case of the my problem from http://www.mathlinks.ro/Forum/viewtopic.php?t=43170