calculus
inequalities
geometry
induction
number theory
IMO

Hi all!
I am particularly interested in how different country conduct their IMO training (and some other issue related to that one) and select their 6 students for the IMO. So please anybody here tell their story about their training program. Especialy I want to know these following points:

1) How close the relation between the math curriculum in high school related to the IMO materials.

2) How to decide 6 people to go to the IMO (describe all the previous preselection).

3) How long and how many time the training taken place ( and I also want to know how to reconciliate this with the obligation as a high school student to attend the school).

4) What is the public (student) opinion in your country about math and how good the average student doing in math.

5) Does teacher (in high school) is capable enough to help you to understand the IMO material.

Maybe that points that I have in my mind right now that I want to discuss. If anyone want to give a comment please address from which country you are and if there are another points that you want to say (out of the 5 points above) please elaborate.

Thanks for your time.

[quote="shobber"]In China, there is a math contest in the middle of September, then the top 6 in every provinces. They can go to CMO, which will take place in the beginning of the next year. Then top students go to TST, there after rounds of tests, 6 of the top are chosen to IMO.[/quote]

Thanks for the post shobber. Can you tell me if there is any training before the TST and for how long.

I'm from Belgium. In Belgium, you have to distinguish the Nothern region (Flanders, Dutch speaking) and the Southern region (Wallony, French speaking). Competition and most trianings are separate, and each region may sends three contestants to the IMO.

I am from Flanders. I will answer the questions for Flanders.

The regional competition is the VWO (Vlaamse Wiskunde Olympiade / Flanders Mathematical Olympiad, cf. [url=http://vwo.be]vwo.be[/url]). The firsts round is multiple choice and is organized in the local schools. The best 10% are selected for the second round, again multiple choice and this time organized per province. The final (with $\pm$ 80 participants) consists of four open questions which have to be answered in three hours.
Each year, the most impressing students ($\pm$ 15) of that year and the past years are selected for a training weekend in March. IMO selecting is based on both the final of the VWO and the performances during this trainig weekend (the IMO participants are always participants of the training weekend).
For the selected IMO participants, there are two more training weekends, one of them together with the participants from Wallony.

Trainings are during the weekend, so there is no problem with school.
Most IMO-participants also trained themselves at home to achieve the selection.

For Olympiad problems, you need to learn certain 'tricks' you don't see at school, but this isn't too much, since the emphasis is on logical thinking, not on knowlegde. Also, much thing you learn at school, for example calculus, are not asked on Olympiads.

Belgium is known for it's excellent math schooling, but scoring good at maths at school doesn't neccesariliy mean that you are good at Olympiads too. Though Belgium is at the world top for math schooling, on the IMO-ranking we are usually about 40th (IMO ranking is strongly influenced by training, and Belgium has little training when you compair with other countries, Flanders has less training than Wallony BTW).

The TST is held in a camp which will last for around half a month.

In the UK, high school mathematics relates little to IMO mathematics - some things are not taught, such as useful inequalities (but I suspect that's the same everywhere), but number theory is not taught at all and geometry education is highly deficient.

Some mathematical competitions bear only an indirect relation to IMO selection. The Intermediate Maths Challenge for people in years 10 and 11 (grades 9 and 10) is held each year (I forget when). It's a multiple choice test; the top-scoring contestants are invited to take an olympiad type test, the IMOK (International Mathematical Olympiad and Kangaroo or something like that; don't ask me why) with five very easy problems. The 40 top scorers are invited to the National Maths Summer School (NMSS) held at Birmingham University in July. This camp is staffed partly by former IMO contestants and an eye is kept out for IMO potential.

In September, around 14 people who have shown promise at the NMSS or in various tests are invited to a training camp at Bath University. Some of these are invited into the IMO squad, filling gaps left by year 13s (grade 12s) leaving for university. The IMO squad consists of about 20 to 25 people, and barring occasional surprises contains everyone who will make the final team. They are each assigned a "mentor" (normally a past IMO contestant) and are given both correspondence work and two sit-at-home IMO-style exams per month.

The first round of the IMO selection, known as the Senior Maths Challenge, is held in November. It's a multiple choice paper with 25 problems, and you get penalised for answering wrongly. The top 800 or so students are invited to compete in the British Mathematical Olympiad Round 1 (BMO1).

This is a 3.5 hour exam with 5 (still rather easy) Olympiad problems. The top 100 (taking into account both BMO1 score and other factors) are invited to take the British Mathematical Olympiad Round 2 (BMO2) in January/February. Before this, however, for the past few years there has been a joint training camp with Hungary, held in Hungary over the New Year.

The BMO2, also 3.5 hours long, contains 4 easy-ish Olympiad problems. After this, the 20 people viewed as most likely to make the IMO team are invited to a training camp at Trinty College, Cambridge in April. During this camp, two TSTs (the First Selection Tests, FSTs) are held. At the end of the camp, eight or nine students, who will form the team and reserves, are chosen. These students are given an intensive correspondence course throughout the rest of April and May. In late May/early June, they attend a final training camp at Oundle School where the NSTs (Next Selection Tests) are held and the team is chosen.

In the week before the IMO itself, the team has a camp in the same timezone as the IMO to acclimatise and for last-minute training.

So... that was long. And took me a long time to write. But I had nothing else to do, so whatever.

I'm from Austria (please do not mix up with Australia). Here you usually have, in order to participate on the mathematical olympiad, attend a special course where you learn the basics of mathematical olympiad like calculating with congruences, induction, mean-ineq. etc. and of course you train a lot (the mathematical knowledge taught at school is very far from beeing sufficient for even half a point at the IMO). Those courses are offered at several schools and every pupil can attend them. They take about 1:40 a week (starting at the beginning of september).

Usually in April a course competition takes place in every course.
The better half of each course ascends to the "Regional Competition" (4 problems, 4 hours) which takes place at the end of April or the beginning of May (altogether there are 3 regions in Austria).

The best of each region then ascend to the National Competition (altogether about 40 participants). In order to prepare the contestants for the (for Austrian's math) 'high level' of this competition a two-week-long preparatory camp takes place. After a week of training, a first-round-competition (4 problems, 4.5 hours) takes place which selects half of the participants for another week of preparation and finally for the TST (2 days; 3 problems and 4.5 hours each day).
Usually 1st - 6th go to the IMO and 7th - 12th participate on a bilateral competition with Poland which includes a single competition and a team competition.

A week before the IMO another 4-day preparation for the IMO-Team takes place.

Thanks a lot for sharing the information shobber, stijn, saul and Yimin Ge.

Anybody else???

In malaysia, not even the top 3 are selected. Don't ask me why, I do not know.

in Hong Kong, we have a preliminary contest in June, then there are three training periods (one lesson per weekend), in the end of each period we have some test. The first one is in late August, the second one is in December (we call it CHKMO), the last one is APMO. Then they have some weighting and have the total score and the top 6 can go to imo. Last year a geometry test was added in October but I'm not sure if this test will be held anymore.

[quote="mecrazywong"][quote="shobber"]In China, there is a math contest in the middle of September, then the top 6 in every provinces. They can go to CMO, which will take place in the beginning of the next year.[/quote]
Actually not necessarily six ;)[/quote]
Really? Well, there are only six selected from Beijing...

in mexico our selection is divided by states first each state meakes a number of tests varying from multiple option questions to normal olympiad standard tests for example in my sate (yucatan where the imo was held) we have an 4 problem a day test wich we have 4 hours and a half to solve. then the fisrt 15 are selected and we have a six month training where the top 6 are selected to go to nationals . in nationals we take a 3 problem a day test with 4 hours and a half to solve there the top 15 are selected and every last week of the month we tarin ans take tsts until the top 6 are selected and thats it.  :D


projective geometry
geometry unsolved
geometry

let (C) be a circle and AB be the diameter of it. point H is on AB. points C,D,E,F are on the circle , in wich,:

angle CHA=angle FHB         and           angle CHD=angle FHE.

let CD and FE meet each other in point G. prove that: GH is prependicular to AB.

Let $C',F'$ be points such that $AB$ is the orthogonal bisector of $CC'$ and $FF'$. by assumption that angles $CHA=FHB$ we conclude that points $C,H,F'$ and $C',H,F$ are colliniar.let $X=CF \cap {AB}$ then we have lines $CF,c'F'$ through $X$ so $X$ is the inverse point of $H$ with respect to the circle. In this way we conclude that $DE$ passes through $X$ too. now $G$ is the intersection of $CD,EF$ so $G$ is on the polar line of $X$. so $GH$ is the polar of $X$ so it's perpendicular to $PO=AB$.

nice solution ;) ali.

according to http://mathlinks.ro/forum/viewtopic.php?t=51865 we have :

let FH intersect (C) in point F'.and let CH intersect (C) in point C'.its clear that FC and F'C' intersect each other in a

point K,on line AB.then the polar of K to (C) ,will goes through H.

let EH intersect (C) in point E'.and let DH intersect (C) in point D'.again its clear why they intersect each other on line

AB.lets name it K'.again we have :the polar of K' to (C)will goes through H.it means that K' should be K,and FC and

ED and  AB are colinear.then in convex and cyclic quadrilaterall DECF,we have : the polar of point K to (C),should

go through G.and it means that GH is prependicular to AB.

It is confusedly. For example, $H\in CF\cap DE\Longrightarrow \emptyset$ ([u]falsely[/u])

I think that the points $C,D,E,F$ belong to the same semiplane w.r.t. the line AB.

wow ,levi ,i couldnt get u sentences.its because of my owfully english.could u plz make them more clear?

$H\in (AB)\cap CF\cap DE,\ G\in CD \cap EF$ and $A,C,D,B,F,E$ in this order on circle $(C)\Longrightarrow \ FALSELY$.

u know,H is in the interior of circle ,on AB,in wich CHA=FHB    and     CHD=FHE.


search
real analysis
real analysis unsolved

$ (x_n)$-sequence of nonzero reals, $ x_n\to\infty$,
there exists $ a\in\mathbb{R}$ s.t. $ x_n \plus{} ax_{n \minus{} 1} \plus{} x_{n \minus{} 2} \equal{} 0$ for all $ n\ge 2$.

Let $ \lambda\in ( \minus{} 1,1)$ be the solution of  $ \lambda^2 \plus{} a\lambda \plus{} 1 \equal{} 0$.

Compute $ \sum_{n \equal{} 0}^{\infty}\frac {1}{x_nx_{n \plus{} 1}}$

Well, from the characteristic equation:

$ r^2\plus{}a r\plus{}1\equal{}0$

we get:

$ r_{1,2}\equal{}\frac{1}{2}\left(\pm \sqrt{a^2\minus{}4}\minus{}a\right)$

So, we have:

$ x_n\equal{}\frac{A}{2^n}\left(\sqrt{a^2\minus{}4}\minus{}a\right)^n\plus{}\frac{B}{(\minus{}2)^n}\left(\sqrt{a^2\minus{}4}\plus{}a\right)^n$

Now, we can write:

$ x_n x_{n\plus{}1}\equal{}\left(A^2\minus{}B^2\right)\frac{\left(\sqrt{a^2\minus{}4}\minus{}a\right)^{2n\plus{}1}}{2^{2n\plus{}1}}$

So we have:

$ \frac{1}{A^2\minus{}B^2} \sum^\infty_{n\equal{}0}\frac{2^{2n\plus{}1}}{\left(\sqrt{a^2\minus{}4}\minus{}a\right)^{2n\plus{}1}}$

That can easily be evaluated (the $ \lambda \in (\minus{}1,1)$ part ensures that it converges) so that we have:

$ \sum ^\infty_{n\equal{}0}\frac{1}{x_n x_{n\plus{}1}}\equal{}\frac{1}{\sqrt{a^2\minus{}4}\left(A^2\minus{}B^2\right)}$

As for A and B, we could try to evaluate them form initial values of $ x_n$ if we knew some.

[quote="milin"]
So, we have:

$ x_n \equal{} \frac {A}{2^n}\left(\sqrt {a^2 \minus{} 4} \minus{} a\right)^n \plus{} \frac {B}{( \minus{} 2)^n}\left(\sqrt {a^2 \minus{} 4} \plus{} a\right)^n$

[/quote]

Could you explain what's going on here?

It's the solution of a linear recursion using the characteristic equation.  You may be familiar with it from the derivation of Binet's formula for the Fibonacci numbers.  Try a forum search, or go read the first chapter or two of generatingfunctionology (free on Herb Wilf's website).

[quote="milin"]
Now, we can write:

$ x_n x_{n \plus{} 1} \equal{} \left(A^2 \minus{} B^2\right)\frac {\left(\sqrt {a^2 \minus{} 4} \minus{} a\right)^{2n \plus{} 1}}{2^{2n \plus{} 1}}$
[/quote]

are you sure this equality holds ?
when substitute $ x_n$ and $ x_{n\plus{}1}$ from your formula it isn't the same :(

now I know the answer to this task: it is $ \frac{1}{x_0(x_1\minus{}\lambda x_0)}$


geometry
rectangle
trigonometry
geometry solved

A point X lies inside or on the boundary of the triangle ABC in which the angle C is a right angle. The points P, Q and R are the perpendicular projections of X onto the sides BC, CA and AB respectively. Prove that if AR*RB = BP*PC + AQ*QC then X is on AB.

We will prove:

[b]Lemma 1.[/b] For every point X inside or on the boundary of the triangle ABC, we have

$AR\cdot RB - \left(BP\cdot PC + AQ\cdot QC\right) = XR^2$.

[i]Proof.[/i] By the Pythagoras theorem in the right-angled triangles ARX and AQX, we have $XR^2+AR^2=AX^2$ and $XQ^2+AQ^2=AX^2$, so that $XR^2+AR^2=XQ^2+AQ^2$. Since the quadrilateral CPXQ has three right angles (in fact, < XPC = 90° and < XQC = 90° since the points P and Q are the perpendicular projections of the point X on the lines BC and CA, and < PCQ = 90° since the triangle ABC is right-angled at C), it is a rectangle, so that XQ = PC. Thus, $XR^2+AR^2=PC^2+AQ^2$. Similarly, $XR^2+BR^2=QC^2+BP^2$. Adding up these two equations, we get

$\left(XR^2+AR^2\right) + \left(XR^2+BR^2\right) = \left(PC^2+AQ^2\right) + \left(QC^2+BP^2\right)$.

After rearranging terms, this becomes

$2\cdot XR^2 + \left(AR^2+BR^2\right) = \left(BP^2+PC^2\right) + \left(AQ^2+QC^2\right)$.

Now, for any two numbers u and v, we have the identity $u^2+v^2=\left(u+v\right)^2-2uv$. Applying this identity to the numbers u = AR and v = RB, we get $AR^2+RB^2 = \left(AR+RB\right)^2-2\cdot AR\cdot RB$. In other words, $AR^2+BR^2 = AB^2-2\cdot AR\cdot RB$. On the other hand, applying the same identity to the numbers u = BP and v = PC, we get $BP^2+PC^2 = \left(BP+PC\right)^2-2\cdot BP\cdot PC = BC^2 - 2\cdot BP\cdot PC$, and applying it to the numbers u = AQ and v = QC, we get $AQ^2+QC^2 = \left(AQ+QC\right)^2-2\cdot AQ\cdot QC = AC^2 - 2\cdot AQ\cdot QC$. Hence, our equation

$2\cdot XR^2 + \left(AR^2+BR^2\right) = \left(BP^2+PC^2\right) + \left(AQ^2+QC^2\right)$

becomes

$2\cdot XR^2 + \left(AB^2 - 2\cdot AR\cdot RB\right) = \left(BC^2 - 2\cdot BP\cdot PC\right) + \left(AC^2 - 2\cdot AQ\cdot QC\right)$,

or, equivalently,

$2\left(XR^2 - AR\cdot RB\right) + AB^2 = -2\left(BP\cdot PC+AQ\cdot QC\right) + \left(BC^2+AC^2\right)$.

But since the triangle ABC is right-angled at C, we have $AB^2=BC^2+AC^2$ by the Pythagoras theorem; thus,

$2\left(XR^2 - AR\cdot RB\right) = -2\left(BP\cdot PC+AQ\cdot QC\right)$,

so that $XR^2 - AR\cdot RB = -\left(BP\cdot PC+AQ\cdot QC\right)$. Consequently, $XR^2 = AR\cdot RB - \left(BP\cdot PC + AQ\cdot QC\right)$, what proves Lemma 1.

Of course, it follows that, if we have $AR\cdot RB = BP\cdot PC + AQ\cdot QC$, so that $AR\cdot RB - \left(BP\cdot PC + AQ\cdot QC\right) = 0$, then we also have $XR^2 = 0$, so that XR = 0. But XR is the distance from the point X to the line AB; thus, XR = 0 means that the point X lies on the line AB. And your problem is solved.

  Darij

Here is my solution.
It is given that $\frac{BP\cdot{RB}}{PC\cdot{AR}}+\frac{AQ\cdot{QC}}{PC\cdot{AR}}=1.(*)$
Denote $\measuredangle{PBX}=\alpha,  \measuredangle{RBX}=\beta,  \measuredangle{QAX}=\gamma,  \measuredangle{XAR}=\phi.$
Now using our definitions in (*) we get $\frac{\cos{\alpha}\cdot{\sin{\gamma}}}{\cos{\beta}\cdot{\cos{\phi}}}+\frac{\cos{\gamma}\cdot{\sin{\alpha}}}{\cos{\beta}\cdot{\cos{\phi}}}=1\Rightarrow\sin(\alpha+\gamma)=\cos{\phi}\cdot{\cos{\beta}}.$
But since $\alpha+\gamma=\frac{\pi}{2}-(\phi+\beta)$ we have $\sin(\alpha+\gamma)=\cos({\phi+\beta})=\cos{\phi}\cdot{\cos{\beta}}-\sin{\phi}\cdot{\sin{\beta}} =\cos{\phi}\cdot{\cos{\beta}}$ thus $\sin{\phi}\cdot{\sin{\beta}}=0$, so $X$ lies on $AB$. QED


Is anyone living in the Waterloo region wanting to come with the WCI math teams to London for the MCTIC competition held by UWO?

MCTIC (Math Challenge Team Invitational Competition) is a very fun and enriching experience. Teams of 9 compete in a variety of competitions including Power and Team Questions (Team of 9 can help each other), Individual Questions and Mental Math (Individual), and Relay Questions (Team of 3).

The date is May 18th - 20th, we leave on the late afternoon of the 18th and get back on the morning of the 20th. The cost is originally $120 but UW is sponsoring$50 per member so that boils downt o $70. The price pays for room and board and all activities on campus (not transportation, however). For transportation we're looking at an additional driver (just one extra is enough) or we'll have to Greyhound which is$11 one-way.

If anyone's interested post here with your name, phone # and email.

For all who's going, have fun and good luck!

I'm definitely going :)

I heard that people from Toronto may join the Waterloo team. Does anyone from Toronto want to come? we are looking for 2 or 3 more people. Alex, can you join?

If anyone wants to come, please contact me or Dwang SOON. MCTIC is this Friday!

No, unfortunately I cannot come. I have too much stuff due -- I have my BC Calc exam, my Chem ISP, my Bio ISP, and like everything. :(
Sorry guys.

Good luck! PWN everyone like last yr. :lol:

boo alex isn't coming? awwwww but yes london will win lol oii prolly not but oh well lol but yes POKER AGAIN WIT WATERLOO PPL lol

[quote="rem"]No, unfortunately I cannot come. I have too much stuff due -- I have my BC Calc exam ... [/quote]

lol @ BC Calc. I thought all AP Exams had to take place at the same time and at the same hour to avoid cheating or what ever. I wrote mine May 9th.

[quote="rem"]My school ordered the wrong exam so it got postponed.[/quote]

 :rotfl:

oh london duzn't get the nice rooms this year i get 2 singles and 4 doubles argh lol

We're mostly set! W00t

Hope we can at least win a book prize  :P  There's really not much hope for first place with our team this year  :(

ya we didnt win but we came second:

1. Windsor 1
2. Waterloo 1
3. Windsor 2 (i think)

Apparently we lost by 6 pts..which sux cuz we couldve got those points in relay..my answer was 9 and i somehow turned my peice of paper :wallbash_red: so the next person thought it was a 6  ..and the last person solved the problem within 4 minutes..so we ended up getting 0 when we mayve got 6..a tie wouldve been interesting..

:O we dint get anything... well i got a pencil lol :D:D:D

LOL, I can't believe how cheap they were being this year =/

Perhaps the Maple 11 drained the resources so they couldn't buy anything else. There was no prizes for 2nd and 3rd place teams, the prizes for the individual parts SUCK (a pencil and a used book, w00t!), bleh.

Congrats Windsor.

i heard the books weren't good either... but i wanted one anyways lol argh 7/8 on individuals i got number 5 wrong. the like 3 odds... 2 evens... probability... answer: 60%... i got 50%... cuz iunno i counted miscounted my O's... but I GOT HIGHEST OVERALL INDIVIDUALS SCORE i think :D:D:D um 17/18?

Everyone in my team who solved number 8 on the individuals used calculus. >_<
I was just wondering if anyone here had non-calculus solutions.

im not sure if my solution used calculus lol cuz i don't know calculus lol but i mite've accidently

I used calculus too on #8 :P

I guess you can use the optimization formula for quadratics on the quartic, it should work because it factors nicely I think.

Calc pwns.

no... the book prizes for perfect individuals were pretty good. Though, I still sold mine to Peter Yang for $\$$7.00  :lol:

[quote="Dwang"]Calc pwns.[/quote]
 :o

you really dont need calc for that problem... you just find the maximum of a quadratic. For people who werent there, the problem was:
 "The graph of $y=ax^{4}+bx^{3}+cx^{2}+dx+e$ is symmetric about the y-axis, reaches a high point of $y=1$ when $x=2$, and has a y-intercept of $-3$. Find all values of $x$ for which $y=0$."

oh and i think the book might not be used... its just that the cover is very worn out.


function
number theory unsolved
number theory

As the title says Number Theory...
So: wrong forum (and it's also wrong located as Solved Problem).


modular arithmetic
decimal representation
combinatorics
Digits
IMO Shortlist

Find the digits left and right of the decimal point in the decimal form of the number \[ (\sqrt{2} + \sqrt{3})^{1980}. \]

We have $( \sqrt 2 + \sqrt 3)^{1980} = ...7,9...$.

Note that $A = ( \sqrt 2 + \sqrt 3)^{1980} = (5 + \sqrt{24})^{990}$.
Let $a = 5 + \sqrt{24}$ and $b = 5 - \sqrt{24}$.
Then, using the binomial expansion, it is easy to verify that $N = A + b^{990} = a^{990} + b^{990}$ is a positive integer.

Moreover $b = \frac {1} {a} < \frac {1} {5}$. It follows that $0 < b < 0,2$, then $0 < b^{990} < 10^{-1}$.
Thus $A = N - b^{990}$ has digit $9$ at the right of the decimal point, and has digit $d$ at the left, with $d = n - 1$ mod[10] where $n$ is the digit of the unity in $N$.

Using binomial expansion, we have :
$N = 2 \prod_{k=0}^{495} 2C_{990}^{2k} 5^{990-2k} 24^k = 2 \times 24^{495}$ mod[10].
Thus $N = 2 \times 4^{495}$ mod[10].
Using that $2^5 = 32 = 2$ mod[10], it follows easily that $N = 8$ mod[10], so that $d = 7$ as claimed.

Pierre.

Let $a_n=(\sqrt3+\sqrt2)^{2n}+(\sqrt3-\sqrt2)^{2n}=(5+2\sqrt6)^n+(5-2\sqrt6)^n$. It satisfies the recurrence $a_n-10a_{n-1}+a_{n-2}=0$, or $a_n=10a_{n-1}-a_{n-2}$, where $a_0=2$ and $a_1=10$. Then the sequence modulo $10$ is $2\to0\to-2\to0\to-2\to\dots$. We know that $a_{990}=(\sqrt3+\sqrt2)^{1980}+(\sqrt3-\sqrt2)^{1980}\equiv-2\pmod{10}$, so its last digit is $8$. Since $(\sqrt3-\sqrt2)^{1980}<<0.1$, the units digit of $(\sqrt2+\sqrt3)^{1980}$ is $7$ and the digit after is $9$.


I'm trying to go into the Introduction to Algebra class right now. I've refreshed numerous times but I can't get in!! The page says that it's loading but it's not!

**I'm following along in the transcript though.

which browser are you using? try using opera.

You may also have a problem with your Java installation -- you can follow the directions on the Classroom page to reinstall Java.

Never mind--I deleted the cookies on my computer twice. Now it works.
Please lock.


Let $ABC$ be an isosceles triangle with $AB=AC$, such that $\angle{BAC}=20^\circ$. Let $D,E$ be points on $(AC), (AB)$, respectively such that $\angle{DBC}=60^\circ$ and $\angle{ECB}=50^\circ$. Find $\angle{BDE}$.

I just found a nice metrico-trigonometric solution, hope you will like it:

First we just observe that $\angle{BDE}=30^\circ$ works, and now to prove it, construct the parallel through $D$ to $BC$, which cuts $AB$ in $F$. And now because $\angle{BDC}=40$, we only must prove that $DE$ bisects $\angle{BDF}$.
This is equivalent to prove that $\frac{EF}{EB}=\frac{FD}{BD}$.
Now we observe that from the parallelism we have that $\frac{FD}{BC}=\frac{AF}{AB}=\frac{BD}{AB}$, because $\angle{BAD}=\angle{ABD}=20$. Therefore $\frac{FD}{BD}=\frac{BC}{AB}$.
So, it remains us to prove that : $\frac{BC}{AB}=\frac{EF}{EB}\iff BC^{2}=AB \cdot EF$, because $\angle{BCE}=\angle{BEC}=50$.

From here, my solution transforms into a trigonometric one:
1) We have that $AB=\frac{BC \cdot \sin{80}}{\sin{20}}$, from the sin law in $\triangle{ABC}$.
2) We have that $EF=AB-BC-BD=BC(\frac{\sin{80}-\sin{20}-2\cos{40}\sin{20}}{\sin{20}})$, again from the sin law in $\triangle{BDC}$.

Therefore, it only remains us to prove that :
$\sin{80}\cdot (\sin{80}-\sin{20}-2\cos{40}\sin{20})=\sin^{2}{20}$

$\iff \sin{80}\cdot [ 2\cos{40}(\sin{40}-\sin{20})-\sin{20}] = \sin^{2}{20}$

$\iff \sin{80}\cdot (4\cos{40}\sin{10}\cos{30}-\sin{20}) = \sin^{2}{20}$
Now using that : $\sin{80}=\cos{10}$:

$\iff 2\sin{20}\cdot (2\cos{40}\cos{30}-\cos{10})=\sin^{2}{20}$

$\iff 2 \cdot (2\cos{40}\cos{30}-2\sin{40}\cos{40})=\sin{20}$

$\iff 4\cos{40}\cdot (\sin{60}-\sin{40}) = \sin{20}$

$\iff 2\cos{40}\cdot \cos{50}= \cos{10}$

$\iff 2\sin{50}\cdot \sin{40}= 2\sin{40}\cos{40}\iff \sin{50}=\cos{40}$, which is true , so the problem is solved, and $\angle{BDE}=30^\circ$.

Hope, you liked it, and I am waiting for your "non-trigonometric" solutions. :)

This a really nice problem.

[color=indigo][b][u]Solution[/u][/b][/color]

Draw $\overline{BC}= \overline{BF}$ with $F \in \overline{AC}$. Then, we're gonna have that $\overline{BF}= \overline{DF}$, then is easy to see that $\triangle BEF$ is equilateral $\implies \overline{EF}= \overline{DF}$. Set $\measuredangle{\text{ }}BDE = x$ finally we get

\begin{eqnarray*}2(x+40^\circ)+40^\circ&=&180^\circ\\x+40^\circ+20^\circ&=&90^\circ\\x&=&30^\circ\end{eqnarray*}

[quote="Virgil Nicula"][color=darkred][b][u][size=150]Extension.[/size][/u][/b] Let $\triangle ABC$ for which $A<30$ and $B=60+A$. Prove that : $\left\{\begin{array}{c}N\in (AC)\ ,\ m(\widehat{ABN})=30\\\\ M\in (AB)\ ,\ m(\widehat{ACM})=30-\frac{A}{2}\end{array}\right\|$ $\implies$ $m(\widehat{CMN})=30$.[/color][/quote]

[hide="[color=red"]Only if you are ... curiously.[/color]]
[color=darkblue][b][u]Proof.[/u][/b] Observe $C=120-2A>60$ and consider the point $G\in (BM)$ for which $m(\widehat{ACG})=60$.

Prove easily that $CB=CG=CN=GN=GM$ and $m(\widehat{CMN})=30$. Indeed, appear three triangles :

an isosceles $\triangle BCG\ (CB=CG)$; an isosceles $\triangle CGM\ (GC=GM)$; an equilateral $\triangle CNG$; an isosceles $\triangle MGN\ (GM=GN)$.[/color][/hide]

[color=darkblue][u]Particular case.[/u] For $A=20$ obtain the above[/color] [color=darkred]Pohoatza's proposed problem[/color].


advanced fields
advanced fields unsolved

Hi!
I've been thinking about this problem for quite a while now. Perhaps somebody
can help...

G_1, G_2 simply connected regions, neither of which is the whole plane.
f one-one analytic mapping of G_1 onto G_2.
a \in G_1, and alpha = f(a).
If h is any one-one analytic mapping mapping of G_1 into G_2 with
h(a)=alpha, show that |h'(a)| <= |f'(a)|.
What can be said if h is not one-one ?

Thanks for any ideas  :)

[quote="ZetaX"]Sorry, but aren't there any conditions missing, because I see no reason why we cannot change $f$ and $h$ to get $|h'(a)|=|f'(a)|$.[/quote]

I think the point here is that f maps G_1 onto G_2 but h maps G_1 only into G_2.
Thanks for your response!

[quote="fedja"][quote="Amar"]
G_1, G_2 simply connected regions, neither of which is the whole plane.
f one-one analytic mapping of G_1 onto G_2.
a \in G_1, and alpha = f(a).
If h is any one-one analytic mapping mapping of G_1 into G_2 with
h(a)=alpha, show that |h'(a)| <= |f'(a)|.
What can be said if h is not one-one ?[/quote]
The same. It is actually just the Schwartz lemma in disguise. The standard formulation of the Schwarz lemma is for the case $G_1=G_2=\mathbb D$, $a=\alpha=0$, but you can easily reduce the general problem to this case by considering two auxiliary conformal mappings $\varphi:\mathbb D\to G_1$ with $\varphi(0)=a$ and $\psi: G_2\to \mathbb D$ with $\psi(\alpha)=0$. Then, for $f$, you have $(\psi\circ f\circ \varphi)'(0)=1$ while the similar expression for $h$ instead of $f$ does not exceed $1$ according to the Schwarz lemma.[/quote]

Hey fedja,

thanks, that works!  :)


Let $ABC$ be a triangle, and let $D$, $E$, $F$ be the points in which the incircle of the triangle touches the sides $BC$, $CA$ and $AB$ respectively. Let $P$ be a point on the side $BC$ and $R$ a point on the segment $AP$ such that $AR=AE=AF$. We denote by $Q$ and $S$ the incircles of the triangles $ABP$ and $ACP$.

Prove that the quadrilateral $DQRS$ is a square if and only if $P$ is the foot of the altitude from $A$ on $BC$.

[hide="Remark"]This problem is slightly connected to [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=344774#p344774]GMB-IMAR 2005, Juniors, Problem 2[/url]
[/hide]


Let $a_{k}$ be the number of pair of natural numbers $(x,\ y)$ such that  $x<y<k<x+y$ for positive integers $k.$
Express $\sum_{k=1}^{2n}a_{k}$ in terms of $n.$

We fix natural number $k$. and count $a_{k}$. if $x<y<k<x+y$
It's need to satisfy $x<k$ and given $x$ satisfying it the number of $y$ must be satisfy $x<y, k-x<y$ and $y<k$.
So , if $x\leq k-x$($x\leq \frac{k}{2}$) then $k-x<y<k$. The number of $y$ is $x-1$.
and if $x>k-x$($x> \frac{k}{2}$) then $x<y<k$. The number of $y$ is $k-x-1$.
If $k$ is odd , we can put $k=2s+1(s\geq 0)$.  If $l$ is a natural number such that $l\leq s$.  About $x=l$, the total number of $y$ is $l-1$, and about $x=k-l$ the total number of $y$ is also $k-(k-l)-1=l-1$. So $a_{k}=2\sum_{l=1}^{s}(l-1)=s(s-1)$.
If $k$ is even, we can put $k=2s$. If $l$ is a natural number such that $l\leq s-1$ then similarly both about $x=l, k-l$ the number of $y$ is $l-1$. And about $k=s$ the number of $y$ is $s-1$. So $a_{k}=2\sum_{l=1}^{s-1}(l-1)+s-1=(s-1)(s-2)+(s-1)=(s-1)^{2}$.
Now the disired sum is
$\sum_{s=0}^{n-1}s(s-1)+\sum_{s=1}^{n}(s-1)^{2}=\sum_{s=0}^{n-1}(s^{2}-s)+\sum_{s=1}^{n}(s-1)^{2}=2\sum_{s=1}^{n}(s-1)^{2}-\sum_{s=0}^{n-1}s$
$=\frac{2(n-1)n(2n-1)}{6}-\frac{n(n-1)}{2}=\frac{n(n-1)(4n-2-3)}{6}=\frac{n(n-1)(4n-5)}{6}$


I need to solve these systems of equations by the addition method:

2x-5y=4
3x-2y=4      the solution in the back of the book is (12/11, -4/11) I cant make my work match the answer.

the next is ,

x/3 + y/6 =1
x/2 - y/4 = 0    the solution is ( 3/2 , 3)

and the last is,

x/3 - y = 2
-x/2 + 3y/2 = -3

Not really pre-olympiad. Try getting started or the middle school forums next time.

GG, he said by the addition method; you used substitution which is not the best method here. And btw, this is more like middle school or getting started exercises.  ;)

Also, let people do their own homework.  If you want to help them, create a problem like theirs and solve it so they have a demonstration and can do it themselves.


inequalities
trigonometry
LaTeX
geometry
3D geometry
geometric transformation

$x\cdot \sqrt{1-y^2}+y\cdot\sqrt{1-x^2}=1$. Find the maximum of $x+y$.

try to use Cauchy's inequality, it's rather easy

Or particulary beautiful trig substitution and jensens

Did it quite hurriedly, is it $\sqrt{2}$ ???

I got the same thing, but I don't know cauchy or jensen's or beautiful trig substitution.

I used the method of Lagrange multipliers to maximize x + y with the big thingy as the constraint curve.

I second seamusoboyle's answer.  This probem is just yelling out "TRIG SUBSTITUTION." :P  You don't need Jensen's btw, but of course it is the simplest way of saying it.

[hide]
$\displaystyle x\cdot \sqrt{1-y^2}+y\cdot\sqrt{1-x^2}=1$
$x= \cos{\theta}, y=\sin{\theta}$
$\cos^3{\theta} + \sin^3{\theta} = 1$
$x^3+y^3= (x+y)(x^2-xy+y^2) = (x+y)(1-xy) = 1$
[/hide]

you can continue from what i did up there.

[quote="xxreddevilzxx"]$\displaystyle x\cdot \sqrt{1-y^2}+y\cdot\sqrt{1-x^2}=1$
$x= \cos{\theta}, y=\sin{\theta}$
$\cos^3{\theta} + \sin^3{\theta} = 1$
$x^3+y^3= (x+y)(x^2-xy+y^2) = 1$

I got to that, but shouldnt there be another condition, if i were to solve?[/quote]
that doesn't look right...
$\displaystyle x\cdot \sqrt{1-y^2}+y\cdot\sqrt{1-x^2}=1$
$x= \cos{\theta}, y=\sin{\theta}$
$\cos^2{\theta}+\sin^2{\theta}=1$
obviously the maximum value is at $\theta=\frac{\pi}{4}$, which implies that $x+y=\frac{2\sqrt{2}}{2}=\sqrt{2}$

oops. didnt sqaure root. sorry. that wouold have been very trivial if i squarerooted!!

Yes, I made the same mistake a reddevilz.

Yeah, that is like 1000 times faster than my method.

[quote="bubala"]oh.  duh...... I was that close!!! stupid me.  Yeah, that is like 1000 times faster than my method.

yeah, reddevilz, I got up to that part in the trig solution.  Look at my post.  I just didn't write it in that understandable of a form.  sorry.[/quote]
no, youre solution is wrong,
you wont get $\cos^3{\theta}+\sin^3{\theta}$ you'll get $\cos^2{\theta}+\sin^2{\theta}$.
you made the same mistake as, xxreddevilzxx

oh, right.  nvm forgot about the sqrt.

thanks.

I see, so cos squared theta plus sin squared theta is always 1, so just pick the maximum of cos theta plus sin theta in general which obviously occurs when theta is pi/4.  Gotcha.

[quote="bubala"]could you elaborate on your beautiful trig substitution?  I would elaborate on my solution, but I'm not that great at LaTeX yet.

I'm assuming it's something like $x = cos (theta)$, $y = sin (theta)$, then $cos cubed (theta) sin cubed (theta) = 1$, then factor that as sum of two perfect cubes, then arriving at $(cos (theta) + sin (theta))(1 - cos (theta) sin(theta))=1$, but I'm not sure where to go from there.[/quote]

whoa...read [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=5261]this[/url] quickly!  :P

[quote="bubala"]thanks joml88.  I know--I need to work on my LaTeX--I haven't really committed to learning it yet.

towersfreak, good point.  Well, my lagrange solution works anyway.  i'd be curious to see a solution with the trig substitution, though.  Or maybe the other solution is valid and someone will prove you wrong, but I doubt it.[/quote]
the above's solutins IS trig substitution...

[quote="towersfreak2006"]Hmm, can you really say $x= \cos{\theta}, y=\sin{\theta}$?
Doesn't that imply $x=\sqrt{1-y^2}$?

Aren't we supposed to be looking more along the lines of $x= \cos{\theta}, y=\cos{\phi}$?[/quote]

I think you are right. However, in this case, the end result is the same. Let $x=\cos{\theta},y=\cos{\phi}$. Then, $\cos{\theta}\sin{\phi}+\cos{\phi}\sin{\theta}=1$, and so by the trigonometric identity, $\sin({\phi+\theta})=1$. Thus, $\phi+\theta=\frac{\pi}{2}$. This is clearly maximized when $\phi=\theta=\frac{\pi}{4}$, and so maximum value is still $\sqrt{2}$.

I think I have the Cauchy solution...

[hide]
$(x^2+y^2)(1-y^2+1-x^2) \ge (x\sqrt{1-y^2}+y\sqrt{1-x^2})^2 = 1$ by Cauchy.

Let $a = x^2+y^2$.

Then $a(2-a) \ge 1 \Rightarrow 0 \ge (a-1)^2 \Rightarrow a=1$.

By Power Mean, $\frac{x+y}{2} \le \sqrt{\frac{x^2+y^2}{2}} = \sqrt{\frac{1}{2}} \Rightarrow x+y \le \sqrt{2}$.[/hide]

So going back to nolachrymose's solution, we have $\sin(\phi+\theta)=1$.  So $\phi+\theta=\pi/2$

We need to show that $\cos(\phi)+\cos(\theta)\leq \sqrt{2}$.  We're working up to finishing with Jensen's inequality, but we need to establish a few things.  First, it's enough to consider $\phi,\theta\in [0,\pi/2]$.

Why is this true?  Well, in the first place, the range of $\phi,\theta$ was $[0,\pi]$ (e.g, up to translation the smallest interval for which $\cos$ attains its (largest) range)  If $\phi\in (\pi/2,\pi]$, then $x<0$, so that $\textstyle \sum_{x,y} x\sqrt{1-y^2}<y\leq 1$, a contradiction of the condition.  So $\phi,\theta\in [0,\pi/2]$.

Next, $\cos$ is concave on $[0,\pi/2]$

By Jensen's

$\cos(\phi)+\cos(\theta)\leq 2\cos(\pi/4)=\sqrt{2}$

and we're done!

Thats the on i had. Nice, i must say.
What about this solution with the LeGrange multipliers?

Ignore solutions with lagrange multipliers unless you know necessary/sufficient conditions for maxima/minima (which are fairly involved for lagrange multipliers)


prove or disprove there are infinitely many primes in the form n^2 +1


inequalities
AMC
USA(J)MO
USAMO
inequalities solved

Well, here's what I did.
I reduced the inequality to the symmetric sum inequality
\[ \sum_{sym} \left[3a^2b^2c^2+2a^5b+a^6+2a^4bc-8a^3b^2c\right] \ge 0\]
I used Schur's inequality and weighted AM-GM four times to arrive at this. Now can sum1 help ( I hope what I got is correct)
Thanks.

Ok, it wasn't easy, but...
We see that if numbers $(a,b,c)$ satisfy the inequality, than so do $(ka, kb, kc)$ for any $k>0$. Hence we can assume that $a+b+c=1.$ The inequality becomes:
\[\sum_{cyc}\frac{(a+1)^2}{2a ^2 + (1-a) ^2}\le8.\]
This means that we must show
\[\sum_{cyc}\frac{a^2+2a+1}{3a ^2-2a+1}\le8.\]
We observe that we have equality when $a=b=c=\frac13.$ So let's show that this is the case when the ewpression is maximal.

Let's show that $\frac{a^2+2a+1}{3a ^2-2a+1}\le4a+\frac43.$ This is equivalent to showing  $a^2+2a+1\le12a^3-4a^2+\frac{4a}{3}+\frac43,$ equivalent to $12a^3-5a^2-\frac{2a}{3}+\frac 13\ge0,$ equivalent to $3(4a+1)\left(a-\frac13\right)^2\ge0,$ which is true.

Now that means that $\sum_{cyc}\frac{a^2+2a+1}{3a^2-2a+1}\le4(a+b+c)+4=8,$ whick concludes the proof.

You might be wondering how I came up with $\frac{a^2+2a+1}{3a^2-2a+1}\le4a+\frac43.$ Well, I searched for a number $k$ such that $\frac{a^2+2a+1}{3a^2-2a+1}\le ka+\frac{8-k}{3}$ would be true for any $a\in (0,1).$ I found $k=4.$

Well I've done this problem about a grand total of 230948023184 times but I've never bothered to write it up so here goes, I guess?

The inequality is homogeneous, so we may set $a+b+c=3.$ Then \[\sum_{\text{cyc}} \frac{(2a+b+c)^2}{2a^2+(b+c)^2}\stackrel{\text{WTS}}{=}\sum_{\text{cyc}} \frac{a^2+6a+9}{3a^2-6a+9} \leq 8.\] Check that $f$ is not convex. By the Tangent Line Trick, it remains to prove \[\frac{a^2+6a+9}{3a^2-6a+9} \le \frac{8}{3}+\frac{4}{3}(a-1)\implies (a-1)^2(4a+3) \geq 0.\] Summing cyclically gives the desired result.

It seems difficult at first, but in the end it will turn out to be amazing
 $$\sum_{cyc} \frac{(2+\frac{b+c}{a})^2}{2+(\frac{b+c}{a})^2} \leq 8$$
by using the substitution of
$$\frac{a+b}{c}=x$$
$$\frac{c+a}{b}=y$$
$$\frac{b+c}{a}=z $$
 we can note that
 $$xyz=x+y+z+2$$
 $$\sum_{cyc} \frac{(2+x)^2}{2+x^2} \leq 8$$
 $$\sum_{cyc} \frac{2x+1}{2+x^2} \leq \frac{5}{2}$$
 $$\sum_{cyc} \frac{(x-1)^2}{2+x^2} \geq \frac{1}{2}$$
 $$\sum_{cyc} \frac{(x-1)^2}{2+x^2} \geq \frac{x+y+z-3}{x^2+y^2+z^2+6}$$
 So, we are left with proving that $$2(x + y + z - 3)^2\geq x^2 + y^2 + z^2 + 6$$. But
this is not difficult. Indeed, this inequality is equivalent to
$$2(x + y + z - 3)^2 \geq (x + y + z)^2 - 2(xy + yz + zx) + 6$$.
Now, from $xyz \geq 8$ (recall who x, y, z are and use the AM-GM inequality three
times), we find that $xy+yz +zx \geq 12$ and $ x +y+z \geq 6 $ (by the same AM-GM
inequality). This shows that it suffices to prove that$ 2(s -3)^2 \geq s^2 - 18$ for all
s > 6, which is equivalent to $$(s -3) (s - 6) \geq 0$$, clearly true. And this difficult
problem is solved


If $a,b,c$ are positive reals and $abc >$max $\left\{ac+bc, ab+cb, ba+ca\right\}$ prove that

$(1+a)^2+(1+b)^2+(1+c)^2 >27$

dividing by abc and setting a -> 1/a the problem is equivalent to :

$a+b < 1$
$b+c < 1$
$c+a < 1$

prove that $\sum (1+1/a)^2 > 27$

We have $a+b+c < 3/2$ so by Cauchy-Schwarz $1/a+1/b+1/c > 6$
Also by convexity of $1/x^2$, $1/a^2 + 1/b^2 > 8$

Summing all that gives wanted result.


function
trigonometry
calculus
calculus computations

Let $ f: \mathbb{R}^2\rightarrow \mathbb{R}$ and define $ g(x,y) \equal{} f(\cos(y),\sin(x))$. Find $ g_{xx}$ and $ g_{yy}$.

Can anyone help me on this? :) I don't understand the notation for $ f$.

It means $ f$ is a function of $ \cos{y}$ and $ \sin{x}$, which are themselves functions of $ x$ and $ y$.  Think of it as $ g\equal{}f(u,v)$, where $ u\equal{}u(y)$ and $ v\equal{}v(x)$.  See the attachment.

In general, if $ g(x,y) = f(u(x,y),v(x,y))$ for differentiable functions $ f,u,$ and $ v$, then the chain rule reads:

$ g_x = \frac{\partial g}{\partial x} = \frac{\partial f}{\partial u} \cdot \frac{\partial u}{\partial x} + \frac{\partial f}{\partial v} \cdot \frac{\partial v}{\partial x}$

Notice for your particular functions, $ u(x,y) = \cos y$ and $ v(x,y) = \sin x$, that $ \frac{\partial u}{\partial x} = 0$ and $ \frac{\partial v}{\partial x} = \cos x$.

Now differentiating again with respect to x gives:
$ g_{xx} = \frac{d}{dx}(g_x) = \frac{d}{dx} [ \frac{\partial f}{\partial u} \cdot \frac{\partial u}{\partial x} + \frac{\partial f}{\partial v} \cdot \frac{\partial v}{\partial x} ] = \frac{d}{dx} [\frac{\partial f}{\partial v}\cdot \cos x ]$

$ = \frac{d}{dx} (\frac{\partial f}{\partial v}) \cdot \cos x + \frac{\partial f}{\partial v} \cdot \frac{d}{dx} (\cos x)$

$ = [\frac{\partial^2 f}{\partial v \partial u} \cdot \frac{\partial u}{\partial x} + \frac{\partial^2 f}{\partial v^2} \cdot \frac{\partial v}{\partial x} ] \cdot \cos x - \frac{\partial f}{\partial v} \cdot \sin x$

$ = \frac{\partial^2 f}{\partial v^2} \cdot \cos^2 x - \frac{\partial f}{\partial v} \cdot \sin x$

Mhmm gauss202, in the first line for  $ g_x$ how come there is no $ \frac{\partial g}{\partial f}$ anywhere> Why can that be omitted? Or that shouldn't be in because you take all of the partials with the next function down with respect to $ x$, and that total sum implies the given?

The chain rule starts from $ f$.  $ g$ is the expression for the complete function you want to differentiate.  Ask yourself if that question makes sense in one variable:  in differentiating $ g(x) \equal{} f(h(x))$, is there a $ \frac{dg}{df}$ expression anywhere?


inequalities
linear algebra
matrix
linear algebra unsolved

Let $ A,B$ be positive semidefinite Hermitian matrices. Prove or disprove $ |tr(A^2B^2AB)|\le tr(A^3B^3)$.
Some conpuster examples do show the validity of the inequality, how to show it then?  :)

$ |tr(A^{2}B^{2}AB)|\le tr(A^{3}B^{3})$
Numerical computations seem to show that the previous result is true.
Moreover it seems also to me that $ tr(A^{2}B^{2}AB)\geq{0}$.
If this last result is true then the first required inequality is equivalent to $ tr(A^2B^2(BA\minus{}AB))\geq{0}$ that is a quite surprising result.

[quote="loup blanc"]$ |tr(A^{2}B^{2}AB)|\le tr(A^{3}B^{3})$
Numerical computations seem to show that the previous result is true.
Moreover it seems also to me that $ tr(A^{2}B^{2}AB)\geq{0}$.
If this last result is true then the first required inequality is equivalent to $ tr(A^2B^2(BA \minus{} AB))\geq{0}$ that is a quite surprising result.[/quote]
Hi, I have examples that $ tr(A^{2}B^{2}AB)\le{0}$.  See for instance the paper 'EIGENVALUES OF WORDS IN TWO POSITIVE DEFINITE
LETTERS'  SIAM J. MATRIX ANAL. APPL. c 2002 Society for Industrial and Applied Mathematics
Vol. 23, No. 4, pp. 916–928


though we have moved to this server for quite awhile now, is it still possible to still use the original black background?

No.  That old black background is history.


$ [k \pi] \leq  k \pi$
$ [ \frac{[k \pi]}{k}] \leq  [ \pi ]$

$ k \pi-1< [k \pi]$
$ [\pi]-1<[ \frac{[k \pi]}{k}]$
then $ [ \frac{[k \pi]}{k}]=3 \forall k \in \mathbb N$
obtaining for that sum $ 2007 \cdot 3$


Let $n \ge 2$ be a positive integer, with divisors \[1=d_{1}< d_{2}< \cdots < d_{k}=n \;.\] Prove that \[d_{1}d_{2}+d_{2}d_{3}+\cdots+d_{k-1}d_{k}\] is always less than $n^{2}$, and determine when it divides $n^{2}$.

[quote="nicetry007"]The sum
$ \sum_{i=1}^{k-1}d_{i}d_{i+1}= d_{k}d_{k-1}+d_{k-1}d_{k-2}+\cdots+d_{2}d_{1}$
                    $ \leq n\cdot \frac{n}{2}+\frac{n}{2}\cdot \frac{n}{3}+\cdots+\frac{n}{k-1}\cdot \frac{n}{k}= n^{2}\left[ 1\cdot\frac{1}{2}+\frac{1}{2}\cdot \frac{1}{3}+\cdots+\frac{1}{k-1}\cdot \frac{1}{k}\right] = n^{2}\left[ 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\cdots+\frac{1}{k-1}-\frac{1}{k}\right] = n^{2}\left(1-\frac{1}{k}\right) < n^{2}$.

Suppose $ n$ is a prime. $ n = 1.p$, $ d_{1}=1$ and $ d_{2}= p$ and $ n^{2}= p^{2}$. The sum $ \sum_{i=1}^{k-1}d_{i}d_{i+1}= 1 \cdot p = p$ and $ p | p^{2}$. Hence, whenever $ n$ is a prime, the sum divides $ n^{2}$.

Suppose $ n$ is composite. This implies $ k \geq 3$. We have $ d_{1}= 1$ and $ d_{2}= p$ where $ p$ is ther smallest prime that divides $ n$.
The sum $ \sum_{i=1}^{k-1}d_{i}d_{i+1}> d_{k}d_{k-1}= n\cdot \frac{n}{p}= \frac{n^{2}}{p}$.

This helps us make the following useful observation.

$ 1 < \frac{n^{2}}{\sum_{i=1}^{k-1}d_{i}d_{i+1}}< \frac{n^{2}}{d_{k}d_{k-1}}= \frac{n^{2}}{\frac{n^{2}}{p}}= p$.[Note the first inequality follows from the first part of the problem.]

If $ \frac{n^{2}}{\sum_{i=1}^{k-1}d_{i}d_{i+1}}$ is an integer, then it should lie between $ 1$ and $ p$. But $ p$ is the smallest prime that divides $ n$. Hence, the ratio  $ \frac{n^{2}}{\sum_{i=1}^{k-1}d_{i}d_{i+1}}$ cannot be an integer.

Hence, the sum  $ \sum_{i=1}^{k-1}d_{i}d_{i+1}| n^{2}$ if and only if $ n$ is a prime.[/quote]


inequalities
calculus
calculus computations

it makes sense for equality. if two things are equal then doing the same operation to the same entities should yield two more equally valid terms.

if 2(3+5) = 16
then (2(3+5)) + 5 = 16 + 5

or
if 2^3 = 4*2
then (2^3)^2 = (4*2)^2

or
if 3/4 < 5
then 2(3/4) < 2*5

but WHY?!

this feels like a dumb question, because (1) I already know how to do the computation, (2) I feel like there is no answer/explanation/proof to it, you just sort of take it as assumption or something like that

What an inequality $ a < b$ means is that $ 0 < b \minus{} a$; in other words, that $ b \minus{} a$ is a positive number.  This immediately tells you that if you add two numbers to both sides of an inequality, it will remain true.  Moreover, the [i]positive numbers are closed under multiplication[/i] (this is a fancy way of saying the product of two positive numbers is positive, and this can be proven from the definition of the real numbers), so if $ b \minus{} a$ is positive then so is $ 2(b \minus{} a)$ or generally $ c(b \minus{} a)$ for any [i]positive[/i] real $ c$.

This is why inequalities have to be reversed when you multiply by negative numbers.


group theory
abstract algebra
superior algebra
superior algebra unsolved

let $ p$ be a prime and $ 1\leq n\leq p\minus{}1$ an integer, and let $ G$ be a p-subgroup of $ S_{np}$ . Prove that $ G$ is commutative

Notice it's enough to show the Sylow p-groups are commutative.  For that, consider the n p-cycles
(1,2,...,p); (p+1,...,2p); ... ( [n-1]p+1,...,np);
and the group they generate.

Steve


abstract algebra
superior algebra
superior algebra unsolved

I guess you can find this one in the archives of the forum (well, just a guess, I'm not sure).

[quote="ma_go"]correct guess...
http://www.mathlinks.ro/Forum/viewtopic.php?t=105693[/quote]
The problem you quoted is quite different from mine.

[quote="cstalg"]Let G be a group such that for all a, b in G we have (ab)^3 = (ba)^3.
Show that G is abelian.[/quote]

Since there are nonabelian group of exponent 3 (the smallest has order 3^3), this proposition is false.

LiMa


Prove that if $ x$ and $ y$ are real numbers with $ x\plus{}y\equal{}1$,

$ x^{m\plus{}1}\sum_{j\equal{}0}^{n}\binom{m\plus{}j}{j}y^j\plus{}y^{n\plus{}1}\sum_{i\equal{}0}^{m}\binom{n\plus{}i}{i}x^i\equal{}1$

for any nonnegative integers $ m$, $ n$.

[hide="The zzz way"]Put in a $ (x\plus{}y)^{m\minus{}i}$ factor in the first sum and a $ (x\plus{}y)^{n\minus{}j}$ factor in the second. After enough simplification you get $ (x\plus{}y)^{m\plus{}n\plus{}1}$ on the left.[/hide]

[hide="The awesome way"]Suppose we are selecting any subset of $ S \equal{} \{0,1,\ldots,m\plus{}n\}$. Each element has probability $ x$ of being selected and $ y$ of not being selected.

The probability of picking any subset is, um, $ 1$. Yay right hand side!

If we end up selecting at least $ m\plus{}1$ of the elements, then case work on which number the $ m\plus{}1$th selected element is. It can be any number from $ m\plus{}n$ to $ m$. Suppose it is element $ m\plus{}j$ for $ 0 \leq j \leq n$. Then of the previous $ m\plus{}j$ elements (remember $ 0$ is in the set), $ j$ of them were not selected. There are $ {m\plus{}j \choose j}$ ways to choose which were selected and which weren't, and the actual selection has probability $ x^{m\plus{}1} y^j$. Combining all of these cases gives the first term.

Otherwise, we end up not selecting at least $ n\plus{}1$ of the elements, and we do the same thing reversed. We get the second term from this.

Finally, there is the issue that $ x,y$ may not be in $ [0,1]$. But the LHS is a polynomial, and we've shown it's equal to $ 1$ on an infinite set. So it's identically $ 1$.[/hide]


I know this has been posted many times but no one seemed to have provided the solution.

The problem is equivalent to finding the greatest natural number $n$ such that $6x+9y+20z=n$ does not have natural solutions $(x,y,z)$.

I managed to show that when $n=43$ no natural solutions exist. But how can I arrive at $43$ before knowing that it is the anwer?

Thanks in advance.

Edit: :wallbash: what am I doing!!! Edited now.

First: 6x+9y=3(2x+3y), which gives solutions for multiples of 3 greater then 1.

The answer must be less then 44 (0, 20, 40) to look at. 1 is 0 mod 3, 1 is 1 mod 3, and 1 is 2 mod 3, and all are at least 3 away from numbers above 44, which means the multiple of 3 necessary is greater then 1.


Hi, I got a question which I need some explainations from the following... thx

Let $L \in R^{n \times n}$ be a nonsingular lower-triangular matrix, then prove that $L^{-1}$ is lower triangular

Here is a proof:

Let $i < j$ and let $L_{ij}$ denote the matrix obtained from $L$ by deleteing row $i$ and column $j$ then

[b]$1.$ $l_{ij}^{-1}= (1/det L)(-1)^{i+j}\times l_{ij}\times det L_{ij}$
$2.$           $= (1/det L)(-1)^{i+j}\times 0 \times det L_{ij}$
$3.$           $=0$[/b]

Thus $L^{-1}$ is lower triangular

Could anyone explain me the line $1$ and line $2$ from the above? What does that mean/ how did we get that?
Thx again!!

The proof you have is relating to the following lemma:
det(L)*I=L*[com(L)]^t (A^t is the matrix with elements a'ij=aji)
but there is an easier solution though just consider u the linear map relating to that matrix L and v the one relating to L^(-1):
you have for ei wich is a vector of the base:u(ei)=Sum l_ki*ek (i<=k) thus ei=Sum l_ki*v(ek ) a simple recursion by starting from i=n will help you to deduce that for each i: v(ei)=Sum f_ki*ek this time f_ki=(l_ki)^-1 and finally you directly conclude that L^-1 is lower triangular :)

So basicly the line $1$ is from some Lemma, right? Is that any better explaination for the line $1$ and how it goes to line $2$, thx alot.  :roll:

If you do speak french (it isn't that necessary as math isn't that relating to a language) :D here's a good link where they explain that lemma (from page 19 to 21):
http://perso.wanadoo.fr/jean-pierre.barani/DETP.PDF
From line one to the line two it's proven that if l_ij is null then the element (l_ij)^(-1) of the matrix L^(-1) is null too which means that for 0<=k<=n (l_kj)^(-1) is null for j<k Thus you conclude that L^(-1) is lower triangulat too.


This will really help my understanding... can someone prove that 4^n - 1 is always a multiple of 3? I did some strange stuff and I know it always works, but can't prove it. I've tried looking in base 3 but I don't know what to do or anything! Thanks very much if someone can show me!

Mike

[hide]Notice that 4^n=2^(2n).  So we can factor the expression as

((2^n)-1)((2^n)+1).  Every group of 3 consecutive numbers has at least one which is a multiple of 3.  The numbers that we have are 2^n-1 and 2^n+1.  2^n would make this a group of 3 consecutive numbers.  One of these is a mult. of 3.  It can't be 2^n because there is no 3 in the prime factorization.  Thus either 2^n-1 or 2^n+1 is a multiple of 3.  [/hide]

[quote="mikewd"]can someone prove that 4^n - 1 is always a multiple of 3?[/quote]

Fine proof by joml.  :)
Here's another, assuming you know the basics of modular arithmetic....

[tex]4^n = 4\times 4\times 4\times 4\times \cdots \times 4[/tex]
[tex]4~$modulo$~3 = 1[/tex]
[tex]4^n = 1\times 1\times 1\times 1\times \cdots \times 1=1~~~(\mod 3)[/tex]
[tex]4^n-1 = 1 - 1 = 0~~~~(\mod 3)[/tex]

That's exactly what I was looking for. Thanks very much!

Yet another proof.  By the formula for geometric series, the sum of the first n powers of 4 is (4n - 1) / 3.  Thus that last expression must be an integer.

In general, if n is an integer, n^x - 1 is divisible by n-1:

n  :equiv: 1 mod n-1
n^x :equiv: 1^x :equiv: 1 mod n-1
n^x - 1 :equiv: 0 mod n-1

Or use binomial theorem:

4^n-1 = (3+1)^n - 1

But I like Ravi's method better.


The 75 dollars cost for a   party was to be shared equally by all those attending, since five more people   attended than was expected , the price per person dropped by 50 cents . How many people attended the party?

Let x the number of people who were supposed to show up. If they had all showed up, each person's fee would be 75/x. However, because 5 more people showed up, each person's fee was 75/(x+5). Thus we can set up an equation: 75/x - 1/2 = 75/(x+5). Solving this equation, we have x + 5 = $ \boxed{30}$.


geometry
circumcircle
inequalities
geometry proposed

Given a traingle $ ABC$ with circumcircle $ (O)$ and three sides $ BC\equal{}a;CA\equal{}b: AB\equal{}c$.Denote by $ A_1;B_1;C_1$ the midpoints of $ BC;CA;AB$,respectively;and $ A_2;B_2;C_2$ are midpoint of the arcs $ BC$ (which does not contain $ A$) ;$ CA$(which does not contain $ B$);$ AB$(which does not contain $ C$).Draw the circles $ (O_1);(O_2)$ and $ (O_3)$ whose diameters are $ A_1A_2;B_1B_2;C_1C_2$,respectively.Prove the inequality:
$ P_{A/(O_1)}\plus{}P_{B/(O_2)}\plus{}P_{C/(O_3)} \geq \frac{(a\plus{}b\plus{}c)^2}{3}$
When does equality occur?

To mod: Please lock this topic because this problem is from Mathematics And Youth Magazine in Viet Nam. The deadline is 15-05-2008.


In fact it can be shown that if we have $x_1\ge x_2\ge \dots \ge x_n\ge 0$, $n\ge 3$ and \[x_1+x_2+\dots +x_n\le 9\\x_1^2+x_2^2+\dots +x_n^2\ge 9\]
then $x_1+x_2+x_3\ge 3$.


Let $ a,b$ and $ c$ be real numbers such that, their product is $ -64$ and:
\begin{eqnarray*}
\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} &=& \dfrac{3}{8}\\
\dfrac{1}{bc}+\dfrac{1}{ac}+\dfrac{1}{ab} &=& -\dfrac{3}{32}
\end{eqnarray*}

If $ P(x)$ is a 3rd degree polinomial with roots $ a,b,c$ and $ p(1)=70$. Find $ p(x)$.

Hide solutions

[quote="tonypr"]Let $ a,b$ and $ c$ be real numbers such that, their product is $ - 64$ and:
\begin{eqnarray*} \dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} & = & \dfrac{3}{8}.............(1) \\
\dfrac{1}{bc} + \dfrac{1}{ac} + \dfrac{1}{ab} & = & - \dfrac{3}{32}...............(2) \end{eqnarray*}

If $ P(x)$ is a 3rd degree polinomial with roots $ a,b,c$ and $ p(1) = 70$. Find $ p(x)$.

Hide solutions[/quote]

[hide="solution"]
Multiply $ (1)$ by $ abc=-64$,  $ ab+bc+ca=-24.............(i)$
Multiply $ (2)$ by $ abc=-64$,  $ a+b+c=6.............(ii)$
$ abc=-64............(iii)$

So the polynomial $ P(x)=k(x^3-6x^2-24x+64)$

$ P(1)=k(35)=70 \Leftrightarrow k=2$

So $ P(x)=2x^3-12x^2-48x+128$
[/hide]

[hide]Writing both equations with common denominators and simplifying, we get

$ bc \plus{} ac \plus{} ab \equal{} \minus{} 24$

$ a \plus{} b \plus{} c \equal{} 6$

This would be very useful if we use Vieta's formulas.

$ p(x) \equal{} a_nx^3 \minus{} 6a_nx^2 \minus{} 24a_nx \plus{} 64a_n$

You can find what $ a_n$since it's given that $ p(1) \equal{} 70$.

$ p(1) \equal{} a_n \minus{} 6a_n \minus{} 24a_n \plus{} 64a_n \equal{} 70$
$ p(1) \equal{} a_n(35) \equal{} 70$
$ a_n \equal{} 2$

So the polynomial is $ \boxed{p(x) \equal{} 2x^3 \minus{} 12x^2 \minus{} 48x \plus{} 128}$[/hide]

A slight variation. Consider a polynomial $ Q(x)$ of third degree whose roots are the reciprocals of $ P(x)$. If $ P(x)\equal{}K(Ax^3\plus{}Bx^2\plus{}Cx\plus{}1)$, then $ Q(x)\equal{}K(x^3\plus{}Cx^2\plus{}Bx\plus{}A)$. From the first equation, $ C\equal{}\frac{\minus{}3}{8}$. From the second, $ D\equal{}\frac{\minus{}3}{32}$, and from $ abc\equal{}\minus{}64 \Rightarrow \frac{1}{abc}\equal{}\frac{\minus{}1}{64}$, so $ A\equal{}\frac{1}{64}$. Now we have $ P(x)\equal{}K(\frac{1}{64}x^3\minus{}\frac{\minus{}3}{32}x^2\minus{}\frac{\minus{}3}{8}x\plus{}1)$. $ P(1)\equal{}70 \Rightarrow K\equal{}128 \Rightarrow P(x)\equal{}2x^3\minus{}12x^2\minus{}48x\plus{}128$.


[quote="guru_bhai"][hide="Hint"]
$ \triangle ABC \sim \triangle DCE$[/hide][/quote]

[hide="Minor quibble..."]
From the texts I've read, the similarity correspondence should be written as $ \triangle ABC \sim \triangle DEC$, such that we can unambiguously declare how the triangles are related. That is, we can instantly gather that $ \angle A \equal{} \angle D, \frac{AC}{BC}\equal{}\frac{DC}{EC}$, etc., even without an accompanying diagram. Granted, since there is a figure here to refer to, there probably would not be any problems with this.
[/hide]

[quote="21 7 15"]clearly,a=4[/quote]
[hide]
you shouldn't assume from the image that the triangles are necessarily congruent

Since $ \triangle ABC \sim \triangle DCE$,

$ \frac {12 \plus{} 18}{6a} \equal{} \frac {3a \plus{} 6a}{18}$

$ a \equal{} \sqrt {10}$

Simple calculations to get $ AB \equal{} 9\sqrt {6}$

Since $ \triangle DBK \sim \triangle ABC$

$ 12\cdot\frac {18}{9\sqrt {6}} \equal{} 4\sqrt {6} \equal{} BK$

$ \frac {12\cdot4\sqrt {6}}{2} \equal{} \boxed{24\sqrt {6}}$

hehe felt like messing around with asymptote.. couldn't figure out how to put the right angle mark on point $ E$, though :huh:
I don't think my code's exactly efficient, either XD

[asy]draw((0,0)--(30,0)--(18,6sqrt(6))--cycle);
draw((12,0)--(12,9sqrt(6))--(30,0));
draw((0,0)--(12,0)--(12,4sqrt(6))--cycle,linewidth(1));
draw((12,0)--(13,0)--(13,1)--(12,1)--cycle,linewidth(.3));
label("$A$",(12,9sqrt(6)),NW);
label("$B$",(12,0),S);
label("$C$",(30,0),SE);
label("$D$",(0,0),SW);
label("$E$",(18,6sqrt(6)),NE);
label("$K$",(12,4sqrt(6)),NW);
label("12",(6,0),S);
label("18",(21,0),S);
label("3$a$",(15,7.5sqrt(6)),NE);
label("6$a$",(24,4sqrt(3)),NE);[/asy]
[/hide]

[quote="21 7 15"]clearly,a=4[/quote]
First, this isn't really a hint if you just state something, without explaining it OR backing it up. Next time, don't assume random stuff thats totally wrong.

[quote="21 7 15"]simply join k to c then you will see that 18=6a,a=3.
By pythagoras,the altitude =s3multiplied by the square root of 35.
In conclusion the area of the small triangle is equal to 6(sqrt35) :wink:[/quote]Wrong, $ KC$ is not an angle bisector.


Hello, i'm new here and I realize of which it is a very useful site. This problem came me to the mind and is less difficult of which it seems.

Probe that prove that in all quadrilateral an ellipse can be inscribed.

First of all, welcome to the forum!

The best argument I can think of is that any quadrilateral can be "stretched" with respect to some line: ie, we can apply a planar mapping which makes all points farther away from some given line by a factor of some number.  Well anyways, we can stretch the quadrilateral so that a circle can be inscribedin it: ie, so that the sums of opposite side lengths are equal.  Then, we inscribe a circle in that, and apply the inverse transformation to get our ellipse (because this transformation preserves intersection points and thereby preserves tangency).  However, I can't formalize that...

[color=darkred]See http://mathworld.wolfram.com/Inellipse.html where you can meet the following sentence :
[b]Newton gave the solution to inscribing an ellipse in a convex quadrilateral[/b] (Dorrie 1965, p. 217). [/color]


conics
parabola
calculus
symmetry
analytic geometry
quadratics

A circle rests in the interior of the parabola with equation $y=x^2$ so that it is tangent to the parabola at two points. How much higher is the center of the circle than the points of tangency?

Don't we need to know the radius of the circle or something?

Let $C (0,a)$ be the center of the given circle and $T (t,t^2)$ be the point of tangency in the first quadrant.

By $\frac{d}{dx}x^2=2x,$ we have $\frac{t}{a-t^2}=2t\Longleftrightarrow a-t^2=\boxed{\frac{1}{2}}.$

Calculus is not necessary.

[hide="hint"]
Equation of the circle
[/hide]

Yes, there are another three solutions including your solution.

But your solution that you are thinking has posiibility to have imcomplete solution.

Please show your solutiuon, chess 64.

[hide]
By symmetry the $x$-coordinate of the center of the circle is 0 so let the circle be centered at $(0,k)$ with radius $r$. The equation of the circle is $x^2+(y-k)^2=r^2$. At the point of tangency, $y=x^2$ so substitute to get \[ x^2 + (x^2-k)^2 = r^2, \qquad \text {or}\qquad x^4-x^2(2k-1)+(k^2-r^2)=0, \] which is a quadratic in $x^2$. There can only be one solution so the discriminant must be equal to 0.
[/hide]

Too lazy to finish... :P

You should give another condition, chess64. :D


algebra
polynomial
inequalities
geometry
blogs
geometric transformation
reflection

Can someone post a solution to these problems as I have no idea how to do most of them!
Thanks in Advance!

Posting the questions would be helpful. :P
Oh, and don't double post...

I have no comment about problem 1 (because I don't really get the problem , it would be helpful if someone put it in other words  :P )
problem 2 : see that if ax^2+bx+c=0 has rational roots then $ \delta \equal{}b^2\minus{}4ac$ has to be a perfect square .
However , since b ,a,c are all odd then $ b^2\minus{}4ac$ gives the remainder is 5 when divided by 8 , Therefore , delta can not be a perfect square which is a contradiction .

problem 3: denote $ X\equal{}\bar{abcdef}$ then $ Y\equal{}\bar{defabc}$ all we have to do is to prove that |X-Y| is divisible by 27
which is clearly true since $ |X\minus{}Y|\equal{}27.37(\bar{abc}\plus{}\bar{def})$

problem 4: Let ABC is a triangle that has 2 equal half-inscribed circle , (I) (I on AB ) and (K) (K on AC)
II' and KK' perpendicular to AC and AB , see that triangle AK'K and AI'I are equal , therefore AI=AK since then AB=AC which is QED

Problem 1:
Verify that the numbers 1, 1, 2, 3, 5, 8 do not contain a triplet which represents the sides of a triangle.

Let the numbers $ 1 \le a_1 \le a_2 \le \cdots \le a_7 \le 10$ be given. If for any numer $ k\in\{0,\dots,5\}$ the relation $ a_k\plus{}a_{k\plus{}1} > a_{k\plus{}2}$ is satisfied, the numbers represent a triangle ($ a_{k\plus{}2} > |a_{k\plus{}1}\minus{}a_k|$ is satisfied by the definition of $ a_i$). Now suppose that $ a_{k\plus{}2} \ge a_{k\plus{}1} \plus{} a_k$ for all $ k$; using $ a_1 , a_2 \ge 1$, we get $ a_7 \ge 13$ which is clearly impossible.

Generalization: If $ F_k$ denotes the kth Fibonacci number ($ F_0 \equal{} F_1 \equal{} 1$), then from the numbers $ 1,\dots,n$ we can choose $ k$ numbers such that no three represent the sides of a triangle iff $ n\ge F_{k\minus{}1}$. (Might be slightly off, but w/e).

@HTA
For number three you lost me after the $ |X \minus{} Y| \equal{} 27.37(\bar{abc} \plus{} \bar{def})$
From what I picked up you are basically saying that if two numbers that are divided by the same number $ N$ and they will have the same remainder $ R$,
then $ |X \minus{} Y|$ must be divisible by $ N$.
A more descriptive and easier to read "idiot version" of your explanation is needed.
Thanks in Advance  :D

ok then, if X and Y have the same remainder S (S>0) when divided by N we may denote X as N*a+S and Y as N*b+S , so |X-Y|=|N(a-b)| , therefore , N | |X-Y|

The formula is supposed to read $ |X-Y| = 27 \cdot 37|\overline{abc}-\overline{def}|$. Recall that $ \overline{abcdef} = 1000\cdot\overline{abc} +\overline{def}$ and that $ 999 = 27\cdot 37$.

I am either not seeing something, or number 2 does not make any sense:
$ x^2 \plus{} 6x \plus{} 9 \equal{} 0$
This is a square polynomial and each of the coefficients is an odd integer.

Edit:  I knew something had to be up, since I already saw that a perfect square was not obtainable in the discriminant, yet I somehow didn't notice the blatant odd coefficent in my polynomial, thank you for pointing it out.

[hide="question 1"]

Let $ x_1,x_2,...,x_7$ be the lengths of the sides, $ x_1 \leq x_2 \leq ... \leq x_7$

Assume the contrary, that the 7 sides work.

By the triangle inequality, we know that $ x_{i} \geq x_{i \minus{} 1} \plus{} x_{i \minus{} 2}, i$ belonging to {3,4,5,6,7}, since otherwise, $ x_{i} < x_{i \minus{} 1} \plus{} x_{i \minus{} 2},$ and since$ x_{i \minus{} 1}\leq x_{i} < x_{i} \plus{} x_{i \minus{} 2},$
$ x_{i \minus{} 2}\leq x_{i} < x_{i} \plus{} x_{i \minus{} 1}$, which makes $ x_{i \minus{} 2},x_{i \minus{} 1},x_{i}$ a triple which form the sides of a triangle

Therefore,

$ x_3 \geq x_2 \plus{} x_1 \geq 2,x_4 \geq x_3 \plus{} x_2 \geq 2 \plus{} 1 \equal{} 3, x_5 \geq x_4 \plus{} x_3\geq 3 \plus{} 2 \equal{} 5, x_6 \geq x_5 \plus{} x_4 \geq 5 \plus{} 3 \equal{} 8, x_7 \geq x_6 \plus{} x_5 \geq 8 \plus{} 5 \equal{} 13$

Therefore, 7 always has a possibility.

For 6, consider 1,1,2,3,5,8

For every 3, the smaller 2 are less than the third, since the smaller 2 sum to less than or equal to the 2 numbers preceding the third, which sum to the third.

$ x_{i} \plus{} x_{j}\leq x_{k \minus{} 2} \plus{} x_{k \minus{} 1} \equal{} x_{k}$ for all i,j,k

[/hide]

[hide="question 2"]

$ (2k \plus{} 1)^2 \equal{} 4k^2 \plus{} 4k \plus{} 1 \equal{} 4(k^2 \plus{} k) \plus{} 1$ and since$ k^2 \plus{} k$ is even, $ 4(k^2 \plus{} k) \plus{} 1$ is congruent to 1 mod 8.

Let a=2x+1, b=2y+1, c=2z+1

The discriminant $ b^2 \minus{} 4ac$ must be a perfect square, and since b^2 is odd, 4ac is even, b^2-4ac is an odd perfect square, so must be congruent to 1 mod 8, as said above.

b^2 is congruent to 1 mod 8 as said above

$ 4ac \equal{} 4(2x \plus{} 1)(2z \plus{} 1) \equal{} 8(2xz \plus{} x \plus{} z) \plus{} 4$ which is congruent to 4 mod 8

1-4=-3 which is congruent to 5 mod 8

but b^2-4ac must be congruent to 1 mod 8, contradiction!

[/hide]

[hide=" question 3"]

Let a be once of the 3 digit numbers, c be the second 3 digit number.

$ 1000a \plus{} c \equal{} (27*37 \plus{} 1)a \plus{} c \equal{} 27*(37a) \plus{} a \plus{} c$ which is congruent to $ a \plus{} c$ mod 27

$ 1000c \plus{} a \equal{} 27*(37c) \plus{} a \plus{} c$ which is congruent to $ a \plus{} c$ mod 27 as well

therefore, 1000a+c is congruent to 1000c+a mod 27, so leave the same remainder upon division.

[/hide]

[hide=" question 4"]

Let the triangle have side lengths a,b,c, opposite vertice A,B,and C respectively.

WLOG, let the semicircles lie on side a and b.

Let the common radius of these 2 circles be r.

Let the center of the circle whose center lies on a have center A1, and the center which lies on b have center B1.

Then, the line A A1 divides the triangle into 2 smaller triangles, which have areas b*r and c*r, using the base height formula for the area of a triangle, and since the radius makes a perpendicular with the sides since they are tangent.

By similar logic, B B2 divides the triangle into 2 smaller triangles with areas a*r+c*r.

Now, ABC area=b*r+c*r and TABC area=a*r+c*r

so b*r+c*r=a*r+c*r

a*r=b*r

a=b

Therefore, the triangle is isoceles.

[/hide]

For number four, I was thinking if you could assume since the two sides are tangent, and they are drawn from the same vertex, you could assume that the two lines would be congruent, which would thus that the two sides are congruent and allowing us to conclude that the side must be isosceles.
(Also I've kind of noticed that pretty much all of the problems want you to not only show your work but "prove" or "show" how it is mathematically correct. I have done some of this in terms of geometric proof but nothing much more than that, is this something that is important in an Olympiad?

[quote="aahmadi"]I am either not seeing something, or number 2 does not make any sense:
$ x^2 \plus{} 6x \plus{} 9 \equal{} 0$
This is a square polynomial and each of the coefficients is an odd integer.[/quote]

6 is not an odd integer

I discuss #2 in some depth in [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=196301]a recent blog post[/url]; it's quite a well-known problem.

An approach to #4:  a half-inscribed circle in triangle $ ABC$ with respect to side $ BC$ is fully inscribed in quadrilateral $ ABA'C$, where $ A'$ is the reflection of $ A$ across $ BC$.

[quote="neurosurgeon1118"]Also I've kind of noticed that pretty much all of the problems want you to not only show your work but "prove" or "show" how it is mathematically correct. I have done some of this in terms of geometric proof but nothing much more than that, is this something that is important in an Olympiad?[/quote]

This is the fundamental thing that distinguishes Olympiad math (and in fact, "real" math) from classroom (high-school) math.


calculus
algebra
polynomial
function
limit
trigonometry
algebra unsolved

I need help proving this.  Sorry if this is the wrong forum, I wasn't quite sure where to post, as this is my first posting of a proof...

Prove: $e^{i\pi}+1=0$.

Prove Euler's formula??

Well you can use calculus:
http://mathworld.wolfram.com/EulerFormula.html

Yes.  Is there a way to do it with only precalculus?  MathWorld is a bit complicated, could someone post as simple a proof as possible?  (I understand De Moivre's and all that stuff...).

It is not exactly a fact, rather its a definition, how ever this fact/definition comes from the thailor series of $e^x$ and $sin x$ and $cos x$, I mean like this :
\[ e^x=x^0+\frac{x^1}{1!}+\frac{x^2}{2!}+\ldots \]
\[ cos x=x^0-\frac{x^2}{2!}+\frac{x^4}{4!}\mp\ldots \]
\[ sin x=\frac{x^1}{1!}-\frac{x^3}{3!}\pm\ldots \]
So if in the series of $e^x$ we put $ix$ instead of $x$ we easily derive at $e^{ix}=cos x+i sin x$ which implies the result.

I think the problem is how to show that the Taylor polynomial of e^x is correct in complex numbers.

Use the following theorem (I will not prove it since the proof is lengthy):

The remainder (error) of an nth degree Taylor polynomial of f(x) centered at c is $\frac{f^{(n+1)}(z)}{(n+1)!}(x-c)^{n+1}$ where z is some value between c and x.

So, if the above expression converges to 0 as n becomes infinite, the Taylor series of that funtion has a remaninder of 0 (i.e. is exact).

$\lim_{n\to\infty}\frac{\frac{d}{dx}\left(e^x\right)}{(n+1)!}(x-c)^{n+1}=\lim_{n\to\infty}\frac{e^x}{(n+1)!}(x-c)^{n+1}=0$
So, the Taylor series for $e^x$ converges to $e^x$ for all x.

Proving the convergance of the Talyor series of sine and cosine to those functions for all x can be done likewise.

The rest has been explained by Nima Ahmadi Pour.


geometry
3D geometry
sphere
tetrahedron
circumcircle
radical axis
geometry unsolved

The inscribed sphere of a tetrahedron $ABCD$ touches $ABC,ABD,ACD$ and $BCD$ at $D_1,C_1,B_1$ and $A_1$ respectively. Consider the plane equidistant from $A$ and plane $B_1C_1D_1$ (parallel to $B_1C_1D_1$) and the three planes defined analogously for the vertices $B,C,D$. Prove that the circumcenter of the tetrahedron formed by these four planes coincides with the circumcenter of tetrahedron of $ABCD$.

$O$ is the circumcenter of $ABCD$ and $\ell_a,\ell_b,\ell_c,\ell_d$ denote the perpendiculars dropped from $O$ to the faces $DBC,$ $DCA,$ $DAB,$ $ABC,$ respectively. Label $\delta_A$ the plane equidistant from $A$ and $B_1C_1D_1$ (parallel to this). Planes $\delta_B,\delta_C,\delta_D$ are denoted analogously. Then $\delta_A,\delta_B,\delta_C$ are the radical planes of the insphere $(I)$ of $ABCD$ and the spheres $(A),$ $(B),$ $(C)$ with zero radii, respectively $\Longrightarrow$  $D' \equiv \delta_A \cap \delta_B \cap \delta_C$ is the radical center of $(A),(B),(C),(I)$ $\Longrightarrow$ $D'$ has equal power with respect to $(A),$ $(B),$ $(C),$ $(I),$ i.e. $AD'=BD'=CD'$ $\Longrightarrow$ $D' \in \ell_d.$ Likewise, $A' \equiv \delta_B \cap \delta_C \cap \delta_D,$ $B' \equiv \delta_C \cap \delta_A \cap \delta_D$ and $C' \equiv \delta_A \cap \delta_B \cap \delta_D$ lie on $\ell_a,\ell_b,\ell_c,$ respectively. Now, tetrahedra $A_1B_1C_1D_1 \sim A'B'C'D'$ are homothetic and $IA_1 \parallel OA',$ $IB_1 \parallel OB',$ $IC_1 \parallel OC',$ $ID_1 \parallel OD'$ means that $I,O$ are homologous points. Thus, since $I$ is the circumcenter of $A_1B_1C_1D_1,$ then $O$ is the circumcenter of $A'B'C'D',$ as desired.

Let $A'B'C'D'$ be the tetrahedron formed by the four planes. Let $I$ be the incenter of the tetrahedron. Note that $D'$ is the radical center of $(A), (B), (C), \omega$, where $\omega$ is the incircle of the tetrahedron and $(P)$ denotes the point-circle at $P$. Similar results hold for $A', B', C'$. Therefore, we know that $C'D'$ is perpendicular to the plane of $ABI$ (Carnot, say). Since $D'A = D'B = D'C$, if we let $O$ be the circumcenter of $ABCD$, then we have that $OD'$ is perpendicular to the plane of $ABC$. Similarly, $OC'$ is perpendicular to the plane of $ABD$. Hence, to show that $OC' = OD',$ we want to show that $C'D'$ makes an equal angle with the planes of $ABC, ABD$. However, as $ABI$ is the "bisector" of the planes of $ABC, ABD$, and $C'D'$ is perpendicular to the plane of $ABI$, we're done.

$\square$


In an arena, each row has 199 seats. One day,
1990 students are coming to attend a soccer match. It is only
known that at most 39 students are from the same school. If
students from the same school must sit in the same row, determine
the minimum number of rows that must be reserved for these
students.

[i]My first attempt [/i]
[hide]
(Up to) 39 students per school
199 seats per row
199/39=5.1025641025641025641025641025641
39*5=195
1990/195=10.205128205128205128205128205128
eleven rows required
 :spider:
[/hide]

[i] My best answer (with help) [/i]
[hide]
What if there were exactly 34 students...?
34 students per school
199 seats per row
199/34= 5.8529411764705882352941176470588
34*5=170
1990/170=11.705882352941176470588235294118
12 rows required... :diablo:
[/hide]

39 is only divisible by 3 (and 13) --not sure how helpful that is. I [i]think[/i] that if some of the schools have different sizes, they just make the packing easier.

A very similar problem (in some sense the dual?) was posted [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1585688#1585688]here[/url], and the argument I gave there shows that 11 is insufficient (as does the second answer of the previous poster).  In this case, we need to show that 1990 students can always be accomodated in 12 rows.  That 13 rows suffices is simple: packing randomly, we can always fit at least $ 199 \minus{} 38 \equal{} 161$ students into each row, and $ 13 \cdot 161 > 1990$.

What is the source for this problem?

How do you know that 12 will [i]not[/i] always work?
Can you give more explanation why 161 is the magic number?

I don't know that 12 will not always work -- indeed, I suspect it does (as I noted in the second sentence of my previous post).  I don't have a proof either that 12 will always work or a proof that 12 will sometimes not work -- I have a proof that 11 will sometimes not work, and a proof that 13 will always work.  Is it clear now?

Hmm... I wonder what apratimdefermat has to say about the proof...?


AMC
AIME
geometry
3D geometry
sphere
Pythagorean Theorem

http://www.artofproblemsolving.com/Wiki/index.php/1998_AIME_Problems/Problem_10

Hmm lets call the circle O_1, O_2,...O_8. and the big circle P.

if we draw triangle O_1 P O_2, dont we have an isoceles triangle with sides 200, 100+r, and 100+r?

but then we can find the altitude of the isosceles triangle because it is half the side of the square that forms the octagon.

So then we could apply pythagorean theorem and solve for r.

Why was my solution wrong?

umm can somebody take a look at this and tell me where i screwed up please?

[quote="unimpossible"]http://www.artofproblemsolving.com/Wiki/index.php/1998_AIME_Problems/Problem_10

but then we can find the altitude of the isosceles triangle because it is half the side of the square that forms the octagon.

[/quote]

Sorry, Im kind of confused but whats that suppose to mean?

Woudnt that isosceles triangle be slanted?  Like it wouldnt be parallel to the octagon so how can you relate it to the octagon?  Since the R=radius of the 9th sphere will be >100 and thus the center will be above the centers of each of the other 8 spheres.

I just briefly read through and I think it's mainly because you are treating spheres as circles....and that height of the isoscles triangle would be in a non-parallel plane to the height of the square so the lenghts can't be the same


AMC
AIME
USA(J)MO
USAMO
AMC 10
AMC 12
probability

If you are in a situation where you have 22 problems answered, and CANNOT do the last 3, do you guess?

If you say no, what if you have 23 right?  The should you guess?

If I have 24, I will definitely guess on the last one.  :P

I'm thinking if you have 23, you should go for the perfect score.  On the other hand, omissions could mean a state high score.  On an AMC-type in-state test (USC) a 10th grader won by getting 24 and omitting 1.  If something like that happens again, omissions will be the key factor.

So when is it a good idea to guess?  I think I would guess if I can eliminate at least 2 answer choices, but I just want to hear your ideas.

I would work like crazy trying to crack the last 2/3 problems, but if i get nothing definite then i would just screw it and leave it blank.

If i got 24 right i would definetly guess since, I must have had a lot of good luck/intuition getting so many right so far, it makes total sense to just go for it using the intuition that got you this far.

Good reasoning fireforce. Haha...
However if I base anything I do on intuition, I will get below a 37.5.  :lol:

depends on how confident you are on your aime score (dont forget usamo index)

I never guess. If I got 24, I would rather stick with the 145.5 than have a 20% chance of 150 and 80% change of 144.

Though I know this doesn't really apply here, I told other people at my school to always answer either 21 or 25 on the AMC10 or 25, 22, 18, or 14 on the AMC12, since these are ALWAYS better than the alternatives (it's not even probability, it's a fact).

AMC10 NEVER APPLIES TO USAMO INDEX. People seem to have a bit more trouble understanding this next one, but freshmen and sophomores will NEVER qualify for the USAMO by index (with the AMC12 score) but not AIME cutoff score. So index never matters for underclassmen.

so if you have done 19 problems, you should guess on a few?

If I absolutely cannot answer a question and i have 24 right, i'll guess it.

If 23 right and i can't figure out either question, I'd leave them blank.  I don't care so much about how well i do in the AMC...I just can't bear missing usamo again this year...and I know how easy it is to screw up the AIME.


Hey all. On amazon.com, I found a book of problems (and solutions) from the "Miklos Schweitzer" contest, which is apparently a take home, 10 day/10 problem contest. How difficult are these problems compared to say, IMO or Putnam? I'm going to be a math major, so would these be a good source of problems to have around?

Thanks!

~Arachnotron

try "Miklos Schweitzer" and the "search" button of the forum ...


LaTeX
inequalities
inequalities unsolved

Let a,b,c are positive real numbers such that abc=1
prove that (a-1+${frac\{1}{b}$ )+(b-1+${frac\{1}{c}$ )+(c-1+${frac\{1}{a}$ ) $\leq$ 1

this problem is out of an ISL or an IMO. so please post your source and try to write clear text...

Naphthalin

[quote="2006"]Let $ a,b,c$ are positive real numbers such that $ abc \equal{} 1,$ prove that

$ \left(a \minus{} 1 \plus{} \frac {1}{b}\right)\left(b \minus{} 1 \plus{} \frac {1}{c}\right)\left(c \minus{} 1 \plus{} \frac {1}{a}\right) \leq 1.$[/quote]This problem is introduced as Example 2.2.5 on page 41 in new book :

Ji Chen, Chao-Cheng Ji, [url=http://kaoshi.book110.cn/book426712/]Algebraic Inequalities (Mathematical Olympiad Proposition people Lecture)[/url], Shanghai Scientific & Technological Educational Publishing House, 2009, 225p. (ISBN 978-7-5428-4848-2/O·613)

Let $ x \equal{} a \plus{} \frac {1}{b},y \equal{} b \plus{} \frac {1}{c},z \equal{} c \plus{} \frac {1}{a}$, then $ abc \equal{} 1\Longleftrightarrow x \plus{} y \plus{} z \plus{} 2 \equal{} xyz.$

More general result is :

If $ x,y,z$ are non-negative numbers such that $ |x \plus{} y \plus{} z \minus{} xyz|\leq 2$, then

$ (x \minus{} 1)(y \minus{} 1)(z \minus{} 1)\leq 1,$

with equality if $ x \equal{} 2,y \equal{} z \equal{} 0.$

See : http://www.mathlinks.ro/viewtopic.php?t=314860


If n, x, y are positive integers with n+x  being a square number and  n+y the next larger square, prove that:

1.   n+xy  and n+x+y +xy are adjacent squares.

2.   n+x+y+xy(2+x+y+xy) is a square

$ n \plus{} x \equal{} k^2$ and $ n \plus{} y \equal{} (k \plus{} 1)^2$ with $ n,x,y \in \mathbb Z$
then $ x \equal{} k^2 \minus{} n$ and $ y \equal{} (k \plus{} 1)^2 \minus{} n$
$ 1)$
$ n \plus{} xy \equal{} n \plus{} (k^2 \minus{} n)((k \plus{} 1)^2 \minus{} n) \equal{} (k^2 \plus{} k \minus{} n)^2$ and
$ n \plus{} x \plus{} y \plus{} xy \equal{} (k^2 \plus{} k \minus{} n)^2 \plus{} (k^2 \minus{} n) \plus{} ((k \plus{} 1)^2\minus{}n) \equal{} ((k^2 \plus{} k \minus{} n) \plus{} 1)^2$

$ 2)$
$ n \plus{} x \plus{} y \plus{} xy(2 \plus{} x \plus{} y \plus{} xy) \equal{} (1 \plus{} k \plus{} 2 k^2 \plus{} 2 k^3 \plus{} k^4 \minus{} 2 n \minus{} 2 k n \minus{} 2 k^2 n \plus{} n^2)^2$
by replacing, expanding and then collecting terms.


$ (h_a)^2\equal{}2(R)^2$ that $ h_a$ is altitude from A and $ R$ is circumradius.$ h_a$ is intersect $ BC$ at $ H$.$ HE,HF$ is perpendicular to $ AB,AC$ ($ E$ is on $ AB$ and $ F$ is on $ AC$) if $ O$ be the circumcenter of $ ABC$ prove that $ E,F,O$ are lies on a line

Since $\angle AEF=\angle AHF=\angle ACB,$ it follows that $EFCB$ is cyclic, i.e $\triangle AFE$ and $\triangle ABC$ are similar $\Longrightarrow$ $EF$ is antiparallel to $BC,$ i.e. $AO \perp EF.$ If $P \equiv AO \cap EF,$ then $E,O,F$ are collinear $\Longleftrightarrow$ $O \equiv P$

$\Longleftrightarrow \frac{R}{AH}=\frac{EF}{BC}=\frac{AF}{AB}=\frac{AH^2}{AC \cdot AB}=\frac{AH}{2R}\Longleftrightarrow AH=\sqrt{2}R.$

The inversion of pole $A$ and power $AH^2$ transforms the circle $\odot (ABC)$ into its diameter perpendicular to $OA$ and, if this diameter intersect $AB$ and $AC$ at $E'$ and $F'$ respectively, we shall get $AE'\cdot AB=AF'\cdot AC = AH^2$, i.e. $HE'\perp AB$ and $HF'\perp AC$, or $E' \equiv E$ and $F'\equiv F$, done.

Best regards,
sunken rock

[quote="sunken rock"] $HE'\perp AB$ and $HF'\perp AC,$ or $E' \equiv E$ and $F'\equiv F,$ done.[/quote]
How does this  lead directly to $E,O,F$ are collinear $\Longleftrightarrow AH=\sqrt{2}R \ ?$

My reasoning: Line $EF$ cuts $(O)$ at double points $U,V$ under the inversion $\Longrightarrow$ $AH=AU=AV$ is the radius of the inversion. Then $AU=AV=AH=AO \sqrt{2}$

To: luisgeometra
1) The problem does not ask 'iff'
2) $AU=AO\sqrt{2}$ iff $UV$ is a diameter of the circle, i.e. $E, O, F$ collinear.

With best regards,
Stan Fulger,
Master Master


function
integration
limit
calculus
real analysis
probability
real analysis unsolved

[quote="Kalle"]Suppose $f: R \to R$ is a nice enough non-negative function such that $\int_{-\infty}^{\infty}f$ exists as a finite value. Can we show that
$\lim_{x\to-\infty}\int_{-\infty}^{x}f = 0$?

Do we need to require anything of $f$?[/quote]
To be completely formal about it, yes you do to require something.

Ignoring the fact that you left the differential out of the integral, the integral $\int_{-\infty}^{\infty}f(u)du$ is improper and should be expressed as: $\lim_{h\to \infty}\int_{-h}^{+h}f(u)du = L$, i.e. said limit exists.

We know then that $\lim_{h\to\infty}F(-h)$, where $F$ is the antiderivative of $f$ exists; let's call this limit $\ell$. And that $\lim_{x\to-\infty}\lim_{h\to\infty}\int_{-h}^{x}f(u)du = \lim_{x\to-\infty}\lim_{h\to\infty}F(x)-F(-h) = \ell-\ell = 0$.

[quote="Kalle"]I was thinking of a more general situation with Lebesgue integrals, $f$ not necessarily having a primitive.

edit: I was wrong about something :)[/quote]
That's okay... I presumed you meant Riemann integrable. I have no idea how to prove it using the Lebesgue integral.

Define $g_{x}(u)=\begin{cases}f(u)&u\le x\\0&u>x\end{cases}$

Then for any sequence $x_{n}\to-\infty,\ g_{x_{n}}\to 0$ pointwise.

$|g_{x}(u)|\le f(u)$ for all $u$ and $f$ is integrable.

That meets the requirements of the Lebesgue Dominated Convergence Theorem. Therefore

$\lim_{n}\int_{-\infty}^{\infty}g_{x_{n}}(u)\,du= \lim_{n}\int_{-\infty}^{x_{n}}f(u)\,du=0$

Since this holds for all such sequences, we have the result you want.

You could relax the condition "$f$ is nonnegative." What you need is that $f\in L^{1}(\mathbb{R}).$ Given that, the same would be true, with essentially the same proof.

If you don't have $f\in L^{1},$ then the integral of $f$ is not defined as a Lebesgue integral. If you try to force it to mean something anyway, then you would have to define what you are talking about as an improper integral, which makes for a different problem.

Kalle's title refers to density functions - by that, I presume he means probability density functions. In that case, the result he is looking for is still true even if the distribution fails to have a density (that is, is not absolutely continuous.)

Suppose $X$ is a random variable. Define the [i]cumulative distribution function[/i] as follows:

$F(x)=P(X\le x).$

Claim: $\lim_{x\to-\infty}F(x)=0.$

A random variable is a measurable real-valued function defined on a probability sample space. Call the space $S$ and use $\omega$ as a generic point in $S.$

As above, consider a sequence $x_{n}$ that decreases to $-\infty.$ Let $E_{n}=\{\omega\in S: X(\omega)\le x_{n}\}.$

Then $S\supseteq E_{1}\supseteq E_{2}\supseteq E_{3}\cdots$ and $\bigcap_{n=1}^{\infty}E_{n}=\emptyset.$

The first two lemmas in measure theory look like this:

Lemma 1: Given an increasing sequence of measurable sets, the limit of the measures is the measure of the union.

Lemma 2: Given a decreasing sequence of measurable sets, for which the first set has finite measure, the limit of the measures is the measure of the intersection.

Lemma 2 applies to our case (We're starting from finite measure because this is, after all, a probability space.) We conclude that

$\lim_{n}F(x_{n})=\lim_{n}P(E_{n})=P(\emptyset)=0.$

Since this holds for all such sequences, we have $\lim_{x\to-\infty}F(x)=0.$

A closely related argument could be used to show that

$\lim_{x\to\infty}F(x)=1.$

Thanks for your help Kent Merryfield. I can't stand the probability course I am taking; it feels like I'm back at high school. Everything is hazily done and very little is proven.


This actually shouldn't be posted here but I think a lot of people in the Getting Started read this than in the Questions/Suggestions Forum. This is just a suggestion:

I have rigorously browsed the past problems in Getting Started and a lot of problems have been left hanging; some have no answers and some have no confirmation if the answers are correct.

Though we cannot avoid those things, if you are giving a problem and you have the answer, kindly post it.  After all, we are looking for solutions.

I have synthesized a lot of problems, and Im going to post it later in pdf form and hopefully can collect/put solutions on it.

Problem Solving Marathons are very good since nobody can post until the right solution/answer is posted.

Im sorry guys, if I feel like Im the moderator. This is just a suggestion, and the moderators can move/remove this post, if they want to. :)

I think for most of the problems without answers, nobody knows the answer. It's not that they just don't want to post it. Those questions should probably be moved Intermediat Topics or something harder.

I think that this is an excellent idea. Thanks for your commitment to the forums.  :)

I'm changing this to an announcement for a while.

Indeed, I agree that they have no answer because they were posted in the wrong forum to begin with.  This is yet another reason it is important that people post in the appropriate forum.

Let me just make a short comment.

You'll see a series of problems in Getting Started, Intermediate, and also AMC forum (not yet but I'll try to divide problems between each forum so one forum doesn't stuck with all posts made by me  :D ). This is a VERY good chance for you to develop your problem solving.

Only warning I give is this:

1. Hide your answer
2. You CANNOT just have "A" or "B" as answer. Such an answer will be deleted and if this is repeated, you're going to receive PM from me.  :)


If $ x_1,\ldots,x_n$ are $ n$ distinct integers and $ s_i \equal{} \sum_{j\neq i} x_j,$ prove that
\[ \sum_{i \equal{} 1}^n \prod_{j\neq i} \frac {s_j}{x_i \minus{} x_j} \equal{} 1.
\]

Noone? Well, then enjoy this
[hide="Hint"]Set $ x \equal{} \sum_{i\equal{}1}^n x_i,$ rewrite the LHS as $ \sum_{i\equal{}1}^n\prod_{j\neq i}\frac{x\minus{}x_j}{x_i\minus{}x_j} \equal{} f(x)$ and prove that $ f(x)\equiv 1.$[/hide]

[u]Solution.[/u] The typical way to approach this problem is to consider the degree of a polynomial $ g(x) \equal{} f(x) \minus{} 1 \equal{} \sum_{i \equal{} 1}^n\prod_{j\neq 1}\frac {x \minus{} x_j}{x_i \minus{} x_j}$, where $ x\equal{}\sum x_i$.  Since $ n > \deg g$ and $ g(x_i) \equal{} 0\ \forall i\in [n]$, we have that $ g(x)\equiv 0$.  Hence, $ f(x)\equiv 1$. $ \Box$


real analysis
function
calculus
calculus computations

Fix b>1.
If m, n, p, q are integers, n>0, q>0, and r=m/n=p/q, prove that

(b^m)^(1/n) = (b^p)^(1/q)

Hence it makes sense to define b^r = (b^m)^(1/n).

Do you understand the definition of the function $ x^{ \frac{1}{q} }$?  Write it out.  What does it tell you about the relationship between the two expressions given?

Hint:  Just use laws of exponents:

eg.
$ (2^4)^{\frac{1}{2}}\equal{}2^\frac{4}{2}\equal{}2^2\equal{}4$.

t0rajir0u,
$ x^{\frac {1}{q} }$ is the unique number $ y > 0$ such that $ y^{q} \equal{} x$.

$ b^{m}$ is the unique number $ x > 0$ such that $ x^{n} \equal{} y$, where $ y > 0$ is the unique number such that $ y^{q} \equal{} b^{p}$.
Not sure where to go with that.

takiguy, if I understand you correctly, the law of exponents you're telling me to use is the one we're trying to prove.

I can prove that $ (a^b)^c \equal{} a^{bc}$ for integers, and this problem is a special case of that but for noninteger rational numbers.  I just can't think of any way to define $ d^{\frac {f}{g}}$ except as representing $ (d^f)^{\frac {1}{g}}$.

Most problems in Rudin work like this:  write out the definitions.  See what they tell you.

In this case, the definitions tell us that there exists a real number $ x > 0$ such that $ b^m \equal{} x^n$ and a real number $ y > 0$ such that $ b^p \equal{} y^q$.  In addition, we are given that $ \frac {m}{n} \equal{} \frac {p}{q} \Leftrightarrow mq \equal{} np$.  This tells us that

$ b^{mq} \equal{} x^{nq} \equal{} b^{np} \equal{} y^{nq}.$

What can we conclude?


Prove that among any three real numbers, no two of which add up to $1$, show that there are two numbers $x,y$ such that $\frac{xy}{x+y-1}$ is not in the interval $(0,1)$

If $0<\frac{xy}{x+y-1}<1$, then we have two cases:

1. $x+y>1$. Then $xy>0$, which gives $x>0,y>0$ (the possibility $x<0,y<0$ doesn't work since we must have $x+y>1$). Also $\frac{xy}{x+y-1}<1\iff (x-1)(y-1)<0$. Therefore $x$ and $y$ must be on the opposite sides of $1$, but among three numbers two must always be on the same side of $1$.

2. $x+y<1$. Then $\frac{xy}{x+y-1}<1\iff (x-1)(y-1)<0$, which gives $x<1, y<1$ (the possibility $x>1,y>1$ doesn't work since we must have $x+y<1$.) Also $\frac{xy}{x+y-1}>0\iff xy<0$. Therefore $x$ and $y$ must be on the opposite sides of $0$, but among three numbers two must always be on the same side of $0$.


how many ways can two of more consecutive positive numbers add up to 2006?
what about 2007?

Well, there will either be only one solution or no solution when you add up a certain number of consecutive positive integers and set it equal to $2007$ (or any integer, for that matter). The question is, then, which numbers give a solution?

[hide="Hint"]
For $2007$, it's equivalent to solving $n(n+1)-k(k+1) = 4014$ for integers $k < n \leq 1004$.[/hide]

man i think we are going to have to factor [factor on a number theory problem...WHAT???]

[hide]
$n(n+1)-k(k+1) = 2j$
$n(n+1)+1/4-k(k+1)-1/4 = 2j$
$(n+1/2)^{2}-(k+1/2)^{2}= 2j$
$(n+k+1)(n-k)=2j$

For 2007:
$(n+k+1)(n-k)=4014$
Since these two factors have different parity, we need to break it down into an odd factor and an even factor, set the first factor equal to the larger, the second to the smaller;

2007,2; 1338,3; 669,6; 446,9; 223,18

First:     n=1004, k=1002
Second: n=670, k=667
Third:    n=337, k=331
Fourth:  n=227, k=218
Fifth:     n=120, k=102

$\boxed{\text{5 ways}}$
[/hide]

[quote="The Zuton Force"]

2007,2; 1338,3; 669,6; 446,9; 223,18
[/quote]

how did u arrive here?
anyway what about 2006? my teacher gave me the factors of 2006 as a hint but i couldn see the link

Umm, I factored 4014 to get that... $4014=(2^{1})(3^{2})(223^{1})$.
Btw, I didn't recognize $1$ as an odd factor because that would imply $n-k=1$, and it would just be one "consecutive" number, 2007.

[hide="2006"]If you do $2006$, its the same except now you have to solve:
$(n+k+1)(n-k)=4012=(2^{2})(17^{1})(59^{1})$

Which is much easier; all the even factors have to be together, so that it results in one even and one odd factor.

$4012=[(2^{2})][(17)(59)]=[(2^{2})(17)][(59)]=[(2^{2})(59)][(17)]$

These factorizations simplify to:
4,1003; 17,236; 59,68

For the first set, you solve $n-k=4\text{ and }n+k+1=1003 \implies n=503, \;k=499$
Similarly,
$n-k=4 \text{ and }n+k+1=1003 \implies n=503, \;k=499$
$n-k=17 \text{ and }n+k+1=236 \implies n=126, \;k=109$
$n-k=59 \text{ and }n+k+1=68 \implies n=63, \;k=4$

So those represent the 3 solutions for $j=2006$[/hide]

By the way, as to which numbers can be written in this way:
[hide="Answer"]
A number can be written as a sum of two or more consecutive integers iff. it can be written in the form $j=\frac{1}{2}(n+k+1)(n-k), \;n-k>1$. That is, $2j$ must have an odd factor greater than $1$. This is true for all numbers except powers of 2.
[/hide]

[quote="The Zuton Force"][hide="Answer"]
A number can be written as a sum of two or more consecutive integers iff. it can be written in the form $j=\frac{1}{2}(n+k+1)(n-k), \;n-k>1$. That is, $2j$ must have an odd factor greater than $1$. This is true for all numbers except powers of 2.
[/hide][/quote]

Be precise; consecutive [b]positive[/b] integers.  If this restriction is removed, any $n$ can be written trivially as $(1-n)+(2-n)+...+0+1+2+...+(n-1)+n$.  And as long as we're discussing positive integers, you need to specify an additional constraint on the odd and even factors of $j$.

The problem becomes much simpler if we restrict it two cases.  One where we consider $2006$ or $2007$ to be the sum of an odd number of consecutive positive integers and one where we consider it to be the sum of an even number of consecutive integers.  In the former case, if we write the integers as $a-j, a-j+1, \ldots, a+j-1, a+j$ then we see the sum is $ka$ where $k$ is an odd positive integer.  In the even case we can write the numbers as $a-j, a-j+1, \ldots, a+j+1$ and the sum is $ka+1$ where $k$ is an even positive integer.  In both cases, we must be careful that our sum starts at a positive integer.

[quote="mna851"]In the even case we can write the numbers as $a-j, a-j+1, \ldots, a+j+1$ and the sum is $ka+1$ where $k$ is an even positive integer.  In both cases, we must be careful that our sum starts at a positive integer.[/quote]

This result is slightly incorrect (it's actually $ka+j+1 = \frac{k}{2}(2a+1)$), and your use of both $k$ and $j$ is confusing, so I will restate it:

- The sum of $2n+1$ consecutive integers with middle term $a$ is $a(2n+1)$.  The first term is $a-n$.
- The sum of $2n$ consecutive integers with middle terms $a, a+1$ is $n(2a+1)$.  The first term is $a-(n-1)$.