Let $ P(n)$ be the product of positive integers smaller than (or equal to) and [b]coprime[/b] to the positive integer $ n$. Prove that

$ P(n) \equal{} n^{\phi(n)}\prod_{d | n}\left(\frac {d!}{d^{d}}\right)^{\mu\left(\frac {n}{d}\right)}$

where $ \mu$ is the famous mobius function.

we take $ f(n) \equal{} \frac {P(n)}{n^{\varphi(n)}},g(n) \equal{} \frac {n!}{n^{n}}$
we have $ S \equal{} \{1/n,2/n,3/n,...,(n \minus{} 1)/n,1\} \equal{} \{\frac {a}{d}: \ d|n,\ 1\le a\le d,gcd(a,d) \equal{} 1\}$
obviously we can find that:
$ i)$ $ \prod_{x\in S}x \equal{} \prod_{i \equal{} 1}^{n}\frac {i}{n} \equal{} \frac {n!}{n^{n}} \equal{} g(n)$
$ ii)$ $ \prod_{x\in S}x \equal{} \prod_{d|n}^{n}\left(\prod_{j < d; gcd(j,d) \equal{} 1}\frac {j}{d}\right) \equal{} \prod_{d|n}\frac {P(d)}{d^{\varphi(d)}} \equal{} \prod_{d|n}f(d)$
then $ \boxed{g(n)\equal{}\prod_{d|n}f(d)\iff f(n)\equal{}\prod_{d|n}g(d)^{\mu(n/d)}}$

finally $ P(n)\equal{}n^{\varphi(n)}f(n)\equal{}...$

Now that's a really beautiful solution! :) Amazing!


dien dan it hs lhp nhi !!!!!!!!!1111
chan qua :D

mình cũng h?c Lê Hồng Phong, nhưng mà ở NAM?ỊNH :D

mà nhi?u hay ít thì có vấn đ? gì đâu

mà rất nhi?u linker Việt Nam đó chứ.okie

chao  :D

TO NGUYENQUOCKHANH
minh se mai mai khong bao gio thi quoc gia!!
vi minh khong hoc o lop chuyen toan cua truong

lhp NAM DINH noi tieng nhieu giai QG ha ?

chuc may man trong ky thi nay :lol:

Hi Cuong.
   Minh cung o Le Hong Phong, tp Ho Chi Minh!
   Uh, dien dan nay it LHP,tpHCM qua'. Ah, ban nay con hoc LHP ko? Neu con, cho minh hoi tham thay Vu Vinh Thai nhe!

Minh cung la hoc sinh Le Hong Phong ne. Sorry tai dao nay ban qua nen it len. De sau nay minh se len thuong xuyen hon. Ma khong biet cuong hoc o lop nao nhi. Con thay Cuong user name la DuyCuong phai khong?
    PS: a, dung de y cai username cua minh nha, do la 1 big mistake, huhu

Chà, Cường trên này nổi nhỉ? Tao nè mày. Mới đăng kí hôm nay. Tự giới thiệu. Mình là Lê Minh Nhật, học chung lớp A2 03-06 với thằng Cường. Còn Nguyễn Hoàng Việt hình như cũng có quen, hồi cấp 2 học Nguyễn Du quận 1, lớp 9/1 phải ko?

Chào các anh , em sắp lên 11CT LHP TPHCM . Thầy Vũ Vĩnh Thái vẫn đang rất "phong độ" anh Gibbernergy ạ :d

Hihi, vào tận đây để mà chat chít lận à, ghê vậy?
Chúc vui vẻ! :lol:

Dân LHP vô đây nhiều wá!!!mình là dân NK mà chả thấy 1 mống nào vào đây mới buồn đây nè!!!hix hix!!! :(

Em hoc lop 10CT LHP, tpHCM ne.
May anh co biet cach nao de dc chon di thi Olympic 30/4 ko?
Giup do em voi.  :D

Học giỏi vào thì được đi thi thôi em ạ. :P

Trên này hình như có Đức Thành cũng tham gia mà, nick là ctlhp thì phải?  :maybe: Mình có bạn làm ở TpHCM nó bảo là ở SG thì LHP còn có tiếng hơn cả NK DHQG :D Chúc mừng! :D

Mình cũng là cựu học sinh chuyên Toán Lê Hồng Phong.
Học chuyên toán LHP chỉ thấy thích thầy Thái thôi, nhiệt tình và sáng tạo........
Các thầy khác thì ôi thôi....tất cả đều chạy theo..............MON...Y

Thay Thai duoc cai that tha va dan da, cai gi cung boc lo qua loi an tieng noi, rat chuan muc! :D

Em la hs moi, mac du hoc lop 10CTi (2008-2009) nhung em rat thich toan, rat mong cac anh giup do... ^^

Nghe nói có thầy Minh mới từ Lam Sơn Thanh Hoá chuyển về dạy tin, nhưng thầy Minh xuất thân từ Toán mà :|

I see that not many people come here to post things. Where do people go? I barely see a new post. This is such a boring section.

Chuyên toán NTH chả có ai vô đây cả, mình mới buồn nè


[quote="tiredepartment"]What do you eat for breakfast before the AIME?[/quote]

Definately a banana w/ orange juice and AAA batteries. :yup:

Why do you need AAA batteries when you can make a battery with the fruit? Haha, but I guess if you really wanted to get energized...

[quote="tiredepartment"]What do you eat for breakfast before the AIME?[/quote]

Definitely previous AIMEs.

don't eat stuff that are notoriously known to cause diarrhea (for yourself of course). And just enough so you dont feel hungry.

Generally, cereal+banana+egg
Perhaps i ate the wrong cereal this time...Normally I eat Honey Bunches of Oats as a rule of thumb before any big thing.  This time I ate Cinnamon Toast Crunch and screwed up.  Well, that teaches us a lesson...

I ate nice Japanese traditional breakfast + some sudoku problems + some geometric figures before the AIME :lol:


Fruit salad can be made with any 3 of these 5 fruits: apples, bananas, grapes, strawberries, and pineapples.  If strawberries and pineapples do not taste good together and grapes and bananas do not look appetizing together how many possible good tasting and appetizing salads are there?

[hide="hint"]Well, if you're having trouble with this, first find out how many different salads you can make, and then find out how many contain the combinations that do not taste good or do not look appetizing. Good luck!!!  :)  :)  :)  [/hide]

[hide]$\frac {5!}{3!2!}$. The ones that don't go well together account for 6 combinations. $4$.[/hide]

[hide][color=indigo]there are 10 combinations total, so when you take out the 6 bad ones there are 4 good salads.[/color][/hide]

[hide]A
Bx
Gx
S*
p*
x's and *'s don't go well with each other.
AB
AG
AS
AP
[b]4[/b][/hide]

[hide]the are 10 possible, but 6 are bad, so 4 good ones are left[/hide]


geometry
circumcircle
trigonometry
IMO Shortlist
Triangle
easy geometry

[quote="orl"]The altitudes through the vertices $ A,B,C$ of an acute-angled triangle $ ABC$ meet the opposite sides at $ D,E, F,$ respectively. The line through $ D$ parallel to $ EF$ meets the lines $ AC$ and $ AB$ at $ Q$ and $ R,$ respectively. The line $ EF$ meets $ BC$ at $ P.$ Prove that the circumcircle of the triangle $ PQR$ passes through the midpoint of $ BC.$[/quote]
[hide="Solution"]
Let $ M$ be the midpoint of $ BC$ and let $ H$ be the orthocenter of $ \triangle ABC$.  Extend $ AH$ to meet the circumcircle of $ \triangle ABC$ at $ X$ and let $ MH$ meet $ AP$ at $ Y$.  Notice that $ \angle BFC \equal{} 90 \equal{} \angle BEC$, so $ BFEC$ is cyclic.  This circle has center $ M$, so it is well-known that $ MH\perp AP$.  Furthermore, from the fact that $ BFEC$ is cyclic, we see that $ \angle RCB \equal{} \angle AFE \equal{} \angle BQR$, so $ BRCQ$ is cyclic, which means that $ BD\cdot DC \equal{} QD\cdot DR$.  Yet, $ ABXC$ is cyclic, so $ BD\cdot DC \equal{} AD\cdot DX$.  From the fact that $ \angle AYM \equal{} 90 \equal{} \angle ADM$, we see that $ AYDM$ is cyclic, so $ \angle PAX \equal{} \angle HMB$.  Yet, it is well-known that $ X$ is the reflection of $ H$ over $ BC$, so $ \angle PMX \equal{} \angle HMB \equal{} \angle PAX$, which gives that $ PAMX$ is cyclic.  Thus, $ PD\cdot DM \equal{} AD\cdot DX \equal{} BD\cdot DC \equal{} QD\cdot DR$, so $ PQMR$ is cyclic, so the circumcircle of $ \triangle PQR$ passes through the midpoint of $ BC$, which is $ M$. [/hide]

Dear Mathlinkers,
this problem has also being proposed at : O.M.  IRAN  1998 and Hubei Math Contest 1994.
Thanks for your nice proof.
I saw another solution based on power in the book of Mohamed Assila.
Sincerely
Jean-Louis

Please post those further solutions you know for this problem in this thread. You are also invited to send your solutions for the other posted IMO Shortlist problems. Thanks.

Dear "Orl",
sorry, I made a confusion in the name. Here is the reference of the book that I have given.
Soulami Tarik Belhaj, Les Olympiades de mathématiques, Ellipses.
I will send my proof that I have in mind.
Sincerely
Jean-Louis

Quatto, what is point T? Do you mean F instead? Also, can you explain why $ MH \perp AP$?

Yes, sorry [b]dgreenb801[/b], $ F$ is the same as $ T$; sorry for the error. [Moderator edit: Error fixed.]

Apply the following lemma to cyclic quadrilateral $ ABCD$ to get that $ MH\perp AP$:

Consider cyclic quadrilateral $ ABCD$ with center $ O$ and intersection of diagonals being $ P$.  Let $ AB$ and $ CD$ intersect at $ X$ and $ AD$ and $ BC$ meet at $ Y$.  Then, $ OP\perp XY$.

This can be proven as follows:

Let the intersection between the circumcircles of $ \triangle ABY$ and $ DCY$ be $ Z$.  Let the feet of the perpendiculars from $ P$ to $ AD$ and $ BC$ be $ Q$ and $ R$ respectively and the midpoints of $ AD$ and $ BC$ be $ M$ and $ N$ respectively.  Now, $ \angle ADZ \equal{} \angle ZCB$ and $ \angle ZBC \equal{} 180 \minus{} \angle ZBY \equal{} 180 \minus{} \angle YAZ \equal{} \angle ZAD$, so $ \triangle ZAD\sim \triangle ZBC$.  Since $ M$ and $ N$ are the respective midpoints of $ AD$ and $ BC$, we see that $ \triangle ZAM\sim \triangle ZBN$, so $ \angle YMZ \equal{} \angle YNZ$, so $ YZNM$ is cyclic.  Since $ O$ is the circumcenter of $ ABCD$, we see that $ \angle YMO \plus{} \angle YNO \equal{} 90 \plus{} 90 \equal{} 180$, so $ YNOM$ is cyclic, so $ YZNOM$ is cyclic.  Thus, $ \angle YZO \equal{} \angle YNO \equal{} 90$, so $ OZ\perp YZ$.  By Miguel's Theorem, we see that $ Y$, $ Z$, and $ X$ are collinear, so $ OZ\perp YX$.  Now, notice that since $ ABCD$ is cyclic, $ \triangle PAD\sim \triangle PBC$, so $ \triangle PAQ\sim \triangle PBR$, so $ \frac {AQ}{BR} \equal{} \frac {AP}{PB} \equal{} \frac {AD}{BC}$, which means that $ \triangle ZAQ\sim \triangle ZBR$, which implies that $ \angle ZQY \equal{} \angle ZRY$, so $ ZRQY$ is cyclic.  Since $ \angle PRY \plus{} \angle PQY \equal{} 90 \plus{} 90 \equal{} 180$, we see that $ YZRPQ$ is cyclic, so $ \angle PZY \equal{} \angle PRY \equal{} 90$, which implies that $ P$ and $ O$ lie on the perpendicular through $ Z$ to $ YX$.  Thus, $ OP\perp YX$.

I think this problem is posted many times on MathLinks, but I can't get the links, so let me present the simplest solution I have in mind (surely that it's not new)  :D

We have quadrilateral $ BCEF$ is cyclic and $ EF\parallel QR$, so the quadrilateral $ BQCR$ is also cyclic. Therefore $ DQ\cdot DR \equal{} DB\cdot DC$ $ (1)$

On the other hand, $ (PBDC)$ is a harmonic division, so $ DB\cdot DC \equal{} DM\cdot DP$ $ (2)$, where $ M$ is the midpoint of $ BC$.

From $ (1)$ and $ (2)$, we conclude that $ M$ lies on the circumcircle of triangle $ PQR$.

[quote="The QuattoMaster 6000"]Consider cyclic quadrilateral $ ABCD$ with center $ O$ and intersection of diagonals being $ P$.  Let $ AB$ and $ CD$ intersect at $ X$ and $ AD$ and $ BC$ meet at $ Y$.  Then, $ OP\perp XY$.[/quote]
We can prove, that O is orthocenter of XYP. We see that $ (A,D;F,Y)\equal{}(B,C;G,Y)\equal{}(D,A;F,Y)$ => (A,D;F,Y)=(B,C;G,Y)=1. Hence PX is polar of Y => $ PX\perp OY$. Analogously $ XO\perp YP$ QED (I think it's known as Brokard's theorem)

I have a simpler way to prove this lemma: it is well known (Pascal to the ‘hexagons’ ABCCDA and ABBCDD) that the tangents to circle (ABCD) at A and C concur on XY, let their common point be K, same the ones at B and D, which concur at L. Then since AP.PC = BP.PD it follows that P belongs to radical axis of the circles (K,KA) and (L,LB). But O belongs to that line too [the tangents from O to both circles are radii of (ABCD)], therefore OP and KL are parallel, but KL and XY are perpendicular.

Best regards,
sunken rock

Lines $ PR$ and $ PQ$ cut circle $ \mathcal{T} \equiv{} \odot (RCQB)$ again at $ X$ and $ Y.$ Notice that $ P$ is radical center of $\mathcal{ T},$ the 9-point circle $(N)$ and the circle with diameter $\overline{BC}.$ Inversion with center $ P$ and power $\overline{ PB} \cdot \overline{PC}$  takes $ D \mapsto M,$ $ R \mapsto X$ and $ Q \mapsto Y.$  In other words, circle $\odot( PQR)$ is taken to the line $ XY.$ But $ (B,C,D,P)=-1$ and $ P \equiv {} QY \cap XR$ is on the polar of the intersection of the diagonals of the quadrilateral $ RXQY$ WRT $\mathcal{ T}$ $\Longrightarrow$ $ XY$ passes through $ D.$ If $ XY$ passes through $ D,$ then its inverse $\odot (PQR)$ passes through the inverse $M$ of $ D$ and the proof is completed.

Let O be the midpoint of BC.

We note that $\angle PFB=\angle AFE=\angle C$

$\angle FPB=\angle B-\angle C$

Applying the law of sines in triangle PFB.

$PB=\frac{BFsinC}{sin(B-C)}=\frac{acosBsinC}{sin(B-C)}$

$PD=PB+BD=\frac{acosBsinC}{sin(B-C)}+ccosB$

$OD=OB-BD=\frac{a}{2}-ccosB$

$\angle QFD=\angle BFD=\angle C$

$\angle FQD=\angle AFE=\angle C$

So $QD=FD=\frac{csin2B}{2sinC}$

Similarly applying the law of sines in triangle DRC.

$DR=\frac{DCsinC}{sinB}=\frac{bsin2C}{2sinB}$

Now points P,Q,O,R lie on a circle

$\Leftrightarrow PD*OD=QD*DR$

$\Leftrightarrow (\frac{acosBsinC}{sin(B-C)}+ccosB)(\frac{a}{2}-ccosB)=\frac{bcsin2Bsin2C}{4sinBsinC}$

$\Leftrightarrow 4sinBcosBsinCcosC=4sinBcosBsinCcosC$

which is true.Hence the proof.

Maths is the doctor of science.

Darn I failed to come up with a synthetic solution, and tried trig bash instead, which I believe is the more natural approach. My solution is probably similar to that posted by [b]sayantanchakraborty[/b], although I didn't read it fully.

[hide=" Diagram"] [img]http://s2.postimg.org/v7qhtxcbt/Untitled.png[/img] [/hide]
[hide=" Solution"]
I shall use the standard notations $BC=a, CA=b, AB=c, \angle BAC = A, \angle ABC= B, \angle BCA= C.$ Let $R$ be the circumradius of $\triangle ABC.$ Let $M$ be the midpoint of $BC.$ WLOG assume $AB<AC$ ($AB \neq AC$ since otherwise $EF \parallel BC$).

Note that $\angle DQA = \angle AFE = \angle QFD,$ so $QF = DF.$ By sine rule on $\triangle DEF,$ we get that $\dfrac{DF}{\sin (\pi - 2B)} = R \implies DF = R \sin 2B,$ where we have used the fact that the radius of the nine point circle is half the circumradius. Similarly we get $DR = R \sin 2C.$

Note that $\angle FPB = \angle ABC - \angle PFB = \angle ABC - \angle DFE = B-C.$ By sine rule on $\triangle DEP$ and using the fact that $DE = R \sin 2C$ (which follows from sine rule in $\triangle DEF$), $\dfrac{R \sin 2C}{\sin (B-C)} = \dfrac{DP}{\sin 2B} \implies DP = \dfrac{R \sin 2C \sin 2B}{\sin (B-C)}.$ Also, note that $MD = \dfrac{a}{2} - c \cos B = R (\sin A - 2 \sin C \cos B).$

We wish to show that $DP \times DM = DP \times DQ,$ from which the result follows by PoP. This is equivalent to showing
\begin{align*} \sin 2B \sin 2C & = (\sin A - 2 \sin C \cos B) \dfrac{\sin 2B \sin 2C}{\sin (B-C)} \\
\iff \sin(B-C) & = \sin (B+C) - 2 \sin C \cos B \\
\iff \sin (B-C)  & = \sin B \cos C + \cos B \sin C - 2 \sin C \cos B \\ & = \sin B \cos C - \sin C \cos B, \end{align*}
which is true. $\blacksquare$

 [/hide]

Dear Mathlinkers,

http://jl.ayme.pagesperso-orange.fr/Docs/Regard%203.pdf  p. 12-13.

Sincerely
Jean-Louis

Let $M$ is midpoint of $BC$
Since: $PQ$ $\parallel$ $EF$ then: $B$, $R$, $C$, $Q$ lie on a circle
So: $\overline{DQ} . \overline{DR} = \overline{DB} . \overline{DC} = \overline{DP} . \overline{DM}$ or $M$ $\in$ $(PQR)$

Let $M$ be a midpoint of $BC$.
[b]Claim: [/b] $BRCQ$ is cyclic.
[b]Proof:[/b]  Note that:
$$ \angle BRQ = \angle PFB = \angle BCQ$$
We will prove that $PRQM$ is cyclic. It is sufficient to show that $PD \cdot DM = RD \cdot DQ$. Since $BRCQ$ is cyclic we have $RD \cdot DQ = BD \cdot DC$. Therefore we are left to prove:
$$ PD \cdot DM =BD \cdot DC$$
Note that:
\begin{align*}
BD \cdot DC =(BM-DM)(MC+DM) =BM^2-DM^2 = PD \cdot DM \\
BM^2=DM(PD+DM)=DM \cdot MP
\end{align*}
Since $(P,D;B,C)=-1$ and $M$ is midpoint of $BC$ by well -known lemma we get that $BM^2 = DM \cdot MP$ and we are done.

See my solution on my Youtube channel here:

[url]https://www.youtube.com/watch?v=CryOVKdVZC0[/url]

[quote=orl]The altitudes through the vertices $ A,B,C$ of an acute-angled triangle $ ABC$ meet the opposite sides at $ D,E, F,$ respectively. The line through $ D$ parallel to $ EF$ meets the lines $ AC$ and $ AB$ at $ Q$ and $ R,$ respectively. The line $ EF$ meets $ BC$ at $ P.$ Prove that the circumcircle of the triangle $ PQR$ passes through the midpoint of $ BC.$[/quote]
I think I have seen this one on the other exam before :huh:
[hide=harmonic]
$M$ is considered as the midpoint of $BC$.
$BCEF$ is a complete quadrilateral so:
\[(CB,DP)=-1
\implies PB.PC=PD.PM \]
\[\implies MD.DP=BD.DC \qquad (1)\]
on the other hand we know that $EF||RQ$ so $RBQC$ is cyclic then:
\[BD.DC=RD.DQ\qquad (2)\]
combining $(1)$ and $(2)$ gives the desired result.
$\boxed{\mathcal{Q.E.D}}$
[/hide]

We have that $\triangle RAQ \sim \triangle FAE \sim \triangle CAB \implies AQ \cdot AC =AR \cdot AB \implies BRCQ$ is cyclic $\implies RD \cdot QD=BD \cdot CD$. Let $X$ be the second intersection of $(AEF)$ and $(ABC)$. Then by the radical center $A,P,X$ are collinear. Also if $N$ is the reflection of $H$ w.r.t. $M$, then $\angle BNC=\angle BHC=180^{\circ}-\angle BAC \implies N \in (ABC)$ and $\angle AXN=\angle ABN=\angle ABC+\angle CBN=\angle ABC+\angle HCB=90^{\circ}=\angle AHN \implies X,H,N$ are collinear. Since $H,M,N$ are collinear we have that $X,H,M$ are collinear. Thus $H$ is the orthocenter of $\triangle PAM$. If $S$ is the reflection of $H$ over $\overline{BC}$, then $\angle BSC=\angle BHC=180^{\circ}-\angle BAC \implies S\in (ABC)$. In a similar way we prove that $S \in (ADM)$. Hence $D$ lies on the radical axis of $(ABC)$ and $(ADM)$, so it has equal powers w.r.t. these circles, which are $BD \cdot CD$ and $PD \cdot MD$, respectively. Thus $RD \cdot QD=BD \cdot CD=PD \cdot MD \implies PRMQ$ is cyclic. Done.

Nice problem for practicing harmonic ratios :D

We easily see that $BQRC$ is a cyclic quadrilateral.
Let $M$ be the midpoint of $BC$.
We have that $-1=(P,D;B,C)=(B,C;P,D)$, this implies that $DM.DP=DB.DC$
But since we have that $BQCR$ is cyclic this implies that $BD.DC=QD.DR$ by PoP, thus we have that $DM.DP=DQ.DR$, by PoP we have that $PQMR$ is cyclic.

$AD,BE,CF$ concure at $H$, thus $(P,D;B,C)=-1\Rightarrow BD\cdot PC=DC\cdot PB\Rightarrow BD\cdot DP+BD\cdot DC=DC\cdot PB\Rightarrow DP(BC-DC)+BD\cdot DC=DC\cdot PB\Rightarrow DP\cdot BM=DC\cdot PB\Rightarrow DP\cdot DM=DC\cdot PB-BD\cdot DP\Rightarrow DP\cdot DM=DC\cdot BD+DC\cdot PB-BD\cdot PC\Rightarrow DP\cdot DM=DC\cdot BD=RD\cdot DQ$. So $PRMQ$ is cyclic.

For storage I guess...

We know that $(P,D;B,C) = -1$ and that $R, Q, B,C$ concyclic. Let $M$ be the midpoint of $BC$.
Then $DM \cdot DP =  DB \cdot DC = DQ \cdot DR$ thus $M \in circ(PQR)$

$M$ is midpoint of  $BC$
$$PD \cdot DM=RD \cdot DQ$$
By sinus Law.

[quote=bora_olmez]For storage I guess...
[/quote]

Can u explain why the condition (P,D;B,C) = -1 implies
DM . DP = DB . DC  ?

We know $BQCR$ is cyclic as $$\angle RQC=\angle FEC=180-\angle CBA=RBC$$ so $DB\cdot DC=DQ\cdot DR.$ Since $(BC;DP)=-1,$ $D^*=P$ with respect to $(BC).$ Hence, letting $M$ be the midpoint of $\overline{BC},$ we see $MD\cdot MP=BM^2.$ Therefore, \begin{align*}DB\cdot DC&=(BM-DM)(CM+DM)=BM^2-DM^2\\&=MP\cdot MD-DM^2=DM(MP-DM)=DM\cdot DP.\end{align*} $\square$

Denote midpoint of $BC$ by $M.$ Since $BRCQ$ are concyclic and $P,D$ are inverse wrt circle with diameter $BC,$ we obtain $$|PD|\cdot |DM|=|BD|\cdot |DC|=|RD|\cdot |DQ|.$$

[hide=storage]Let $M$ be the mid-point of $\overline{BC}$.
Note that since $QR\parallel EF$, and line $EF$ is antiparallel to line $BC$, we conclude that $B,R,Q,C$ are concyclic. So, $$DB\cdot DC=DQ\cdot DR \ \ (1)$$
Moreover, since $(P,D;B,C)=-1$, a few length computations gives $$DM\cdot DP=DB\cdot DC \ \ (2)$$
Combining $(1)$ and $(2)$, we get $$DQ\cdot DR=DM\cdot DP,$$ and the result follows. $\square$[/hide]


linear algebra
matrix
real analysis
real analysis unsolved

Let $ A\equal{}\frac{1}{\sqrt{5}}\begin{pmatrix} 1 & 2 \\ \minus{}2 & 1 \end{pmatrix}$ and $ x: \mathbb{R}\to\mathbb{R}^2$ be a function.

Find the solution of $ x'(t)\equal{}Ax(t)$ with initial value $ x(0)\equal{}(1,\minus{}1)$.

$ x(t)\equal{}e^{tA}x(0)$. [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=177357]How to find $ e^{tA}$[/url].

In addition to the method in the link, you may use Laplace transformation to find $ e^{\bb{A}t}$. See an example here: [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=177356[/url]


If $ a$ and $ b$ are roots of $ x^4 \plus{} x^3 \minus{} 1$, then prove that $ ab$ is a root of $ x^6 \plus{} x^4 \plus{} x^3 \minus{} x^2 \minus{} 1.$

[hide="Solution"]The problem must work for any choice of two roots. Hence, we will show that there exists numbers $ (p,q,r,s)$ such that the two polynomials are $ (x \minus{} p)(x \minus{} q)(x \minus{} r)(x \minus{} s)$ and $ (x \minus{} pq)(x \minus{} pr)(x \minus{} ps)(x \minus{} qr)(x \minus{} qs)(x \minus{} rs)$.

The coefficients from the first polynomial require that:

$ x^3 : \qquad a \equal{} p \plus{} q \plus{} r \plus{} s \equal{} \minus{} 1$
$ x^2 : \qquad b \equal{} pq \plus{} pr \plus{} ps \plus{} qr \plus{} qs \equal{} 0$
$ x^1 : \qquad c \equal{} pqr \plus{} pqs \plus{} prs \plus{} qrs \equal{} 0$
$ x^0 : \qquad d \equal{} pqrs \equal{} \minus{} 1$

The coefficients from the second polynomial require that:

$ x^5 : \qquad b \equal{} 0$
$ x^4 : \qquad ac \minus{} d \equal{} 1$
$ x^3 : \qquad c^2 \plus{} a^2d \minus{} 2bd \equal{} \minus{} 1$
$ x^2 : \qquad acd \minus{} d^2 \equal{} \minus{} 1$
$ x^1 : \qquad bd^2 \equal{} 0$
$ x^0 : \qquad d^3 \equal{} \minus{} 1$

Where the above conditions are derived by writing the coefficients in terms of symmetric sums.

Plugging in the values of $ A,B,C,D$ found from the first polynomial show that they are consistent.[/hide]

I hope to see a faster solution

hello, do you mean with the roots $ a$ and $ b$ the real roots of the equation $ x^4 \plus{} x^3 \minus{} 1 \equal{} 0$?
Sonnhard.

[b]Restatement:[/b] Given that $ a^4\plus{}a^3\equal{}1$ and $ b^4\plus{}b^3\equal{}1$ for $ a,b \in \mathbb{C}$, show that $ (ab)^6\plus{}(ab)^4\plus{}(ab)^3\minus{}(ab)^2\equal{}1$

We must also have that $ a\ne b$. I don't think that restatement helps.

Solution:

Simply consider $ x^4\plus{}x^3\minus{}1\equal{}(x^2\plus{}ax\plus{}b)(x^2\plus{}cx\plus{}d)$. Expand, compare coefficients, and solve for $ b$.

$ x^4\plus{}(a\plus{}c)x^3\plus{}(d\plus{}ac\plus{}b)x^2\plus{}(ad\plus{}bc)x\plus{}bd$

Then $ a\plus{}c\equal{}1$, $ b\plus{}d\plus{}ac\equal{}0$, $ ad\plus{}bc\equal{}0$, $ bd\equal{}\minus{}1$.

Firstly, $ d\equal{}\minus{}b^{\minus{}1}$.
So $ ad\plus{}bc\equal{}0\implies b^2c\equal{}a$.

So $ 1\equal{}a\plus{}c\implies 1\equal{}b^2c\plus{}c\implies c\equal{}(b^2\plus{}1)^{\minus{}1}$

So $ a\equal{}b^2c\equal{}b^2\cdot (b^2\plus{}1)^{\minus{}1}$.

Now $ b\plus{}d\plus{}ac\equal{}0\implies b\minus{}b^{\minus{}1}\plus{}b^2\cdot (b^2\plus{}1)^{\minus{}2}\equal{}0$.

Clearing the denominators gives us the desired polynomial!

[quote="TZF"]$ x^5 : \qquad b \equal{} 0$
$ x^4 : \qquad ac \minus{} d \equal{} 1$
$ x^3 : \qquad c^2 \plus{} a^2d \minus{} 2bd \equal{} \minus{} 1$
$ x^2 : \qquad acd \minus{} d^2 \equal{} \minus{} 1$
$ x^1 : \qquad bd^2 \equal{} 0$
$ x^0 : \qquad d^3 \equal{} \minus{} 1$[/quote]
Is there a fast way to derive this?


LaTeX
inequalities
induction
number theory proposed
number theory

Let a,b be 2 positive integers such that s(an)=s(bn) for all positive integers n (s(x) is the sum of digits of x). Prove that a/b=10^k for some integer k.

Oh...it was an open question by Harazi not so long ago. Does it means that you found a solution?
That's nice!

Pierre.

Uups... I didn't know Harazi posted it as an open question. That's a bit strange because we solved it a few months ago. Anyway, if I (actually we) found a solution, it means that many others can. So... have fun!

I think the unsolved problem was a different one, involving an and bn (if I remember correctly). I think it asked to show the same thing, given that S(an)=S(bn) for sufficiently large n.

Harazi can probably shed some light on this issue :).

You are right Grobber, Imisread the statement :

http://www.mathlinks.ro/viewtopic.php?t=1684

But it is also a nice problem.

Pierre.

This one, posted by dZeta is easier than the open question posted long time ago. But wait, this doesn't mean that we didn't spend lots of time on it. But believe me, it worths.

We may assume a and b aren't divisible by 10 (otherwise we just work with a'=a/10 and we repeat this until there's no 10 left). If we look a bit harder we can see that we may assume (a,10)=1 because if 2|a, for example, we work with a'=5a and b'=5b and then we work with a"=5a/10 and b' etc. In the same way we show that we may assume (a,10)=(b,10)=1.

Now assume further a>b. (a,10)=1 means that 10phi(a)=1 (mod a), so there is a multiple of a , ka=10t-1=99..9. kb<ka, so kb has fewer digits than ka, and if it has the same number of digits, they can't all be 9 because otherwise we would have a=bS(kb), which contradicts the initial conditions, so our assumption that a>b was wrong. We treat the case a<b in the same manner to get the conclusion a=b (this conclusion holds for "reduced" a and b, that is a and b dividd by 10 as many times as we need so that 10 doesn't divide any of them).

Since S(x)=S(10tx) for all t>=0 we deduce that a/b=10t with integer t.

Terribly sorry about that! What was I thinking? I just meant (a,10)=(b,10)=1. I don't need (a,b)=1. It's been edited. Is the rest Ok?

Sorry, but I don't think we can consider (a,10)=(b,10)=1.

Hmm.. I see what you mean. Do you at least think we can assume (a,10)=1? I'm pretty sure about this one, but this wouldn't be enough to solve the problem as quickly as I wanted to :). However, I think we can build an argument starting from (a,10)=1.

To show that we may assume (a,10)=1: We've seen that we can take a s.t. 10 doesn't divide it, and let's assume a=2tk, with odd k. Then we work with th numbers a'=5ta, b'=5tb, and then with a"=a'/10t=k, b"=b'. If 5 divides k we repeat the process, going to a"' and so on, until we reach a(h) which isn't divisible with 2 or 5. Howwever, this will give b(h) some 5 and 2 factors, We can divide it by 10 until it only has 2 factors or only 5 factors, but we haven't reached (a,10)=(b,10)=1, just (a,10)=1. At this stage, b can be much greater than a, so my argument doesn't work anymore. Maybe something else does.. I'll think about it some more.

There was a little error in my previous message: I assumed 10 doesn't divide a, so if a=2tk then k can't be divisible with 5 (it wasn't that important anyway).

I've been thinking about it for a long time, but I haven't been able to finish it in the case when the greatest among a and b isn't coprime with 10. I think I can reduce it to this:

Let b a number not equal to 99..9 with n 9's. Then the terms of the sequence ak=(1+10n+102n+..+10kn)b can't have a bounded number of digits different from 9. I'm sure it's true, but I can't prove it (to make it clearer: there is no N s.t. if 9(k) is the number of digits of ak which are different from 9 then 9(k)<=N for all k>=0).

It probably doesn't yield a nice solution, but I've been thinking about it for so long (or so it seems) that I just had to get it out of my system :D.

Sorry! Looks like I'm wrong again! :( The "lemma" in my previous post is wrong, and a counterexample is: n=1, b=45<>9 and N=2. All the terms ak are of the type 499..95.

This seems to be a dead-end  :?

P.S. Really sorry for all the useless posts on this topic, but I keep making mistakes and I have to correct them and so on. I suppose you people could erase these..

Ok, let's solve this problem. Very important: the following solution is the opera of my friend Adrian (Dzeta), only the last argument (which is trivial) being mine:
   I will try to write it in Latex:
    Let us take a natural t such that [tex] 10^t>a [/tex]. For any [tex] k\geq t [/tex] there is a unique number [tex] n_k [/tex] such that [tex] 10^k\leq a\cdot n_k<10^k+a [/tex]. Thus, [tex] 10^k\cdot\frac {b}{a}\leq b\cdot n_k< 10^k\cdot\frac{b}{a}+b [/tex]. Since [tex] s(a\cdot n_k) [/tex] is bounded, so is
 [tex] s(b\cdot n_k) [/tex] and from the above inequalities we find that this implies that [tex]\frac{b}{a}[/tex] has a finite number of zeroes after the comma. Exactly the same it follows that [tex]\frac{a}{b} [/tex] has a finite number of zeroes after the comma. Thus, we can find (m,10)=1 such that [tex] a=2^x\cdot 5^y\cdot m [/tex] and [tex] b=2^z\cdot 5^u\cdot m [/tex]. Now we take [tex] c>max(x,y) [/tex] and [tex] N=2^{c-x}\cdot 5^{c-y}\cdot n [/tex]. Since [tex] s(aN)=s(bN)[/tex] we find that [tex] s(mn)=s(mn\cdot 2^{z+c-x}\cdot 5^{u+c-y}) [/tex] for any n. We are done if [tex] z-x=u-y [/tex]. Otherwise, we may asume that [tex] z-x>u-y [/tex] (in the other case we multiply by a very large power of 2) . Let [tex] p=z-x+y-u[/tex]. We have [tex] s(mn)=s(mn\cdot 2^p) [/tex] for any n and an easy induction shows that [tex] s(m)=s(m\cdot 2^{kp}) [/tex] for any k. But by an easy application of Kronecker theorem we can show that for any s there is a number k such that [tex] m\cdot 2^{kp} [/tex] begins with 9,9,...,9 (k times), contradiction. Too bad we are too young to propose this problem to a TST. I think it's very good for a TST, but now...

And one more thing: do not trust what Dzeta says: if he made the problem, then lots of people can. Believe me, he's my friend and I know what he is capable of and I know lots of people can't do this problem. I couldn't solve it and I'm not alone.


Let $ x,y,z$ be real numbers . Prove that $(x^3\plus{}y^3\plus{}z^3)^2\plus{}3(xyz)^2\ge 4(x^3y^3\plus{}y^3z^3\plus{}z^3x^3)$ .

Determine the cases of equality .

[quote="Mateescu Constantin"]Let $ x,y,z$ be real numbers . Prove that $ x^3 \plus{} y^3 \plus{} z^3 \plus{} 3(xyz)^3\ge 4(x^3y^3 \plus{} y^3z^3 \plus{} z^3x^3)$ .

Determine the cases of equality .[/quote]
Try $ x \equal{} \minus{} 1$ and $ y \equal{} z \equal{} 0$ or $ x \equal{} y \equal{} z\equal{}1.$ :wink:

I'm sorry, the correct statement was this :

If $x,y,z\in\mathbb{R}$ then : $\boxed{\ (x^3+y^3+z^3)^2+3(xyz)^2\ge 4(x^3y^3+y^3z^3+z^3x^3)\ }$

[quote="Mateescu Constantin"]I'm sorry, the correct enunciation was this :

If $x,y,z\in\mathbb{R}$ then : $\boxed{\ (x^3+y^3+z^3)^2+3(xyz)^2\ge 4(x^3y^3+y^3z^3+z^3x^3)\ }$[/quote]
The inequality is following to:
$ x^6+y^6+z^6+3x^2y^2z^2 \ge 2(x^3y^3+y^3z^3+z^3x^3) $
Setting $ a=|x^3|,b=|y^3|,c=|z^3| $
By AM-GM:  $ a+b+c \ge 3\sqrt[3]{abc} \Rightarrow 3x^2y^2z^2=3\sqrt[3]{a^2b^2c^2} \ge \frac{9abc}{a+b+c} $
Using Schur ineq we have: $ a^2+b^2+c^2 + \frac{9abc}{a+b+c} \ge 2(ab+bc+ca) \ge 2(x^3y^3+y^3z^3+z^3x^3) $
Q.E.D


[quote="VIMF"]let $ a,b,c > 0$. Prove that

$ \sqrt {\left( {a \plus{} b \plus{} c} \right)\left( {\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}} \right)} \ge 1 \plus{} \sqrt [3]{{5 \plus{} \sqrt {\left( {{a^3} \plus{} {b^3} \plus{} {c^3}} \right)\left( {\frac {1}{{{a^3}}} \plus{} \frac {1}{{{b^3}}} \plus{} \frac {1}{{{c^3}}}} \right)} }}$[/quote]
For the proof we need to check only one case: $ b\equal{}c\equal{}1.$ :wink:

[quote="VIMF"][quote="arqady"][quote="VIMF"]let $ a,b,c > 0$. Prove that

$ \sqrt {\left( {a \plus{} b \plus{} c} \right)\left( {\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}} \right)} \ge 1 \plus{} \sqrt [3]{{5 \plus{} \sqrt {\left( {{a^3} \plus{} {b^3} \plus{} {c^3}} \right)\left( {\frac {1}{{{a^3}}} \plus{} \frac {1}{{{b^3}}} \plus{} \frac {1}{{{c^3}}}} \right)} }}$[/quote]
[/quote][/quote]
i think that your problem is inspired from prof Cirtoaje's one(GM-B 5 2004)

[quote="VIMF"][quote="arqady"][quote="VIMF"]let $ a,b,c > 0$. Prove that

$ \sqrt {\left( {a \plus{} b \plus{} c} \right)\left( {\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}} \right)} \ge 1 \plus{} \sqrt [3]{{5 \plus{} \sqrt {\left( {{a^3} \plus{} {b^3} \plus{} {c^3}} \right)\left( {\frac {1}{{{a^3}}} \plus{} \frac {1}{{{b^3}}} \plus{} \frac {1}{{{c^3}}}} \right)} }}$[/quote]
For the proof we need to check only one case: $ b \equal{} c \equal{} 1.$ :wink:[/quote]

I don't understand
can you write your solution?
thanks[/quote]
Let $ a \plus{} b \plus{} c \equal{} 3u,$ $ ab \plus{} ac \plus{} bc \equal{} 3v^2$ and $ abc \equal{} w^3.$
 Hence, our inequality is equivalent to $ f(w^3)\leq3\sqrt {uv^2},$ where $ f$ is increasing function.
Thus, $ f$ gets a maximal value when $ w^3$ gets a maximal value, which happens when two numbers from $ \{a,b,c\}$ are equal.
Id est, it remains to check the case $ b \equal{} c \equal{} 1,$ which gives that your inequality is true.

[quote="arqady"][/quote][quote="VIMF"][quote="arqady"][quote="VIMF"]let $ a,b,c > 0$. Prove that

$ \sqrt {\left( {a + b + c} \right)\left( {\frac {1}{a} + \frac {1}{b} + \frac {1}{c}} \right)} \ge 1 + \sqrt [3]{{5 + \sqrt {\left( {{a^3} + {b^3} + {c^3}} \right)\left( {\frac {1}{{{a^3}}} + \frac {1}{{{b^3}}} + \frac {1}{{{c^3}}}} \right)} }}$[/quote]
For the proof we need to check only one case: $ b = c = 1.$ :wink:[/quote]

I don't understand
can you write your solution?
thanks[/quote][quote="arqady"]
Let $ a + b + c = 3u,$ $ ab + ac + bc = 3v^2$ and $ abc = w^3.$
 Hence, our inequality is equivalent to $ f(w^3)\leq3\sqrt {uv^2},$ where $ f$ is increasing function.
Thus, $ f$ gets a maximal value when $ w^3$ gets a maximal value, which happens when two numbers from $ \{a,b,c\}$ are equal.
Id est, it remains to check the case $ b = c = 1,$ which gives that your inequality is true.[/quote]
I do not like this solution
[url=http://ddbdt.tk/forum/ca-c-cua-c-thi-trasn-vimf/136-ae-thi-cha-nh-tha-c-cua-c-thi-vicfj.html]see more[/url]
this is my solution by Cauchy-Schwarz[/url]

Nice proof, VIMF! :lol:
But I think, your method does not help for the following inequality.
Let $ a,$ $ b$ and $ c$ are positive numbers. Prove that:
\[ \sqrt {\left( {a \plus{} b \plus{} c} \right)\left( {\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}} \right)} \ge 1 \plus{} \sqrt [3]{{4.1 \plus{} 1.3\sqrt {\left( {{a^3} \plus{} {b^3} \plus{} {c^3}} \right)\left( {\frac {1}{{{a^3}}} \plus{} \frac {1}{{{b^3}}} \plus{} \frac {1}{{{c^3}}}} \right)} }}\]

[quote="arqady"]
Let $ a,$ $ b$ and $ c$ are positive numbers. Prove that:
\[ \sqrt {\left( {a \plus{} b \plus{} c} \right)\left( {\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}} \right)} \ge 1 \plus{} \sqrt [3]{{4.1 \plus{} 1.3\sqrt {\left( {{a^3} \plus{} {b^3} \plus{} {c^3}} \right)\left( {\frac {1}{{{a^3}}} \plus{} \frac {1}{{{b^3}}} \plus{} \frac {1}{{{c^3}}}} \right)} }}\]
[/quote]
It's wrong. I am sorry.


Given a positive integer $ n$, find the smallest possible $ t \equal{} t(n)$ such that for all real numbers $ x_1, x_2, . . . , x_n$ we have
$ \sum_{k\equal{}1}^n(x_1 \plus{} x_2 \plus{} ... \plus{} x_k)^2 \leq t(x_1^2\plus{} x_2^2 \plus{}... \plus{} x_k^2)$.

On the right hand side of your inequality, do you sum up over all $ k\equal{}1,...,n$ or should the last $ k$ be a $ n$?

It's probably
$ (x_1\plus{}x_2\plus{}\ldots \plus{}x_n)^2\le t(x_1^2\plus{}x_2^2\plus{}\ldots \plus{}x_n^2)$

[quote="jbmorgan"][quote="Bugi"]It's probably
$ (x_1 \plus{} x_2 \plus{} \ldots \plus{} x_n)^2\le t(x_1^2 \plus{} x_2^2 \plus{} \ldots \plus{} x_n^2)$[/quote]

Yep, that's what I want...[/quote]

If that what you want then the best is $ t\equal{}n$. But I think it should be
\[ x_1^2  \plus{} \left( {x_1  \plus{} x_2 } \right)^2  \plus{} ... \plus{} \left( {x_1  \plus{} ... \plus{} x_n } \right)^2  \le k\left( {x_1^2  \plus{} x_2^2  \plus{} ... \plus{} x_n^2 } \right)
\]
 :)
Then it's more interest.  :)

[quote="jbmorgan"][quote="Bugi"]It's probably
$ (x_1 \plus{} x_2 \plus{} \ldots \plus{} x_n)^2\le t(x_1^2 \plus{} x_2^2 \plus{} \ldots \plus{} x_n^2)$[/quote]

Yep, that's what I want...[/quote]

[quote="jbmorgan"][quote="Honey_S"] But I think it should be
\[ x_1^2 \plus{} \left( {x_1 \plus{} x_2 } \right)^2 \plus{} ... \plus{} \left( {x_1 \plus{} ... \plus{} x_n } \right)^2 \le k\left( {x_1^2 \plus{} x_2^2 \plus{} ... \plus{} x_n^2 } \right)
\]
:)
Then it's more interest.  :)[/quote]

Yeah, yeah, of course I mean this one...[/quote]

Wait,wait,wait which is the actual?

I'm sorry. Only one small typo in the original.

Here's the actual:

Given a positive integer $ n$, find the smallest possible $ t \equal{} t(n)$ such that for all real numbers $ x_1, x_2, . . . , x_n$ we have
$ \sum_{k \equal{} 1}^n(x_1 \plus{} x_2 \plus{} ... \plus{} x_k)^2 \leq t(x_1^2 \plus{} x_2^2 \plus{} ... \plus{} x_n^2)$


Given triangle AXY, four points are spaced evenly along XY and the segments connecting them to A are drawn. The midpoint of AY is Z and XZ is drawn. The four points of intersection are B C D and E respectively from X to Z. Find the ratio CD/DE.

My instinct is to go with mass points, but I'm not too sure how to assign the weights.

hey everyone, i am reviving this topic because it is a mass points problem.

But I don't know how to do it.
Can someone help?

I fail at mass points, so here's a non mass point solution...

[hide="Solution"]
Without loss of generality, let $ AZ\equal{}1$, so $ ZY\equal{}1$ as well. Draw a line $ \overline{EE'}$ parallel to $ \overline{XZ}$ such that $ E'$ is on line $ \overline{AY}$. If we let $ XZ\equal{}x$, then since the points drawn were equally spaced into $ 5$ regions, $ EE'\equal{}\frac x5$. Also, $ E'Y\equal{}\frac15\implies ZE'\equal{}\frac45$. Now, since $ \triangle AEZ\sim\triangle AEE'$, $ EZ\equal{}\frac x5\left(\frac1{1\plus{}\frac45}\right)\equal{}\frac x9$. Proceed in a similar manner to find that $ DZ\equal{}DE\plus{}EZ\equal{}\frac x4$ and $ CZ\equal{}CD\plus{}DE\plus{}EZ\equal{}\frac37x$, so $ CD\equal{}\frac37x\minus{}\frac x4\equal{}\frac5{28}x$ and $ DE\equal{}\frac x4\minus{}\frac x9\equal{}\frac5{36}x$, and the desired ratio is $ \frac{CD}{DE}\equal{}\frac{\frac5{28}x}{\frac5{36}x}\equal{}\frac{36}{28}\equal{}\boxed{\frac97}$.
[/hide]

[quote="Smartguy"]why won't anyone respond to mass point threads? is it not popular??? :mad:[/quote]  This is not your personal problem-solving service.  Your two posts were less than 24 hours apart.  Have some patience!

Here's a solution with barycentric coordinates, and I think (?) any such proof can be translated into mass points without too much difficulty:
[hide]$ A \equal{} (1, 0, 0)$, $ X \equal{} (0, 1, 0)$, $ Y \equal{} (0, 0, 1)$.  Then $ Z \equal{} (1/2, 0, 1/2)$.  $ C$ lies on the line joining $ A$ to $ (0, 3/5, 2/5)$, $ D$ lies on the line joining $ A$ to $ (0, 2/5, 3/5)$ and $ E$ lies on the line joining $ A$ to $ (0, 1/5, 4/5)$.  It follows that every point on $ XZ$ is of the form $ (t, 1 \minus{} 2t, t)$, every point on $ AC$ is of the form $ (1 \minus{} 5s, 3s, 2s)$, and so on.  This gives us some quick linear equations to solve, and we find that $ C \equal{} (2/7, 3/7, 2/7)$, $ D \equal{} (3/8, 1/4, 3/8)$ and $ E \equal{} (4/9, 1/9, 4/9)$.  Then $ \frac{CD}{DE} \equal{} \frac{3/8 \minus{} 2/7}{4/9 \minus{} 3/8} \equal{} \frac{9}{7}$, which luckily is the same as math154's answer.[/hide]


calculus
inequalities solved
inequalities

If $a \geq b \geq c>0$ then we have:

$ \sum \frac{a^3b}{a^3+b^3} \geq \sum \frac{ab^3}{a^3+b^3}$.

Multiplying (a^3+b^3)(b^3+c^3)(c^3+a^3),we have

 \sum a^6b^4 >=  \sum a^4b^6 (using force)

It's trivial by (x-y)(x-z)(y-z)(xy+yz+zx)>=0  <==>   \sum x^3y^2>= \sum x^2y^3 when x>=y>=z>0.

Thanks.


prove that we always choose 9 from 17 unexpected numbers satisfy to total them that I have a sum which divide 9=0

[quote="ZetaX"]- bad title.
- avatar stealing.
- see http://www.mathlinks.ro/Forum/viewtopic.php?t=57546 , my post.[/quote]
avatar stealing ??? it's a joke ,you think avatar has copyright???
 bad title ! maybe
thanks for your link

Yes, avatars have, if selfmade, a copyright on it. The owner could even sue you if he wants.

Problem follow posted before on this forum

prove that we always choose $n$ from $2n-1$ unexpected numbers satisfy to total them that I have a sum which divisble $n$.

Sorry , because I not see that link  :oops: . I know this assertion, but not know its prove. Thanks!  :wink:


A rod of length 100 meters is inclined with a wall. It is sliding in such a manner that the horizontal end slides on the floor in the velocity of 10 m./s. What is the rate of sliding of the middle point of the rod at the inclination of pi/6 with the horizontal floor?

100 meters? Seriously?  :o

The midpoint moves horizontally at half the rate of the bottom end (which makes $ 10/2\equal{}5$ m/s). It also move vertically at half the rate of the top end. So you need to find the velocity of the top end, which is probably done in your textbook. The final step is to use the Pythagorean theorem.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=125044]Older topic with the same title (but a harder problem)[/url]

$ x^2\plus{}y^2\equal{}100^2$
$ 2x\frac{dx}{dt}\plus{}2y\frac{dy}{dt}\equal{}0$

when their angle is $ \frac{\pi}{6}$
$ {dx}{dt}\equal{}10$, $ x\equal{}50\sqrt 3$, $ y\equal{}50$

so $ \frac{dy}{dt}\equal{}\frac{\minus{}x\frac{dx}{dt}}{y}\equal{}\minus{}10\sqrt 3$

giving us midpoint speeds of $ (5,\minus{}5\sqrt 3 )$

cool part, final answer = 10


algebra
polynomial
LaTeX
inequalities
geometry
functional equation
algebra unsolved

Find all polynomial solutions of the functional equation f(x)f(x+1)=f(x2+x+1). Note:x2 means x square. I don't know how to type x square.

[quote="nealth"]Find all polynomial solutions of the functional equation$ f(x)f(x+1)=f(x^2+x+1)$. [/quote]
if you want to know how can use you should find one of [b]valentin veronica [/b] post because in his signeture there is the LATEX pdf form and you can download it easily.

there is another thing;is $f:R\rightarrow R$ or not because when you say nothing about this it means $f:R\rightarrow R$.

i think $f(x)=1$or $f(x)=0$ because:
suppose that $f(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0$we have
 $f(x).f(x+1)=f(x^2+x+1) $ $\rightarrow$
 $(a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0)$
$\times (a_n(x+1)^n+a_{n-1}(x+1)^{n-1}+...+a_1(x+1)a_0$=
$a_n(x^2+x+1)^n+a_{n-1}(x^2+x+1)^{n-1}+...+a_1(x^2+x+1)+a_0$ $\rightarrow$
the maximum degree in the left of equality is $3n$(it belongs to $(a_n)^2x^nx^{2n}$)
 but the maximum degree in the right of this equality is $2n$(it belongs to $a_nx^{2n}$)
 $\rightarrow $ we supposed that $a_n$ is not $0$ $\rightarrow 3n=2n \rightarrow n=0 \rightarrow f(x)=a_0$ and we have $f(0).f(1)=f(1) \rightarrow f(1)=0$ or
$f(0)=1$
we R done. :D

My solution:
Suppose it's nonzero polynomial.And f(x)=1 does not always hold.
we have $f(x)f(x-1)=f(x^2-x+1)$
it's obvious that the lead coefficient is 1.
And if x is a root, then $x^2+x+1, x^2-x+1$ are roots also. Since the number of roots is finite, if a root x belongs to a set I, the I is closed to $g(x)=x^2+x+1,h(x)=x^2-x+1$
Let z be the complex root with the largest |z|, , and z is not 0 obviously. If $|z^2+1|$ is not 0,
then either $|z^2+z+1|>|z|$ or $|z^2-z+1|>|z|$ holds.
so $z^2+1=0$ what's more the polynomial is real so (x+i) and (x-i) must appear in pairs.
Let $f(x)=(x^2+1)^nr(x)$ which (x^2+1) doesn't divide r(x).
We also have $r(x)r(x+1)=r(x^2+x+1)$ so r(x)=1
So the only solution are f(x)=0 or $(x^2+1)^n$.

your solutions is very nice but i think you make it so hard and you can solve it easier.and there is another thing don't ever say that something in math is obvious because may be you think it is obvious but it is not,you should prove all your steps. :D

[quote="jensen"]...and there is another thing don't ever say that something in math is obvious because may be you think it is obvious but it is not,you should prove all your steps. :D[/quote]
Well I was in a hurry to type my solution. I just say "obvious" to make others think a little and also save time. However, thanks to your suggestion.
And have you noticed that your last solution is wrong? The degrees are both 2n

[quote="jensen"]i think you are wrong because you said that both of them are 2n but the maximum degree of the left side is $x^{2n} \times x^n$ and it is equal to $x^{3n}$.so what do you mean that both are 2n?[/quote]
Maybe thinking a little before posting is not such a bad idea ;).

yes you are correct i am so sorry lets have another solution:
i will correct my solution as soon as i can.

[quote="jensen"]
i will correct my solution as soon as i can.[/quote]
We're still waiting :)

[quote="Arne"][quote="jensen"]
if you want to know how can use you should find one of [b]valentin veronica [/b] post because in his signeture there is the LATEX pdf form and you can download it easily.[/quote]

When did Valentin change his surname?[/quote]I will pretend this is (yet another) just a mistake from our Iranian friends ...  :ninja:

Jensen ,behave yourself.Look at your posts in Geometry part too,Please stop making useless posts.
If you want to tell someone about latex ,just Pm him ,there is no need of a public post
Dont make such a arguments about the others solution(This forum has its moderators)
We have already made enough problems here so there is no need of a new one.
i am sure you didnt mean anything by Using vornica beside of vornicu ,but the correct name is[b]Mr.Valentin Vornicu

[quote]I will pretend  this is (yet another) just a mistake from our Iranian friends[/quote]
I can assure you ,Mr.Vornicu, that this is just a simple mistake made by jensen ,He must be carefull when using the others names but this is a mistake ,not an insult.


calculus
integration
trigonometry
calculus computations

calculate the following integral:

INTEGRAL from 0 to pi/2 of

[((sinx)^2 * cos(x))/(sinx+cosx)]dx

regards, TOM.

Let $I = \int_0^\frac{\pi}{2}\frac{\sin^2x\cos x}{\sin x+\cos x}dx$
Variable substition: $y=\frac{\pi}{2}-x$
$I = \int_0^\frac{\pi}{2}\frac{\sin y\cos^2 y}{\sin y+\cos y}dy$
Thus
$2I = \int_0^\frac{\pi}{2}\frac{\sin^2x\cos x}{\sin x+\cos x}dx + \int_0^\frac{\pi}{2}\frac{\sin y\cos^2 y}{\sin y+\cos y}dy = \int_0^\frac{\pi}{2} \sin x\cos x\ dx = \frac{1}{2}$

$I = \frac 14$.


[quote="Interval"]What is e^pi times pi^e?[/quote]
What is the point of posting this in two forums. I'm like 99% sure that the answer isn't anything special.

[quote="t0rajir0u"]Something rather ugly and pointless.

A more interesting question:  Which is larger?[/quote]
[hide]Taking the ln of both sides we compare now
$\pi$ and $e \ln \pi$ . The ln of pi is like 1.1-1.3 so $e^{\pi}$ is greater.[/hide]
EDIT: What the heck, I guess I didn't click hide and my latex is messed up. Also, to torajirou, that was an approximation by hand, it would obviously be trivialized by a calculator... I guess I'm missing some obvious method, I wish I were better at math...

Of course I meant, "reason without a calculator."  (Two points:  One, if $\ln \pi = 1.3$ (which it does not) then your logic fails.  Two, if you were allowed to evaluate $\ln \pi$ to solve this problem, then why not evaluate $e^{\pi}, \pi^{e}$?  Of course you don't because you would trivialize the problem.)

The theoretical answer is really very interesting.  Won't someone try?

If you expand out $e^{\pi}$ into a series, you can estimate $\pi^{e}$ with $\pi^{3}$ and play round with it. I havnt actually done this, just hypothising a bit.

If $F(x)=x^{-1}\text{ ln }x$ on $x\in ]0,\infty[$, we have $F'(x)=x^{-2}(1-\text{ln }x)$, which is positive if $x<e$ and negative if $x>e$. Therefore $F$ is increasing if $x<e$ and decreasing if $x>e$. Therefore $F$ gets its maximum value at the point $e$, so $e^{-1}\text{ ln }e>\pi^{-1}\text{ ln }\pi$ and thus $e^\pi>\pi^{e}$.


I have 30 coins consisting of nickels and quarters.  The total value of the coins is $\$4.10.  How many of each kind do I have?

[hide="da answer"]
i set up a system of equations.
They are:
n+q=30
and...
5n+25q=410
solve for n.
n=30-q
substitute...
5(30-q)+25q=410
150-5q+25q=410
20q=260
q=13
n+q=30
13+n=30
n=17
[b]there are 13 quarters and 17 nickels[/b][/hide]

[hide] ans: 13 quarters and 17 nickels
[/hide]

[hide]let n be the number of nickels and q be the number of quarters, so

5n + 25q = 410
n + q = 30

solve it from there.
[color=red]13 quarters, 17 nickels [/color][/hide]


Evaluate $ \allowbreak \int_{0}^{10\pi }\frac{\cos 6x\cos 7x\cos 8x\cos 9x}{1\plus{}e^{2\sin ^{3}4x}}\,dx\allowbreak$

$ I\equal{}\int_{0}^{10\pi}\frac{cos6xcos7xcos8xcos9x}{1\plus{}e^{2sin^3 4x}} dx$

Replace $ x$ by $ 10\pi\minus{}x$

$ I\equal{}\int_{0}^{10\pi}\frac{cos6xcos7xcos8xcos9x}{1\plus{}e^{\minus{}2sin^3 4x}} dx$

On adding

$ 2I\equal{}\int_{0}^{10\pi} cos6xcos7xcos8xcos9x dx$

You may be able to tackle it now


Every card in a deck has a picture of one shape - circle, square, or triangle, which is painted in one of the three colors - red, blue, or green.  Furthermore, each color is applied in one of three shades - light, medium, or dark.  The deck has 27 cards, with every shape-color-shade combination represented.  A set of three cards from the deck is called complementary if all of the following statements are true:

i. Either each of the three cards has a different shape or all three of the card have the same shape.
ii. Either each of the three cards has a different color or all three of the cards have the same color.
iii. Either each of the three cards has a different shade or all three of the cards have the same shade.

How many different complementary three-card sets are there?

[hide]Case 1: All three attributes are different.  There are 27 ways to choose the first card, 8 ways to choose the second card, and 1 way to choose the third card.  There are $\frac{27\times 8}{6}=36$ possibilities for this case.

Case 2: Two attributes are different, and one is the same.  There are 3 ways to choose which attributes will be different.  There are 9 ways to choose the first card, 4 ways to choose the second card, and 1 way to choose the third card.  There are $\frac{3\times 9\times 4}{6}=18$ possibilities for this case.

Case 3: One attribute is different, and two are the same.  There are 3 ways to choose which attribute will be different.  There are 3 ways to choose the first card, 2 ways to choose the second card, and 1 way to choose the third card.  There are $\frac{3\times 3\times 2}{6}=3$ possibilities for this case.

Case 4: All three attributes are the same.  Since there are no identical cards, there are no possibilities for this case.

In total there are $36+18+3=57$ sets.[/hide]

[hide]

We can split it into 4 different cases:

Case 1: All 3 cards the same - not possible since we only have 27 distinct cards
Case 2: 2 same attribute, one is different: $\binom{3}{1}\binom{3}{1}$ and we multiply 3 because of symmetry so $27$ for this case
case 3: 1 same attribute, 2 different: we first assign the one fixed attribute $\binom{3}{1}$. Then, the 2nd attribute is just assigned because it makes no difference because the first attribute is the same. The third attribute can be distributed in $3!$ ways so there are a total $(3)(3!)(3) =54$ ways once you account for symmetry.
case 4: all attributes different: assign the first attribute randomly the next 2 can be assigned as: $3!$ and $3!$. There's not symmetry this time because we are already accounting for distinctness through the factorials: so $36$ in this case.

Add it all up and we have $36 + 54 + 27 = 117$[/hide]

[quote="me@home"][hide="hmmm.... this seems to be wrong but here is what I keep getting:"]
Let's say we have picked two cards. We now compare their attributes to decide how we can pick the third card to make a complement set.
For each of the three attributes, should the two values be the same we have one option - choose a card with the same value for that attribute. Furthermore, should the two be different there is only one option- choose the only value that is remaining. In this way, every two card pick corresponds to exactly one set, for a total of $\binom{27}{2}=27*13=\boxed{351}$ possibilities.[/hide]
somewhere along the way I had to divide a three, but I don't see where... can someone help?[/quote]

Every two cards chosen correspond to exactly one other card to complete the set. However, any pair of that set of three will correspond to the same set. There are $\displaystyle{3\choose 2}$ ways of choosing a pair out of three cards, so your method overcounts by a factor of three. Cool solution  :D

thx... its funny, this solution was the first thought that came to mind when I first saw the problem, yet I usually only see tedious solutions...
have you ever played the card game Set that this problem was modeled after?

I don't really understand ultramanzzy's solution  :(
Also, what's the actual answer? 57 or 117?

ultramanzzy's "solution" is meaningless.  agolsme's post or the combination of the two posts that follow it give the correct answer.

Here's my solution:
[hide]
Case 1: The cards share no properties.
Using case 1 of matt276eagles's solution, there are $ \boxed{36}$ possibilities here.

Case 2: The cards share at least one property.
Envision each card as a point $ (x,y,z)$ in 3D space, where $ x,y,z\in\{1,2,3\}$, where the x-, y-, and z-coordinates of each point correspond to the card's color, shade, and shape respectively.
If the cards share a property, then the points lie in a plane $ P$ parallel to one of the planes $ x\equal{}0$, $ y\equal{}0$, or $ z\equal{}0$.  There are twelve possible ways to choose three points in $ P$ such that they correspond to a complementary set of cards:

[asy]path P = (1,1)--(1,-1)--(-1,-1)--(-1,1)--(1,1)--(0,1)--(0,-1)--(0,0)--(1,0)--(-1,0);
for (int i=0; i<4; i+=1)
{
  for (int j=0; j<3; j+=1)
  {
    draw(shift((3*i+1,3*j+1))*P);
    for (int a=0; a<3; a+=1)
    {
      for (int b=0; b<3; b+=1)
      {
        dot((3*i+a,3*j+b),gray(0.7));
      }
    }
  }
}
dot((0,6)); dot((0,7)); dot((0,8));
dot((1,3)); dot((1,4)); dot((1,5));
dot((2,0)); dot((2,1)); dot((2,2));
dot((3,8)); dot((3,4)); dot((3,0));
dot((4,8)); dot((4,4)); dot((4,0));
dot((5,8)); dot((5,4)); dot((5,0));
dot((6,6)); dot((7,7)); dot((8,8));
dot((9,8)); dot((10,7)); dot((11,6));
dot((6,5)); dot((7,3)); dot((8,4));
dot((6,0)); dot((7,2)); dot((8,1));
dot((11,5)); dot((10,3)); dot((9,4));
dot((11,0)); dot((10,2)); dot((9,1));[/asy]

Since there are 9 possible values for $ P$, we have $ 12\cdot 9\equal{}108$ sets.  But we're overcounting, since this counts the cases in which we have two shared properties twice.
There are 3 ways to choose which property is not shared, and there are $ 3^2\equal{}9$ ways to choose the values for the properties that are shared.  So there are $ 27$ sets that have two shared properties, so case 2 contains $ 108\minus{}27\equal{}\boxed{81}$ possibilities.
(We need not worry about cases where all three properties are shared, since that means all the cards are the same.)

Therefore the answer is $ 36\plus{}81\equal{}\boxed{117}$.
[/hide]

Hmm could someone please point out what's wrong with this using PIE? I couldn't find the flaw... :maybe:

[hide]We could call the restrictions $ i_1,i_2,ii_1,ii_2,iii_1,iii_2$ with $ _1$ corresponding to DIFFERENT and $ _2$ corresponding to SAME as in $ i,ii,iii$ in the problem.

Then let $ x_1y_1$ denote the cardinality of the intersection of $ x_1,y_1$ with $ x,y\in \{i,ii,iii\}$ and $ x\neq y$.

Then the problem is equivalent to

$ \sum x_1 \plus{} \sum x_2 \minus{}\sum x_1y_1 \minus{}\sum x_1y_2 \minus{}\sum x_2y_2 \plus{}\sum x_1y_1z_1 \plus{}\sum x_1y_1z_2 \plus{}\sum x_1y_2z_2\plus{}\sum x_2y_2z_2$

(note that $ x_1x_2\equal{}0$ since we cannot have an element that is both DIFFERENT and SAME, so the rest sums of 4 or more variables are $ 0$)

$ x_1\equal{}3^6$; $ \sum x_1 \equal{} 3x_1\equal{}3(3^6)$ (since $ x_1\in\{i_1,ii_1,iii_1\}$)

$ x_2\equal{}3*3^6$; $ \sum x_2\equal{}3x_2\equal{}3(3*3^6)$

$ \sum x_1y_1\equal{}3(3!*3^3)$

$ \sum x_1y_2\equal{}3*2(3*3^3)$

$ \sum x_2y_2\equal{}3(3*3)$

$ \sum x_1y_1z_1\equal{}1(3!*3!)$

$ \sum x_1y_1z_2\equal{}3(3!*3)$

$ \sum x_1y_2z_2\equal{}3(3*3)$

$ \sum x_2y_2z_2\equal{}1(3*3*3)$

But doing the operations with the known values won't yield the right answer... :maybe:  :huh:[/hide]

The number of triples of cards of all the same shape is not $ 3 \cdot 3^6$ (and I really wish that you'd begin using \cdot for multiplication) -- you've counted in that expression triples that contain the same card three times, and it's not clear to me that these impossible triples ever get filtered out.  (There may be other errors as well.)

Yes, you are right! I always make this kind of dumb mistakes whenever there are so many cases!  :o  :(
anways what are the impossible ones you were referring (i guess you meant not those of both DIFFERENT and SAME since my method didn't include those - all $ 0$ as i said earlier)...
BTW nice solution me@home!!!! :)

Associate each card with a triple $ (a_1,a_2,a_3)$ where $ a_i\in \{0,1,2\}$ and $ a_1$ is the color, $ a_2$ is the shape, and $ a_3$ is the shade.

Three cards corresponds to three triples. The conditions assert that either each component is all the same or all different.

Count: all ordered sets of cards that satisfy the all equal or all different component criterion: consider component one for the three cards, there is $ 9$ possibilities  (the permutations of $ 0$,$ 1$, and $ 2$ or all equal). So we'd have $ 9^3$ possibilities.

Count: all ordered sets of distinct cards with all equal or different components:

If the cards are not distinct, then there are either two that are equal or three that are equal. But if two triples are equal, they are all equal. So we just need to consider when they are all equal. This is just the number of cards or $ 27$. So $ 9^3\minus{}27$.

Count: all unordered sets of distinct cards with all equal or all different components: Using what we had before, we counted all ordered sets of distinct cards that satisfied the all equal or all different for each component. So for any triple of triples, we can permute those cards in exactly $ 3!\equal{}6$ ways to get another counted item. So our answer is $ \frac{9^3\minus{}27}{6}\equal{}117$.

I did $\frac{27*26}{6} $

6 ways to order each set of 3.

Sorry for the huge bump, but I'm having trouble understanding cases 2 and 3 of agolsme's solution.


[quote="mathswizard1992"]If in a group of hens and cows ,the number of legs is twice the no. of heads in the group . Find the no. of each type of animals......
Give reason also.......[/quote]
[hide="Solution"]
Each cow has only 1 head; likewise, each hen has only 1 head.

This implies totalHeads = cows + hens.

Each cow has 4 legs; each hen has 2 legs.

Implying  totalLegs = 4 x cows + 2 x hens.

totalLegs = 2 x totalHeads --> 4 x cows + 2 x hens = 2 x (cows + hens).

Therefore, 4 x cows + 2 x hens = 2 x cows + 2 x hens.

4 x cows + 2 x hens - 2 x hens = 2 x cows + 2 x hens - 2 x hens.

--> 4 x cows = 2 x cows.

--> 4 x cows  - 2 x cows = 2 x cows - 2 x cows.

--> 2 x cows = 0.

--> 2 x cows/2 = 0/2.

Therefore, cows = 0.

This means the entire group is hens, and only hens. Of course that makes intuitive sense because hens have exactly twice as many legs as heads; whereas, cows have four times as many legs as heads.

Since we were not told how many animals there were to begin with, we don't how many hens there are.[/hide]

[hide="answer"]I think that the answer would be that there would be nothing but hens, because the numbers of legs is twice as much as the number of heads, and that a hen also has twice as many legs as heads, therefore there must be only hens.[/hide]


A rubber ball is dropped from a 100 ft tall building. Each time it bounces, it rises to threequarters
its previous height. So, after its first bounce it rises to 75 ft, and after its second
bounce it rises to 3/4 of 75 ft, and so on forever. What is the total distance the ball travels?

This is what I did:
Sum of an Infinite Geometric Sequence= a/1-r
So 100/ 1-(3/4) = 100/(1/4)
So sum = 400

BUT THE ANSWER KEY SAYS 700!!!

Can someone tell me if I am right, or not please.

Thanks!

Click here --> [hide=" :D  :D  :D "]First off, after the ball bounces, it has to fall back down. So double 400. However, the first time, the ball does not go up 100. So you should subtract 100. Thus the answer is 700.[/hide]


vector
geometry
linear algebra
number theory
number theory proposed

Given a set S of 11 natural numbers. Prove that this set S has a non-empty subset T such that if you take all the numbers from T and throw all of their digits in a bin, then each digit occurs in this bin an even number of times.

  Darij

If it's not number theory, then why do you post it in number theory? :D

It's a combinatorics problem; there are ten digits, so then the problem can be stated as follows (it paractically solves itself :)):

Given $11$ vectors, each one with $10$ components in $\mathbb Z_2$, show that they do not form an independent set over $\mathbb Z_2$. This is, of course, obvious, since the $\mathbb Z_2$ vector space we're working with has dimension $10$.

Indeed, Grobber, this was my solution too. And I actually wanted to post it into "Algebra" first (it's actually a linear algebra problem), but then I thought that if I post it into "Algebra" then it would be too simple...

  Darij

Peter : I think it is a kind of private joke with Darij (check some threads in algebra and geometry...)

Pierre.

[quote="pbornsztein"]Peter : I think it is a kind of private joke with Darij (check some threads in algebra and geometry...)[/quote]

Actually, if it would be a private joke with me, then I, at least, should be able to understand it?... or am I too dumb now?... or is 2:00 AM too late for doing maths?...

Ah yes, and note that I hate problems where numbers are represented in the decimal system, like number rebuses, IMO 2004/B3 etc.. Hence it is clear that when I post a problem about digits, then it's not really about digits, like this one  :D

  Darij

http://www.mathlinks.ro/Forum/viewtopic.php?t=15123&highlight=combinatorical+tools

and

http://www.mathlinks.ro/Forum/viewtopic.php?t=15139

Pierre.

Ah, you meant these twos... okay, as I have said, I posted the problem into Number Theory in order not to make it too simple. I didn't actually think about my promise "Next time I will post a geometric problem into the number theory section and it will be solved using number theory...", and of [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=6081]a similar expression of mine in the Geometry forum[/url]; actually, the problem above doesn't seem to be related to geometry...

  Darij

Ya I didn't make the link... I seem to remember this one as an older topic than the other two... for some reason :?

Anyway very nice problem! :)


[ A partition of positive integer m into n components is any sequence a1,...,an  of positive integers such that a1+...+an=m and a1<=a2<=...<=an]

[ The lexicographic order is defined as follows: sequence a1,...,an  comes before b1,...,bn iff there exists such an integer i,1<=i<=n, that aj=bj  for all j, 1<= j< i, and ai< bi.]

We are given integer m and we want to determine the partition, which occupies the k-th position in the lexicographic order of all partitions of m into n components - how to do that??

For example:
m=9
n=4
k=3

So partition k=1 is [1 1 1 6], k=2 [1 1 2 5] and k=3 [1 1 3 4]

But how to find partition when, for example m=220, n=10 and k=1000000000??

This can be done fairly easily if you know the numbers P(m,n) of all partitions
of m into n components. If you know all such numbers, then to determine the
first component of the N-th partition, compare N to P(m-1,n-1). If N does not exceed P(m-1,n-1), then the first component is 1. If N is greater than P(m-1,n-1), then you are looking at partitions of m into n components starting with 2 or higher. The number of partitions of m into n components starting with 2 is the same as the number of partitions of m-n into n components starting with 1 (subtract 1 from each component), that is P(m-n-1,n-1). Thus, the first component is 2 if P(m-1,n-1)<N<=P(m-1,n-1)+P(m-n-1,n-1). Similarly, it is 3 if
P(m-1,n-1)+P(m-n-1,n-1)<N<=P(m-1,n-1)+P(m-n-1,n-1)+P(m-2n-1,n-1) and so on.
Once you have determined that the first component is k, you reduced the problem to  finding the N'=N-P(m-1,n-1)-...-P(m-(k-2)n-1,n-1)-th partition of m-k into n-1 components with the smallest component of size at least k. Subtracting k-1 from
each component, you see that it is equivalent to the problem of finding the N'-th
partition of m-k-(n-1)(k-1) into n-1 components.

This is still quite a work but it is easily doable on any decent computer. This leaves unanswered just one question: how to find P(m,n) fast? For n=1 and n=2 the answer is obvious: P(m,1)=1 (m>=1); P(m,2)=[m/2] ([x] stands for the integer part of x).  Also P(m,n)=0 if m<n and P(n,n)=1.

The other values can be found from the relation P(m,n)=P(m-1,n-1)+P(m-n,n). This will force you to fill out a 220 by 10 table first to do your particular example but, again, any decent computer will manage 2200 numbers without a big problem.  Hope this helps.

fedja

Everything seems to be OK, but I can't understand how can I count N'... When m=9, n=4 and k=3 (in your algorithm it's N), P(8,3)=5 and k (the first element) is equal to 1, so N'=N-P(m-1,n-1)=3-5=-2... What should I do then?? Maybe I'm just doing something wrong...

Can You explain me how to count N', please....

The sum $P(m-1,n-1)+...+P(m-(k-2)n-1,n-1)$ one should subtract from $N$ to
find $N'$ is a bit confusing thing for small $k$: for $k=1$ it is empty: if you write it in the sigma notation, you will have
$\sum_{j=0}^{k-2} P(m-jn-1,n-1)$ and for $k=1$ the range of summation will be
$0\leq j\leq -1$, i.e., empty.

So, for $k=1$, we have $N'=N$, for $k=2$ (not for $k=1$!), we have $N'=N-P(m-1,n-1)$, for $k=3$ and higher the sum makes perfect sense as written.
Sorry for not writing it clearly the first time. :)

Ok - I understand that. But now I have another question:

When I find first element k of n-elements partition of integer m, then I'm looking for second one, but before doing it I have do decrement m by k (m=m-k) and decrement n by 1 (n=n-1). You wrote that next element has to be larger or equal than previous one, so what I should check?? I mean in first iteration I have to check if N<=P(m-1,n-1), if not then P(m-1,n-1)<N<=P(m-1,n-1)P(m-n-1,n-1) and so on... But in each next when k has to be larger or equal then previous one what condition I should start with?? When I'm starting always with the same condition (N<=P(m-1,n-1)) I don't know what k to take as "solution" in iteration, and usualy I get wrong answer. :(

I'll start with repeating what I wrote originally (just to make sure you paid attention to that part)

Once you have determined that the first component is k, you reduced the problem to finding the N'=N-P(m-1,n-1)-...-P(m-(k-2)n-1,n-1)-th partition of m-k into n-1 components with the smallest component of size at least k. Subtracting k-1 from
each component, you see that it is equivalent to the problem of finding the N'-th
partition of m-k-(n-1)(k-1) into n-1 components.

In the algorithm it means that you do things recursively: once you determined the first component k, you call your algorithm with parameters m-k-(n-1)(k-1),n-1, N' in place of m,n,N and then add k-1 to every element of the resulting partition to get the correct tail.

Let's run an example with m=12, n=3, N=8.
Since P(11,2)=5, P(8,2)=4, we see that k=2, N'=3
Now we would like to search the 3-d partition among the partitions
of 12-2=10 into 2 components starting with 2 or a higher number.
So, actually, we are interested in partitions 2+8 ; 3+7; 4+6; 5+5, the third one being the one we need. The trick is to think of them as of arbitrary partitions of 8 into 2 parts to which 1 is added in each component.
So, instead of 2+8=10, you think of 1+7=8, instead of 3+7=10, you think of 2+6=8 and so on. Finding the third partition of 8 among all partitions into 2 components gives you 3+5. You add 1 back and get the tail 4+6 and the final answer 2+4+6.
 :)

Thank You very much - I've made it - now everything is OK... :)


There are 4 prisnors lined up in a row they are wearing black, white, black, white hats on the 1, 2, 3, and 4 man respectively. The 1st man is facing a wall and the rest can only look at the hat the man in front of them is wearing. The jail officer goes down the row starting with the 1st man. If they guess the right hat they are wearing they can go if they don't he/she will die. which man can be sure of which hat they are wearing. See if you can figure this out  :P
 Answer  :downto:
[hide]The 3rd man because the 1st man will answer and if he is wrong the 3rd man will know that he is wearing the opposite since there are only two colors. The 2nd man (right or wrong) will be wearing the hat that he said if right and opposite if wrong. Then the 3rd man will know there are two hats left one black and one white which the man in front of him is wearing[/hide]

wow. that is quite a diagram. Although ive solved it, ill let others to figure out.


Find all natural numbers which can be written in the form
$\frac{(a+b+c)^{2}}{abc}$, where $a,b,c\in N$

And again a well-known idea.

If $n$ satisfies the condition then consider its representation with the smallest possible sum $a+b+c$.
Suppose $a\leq b\leq c$. Consider the equation $x^{2}-(nab-2a-2b)x+(a+b)^{2}=0$. It has two natural solutions: $c$ and $\frac{(a+b)^{2}}{c}$, and their sum is $nab-2a-2b$. Note that $c+\frac{(a+b)^{2}}{c}\leq b+\frac{(a+b)^{2}}{b}$, since $c<a+b$ (otherwise the triple $(a,b,\frac{(a+b)^{2}}{c})$ have lesser sum than $a+b+c$). Thus \[nab-2a-2b \leq b+\frac{(a+b)^{2}}{b}=2b+2a+\frac{a^{2}}{b}.\] It follows $nab\leq 4(a+b)+\frac{a^{2}}{b}\leq 9b$, hence $n\leq 9$. And if $a>1$ then $n\leq 4$.

I know example for $n=1,2,3,4,6,9$.

[quote="Myth"]Note that $c+\frac{(a+b)^{2}}{c}\leq b+\frac{(a+b)^{2}}{b}$. [/quote]

This would require $(a+b)^{2}\geq bc$, but is it so?

Really, I forget to mention one thing in the proof. I have edited it.


I recall i once saw someone making a square around a text, a textbox if you like. Is that made in $\LaTeX$, and if yes, how to do so?

Thx in advance for your help..

Like $\boxed{this?}$ \boxed{math mode stuff}


induction
number theory unsolved
number theory

I solve this question before .
First try to prove $a=b$ is the only case .(Set $a=\text{min}\{a,b\}$ so $c\geq a$ .Then case work $c\geq b$ and $b>c$ which you found no solution )
then from $a!(a!-2)=c!$ notice that $gcd(a!,a!-2)=2$ . From here we can say that $a+1\leq c\leq a+3$ .Then just case work out .

By the way , (2,2,0) is not the solution since $0!=1$ :wink:

WLOG, $a \geq b$. Since $b! | LHS$, then $b! | RHS$ and $c \geq b$
Then, the equation becomes $a! = \frac{a!}{b!}+\frac{c!}{b!}+1$

i) $c = b, a > b$ yields $a! \cdot b! = a!+2 \cdot b! \Rightarrow a! | 2b!$
-) $a = b$ gives $a! (a!-3) = 0$ which produces no solution
-) $a > b \geq 1$ gives $\frac{a!}{b!}| 2 \Rightarrow a! = 2b! \Rightarrow 2b! = 4 \Rightarrow a! = 4$ which also produces no solution.

ii) $c > b$
-) $a \geq c > b$. Since $\frac{c!}{b!}| LHS \Rightarrow \frac{c!}{b!}| RHS \Rightarrow \frac{c!}{b!}| 1$ which is impossible because $\frac{c!}{b!}\geq b+1 \geq 2$
-) $c > a \geq b$. Since $\frac{a!}{b!}| LHS \Rightarrow \frac{a}{b!}| RHS \Rightarrow \frac{a!}{b!}| 1$
If $a > b$, we come to same conclusion as the preceding case. Thus, $a = b$ yields $c! = a! \cdot (a!-2)$ with $a > 2$. Let $p$ be the highest power of $2$ divides $a$, then $p+1$ is highest power of $2$ divides $RHS$.

--) $a$ even $\Rightarrow c = a+2$ or $c = a+3$.
---) $c = a+2 \Rightarrow (a+2)(a+1) = a!-2$. For $a \geq 4$, by induction, we can prove that $a!-2 > (a+2)(a+1)$. It follows that there's no solution for this case.
---) $c = a+3 \Rightarrow (a+3)(a+2)(a+1) = a!-2$ For $a \geq 6$, by induction, we can prove that $a!-2 > (a+3)(a+2)(a+1)$. Since $a = 4$ doesn't produce any solutions then it follows that there's no solution for this case.

--) $a$ odd $\Rightarrow c = a+1 \Rightarrow a+3 = a! \Rightarrow a | 3$. This gives the solution : $(a,b,c) = (3,3,4)$

Therefore, the only solution is : $(a,b,c) = (3,3,4)$

[quote="nayel"][quote="kathi;)"]$a!b!=a!+b!+c! \\ c!=(a!-1)(b!-1)-1$

for a=b, the only solutions are (2,2,0) and (3,3,4)
because
$c!=(a!-1)^{2}-1 \\ a!(a!-2)=c!$
Case 1:
$a!(a!-2)=0 \Rightarrow a!-2=0 \Rightarrow a=2$

Case 2:
$a!(a!-2)=c! \\ a!-2=(a+1) \\ a!-a=3$
with the only solution a=3

Case 3:
$(a!-2)=\frac{c!}{a!}=(a+1)(a+2)...(c-1)c$
This equation has no solution, because $a!-2=2p$ with p prime.
Proof: $a!-2=2(3*4*5*...*(a-1)*a-1)$ so the right side is 2*p (because $3*4*5*...*(a-1)*a-1$ isn't divisible by any prime q<a)
wlog you can say a<b, but these cases i have to check first ;)[/quote]

what if $3*4*5*...*(a-1)*a-1$ is divisible by prime $q>a$?[/quote]

hm? thats not possible, because if it would be divisible by a prime q>a, the other factor has to be smaller!

(now i'm at home again..... yes of course it was my fault, i thought that moment 0!=0 (i had to hurry up  :blush: ))

[quote="kathi;)"][quote="nayel"][quote="kathi;)"]$a!b!=a!+b!+c! \\ c!=(a!-1)(b!-1)-1$

for a=b, the only solutions are (2,2,0) and (3,3,4)
because
$c!=(a!-1)^{2}-1 \\ a!(a!-2)=c!$
Case 1:
$a!(a!-2)=0 \Rightarrow a!-2=0 \Rightarrow a=2$

Case 2:
$a!(a!-2)=c! \\ a!-2=(a+1) \\ a!-a=3$
with the only solution a=3

Case 3:
$(a!-2)=\frac{c!}{a!}=(a+1)(a+2)...(c-1)c$
This equation has no solution, because $a!-2=2p$ with p prime.
Proof: $a!-2=2(3*4*5*...*(a-1)*a-1)$ so the right side is 2*p (because $3*4*5*...*(a-1)*a-1$ isn't divisible by any prime q<a)
wlog you can say a<b, but these cases i have to check first ;)[/quote]

what if $3*4*5*...*(a-1)*a-1$ is divisible by prime $q>a$?[/quote]

hm? thats not possible, because if it would be divisible by a prime q>a, the other factor has to be smaller!

[/quote]

so you are saying that $\forall a\in \mathbb N$ we have $\frac{a!-2}{2}$ is a prime number. so you have invented a formula which gives only prime numbers. so why not publish it in a journal? :huh:

[quote="nayel"]find all $a,b,c\in \mathbb N$ such that $a!b!=a!+b!+c!$.[/quote]

Posted before.
Have a look here :
http://www.mathlinks.ro/Forum/viewtopic.php?t=116937

Pierre.

[quote="kathi;)"][quote="nayel"]
so you are saying that $\forall a\in \mathbb N$ we have $\frac{a!-2}{2}$ is a prime number. so you have invented a formula which gives only prime numbers. so why not publish it in a journal? :huh:[/quote]

(sorry that i didn't answer so long!)

well... yes in that moment, i found out that these numbers have to be prime... i hope that my proof is right ;)
why should i publish it?[/quote]

because if my memory doesn't fail me upto this day noone has ever come up with a formula that gives only prime numbers. :wink:

Yup, youre rite.. There's none at all... Except the Riemann's Hypothesis, if someone has proved it..  :oops:

okey can you explain [b]why[/b] they give only prime numbers and why can't they have a divisor >a?(or $p_{k}$)

This problem was British MO Round 1, 2002/5

yes, ill explain it.

first, to answer some pm-questions:

i "found" it myself, but "found" is not really the right word. i tried the proof for a=b and i had the equation
$a!(a!-2)=c!$, so i had to look how this can be. of course, it can be when $a!-2=c!=0$ (but i forgot that $0!=1$) and when $a!-2=(a+1)(a+2)...(c-1)c$ but with trial and error, i found out that this may only be one factor because it is a prime number. this i had to prove^^ so my proof follows here:
$a!-2=2(\frac{a!}{2}-1)=2(3*4*5*...*a-1)$
$3*4*5*...*a$ is an even number, divisble by all natural numbers $n\leq a$ (exception: $a=3, ~\frac{3!}{2}=3$ which is prime) . so when you have $(3*4*5*...*a-1)$, its an odd number, so it's not divisible by $2n$. it can't be divisible by any odd number $u$ which is smaller than $a$, because $(a!-1)\equiv u-1 (mod u)$. when it has no primefactor that is smaller than $a$, there can't be a primfactor $p>a$, because the other factor would then be .................. - hm here is my mistake: i started to look for primefactors of $a$, and no longer of $a!$
example: $\frac{15!-2}{2}=353 \cdot 1852229983$
so it's unfortunately
[b]not[/b] a formular that gives only primes  :(
;)

Kathrin


algebra
polynomial
geometry
3D geometry
complex numbers

Let $f(z)$ be a polynomial over complex numbers, If $f(z)$ is divided by $z^2 + z + 1$ , $z-1$ , remainders are respectively $2 \omega z + \omega$ , $5 - \omega$ ; where $\omega$ is nonreal cube root of unity. Find the remainder when $f(z)$ id divided by $z^3 - 1$

Put $f(z) = (z^3 - 1)g(z) + az^2 + bz + c$.  Then trying $z=w, z=1$, we get $aw^2 + bw + c = 2w^2 + w$ and $a+b+c = 5-w$.

Try a = 2.  Then we solve $bw + c = w$ and $b+c = 3 - w$.   So $b = (w-c)/w$ and $c = (2 -w)(1 - 1/w)$.


Exatly problem:
Find all polynomial with real coefficients $ P(x)$ such that $ P(P(x) \plus{} x) \equal{} P(x)P(x \plus{} 1)$ for all $ x$.

If $ P$ has degree $ n>1$ then so has $ P((x)\plus{}x$ thus equaling the degrees of the RHS and the LHS, we deduce that $ n^2 \equal{} n\plus{}1$, which is impossible for an integer.
Thus, $ P(x)\equal{}ax\plus{}b$ for some constants $ a,b$. Direct checking leads to $ P(x)\equal{}0$.

Pierre.

See this:

http://www.mathlinks.ro/viewtopic.php?t=179310


function
algebra
domain
real analysis
real analysis theorems

What is/are the critical points of the function

$ f: R\rightarrow R$

$ f(x)\equal{}2$

Are all points in the domain of f critical points.So what is so special about the critical points in this example.

I'd call all of these points "non-isolated local maxima." And also non-isolated local minima.

Yes, they're all critical points. Of course, we just need to exercise some common sense - we do know how constant functions act.

Non isolated local maxima/minima.

Why so.Is it because you normally locate  points of local maxima and points of local minima distinctly on graph whereas in this situation they appear to be clubbed together.

I think one may also call them degenerated points of local maxima/minima.


calculus
real analysis
real analysis solved

[quote="Sporally"]Our new subject: Diffentiate calculus.

1) I've made this one myself and would first of like somebody to confirm my result:

$f(x)=(x-1)*(x^2+x+1)$

$f'(x)=3x^2$

2) But this one is out of my reach of what i can do. I've tried doing something with it, but i can't say i feel like i'm doing it right:

$f(x)=(5x^2+2)*(4\sqrt{x}+1)$

$f'(x)=(5x^2+2)'*(4\sqrt{x}+1)+(4\sqrt{x}+1)*(5x^2+2)$

$f'(x)=10x*(4\sqrt{x}+1)+\frac{4}{2\sqrt{x}}*5x^2+2)$

So how do i (in case i'm doing it right until now) continue from here and end this diffentiation? And i would really appriciate you to show me the final result. I appriciate hints that can help me forward, but i need the result tomorrow in case the hints aren't enough to end the diffentiation myself. A step by step gothrough is my favorite - easy to understand if the steps aren't too great and you get the result at the same time.

Thx in advance.[/quote]
[hide]
well your first one is correct. The second one I kind of made up my own rules while doing the problem (AKA I can't do it  :roll: ) and got $f'(x)=10x(4\sqrt{x}+1)$[/hide]

This should be in the calculus section if you want to get fast and good responses.
The first one is correct, because you know that $f(x)=x^3-1\implies f'(x)=3x^2$
The second one can be done as follows.
$\\f(x)=(5x^2+2)(4\sqrt{x}+1)\\\\ =20x^2\sqrt{x}+5x^2+8\sqrt{x}+2\\\\ =20x^{2+\frac{1}{2}}+5x^2+8x^{\frac{1}{2}}+2\\\\ =20x^{\frac{5}{2}}+5x^2+8x^{\frac{1}{2}}+2\\\\ \implies f'(x)=\frac{5}{2}\cdot 20 x^{\frac{3}{2}}+10x+\frac{1}{2}\cdot 8x^{-\frac{1}{2}}\\\\ =50\sqrt{x^3}+10x+\frac{4}{\sqrt{x}}$

Masoud Zargar

Ok, so you want to make a major rewriting of $f(x)$. That was what i also wanted to do first of, but on the second hand thought my guess would be it was easier to diffentiate to begin with. Thx for your replies. This help should also settle the answer for many other question on diffentiate that might occur during this chapter. Your way of doing it had many elements.

[quote]This should be in the calculus section if you want to get fast and good responses. [/quote]
This should have started out in "Calculus Computations and Tutorials." Since it's solved now, I won't move it, but for future reference, that's where you should start things like this (and where moderators for other sections should move such things.)

OK, just wouldn't call this a collage topic, but then again, the american and danish school systems are equal. This is a high school topic in Denmark.

And a lot of the students who ask these questions are U.S. high school students (taking what we call "AP Calculus".) Still, Calculus questions belong in the calculus section.

OK, that way. Maybe someone could verify these results have a found. Personally i don't believe my text book is explaining every aspect of diffentiation. Quite a lot tasks, hope you don't mind:

1) $f(x)=\frac{1}{3}x^3 \implies f'(x)=3x^2$

2) $f(x)=\frac{1}{x^2} \implies f'(x)=-\frac{2x}{x^4}$

3) $f(x)=\frac{x^2-x-2}{x} \implies f'(x)=1+\frac{2}{x^2}$

4) $f(u)=\frac{12u^2+3x-8}{3u} \implies f'(u)=4+\frac{8}{3u^2}$

5) $f(t)=\frac{\sqrt{t}}{t-1} \implies f'(t)=-\frac\frac{1}{2\sqrt{x}}{t-1}^2$

6) $f(x)=\frac{(x+1)(1-x^2)}{3+x^3} \implies f'(x)=\frac{2x-1}{3x^2}$

7) $f(x)=x^4\sqrt{x} \implies f'(x)=4x^3\frac{1}{2\sqrt{x}}$

8) $f(x)=x^3+3x^5 \implies f'(x)=3x^2+15x^4$

9) $f(x)=\frac{3}{x} \implies f'(x)=-\frac{3}{x^2}$

10) $f(x)=x^5-5x^2+\frac{1}{x} \implies f'(x)=5x^4-10x-\frac{1}{x^2}$

11) $f(x)=\frac{x^5}{5}-\frac{1}{2}*x^-1 \implies f'(x)=x^4+\frac{0,5}{x^2}$

12) $f(x)=3\sqrt{x}-x^-1 \implies f'(x)=\frac{3}{2\sqrt{x}}+\frac{1}{x^2}$

Hope i didn't write something wrong :)

Plz reply if i have done something wrong (which i believe i have done at least one time) or if you think i could shorten it further.


AMC USA(J)MO USAMO calculus derivative limit blogs

Why a=b=0
If a=b => a.b=a.a=b.b
=>a.a-a.b=a.a-b.b
=>a(a-b)=(a-b)(a+b)
=>a=a-b =>a=b=0!!!!
Do you think there is any wrong thing :maybe:  :maybe:
 :D

Dividing by 0 is not acceptable in mathematics. But we didn't even know that a=b=0 until we derived it from that equation. This is sort of like "which came first: the chicken or the egg": it can't be answered.

If a=b, then you cannot divide by (a-b). It is dividing by 0.

[quote="dafjbomb"]lol if this wasnt on a websites filled with mathletes, it would baffle people for atleast a few hours.

if this was at my school no one would find the error in the solution. (other than me of course!)[/quote]

Hm, yeah.  On the first day, my Alg II teacher tried to "stump" us with this.  I had it in like 15 seconds, but I was waiting for the period to end before I told of it.

you can also say :

$ a^2\minus{}a^2\equal{}a^2\minus{}a^2$

$ (a\plus{}a)(a\minus{}a)\equal{}a(a\minus{}a)$

cancel out the a-a

$ (a\plus{}a)\equal{}a$

$ 2a\equal{}a$

$ 1\equal{}2$

but of course, thats also dividing by zero...so yeah.

...which probably didn't work because the derivative didn't exist? :P  :wink:

[quote="AIME15"]Meh.  At math camp, james4l was talking with another usamo-er, and there was some calc proof that 1=2...something that started with x^2=x+x+x+x...x times, then with taking the derivative of something...[/quote]

I think that is;
let $ x^2 \equal{} x \plus{} x \plus{} x \plus{} x...$ ($ x$ times)

Taking the derivative we get: $ 1 \plus{} 1 \plus{} 1 \plus{} 1...$($ x$ times) $ \equal{} x$

However the derivative of $ x^2$ is $ 2x$

Where is the fallacy?

[hide="solution"]
The fallacy is that the number of $ x$'s is always changing and therefore the derivative falls apart.

Consider derivatives by first pricipals
The derivative of $ x^2 \equal{} \lim_{x \to \infty} \dfrac{(x\plus{}h)^2\minus{}x^2}{h}\equal{}2x$

However by writing $ x^2$ as the sum of $ x$'s we are effectivly evaluating

$ \lim_{x \to \infty} \dfrac{x(x\plus{}h)\minus{}x^2}{h} \equal{} x$

[/hide]

i love how a simple trick turned into derivatives...

 :rotfl:

Here is another bogus proof that two numbers that aren't equal are:
\begin{eqnarray}
1=1\\
1=\sqrt{1}\\
1=\sqrt{(-1)(-1)}\\
1=\sqrt{-1}^2\\
1=-1\\
\end{eqnarray}
No one in my math class with the exception of me and and another AoPSer was able to find the flaw (partly because none of the other students had more than a superficial knowledge of $ i$)

That's because the square root of anything is positive.

[quote="Discrete_Math"]Here is another bogus proof that two numbers that aren't equal are:
\begin{align*} 1 & = 1 \\
1 & = \sqrt {1} \\
1 & = \sqrt {( - 1)( - 1)} \\
1 & = \sqrt { - 1}^2 \\
1 & = - 1 \end{align*}
No one in my math class with the exception of me and and another AoPSer was able to find the flaw (partly because none of the other students had more than a superficial knowledge of $ i$)[/quote]

FIxed ur tex.

I still like the one found in Calvin and Hobbes better:

Calvin: Help me with this homework ok? What's 6+3?
Hobbes: 6+3 eh? Well this one is a bit tricky. First we call the answer y as in "Y do we care?" Now y may be a sqaure number, so we'll draw a square and make this side 6 and that side 3. Then we'll measure the diagonal.
Calvin: I don't remember the teacher explaining it like this.
Hobbes: She doesn't know higher math. When you deal with high numbers, you need higher math.
Calvin: But this diagonal is just a little under two.
Hobbes: Ok, here, I'll draw a bigger square.
:rotfl:

When I Googled this, it showed up on 7h3.D3m0n.117's blog, too.  :o

I live in Oklahoma, and I went to a math camp at a math and science high school and they taught us this:

A = B

A^2 = AB

A^2-B^2 = AB-B^2

(A-B) (A+B) = B(A-B)

A+B = B

?

 :D

[quote="AIME15"]In these "proofs", the error is usually a division by $ 0$ somewhere.[/quote]

Or an illegal square root.

In this case its division by A-B, which, as A=B, is 0.

I think I prefer paradoxes over bogus proofs. Although bogus proofs are fun to read, paradoxes are fun to think about. Okay, I'm bored.  :rotfl:

But the correct answer is
If a=b =>a-b=0
so (a+b)(a-b)=a(a-b)=0
and we can't say a+b=a
I think it very easy


Mrs Li has 2m of cake frill. She uses it for trimming a square cake of side 20cm and the circumference of a round cake of diameter 28cm. How muvch of the cake trimming is left over? ( take pi=22/7)


I was so close to making the AIME!!!  :read:  :wallbash_red:  :wallbash:  :(  :mad:  I took the AMC 12B I guess I shouldnt be mad cause I'm only in 5th/6th grade, but if I had gotten 2 more probs right I would have gone to the AIME.  I am so close!!!  What is your best advice, if you are as close as I was to making the AIME, how do you get TO making the AIME?  Any specific classes/books?  Once your sure you'll make the AIME hwo do you prepare for it?

The AMC 8 (i got a 21 or 22, i'm not sure, my coach never toald me, i just know i was second in my school  :winner_second: ) and AMC12 (got a 89 cause I guessed trying to get into AIME, since i wasn't taking it with my school, otherwise would have got a 94) are way easier than the AMC 10.

WHy isn't the poll open yet?????

its open now, try to keep discussion in there

for classes/books:

AoPS Vol 1 is top of the line for AMC level
of course, use the Contests/Resources section and practice lots and lots of old tests
doing Mathcounts should help on the AMC too, and Mathcounts is much more targeted towards people your age

you're barely in middle school, don't get so stressed out, heh

my coach said i couldn't do math counts cause i was too young  :(  :mad: .  I don't like the buzzer thing any way, but the MAndelbrot rules!!!

If you didn't know my grade, would you recomend All of Voume 1 and 2, or what, and what books other than that, what AoPS courses,the reason i am asking is that I though i would be forever away from gettin to AIME now that i am SO close, i want to get to AIME.  :mad:  :lol:  :showoff:  :weightlift:

You can do mathcounts, you just can't do it for more than 3 years. Just think you great you'll be next year when you can be!

And really, don't complain about getting a 86 or such in the 5th grade on the AMC12!!!. Most of us didn't even know about the AMC in 5th grade, much less almost made the AIME... For books, study AoPS1. You'll be really good after that. When you can answer most of the problems in AoPS1 quickly, you will be ready to do Mathcounts (and do well!). AoPS2 comes after AoPS1; it's good for advanced AMC stuff, Mandelbrot, etc. (You'll be ready for the Comprehensive level of high school competitions after you've done a good deal of AoPS2.)

Have fun! You'll be beating most of us in 3 or 4 years.

(Are you perchance related to Nathan Benjamin? He's another one of the really exceptional AMC prodigies.)

Billy

Hey, you beat several junior and seniors from my school. If you are barely missing it now, you'll be flying into MOP with ease when you're in HS


Could you give me some explains of Parzen windows method to estimate densities or some links that can help me understand more about it.

 :oops:

After finding this site, i think i have found an easy explain.

[url]http://www.personal.rdg.ac.uk/~sis01xh/teaching/CY2D2/Pattern%20Recognition%20Lecture%20Notes.htm[/url]

Slide : Parzen windows

Hope it helps.


analytic geometry
geometry
perpendicular bisector

I am looking for someone to solve the questions below using steps as you go through the lesson. I would like to learn how to solve similar questions on my own. Ready?

1) How many points are equidistant from point R(3,1)
and S(3, 6) and also 4 units x-axis?

2) How many points are equidistant from point R(3,4)
and S(3, 8) and also 6 units y-axis?

Consider the locus of the points, which are equidistant from the two points you wrote, R and S. These line will intersect somewhere the x-axis and the y-axis, so there will be two points for each exercise you proposed, which are also equidistant (4 or 6 units) from the x-axis or the y-axis, always depending on what's the exercise.
Then, you just have to find the equation of the locus and
1) consider the two points with $y=\pm 4$;
2) consider the two points with $x=\pm 6$.
In both cases the points you're looking for are two, maybe coincident (if their distance from an axis must be 0). If the locus is parallel to the axis you're referring to, then you have infinite points with fixed distance from it... well, these are extreme cases...  :wink:

This holds if I understood well this thing of the distance from the axis, I can't guarantee! Anyway, I hope I've been clear!  :)

I understood your explanation but what did you mean by extreme cases?

Also, do I need to graph the given points for R and S to see what is going on?

Also, I noticed that both x coordinates are the same for BOTH questions.  See it?

What is each coordinate is DIFFERENT?

What should I do if that is the case?

What if the each coordinate is a fraction, decimal, pi or e?

How do I handle such questions?

1) Extreme cases, because among all the possible straight lines, the two axes are exceptions! Also, if you have to find points with distance 0 from an axis, that's an exception, among all the possible values of that distance! Do you understand? It's really banal, perhaps I'm not able to write it in a good way!...

2) If you want to draw a picture, then it's better to include the points R and S, even if you can solve the problem without drawing any graph... it's easy to see that all of these problems are really the same, what's changing are just the numbers involved!

3) I see... that doesn't change the matter!

4), 5), ... I invite you to solve the problem without any number, just with letters... in a few words, I think it would be a good thing trying to generalize it...
Call the point R $(x_{1},y_{1})$ and the point S $(x_{2},y_{2})$ and the distance from the axis you want to consider $d$. Then solve and see what comes out: you'll find out that even with the worst values of the coordinates the question is simple... If you have other problems, please post your mathematical reasoning, so that it'll be simpler to answer questions! Bye!  :)

P.S.: By the way, the answer to your original question is 2 points (at most coincident), as you should already have understood, if the perpendicular bisector of segment AB isn't parallel to the axis you're considering!


Find all positive integrers $k$ such that the result of multiplying the digits of $k$ is $\frac{25}{8}k-211$

Ummm.....  :roll:
[hide]So k must be divisible by eight for 25/8 to be an integer. Now, 25k/8 must be odd, so k isn't divisible by 16.  Now, let k=8n, so n is odd. Now, 25k/8=25n, so the last 2 digits must be 25 or 75.  Subtract 11, and so the last 2 digits of k must be 14 or 64, so now, since k is a multiple of 8, the last 3 digits must be divisible by 8.
Case 1, the last 2 digits are 64.  by contradiction, so assume true
now, since the product of the digits is divisible by 8, but not 16, the hundred's digit must be odd, but 164, 364, 564, 764, and 964 are all not divisible by 8. So =><=
Case 2, the last 2 digits are 14,
this isn't even divisible by 4, so it isn;t divisible by 8 =><=
[/hide]

Bad proof, but I didn't want to write it up.   :D

EDIT: oops nvm, something [i]seems[/i] wrong. :(  :oops:

I don't remember the answer, but I think that there are solutions.

[hide]$72$ and $88$ are two solutions.  I think they're the only two, but I'm not sure.

$8|k$, but $\frac{25}{8}k-211$ must be odd, since the units digit of $k$ will be even and so the product of all its digits will also be even.  Therefore, $k=16n+8$ for some nonnegative integer $n$, and $\frac{25}{8}k-211=50n-186$, so the product of $k$'s digits must be congruent to $14\mod50$.[/hide]

So, Jose, what's the source of the question?  Do you know what the answer is (and just as importantly, how do find it)?  Thanks!


The sequences $(a_n)$ , $(b_n)$ are defined as follows:
$a_n=1+\frac{n(n+1)}{1+n^2}+........+\frac{n^n(n^n+1)}{1+n^{2n}}$
$b_n=(\frac{a_n}{1+n})^{\frac{1}{n(1+n)}}$ for n>0.
find $\lim_{n\rightarrow+\infty} b_n$

If i'm not wrong, using Stolz,
$\lim_{n\to\infty} \frac{a_n}{1+n} = \lim_{n\to\infty} (a_n-a_{n-1})=1$,
and then
$\lim_{n\to\infty} b_n = 1^0 = 1$.


I know now that it is not legal to distribute old national tests for legal reasons, but I think that some student should explain to MATHCOUNTS that we should be allowed to distribute them as long as we don't get money or we cannot purchase them. I have a few reasons.

There is no where to obtain the tests at all except your coach and that is only if they were at the competition so that gives an unfair advantage to students that have coaches that werent at the competition.

There is no reason really not to. MATHCOUNTS doesn't lose any money because they are not sold and they are not being asked to distribute them.

I think that with this information, MATHCOUNTS might consider no longer considering sharing the tests a copyright infringement.

I would write the letter but I am a horrible persuasive writer :(

not the ones before 1999. We wouldn't be able to distribute those, as they are copyrighted.

I think you have to purchase most of them.  I don't know about earlier rounds, but my dad got them somehow.  I'll ask my dad.

The ones before 1999 are not sold anywhere and I don't know if MATHCOUNTS even prints them. if they don't print them, then its not a copyright violation.

Your dad was probably at previous competitions or bought the pre-1999 ones when they were available.

My mathcounts teacher has the tests for 1995-1998.  I'll ask her how she got them.

:bomb: This topic is about if someone would write a letter to MATHCOUNTS. Math92, your teacher and every other person who has them legally either bought them online when they were available or was at the competiton (or got them illegally :D ). Obviously, they are no longer sold and I my coach wasn't the competitons, so to get them, MATHCOUNTS would have to make it legal to distribute them.

Maybe Mathcounts just doesn't want rich students to get too much of an advantage. :D

This really belongs in the Round Table or something. I think.

Why would it belong in the round table :huh:  :huuh:

Seriously, would someone write a letter?


calculus
integration
logarithms
calculus computations

The answer is $\ln 2$, this question has been discussed here: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=32894[/url].

nice proof in this topic. Mine was ugly ..

Could you post yours anyway?  I like to see different solutions to a problem.

Thanks.


integration
calculus
real analysis
real analysis unsolved

Assume that $f(x)$ is continuous on$[0;1]$ and$\int_{0}^{1}xf(x)dx=0$.Show that there is $p$ s.t: $0<p<1$ and$\int_{0}^{p}f(x)dx=0$. :)

$f(x)=1-\frac{3}{2}x$ is a counterexample to what you wrote. What you, probably, meant was $\int_{p}^{1}f(x)\,dx=0$. Have you ever heard about integration by parts? ;)


Μια ανισότητα για τους λάτρεις του είδους ...

Αν x,y,z > 0,  τότε δείξτε ότι $ 2\sqrt{x\plus{}y\plus{}z}\plus{}\sqrt{\frac{1}{x^2}\plus{}\frac{1}{y^2}\plus{}\frac{1}{z^2}}\ge 3\sqrt{3}$

 :)

Θέτω:

$ x \equal{} \frac {1}{a} , y \equal{} \frac {1}{b}, z \equal{} \frac {1}{c}$

και έχουμε:

$ 2\sqrt {\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}} \plus{} \sqrt {a^2 \plus{} b^2 \plus{} c^2}\geq 2\sqrt {\frac {ab \plus{} bc \plus{} ca}{abc}} \plus{} \sqrt {ab \plus{} bc \plus{} ca} \equal{} \sqrt {\frac {ab \plus{} bc \plus{} ca}{abc}} \plus{} \sqrt {\frac {ab \plus{} bc \plus{} ca}{abc}} \plus{} \sqrt {ab \plus{} bc \plus{} ca}\geq3\sqrt [3]{\sqrt {\frac {ab \plus{} bc \plus{} ca}{abc}}\cdot\sqrt {\frac {ab \plus{} bc \plus{} ca}{abc}}\cdot\sqrt {ab \plus{} bc \plus{} ca}} \equal{} 3\sqrt [6]{\frac {(ab \plus{} bc \plus{} ca)^3}{(abc)^2}}\geq 3\sqrt [6]{\frac {[3\sqrt [3]{(abc)^2}]^3}{(abc)^2}} \equal{} 3\sqrt {3}$

Sorry για το κακό layout και για τυχόν λαθάκια στις πράξεις. Είναι περασμένες 3 ..    :oops:

Ωραίος ρε Στέλιο ...  :wink:

Κανένας άλλος τρόπος υπ΄όψη ?

Ως γνήσιος λάτρης θα βάλλω τη λύση μου αν και το σκεπτικό της είναι ίδιο με του Στέλιου πάνω κάτω .

Έχουμε
$ LHS\geq 2\sqrt {a \plus{} b \plus{} c} \plus{}\frac{1}{ \sqrt {3}}(\frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z})\geq 3\sqrt [3]{\frac{1}{\sqrt {3}}(x \plus{} y \plus{} z)(\frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z})}\geq 3\sqrt {3}$

afou $ (a \plus{} b \plus{} c)(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})\geq 9$


A circle with radius 4 cm is tangent to three sides of a rectangle, as shown. The area of the rectangle is twice the area of the circle. What is the length of the longer side of the rectangle, in centimeters? Express your answer in terms of $ \pi$.

[asy]import graph;
draw((0,0)--(30,0)--(30,20)--(0,20)--cycle);
draw(Circle((10,10),10));[/asy]

Rectangle: 2ra
Circle: $ r^2\pi$
$ a\equal{}r\pi$
$ a\equal{}4\pi$

I feel Bugi jumped a few steps. Remember, people asked these questions because they had trouble!

Let's write out what we know, calling the radius of the circle $ r$, and the sides of the rectangle $ w$ and $ h$ where $ w$ is the longer side.

We know:

Area of the circle: $ A_c\equal{}\pi r^2$

Area of the rectangle: $ A_r\equal{}2A_c\equal{}2\pi r^2$ [b]and it can also be written as[/b] $ A_r\equal{}hw$

Thus by substitution $ 2 \pi r^2\equal{}hw$

This doesn't help us. What else do we know? We know $ r\equal{}4$! Let's use that!

$ 2 \pi r^2\equal{}2 \pi 4^2\equal{}32\pi\equal{}hw$

Ok. Now what? Let's look at the diagram. We quickly realize that $ h$ is the same lenght as the diameter of the circle! Yay! Now let's use that $ h\equal{}2r\equal{}8$

$ 32\pi\equal{}hw$

Substituting in our new value of h,

$ 32\pi\equal{}8w$ and dividing by 8 gets our desired answer of

$ w\equal{}\boxed{4\pi}$ as Bugi said.