geometry
geometric transformation
reflection
Euler
circumcircle
INMO
perpendicular bisector

Let $ R$ the center of the nine points circle. Let $ Q_1$, $ P_1$ and $ R_1$ the projections of $ Q$, $ P$ and $ R$ on the altitude $ AK$. Then for do the problem we have to prove that $ Q_1$ is the midpoint of $ AK$. But is easy to prove that $ Q_1P_1\equal{}2P_1R_1$.

So:

$ Q_1K\equal{}Q_1P_1\plus{}P_1K\equal{}2P_1R_1\plus{}P_1K\equal{}2(R_1K\minus{}P_1K)\plus{}P_1K\equal{}2R_1K\minus{}P_1K\equal{}\frac{1}{2}AH\plus{}HK\minus{}P_1K\equal{}\frac{1}{2}AK$ as claimed.

[hide="Soution 1 (own)"]
Due to a plethora of points I'll refer to...I've just indicated them in the diagram...(Note that I've drawn $ PS \perp HK$ and then extended it to $ R$ on $ BH$.)

Now What I wrote was that I first made a lemma stating that If $ R$ is the midpoint of $ BH$, and $ E$ is the midpoint of $ AC$, then $ RE$ is a diameter of the nine - point circle...and then proved it in the same style as PCM...

Now, coming back to the question....(I'm feeling too lazy so I'm just saying in brief....)
Now, since $ P$ is the circumcentre of $ \triangle OKH$,
Therefore, $ S$ is the midpoint of $ HK$, combining this with $ RS \parallel{} BK$ gives $ \boxed{R \ is \ the \ Midpoint \ Of \ BH}$.
Now, since $ P$ is the circumcentre of $ \triangle OKH$, we also have $ N$ is the midpoint of $ OH \Rightarrow N$ is the nine - point centre.
So in $ \triangle s RNP, ENQ$ we have $ PN \equal{} NQ$(By reflection), $ EN \equal{} NR$(By lemma), $ \angle RNP \equal{} \angle QNE$(since $ \overline{RNE}$ is a straight line by lemma)
Therefore, $ \triangle RNP \equiv \triangle ENQ$ (By SAS)
and therefore, $ EQ \parallel{} RP \parallel{} BC \parallel{} DE$(First two we have proved by using the congruency, next two already proved, and next two follow from midpoint theorem)
So, $ EQ$ and $ DE$ coincide and thus the result follows..[/hide]

[hide=" Solution 2 (Due to Akashnil)"]
Point-reflect (half-turn) everything about midpoint $ N$ of $ OH$, the nine-point center. points lying on the nine-point circle will be mapped to diametrically opposite points. Hence the mid points of sides $ AB,AC$ maps respectively to midpoints of $ CH,BH$. the line joining them is the perpendicular bisector of $ HK$ as well, and that contains $ P$, the image of $ Q$ under the point-reflection. :)[/hide]

i have a very easy solution using pure geometry

 Let ABC be an acute-angled triangle with altitude AK. Let H be its ortho-centre and O be its circum-centre. Suppose KOH is an acute-angled triangle and P its circum-centre. Let Q be the reflection of P in the line HO.
it is very easy to prove that
 quadrilateral HPOQ is a rhombus.

Thus now consider the \triangle HKO
in this triangle
$ \angle HPO \equal{} 2\angle HKO$or $ 1/2\angle HPO \equal{} \angle HKO \equal{} \angle PQO    \minus{}\minus{}\minus{}\minus{}\minus{}\minus{}\minus{}\minus{}\minus{}\minus{}\minus{}\minus{}\minus{}\minus{}\minus{}\minus{}\minus{}\minus{}(1)$
since P is the circumcenter and if we draw the circle we will get that \angle HPO is subtended at the center and \angle HKO is subtended at any other point by the same arc .

now we know that AK\perp BC since it is the altitude and $ UN\perp BC$ thus $ UN \parallel AK$.
hence$ \angle HKO \equal{} \angle KOT$

but $ 1/2\angle HKO \equal{} \angle HPO$
hence $ 1/2\angle HPO \equal{} \angle KON$
since HPOQ is a rhombus ,
$ \angle HQP \equal{} \angle PQO$

we get $ \angle PQO \equal{} \angle KON$  OR  $ \angle MQO \equal{} \angle KOT$
consider$ \triangle KOT$and$ \triangle OQM$
$ \angle MQO \equal{} \angle KON$
and $ \angle QMO \equal{} \angle OTK$
thus $ \triangle KOT$  and $ \triangle OQM$ are similar so we get
$ \angle OKT\equal{} \angle QOM$

now since UT is a straight  line
 $ \angle QOX \plus{} \angle QOM \plus{} \angle HOK  \plus{}  \angle KOT\equal{}180$
  $ \angle QOX \plus{} \angle OKT \plus{} \angle HOK  \plus{}  \angle KOT\equal{}180$  since $ \angle OKT\equal{} \angle QOM$
   $ \angle QOX \plus{}  \angle HOK  \plus{}90\equal{}180$
  $ \angle QOX \plus{}  \angle HOK\equal{}90$
but $ \angle QOX\equal{}90\minus{}\angle OQX$
  thus $ \angle OQX\equal{} \angle HOK$  $ \minus{}\minus{}\minus{}\minus{}\minus{}\minus{}\minus{}\minus{}\minus{}\minus{}\minus{}\minus{}\minus{}\minus{}(2)$
 now consider $ \triangle HLN$and $ \triangle NMQ$
  $ \angle HLN\equal{} \angle NMQ$
  and $ \angle HNL\equal{} \angle NQM$
 thus $ \triangle HLN$ and $ \triangle NMQ$ are similar.
 hence $ \angle LHN\equal{}\angle NQM$
 but $ \angle LHN and \angle KOH$ are same.
hence we get $ \angle KOH\equal{} \angle NQM\equal{}\angle RQP$
we know that in a triangle sum of all angles is 180.
Thus in $ \triangle KOH$,

$ \angle HKO \plus{} \angle KOH \plus{} \angle KHO\equal{}180.$
or    $ \angle PQO \plus{} \angle RQP \plus{} \angle OQX \equal{}180$ from (1) and (2),
this implies that RQX is a straight line or point Q lies  on RS.

[quote="keyree10"]Let $ ABC$ be an acute-angled triangle with altitude $ AK$. Let $ H$ be its ortho-centre and $ O$ be its circum-centre. Suppose $ KOH$ is an acute-angled triangle and $ P$ its circum-centre. Let $ Q$ be the reflection of $ P$ in the line $ HO$. Show that $ Q$ lies on the line joining the mid-points of $ AB$ and $ AC$.[/quote]
Let $ a,b,c,h,p,q,z_O,k$ be the complex numbers corresponding to the points $ A,B,C,H,P,Q,O,K$ in that order.

Wlog assume that $ z_O \equal{} 0, |a| \equal{} |b| \equal{} |c| \equal{} 1$. Then $ h \equal{} 3(g \minus{} z_O) \plus{} z_O \equal{} a \plus{} b \plus{} c$. $ k \equal{} \frac {1}{2} \left ( a \plus{} b \plus{} c \minus{} \overline{a}bc \right )$ is easy to verify.

A point $ Z(z)$ lies on the line joining the midpoints of $ AB$ and $ AC$ iff $ \frac {z \minus{} \frac {a \plus{} b}{2}}{\overline{z \minus{} \frac {a \plus{} b}{2}}} \equal{} \frac {\frac {a \plus{} c}{2} \minus{} \frac {a \plus{} b}{2}}{\overline{\frac {a \plus{} c}{2} \minus{} \frac {a \plus{} b}{2}}} \iff$
$ 2z \plus{} 2bc\overline{z} \equal{} a \plus{} b \plus{} c \plus{} \overline{a}bc$. $ bc \equal{} \frac {h \minus{} k}{\overline{h \minus{} k}}$ and $ a \plus{} b \plus{} c \plus{} \overline{a}bc \equal{} 2h \minus{} 2k$, so:
$ \iff z \plus{} \frac {h \minus{} k}{\overline{h \minus{} k}}\overline{z} \equal{} h \minus{} k$

$ P$ is the orthocenter of $ \triangle OKH$ so:
$ p \equal{} \frac {hk(\overline{h \minus{} k})}{\overline{h}k \minus{} h\overline{k}}$

$ Q$ is obtained by the reflection of $ P$ in $ OH$, so:
$ q \equal{} \overline{p} \frac {h}{\overline{h}} \equal{} \frac { \minus{} h\overline{k}(h \minus{} k)}{\overline{h}k \minus{} h\overline{k}}$

Inserting it is obvious that:
$ q \plus{} \frac {h \minus{} k}{\overline{h \minus{} k}}\overline{q} \equal{} (h \minus{} k) \frac {\overline{h}k \minus{} h\overline{k}}{\overline{h}k \minus{} h\overline{k}} \equal{} h \minus{} k$, and therefore $ Q$ lies on the line joining the midpoints of $ AB$ and $ AC$, and we are done :)

[quote="keyree10"]Let $ ABC$ be an acute-angled triangle with altitude $ AK$. Let $ H$ be its ortho-centre and $ O$ be its circum-centre. Suppose $ KOH$ is an acute-angled triangle and $ P$ its circum-centre. Let $ Q$ be the reflection of $ P$ in the line $ HO$. Show that $ Q$ lies on the line joining the mid-points of $ AB$ and $ AC$.[/quote]
Nice Problem!
Let the midpoints of $ AB,CD$ be $ Q,R$ respectively, and let the reflection of $ P$ wrt $ OH$ be $ P'$.
We consider a half turn wrt $ T$, the midpoint of $ OH$ (and the center of the nine point circle of $ \triangle ABC$). This maps $ P' \to P$, $ QR \to Q'R'$.
It is enough to prove that $ P$ lies on $ Q'R'$.
We know that $ Q'R' \parallel CB$. So $ Q'R'$ intersect $ HK$ at the midpoint, and also $ HK \perp Q'R'$. So  $ P$ lies on $ Q'R'$.

[quote="groshanlal"]By the way guys, how many questions do you think one needs to do or the minimum marks required to get through the INMO?[/quote]
Though I have not ever took part in InMO (I am from Bangladesh), I believe that it depends on many facts, such as: the quality of the contestants, difficulty level of the problems etc. In general, in Olympiads the contestants who solve half of the problem gets some sort of awards. I don't know if it true for InMO.  :P

My goodness - Dr Rijul has posted his solution already. :lol:
[hide="My Solution:"]
[img]http://totalgadha.com/rmo/wp-content/uploads/2010/01/INMO-2010-5s2.PNG[/img][/hide]

we have not used that triangle is acute in rijul's solution

Let E be midpoint of AC and D be midpoint of BH. Note that BH = 2ON so DH = ON and we also have DH || ON because they're both perpendicular to AC so DHEO is parallelogram. Let S be midpoint of OH we know S is midpoint of ED and OH and PQ so DQEP is parallelogram so EQ || PD || BC so Q lies on the line made of midpoints of AC and AB.
we're Done.


For how many positive integers $ n$ between 1 and 2007 is $ 6\cdot n!(n\minus{}3)!$ a factor of $ 2n!$?

[hide="hint"]Consider $ \frac{2n!}{6\cdot n!(n\minus{}3)!}$.[/hide]
[hide="hint 2"]Multiply top and bottom by $ n!$. Note that $ 6 \equal{} 3!$.[/hide]
[hide="hint 3"]Write as the product of binomial coefficients.[/hide]

[quote="nhansen"]Nice problem!  But I think you've mean "factor", not multiple?[/quote]

oh yeah, i'll edit that


geometric series
real analysis
real analysis unsolved

Can someone help me with this problem?
Let r>0 be a real number. Find an example of a power series $ \sum c_nx^n$ with radius convergence r such that:
1) The power series converges for x= r but diverges for x=-r . What about converges for x=-r and diverges for x=r?
2) The power series converges for neither x=r nor x=-r

[quote="t0rajir0u"]1)  $ c_n \equal{} \frac {( \minus{} 1)^n}{r^n n}$ and $ c_n \equal{} \frac {1}{r^n n}$.

2)  $ c_n \equal{} \frac {1}{r^n}$.[/quote]

Thank you very much for your help. Can you help me how to construct those power series? I want to be able to do it too.

One way to interpret the first question is that $ c_n r^n$ is a series whose sum diverges but whose alternating sum converges.  This is true of any monotonically decreasing sequence that diverges but passes the $ n^{th}$-term test, the most well-known example of which is the harmonic series.

My answer to the second question is motivated by the fact that a geometric series sum $ 1 \plus{} x \plus{} x^2 \plus{} ... \equal{} \frac{1}{1 \minus{} x}$ is valid precisely for $ |x| < 1$.

1)  $ c_n \equal{} \frac {( \minus{} 1)^n}{r^n n}$ and $ c_n \equal{} \frac {1}{r^n n}$.

2)  $ c_n \equal{} \frac {1}{r^n}$

I have to multiply these with $ x^n$ right because otherwise, they are not power series. But then I think you meant $ r^n$ is the same as $ x^n$, just different notation from my question. If that is the case, then those series are not power series?

I mean exactly what is written.  $ r$ is a constant (the given radius of convergence) and the relevant series is $ \sum c_n x^n$.


inequalities
geometry
inequalities proposed

$\triangle ABC,\ \triangle XYZ,\ S=\sigma [ABC],\ T=\sigma [XYZ]\Longrightarrow$ \[ \overline {\underline {\left|\ 2\sum a^2\sum x^2\ge 3\sum a^2x^2+48ST\ \right| }}  \] [u]Particular cases:[/u]

$\blacksquare\ 1.\ (x=y=z=1)\Longrightarrow a^2+b^2+c^2\ge 4\sqrt 3\cdot S$ (the [b]Weitzenback's inequality[/b]).

$\blacksquare\ 2.\ (x=\sqrt a,\ y=\sqrt b,\ z=\sqrt c)\Longrightarrow 2\sum a^2\sum a\ge 3\sum a^3+24S\sqrt{r(4R+r)}$

$\blacksquare\ 3.\ \ldots \ldots\ a.s.o$

(you can try and other particular cases, for example, $x=\sqrt{a(s-a)},\ldots$).

[u]Remark.[/u] The inequality $(2)$ is equivalently with the framed relation

$\left| \overline {\underline {\ ab+bc+ca\ge 2r(R-2r)+12r\sqrt{r(4R+r)}\ }}\right| \ge \max \left\{ 12r\sqrt{r(4R+r)},2r(R+16r)\right \}.$

[u]Atention.[/u] $\sigma [XYZ]$ - the area of the triangle $XYZ$.

Some noninteresing inequalities ? Really is it ?


inequalities
combinatorics unsolved
combinatorics

In a school there are $ 2008$ students. Students are members of certain committees. A committee has at most $ 1004$ members and every two students join a common committee.
(i) Determine the smallest possible number of committees in the school.
(ii) If it is further required that the union of any two committees consists of at most $ 1800$ students, will your answer in (i) still hold?

Here is my solution for the first part of the problem: we shall prove that the number is 6. Let's consider the situation with 5 committees. If every student is a mamber of at least 3 committees the we will have the inequality 3.2008<5.1004 since in each committeee there are al most 1004 members. This gives 6<5 - contradiction. So the number of committees is at least 6. Here is an example for 6 committees: devide the 2008 students in 4 groups of 502 students each - lets say they are A, B, C, D  with |A|=|B|=|C|=|D|=502. For the 6 committees take (A,B); (A,C); (A,D); (B,C); (B,D); (C,D). It's easy to check that the conditions are OK.

And could anybody solve the ii) part  :?:

Hey could anybosy post some ideas on ii???? Its exceedingly hard!

For part (ii), $6$ committees is enough. Split the students into $8$ equal groups $A_1,\dots,A_8$. Then form the six committees: $$\begin{cases}A_1,A_2,A_3,A_4\\A_1,A_5,A_6,A_7\\A_1,A_2,A_7,A_8\\A_2,A_5,A_6,A_8\\A_3,A_4,A_5,A_8\\A_3,A_4,A_6,A_7\end{cases}$$ The union of two committees has at most $7\times251=1757<1800$ members, so we're done.


1. Alice attends a college in which classes meet only once per week. There are 30 non overlapping classes to choose from, and 6 classes per day for Mon-Fri. Alice decides to register at random for 7 classes. What is the probability she will have class every day of the week? (M-F)

2. A fair coin is flipped 3 times, and the result is written on a slip of  paper and thrown into a hat.
(a) Find the probabilility all 3 tosses were H, given at least two of them were H.
(b) Two of the slips of paper are drawn at random and both say H. Given this, find the probabililty all 3 heads are H.
(c) Explain the difference or equivalence of (a) and (b)

3. You are going to play 2 games of chess against an opponent who's equally like to be a master, an intermediate, or a beginner. Your P(winning) any single game is 90%, 50%, 30%, against beg, int, master, respectively.
(a) What is P(winning first game)
(b) You won game 1! Now, what's P(winning game 2)? Assume outcomes are independent
(c)You won game 2! What are the conditional probabilities that that ur opponent is a beg? int? master?

4. Two cards are randomly drawn w/o replacemtn from a standard deck. Like B = both cards are aces. Let A_s = ace of spades is chose. Let A = at least one card is an ace. Find
(a) P(B|A_s)
(b) P(B|A)

5. Suppose we have 10 coins such that if the i-th coin is flipped, heads will appear with probability i/10. When one of the coins is randomly selected and flipped, it shows heads. What is the conditional probability that it was the fifth coin?

6. A stock market investor owns shares in a stock whose present value is 25. She has decided taht she must sell her stock if it either goes down to 10 or up to 40. If each change of price is either up 1 point with Probability .55 or down 1 point with probability .45, and the successive changes are independet, what is the probability the ivnestor retires a winner? (i.e. sells at 40)

7. Let A, B, E, F be events, with A,B independent and P(A^B)>0.
(a) Are A and B necessarily conditionally independent given E?
(b)Suppose that P(A|E)>P(A), and P(A|F)>P(A), i.e. E and F separately provide evidence in favor of A. Is it necessarily true that P(A|E^F)>P(A)?

btw, i fizsle, i believe ur #1 is incorrect

So whats the answer for #1 ???
I think it could be (6^5 x 25C2)/30C7  :maybe:

everything else ok??? here is 4 - 6... i find 7 a bit complex..will try
[hide]
4) a) 3/51 b) 1/5

5) 1/11

6)(11/20)^15 - Not sure about this one too... :D
[/hide]

Fizsle, your first answer for number one was pretty close - you may have missed one case and that threw your answer off a little.

yea, and fizsle, your #2 is incorrect as well. Try the ocnditional probs again

here is the answer for 1... i think i got it this time
2 cases are that the additional 2 clases be on the same day or on different days

[6 x (6^5) x 5c2]/30c7

Which one in #2 is wrong??? a or b??

hmm still not quite.

Case 1: 2 extra classe are on the same day. Then, we have 5 ways to order the # of classes, e.g. (M,T,W,T,F) = (1,1,1,1,3), (1,1,1,3,1,), etc. For each of those orderings, we have 6^4 ways to choose classes for the "1" days and 6C3 ways to choose classes for the "3" day. So case 1 has 5*6^4*(6C3)

Case 2: (M,T,W,T,F) = (1,1,1,2,2), etc. There are 10 ways to permutate that. Each way has 6^3 ways to choose classes for the "1" days and 6C2 ways to choose classes for each of the "2" days.
So case 2 yields 10*6^3*(6C2)^2

So total is 114/377. I think. Somone check me ont hat.

In addition, I believe 2b, 4b, 5, and 6 are all incorrect

I don't think we need to permutate for #1... we just need selection...

As for the rest...dont you have the answers??? I'm now more confused than ever!lol  :huh:

Somehow i think 4b is right

P(B/A)  = (probability both are aces)/(probability of 1 card being an ace)

As for five the same thing applies i believe...

6 i'm not sure...

Anyone else wants to give a try and tell me..???

[quote="darkquantum"]
6. A stock market investor owns shares in a stock whose present value is 25. She has decided taht she must sell her stock if it either goes down to 10 or up to 40. If each change of price is either up 1 point with Probability .55 or down 1 point with probability .45, and the successive changes are independet, what is the probability the ivnestor retires a winner? (i.e. sells at 40)
[/quote]

Solution:

[hide]
The optimum best for her getting to 40 fastest has a (.55)^15 chance of happening.

The other ones are (.55^16)(.45)(15)
Next is: (.55^17)(.45^2)(16x15)
Next is: (.55^18)(.45^3)(17x16x15)

and so on

up to: (.45^14)(.55^29)(28x27x26x25...x15)

So, we have equation to get to 40 is:

(.55^(a+15))(.45^a)((a+14)!/(14)!)

Take the sum from a=0 to a=14 and you get an answer.[/hide]

wow, hmm, interesting, i'd just use a recursion...idk if that gets the same answer, lol, too tired.

i got the answer to #6 is about 95.3%.

As to fizsle, um, P(B|A) isn't P(B)/P(A). It's the probability of B given A, which is P(B and A)/P(A)

[quote="darkquantum"]wow, hmm, interesting, i'd just use a recursion...idk if that gets the same answer, lol, too tired.

i got the answer to #6 is about 95.3%.
[/quote]

yea doing the sum, i got that

what is a recursion tho?


calculus integration geometry quadratics function trigonometry algebra

Here's a plot of the region I did with Maple v10:

[url=http://www.uploadfile.info][img]http://www.uploadfile.info/uploads/e185927597.jpg[/img][/url]

[quote="kunny"]Find the area of the figure surrounded by the curve $2(x^{2}+1)y^{2}+8x^{2}y+x^{4}+4x^{2}-1=0.$[/quote]

Since the given equation is quadratic in y, we have by the quadratic formula

$y=\frac{-8x^{2}\pm\sqrt{64x^{4}-8(x^{2}+1)(x^{4}+4x^{2}-1)}}{4(x^{2}+1)}$
$=\frac{-4x^{2}\pm\sqrt{6x^{4}-6x^{2}-2x^{6}+2}}{2(x^{2}+1)}$
$=\frac{-4x^{2}\pm\sqrt{2(1-x^{2})^{3}}}{2(x^{2}+1)}$
$=\frac{-4x^{2}\pm(1-x^{2})\sqrt{2(1-x^{2})}}{2(x^{2}+1)}$

It is evident from the graph that the region is best described in the form $g(x)\leq y\leq f(x)$ where $-1\leq x\leq 1$ and, as the the term $(1-x^{2})\sqrt{2(1-x^{2})}$ is always positive for $-1\leq x\leq 1$, it follows that taking the plus sign gives the larger value and hence the function which describes the upper, "bell-shaped" curve; and the choice of the minus sign gives then the lower curve which connects the two points (cusps). The area bounded by the given equation is then given by

$\int_{-1}^{1}\left(\frac{-4x^{2}+(1-x^{2})\sqrt{2(1-x^{2})}}{2(x^{2}+1)}-\frac{-4x^{2}-(1-x^{2})\sqrt{2(1-x^{2})}}{2(x^{2}+1)}\right) dx$
$=\int_{-1}^{1}\frac{(1-x^{2})\sqrt{2(1-x^{2})}}{x^{2}+1}\, dx$

taking a break... the eval of this is escaping me: but thanks be to one of you mathlinker's for finishing it ;)

    -Ben

[quote="Xevarion"]Maple gives the same answer as you. Maybe quickmath is wrong?[/quote]

Maple, http://www.quickmath.com, and FireyHydra give equivalent answers, and thank you for finishing the problem FireyHydra.

  -Ben

Your answer is correct,　FieryHydra. :)

$\int_{0}^{1}\frac{(1-x^{2})^{\frac{3}{2}}}{1+x^{2}}\ dx$ Let $x=\sin \theta$

$=\int_{0}^{\frac{\pi}{2}}\frac{\cos^{3}\theta}{1+\sin^{2}\theta}\cos \theta\ d\theta=\int_{0}^{\frac{\pi}{2}}\frac{(1-\sin^{2}\theta)^{2}}{1+\sin^{2}\theta}\ d\theta=\int_{0}^{\frac{\pi}{2}}\left(\sin^{2}\theta-3+\frac{4}{1+\sin^{2}\theta}\right) d\theta$

$=\int_{0}^{\frac{\pi}{2}}\left(\frac{1-\cos 2 \theta}{2}-3+\frac{4}{1+\sin^{2}\theta }\right) d\theta$

$=\left[-\frac{1}{4}\sin 2\theta-\frac{5}{2}\theta \right]_{0}^{\frac{\pi }{2}}+4\cdot \frac{\sqrt{2}}{4}\pi\ \because \int_{0}^{\frac{\pi}{2}}\frac{dx}{1+\sin^{2}x}\ dx=\frac{\sqrt{2}}{4}\pi,$ which is derived by [b]FieryHydra[/b].

$=\left(\sqrt{2}-\frac{5}{4}\right)\pi.$

Thus as [b]benorin[/b] derived, the desired area is

$\int_{-1}^{1}\frac{\sqrt{2}(1-x^{2})^{\frac{3}{2}}}{x^{2}+1}dx=2\sqrt{2}\int_{0}^{1}\frac{\sqrt{2}(1-x^{2})^{\frac{3}{2}}}{x^{2}+1}dx=2\sqrt{2}\left(\sqrt{2}-\frac{5}{4}\right)\pi =\left(4-\frac{5\sqrt{2}}{2}\right)\pi .$ which is derived by [b]FieryHydra.[/b]


Define $ a \Delta b \equal{} a^2 \minus{}b$.  What is the value of $ (2^{4 \Delta13})\Delta(3^{3\Delta5})$

Simply plug in the numbers.

\[ 4 \Delta 13 \equal{} 4^2 \minus{} 13 \equal{} 3 \]

\[ 2^3 \equal{} 8 \]

\[ 3 \Delta 5 \equal{} 3^2 \minus{} 5 \equal{} 4 \]

\[ 3^4 \equal{} 81 \]

\[ 8 \Delta 81 \equal{} 8^2 \minus{} 81 \equal{} \boxed{\minus{}17} \]


$f_{n}(x)$ uniform converge to $f(x)$,and for every fixed n,then $f_{n}(x)$ is bounded,can we prove that $f(x)$ is bounded?

Thanks for detail.
I am sorry if it has been posted :)

there exist $N$ such that $\sup |f_{N}-f| \le 1$, so $\sup |f| \le \sup|f_{N}|+1 < \infty$...
is there a trick? :huh:


Let x,y,z be real number,and $ x^2\plus{}y^2\plus{}z^2\equal{}9$,prove taht:$ 2(x\plus{}y\plus{}z)\minus{}xyz \leq 10$

This problem was used in Vietnam National Olympiad 2002, not in Vietnam TST.

And there is the similar problem of Poland Olympiad: if $ x^2 \plus{} y^2 \plus{} z^2 \equal{} 2$ then $ x \plus{} y \plus{} z \le 2 \plus{} xyz.$

This problems posted before in Mathlinks, but I provide one approach here:

Use Cauchy -
      $ (2x \plus{} 2y \plus{} 2z \minus{} xyz)^2 \equal{} [(x \plus{} y).2 \plus{} z(2 \minus{} xy)]^2 \le [(x \plus{} y)^2 \plus{} z^2][4 \plus{} 4 \minus{} 4xy \plus{} x^2y^2] \equal{} (9 \plus{} 2xy)(8 \minus{} 4xy \plus{} x^2y^2)$
It is enough to show that
      $ (9 \plus{} 2xy)(8 \minus{} 4xy \plus{} x^2y^2) \le 100$
which is equivelent to $ (7 \minus{} 2xy)(xy \plus{} 2)^2 \ge 0$

The last one holds, because we can choose z with |z| = max{|x|, |y|, |z|} then $ x^2 \plus{} y^2 \le 6$ => $ xy \le 3$.

The equality holds when xy+2 = 0 and (x+y):z = 2: (2-xy) => (x, y, z) = (-1, 2, 2) (or its permutations).

from schur we have
$ r\geq \frac {p(4q \minus{} p^2)}{9} \equal{} \frac {p(p^2 \minus{} 18)}{9}$ since $ p^2 \minus{} 9 \equal{} 2q$
so$ 10 \plus{} r\geq 10 \plus{} \frac {p(p^2 \minus{} 18)}{9}$
it remain to prove
$ f(p) \equal{} p^3 \minus{} 36p \plus{} 90\geq 0$ where $ 3\sqrt {3}\geq p$
but from $ f'(p) \equal{} 3p^2 \minus{} 36$ the minimum of $ f(p)$ happen at $ p \equal{} 2\sqrt {3}$ but we can check it easily that $ f(2\sqrt {3}) > 0$
so the inequality is proven


prove that riemannian 2 dimensional surface is conformal to the plane.

That's not quite right. See the [url=http://en.wikipedia.org/wiki/Uniformization_theorem]Uniformization Theorem[/url]


[quote="shyong"]For non-negative real $a,b,c,d$ with $a+b+c+d=1$ , prove that

$8(a^3+b^3+c^3+d^3)+3(a^2+b^2+c^2+d^2)+30(abc+abd+acd+bcd)\geq 3$[/quote]
Yes this is true. I have not found the nice proof of your problem.
 shyong, whether is there  a nice proof? Thank you!

yes , i have a nice prove . The hint is try to write it as $\sum pS_1+qS_2+rS_3$ where $S_n=a^n+b^n+c^n+d^n$ ;)

[quote="arqady"]By your hint:
$\sum_{cyc}(8a^3+3a^2+30abc)\geq 3\Leftrightarrow\sum_{cyc}a(3a-1)^2\geq0.$ This is very nice! :lol:[/quote]

yah , you are right  :D

By using similar  method , we can prove that for

$a_1,a_2,\cdots ,a_n,a_{n+1}\geq 0$ , $n\geq 2\in\mathbb{N}$ , $p\in\mathbb{R}$ and

$a_1+a_2+\cdots +a_n+a_{n+1}=1$ , then

$(n^2-2p)(a_1^3+a_2^3+\cdots +a_{n+1}^3)+6p\sum_{i\neq j\neq k} a_ia_ja_k\geq (2n-3p)(a_1^2+a_2^2+\cdots +a_{n+1}^2)+(p-1)$


hello, if you mean $ 2^{3^4}\equal{}2417851639229258349412352$ then the answer is a).
Sonnhard.

[quote="SuperNerd123"]Actually the answer is c) because a) and b) are equal to eache other and so are d) and e)[/quote]

that's actually incorrect

let's figure out what the expressions are equal to

a)$ 2^{3^4}\equal{}2^{(3^4)}\equal{}2^{81}$
b)$ 2^{4^3}\equal{}2^{(4^3)}\equal{}2^{64}$
c)$ 3^{4^2}\equal{}3^{(4^2)}\equal{}3^{16}$
d)$ 4^{3^2}\equal{}4^{(3^2)}\equal{}4^{9}$
e)$ 4^{2^3}\equal{}4^{(2^3)}\equal{}4^{8}$

obviously, the greatest should be $ \boxed{\text{a}}$

btw how do u do the double power?(in Latex obviously)

[quote="SuperNerd123"]Actually the answer is c) because a) and b) are equal to eache other and so are d) and e)[/quote]

Remember that you take care of the inner exponents first.

idk...depends where the praenthesis are at


What's the average of 2^1999 and 2^2001?

[quote="isabella2296"]The number between them is $ \boxed{2^{2000}}$.[/quote]

is $ 2^3$ between $ 2^2$ and $ 2^4$?

i think buzzer is correct
i'm not sure though

Oops, stupid moment of the day.  :oops:

I believe buzzer is correct.

[hide]We wish to find

$ \frac{2^{1999}\plus{}2^{2001}}{2}$.

Factoring the $ 2^{1999}$ out of the numerator

$ \frac{2^{1999}(1\plus{}2^2)}{2}$

Simplifying $ 1\plus{}2^2$

$ \frac{2^{1999}(5)}{2}$

Dividing by 2

$ \boxed{2^{1998}\times5}$

[/hide]

Double checking. Yes, buzzer is correct

Isabella would be correct if you had asked for the [b]geometric mean.[/b]

Okay, this is getting spammy 'cause people keep posting the same solution.

Also to get $ 5\cdot2^{1998}$, you need to use \cdot.

It is a $ \text{\LaTeX}$ command.

D3m0n Shad0w, robinhe, please lock this thread - it's getting out of hand.


hi,

Could you please show me step by step how to find formula for the nth term for this sequence 5,13,25,41,61,85

I can do the easy ones but found this one hard.

Thank YOu.

Charikaar

Do you want a closed form or a rule?
The rule is the following: each time increment increase by 4.

Closed Form:

For $f(n)$, where $n$, is the $n$th number of the set, a closed form would be:

$1 + \displaystyle 4\left(\frac{n(n+1)}{2}\right)$

I think.

[quote="Myth"]Maybe you mean $1+2n(n+1)$ ? ;)[/quote]

Bleh... Yeah you're right Myth. My mistake... :D  :blush:

When I posted that I was thinking along the lines of:

$1 + 4 \displaystyle\sum_{i=1}^n {i}$

At the risk of making  this  a little more advanced than "getting started section"  I will make some comments here. I think these comments  are important enough that all serious young  matheletes should think about it.

To put it  mathematically,  IMO, "finding formula for next term"   do not make much sense.  The pattern one sees in any such finite sequence  is [b]in the eye[/b] of the of the person who looks at the sequence and  [i]any[/i] answer you give runs the risk of being  [b]subjective.[/b]

Edited later: Rest of the comments  follows:

[hide]Yes I know these type   of problems are very common and people think that  there is  only one 'right' answer.  I have seen  such problems  too many times in contests . (And yes, some times even my comments ended up in controvercy :))

Point is, unless you tell us the rule,   one can see MANY patterns  not JUST one..  ( sometimes, there a  is difference between  a "riddle" and a  math question.  Unlike a riddle, a math question has to be well defined)

Okay look at the  sequence given here    $ 5,13,25,41,61,85 $  - I can see many patterns.

[b] Pattern A:[/b]    One noticed by Myth above...
[b] Pattern B: [/b]   Each is sum of two consecutive squares .. ($ 1^2+2^2 = 5, 2^2+3^2 = 13, 3^2+4^2 = 25 $ etc  .. a nice sequence  you would agree   [hide](BTW this just happens to  be  the same sequence,  but that is not the point here.. I was just trying to show that there is another nice  way of looking at the  pattern here)
[/hide]
[b]Pattern C:[/b]  Pythogorian Triplets :  Hypotenuse  of the smallest right angle  triangle one can form when  one of the other other side is (2n+1) eg $ (3,5,7,9,11..etc..) $

The Pythogorian triplets are: $(3,4,5),(5,12,13),(7,24,25),(9,40,41),(11,60,61).(13,84,85),(15,8,17) $

     Here  the terms you get it   $5,13,25,41,61,85 $ but the next term is $17$ and not $113$

Okay some one may say: "But Pattern C" is more complicated etc.. If simplicity  is what you want,  you can select the famous rule given,  once by Scientific American's Gardner  --- Every term is 1 million when n>6!  IOW the  pattern goes  $5,13,25,41,61,85,1000000,1000000,1000000, ..$

I would end these  commenst with one  last example . In  national  mathcounts (1998 ?)  in the masters round had a problem regarding some graphs .. the first few values were coming as   $1,2,4,8,16 $ but the next value was  not $32$ but rather $31$   and those who "guessed" the pattern as $2^n$ were wrong. [/hide]

Hope this is helpful.  .. (May be ,  if there is interest, one can open a thread in "other problems solving topics)

A traditional question to ask in a New York City math team learning sequences and series:

What is the next term in the sequence which begins 103, 96, 86, 81, 72, 59, 50, 42?

The answer, of course, is [hide][url=http://www.mta.info/nyct/service/aline.htm]Penn Station[/url], if you happen to be going downtown on the A train.[/hide]


farbod_mp gave me a nice inequality that i thought should be posted:

$ x,y,z > 1$ are real numbers, and $ \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z} \equal{} 2$
show that:
$ \sqrt {x \plus{} y \plus{} z} >\equal{} \sqrt {x \minus{} 1} \plus{} \sqrt {y \minus{} 1} \plus{} \sqrt {z \minus{} 1}$

Since $ \frac{1}{x} \plus{} \frac{1}{y} \plus{} \frac{1}{z} \equal{} 2 \Rightarrow (1\plus{}1\plus{}1)\minus{}(\frac{1}{x} \plus{} \frac{1}{y} \plus{} \frac{1}{z}) \equal{} 1$

Hence $ \frac{x\minus{}1}{x} \plus{} \frac{y\minus{}1}{y} \plus{} \frac{z\minus{}1}{z} \equal{}1$

By Cauchy-Schwarz Inequality:

$ (\frac{x\minus{}1}{x} \plus{} \frac{y\minus{}1}{y} \plus{} \frac{z\minus{}1}{z})(x\plus{}y\plus{}z) \geq (\sqrt{x\minus{}1} \plus{} \sqrt{y\minus{}1} \plus{} \sqrt{z\minus{}1})^{2}$

Thus:

$ \sqrt{x\minus{}1} \plus{} \sqrt{y\minus{}1} \plus{} \sqrt{z\minus{}1} \leq \sqrt{x\plus{}y\plus{}z}$ as desired.

$ \mathbb{Q.E.D}$

It's Iran 1998
from $ \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z} \equal{} 2 \therefore \frac {x \minus{} 1}{x} \plus{} \frac {y \minus{} 1}{y} \plus{} \frac {z \minus{} 1}{z} \equal{} 1$
By Cauchy-Schwarz; $ (\frac {x \minus{} 1}{x} \plus{} \frac {y \minus{} 1}{y} \plus{} \frac {z \minus{} 1}{z})(x \plus{} y \plus{} z) \geq (\sqrt {x \minus{} 1} \plus{} \sqrt {y \minus{} 1} \plus{} \sqrt {z \minus{} 1})^2$
$ \therefore \sqrt {x \plus{} y \plus{} z} \geq \sqrt {x \minus{} 1} \plus{} \sqrt {y \minus{} 1} \plus{} \sqrt {z \minus{} 1}$


Let $ F$ be the set of all numbers of the form $ x \plus{} y \sqrt{2}$, where $ x$ and $ y$ are rational numbers. Show that $ F$ has all the properties of an ordered field but does not have the completeness property.

I don't understand why this field doesn't have the completeness property, which states that all nonempty sets of real numbers have a least upper bound... or does $ F$ not satisfy the completeness requirement since $ \sup F \equal{} \infty$?

[quote="JoeBlow"]Let $ x_n \equal{} \left ( 1 \plus{} \frac {1}{n}\right)^n$.  Clearly $ x_n \in F$ for each $ n\in \mathbb{N}$, but what is $ \sup_{n\in \mathbb{N}} \lbrace x_n \rbrace$?  Is it in $ F$?[/quote]

This doesn't represent the set of ALL numbers of the form $ x\plus{}y\sqrt{2}$ though...

You're right that completeness is a "for all" property, but to show that something is not complete all you need is a single instance that it doesn't exist. Since each of the stated $ x_n$ are rational, it is of the form $ x \plus{} y\sqrt {2}$ (the $ y$ is just $ 0$ in each case). Now you have a subset with no sup (the sup is irrational and not of that form, which may take a short proof in and of itself).


Hello .. can you help me whith this ?

Let the function f(x) = ax^3+bx^2+cx+d. If we know that f΄(x) have positive minimum value, is there any way to find
the sign of a,b,c,d ?
I think it is easy to find that  a,c>0.  What about  b and d ?

Thanks :)

It is $ f'(x)=3ax^2+2bx+c$ and if we know that $ f΄(x)$ has positive minimum value, then $ f'(x)$ has no roots which means that $ (2b)^2-4(3a)<0$

Since $ d$ does not show up in $ f'$, we can't find anything about it.
The coefficient $ b$ can have either sign as well. For example, $ f'$ could be $ x^2\plus{}x\plus{}1$ or $ x^2\minus{}x\plus{}1$.

i have a similar question to this
"Find a, b, c, d such that the cubic function f(x)=ax^3+bx^2+cx+d has horizontal tangent lines at (-2,-31) and (4,77)

This can be expressed by four equations: $ f'(\minus{}2)\equal{}0$, $ f'(4)\equal{}0$, $ f(\minus{}2)\equal{}\minus{}31$, $ f(4)\equal{}77$. Write them out in terms of $ a,b,c,d$: it's a linear system.

[quote="gregg"]It is $ f'(x) = 3ax^2 + 2bx + c$ and if we know that $ f΄(x)$ has positive minimum value, then $ f'(x)$ has no roots which means that $ (2b)^2 - 4(3a) < 0$[/quote]

Ι made the same things and i find that  a >0 and c > 0 . I think that b and d can have either signs.  Thanks  :)


geometry
topology
advanced fields
advanced fields theorems

I'm a total noob when it comes to Minkowski's stuff. Just wondering if there's any link between his geometry of space and his geometry of numbers. Thanks!

The MNG studies bodies in the context of n-dimensional space whereas the MSG involves the use of the space dimensions and time dimension to build manifold, that is currently claimed gradient-like (use of heat equations and Ricci (or similar) flow). [b]The former lies foundations for its application, ie. the latter[/b]; still, the MSG uses four dimensions only and this is only a proposition.

From the perspective of mathematics, the MNG is used in different areas of problem solving.


[quote="Altheman"][quote="shyong"][quote="t0rajir0u"]Show that $x^x \ge \left( \frac{x+1}{2} \right)^{x+1}$ for $x > 0$.[/quote]

[hide]
AM-GM

$\frac{x+1+1\cdots +1}{x+1}\geq x^{\frac{1}{x+1}}$    ($x$ times $1$)
Simplify and done .

[/hide][/quote]

clever proof but i think more needs to be done since x is not always a positive integer[/quote]

I don't know if I am mistaken but at the end of the problem the condition says $x>0$

Actually Altheman wants to argue about the " integer " but think about the form of weight AM-GM

$w_1x+w_2(1)\geq x^{w_1}1^{w_2}$

where $w_1=\frac{1}{x+1}$ , $w_2=\frac{x}{x+1}$ .So you dont need the $x$ to be positive " integer " .
:)

Is it ok ?

Could this problem be solved using mathematical induction? I tried to do that, but I only got so far.

No, since $x$ can be any positive real number.

:blush: Oh yeah. For some reason I got it into my head it was a sequence.

[quote="Lovasz"]$f(t)=t^t$ is strictly decreasing function.[/quote]

Sure??? decreasing?? for what $t$???

And how are you going to use that fact, if it is true?


At a party there are $n$ women and $n$ men. Each woman likes $r$ of the men, and each man likes $s$ of then women. For which $r$ and $s$ must there be a man and a woman who like each other?

I'm thinking that either:
a. Each man likes $s$ women
b.  It's supposed to say, for what $n$ and $r$

If each of the boys select $s$ girls then, by the pidgeonhole principle, there exists a girl that is selected at least $s$ times. Now, this girl in particular has to select $r$ boys. To be forced to select a boy that selected her, what it’s needed is that the ammount of guys that didn’t select her is less than $r$.
In particular:
$$r > n-s$$
Then we can conclude that the values of $r$ and $s$ that satisfy the conditions are the ones that satisfy:
$$r+s>n$$
(For $s$, $r$ $\geq 1$ and $\leq$ $n$)

@above, bumping old threads does not do any harm. You are allowed to bump threads as long as you are adding a solution or useful information to the thread. :)


quadratics
algebra
quadratic formula
special factorizations

Sample probs of what is expecting this week...I did then already, but since i cant check them i would like to see what you guys come up with..Thank You

1. Solve by factoring:
x2 – 9x = -8

2. Solve using the square root property:
2x2 – 5 = 93

3. Solve using the square root property:
(x + 4)2 = 81

4. Solve by completing the square:
x2 + 6x + 2 = 0

5. Solve using the quadratic formula:
x2 – 3x = -6x – 1

6. Solve using the quadratic formula:
x2 – 10x – 1 = -10

[quote="Flash8o"]1. Solve by factoring:
x2 – 9x = -8 [/quote]

[hide]
$ x^{2}\minus{}9x \equal{}\minus{}8$

$ x^{2}\minus{}9x\plus{}8 \equal{} 0$

$ (x\minus{}8)(x\minus{}1) \equal{} 0$

$ \boxed{x \equal{} 8}$

$ \boxed{x \equal{} 1}$
[/hide]

[quote="Flash8o"]2. Solve using the square root property:
2x2 – 5 = 93 [/quote]

[hide]
$ 2x^{2}\minus{}5 \equal{} 93$

$ 2x^{2}\equal{} 98$

$ x^{2}\equal{} 49$

$ \boxed{x \equal{}\pm 7}$
[/hide]

[quote="Flash8o"]3. Solve using the square root property:
(x + 4)2 = 81 [/quote]

[hide]
$ (x\plus{}4)^{2}\equal{} 81$

$ x\plus{}4 \equal{}\pm 9$

$ x \equal{}\minus{}4\pm 9$

$ \boxed{x \equal{} 5}$

$ x \equal{}\minus{}13$
[/hide]

[quote="Flash8o"]4. Solve by completing the square:
x2 + 6x + 2 = 0 [/quote]

[hide]
$ x^{2}\plus{}6x\plus{}2 \equal{} 0$

$ x^{2}\plus{}6x \equal{}\minus{}2$

$ x^{2}\plus{}6x\plus{}9 \equal{} 7$

$ (x\plus{}3)^{2}\equal{} 7$

$ x\plus{}3 \equal{}\pm\sqrt{7}$

$ \boxed{x \equal{}\minus{}3\pm\sqrt{7}}$
[/hide]

[quote="Flash8o"]5. Solve using the quadratic formula:
x2 – 3x = -6x – 1 [/quote]

[hide]
$ x^{2}\minus{}3x \equal{}\minus{}6x\minus{}1$

$ x^{2}\plus{}3x\plus{}1$

$ x \equal{}\frac{\minus{}3\pm\sqrt{3^{2}\minus{}4}}{2}$

$ \boxed{x \equal{}\frac{\minus{}3\pm\sqrt{5}}{2}}$
[/hide]

[quote="Flash8o"]6. Solve using the quadratic formula:
x2 – 10x – 1 = -10[/quote]

[hide]
$ x^{2}\minus{}10x\plus{}9 \equal{} 0$

$ x \equal{}\frac{10\pm\sqrt{\minus{}10^{2}\minus{}36}}{2}$

$ x \equal{}\frac{10\pm 8}{2}$

$ \boxed{x \equal{} 9}$

$ \boxed{x \equal{} 1}$
[/hide]

The ones with the Quadratic formula might be wrong...

Nope, you seem to have gotten all of the correct solutions for these problems. :) You could've checked your answer to number 6 by factoring.
[hide]$ x^{2}\minus{}10x\minus{}1 \equal{}\minus{}10\implies x^{2}\minus{}10x\plus{}9 \equal{} 0\implies (x\minus{}9)(x\minus{}1) \equal{} 0\implies\boxed{x \equal{} 9, 1}$[/hide]


ok i didn't do so hot on the AMC A (partly because i had a stomach ache)...but my school is offering the AMC B too. Suppose I do 1-2 Practice AMCs a day, is that enough to improve my score by say....15-20 points? the reason i'm not sure if its worth it is because i have SATs march 2nd, and im not too sure if i want to sacrifice SAT studying for AMC studying

[quote="hcheng501"]ok i didn't do so hot on the AMC A (partly because i had a stomach ache)...but my school is offering the AMC B too. Suppose I do 1-2 Practice AMCs a day, is that enough to improve my score by say....10 points?[/quote]

Sleeping well is enough to improve your AMC score by 10 points. But practicing can't hurt. And besides, what would you do if someone said no? Not try? :huh:

1-2 AMC's a day is a lot. Make sure you don't get burned out. Also, simply doing the tests is not enough to improve dramatically. It's important to go back over the test once you finish and review concepts that you didn't understand. AoPS Vol. 2 would be helpful, but I don't know if it would arrive in time. Consider picking one topic, say combinatorics, and doing lots of those problems that people post on the Forum.

Click "Intermediate Forum" and solve every problem you see.

A lot of improving your score is whether or not you have a good day... 15-20 points is only 3 questions more, which could be just watching out for unnecessary errors.

For the questions you don't know how to do, after you spend some time on it, read the solution and try to recreate it after a couple days, to prove your understanding. Ideally you should be able to explain the solution to someone else without peeking in the solution manual.

I say instead of taking bunch of AMC's and making yourself worned out, why don't you attack last 5 problems from past years and see how fast you can get them (that is, if you can).

Many of AMC problems use similar concepts but in different aspects and wording. Of course, problems ARE different but if you do tons of geometry, you should do well on geometry type. Same thing goes for others although counting, at least for me, is always difficult subject.

bump

i have AoPS 1 and 2, borrowed from a friend whos now at MIT. my biggest weakness is probably geometry, but there seems to be fewer and fewer geometry problems on the AMC these past few years.

aside from that, my other flaw in preparing was the short time allowed for the AMCs. i was able to solve many of the problems i left blank after the contest was over and i had time to think about them without pressure.

p.s. 1 down, AMC marathon this weekend. anybody think i should change my strategy?

you didn't have enough time because you couldn't think of the way to solve the problem or you didn't have enough time because you solved most of the problems using the most complicated and time-consuming methods?


limit
floor function
algebra proposed
algebra

[quote="Nguyen Van Linh"]Let a real $ a\in (0;1)$. Determine the limit $ \lim_{n\to \infty} [n.a^{n \plus{} 1} \minus{} (n \plus{} 1)a^n]$[/quote]

$ a\in(0,1)$ $ \implies$  $ \lim_{n\to \infty} n.a^{n \plus{} 1}\equal{}0$ and  $ \lim_{n\to \infty} (n \plus{} 1)a^n\equal{}0$ and $ n.a^{n \plus{} 1} < (n \plus{} 1)a^n$. Then :

1) if $ [...]$ in your question are just parenthesis, then the answer is  :  $ \lim_{n\to \infty} (n.a^{n \plus{} 1} \minus{} (n \plus{} 1)a^n)\equal{}0$

2) if $ [...]$ in your question is "integer part of ...", then the answer is  :  $ \lim_{n\to \infty}\lfloor n.a^{n \plus{} 1} \minus{} (n \plus{} 1)a^n\rfloor\equal{}\minus{}1$


I found this problem in a magazine and solved it. I thought it was an interesting problem.

There are 2000 white balls in a box. There are also unlimited supplies of white, green, and red balls, initially outside the box. During each turn, we can replace two balls in the box with one or two balls as follows: Two whites with a green, two greens with a white and a red, a white and green with a red, or a green and red with a white.

(a) After finitely many of the above operations there are three balls left in the box. Prove that at least one of them is a green ball.

(b) Is it possible after finitely many operations to have only one ball left in the box?

I think I can solve this.
Suppose we do the operations a, b, c and d times in that order. Then we have 2000+2a-b+c-d whites, a-2b-c-d greens, -b-c+d reds. Take that first equation, subtract twice the second and add three times the third to get W-2G+3R = 4d+2000. W+G+R = 3 and G=0 (assume this) then we get W+3R = 4d+2000 and W+R=3, so 2R = 1997 and this is impossible. So at least one is green.
If we had one left at the end, then we still must have W-2G+3R = 4d+2000, which is divisible by 4. However by letting one of the three variables equalling one and the other two 0, we can't get a multiple of 4, and thus we cannot end with only a single ball.

Interesting -- I like that method. Here's what I did:

Let each white ball be 1, a green be 2, and a red be 3. Then we take the sum modulo 4. We start with 1*2000 :equiv: 0 (mod 4). The operations we have don't change the sum. Then at the end, the only ways to get 0 (mod 4) with three balls are WWG and GGR, each of which has at least one green.

For the second part, we want to get just one ball left, but that ball must be 1, 2, or 3 (mod 4), but the sum is 0 (mod 4), which is impossible.

I'm not quite sure how I'd write this solution up formally. Can anyone help please? Thanks.

I think our proofs are actually exactly the same. We both get the sum W-2G+3R is 0 mod 4 (or W+2G+3R, same thing) and that solves it pretty easily. Coming up with the W+2G+3R is probably the hardest bit.

Hm I see what you mean. If it were on an olympiad (which it actually was), would it be possible to write it up something like the way I did it and still get 7 points, or would it be necessary to write it up as you did?

Your way would be perfectly fine..  just saying that 1*W + 2*G + 3*R is always 0 mod 4, probably would be a good idea to say why each operation doesn't change it, and then why a green must be in it to make it 0 mod 4 for three balls. But it would be perfectly fine.

I did it essentially the same as TripleM. Your way is really nice though Zeta. Nice problem.

Speaking of this problem Zeta, how did you think of this solution. Is this a general technique that you use on problems like this?

The first time I saw a problem like this was on the USAMTS. The problem had to do with proving that a certain word was not part of a language with certain rules about the formation of words. You had to prove that ``baabaabaa'' was not a word in this language. It turned out that it was impossible to have 0(mod 3) a's. Just to be different from what I thought everyone else would do, I let a=e^(2i:pi:/9), which essentially reduces to the same argument, but I like it more because it involves complex numbers. I basically figured from this problem when I looked at it that there would be some invariant based on the sum (or product if I choose complex numbers). So I let white be 1 and then just figured out what the others should be modulo four for the sum to be invariant. Now that invariant method is the first thing I think of when I look at a problem like this.


function
search
number theory unsolved
number theory

Find all surjective function $f : N$ → $N$ satisfying the condition
                $m|n$ ↔ $f(m)|f(n)$        $\large m,n \in N$.

Clearly you didn't even try to search for it + bad title.

http://www.mathlinks.ro/Forum/viewtopic.php?t=135914


Calculate the accelaration of the centre of mass. The figure is attached. Both the blocks are of equal mass. No friction.

I calculated the accleration of the blocks. But how to find the accelaration of centre of mass of the blocks.

$ \sum F_{ext} \equal{} M \cdot a_{CM}$.

I did it in the following way:

Let the accelaration of the blocks is a.
The angle between the directions of accelaration of the bodies is 90 degrees. So the resultant is a*sqrt(2).

Is it correct?

Calculate a_CM with the expression I gave you above. If it is equal to your answer then it is correct. However, at first sight your answer does not seem to be right.


Denote $ M\equal{}${$ \minus{}1,0,1$}.How many $ (a_1,a_2,...,a_n)$ are there satisfying the following conditions:
i)$ a_i \in M$ for $ i\equal{}1,2,3,...,n$
ii)$ (a_i\minus{}a_{i\plus{}1})\in M$ for $ i\equal{}1,2,3,...,n$

[quote="SUPERMAN2"]Denote $ M \equal{}${$ \minus{} 1,0,1$}.How many $ (a_1,a_2,...,a_n)$ are there satisfying the following conditions:
i)$ a_i \in M$ for $ i \equal{} 1,2,3,...,n$
ii)$ (a_i \minus{} a_{i \plus{} 1})\in M$ for $ i \equal{} 1,2,3,...,n$[/quote]

Let $ P_{(n,k)}$ be the number of $ (a_1,a_2,...,a_n)$ satisfies the condition with $ a_1 \equal{} k$ and let $ Sum_n \equal{} P_{(n, \minus{} 1)} \plus{} P_{(n,0)} \plus{} P_{(n,1)}$

Then $ P_{(n \plus{} 1,0)} \equal{} Sum_n$ and
$ Sum_{n \plus{} 1} \equal{} (P_{(n, \minus{} 1)} \plus{} P_{(n,0)}) \plus{} Sum_n \plus{} (P_{(n,0)} \plus{} P_{(n,1)}) \equal{} 2 Sum_n \plus{} P_{(n,0)} \equal{} 2 Sum_n \plus{} Sum_{n \minus{} 1}$

Now $ Sum_1 \equal{} 3$ and $ Sum_2 \equal{} 7$ .

I am too lazy to find the general formula, so that's it. :sleeping:


Someone made a post but there's no response....so I'll say it here again.

Anyway to download the cse5 package??? That really helps because I'm to lazy to type the long-winded method names... :rotfl:

And also, anyway to type up the code in asymptote like a text editor, then run the whole code at one time??

I found this URL in a previous thread http://www.artofproblemsolving.com/Forum/latexrender/cse5.asy (right-click and choose Save Target As (IE)/Save Link As (Firefox) but I agree it needs to be highlighted somewhere. Fedja can I suggest this link is added to the beginning of your [i]Description of the package cse5[/i] thread?

All right, I added the link to the beginning of the description. By the way, since I intend to update it to the new version by the end of July, any suggestion about what other functions you want to see there are welcome. Right now I'm thinking of upgrading MP to make it possible to label in color, adding a MarcCurve (MC) function for labeling curves and WayPoint (WP) that combines waypoint and midpoint from the olympiad package. I'm also thinking of adding a MarkAngle (MA) function that would create an arc, fill the sector if asked, and put the label near the middle of that arc. The question with the last one is what is the best way to do it (I mean, what the parameters of the function should be). Once it is done in some way, there is no road back because, since the package is for public use, the backward compatibility should be complete and the meaningful 2/3-letter combinations are not that plentiful :P


trigonometry
algebra
polynomial
calculus
calculus computations

For the approximations

sin (x) = x

sin (x) = x - (x^3)/6

What is the range of x-values for which each of these approximations is correct to within 1%?

I am guessing this has to do with Linearization / Taylor Polynomials?

Help would be Aprreciated. Thanks

I do not quite grasp what that does/mean


algebra
polynomial
integration
college contests

Let be given a polynomial $ P(x) $ satisfying condition  $ P(a) = P(b) $ where $ a < b$. Denote $ M = max_{a\leq x \leq b} |P''(x)| $. Prove that

$a) \int^b_a P\"(x)(x-a)(x-b)dx = 2\int^b_a P(x)dx $
$b) |\int^b_a P(x)dx| \leq \frac 1{12}M(b-a)^3$

You are right, Grobber, thank you and congratulations. I will correct it.

[quote="grobber"]To get (a) out of the way: we integrate by part and we get

$\int_a^bP\"(x)(x-a)(x-b)dx=-\int_a^bP'(x)(2x-a-b)dx=2\int_a^bP(x)dx$.

For (b), is it possible that you meant $M=max_{a\leq x\leq b}|P\"(X)|$ instead of $P'(x)$?. It's easy to solve it in these conditions by using (a), but maybe I'm wrong. Here's what I did:

$2|\int_a^bP(x)dx|=|\int_a^bP\"(x)(x-a)(x-b)dx|\leq |M\int_a^b(x-a)(x-b)dx|=\nolinebreak M\frac{(b-a)^3}{6}$, which is exactly what we want. That's what makes me think that maybe there's a mistake in the text you've posted.[/quote]

Isn't it also necessary to mention that M is positive?


calculus
real analysis
real analysis solved

Determine f:[a,b]->R, of class C1, s.t. f(a)=0 and there is a k>0 s.t.   |f'(x)|<= k|f(x)|.

Let F(x)=(f(x)) is positive C1, F(a)=0,
F'(x)=2f'(x)f(x)  \leq 2|f'(x)||f(x)|  \leq 2kf(x)
F'(x) \leq 2kF(x)
We consider g(x)=F(x).exp(-2kx)
g'(x)=exp(-2kx)(F'(x)-2kF(x)) \leq 0
g is positive continous decreasing g(a)=0=> g = 0 =>F=0
f=0

Moubinool

grobber please post the sources of the problems. this problem is from IMC 199x, x \in {4,5} :) (don't know which exactly).

Oh, ho, ho, it was certainly not given for the first time at that IMC. This problem is quite old and famous.

Yeah, i've seen it like a hundred times in all sorts of old calculus books.


Prove that
$4b^3+4b+1=(2a-1)^2$, $a,b \in \mathbb{N} \cup \left\{0\right\}$
has only solutions b=0, a=0; b=0, a=1; b=1, a=2; b=3, a=6.

nice problem from friedrich. i'm already very interestested whether he will get to the training camps this year again :lol: :lol: . you have any ideas how he did, idril? btw, who are you? do i know you?

so, to the problem:

we have
$(b^2+1)b=(a-1)a$.
if $b=0$, then $a=0$ or $a=1$.
if $a=0$ or $a=1$, then $b=0$.
else we immediately get $b|a$ or $b|a-1$.
first case: $a=kb$. we have $k\geq 1$. then $b^2+1=k^2b-k$. take the discriminant to get that $k^4-4k-4$ is a square. now $(k^2-1)^2<k^4-4k-4<(k^2)^2$ for $k\geq 3$. thus $k=1$, giving no solution, or $k=2$, giving $b=1$, $a=2$ or $b=3$, $a=6$.
second case: $a=kb+1$. then $b^2+1=k^2b+k$.  the discriminant now is $k^4+4k-4$. but $(k^2)^2<k^4+4k-4<(k^2+1)^2$ for $k\geq 2$, thus $k=1$ giving $b=1$, $a=2$.

Peter

i dont understand why $b|a$ or $b|a-1$? can you explain it please.

... it's wrong, and i already said that before, but that post was moved since the discussion here got quite away from the problem. actually, this problem may be solved by some brutal methods for elliptic curves; this doesn't get quite nice, though.

yes, my post was here before the discussion an i have already see your reply. Thanks


[quote="gorrina"]can you prove it?

$ 1 \leq (\frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z})(x^2(x^2 \plus{} 2y^2 \plus{} 2z^2) \plus{} 4y^2z^2)$, for $ x \plus{} y \plus{} z \equal{} 1$ and $ x, y, z$ are positive real numbers.[/quote]
It's wrong. Try $ x\equal{}y\equal{}0.1$ and $ z\equal{}0.8.$

[quote="gorrina"]what about :

$ 9 \leq (\frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z})^2(x^2(x^2 \plus{} 2y^2 \plus{} 2z^2) \plus{} 4y^2z^2)$, for $ x \plus{} y \plus{} z \equal{} 1$ and $ x, y, z$ are positive real numbers.[/quote]
It's obviously true because $ (xy\plus{}xz\plus{}yz)^4\geq9x^2y^2z^2(x\plus{}y\plus{}z)^2$ is true.


This is not the wrong forum; the nicest solution to this problem involves matrices I think :)

If $ a,b,c$ are coprime integers, show that
    \[ r\equal{}\minus{}3a\plus{}3b\plus{}2c, s\equal{}5a\plus{}4b\minus{}6c,t\equal{}\minus{}7a\plus{}6b\plus{}5c
    \]are also coprime integers.

Source: an old Linear Algebra assignment.

My solution

    We see that
    \[ \begin{bmatrix}
          r \\
          s \\
          t
       \end{bmatrix}\equal{}
          \begin{bmatrix}
             \minus{}3 & 3 & 2 \\
             5 & 4 & \minus{}6 \\
             \minus{}7 & 6 & 5
          \end{bmatrix}\begin{bmatrix}
          a \\ b \\ c
          \end{bmatrix}
    \]
    Let us take the determinant of the coefficient matrix, let's call it $ C$. It's a $ 3\times3$, so we can compute it by hand:
    \[ \det{C}\equal{}(\minus{}3)(4)(5)\plus{}(3)(\minus{}6)(\minus{}7)\plus{}(2)(5)(6)\minus{}(\minus{}7)(4)(2)\minus{}(6)(\minus{}6)(\minus{}3)\minus{}(5)(5)(3)\equal{}\minus{}60\plus{}126\plus{}60\plus{}56\minus{}108\minus{}75\equal{}\minus{}1
    \]

We know that if the determinant of a matrix with integer entries is $ \pm1$, then its inverse also has integer entries. (If you didn't know that, look at this hidden section:
[hide]
[b]Suppose $ A\in M_n(\mathbb Q)$, that $ A$ is invertible, and that the entries of $ A$ are integers. Prove that $ A^{\minus{}1}$ has integer entries if and only if $ \det A\equal{}\pm1$.[/b]

Since $ A$ is invertible, we know we definitely have $ A^{\minus{}1}$ such that
\[ \det{(AA^{\minus{}1})}\equal{}\det{A}\det{A^{\minus{}1}}\equal{}\det{I}\equal{}1\implies\det{A}\equal{}\frac{1}{\det{A^{\minus{}1}}}
\]
We know that $ A$ is composed entirely of integers, so its determinant is also an integer. Thus, if $ A^{\minus{}1}$ has all integer entries, its determinant is also an integer; but the only integer such that $ \frac{1}{\det{A^{\minus{}1}}}$ is also an integer is $ \pm1$. So then $ \det{A}\equal{}\frac1{\pm1}\equal{}\pm1$. Thus we have proven one side of the implication; it remains to prove the converse.

Next we consider the classical adjoint of $ A$. Since the fields of $ A$ are all integers, then the determinant of each minor matrix will also be an integer, so $ A^{\mathrm{adj}}$ has all integer entries as well. Then, by the formula for the inverse of a matrix according to its classical adjoint, we have
\[ A^{\minus{}1}\equal{}\frac{1}{\det{A}}A^{\mathrm{adj}}
\]
Thus, if $ \det{A}\equal{}\pm1$, then $ A^{\minus{}1}\equal{}\pm A^{\mathrm{adj}}$. Obviously this preserves the integers of $ A^{\mathrm{adj}}$, so $ A^{\minus{}1}$ has integer entries as well.
[/hide]
So we see that $ \det C\equal{}\minus{}1$. By the result in the hidden section above, this means that the inverse of $ A$ has entirely integer entries. So there is some matrix $ A^{\minus{}1}\in M_3(\mathbb Z)$ such that
    \[ A^{\minus{}1}\begin{bmatrix} r \\ s \\ t\\\end{bmatrix}\equal{}\begin{bmatrix}a \\ b\\ c\end{bmatrix}
    \]

    Let us assume that $ r,s,t$ are not coprime, and seek a contradiction. If they are not coprime, then there is some $ d$ that divides all of them. We can take it out as a common factor, and we will have
    \[ dA^{\minus{}1}\begin{bmatrix} \frac rd \\ \frac sd \\ \frac td\\\end{bmatrix}\equal{}\begin{bmatrix}a \\ b\\ c\end{bmatrix}
    \]
    Since $ d>1$ is a common divisor, every entry in the new column matrix is an integer. Moreover, we know that $ A^{\minus{}1}$ also has only integer entries. However, if we divide both sides of this equation by $ d$, we see that the right column vector must contain a non-integer, since $ d$ cannot have been a common factor of $ a,b,c$ because they were coprime. Thus we have two integer matrices multiplying to give a non-integer matrix. This is a contradiction, so our assumption that $ d\not\equal{}1$ must have been false. Therefore, $ r,s,t$ are also coprime.

I think after you find that inverse, you can alternatively show that some linear combination of r s and t gives 1.


advanced fields
advanced fields unsolved

Let $ H$ be a Hilbert space and $ T \colon H \to H$ be a self-adjoint operator. Must $ T$ have a closed range?

Certainly not. Let $ e_n$ be the orthonormal basis. Take $ Te_n\equal{}\frac 1n e_n$. Then the range of $ T$ is dense in $ H$ but not the entire $ H$.


search
inequalities
inequalities proposed

Let $ a,b,c >0$. Prove that !
$ \sum \frac{1}{a(1\plus{}b)} \ge \frac{3}{abc\plus{}1}$

[quote="bigbang195"][quote="bigbang195"]Let $ a,b,c > 0$. Prove that !
$ \sum \frac {1}{a(1 \plus{} b)} \ge \frac {3}{abc \plus{} 1}$[/quote]

 :idea:  :idea:  :idea:[/quote]

Be more patient and try to search for such well-known problems. This is famous as BMO 2006, but it had appeared in other contests before.

We have,
  \[ \frac{{abc \plus{} 1}}{{a(1 \plus{} b)}} \plus{} 1 \equal{} \frac{{abc \plus{} 1 \plus{} a \plus{} ab}}{{a(1 \plus{} b)}} \equal{} \frac{{ab(c \plus{} 1) \plus{} 1 \plus{} a}}{{a(1 \plus{} b)}} \equal{} \frac{{b(c \plus{} 1)}}{{1 \plus{} b}} \plus{} \frac{{1 \plus{} a}}{{a(1 \plus{} b)}}\]

Similarly, we deduce
 \[ 3 \plus{} \sum {\frac{1}{{a(1 \plus{} b)}}}  \equal{} \frac{{b(c \plus{} 1)}}{{1 \plus{} b}} \plus{} \frac{{1 \plus{} a}}{{a(1 \plus{} b)}} \plus{} \frac{{c(1 \plus{} a)}}{{1 \plus{} c}} \plus{} \frac{{1 \plus{} b}}{{b(1 \plus{} c)}} \plus{} \frac{{a(b \plus{} 1)}}{{1 \plus{} a}} \plus{} \frac{{1 \plus{} c}}{{c(1 \plus{} a)}}\]

Using the AM-GM Inequality. we also have
 \[ \frac{{b(c \plus{} 1)}}{{1 \plus{} b}} \plus{} \frac{{1 \plus{} a}}{{a(1 \plus{} b)}} \plus{} \frac{{c(1 \plus{} a)}}{{1 \plus{} c}} \plus{} \frac{{1 \plus{} b}}{{b(1 \plus{} c)}} \plus{} \frac{{a(b \plus{} 1)}}{{1 \plus{} a}} \plus{} \frac{{1 \plus{} c}}{{c(1 \plus{} a)}} \ge 6\]

          So the Inequality is proved

This is very nice inequality. But no new problem. Balkan MO 2006, Problem1.


quadratics
function
algebra
polynomial
algebra unsolved

Let $ x$ be a real number and $ n$ a positive integer such that $ x \plus{} x^n$ and $ x \plus{} x^{n \plus{} 1}$ are both rational numbers. Prove that $ x$ is a rational number as well.

Edit: Sorry, this should have been posted in Algebra Proposed & Own Problems. I have a solution, but I'm interested to see others.

The official solution: let $ x\plus{}x^n\equal{}a$ and $ x\plus{}x^{n\plus{}1}\equal{}b$. Then $ b\minus{}x\equal{}x(a\minus{}x)$, hence $ x$ is the root of a quadratic equation with rational coefficients. Therefore, $ x\equal{}u\plus{}\sqrt{v}$ or $ x\equal{}u\minus{}\sqrt{v}$ for some rationals $ u,v$. If $ \sqrt{v}$ is not rational, we obtain a contradiction by considering the function $ f(x)\equal{}x\plus{}x^k$, with $ k\equal{}n$ or $ k\equal{}n\plus{}1$,  such that $ k$ is odd. The function is increasing, so we cannot have $ f(u\plus{}\sqrt{v})\equal{}f(u\minus{}\sqrt{v})$.
However, for the nine graders, the fact that if a polynomial with rational coefficients having the root $ u\plus{}\sqrt{v}$ also has the root $ u\minus{}\sqrt{v}$ is not common knowledge.
I have a proof which does not use this result but, still, I'm waiting for your feedback.

[quote="enescu"] The function is increasing, so we cannot have $ f(u \plus{} \sqrt {v}) \equal{} f(u \minus{} \sqrt {v})$.
[/quote]

I'm confused... how do you go from $ u\minus{}\sqrt{v}$ being a root of the quadratic to $ f(u\plus{}\sqrt{v})\equal{}f(u\minus{}\sqrt{v})$?

Now, $ x$ is a real number which satisfies both equalities $ x\plus{}x^n\equal{}0,x\plus{}x^{n\plus{}1}\equal{}0$. An algebraic manipulation shows that the only possible candidates are $ u\plus{}\sqrt{v}$ and $ u\minus{}\sqrt{v}$. So, one of these two numbers must satisfy both equalities. As I pointed above, the other one must satisfy both equalities, as well.

Sorry if I'm missing something obvious, but does $ u\minus{}\sqrt{v}$ satisfying the quadratic equation mean that it satisfies the two original equations?

Think about the problem rephrased as follows: the polynomials $ x\plus{}x^n\minus{}a$ and $ x\plus{}x^{n\plus{}1}\minus{}b$ have a common root. Prove that this root is also one of the roots of the quadratic equation $ x^2\minus{}(a\plus{}1)x\plus{}b\equal{}0$.

So, one of the numbers $ u\plus{}\sqrt{v}, u\minus{}\sqrt{v}$ satisfies both equations $ x\plus{}x^n\equal{}a,x\plus{}x^{n\plus{}1}\equal{}b$.
However, if $ \sqrt{v}$ is irrational,  since the equations have [b]rational[/b] coefficients, if one of $ u\plus{}\sqrt{v}, u\minus{}\sqrt{v}$ is a root, the other one is a root, as well.
Hence, both numbers $ u\plus{}\sqrt{v}, u\minus{}\sqrt{v}$ satisfy both equations $ x\plus{}x^n\equal{}a,x\plus{}x^{n\plus{}1}\equal{}b$.


$ {PQR}$ is an equilateral triangle. The point $ {U}$ is the midpoint of $ {PR}$. Points $ {T}$ and $ {S}$ divide $ {QP}$ and $ {QR}$ in the ratio $ {1: 2}$. The point of intersection of $ {PS}$, $ {RT}$ and $ {QU}$ is $ {X}$. If the area of $ \triangle{QSX}$ is $ {1}$ square unit, what is the area, in square units, of $ \triangle{PQR}$

[/img]

[quote="stevenmeow"]is point t closer to point q or point p?
similarly, is point s closer to point q or point r?[/quote]

There was a diagram with the question, and i have it, but how do i put it on?

follow the directions here
"Insert image: [img]http://image_url[/img]  (alt+p)"
they are in the small blue bar above the text box under the 'toolbar' when you hover over the Img button

alternatively, you could add an attachment if it is saved as a file on your computer

[hide="Big Hint"]Common altitudes.[/hide]
[hide="Solution"]Note that $ \triangle QTX\cong\triangle QSX$, so the area of $ \triangle QTX$ is also 1.  Also note that $ \triangle QSX$ and $ \triangle SRX$ share a common altitude dropped down to $ \overline{QR}$, so the area of $ \triangle SRX$ is 2.  Thus the area of $ \triangle QRT$ is 4.  Finally, $ \triangle QRT$ and $ \triangle QRP$ share a common altitude dropped down to $ \overline{PQ}$, so the area of $ \triangle QRP$ (a.k.a. $ \triangle PQR$) is $ \boxed{12}$.[/hide]


Compute the number of ordered triplets of real numbers $ (a, b, c), a \le b \le c$ for which $ a^2 \plus{} b^2 \equal{} 410$, $ b^2 \plus{} c^2 \equal{} 490$, and $ a^2 \plus{} c^2$ is a positive integer.

There are no solutions I believe.
The only solutions for the first one are (7,19) (11,17) where a < b.
The only solution for the second one is (7,21) where b < c.

therefore, the values of A are 7 or 11. The values of B are 19,17, or 21. the value of c is 21.
Therefore there are no solutions.

I think that should be right.
I may be wrong though. I'm new at this.

Nobody said $ a,b$ and $ c$ are integers... :P

Suppose $ a^2 \plus{} c^2 \equal{} k$. Summing the equations and dividing by two, we see that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 450 \plus{} \frac{k}{2}$. Thus, $ a^2 \equal{} \frac{k}{2} \minus{} 40$, $ b^2 \equal{} 450 \minus{} \frac{k}{2}$, and $ c^2 \equal{} 40 \plus{} \frac{k}{2}$. Since $ a \leq b \leq c$, we see that $ a^2 \leq b^2 \leq c^2$. It thus sufficies to find the number of positive integers $ k$ such that $ k \minus{} 80 \leq 900 \minus{} k \leq 80 \plus{} k$.

$ k \minus{} 80 \leq 900 \minus{} k \to k \leq 490$; $ 900 \minus{} k \leq 80 \plus{} k \to k \geq 410$. Thus, $ k \in [410, 490]$, so there are $ \boxed{91}$ such $ k$ satisfying the desird property.

Hope I didn't do anything wrong.

This answer is incorrect, remember that the numbers may be negative.

[quote="Zhero"]Suppose $ a^2 \plus{} c^2 \equal{} k$. Summing the equations and dividing by two, we see that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 450 \plus{} \frac {k}{2}$. Thus, $ a^2 \equal{} \frac {k}{2} \minus{} 40$, $ b^2 \equal{} 450 \minus{} \frac {k}{2}$, and $ c^2 \equal{} 40 \plus{} \frac {k}{2}$. Since $ a \leq b \leq c$, we see that $ a^2 \leq b^2 \leq c^2$. It thus sufficies to find the number of positive integers $ k$ such that $ k \minus{} 80 \leq 900 \minus{} k \leq 80 \plus{} k$.

$ k \minus{} 80 \leq 900 \minus{} k \to k \leq 490$; $ 900 \minus{} k \leq 80 \plus{} k \to k \geq 410$. Thus, $ k \in [410, 490]$, so there are $ \boxed{91}$ such $ k$ satisfying the desird property.

Hope I didn't do anything wrong.[/quote]

Because $ a,b,c$ are real numbers, I think this part
$ a \leq b \leq c$, we see that $ a^2 \leq b^2 \leq c^2$
isn't true
For example

$ \minus{}3<\minus{}2<\minus{}1$, but $ 9>4>1$
Sorry,if I'm wrong :)

Yes, that's true.
[hide="Hint"]Consider the four cases:
$ 1)$ $ a, b, c \ge 0$
$ 2)$ $ b, c \ge 0, a < 0$
$ 3)$ $ c \ge 0, a, b < 0$
$ 4)$ $ a, b, c < 0$[/hide]

Another thing to point out... the problem asks to find the number of triples, not the number of possible values of $ k$.

[hide="Continuing from..."]Denote $ a^2 + c^2 = X$. So after some manipulations we have
\begin{eqnarray*} 2a^2 & = & X - 80 \\
2b^2 & = & 900 - X \\
2c^2 & = & 80 + X \end{eqnarray*}
that forces $ 80 \le X \le 900$ for each square to be nonnegative.

Now we go on to find $ a \le b \le c$ for each X. Let's compare $ |a|$, $ |b|$, and $ |c|$, particularly since we have squares in our equations. Since $ X - 80 < 80 + X \iff a^2 < c^2$, we immediately have that $ |a| < |c|$ and can consider each of the three possibilities:

[b]1. [/b]$ 0 \le |b| < |a| < |c|$ for $ 900 - X < X - 80$, i.e. $ 490 < X \le 900$
For each $ X$ we can find a triple $ (a,b,c)$ with $ a < b \le 0 < c$ such that $ a \le b \le c$.

[b]2. [/b]$ 0 \le |a| \le |b| \le |c|$ for $ X - 80 \le 900 - X \le X + 80$, i.e. $ 410 \le X \le 490$
For each $ X$ we can find two triples $ (a,b,c)$, one with $ 0 < a \le b < c$ and another with $ a < 0 < b \le c$, each such that $ a \le b \le c$

[b]3. [/b]$ 0 \le |a| < |c| < |b|$ for $ 80 \le X < 410$.
No triple $ (a,b,c)$ such that $ a \le b \le c$ is possible.

So $ 410 + 2 \cdot 81 = 572$ triples.
[/hide]

I might have missed a few of them.

That answer is still incorrect.  Does anybody want to post a solution?

If $ a^2 \plus{} c^2 \equal{} n$, then $ b^2 \equal{} 450 \minus{} {n\over 2}\implies a^2 \equal{} {n\over 2} \minus{} 40\land c^2 \equal{} {n\over 2} \plus{} 40$

By non-negativity of the squares, we get $ 80\leqslant n\leqslant 900$.

For each of $ a^2, b^2, c^2$, the square root of the corresponding expression can take either positive or negative value.

1. $ \plus{} \sqrt {a^2}\leqslant \plus{} \sqrt {b^2}\leqslant \plus{} \sqrt {c^2}$. That gives $ {n\over 2} \minus{} 40\leqslant 450 \minus{} {n\over 2}\iff n\leqslant 490$ and $ 450 \minus{} {n\over 2}\leqslant {n\over 2} \plus{} 40\iff n\geqslant 410$, hence $ 410\leqslant n\leqslant 490$, therefore $ 81$ values for $ n$ i.e. $ 81$ triplets.

2. $ \minus{} \sqrt {a^2}\leqslant \plus{} \sqrt {b^2}\leqslant \plus{} \sqrt {c^2}\iff 450 \minus{} {n\over 2}\leqslant {n\over 2} \plus{} 40\iff n\geqslant 410$. Since $ n\leqslant 900$, here we have $ 491$ triplets.

3. $ \minus{} \sqrt {a^2}\leqslant \minus{} \sqrt {b^2}\leqslant \plus{} \sqrt {c^2}\iff \plus{} \sqrt {b^2}\leqslant \plus{} \sqrt {a^2}\iff 450 \minus{} {n\over 2}\leqslant {n\over 2} \minus{} 40$. That yields $ n\geqslant 490$, which combined with $ n\leqslant 900$ gives $ 411$ triplets.

4. $ \minus{} \sqrt {a^2}\leqslant \minus{} \sqrt {b^2}\leqslant \minus{} \sqrt {c^2}\iff \plus{} \sqrt {c^2}\leqslant \plus{} \sqrt {b^2}\leqslant \plus{} \sqrt {a^2}$. That yields $ {n\over 2} \plus{} 40\leqslant 450 \minus{} {n\over 2}\leqslant {n\over 2} \minus{} 40$. That case is impossible, hence $ 0$ triplets.

Therefore total is $ 81 \plus{} 491 \plus{} 411 \plus{} 0 \equal{} 983$ triplets.


calculus
integration
logarithms
calculus computations

can you solve :|

$ \int_{0}^{\infty}\;\;\frac{x\cdot\ln x}{e^{2\pi x}\minus{}1}\;\;\textbf dx$

Oh, I found this: http://mathworld.wolfram.com/Glaisher-KinkelinConstant.html

[quote="sos440"]Is there anyone who can simplify $ \frac{1}{24}\left( 1\minus{}\gamma\minus{}\ln(2\pi)\plus{}\frac{\zeta'(2)}{\zeta(2)}\right)$ ?[/quote]

wow !  :| did you actually derive this answer, 'cause if you did, then it is also
[b]EXACTLY [/b]what i derived in class!

really cool if you did it by hand...   :coolspeak: took me well over 3 pages
to get to it using complex analysis...

Using the formula:

$ \int_{0}^{\infty}\frac{t^{x\minus{}1}}{e^{t}\minus{}1}\,dt \equal{}\zeta(x)\Gamma(x)$

and differentiating under the integral sign with respect to $ x$ we have:

$ \int_{0}^{\infty}\frac{t^{x\minus{}1}\ln(t)}{e^{t}\minus{}1}\,dt \equal{}\zeta'(x)\Gamma(x)\plus{}\zeta(x)\Gamma'(x)$

Using $ \Gamma'(2) \equal{} 1\minus{}\gamma$ and $ \zeta(2) \equal{}\frac{\pi^{2}}{6}$ we have:

$ \int_{0}^{\infty}\frac{t\ln(t)}{e^{2\pi t}\minus{}1}\,dt \equal{}\frac{1}{4\pi^{2}}\int_{0}^{\infty}\frac{t\ln(\frac{t}{2\pi})}{e^{t}\minus{}1}\,dt$
$ \equal{}\frac{1}{4\pi^{2}}\int_{0}^{\infty}\frac{t\ln(t)}{e^{t}\minus{}1}\,dt\minus{}\frac{1}{4\pi^{2}}\ln(2\pi)\int_{0}^{\infty}\frac{t}{e^{t}\minus{}1}\,dt$
$ \equal{}\frac{1}{4\pi^{2}}\left(\zeta'(2)\plus{}\zeta(2)\minus{}\gamma\zeta(2)\right)\minus{}\ln(2\pi)\zeta(2)\frac{1}{4\pi^{2}}$
$ \equal{}\frac{1}{24}\left(1\minus{}\gamma\minus{}\ln(2\pi)\plus{}\frac{\zeta'(2)}{\zeta(2)}\right)$


function
geometry
rhombus
geometric transformation
rotation
perpendicular bisector

[quote="perfect_radio"][quote="ehsan2004"]Prove that $f(x)=0$ for all reals $x$.[/quote]

You wanted to say "$f(A)=0$ for all points $A$", right?[/quote]

excuse me, my meant was $f(x)\equiv 0$

Take $A \neq B$. Let $\ell$ be the perpendicular bisector of $AB$. Construct a rhombus $ACBD$, with $C,D \in \ell$ and $\measuredangle DAC = \measuredangle DBC = \dfrac{\pi}{3}$. This yields $f(A)+f(C)+f(D)=0=f(B)+f(C)+f(D)$, so $f(A)=f(B)=t$, $\forall A \neq B$.

Therefore $nt=0$, so $t=0$.

Have I done something wrong? It looks too good to be true  :blush: (because I used the condition given only for $n=3$)

[quote="enescu"]Actually, $n$ is fixed in the original statement.[/quote] Oops... sorry :( . do you know the solution for $n \geq 4$?

Yes. Let $A$ be an arbitrary point. Consider a regular $n-$gon $AA_{1}A_{2}\ldots A_{n-1}.$ Let $k$ be an integer, $0\leq k\leq n-1.$ A rotation with center $A$ of angle $\dfrac{2k\pi}{n}$ sends the polygon $AA_{1}A_{2}\ldots A_{n-1}$ to $A_{k0}A_{k1}\ldots A_{k,n-1},$ where $A_{k0}=A$ and $A_{ki}$ is the image of $A_{i}$, for all $i=1,2,\ldots,n-1.$

From the condition of the statement, we have
\[ \sum_{k=0}^{n-1} \sum_{i=0}^{n-1}{f(A_{ki})}=0.  \]
Observe that in the sum the number $f(A)$ appears $n$ times, therefore
\[ nf(A)+\sum_{k=0}^{n-1} \sum_{i=1}^{n-1}{f(A_{ki})}=0.  \]
On the other hand, we have
\[ \sum_{k=0}^{n-1} \sum_{i=1}^{n-1}{f(A_{ki})}=\sum_{i=1}^{n-1} \sum_{k=0}^{n-1}{f(A_{ki})}=0,  \]
since the polygons $A_{0i}A_{1i}\ldots A_{n-1,i}$ are all regular $n-$gons. From the two equalities above we deduce $f(A)=0,$ hence $f$ is the zero function.

We may assume \(n\) is even, since for \(n\) odd, the sum of the vertices in any \(2n\)-gon is zero.

Now let \(A_1\cdots A_n\) be a regular \(n\)-gon. For each \(i\) and \(j\), let \(M_{ij}\) be the midpoint of \(\overline{A_iA_j}\) (so in particular, \(M_iM_i=A_i\)), and let \(O\) be the center of the \(n\)-gon.

We know since \(M_{i1}M_{i2}\cdots M_{in}\) and \(M_{1,1+i}M_{2,2+i}\cdots M_{n,n+i}\) are regular \(n\)-gons that \begin{align*}     0=\sum_i\sum_jf(M_{ij})     =n\cdot f(O)+\sum_j\sum_{\substack{i<n\\ i\ne n/2}}f(M_{j,j+i})     &=n\cdot f(O) \end{align*}

[quote=ehsan2004]Let $ f: \mathbb{R}^2 \rightarrow \mathbb{R} $ be a function such that for every regular $ n $-gon $ A_1A_2 \ldots A_n $ we have $ f(A_1)+f(A_2)+\cdots +f(A_n)=0 $. Prove that $ f(x)=0 $ for all reals $ x $.[/quote]

The claim for just $n=4$:
[url]https://aops.com/community/p1703551[/url]


Let $ a$, $ b$, $ c$, $ d$ be positive real numbers such that $ abcd \equal{} 1$ and $ a \plus{} b \plus{} c \plus{} d > \dfrac{a}{b} \plus{} \dfrac{b}{c} \plus{} \dfrac{c}{d} \plus{} \dfrac{d}{a}$. Prove that
\[ a \plus{} b \plus{} c \plus{} d < \dfrac{b}{a} \plus{} \dfrac{c}{b} \plus{} \dfrac{d}{c} \plus{} \dfrac{a}{d}\]
[i]Proposed by Pavel Novotný, Slovakia[/i]

[quote="April"]Let $ a$, $ b$, $ c$, $ d$ be positive real numbers such that $ abcd \equal{} 1$ and $ a \plus{} b \plus{} c \plus{} d > \dfrac{a}{b} \plus{} \dfrac{b}{c} \plus{} \dfrac{c}{d} \plus{} \dfrac{d}{a}$. Prove that
\[ a \plus{} b \plus{} c \plus{} d < \dfrac{b}{a} \plus{} \dfrac{c}{b} \plus{} \dfrac{d}{c} \plus{} \dfrac{a}{d}\]
[/quote]
Very Easy !

We know that:

$ (a \plus{} b \plus{} c \plus{} d).(ab \plus{} bc \plus{} cd \plus{} da) > (\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{d} \plus{} \frac {d}{a}).(ab \plus{} bc \plus{} cd \plus{} da) \geq (a \plus{} b \plus{} c \plus{} d)^2$

It follows that $ ab \plus{} bc \plus{} cd \plus{} da > a \plus{} b \plus{} c \plus{} d$

It suffices to prove that:

$ \frac {b}{a} \plus{} \frac {c}{b} \plus{} \frac {d}{c} \plus{} \frac {a}{d} \geq ab \plus{} bc \plus{} cd \plus{} da$

But:

$ (\frac {b}{a} \plus{} \frac {c}{d}) \plus{} (\frac {c}{b} \plus{} \frac {d}{a}) \plus{} (\frac {d}{c} \plus{} \frac {a}{b}) \plus{} (\frac {a}{d} \plus{} \frac {b}{c}) \geq 2.(ab \plus{} bc \plus{} cd \plus{} da) > (ab \plus{} bc \plus{} cd \plus{} da) \plus{} (a \plus{} b \plus{} c \plus{} d) > (ab \plus{} bc \plus{} cd \plus{} da) \plus{} \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{d} \plus{} \frac {d}{a}$

and the result follows immediately.

[quote="April"]Let $ a$, $ b$, $ c$, $ d$ be positive real numbers such that $ abcd \equal{} 1$ and $ a \plus{} b \plus{} c \plus{} d > \dfrac{a}{b} \plus{} \dfrac{b}{c} \plus{} \dfrac{c}{d} \plus{} \dfrac{d}{a}$. Prove that
\[ a \plus{} b \plus{} c \plus{} d < \dfrac{b}{a} \plus{} \dfrac{c}{b} \plus{} \dfrac{d}{c} \plus{} \dfrac{a}{d}\]
[/quote]
It was also a problem in Russia Winter TST 2008. Russia_Man posted it but received no replies.
http://www.mathlinks.ro/viewtopic.php?search_id=1658696324&t=250987.
rachid's solution is very nice.

By Cauchy Schwarz Inequality, we have that
\[ a \plus{} b \plus{} c \plus{} d > \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{d} \plus{} \frac {d}{a} \equal{} \frac {a^2}{ab} \plus{} \frac {b^2}{bc} \plus{} \frac {c^2}{cd} \plus{} \frac {d^2}{da} \ge \frac {(a \plus{} b \plus{} c \plus{} d)^2}{ab \plus{} bc \plus{} cd \plus{} da}\]
so we have that
\[ ab \plus{} bc \plus{} cd \plus{} da > a \plus{} b \plus{} c \plus{} d.\]
Now, by Cauchy Schwarz Inequality, we also have that
\[ \left(\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{d} \plus{} \frac {d}{a}\right)\left(\frac {d}{c} \plus{} \frac {a}{d} \plus{} \frac {b}{a} \plus{} \frac {c}{b}\right) \ge \left(\sqrt {\frac {ad}{bc}} \plus{} \sqrt {\frac {ab}{cd}} \plus{} \sqrt {\frac {bc}{da}} \plus{} \sqrt {\frac {cd}{ab}}\right)^2 \equal{} (ad \plus{} ab \plus{} bc \plus{} cd)^2 > (a \plus{} b \plus{} c \plus{} d)^2\]
and since $ a \plus{} b \plus{} c \plus{} d > \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{d} \plus{} \frac {d}{a}$, then we must have $ a \plus{} b \plus{} c \plus{} d < \frac {b}{a} \plus{} \frac {c}{b} \plus{} \frac {d}{c} \plus{} \frac {a}{d}.$ QED

By AM-GM, $ 4a \equal{} 4\sqrt [4]{\frac {a}{d}\frac {a}{b}\frac {a}{b}\frac {b}{c}}\leq\frac {a}{d} \plus{} 2\frac {a}{b} \plus{} \frac {b}{c}$, and similarly for its cyclic permutations.  Adding all of them up,
\[ 4(a \plus{} b \plus{} c \plus{} d)\leq3\left(\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{d} \plus{} \frac {d}{a}\right) \plus{} \frac {b}{a} \plus{} \frac {c}{b} \plus{} \frac {d}{c} \plus{} \frac {a}{d} < 3(a \plus{} b \plus{} c \plus{} d) \plus{} \frac {b}{a} \plus{} \frac {c}{b} \plus{} \frac {d}{c} \plus{} \frac {a}{d}.\]

[quote="April"]Let $ a$, $ b$, $ c$, $ d$ be positive real numbers such that $ abcd \equal{} 1$ and $ a \plus{} b \plus{} c \plus{} d > \dfrac{a}{b} \plus{} \dfrac{b}{c} \plus{} \dfrac{c}{d} \plus{} \dfrac{d}{a}$. Prove that
\[ a \plus{} b \plus{} c \plus{} d < \dfrac{b}{a} \plus{} \dfrac{c}{b} \plus{} \dfrac{d}{c} \plus{} \dfrac{a}{d}\]
[i]Proposed by Slovakia[/i][/quote]
$ \frac{a}{b}\plus{}\frac{b}{a} \ge a\plus{}b \iff (a\minus{}b)^2(a\plus{}b) \ge 0$, and so on. So
$ \dfrac{a}{b} \plus{} \dfrac{b}{c} \plus{} \dfrac{c}{d} \plus{} \dfrac{d}{a} \plus{} \dfrac{b}{a} \plus{} \dfrac{c}{b} \plus{} \dfrac{d}{c} \plus{} \dfrac{a}{d} \ge 2(a\plus{}b\plus{}c\plus{}d)$ from which the problem follows.

[quote="Mathias_DK"][quote="April"]Let $ a$, $ b$, $ c$, $ d$ be positive real numbers such that $ abcd \equal{} 1$ and $ a \plus{} b \plus{} c \plus{} d > \dfrac{a}{b} \plus{} \dfrac{b}{c} \plus{} \dfrac{c}{d} \plus{} \dfrac{d}{a}$. Prove that
\[ a \plus{} b \plus{} c \plus{} d < \dfrac{b}{a} \plus{} \dfrac{c}{b} \plus{} \dfrac{d}{c} \plus{} \dfrac{a}{d}\]
[i]Proposed by Slovakia[/i][/quote]
$ \frac {a}{b} \plus{} \frac {b}{a} \ge a \plus{} b \iff (a \minus{} b)^2(a \plus{} b) \ge 0$, [/quote]
It 's wrong.Try $ a\equal{}1;b\equal{}2$
The solution of daniel 73 very nice :)

[quote="quykhtn-qa1"][quote="Mathias_DK"][quote="April"]Let $ a$, $ b$, $ c$, $ d$ be positive real numbers such that $ abcd \equal{} 1$ and $ a \plus{} b \plus{} c \plus{} d > \dfrac{a}{b} \plus{} \dfrac{b}{c} \plus{} \dfrac{c}{d} \plus{} \dfrac{d}{a}$. Prove that
\[ a \plus{} b \plus{} c \plus{} d < \dfrac{b}{a} \plus{} \dfrac{c}{b} \plus{} \dfrac{d}{c} \plus{} \dfrac{a}{d}\]
[i]Proposed by Slovakia[/i][/quote]
$ \frac {a}{b} \plus{} \frac {b}{a} \ge a \plus{} b \iff (a \minus{} b)^2(a \plus{} b) \ge 0$, [/quote]
It 's wrong.Try $ a \equal{} 1;b \equal{} 2$
The solution of daniel 73 very nice :)[/quote]
Oh yes, of course i'm wrong :blush: (The idea is the same though)

WLOG $ac\ge1$. Suppose for the sake of contradiction that $a+b+c+\frac{1}{abc}>\frac{a}{b}+\frac{b}{c}+abc^2+\frac{1}{a^2bc}$ and $a+b+c+\frac{1}{abc}\ge\frac{b}{a}+\frac{c}{b}+\frac{1}{abc^2}+a^2bc$. Expanding, we get that
\[(ac^3-c+1)(ab)^2-ac(a+c)(ab)+(a^3c-a+1)<0\]and
\[(a^3c-a+1)(bc)^2-ac(a+c)(bc)+(ac^3-c+1)\le0.\]Since $ac\ge1$, $ac^3-c+1\ge c^2-c+1>0$ and $a^3c-a+1\ge a^2-a+1>0$. Consider
\[f(x)=(ac^3-c+1)x^2-ac(a+c)x+(a^3c-a+1)=(ac^3-c+1)(x-r)(x-s),\]which must have real roots $r<s$ since $\lim_{x\to\infty}f(x)=\infty$ and $f(ab),f(1/bc)\le0$. Thus $r<ab<s$ and $r\le \frac{1}{bc}\le s$, whence
\[r^2 < \frac{a}{c} < s^2\implies f(\sqrt{a/c})<0,\]i.e. letting $u=a+c$ and $v=ac\ge1$ and noting that $u\ge2\sqrt{v}$,
\[2v>u(v^2-v^{3/2}+1)\ge2\sqrt{v}(v^2-v^{3/2}+1)\implies (v^{3/2}-1)(v^{1/2}-1)<0,\]a clear contradiction.

[quote="April"]Let $ a$, $ b$, $ c$, $ d$ be positive real numbers such that $ abcd = 1$ and $ a + b + c + d > \dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{d} + \dfrac{d}{a}$. Prove that
\[ a + b + c + d < \dfrac{b}{a} + \dfrac{c}{b} + \dfrac{d}{c} + \dfrac{a}{d}\]
[i]Proposed by Pavel Novotný, Slovakia[/i][/quote]

Suppose, for the sake of contradiction, that \[{a + b + c + d \ge \frac{b}{a} + \frac{c}{b} + \frac{d}{c} + \frac{a}{d}. \quad \text(1)}.\] Then, by multiplying the given inequality by $(1)^3$, it follows that \[\left(a + b + c + d\right)^4 > \left(\frac{a}{b} + \frac{b}{c} + \frac{c}{d} + \frac{d}{a}\right)\left(\frac{b}{a} + \frac{c}{b} + \frac{d}{c} + \frac{a}{d}\right)^3. \quad \text{(2)}\] However, this is false, since, by Holder's Inequality, we have that
\begin{align*}
\left(\frac{a}{b} + \frac{b}{c} + \frac{c}{d} + \frac{d}{a}\right)\left(\frac{b}{a} + \frac{c}{b} + \frac{d}{c} + \frac{a}{d}\right)^3 \\
&= \left(\frac{a}{b} + \frac{b}{c} + \frac{c}{d} + \frac{d}{a}\right)\left(\frac{d}{c} + \frac{a}{d} + \frac{b}{a} + \frac{c}{b}\right)\left(\frac{a}{d} + \frac{b}{a} + \frac{c}{b} + \frac{d}{c}\right)\left(\frac{a}{d} + \frac{b}{a} + \frac{c}{b} + \frac{d}{c}\right) \\
&\ge \left(\sqrt[4]{\frac{a^3d}{bcd^2}} + \sqrt[4]{\frac{b^3a}{cda^2}} + \sqrt[4]{\frac{c^3b}{dab^2}} + \sqrt[4]{\frac{d^3c}{abc^2}}\right)^4 \\
&= \left(a + b + c + d\right)^4.
\end{align*}
Therefore, we conclude that $(2)$ was false, and hence our assumption was flawed. Thus, we see that $\;$ $a + b + c + d < \frac{b}{a} + \frac{c}{b} + \frac{d}{c} + \frac{a}{d}$ $\;$ as desired. $\blacksquare$

Isn't this just holder:
$(\frac{b}{a} +...+\frac{a}{d})^3\cdot(\frac{a}{b}+...)(a+b+c+d) \geq (a+b+c+d)^6$
Which is by easy manipulation :$(\frac{b}{a}+...)>a+b+c+d$
PS. I wrote this on phone so thatsbwhy I was lazy about fractions

Assume not : $ a+b+c+d \geq \frac{b}{a}+\frac{c}{b}+\frac{d}{c}+\frac{a}{d} $
$ \Rightarrow a+b+c+d>\sqrt{(\frac{b}{a}+\frac{c}{b}+\frac{d}{c}+\frac{a}{d})(\frac{c}{d}+\frac{d}{a}+\frac{a}{b}+\frac{b}{c})} \geq (a+c)(b+d) $
On the other hands :
$ (a+b+c+d)(ab+bc+cd+da) \geq (ab+bc+cd+da)(\frac{b}{a}+\frac{c}{b}+\frac{d}{c}+\frac{a}{d}) \geq (a+b+c+d)^2 $
$ \Rightarrow ab+bc+cd+da \geq a+b+c+d \rightarrow\leftarrow $
done!

Solved with [b]william122[/b]  :love:

Use $\sum$ to denote cyclic sums. The problem follows from the following chain of inequalities:
$$\sqrt{\sum \frac{a}{b}\sum \frac{d}{c}} \ge \sum \sqrt{\frac{ad}{bc}} = \sum ad \ge \frac{(\sum a)^2}{\sum \frac{d}{a}} > \sum a.$$

Define the following quantities :=

$$S_1 = \frac ab +\frac bc +\frac cd +\frac da$$$$S_2 = \frac ba +\frac cb +\frac dc +\frac ad$$$$W =a+b+c+d$$.

We claim that $3S_1+S_2 \geq 4W$ .
This definitely finishes as then $S_2 \geq W + \underbrace{3(W-S_1)}_{>0} >W$

To prove the inequality we note that$$2 \cdot \frac ab + \frac bc + \frac ad \overset{\text{AM-GM}}{\geq} 4\sqrt [4]{\frac {a}{d}\cdot \frac {a^2}{b^2} \cdot \frac {b}{c}} \geq 4a$$.

Cyclically summing up we are done . $\blacksquare$

[hide=Motivation]So starting out , the problem looks very AM-GM able . First the presence of too many terms like $\frac ab$  made me keep $a,b,c,d$ as it is , without substituting $a=\frac xy$ etc . Next I defined $S_1$ , $S_2$ and $W$ as above . I always felt that we should be able to prove something like $xS_1+yS_2 \geq (x+y)W$ . Now since this is a 4-variable inequality , I try setting $x+y=4$ . So we check the cases $(x,y)=(1,3),(2,2),(3,1)$ and eventually the last one works out  :P

[/hide]

Posting for storage. Solved with [b]eisirrational[/b], [b]goodbear[/b], and [b]nukelauncher[/b].

First note $\sum a > \sum a/b \geq (a+b+c+d)^2/(ab+bc+cd+da)$ or $a+b+c+d < ab+bc+cd+da$. Then since
\[(\sum a/b)(\sum d/c) \geq (ad+ba+cb+dc)^2 > (a+b+c+d)^2,\]
the two factors on the LHS cannot both be $\leq a+b+c+d$.

The idea is to show
\[(a+b+c+d)^2\le (\tfrac ab+\tfrac bc+\tfrac cd+\tfrac da)(\tfrac ba+\tfrac cb+\tfrac dc+\tfrac ad).\]
Expand
\[(a+b+c+d)^2=\frac{(a+b+c+d)^2}{\sqrt{abcd}}=\sum_{\mbox{cyc}}\sqrt{\frac{a^3}{bcd}}+\frac{1}{2}\sum_{\mbox{sym}}\sqrt{\frac{ab}{cd}}.\]
Expand
\[(\tfrac ab+\tfrac bc+\tfrac cd+\tfrac da)(\tfrac ba+\tfrac cb+\tfrac dc+\tfrac ad)=4+\sum_{\mbox{cyc}} (\tfrac{a^2}{bd}+\tfrac{bd}{a^2})+\sum_{\mbox{cyc}}\frac{ab}{cd}.\]
Write
\[1+\frac{ab}{cd}\ge 2\sqrt{\frac{ab}{cd}}\]
and symmetric results. Then it suffices to show
\[\sum_{\mbox{cyc}} (\tfrac{a^2}{bd}+\tfrac{bd}{a^2})\ge \sum_{\mbox{cyc}}\sqrt{\frac{a^3}{bcd}}+2\sqrt{\frac{ac}{bd}}+2\sqrt{\frac{bd}{ac}}.\]
Observe that
\[\sum_{\mbox{cyc}}\sqrt{\frac{a^3}{bcd}}+2\sqrt{\frac{ac}{bd}}+2\sqrt{\frac{bd}{ac}}\le \frac{1}{2}\sum_{\mbox{cyc}}(\tfrac{a^2}{bd}+\tfrac{a}{c})+2\sqrt{\frac{ac}{bd}}+2\sqrt{\frac{bd}{ac}}.\]
Additionally, observe the AM-GM powered results of
\begin{align*}
    \sum_{\mbox{cyc}}\left(\frac{a^2}{bd}+\frac{bd}{c^2}\right)&\ge 2\sum_{\mbox{cyc}}\frac{a}{c}\\
    \sum_{\mbox{cyc}}\frac{bd}{a^2}&\ge 2\sqrt{\frac{bd}{ac}}+2\sqrt{\frac{ac}{bd}}\\
    \sum_{\mbox{cyc}}\frac{a^2}{bd}&\ge 2\sqrt{\frac{bd}{ac}}+2\sqrt{\frac{ac}{bd}}.\\
\end{align*}
Sum the first inequality multiplied by $\tfrac 14$, the second inequality multiplied by $\tfrac 34$, and the third inequality multiplied by $\tfrac 14$ to conclude the result.

It is given to us that $$a+b+c+d> \frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}.$$
Assume for the sake of contradiction that $$a+b+c+d\geq \frac{b}{a}+\frac{c}{b}+\frac{d}{c}+\frac{a}{d}.$$ Note that by our assumption$,$ $$(a+b+c+d)^2\bigg(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\bigg)>\bigg(\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}\bigg)\bigg(\frac{a}{d}+\frac{b}{a}+\frac{c}{b}+\frac{d}{c}\bigg)\bigg(\frac{1}{c}+\frac{1}{d}+\frac{1}{a}+\frac{1}{b}\bigg)=:P.$$

But by $\text{Hölder's inequality},$ $$P\geq \bigg(\sqrt[3]{\frac{a^2}{bdc}}+\sqrt[3]{\frac{b^2}{cad}}+\sqrt[3]{\frac{c^2}{dba}}+\sqrt[3]{\frac{d^2}{acb}}\bigg)^3=(a+b+c+d)^3,$$ so it follows that $$\fbox{$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}> a+b+c+d.$}$$

This time notice that $$(a+b+c+d)^3>(a+b+c+d)\bigg(\frac{b}{c}+\frac{c}{d}+\frac{d}{a}+\frac{a}{b}\bigg)\bigg(\frac{d}{c}+\frac{a}{d}+\frac{b}{a}+\frac{c}{b}\bigg)=:Q.$$

With the use of $\text{Hölder's inequality},$ $$Q\geq \bigg(\sqrt[3]{\frac{abd}{c^2}}+\sqrt[3]{\frac{bca}{d^2}}+\sqrt[3]{\frac{cdb}{a^2}}+\sqrt[3]{\frac{dac}{b^2}}\bigg)^3=\bigg(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\bigg)^3,$$ from where it follows that $$\fbox{$a+b+c+d>\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}$},$$ which was the desired contradiction$.$ $\blacksquare$

By Cauchy-Schwarz Inequality, \[(ad+ab+bc+cd)\left(\frac{a}{d}+\frac{b}{a}+\frac{c}{b}+\frac{d}{c}\right)\geqslant (a+b+c+d)^2\] so $ad+ab+bc+cd\geqslant \frac{(a+b+c+d)^2}{\frac{a}{d}+\frac{b}{a}+\frac{c}{b}+\frac{d}{c}}.$ Then
\begin{align*}
(a+b+c+d)\left(\frac{b}{a}+\frac{c}{b}+\frac{d}{c}+\frac{a}{d}\right)>& \left(\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}\right)\left(\frac{b}{a}+\frac{c}{b}+\frac{d}{c}+\frac{a}{d}\right) \\ \geqslant& \left(\sqrt{\frac{ad}{bc}}+\sqrt{\frac{ab}{cd}}+\sqrt{\frac{bc}{ad}}+\sqrt{\frac{cd}{ab}}\right)^2 \\ =& (ad+ab+bc+cd)^2 \\ \geqslant& \frac{(a+b+c+d)^4}{\left(\frac{a}{d}+\frac{b}{a}+\frac{c}{b}+\frac{d}{c}\right)^2}.
\end{align*}
Thus, \[ a + b + c + d < \dfrac{b}{a} + \dfrac{c}{b} + \dfrac{d}{c} + \dfrac{a}{d}. \quad \blacksquare\]

Solved with Max Lu and Ryan Yang.

First note by Cauchy-Schwarz that \begin{align*}     (a+b+c+d)(ab+bc+cd+da)     &>\left(\frac ab+\frac bc+\frac cd+\frac da\right)(ab+bc+cd+da)\\     &\ge(a+b+c+d)^2, \end{align*} implying that \(ab+bc+cd+da>a+b+c+d\).

It follows, again by Cauchy-Schwarz, that \begin{align*}     (a+b+c+d)     \left(\frac ba+\frac cb+\frac dc+\frac ad\right)     &>     \left(\frac ab+\frac bc+\frac cd+\frac da\right)     \left(\frac dc+\frac ad+\frac ba+\frac cb\right)\\     &\ge\left(\sqrt{\frac{da}{bc}}+\sqrt{\frac{ab}{cd}}+\sqrt{\frac{bc}{da}}+\sqrt{\frac{cd}{ab}}\right)^2\\     &=(da+ab+bc+cd)^2     \ge(a+b+c+d)^2, \end{align*} implying \(\frac ba+\frac cb+\frac dc+\frac ad>a+b+c+d\) as desired.

[quote=Stephen]
$$(\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a})+(\frac{b}{a}+\frac{c}{b}+\frac{d}{c}+\frac{a}{d})
=(a+c)(\frac{1}{b}+\frac{1}{d})+(b+d)(\frac{1}{a}+\frac{1}{c})
$$$$=(a+c)(b+d)(\frac{1}{ac}+\frac{1}{bd})
=(a+c)(b+d)(ac+bd)=(a+b+c+d)(\frac{1}{\frac{1}{a+c}+\frac{1}{b+d}})(ac+bd)
$$$$\ge (a+b+c+d)(\frac{1}{\frac{1}{2\sqrt{ac}}+\frac{1}{2\sqrt{bd}}})(ac+bd)
\ge 2(a+b+c+d)$$ [/quote]
[url]https://artofproblemsolving.com/community/c6h1280948p23908388[/url]
Let $a,b,c,d>0.$Prove that
$$ \dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{d} + \dfrac{d}{a} + \dfrac{b}{a} + \dfrac{c}{b} + \dfrac{d}{c} + \dfrac{a}{d} \geq 2\sqrt{\left( \dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{d} + \dfrac{d}{a}\right )\left( \dfrac{b}{a} + \dfrac{c}{b} + \dfrac{d}{c} + \dfrac{a}{d}\right)}$$
$$\geq \sqrt{2\left( \dfrac{1}{ac} + \dfrac{1}{bd}\right )}\\\ (a+b+c+d)\geq \frac{2(a+b+c+d)}{\sqrt[4]{abcd}}$$
[size=50](SXTX   (11)2021)[/size]

oops

Suppose that $a+b+c+d>\frac ab+\frac bc+\frac cd+\frac da$ and $a+b+c+d\geq\frac ba+\frac cb+\frac dc+\frac ad$. Let $a=\frac wx$, $b=\frac xy$, $c=\frac yz$, and $d=\frac zw$. Then, we have
\begin{align*}
\frac wx+\frac xy+\frac yz+\frac zw&>wy\left(\frac1{x^2}+\frac1{z^2}\right)+xz\left(\frac1{y^2}+\frac1{w^2}\right)\\
\frac wx+\frac xy+\frac yz+\frac zw&\geq\frac{x^2+z^2}{wy}+\frac{w^2+y^2}{xz},
\end{align*}
so
\begin{align*}
w^2yz+x^2zw+y^2wx+z^2xy&>\frac{w^2y^2z}x+\frac{w^2xy^2}z+\frac{wx^2z^2}y+\frac{x^2yz^2}w\\
w^2yz+x^2zw+y^2wx+z^2xy&\geq x^3z+xz^3+w^3y+wy^3.
\end{align*}

However, by the AM-GM Inequality, we have
\begin{align*}
&w^2yz+x^2zw+y^2wx+z^2xy\\
>&\frac34(x^3z+xz^3+w^3y+wy^3)+\frac14\left(\frac{w^2y^2z}x+\frac{w^2xy^2}z+\frac{wx^2z^2}y+\frac{x^2yz^2}w\right)\\
=&\left(\frac12w^3y+\frac14z^3x+\frac14\frac{y^2zw^2}x\right)+\left(\frac12x^3z+\frac14w^3y+\frac14\frac{z^2wx^2}y\right)+\left(\frac12y^3w+\frac14x^3z+\frac14\frac{w^2xy^2}z\right)+\left(\frac12z^3x+\frac14y^3w+\frac14\frac{x^2yz^2}w\right)\\
\geq&w^2yz+x^2zw+y^2wx+z^2xy.
\end{align*}

This is a contradiction. Therefore, it is impossible for $a+b+c+d\geq\frac ba+\frac cb+\frac dc+\frac ad$, so we must have $a+b+c+d<\frac ba+\frac cb+\frac dc+\frac ad$.

Note that \[(a+b+c+d)(ab+bc+cd+da)> \left(\dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{d} + \dfrac{d}{a}\right)(ab+bc+cd+da)\ge (a+b+c+d)^2\] by Cauchy-Schwarz. Thus, $ab+bc+cd+da>a+b+c+d.$ It remains to show that $ab+bc+cd+da\le \dfrac{b}{a} + \dfrac{c}{b} + \dfrac{d}{c} + \dfrac{a}{d}.$ Now, note that \[\left(\dfrac{b}{a} + \dfrac{c}{b} + \dfrac{d}{c} + \dfrac{a}{d}\right)\left(\dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{d} + \dfrac{d}{a}\right)\ge \left(\sqrt{\frac{bc}{ad}}+\sqrt{\frac{cd}{ab}}+\sqrt{\frac{da}{bc}}+\sqrt{\frac{ab}{cd}}\right)^2\]
We have \[\left(\sqrt{\frac{bc}{ad}}+\sqrt{\frac{cd}{ab}}+\sqrt{\frac{da}{bc}}+\sqrt{\frac{ab}{cd}}\right)^2=\left(\frac{bc+da}{\sqrt{abcd}}+\frac{cd+ab}{\sqrt{abcd}}\right)^2=(ab+bc+cd+da)^2>(a+b+c+d)^2.\] The result, thus follows.

By C-S: $$\left(\sum_{\text{cyc}} \frac{a}{b}\right )\left (\sum_{\text{cyc}} \frac{b}{a}\right )\geqslant \left (\sum_{\text{cyc}} \sqrt{\frac{ad}{bc}}\right )  \left (\sum_{\text{cyc}} \sqrt{\frac{ad}{bc}}\right )= \left (\sum_{\text{cyc}} ad\right ) \left (\sum_{\text{cyc}} ad\right )\geqslant \left (\frac{\left (a+b+c+d\right )^2}{\sum ab^{-1}} \right)^2 > \left (\sum_{\text{cyc}} a\right)  \left (\sum_{\text{cyc}} a\right)$$

Note the inequality
$$\frac{3}{8}\cdot \frac{a}{b}+\frac{3}{8}\cdot \frac{a}{d}+\frac{1}{8}\cdot \frac{b}{c}+\frac{1}{8}\cdot \frac{d}{c} \geq \frac{a}{\sqrt[4]{abcd}}=a$$
by weighted AM-GM. Summing cyclically and doubling, it follows that
$$\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}+\frac{b}{a}+\frac{c}{b}+\frac{d}{c}+\frac{a}{d}\geq 2(a+b+c+d),$$
so if $\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}<a+b+c+d$ we must have $\frac{b}{a}+\frac{c}{b}+\frac{d}{c}+\frac{a}{d}>a+b+c+d$. $\blacksquare$

Got a hint to use weighted AM-GM why did I not think of that i guess im rusty :(

We have $$4a=4\sqrt[4]{\frac{a^4}{abcd}}=4\sqrt[4]{\frac{a^3}{bcd}}\le\frac ad+\frac bc+2\frac ab\implies \sum_{cyc}4a\le\sum_{cyc}\frac ad+2\frac ab+\frac bc=\sum_{cyc}3\frac ab+\frac ba<\sum_{cyc}3a+\frac ba,$$ which readily implies the result. $\blacksquare$

Note that
\[ \frac{3}{8} \cdot \frac{a}{b} + \frac{3}{8}\cdot \frac{a}{d} + \frac{1}{8}\cdot \frac{d}{c} + \frac{1}{8}\cdot \frac{b}{c} \geq a^{\frac{3}{4}}b^{-\frac{1}{4}}c^{-\frac{1}{4}}d^{-\frac{1}{4}} = a \]
So summing cyclically yields that
\[ \frac{1}{2}\left( \frac{b}{a} + \frac{c}{b} + \frac{d}{c} + \frac{a}{d}\right) + \frac{1}{2}\left( \frac{a}{b} + \frac{b}{c} + \frac{c}{d} + \frac{d}{a}\right) \geq a + b + c + d \]
Using the given condition implies that $\frac{b}{a} + \frac{c}{b} + \frac{d}{c} + \frac{a}{d} \geq a + b +c + d$. $\blacksquare$

[b]Generalization 1[/b]
Let $a_{1},a_{2},\cdots,a_{n}$ be positive reals such that $\prod{a_1}=1$ and
$$\sum_{cyc}{a_{1}}> \sum_{cyc-j}{\dfrac{a_{j}}{a_{j+1}}}$$
equalities hold. Then prove that

$$\sum_{cyc}{a_1}< \sum_{cyc-j}{\dfrac{a_{j}^2a_{j+1}a_{j+2}}{\prod{a_1}}}$$

Let $S_1 = \dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{d} + \dfrac{d}{a}$ and $S_2 = \dfrac{b}{a} + \dfrac{c}{b} + \dfrac{d}{c} + \dfrac{a}{d}$

Note that $S_1 = \frac{ad+bc}{bd}+\frac{ab+cd}{ac}$ and $S_2 = \frac{bc+ad}{ac}+\frac{cd+ab}{bd}$. So, we get that $$S_1+S_2 = (b+d)(a+c)\left(\frac{1}{ac}+\frac{1}{bd}\right) \geq 2(b+d)(a+c)$$

Now, by Cauchy Schwarz inequality,
$$\left(\sqrt{\frac{a}{c}}+\sqrt{\frac{b^2}{ac}}+\sqrt{\frac{c}{a}}+\sqrt{\frac{d^2}{ac}}\right)^2 \leq \left(\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}\right)\left(\frac{b}{c}+\frac{b}{a}+\frac{d}{a}+\frac{d}{c}\right) = S_1\frac{(b+d)(a+c)}{ac}$$
multiplying by $ac$ we get that
$$(a+b+c+d)^2 \leq (b+d)(a+c)S_1$$
Similarly, $(a+b+c+d)^2 \leq (b+d)(a+c)S_2$, and taking the mean of the $2$ inequalities we get that
$$(a+b+c+d)^2 \leq (b+d)(a+c)\left(\frac{S_1+S_2}{2}\right) \leq \left(\frac{S_1+S_2}{2}\right)^2$$

implying that if $S_1 < a+b+c+d$ then $S_2 > a+b+c+d$

If we show $2(a+b+c+d)\le \frac ab+\frac bc+\frac cd+\frac da+\frac ba+\frac cb+\frac dc+\frac ad$ we would be done. This rearranges to $(a+c)\left(2-\frac1b-\frac1d\right)+(b+d)\left(2-\frac1a-\frac1c\right)\le 0.$ Dividing by $(a+c)(b+d)$ turns this into $\frac2{b+d}+\frac2{a+c}\le\frac1{ac}+\frac1{bd}.$ However notice that replacing $a,c$ by $\sqrt{ac},\sqrt{ac}$ increases the left side so assuming $a=c,b=d$ makes it $\frac1{a^2}+\frac1{b^2}\ge\frac1a+\frac1b$ for positive reals $ab=1.$ Squaring and homogenizing this becomes $\frac b{a^3}+\frac a{b^3}\ge\frac1{b^2}+\frac1{a^2}.$ This following by summing AM-GM $\frac34\frac b{a^3}+\frac14\frac a{b^3}\ge\sqrt[4]{\left(\frac b{a^3}\right)^3\frac a{b^3}}=\frac1{b^2},\frac14\frac b{a^3}+\frac34\frac a{b^3}\ge\sqrt[4]{\frac b{a^3}\left(\frac a{b^3}\right)^3}=\frac1{a^2}.$


Using projective transformations prove the Pascal theorem (also find where the Pascal line intersects the circle).

so many times in so many books... ;)
[b]$A,B,C,D,E,F$ are $6$ points on a circle prove that the intersections between lines $AB,DE$ and $EF,BC$ and $CD,FA$ are collinear[/b]
[b][u]solution:[/b][/u]
determine points $K,L,M,M',G,H$ such that:
$K=(DE,AB)$
$L=(EF,BC)$
$M=(FA,CD)$
$M'=(KL,CD)$
$G=(CD,AB)$
$H=(DE,BC)$
now we have to prove that $M,M'$ are logic
the projection from $CD$ from $A$ to circle $S$ transformates points $G,C,D,M$ into $B,C,D,F$
and the projection from these points from $E$ to line $BC$ transformates them into $B,C,H,L$
and the projection from these points from $K$ to line $CD$ transformates them into $G,C,D,M'$
and from a famous problem we get that $M,M'$ are logic...
got it?? :D

note that in the above solution $M=(AB,CD)$ means that point $M$ is the intersection between lines $AB,CD$


Let $f(x)$ be a polynomial with integer coefficients, satisfying $f(0)=11$, $f(x_{1})=f(x_{2})=......=f(x_{n})= 2002$, for some distinct integers $x_{1},x_{2},.....,x_{n}$. Find the maximal value of $n$.

Do you have to double the number of factors to include for negative integers?

[quote="BanishedTraitor"][quote="t0rajir0u"][hide="Solution"] $a-b | f(a)-f(b) \implies x_{i}| (2002-11) = 1991, i = 1, 2, ... n$.  We have $1991 = 11 \cdot 181$ which has $\boxed{8}$ distinct integral factors. [/hide][/quote]

Look more closely at your factorisation.[/quote]

I fail to see the problem.

Well your answer is not correct. And when you do find a maximum value for $n$ you should provide an example just to make sure the maximum works.

[hide="Outline Of What I Did"] Let $g(x)= f(x)-2002$ and analyse the case when $x=0$. Then notice for all values of $x$ such that $f(x)=2002$, these values become the root of $g(x)$.[/hide]

Oh.  Point taken.

[hide="Then..."] $g(x) = f(x) (x-x_{1})(x-x_{2})... (x-x_{n})$
$g(0) = f(0) (-x_{1})(-x_{2})...(-x_{n}) =-1991$

We want to write $-1991$ as the product of a maximal number of distinct integers.  Well,

$-1991 = 1 \cdot-1 \cdot 11 \cdot 181$

Is the best we can do, so the maximal value of $n$ is $\boxed{4}$. [/hide]