Suppose you have three squares where a side from each square forms a triangle.  The areas of the squares are 388, 153, and 61 units :^2: .  Without using a calculator, what is the area of the triangle?

Who is Henry Dudeney?  Is that you?  A nice phrasing of the question, I think.

He was one of the greatest puzzlists of all time who found a quick way to solve this problem without a calculator or complex math.

hmm.  go figure.  so I suppose the heron's solution is not the way to go, then?

This problem was on Discovery Quest 2003. Unfortunately, they gave a really big hint that made it way too easy. It's a neat problem.

I can't believe I gave it away! AAARRRGGGHHH! :-x   :-x   :oops:  :oops:
I almost feel like changing my name from frodo to Freud-o.

How did you give it away? Someone needs to solve this problem. It's too good of a problem to ignore. Incidentally, on Discovery Quest, it said that you couldn't use Heron's Theorem or else you'd get 0 points or something like that.

i suppose the area is an integer?  what is it

I'd prefer to see someone solve this problem.

[quote="ComplexZeta"] How did you give it away?[/quote]

OOHHHH! :shock:  :oops:   I thought you were talking about me for second, but I did not read it carefully.  You meant the Discovery Quest 2003 gave it away, not me.  I understand now. :wink:  I think I should be considered Freud-o.

lets see, connecting the square centers, you get a equilateral triangle, connecting the opposite vertices with the opposite square centers gets you some famous point of concurrency which i dont know about...

Here's a little hint, or more of a clarification:
No square overlaps another at all.

somehow find the area of hexagon, then subtract the area of squares and divide by 4??  i just want to know if this generalizes or not
and does the point of concurrency have a name?

The point of concurrency if you put triangles on the sides is called the Fermat point, and is also a distance-minimizing point.  I couldn't find any information on what happens when you use squares.

i dont understand, i do not need a calculator and i can calculate the area using heron's or cosine law with relative ease... why am i NOT ALLOWED to use that??

You aren't ``allowed'' to use those because there's a much more elegant solution.

I'm thinking you all need a picture to help you.  Can we post pictures on this site?

Okay, I have a start -- I know what's special about those numbers.  I'll keep working on it.

Nice, I got it!  That was tough -- I think it's difficult to give a good hint for that question.  The answer is [hide]21[/hide], right?

I'm not sure I would call that "easier" -- to test whether it satisfies what it needs to satisfy takes almost as long as plugging into Herons would, I would think.

I actually conciously didn't put it up -- I thought we all might taunt bugzpodder a little bit first  :wink:

Here does this help? I dunno if this picture is any good but....
[img]http://bwsolo.com/triangle.jpg[/img]

Actually, your picture is pretty inaccurate -- the triangle is nearly degenerate.

Here's my solution, but bugzpodder, I think you should keep working on it until you get it.  It's sort of a cheap trick, and it's more fun to get it on your own:

[hide]388 = 18^2 + 8^2

61 = 6^2 + 5^2

153 = 12^2 + 3^2

18 = 6 + 12

8 = 5 + 3

Thus, construct a right triangle with legs of 8 and 18.  Construct a 6x3 rectangle, whose side of length 6 lies within the side of length 18 of the triangle and whose side of length 3 lies within the side of length 8 of the triangle.  Thus, the rectangle and the triangle share a right angle.  Connect the fourth vertex of the rectangle to the two non-right angles of the triangle.  This gives us a setup which includes our given triangle, as verified by the Pythagorean theorem.  Then add and subtract areas to get 21 as the final answer.[/hide]

wow, thats sweet JBL :shock:
nicely done

Good work Joel. That's essentially the solution that I gave, except that after I put the triangle on the plane with vertices at (0,0), (8,18), and (3,12), I took a determinant to find the area. The ``hint'' was 388=82+182.

woh, i am confused, how does 8=5+3 have anything to do with 8^2, 5^2 and 3^2?

Draw the picture as I described.  Hopefully they you'll see.  If someone could post it, that would be nice.

yes, drawing it did help, that is a nice solution. thanks

[quote="JBL"]388 = 18^2 + 8^2
61 = 6^2 + 5^2
153 = 12^2 + 3^2
18 = 6 + 12
8 = 5 + 3
Thus, construct a right triangle with legs of 8 and 18.  Construct a 6x5 rectangle, whose side of length 6 lies within the side of length 18 of the triangle and whose side of length 5 lies within the side of length 8 of the triangle.  Thus, the rectangle and the triangle share a right angle.  Connect the fourth vertex of the rectangle to the two non-right angles of the triangle.  This gives us a setup which includes our given triangle, as verified by the Pythagorean theorem.  Then add and subtract areas to get 21 as the final answer.[/quote]

wait a minute shouldn't you draw a 6*3 rectangle?

Yes.  Good catch.  Gone back and edited.


Prove that the number of triples $ (A, B, C)$ where $ A, B, C$ are subsets of $ {1, 2, ..., n}$ such that $ A \cap B \cap C  \equal{} \phi, A \cap B \ne \phi, B \cap C\ne \phi$ is $ 7^n \minus{} 2.6^n \plus{} 5^n$.


If I run at 15 miles per hour to a rock 365.767 yards away, and then back at 12 miles per hour, what was my average speed for the whole trip?

well, if we approximate the distance as 1/5 mile, i think the average speed is 13 and 1/3mph. the distance is 2/5 miles total and it would take 9/5 minutes to go that distance. so, after applying the d=rt formula, we should get 13 and 1/3.

That's correct, but there's an easier way to do this (your's was pretty easy though), see if you or someone else can find it.

but, it doesn't take an equal amount of time to get to and from the rock.

so, what's the easier way to solve the problem 236factorial? :)

The average speed of a trip to and back is $\frac{2}{\frac{1}{a}+\frac{1}{b}}$ where a and b are the speeds.

So $\frac{1}{15}+\frac{1}{12}=\frac{9}{60}$.

$\frac{2}{\frac{9}{60}}=\frac{120}{9}=13\frac{1}{3}$

if you're slow at fractions, this is not recommended  :D


abstract algebra
vector
linear algebra
matrix

1) Find all linear transformations $T_{A} \in L(V_{2}(R),V_{2}(R))$  such that $KerT_{A}=ImT_{A}$
2) Proof that $ImT \subset KerT \iff T^2= \theta$

To the point 1)  I have that $dimKerT_{A}=dimImT_{A}=1$. But what's next?

Thanks a lot for any help.

Choose a basis for the space: The first vector generates the kernel and the image, while the second is an arbitrary vector in the rest of the space. What does the matrix of $T$ look like in this basis?

Once you have that, note that a change of basis is a similarity transformation.

How does the vector that generates the basis look like? Is there only one vector or infinite number of similar vectors?


logarithms
limit
real analysis
real analysis unsolved

Prove that for every $ n\in \mathbb{N}$ there exists unique $ t(n) > 0$ such that $ (t(n) \minus{} 1)\ln t(n) \equal{} n.$ Calculate $ \mathop{\lim}_{n\to\infty}\left(t(n)\frac {\ln n}{n}\right).$

Consider the equation :
$ (x \minus{} 1)\ln x \equal{} n$
We will prove this equation has solution for all n.
$ f(x) \equal{} (x \minus{} 1)\ln{x}$
$ \lim_{x\to 1}f(x) \equal{} 0$
$ \lim_{x\to\infty} \equal{} \plus{} \infty$
$ f(x)$ increasing so it has unique solution .
So the equation $ f(x) \equal{} n$ has solution.
Easy to check $ x_n\to\infty$
From condition we have :
$ n \equal{} (x \minus{} 1)\ln{x}$
$ \ln{n} \equal{} \ln{((x \minus{} 1)\ln{x})}$
So $ \lim_{n\to\infty}\frac {t_n \ln n}{n} \equal{} \lim_{n\to\infty} \frac {x_n \ln{(x_n \minus{} 1)\ln{x_n})}}{(x_n \minus{} 1) \ln{x_n}} \equal{} 1$

[quote="TTsphn"]
$ f(x)$ [b]increasing[/b] so it has unique solution .
[/quote]
Why?  :maybe:

I am sorry.
$ f'(x)>0,\forall x>1$
$ f'(x) < 0,\forall x<1$
Consider sign of $ f(x)$.


Hi all!
Our high school's math club is trying to hold a math contest for
local elementary school kids. We haven't organized & planned anything yet. and we just have an idea.
Anyone have the experience of similar kind??

The hardest thing is to make the contest fair and public.. I think...
How should we advertise this to the kids...
Several of our math teachers agreed to supervise for us.

Thanks!!

Any ideas & opinions & suggestions will help.....

1) Tell your elementary teachers (the one that still remember you) about the contest. And tell them that it is extremely educational (it also helps if there is a prize).

2) Make sure your elementary school is "math worthy" which means a good number of them actually like doing math. It won't help to have contestants that hate the contest  :wink: .

Definitely offer prizes. I am sure you can get some money from your school in order to buy some decent awards.

You can probably go to your school district's web site (and those of surrounding school districts) and find the names of the math teachers or math department heads at all of the elementary schools. Send them all nice letters or emails inviting them and their students to the contest. (You might want to let one of your faculty advisors look over your letter before you send it off.)

Make sure you plan the contest with a reasonable amount of detail before you send the letter. If you appear well-organized, teachers will be more likely to want to put in the time and effort to send their kids to your contest.

You should make sure it's scheduled after school or on a day when there's spare time for elementary school students....

Also, make sure the problems are easy enough. Not all elementary school students are as adept at math as most AoPSers were in elementary school. We don't don't want them doing olympiad-level problems.  (Seriously, though don't include anything involving even the simplest algebra on the test. The toughest thing should be exponents and fractions)

And don't forget to mark all the answers wrong if they don't have QED on the end.

To be perfectly honest, I think math contests are not that well suited for elementary schoolers. Think about what they know at that point in time: they've memorized a few arithmetic tables, and barely know any applications.

Most of them don't know what the distributive property is, or how to solve 2x+1=5. Although, if you want to put algebra, make sure to use question marks so that they understand that it's variable.

Very simple geometry... like finding an area of something with all right angles.

etc.

[b]Logic puzzles would be much more similar to problem solving at that level than any math question could be.[/b]

But, seriously, won't you be better off catering to middle schoolers? It's difficult to do problem solving without tools. And if you water it down, you just might give the wrong impression of problem solving.

The elementary school children would not know a lot of math but wouldn't the presence of a competition encourage them to learn more? Sure, the questions might be easy, but they shouldn't just all be arithmetic. The idea is to expand their idea of mathematics to problem solving, it's not just about "finding $ x$" or "why $ 1\plus{}1 \equal{} 2$."

[quote="The Zuton Force"]
Most of them don't know what the distributive property is, or how to solve 2x+1=5. Although, if you want to put algebra, make sure to use question marks so that they understand that it's variable.[/quote]

I'm certain only about 1 in 10,000 elementary school children would know how to solve even the simplest algebra... if you really want to, use it only as bonus.

Simple geometry is fine, $ A\equal{}s^{2}$ is known to many elementary school children in one form or another... some [b]very[/b] simple number theory would be good, like dealing with evens/odds and what they are divisible by.

One other thing: patterns. Display a series of numbers and ask them to describe the pattern.

We're doing patterns right now in geometry :roll:
But yeah, I guess you can post simple patterns, like multiplying by 2, adding some number,etc.

The Continental Math League conducts math contests for kids as young as 2nd grade.  For sample problems, see
[url]http://www.continentalmathematicsleague.com/cml.html[/url].
Obviously the problems will be simple for us folks on AoPS.  But they are still interesting for little kids to think about.

there's also moems, which has an elementary school division

http://www.moems.org/sample.htm

One thing that's essential is to give them either
1) Not many probs
2) A lot of time
and expect them to finish all or
1) Many probs
2) Barely any time
for pressure finish.


Ai giúp mình dịch bài này ra Tiếng việt với:
[b]Problem:[/b]Do there exist 16 three digit numbers, using only three different digits in all, so that the all numbers give different residues when divided by 16?

[b]Solution:[/b]The answer is no.

Suppose the answer would be affirmative. Obviuously the three digits cannot be all of the same parity, because we would not be able to cover all the residues modulo 2 :)

Thus there exists at least one even and one odd digit. Suppose WLOG that we have one odd digit and two even digits (the other case can be treated in the same way, replacing the word even with the word odd everywhere). Thus the 8 odd residues modulo 16 must be covered by odd numbers, that is the last digit is the odd one. But there are 9 ways to choose the first two digits, and out of those at least 4 must give odd residue when divided by 8, because the number $\overline{xy}\cdot 10$ ( $\overline{xyz}=\overline{xy}\cdot 10+z\ \Rightarrow\ \overline{xy}\cdot 10$ ) must pass through all the even residues modulo 16.

The last 4 numbers which give odd residue when divided by 8 must be formed having the last digit y, and odd number, but there are only 3 such numbers, which leads to a contradiction.

Thus the problem is solved.

Có tồn tại hay không 16 số có ba chữ số, chỉ sử dụng 3 chữ số phân biệt, sao cho chúng để lại các số dư khác nhau khi chia cho 16?

De bai thi em dich duoc,con loi giai thi sao?

Đề bài cũng có vấn đề thì phải? Nên anh không dịch tiếp! Có nghĩa là 16 số đó đều có tập các chữ số bằng nhau à? Ví dụ {a,b,c} thì chỉ có tối đa 3!=6 số thôi chứ nhỉ? Hay nó cho cả lặp nữa à? Chú xem lại cái!

Đây là bài 4 2nd Junior Balkan 1998  mà!

Vậy là có thể lặp.

Các chữ số không thể cùng tính chẵn lẻ. Gỉa sử có 1 lẻ và 2 chẵn, do đó 8 số dư lẻ phải sinh bởi các số lẻ, đó là các số có chữ số cuối cùng là số lẻ nói trên. Nhưng có đúng 9=3x3 cách chọn 2 chữ số đầu, quan tâm đến các số có hai chữ số là hai chữ số đầu tiên của các số trên là xong! Dễ chưa em? :D

[quote="chien than"]Vậy anh dịch lời giải hộ em đi![/quote]
Xong rồi đấy em :)

Anh nói dễ hiểu 1 tí được không?
Em mới học THCS mà!


prove that: ${{p-1}\choose{a}}\equiv(-1)^{a}modp$
for p any prime.

I assume by $p$ you mean a prime.
[hide="dont know if this work"]
Let's do induction - if $a$ is $0$, we have $1^{0}\equiv (-1)^{0}(\bmod\ p)$, which is obviously true.

Assume it works for some $a$.  Then, to make the choosing thing equal to $\binom{p-1}{a+1}$, we have to first divide by $a+1$, and then multiply by $p-a-1$.  let's say $\frac{\binom{p-1}{a}}{a+1}= k$ for some integer $k$.  Then we must have $k(a+1) \equiv (-1)^{a}(\bmod\ p)$ by assumption.  Therefore, $k$ is the multiplicative inverse of $a+1$ mod $p$.  Now when we multiply by $p-a-1$, note that:

$k(p-a-1) \equiv k(-a-1) \equiv-k(a+1) \equiv-(-1)^{a}\equiv (-1)^{a+1}(\bmod\ p)$

Because $k$ is the inverse of $a+1$.  Now by induction it'll work for all $a$, and $p$ can be any prime.

Will this argument work??? (the division to the mod part I'm unsure about...)
[/hide]

[quote="pkerichang"]Will this argument work??? (the division to the mod part I'm unsure about...)[/quote]

Yes.  It works because the integers mod p form a field, so you know that you can always divide (except by zero) and get a meaningful result.


AMC
USA(J)MO
USAMO
AMC 10
AMC 10 A
trigonometry
AIME

Sign up for the mock competition here. The link to the competition is http://www.artofproblemsolving.com/Forum/viewtopic.php?t=125028.

Due December 25. :wink:

ALL OF THE QUESTIONS HAVE BEEN VERIFIED AND THE CHOICES HAVE THE ANSWERS. I CANNOT GIVE ANY HELP REGARDING THE QUESTIONS. SORRY.

I'll sign up. Do we pm the answers to you, right? And why do we need solutions for a multiple choice text???

yes. please send solutions for MCQ questions.

Pm me the answers w/solutions.

PDF/DVI format preferred for solutions.

[u]Sign up list:[/u]
[list]mustafa
tim1234133
Go Around the Tree
Xantos C. Guin[/list]

AMC doesn't expect solutions until the USAMO!  Why do you expect them for an AMC 10?  Also note that in order to give an accurate representation of the score, you should award 6 points for a correct answer and [i][b]1.5[/b][/i] points for a blank.  (They changed the scoring for this year)  Also, what are you going to do about time?  It doesn't make sense to use AMC's 75 minute rule, while requiring full solutions.  Also, you may want to change your signature to something that the forum will display.

/me volunteers, as this is my 4th year of taking the AMC10, and I'd like to do tolerably on it.

i dont think you're given enough information on number 12...

[u]List (Updated):[/u]
[list]
mustafa
tim1234133$\sqrt$
Go Around the Tree
Xantos C. Guin
NeverOddOrEven
calc ruz
roadnottaken?
[/list]

A $\sqrt$ has been put next to your name I have received your answer.

[quote="anirudh"]The solutions don't count in the 75 minutes.

@NeverOddorEven: 12 has enough info[/quote]
So what happens if one finds corrections in their solution typing up?

[quote="now a ranger"]Anirudh, you should maybe trust RnWAng's words.

Does she want to make a false impression? nah doubt it  :lol:[/quote]

I took the score down.

Happy

[quote="now a ranger"]Anirudh, you should maybe trust RnWAng's words.

Does [b]she[/b] want to make a false impression? nah doubt it  :lol:[/quote]

1. If that "she" is referring to me, wrong gender.

2. I seriously did not send anything to you anirudh.

[quote="anirudh"][quote="rnwang2"][quote="now a ranger"]Anirudh, you should maybe trust RnWAng's words.

Does [b]she[/b] want to make a false impression? nah doubt it  :lol:[/quote]

1. If that "she" is referring to me, wrong gender.

2. Why do people always have to use such vague pronouns?

3. I seriously did not send anything to you anirudh.[/quote]

Exactly. I took you down.

(even though I know you sent me B B B ... under blah :P )[/quote]

That's preposterous. I refuse to engage in any more conversation on this issue.

[quote="anirudh"][quote="rnwang2"][quote="now a ranger"]Anirudh, you should maybe trust RnWAng's words.

Does [b]she[/b] want to make a false impression? nah doubt it  :lol:[/quote]

1. If that "she" is referring to me, wrong gender.

2. Why do people always have to use such vague pronouns?

3. I seriously did not send anything to you anirudh.[/quote]

Exactly. I took you down.

(even though I know you sent me B B B ... under blah :P )[/quote]
[quote="calc rulz mock amc guidelines thread"]PS: Also note that I agree that the issue of people submitting bogus answers is a separate issue. And I must admit that I did submit one set of B answers to Anirudh under the username blah Blush , although I was just trying out the form because I had never seen and automatic-submit answer form Mr. Green (If you want you can put a 60-something next to my username on the scores list and let me hang my head in shame Razz[/quote]

[quote="now a ranger"]maybe anirudh shouldn't be lazy and grade the answers when they are pmed. you have a record for that too, and people wouldn't be phony.[/quote]

Dude...I had all the answers except rnwang's. Except then I got the problem of grading rnwang's even though it had a phony entry into the website. And by grading the B's, I got somewhere around 120, because half of the boxes were blank and he got like 1.5 pts for each blank box. So the all 'B' was the one that got around 120

[quote="rnwang2"][quote="anirudh"][quote="now a ranger"]maybe anirudh shouldn't be lazy and grade the answers when they are pmed. you have a record for that too, and people wouldn't be phony.[/quote]

Dude...I had all the answers except rnwang's. Except then I got the problem of grading rnwang's even though it had a phony entry into the website. And by grading the B's, I got somewhere around 120, because half of the boxes were blank and he got like 1.5 pts for each blank box. [b]So the all 'B' was the one that got around 120[/b][/quote]

Ahem....so [i]who[/i] was the all "B" one?[/quote]

OH VERY SORRY. I THOUGHT YOU WERE CALC_RUZ :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:

[quote] :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush: [/quote]

Oops sorry Rnwang2, I wasn't exactly paying attention to this topic/discussion  :blush:

This mock AMC was kind of pointless.

You just copied many of the questions on old AHSMEs or amc 10s or 12s (exact same questions-like your number 16,18,25)

and when they wernet the same, the question was phrased exactly the same except the numbers were just changed a bit (like 19,21,22,23,24)

ALl the hard questions were almost or exact replicas of old AHSME questions..

and 20 is never going to be a question.. thats tested on basic knowledge not any fundamental math skills its one step either you do or you dont...

Next time, dont copy questions fomr old ashme and call it a mock....and fix your typos...

-Sorry-

I didn't notice the date the last post was posted. It was last year!

Do not revive the old thread with pointless stuffs like this.  :mad:

This is not just to ThetaPi and vallon22. It's for everyone (in this case, those two).


Four persons are standing in front of a bridge spanning a canyon. The bridge can carry only two persons at a time. It is night time so they have to use a lantern to light their path, and they have only one lantern. Whenever two persons have crossed the bridge one person must carry back the lantern. They take $5$, $10$, $20$ and $25$ minutes respectively to cross the bridge. Two persons walking together need the time of the slower one. What is the minimum time for the group to cross the bridge?

[hide]

If I'm thinking right, we want the $5$ minute person walking back each time. It will take two full trips for two of the slower people to make it, and then a half trip for the 5 to walk back with the remaining person. $10 + 20 + 25 + 2(5) = \boxed{65}$

[/hide]

Actually, you can do it in 60 minutes.

[hide]1) $10$ and $5$ walk across together. Total time: $10$ minutes
2) $5$ comes back. Total time: $15$ minutes
3) $25$ and $20$ walk across together. Total time: $40$ minutes
4) $10$ comes back. Total time: $50$ minutes.
5) $10$ and $5$ walk across together, for a total time of $\boxed{60}$ minutes.[/hide]

I think I saw this before as an Indiana Jones problem.


Two  triang1es  are similar. The lengths of the sides of the smaller triangle are 3, 5, and 6, and the length of the longest side of the larger triangle is 18. What is the perimeter of the larger triangle?

[hide]The longest side of the small triangle is 6 so the ratio of their sides is $3: 1$. So the other two sides of the big triangle is 15 and 9. So the perimeter would be 42.[/hide]

actually, you don't have to find the sides of the bigger triangle individually (not saying that's hard or anything). the ratio of the perimeters is the ratio of any corresponding sides, so 18:6=3:1 is the ratio of corresponding sides 14 is the perimeter of the smaller triangle so 42(=14*3) is the perimeter of the larger.

can you find the ratio of the area of the smaller to the area of the larger?

[hide]
The ratio of the areas would simply be $1^{2}: 3^{2}: : 1: 9$
[/hide]

It says to find the ratio of the smaller triangle first so it would be $1: 9$.

For any similar shapes, and the ratio of the sides is $x: y$, then,
Ratio of Perimeter = $x: y$
Ratio of Area = $x^{2}: y^{2}$
Ratio of Volume = $x^{3}: y^{3}$

[quote="Interval"]So, is the answer 1 : 9?[/quote]
The answer is 1:9 for the area and 1:3 for the perimeter

Yeah, because if you plug in x=1 and y=3 into my equation, you will get that.


function
trigonometry
algebra unsolved
algebra

Prove that if $f$ is a non-constant real-valued function such that for all real $x$, $f(x+1) + f(x-1) = \sqrt{3} f(x)$, then $f$ is periodic. What is the smallest $p$, $p > 0$ such that $f(x+p) = f(x)$ for all $x$?

By some calculations, it seems that $f(x+6) = f(x)$

Can you post little bit more details on how you achieved that value?

How do you prove that 12 is the smallest p?

In fact, it is not always.  We can let $f(x)=\cos\left(\frac{11\pi x}{6}\right)$.  This has period $\frac{12}{11}$, but

$ f(x+1)+f(x-1)=\cos\left(\frac{11\pi (x+1)}{6}\right)+\cos\left(\frac{11\pi (x-1)}{6}\right) $

$ =\left(\cos\left(\frac{11\pi x}{6}\right)\cos\left(\frac{11\pi}{6}\right)-\sin\left(\frac{11\pi x}{6}\right)\sin\left(\frac{11\pi}{6}\right)\right) $

$ =2\cos\left(\frac{11\pi x}{6}\right)\cos\left(\frac{11\pi}{6}\right)=\sqrt{3}f(x) $

So it's not true.  But every function has period 12, and 12 is the minimum common period because we can let $ f(x)=\cos\left(\frac{\pi x}{6}\right) $.  Then

$ f(x+1)+f(x-1)=\cos\left(\frac{\pi (x+1)}{6}\right)+\cos\left(\frac{\pi (x-1)}{6}\right) $

$ =\left(\cos\left(\frac{\pi x}{6}\right)\cos\left(\frac{\pi}{6}\right)-\sin\left(\frac{\pi x}{6}\right)\sin\left(\frac{\pi}{6}\right)\right) $

$ =2\cos\left(\frac{\pi x}{6}\right)\cos\left(\frac{\pi}{6}\right)=\sqrt{3}f(x) $.

And the period of $\cos\left(\frac{\pi x}{6}\right)$ is 12 and not less.  So 12 might not be the least period for some function $f$ that satisfies the equation, but 12 is the least period that works for every function that satisfies the equation.


Prove $ 2^{349} \equal{} 2[7]$

what's the latex symbol for congurence?

\equiv is $ \equiv$ and you can use 2\bmod 7: $ 2\bmod 7$.

[hide="solution"]$ 2^3\equal{}8\equiv 1\bmod 7$, thus the powers repeat mod 7 every three powers. Thus, $ 349\minus{}3\cdot \lfloor \dfrac{349}{3}\rfloor\equal{}1$, and $ 2^1\equiv 2\bmod 7$. Thus $ 2^{349}\equiv 2\bmod 7$.[/hide]

i can!
$ 2^3\equiv 1(7)\Rightarrow 2^{348}\equiv 1(7)\Rightarrow 2^{349}\equiv 2(7)$
xDDDDDDDDDDD


abstract algebra
group theory
modular arithmetic
vector
invariant
superior algebra
superior algebra unsolved

Prove that a finite abelian group having more than one element of order 2  then the product of all its elements is the identity element. (For example, the Klein 4-group)

Since the group is abelian, let's write the group operation as $ \plus{}$ and the identity as $ 0.$ Suppose we have $ a$ and $ b$ such that $ 2a\equal{}2b\equal{}0,a\ne 0,b\ne 0,$ and $ a\ne b.$ Let $ H\equal{}\{0,a,b,a\plus{}b\}.$ By direct computation, $ H$ is a subgroup of $ G,$ and since this is an abelian group, it is a normal subgroup of $ G.$

We can then write $ G\equal{}H\oplus (G/H).$ Consider one particular coset of $ H$ in $ G.$ We can find a $ v\in G$ such that we can write this coset as $ H\plus{}v.$

Compute $ \sum_{x\in H\plus{}v}x\equal{}v\plus{}(a\plus{}v)\plus{}(b\plus{}v)\plus{}(a\plus{}b\plus{}v)\equal{}4v.$

There are two possibilities. If $ v$ has order $ 1$ or $ 2,$ then $ 4v\equal{}0.$ Else, $ v$ has an inverse $ \minus{}v$ such that $ v\ne\minus{}v.$

Let's split that into two more cases. Either $ v$ and $ \minus{}v$ belong to the same coset of $ H,$ or they belong to different cosets. If $ v\equiv\minus{}v\pmod{H},$ then $ 2v\in H.$ But if $ 2v\in H,$ then $ 4v\equal{}0.$

Else, $ v$ and $ \minus{}v$ belong to separate cosets of $ H.$

So we take the sum $ \sum_{x\in G}x$ and break it up into cosets. For each coset, the sum is $ 4v.$ Either $ 4v\equal{}0$ already, or $ 4v$ can be paired with $ 4(\minus{}v)$ which comes from a separate coset. Those two terms combine to add up to zero, and the whole sum is zero.

Essentially, the argument works by finding a copy of the Klein 4-group embedded in $ G.$

Kent, this is the solution I was looking for. I almost had it,  but needed the idea that one needs to consider the product on each coset and breakdown the cases accordingly. This is the solution Herstein had and intended for this problem. It appears in the section on Lagrange's Theorem and Cosets. Factor groups are not yet discussed, let alone vector spaces or automorphisms.

I am intrigue by the other solution, but fail to see why an element invariant under all automorphisms of a vector space of dimension at least 2 must be the identity element....

Bret

There's a slightly more general way of stating this result: in a finite abelian group, the product (or sum, or whatever we want to call the group operation) of all of the elements of the group is either the unique element of order 2, if there is such a thing, or the identity, if there is not a unique element of order 2.


Δίνονται $ 7$ πραγματικοί αριθμοί. Να δειχτεί ότι ανάμεσά τους υπάρχουν $ 2$, έστω οι $ x,y$, ώστε $ 0\leq \frac {x \minus{} y}{1 \plus{} xy}\leq \frac {1}{\sqrt {3}}$.

[hide]Εστω $ x_1, x_2,... x_7$ οι 7 αριθμοι. Θετουμε $ x_i \equal{} \tan{y_i}, y_i \in ( \minus{} \frac {\pi}{2}, \frac {\pi}{2})$, για $ i \equal{} 1,2,3,4,5,6,7$. Τοτε αρκει $ 0 \leq \tan{(y_i \minus{} y_j)} \leq \tan{\frac {\pi}{6}}$, η ισοδυναμα (επηδη $ \tan{x}:$ γνησιως αυξουσα) $ 0 \leq y_i \minus{} y_j \leq \frac {\pi}{6}$ για καποιους δεικτες $ i, j \in \{1,2,3,4,5,6,7\}$. Εστω ΧΒΤΓ $ y_i \geq y_j$ για $ i > j$, τοτε οι διαφορες: $ y_{i \plus{} 1} \minus{} y_i, i \in \{1,2,3,4,5,6\}$ ειναι θετικες και αν υποθεσουμε οτι ολες ειναι μεγαλυτερες απο $ \frac {\pi}{6}$ τοτε το αθροισμα τους θα ειναι: $ y_7 \minus{} y_1 > \frac {7\pi}{6} > \pi$ το οποιο ειναι ατοπο αφου οι αριθμοι $ y_i$ ανηκουν σε ανοιχτο διαστημα ευρους $ \pi$. Αρα ισχυει το ζητουμενο.[/hide]


LaTeX
inequalities unsolved
inequalities

Let $ABCD$ be a quadrilateral inscribed in a circle of radius $R$ and circumscribed in a circle of radius $r$.  If $s$ is the semiperimeter we have

$ s \leq 2R +(4-2 \sqrt{2})r $

with equality if and only if the quadrilateral is a square

I found an easy way to crack problems which you proposed about qadrilateral which is inscibed in a circle with radius $R$ and circumscribed in a circle of radius $r$.I think it will interest you.

Yes, I am really interested in it   ;)

Thank you very much :lol:

Sorry but I am novice in LaTex and it takes a long  time of me to type the solution.
Anyway tomorow I will send you attached MS-Word file with solution  :blush:


ai , bi , ci>0  prove  that
(Σai*bi*ci) 2  \leq (Σai) 2 (Σbi) 2 (Σci) 2


Let $m, n \ge 2$ be positive integers, and let $a_{1}, a_{2}, \cdots,a_{n}$ be integers, none of which is a multiple of $m^{n-1}$. Show that there exist integers $e_{1}, e_{2}, \cdots, e_{n}$, not all zero, with $\vert e_i \vert<m$ for all $i$, such that $e_{1}a_{1}+e_{2}a_{2}+ \cdots +e_{n}a_{n}$ is a multiple of $m^n$.

Consider the set: $ A = \{e_1a_1  +  e_2a_2  + ...  +  e_na_n: 0 \leq e_1, e_2, ..., e_n \leq m - 1\}$. It clearly consists of $ m^n$ elements. If some two of them would be congruent mod $ m^n$, i.e.
$ e_1a_1  +  ...  +  e_na_n \equiv e'_1a_1  +  ...  +  e'_na_n \pmod{m^n}$
then it would follow that
$ (e_1  -  e'_1)a_1  +  ...  +  (e_n  -  e'_n)a_n \equiv 0 \mod{m^n}$
and the numbers $ e_1  -  e'_1$, ..., $ e_n  -  e'_n$ satisfy given conditions.

So suppose that all numbers from the set $ A$ give different residues, which means that they give all residues. Consider $ \omega$ the primitive root of $ 1$ of degree $ m^n$-th. Since
$ 1  +  \omega^{1}  +  \omega^{2}  +  ...  +  \omega^{m^{n} - 1}  =  0$
we also have $ \sum_{k \in A} \omega^k  =  0$.
But on the other hand it's not hard to see that
$ \sum_{k \in A} \omega^k  =  \prod_{i = 1}^{n} (1 + \omega^{a_i} + \omega^{2a_i} + ... + \omega^{(m - 1)a_i}).$
So it follows that for some $ i$:
$ 0  =  1 + \omega^{a_i} + \omega^{2a_i} + ... + \omega^{(m - 1)a_i}  =  \frac{\omega^{ma_i} - 1}{\omega - 1}$
and it's a contradiction since $ m^{n - 1}$ doesn't divide $ a_i$.

Let $ B$ be the set of all $ n$-tuples $ b\equal{}(b_1, b_2, \dots,
b_n)$, where each $ b_i$ is an integer with $ 0\le i<m$. For $ b\in B$,
define $ f(b)\equal{}b_1a_1\plus{}b_2a_2\plus{}\dots\plus{}b_na_n$. As in the above solution, suppose $ f(b)$ gives a complete residue class. Then

\[ (0\plus{}1\plus{}2\plus{}\cdots\plus{}m\minus{}1)(a_1\plus{}a_2\plus{}\cdots\plus{}a_n)\equal{}\displaystyle\sum_{b\in B}{f(b)}\equiv 0\plus{}1\plus{}2\plus{}\cdots\plus{}m^{n}\minus{}1\pmod {m^n}\]
We have
\[ (a_1\plus{}a_2\plus{}\cdots\plus{}a_n)\dfrac{m(m\minus{}1)}{2}\equiv \dfrac{m^{n}(m^{n}\minus{}1)}{2}\pmod {m^n}\]
which means $ (a_1\plus{}a_2\plus{}\cdots\plus{}a_n)\equiv 0\pmod {m^{n\minus{}1}}$. Now by
the assumption there exist a tuple $ b$ such that $ f(b)\equiv
\minus{}(a_1\plus{}a_2\plus{}\cdots\plus{}a_n)\pmod {m^n}$.

Taking $ e_i\equal{}b_i\plus{}1\minus{}m$ finishes the problem.

[quote="nealth"]Then
\[ (0 \plus{} 1 \plus{} 2 \plus{} \cdots \plus{} m \minus{} 1)(a_1 \plus{} a_2 \plus{} \cdots \plus{} a_n) \equal{} \displaystyle\sum_{b\in B}{f(b)}\equiv 0 \plus{} 1 \plus{} 2 \plus{} \cdots \plus{} m^{n} \minus{} 1\pmod {m^n}
\]
[/quote]

Why was this?

Additionally, where did you use the that ai are not multiples of $ m^{n\minus{}1}$? Notice that the problem is not true without this condition.

Imo short list 2002 n6
Complex combinatorics


trigonometry
geometry
trig identities
Law of Sines

Prove that in a triangle $ ABC$,

$ \frac{\tan \frac{B\minus{}C}{2}}{\tan \frac{B\plus{}C}{2}} \equal{} \frac{b\minus{}c}{b\plus{}c}$.

Also known as the Law of Tangents.

[hide="Hint"]Write $ \tan$ in terms of $ \sin$ and $ \cos$. Does the form of the LHS remind you of a particular identity?[/hide]

[hide="Solution"]Look at the LHS:

$ \frac{\tan\frac{B\minus{}C}{2}}{\tan\frac{B\plus{}C}{2}}\equal{}\frac{2\sin\frac{B\minus{}C}{2}\cos\frac{B\plus{}C}{2}}{2\sin\frac{B\plus{}C}{2}\cos\frac{B\minus{}C}{2}}$

Now apply the sum-to-product identity backwards.

$ \text{LHS}\equal{}\frac{\sin B\minus{}\sin C}{\sin B\plus{}\sin C}$

Multiply by $ \frac{bc}{bc}$

$ \equal{}\frac{b(c\sin B)\minus{}c(b\sin C)}{b(c\sin B)\plus{}c(b\sin C)}$

From the Law of Sines, $ \frac{\sin C}{c}\equal{}\frac{\sin B}{b}\implies b\sin C\equal{}c\sin B\equal{}x$

We finally reduce it to $ \frac{bx\minus{}cx}{bx\plus{}cx}\equal{}\frac{b\minus{}c}{b\plus{}c}$ $ \Box$[/hide]

From [b]Extended Law of Sine[/b]:

\[ \frac{a}{\sin\alpha}\equal{}\frac{b}{\sin\beta}\equal{}\frac{c}{\sin\gamma}\equal{}2R\]
we obtain:

$ a\equal{}2R\cdot \sin \alpha$

$ b\equal{}2R\cdot \sin \beta$

$ c\equal{}2R \cdot \sin \gamma$

Applying the [b]sum-to-product[/b] formulas the result follows.

\[ \frac{b\minus{}c}{b\plus{}c}\equal{}\frac{2R\cdot (\sin \beta \minus{}\sin \gamma)}{2R\cdot (\sin \beta\plus{}\sin\gamma)}\equal{}\frac{2\cdot \cos \frac{\alpha\plus{}\beta}{2}\cdot \sin\frac{\alpha\minus{}\beta}{2}}{2\cdot \sin \frac{\alpha\plus{}\beta}{2} \cdot \cos \frac{\alpha\minus{}\beta}{2}}\equal{}\frac{\tan\frac{\alpha\minus{}\beta}{2}}{\tan\frac{\alpha\plus{}\beta}{2}}\]

[quote="Obel1x"][hide="The solution"]From [b]Extended Law of Sine[/b]:
\[ \frac {a}{\sin\alpha} \equal{} \frac {b}{\sin\beta} \equal{} \frac {c}{\sin\gamma} \equal{} 2R\]
we obtain:

$ a \equal{} 2R\cdot \sin \alpha$

$ b \equal{} 2R\cdot \sin \beta$

$ c \equal{} 2R \cdot \sin \gamma$

Applying the [b]sum-to-product[/b] formulas the result follows.
\[ \frac {b \minus{} c}{b \plus{} c} \equal{} \frac {2R\cdot (\sin \beta \minus{} \sin \gamma)}{2R\cdot (\sin \beta \plus{} \sin\gamma)} \equal{} \frac {2\cdot \cos \frac {\alpha \plus{} \beta}{2}\cdot \sin\frac {\alpha \minus{} \beta}{2}}{2\cdot \sin \frac {\alpha \plus{} \beta}{2} \cdot \cos \frac {\alpha \minus{} \beta}{2}} \equal{} \frac {\tan\frac {\alpha \minus{} \beta}{2}}{\tan\frac {\alpha \plus{} \beta}{2}}\][/hide]
[/quote]

This is actually equivalent to my solution, except that I went backwards. You don't [i]really[/i] need the extended sine law (the fact that they are equal to $ 2R$ is irrelevant)


For all $ a,b,c \ge 0$ prove the following:

 \[ \sqrt[4]{9(a\plus{}b)(a\plus{}b\plus{}c)} \ge \sqrt[4]{4b(a\plus{}c)} \plus{} \sqrt[4]{ab}\]

By Hölder we have:
\[ \sqrt[4]{(2\plus{}1)(2\plus{}1)(b\plus{}a)((a\plus{}c)\plus{}b)} \ge \sqrt[4]{4b(a\plus{}c)} \plus{} \sqrt[4]{ab},\]
as desired. Equality holds iff $ (a,b,c)\equal{}(t, 2t, 3t)$.


Let $ A\equal{}\begin{pmatrix}636287&\minus{}355892&160808\\390432&\minus{}734029&1289016\\\minus{}22400&749368&u\end{pmatrix}$ where $ u\in{\mathbb{Z}}$ s.t. $ |u|\leq{10^5}$.
Find $ u$ s.t. the equation $ X^3\plus{}X\plus{}I\equal{}A$ admits at least one solution in $ \mathcal{M}_3(\mathbb{Z})$.
With this (or these) $ u$-value solve for $ X$.

A hint:
The following program (time: 12') gives the value(s) of $u$ -with Maple (for example)-:

For u from -10^5 to 10^5 do
if P(u)=false then print(u); fi:
od:

P() is a property; how do you choose it ?


function
calculus
derivative
linear algebra
matrix
real analysis
real analysis theorems

I know this may not be the right place for this question, but anyway: how do I prove for a function defined on 2 variables that is convex/concave?Does fixing one of them and taking second derivative for the other variable, then fixing the other and checking the second derivative of the second one work?

You can check a function f with two variables that is convex or not by the way affirm that matrix Hessen is positive define.


ok , i hope that i didn't makes idiot  mistakes :D

this is my solution:

using cauchy_schwart ineq we have:
$\sqrt{(5b^2-bc+5c^2)(a^2)}$
$\frac{(a^2+b^2+c^2)(a+b+c)^2}{10((ab)^2+(bc)^2+(ac)^2)-abc(a+b+c)}\geq1$

after some computation:

$(a^2+b^2+c^2)^2+2(ab+bc+ca)(a^2+b^2+c^2)+abc(a+b+c)\geq10((ab)^2+(bc)^2+(ca)^2)$

expanding and compute we have:

$a^4+b^4+c^4+2\sum ab(a^2+b^2)+3abc(a+b+c)\geq8((ab)^2+(bc)^2+(ca)^2)$

using cosi again:

our ineq remain:

$a^4+b^4+c^4+3abc(a+b+c)\geq4((ab)^2+(bc)^2+(ac)^2)$

the last , using VORNICU_SCHUR ineq:

$(a^2-b^2-c^2)(a-b)(a-c)+(b^2-c^2-a^2)(b-a)(b-c)+(c^2-b^2-a^2)(c-a)(c-b)\geq0$

expanding it we get:

$a^4+b^4+c^4+3abc(a+b+c)\geq2((ab)^2+(bc)^2+(ca)^2)+\sum ab(a^2+b^2)$

ok , only using cosi we get our proof :D  :D

something tell me that you also have a generalization of the ineq above,vasc :lol:

also ,many thanks if you kindly show us your soluti ;) on,vasc :)

I think that to apply Valenticu-Shur ineq

http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=vornicu&t=24555

you must have $(a^2-b^2-c^2),(b^2-c^2-a^2),(c^2-a^2-b^2)$ positive

Cuong,this inequality:
[tex] a^4+b^4+c^4+2[ab(a^2+b^2)+bc(b^2+c^2)+ca(c^2+a^2)]+3abc(a+b+c)\geq\ 8[(ab)^2+(bc)^2+(ca)^2][/tex] is not true.See case a->0;b=c

Is this true ?

$\displaystyle a^4+b^4+c^4+3abc(a+b+c)\geq4((ab)^2+(bc)^2+(ac)^2) $

ohhhhhhhh so terrible

it's wrong :(

but i found that using cauchy_schwart  is good ideas

this one may be true , (i don't have time to confirm it  :D )

using cauchy with this number $\sqrt{(5b^2-bc+5c^2)a^4}$

i wait for you ,vasc :D

please , give us a hint  or is it your own ineq?? :)

@thanks spdf and manlio  :blush:  :blush:

ok, our ineq is equivalent to:

$\sum\frac{5a^2+bc}{5b^2-bc+5c^2}\geq2$

using cauchy_schwart ineq:

$\sqrt{(5b^2-bc+5c^2)(5a^2+bc)}$

$\frac{(5(a^2+b^2+c^2)+ab+bc+ca)^2}{\sum (5b^2-bc+5c^2)(5a^2+bc)}\geq2$

the last is equyvalent to:

$25(a^4+b^4+c^4)+22abc(a+b+c)\geq47((ab)^2+(bc)^2+(ca)^2)$

using$ 3(a^4+b^4+c^4)\geq3((ab)^2+(bc)^2+(ca)^2)$

and schur to complete our proof ;)  ;)

thanks for your hint , vasc :lol:

[quote="Vasc"]We write the ineq in the standard form:

$\displaystyle \sum \frac{5a^2+bc}{5b^2-bc+5c^2} \ge 2$ :)[/quote]

how to think of the standard form?
 :?

i also have a question like siuhochung:
 :D

dear manlio:

i just found one of your posts and see this ineq:

$\frac{a^2+2bc}{b^2+2ac}+\frac{b^2+2ac}{a^2+c^2}+\frac{c^2+ab}{a^2+b^2}\geq3$

you said that it can be solved by muirhead
can you give me a link of this ineqsolved by muirhead??/ :)

many thanks :lol:

Sorry, I don't remember that. Can you please show me the link?

For Siuhochung.
Because the general standard ineq

$\displaystyle \displaystyle \frac{2a^2+qbc}{b^2+pbc+c^2}+\frac{2b^2+qca}{c^2+pca+a^2}+\frac{2c^2+qab}{a^2+pab+b^2} \ge \frac{3(2+q)}{2+p}$

is valid for $p>-2$ and $q\ge 2p+1$. However I haven't a solution for $p<-1$.

For Cuong.
The Cauchy-Schwarz way works for our ineq, but doesn't work for

$\displaystyle \displaystyle \frac{1}{4b^2-bc+4c^2}+\frac{1}{4c^2-ca+4a^2}+\frac{1}{4a^2-ab+4b^2} \ge \frac{9}{7(a^2+b^2+c^2)}$
 ;)

very hard ineq indeed

 :D

i transform it into a standard form , but i can't solve it

please give me a hint, vasc :lol:

sory , if it's your secret ;)

Here is a hint. I wrote the inequality in the form
$A(b-c)^2+B(c-a)^2+C(a-b)^2 \ge 0$.
This way is your speciality. :lol:

Every inequality of this kind can be solved by one that way! :)

Ok, i have a proof for it ,i use Vasc's sugget . :lol:

Reyes, can you post your solution, please?

Thank you very much.

[quote=Vasc]
$\displaystyle \displaystyle \frac{2a^2+qbc}{b^2+pbc+c^2}+\frac{2b^2+qca}{c^2+pca+a^2}+\frac{2c^2+qab}{a^2+pab+b^2} \ge \frac{3(2+q)}{2+p}$

is valid for $p>-2$ and $q\ge 2p+1$. [/quote]
It should be $p>-2$ and $q\leq2p+1$.

[quote=manlio]For $a,b,c$ positive reals prove that

$\displaystyle \sum \frac{1}{5b^2-bc+5c^2} \ge \frac{1}{a^2+b^2+c^2} $

[quote]If you solve this challenge you will have like deserved prize Vasc solution for

$\displaystyle \frac{2a^2+qbc}{b^2+pbc+c^2}+\frac{2b^2+qca}{c^2+pca+a^2}+\frac{2c^2+qab}{a^2+pab+b^2} \ge \frac{3(2+q)}{2+p}$  ;)

Good work  :)[/quote]

This is no little. :lol:[/quote]

Stronger is:
$a,b,c>0$
${\frac {{a}^{2}+{b}^{2}+{c}^{2}}{4\,{b}^{2}-bc+4\,{c}^{2}}}+{\frac {{a
}^{2}+{b}^{2}+{c}^{2}}{4\,{c}^{2}-ac+4\,{a}^{2}}}+{\frac {{a}^{2}+{b}^
{2}+{c}^{2}}{4\,{a}^{2}-ab+4\,{b}^{2}}}\geq{\frac {9}{7}}$


$ 2x^{3}\minus{}5x^{2}\plus{}4x\plus{}6\equal{}0$ has roots $ \alpha,\beta,\gamma$. Find equations with roots

(A)$ \alpha\plus{}1,\beta\plus{}1,\gamma\plus{}1$

Why can this be written as $ y\equal{}x\plus{}1$
$ x\equal{}y\minus{}1$ then sub into eqn and simplify??

$ 2x^{3}\minus{}5x^{2}\plus{}4x\plus{}6\equal{}0$ has roots $ \alpha,\beta,\gamma$. Find equations with roots
(B)And how do I solve this?
$ \alpha\beta,\beta\gamma,\gamma\alpha$

Well once you make the substitution $ y\equal{}x\plus{}1$ and obtained an polynomial in terms of $ y$ then you can re-substitute $ y\equal{}x\plus{}1$ and you will see that the new polynomial has zeroes at $ \alpha\plus{}1,\beta\plus{}1,\gamma\plus{}1$.

For the second question notice that $ \alpha\beta,\beta\gamma,\gamma\alpha$ are really just $ \frac{\minus{}3}{\gamma},\frac{\minus{}3}{\alpha},\frac{\minus{}3}{\gamma}$ respectively.

[hide="a hint"]Viete[/hide]
(A)[hide="a solution"]We have: (let $ \alpha\equal{}a,\beta\equal{}b,\gamma\equal{}c$)
$ a\plus{}b\plus{}c\equal{}\frac{5}{2}\\ ab\plus{}bc\plus{}ac\equal{}2\\ abc\equal{}\minus{}3$
hence
$ (a\plus{}1)\plus{}(b\plus{}1)\plus{}(c\plus{}1)\equal{}\frac{11}{2}\\ (a\plus{}1)(b\plus{}1)\plus{}(a\plus{}1)(c\plus{}1)\plus{}(b\plus{}1)(c\plus{}1)\equal{}ab\plus{}bc\plus{}ac\plus{}2(a\plus{}b\plus{}c)\plus{}3\equal{}2\plus{}2\cdot\frac{5}{2}\plus{}3\equal{}10\\ (a\plus{}1)(b\plus{}1)(c\plus{}1)\equal{}abc\plus{}(ab\plus{}bc\plus{}ac)\plus{}(a\plus{}b\plus{}c)\plus{}1\equal{}\minus{}3\plus{}2\plus{}\frac{5}{2}\plus{}1\equal{}\frac{5}{2}$
hence we have our equation:
$ x^{3}\minus{}\frac{11}{2}x^{2}\plus{}10x\minus{}\frac{5}{2}\equal{}0$[/hide]

i dont get anything from this post im dmb can someone explain plz what this post is about ?


Just curious if anybody on here is attempting [url=http://infinitesummer.org/]Infinite Summer[/url]. I read [i]Infinite Jest[/i] a few years ago and probably won't re-read it at this point in my life. But it is a truly amazing work of literature. It contains quite a bit of math by someone who knew an amazing amount for his given profession.

That's a lot of pages... Does it ever get dull?

David Foster Wallace also wrote [url=http://en.wikipedia.org/wiki/Everything_and_More]Everything and More[/url], a book about infinity.  Anyways I have a copy of Infinite Jest, I've been meaning to read it for awhile, it's sitting on my bedside table right now, this Infinite Summer thing sounds intriguing and I may give it a go.

Dull is not quite the word. The pacing and plot action are really slow in some parts, but it is very rarely dull in my mind.

I've been meaning to read [i]Infinite Jest[/i] for... well I haven't gotten to it yet. I want to build up to it. I've read his essay collection [i]Consider the Lobster[/i]$ .^1$ and am planning on finding his collection of short stories [i]Oblivion[/i], which I am told is quite good. Oddly enough, I saw [i]Everything and More: a Compact History of Infinity[/i] at the book store today and thumbed through it. It's a terrible loss that he died.

*** Interpolation (ok, really an edit) ***
DFW's speech at a college commencement thing "This is Water" served as the inspiration for my own valedictorian address. Simply beautiful. Thankfully a text of it was still up on the internet when I looked for it, but now you can buy it in book form$ .^2$

$ .^1$[size=75] Of which [i]Authority and American Usage, or Politics and Language is Redundant[/i] was particularly brilliant, among other book reviews/philosophical interpolations.
$ .^2$ worth it. [/size]


A whole number is called an \textit{increasing number} if each digit in the number is greater than the digit to its left; for example, 2359. How many increasing numbers are there between 5000 and 10000?

working systematically from smallest to largest:

5678
5679
5689
5789
6789

= $ \boxed{5}$


A four-digit integer m and the four-digit integer obtained by reversing the order of the digits of m are both divisible by 45. If m is divisible by 7, what is the greatest possible value of m?

[hide="  Not sure if this is the greatest value"] Both integers are divisible by $ 45 \equal{} 5 \times 9$ . Let $ a,b,c,d$ be the digits of m in that order, then we have $ a\plus{}b\plus{}c\plus{}d \equal{} 9k$ and $ a,d \neq 0$. Therefore it's $ a \equal{} d \equal{} 5$, because m's last digit is d (which follows it is either $ 0$ or $ 5$) and the same thing applies to a if the order of the digits is reversed. So $ 10 \plus{} b \plus{} c \equal{} 9k$. Because $ 0 \leq b,c \leq 9$ then $ b\plus{}c \equal{} 8$ or $ b\plus{}c\equal{}17$ which the latter is only possible for $ b\equal{}9,c\equal{}8$ or $ c\equal{}9, b\equal{}8$. $ 5895$ is not divisible by 7 and $ 5985$ is. There are no other values of b,c greater than the pair (9,8) that fullfil b+c = 8 (The pairs in that case are $ (8,0),(7,1),(2,6),(3,5),(4,4),(5,3),(6,2),(7,1),(8,0)$ and the resulting numbers are $ 5805, 5715,5625,5535,5445,5355,5265,5175,5085$, of which only $ 5355$ is divisible by 7.

So the greatest value of m is $ m\equal{}5985$? [/hide]

[hide="answer"]5985.[/hide]
[hide="how I got it"]I figured both the first and last digits must be 5, since it is divisible by 5 and it has to be the largest value possible. Then I got that the sum of 5, 5, and the middle digits has to be divisible by three. 5115 didn't work; 5145 did, as it has to be divisible by seven. Then I kept adding 210, to make sure it stayed divisible by 3 and 7, until I got to 5985, which was as high as I could go without changing the 1000s digit to 6.[/hide]


The length of the year on the planet Mars is
exactly 697 days. If Mars has a calendar with a
12-day week, and year 0 begins on the first day of
the week, what is the next year which will begin
on the first day of the week?

$ 697\equiv1\pmod{12}$, so every next year begins with a day of the week later.
Answer: 12th year.

In general, the number of days in a week determines "what is the next year which will begin on the first day of the week?"

...am I right or am I right? (or am I wrong)

[hide=Alcumus Solution:]Since $697 = 12 \cdot 58 + 1$, each Martian year consists of 58 weeks and a day. Therefore, for every year that passes, the first day of the year shifts to the next day of the week. Since year 0 begins on the first day, year 1 begins on the second day, then year 2 begins on the third day, and so on. A week consists of 12 days, so the next year that begins on the first day again will be year $\boxed{12}$.[/hide]

Sorry, I don't think copying and pasting Alcumus solutions is the right way to go about posting solutions in this forum.

since $697 \equiv 1 [12]$ and 1 is relatively prime to 12, it takes 12 years for the residues to cycle around, so the answer is 12


function
limit
integration
LaTeX
trigonometry
Asymptote
real analysis

[quote="Cezar Lupu"]Let $ a(.), b(.): (0,\infty)\to\mathbb{R}$ two continuous functions satisfying the following differential equation:
\[ \frac{dx}{dt}\equal{}\minus{}a(t)x\plus{}b(t).(*)\]
If $ a(t) > c > 0,\forall t\in I$ and $ \lim_{t\to\infty}b(t) \equal{} 0$, then prove that $ \lim_{t\to\infty}\varphi(t) \equal{} 0$, for all $ \varphi(.)$ solution of the $ (*)$ differential equation.[/quote]

$ x(t)\equal{}\biggl(C\plus{}\int_{t_{0}}^{t}b(\tau) e^{\int_{t_{0}}^{\tau}a(s)ds}\biggr)e^{\minus{}\int_{t_{0}}^{t}a(\tau)\,d\tau}$.

Note that $ \biggl|Ce^{\minus{}\int_{t_{0}}^{t}a(\tau)\,d\tau}\biggr|\leqslant C_{1}e^{\minus{}ct}\to 0$.

If $ \biggl|\int_{t_{0}}^{t}b(\tau) e^{\int_{t_{0}}^{\tau}a(s)ds}\biggr|\leqslant M$ then all right.

If $ \int_{t_{0}}^{t}b(\tau) e^{\int_{t_{0}}^{\tau}a(s)ds}\to\infty$, we use Lopitale rule:

$ \lim_{t\to\infty}\frac{\int_{t_{0}}^{t}b(\tau) e^{\int_{t_{0}}^{\tau}a(s)ds}}{e^{\int_{t_{0}}^{t}a(\tau)\,d\tau}}\equal{}\lim_{t\to\infty}\frac{b(t) e^{\int_{t_{0}}^{t}a(\tau)d\tau}}{a(t) e^{\int_{t_{0}}^{t}a(\tau)\,d\tau}}\equal{}\lim_{t\to\infty}\frac{b(t)}{a(t)}\equal{}0$.

[quote="Kakael"]Sorry, but i didn´t find another place to ask for help. I´m new here and I would like to know how can i post, how can i write in mathematical notation, is there a program? Thanks![/quote]
You must just click on "Post new topic" in the left corner.If you want to use LaTeX just put dollar signs and between them write the things you need. :wink:
For example:
[code]
$a+b+c=1$.
[/code]
will be
$ a\plus{}b\plus{}c\equal{}1$.
For more information about LaTeX click on the LaTeX Help :wink:

You can use latex here. Any ratex expression must lie between two dollar signs.

For example, if you write down

$\lim_{x \to \infty} \frac{e^{-x}}{1-2\sin \theta x + x^{2}} = 0$

and post it, then you'll see

$ \lim_{x\to\infty}\frac{e^{\minus{}x}}{1\minus{}2\sin\theta x\plus{}x^{2}}\equal{} 0$.

For more detail about ratex, you can refer to these:

[1] [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php]About LaTeX[/url]
[2] [url=http://www.mathlinks.ro/Forum/index.php?f=123]LaTeX and Asymptote[/url]
[3] [url=http://tug.ctan.org/tex-archive/info/math/voss/mathmode/Mathmode.pdf]Math mode - v.2.27[/url]

Thanks a lot!!! I´ll learn how to use latex then a I post a topic.


Hey if anyone has gotten to this problem and has solved it under the new 1 second time limit can you help me?
I have done all the optimizations possible except making linked lists for the strings (I use character arrays). My program times out on test case #8. I think it takes around 4 seconds on that case. What else can I do to get it to run under 1 second besides making the linked lists or is that the only option?

My algorithm is a complete search using dfs and it stops the recursion function on impossible strings.

[edit] If anyone has done this problem already, can you just tell me your runtime for case #8? Right now I just need to know if its possible to get it in under 1 sec.

Test 1 OK [0 secs]
Test 2 OK [0 secs]
Test 3 OK [0 secs]
Test 4 OK [0 secs]
Test 5 OK [0 secs]
Test 6 OK [0.01 secs]
Test 7 OK [0 secs]
Test 8 OK [0.24 secs]
Test 9 OK [0.02 secs]
Test 10 OK [0.05 secs]

Try more optimizations. There's a forum on Usaco

wow! that's impressive! i am now in probably 5.4.. anyhow can u plz send ur code kazenokizu? (r u japanese?)

Mahbub


geometry
search
analytic geometry
vector
ratio
college
similar triangles

Supposedly, if I am to understand what mass point geometry is, I first need to understand baricentrics (is that how you spell it? MY inability to spell "baricentrics" only proves my paucity of knowledge in that area).

So, any help on baricentrics will be greatly appreciated.

It is actually spelled "barycentrics", but, like you, I have little knowledge in the field. Perhaps a google search would help.

what is the connection between barycenters (aka centroids, center of gravity for objects) and "mass point geometry" (not that i know what mass point geometry is)?

did you google it? or perhaps mathworld it?

well, i googled and mathworlded it, and i kno wat it is

but i don't ahve a good understanding of its applications and uses...

well, i dont either...so i guess you'll have to wait for complexzeta and triplem and everyone else..

To be frank, I had never heard of mass point geometry until Joel brought it up. But I Googled ``mass point geometry'' and figured it out. Briefly, mass point geometry is a way of applying basic physical principles to geometry problems. (For example, we can say point P has a mass of 3, whereas point Q has a mass of 2, and point R has a mass of 7.) The first example of the use of mass point geometry that springs to my mind is the centroid of a triangle, which is also the center of mass. To prove this, we can say that in triangle ABC, A, B, and C are points of mass 1. Then the center of mass of AB is the midpoint of AB, the center of mass of BC is the midpoint of BC, and the center of mass of CA is the midpoint of CA. Then the center of mass of triangle ABC lies on the line connecting A to the midpoint of BC, and so on symmetrically. But that's the definition of the centroid.

[quote="darkquantum"]any help on baricentrics will be greatly appreciated.[/quote]

ok, well as someone already pointed out it's 'barycentrics', though its from the greek meaning 'weight' so difference in spelling dont really matter too much.

anyway, ive not heard much about this mass point thing but i've done quite a bit on barycentric coordinates, also known as areal coordinates.  basically, given a triangle ABC you can uniquely define any point in the plane of the triangle with three coordinates.

the coordinates of A, B, C would be (0,0,1), (0,1,0) and (1,0,0) in some order.  now the nice thing is that this is just vectors really, so if we want to express any other point on that plane we can do it by a linear combination of [u]a[/u], [u]b[/u] and [u]c[/u], which are conveniently normalised to 1.  hence the sum of the coordinates of any point will always be 1.

trivially, the coordinates of the bases of the medians are (1/2, 1/2, 0), (1/2, 0, 1/2) and (0, 1/2, 1/2), and any other point on the line (say AB) has coordinates (0, x, 1-x).

any line dividing the triangle in a specific ratio m/n through a vertex is expressed as (1-x, xm/n, x(1-m)/n) where x varies.  check that these add up to 1 and try a median.

the centroid is obviously (1/3,1/3,1/3) in this coordinate system.

the major advantages are:

1. there is no origin.  everything is just relative to the triangle, so you dont have any asymmetries except where they are in the problem.

2. its very easy to express divisions along lines, so problems with ratios of distances and areas are easy

(oh, i forgot to mention why its also called areal coordinates: the coordinates of a point P are in the same ratio as the areas of the three small triangles APB, BPC, CPA)

3. there are nice expressions for most of the triangle centres (figure them out for yourself)

4. the point P(x,y,z), x+y+z=1, is the geometric centroid of a triangle with masses x,y,z placed at the three vertices.  (this may be the origin of the phrase 'mass point geometry'?)

basically if youre having trouble understanding it just think of it as vectors.

if you want examples of how it can be used, try proving ceva's theorem and menelaus' theorem using barycentric coordinates.

the point in Ceva's thm (lets call it P) may also lie outside the triangle.  in a pure geometrical proof you would have to be very careful and maybe consider these cases separately, but using these coordinates its evident that the coordinates of P may be negative and the proof, as long as you dont do anything too complicated, will hold equally well for all points in the plane, not just in the interior of the triangle.

thats also a hint for menelaus: remember that the point dividing an edge externally is going to have at least one negative coordinate.

Have you checked out the lecture at the Berkeley Math Circle?

http://mathcircle.berkeley.edu/mpgeo.pdf

Mass point geometry can be used almost anywhere where you need to find the ratio of a line length to another in a triangle; however, you can solve that same question with vertexes, similar triangles, and the sort.


Given two sets $A$ and $B$ with $|A|=m$ and $|B|=n$ and $m>n$. What is the number of mappings $f: A \rightarrow B$ which are "on" (for every element $y$ of $B$ there is $x\in A$ s.t. $f(x)=y$).

Do you mean that $f$ is surjective?
In that case, the desired number is $n!S(m,n)$ where $S(m,n)$ is the stirling number of second kind :
http://mathworld.wolfram.com/StirlingNumberoftheSecondKind.html

Pierre.


Given the area of triangle A.
How can i find the three sides of that triangle?

You can't.

The area of a triangle does not uniquely determine its side lengths.

For example, a right triangle with legs of length 12,1 and 6,2 have the same area. All their side lengths are different.

You can't do that like Zhero said. But if you want to find the area of the triangle using the side lengths, use Heron's formula.
http://mathworld.wolfram.com/HeronsFormula.html


While i were trying to prove APMO inequality,
i found this inequality:
let a,b,c>0 be positive reals.
prove:
(b+c)^1/2*( (a+b)^1/2 + (a+c)^1/2 ) >= (b+c)/2 +(ab)^1/2+(ac)^1/2

try to solv ;-)

Have you tried to use AM-GM and then  use Cauchy-Schwartz? :D

cheers!

An old  problem but I'll try to give some solution!
$\sqrt{a+b}\sqrt{b+c}\geq(b+\sqrt{ac})$

By the same way we get $\sqrt{a+c}\sqrt{b+c}\geq(c+\sqrt{ab})$

adding we have $LS\geq b+c+\sqrt{ac}+\sqrt{ab}\geq \frac{b+c}{2} +(ab)^1/2+(ac)^1/2$


function
trigonometry
real analysis
real analysis theorems

How do you generally approach the differential equations containing composite functions? E.g. $2f(x)f'(x)=f(2x)$ (both $f(x)=x$ and $f(x)=\sin x$ are the solutions)

Do such equations arise naturally?  :huh:

I doubt that there is a reasonable description of all solutions of this equation. Let $f$ be a $C^{\infty}$-smooth function on the interval $[1,2]$ such that $f=0$ outside of $(1+\epsilon,2-\epsilon)$. Extend its definition to $[2,4]$ by the formula $f(2x)=2f(x)f'(x)$. Then extend to $[4,8]$, etc. You can also extend $f$ to $[1/2,1]$, etc, by solving an ODE.


geometry
circumcircle
inequalities unsolved
inequalities

Let $P$ be an interior point in triangle $ABC$ and let $r_{1}$ , $r_{2}$ and $r_{3}$ denote the distance between $P$ to the triangle sides with side's length : $a_{1}$, $a_{2}$ dan $a_{3}$, respectively.
Let $R$ is the circumradius of $ABC$, and $0<a<1$ is real.
If $b = \frac{2a}{1-a}$. Prove that

$r_{1}^{a}+r_{2}^{a}+r_{3}^{a}\leq \frac{1}{(2R)^{a}}(a_{1}^{b}+a_{2}^{b}+a_{3}^{b})^{1-a}$


Given 3 masses, $m$ each, hanging from strings of length $r < L$ on the following pendulum:

 __________$P$__________
$|\ \ \ \ \ \ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |$
$|\ \ \ \ \ \ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |$
$|\ \ \ \ \ \ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |$
$0\ \ \ \ \ \ \ \ \ \ \ \ \ 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0$

Where $P$ is the pivot point, and the horizontal bar at the top has length $2L$ and mass $M$, with $P$ at its midpoint.

Find the position of each bob as a function of time, and calculate the period of the device (time for all 3 bobs to be in the original places again).

[hide="hint 1"]You can assume $\theta$ is small enough that $\sin \theta \approx \theta$, for any of the angles.[/hide]

[hide="hint 2"]If the periods are $T_{1}, T_{2}, T_{3}$, the period of the whole system is $lcm(T_{1}, T_{2}, T_{3})$.[/hide]

The pendulum would just stay there unless you changed it from its equilibrium position.  How do you change it? Rotate the mass at the end of the strings away by some angle $\theta$?

Did you write this problem yourself?

Well, let $\phi$ be the displaced angle of the horizontal bar (counterclockwise = +).

And let $\theta_{1}, \theta_{2}, \theta_{3}$ by the angles of the pendulums from left to right.

So starting it from an angle $\phi_{0}\neq 0$.

Yes, I also came up with this problem myself.

I had a sudden urge to write a bunch of unusual pendulum problems.


It's been posted many times and it's not really appropriate for HSB.

I think it's nice problem. But it's difficult, do you think so?


geometry
analytic geometry
graphing lines
slope

ABC is a triangle: A=(0,0), B=(36,15) and both the coordinates of C are integers. What is the minimum area $\triangle ABC$ can have?
(A) 1/2
(B) 1
(C) 3/2
(D) 13/2
(E) there is no minimum

[hide]C=(x,y)

$[ABC]=\frac{3|5x-12y|}{2}$

No isn't the answer zero? or are degenerate triangles not allowed?

If they're not, then the answer would be $\frac{3}{2}$[/hide]

[hide="Solution"]
Let $C=(x,y)$. Then use the shoelace theorem to get $\triangle = \left| \frac{ (0\cdot 15+36 \cdot y+x \cdot 0)-(0 \cdot 36+15 \cdot x+y \cdot 0) }{2}\right| = 3 \cdot \left| \frac{12y-5x}{2}\right|$. I'm going to assume that we cannot have a degenerate triangle (which would have $C$ along the line $12x=5y$); then with Bézout's Identity we see that for some lattice point $(x,y)$ we'll have $12y-5x=\gcd(5,12)=1$. Thus, the minimum area of the triangle will be $\triangle=3 \cdot \left| \frac{1}{2}\right| = \boxed{\frac{3}{2}\Longrightarrow C}$[/hide]

I'd say $C$.

[hide]
The slope of the line is 36/15 = 12/5, so we have a 5,12,13 right triangle.

The base of the triangle is 3*13 = 39.

The minimum vertical distance from a multiple of 1/12 to an integer is 1/12 (attained at x=5). So, the minimum height is 12/13 * 1/12 = 1/2 * 1/13 * 39 = 3/2

[/hide]

[hide="Another method"] There are no interior lattice points in $\triangle ABC$ (or else we could use them to form a smaller triangle), so by Pick's Theorem

$[ \triangle ABC ] = \frac{5}{2}+0-1 = \boxed{ \frac{3}{2}}$

Where the border points are $(0, 0), (12, 5), (24, 10), (36, 15), C$.  (I am assuming that the minimal case for $\triangle ABC$ is such that there are no border points on $AC$ or $BC$, which is not entirely rigorous, but it's a safe assumption for a contest problem, especially considering the only answer larger than $\frac{3}{2}$ is [b]much[/b] larger than $\frac{3}{2}$ :P ) [/hide]

[hide]Letting C=(x,y) and using Shoelace,
$A=\frac{|36y-15x|}{2}=\frac{3|12y-5x|}{2}$. Minimum is $\boxed{C: 3/2}$.[/hide]


I remember seeing this in this thread a while back, but I can't seem to find it using the search function. Can anyone link me?

Thanks

it seems i can't find it either. but it's not so hard. here are two approaches:

[hide="first approach"]show that if $ d$ is the distance between the incenter and the circumcenter of a triangle then $ d^{2}=R^{2}-2Rr$.[/hide]
[hide="second approach"]it's well known that the nine point circle has radius $ \frac{R}{2}$. and it's not difficult to see that the nine point circle has radius $ \ge r$. thus $ \frac{R}{2}\ge r$.[/hide]

I don't know if this is the thread you are referring to, but I posted this problem recently enough
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1998973251&t=140629


The product $N$ of three positive integers is $6$ times their sum, and one of the integers is the sum of the other two. Find the sum of all possible values of $N.$

[hide="Generalizable solution"]
[size=117]
Call the numbers $a,b,$ and $c$, set up so that $a=b+c$, and $c$ is the smallest. The product, $N$, equals $abc$ and the sum ("$S$") equals $a+b+c$. Therefore, \[N=6S\iff abc=6(a+b+c)\iff (b+c)bc=6(2a)\iff 12=bc^{*}\]
$*$This last step is valid since $b,c$ are positive integers, the quantity $b+c\not =0$.

Since $bc$ is a constant ($12$, clearly), \[\sum_{all}N=\sum_{all}(b+c)bc=12\left(\sum_{all}b+c\right)\]
We know that $c\leq b$. Since $12$ isn't a perfect square, $b\not =c\implies c<b$. Thus \[\left(\sum_{all}b+c\right)=\sigma(12)\], where the $\sigma$ function denotes the sum of the divisors of $12$.

Now, factor $12: \ 2^{2}\cdot 3$. Using well known properties of the $\sigma$ function (easy to prove and see [url=http://en.wikipedia.org/wiki/Divisor_function]here[/url] for reference), $\sigma(12)=(1+2+2^{2})(1+3)=28$. Thus \[\sum_{all}N=12\sum_{all}b+c=12\sigma(12)=12(28)=\boxed{336}\]

Yes, SM4RT, your answer is correct. This is an alternate to summing up after casework.[/size]
[/hide]

Now solve: $N=6000$ times their sum... (if you read my solution it should take seconds)


Find all $P \in \mathbb{R}[X{\text{]}}$ such that : \[ {\text{(X + 3)P(X) = XP(X + 1)}}  \]

P(x) is divisible by x so P(x)=xQ(x) substitute and get (x+3)Q(x)=(x+1)Q(x+1)
then Q(x)is divisible by x+1 so we can write Q(x)=(x+1)R(x)
substitute and get (x+3)R(x)=(x+2)R(x+1)then R(x)=(x+2)F(x)and finally
we get F(x)=F(x+1) => F(x)=a with $a$ in R
so P(x)=ax(x+1)(x+2)

this is my first post on Mathlink so .. :?
my solution is quite easy , as u have P(x+1)/P(x)=x+3/x so P(x)/P(x-1)=x+2/x-1 & so on ...
therfore ,
P(x+1)/P(1)=x+3!/x!=(x+1)(x+2)(x+3)
then P(x)=ax(x+1)(x+2)
                                                         chevalir3as


I am sorry not to start with some solved problems as my first post but  i think i should wait a bit longer before posting solutions.Still i found this very interesting, hope you like it:

From a strictly mathematical viewpoint it goes like this:
 What makes 100%?
 What does it mean to give MORE than 100%?
 Ever wonder about those people who say they are giving more than 100%?
 We have all been to those meetings where someone wants you to give over 100%.
 How about achieving 103%?
 Here's a little mathematical formula that might help you answer these questions:
 If:
 A B C D E F G H I J K L M N O P Q R S T U V W X Y
 is represented as:
 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26,
 then:
 H-A-R-D-W-O-R-K
 8+1+18+4+23+15+18+11 = 98%
 K-N-O-W-L-E-D-G-E
 11+14+15+23+12+5+4+7+5 = 96%
 A-T-T-I-T-U-D-E
 1+20+20+9+20+21+4+5 = 100%
 B-U-L-L-S-H-I-T
 21+12+12+19+8+9+20 = 103%
 and, look how far ass kissing will take you:
 A-S-S--K-I-S-S-I-N-G
 1+19+19+11+9+19+19+9+14+7 = 118%
 So, one can then conclude with mathematical certainty that:
 While Hardwork and Knowledge will get you close, and Attitude will get you there, Bullshit and Ass Kissing will put you over the top!

The person who wrote this was definetly narrow minded,what do you think?

[quote="Ran"]The person who wrote this was definetly narrow minded, what do you think?[/quote]
I am thinking exactly the opposite. This is supposed to be a joke, and the guy who made it was pretty clever to add up the letters of certain words to obtain a good combination (one with a meaning what so ever).

Then again, a computer program can be made to compute the %-ness of each word, and by testing it versus an english dictionary you can extract lots of metaphoras :D.

okay. that's in English.
what about other languages?

oh,i thought asskicking was supposed to get u farther than asskissing :D


1 ) DAR MOSALASE [u]ABC [/u]DARIM[u] A=90 [/u]VA[u] B<C[/u] . MOMAS DAR [u]A[/u] BAR DAYEREYE MOHITI MOSALAS YANI [u]W[/u]  KHATE [u]BC[/u] RA DAR [u]D[/u] GHAT MIKONAD . [u]E[/u] RA GHARINEYE [u]A[/u] NESBAT BE ZEL'E [u]BC[/u] VA [u]X[/u] RA PAYE AMODE AZ [u]A[/u] BAR [u]BE[/u] VA [u]Y[/u] RA VASATE [u]AX[/u] BEGIRID . FARZE KONID KAHTE [u]BY[/u] DAYEREYE [u]W[/u] RA DAR [/u]Z HAM KHAT KONAD . SABET KONID KHATE [u]BD[/u] BAR DAYEREYE MOHITI MOSALASE [u]ADZ[/u] MOMAS AST .

2) CHAHAR NOGHTEYE [u]A , B, C ,D[/u] TORI ROYE DAYERE ENTEKHAB SHODEHAND KE [u]BCD[/u] GHAEMOLZAVIYE NIST  VA AMODMONESEFHAYE [u]AB[/u] VA [u]AC[/u] KHATE [u]AD[/u] RA DAR NOGHATE[u] W[/u] VA [u]V[/u] GHAT MIKONAND VA HAMCHENIN KHOTOT [u]CT[/u] VA [u]BW[/u] YEKDIGAR RA DAR NOGHTEYE [u]T [/u] GHAT MIKONAND .SABET KONID TOLE YEKI AZ PAREKHATHAYE  BARABARE HASELE JAME TOLE 2TAYE DIGAR AST .

[quote="SoNYDvDPro"]1 ) DAR MOSALASE [u]ABC [/u]DARIM[u] A=90 [/u]VA[u] B<C[/u] . MOMAS DAR [u]A[/u] BAR DAYEREYE MOHITI MOSALAS YANI [u]W[/u]  KHATE [u]BC[/u] RA DAR [u]D[/u] GHAT MIKONAD . [u]E[/u] RA GHARINEYE [u]A[/u] NESBAT BE ZEL'E [u]BC[/u] VA [u]X[/u] RA PAYE AMODE AZ [u]A[/u] BAR [u]BE[/u] VA [u]Y[/u] RA VASATE [u]AX[/u] BEGIRID . FARZE KONID KAHTE [u]BY[/u] DAYEREYE [u]W[/u] RA DAR [/u]Z HAM KHAT KONAD . SABET KONID KHATE [u]BD[/u] BAR DAYEREYE MOHITI MOSALASE [u]ADZ[/u] MOMAS AST .[/quote]

mahalle talaghie $ AE,BC$ ra $ H$ va mahalle talaghie $ AZ,BC$ ra $ K$ benamid,khob kafie sabet konim $ KD^2\equal{}KZ.KA$ dge...

khob midanim 4zelie $ AZBE$ mohatist pas:

$ \angle ZBX\equal{}\angle ZAH$

hamchenin dar mosalase $ \triangle AEX$,noghate $ Y,H$ be tartib osate azlae $ AX,AE$ hastand,pas tebghe tales $ HY\parallel EX$ pas:

$ \angle HYB\equal{}\angle YBX\equal{}\angle ZAH$

pas 4zelie $ YZAH$ mohatie dar natije:

$ \angle HYA\equal{}\angle HZA\equal{}90^\circ$

amma az tarafi $ \angle AHB\equal{}90^\circ$ pas $ \angle ZHB\equal{}\angle ZAH$ pas $ \triangle KZH\sim\triangle KAH$ pas $ KH^2\equal{}KZ.KA\equal{}KB.KC$

khob pas tebghe oon chizi ke avvale esbat goftim kafie sabet konim $ KH\equal{}KD$ pas age gharine $ H$ nesbat be $ K$ ra $ D'$ benamim kafie sabet konim $ D\equiv D'$...

amma tebghe rabete nioton darim $ \boxed{(D'H,BC)\equal{}\minus{}1} (I)$

amma az tarafi midoonim $ \boxed{(DH,BC)\equal{}\minus{}1} (II)$

hala ba tavajoh be $ (I),(II)$ natije mishe $ D\equiv D'$ yeni hamoon chizi ke mikhastim pas $ BD$ bar dayere mohitie $ \triangle ADZ$ momase...

[quote="SoNYDvDPro"]2) CHAHAR NOGHTEYE [u]A , B, C ,D[/u] TORI ROYE DAYERE ENTEKHAB SHODEHAND KE [u]BCD[/u] GHAEMOLZAVIYE NIST  VA AMODMONESEFHAYE [u]AB[/u] VA [u]AC[/u] KHATE [u]AD[/u] RA DAR NOGHATE[u] W[/u] VA [u]V[/u] GHAT MIKONAND VA HAMCHENIN KHOTOT [u]CT[/u] VA [u]BW[/u] YEKDIGAR RA DAR NOGHTEYE [u]T [/u] GHAT MIKONAND .SABET KONID TOLE YEKI AZ PAREKHATHAYE  BARABARE HASELE JAME TOLE 2TAYE DIGAR AST .[/quote]
fek konam manzooret [b]$ CV$,$ BW$ hamdigar ra dar $ T$ ghat konanado az beine $ AD,BT,CT$ toole yeki barabare jame dotaye digas[/b]...
khob fek konam majboor bashim parekhattaro jahatdar begirim:

sabet mikonam $ AD\equal{}BT\plus{}CT$...

mahalle talaghie $ BT$ ba dayeraro $ K$ benamid,khob hala chon $ W$ rooye amood monassefe $ AB$ gharar dare pas:

$ WA\equal{}WB$ yeni $ \boxed{\angle WAB\equal{}\angle WBA} (I)$...

hamchenin do zavie $ \angle DAK$ va $ \angle DBK$ do zavie mohati roobe yek kamanan pas:

$ \boxed{\angle DAK\equal{}\angle DBK} (II)$...

hala az $ (I),(II)$ natije mishe:

$ \angle DBA\equal{}\angle KAB\Rightarrow \boxed{AD\equal{}BK}$

pas kafie sabet konim:

$ BK\equal{}BT\plus{}CT$

$ \Rightarrow BT\plus{}TK\equal{}BT\plus{}CT$

pas kafie sabet konim $ TK\equal{}CT$ ya be ebarati $ \angle TKC\equal{}\angle TCK$

khob darim:

$ \angle WAB\equal{}\angle WBA\equal{}\angle KCA\Rightarrow\boxed{\angle DAB\equal{}\angle KCA} (1)$...

khob hamchenin midoonim noghte $ V$ rooye amood monassefe $ AC$ gharar dare pas:

$ \boxed{\angle VAC\equal{}\angle VCA} (2)$

hamchenin do zavie $ \angle CKB$ va $ \angle CAB$ zavie mohati rooberoo be yek kamanan pas ba ham barabaran yeni:

$ \boxed{\angle CKT\equal{}\angle CAB} (3)$

khob hala az $ (1),(2),(3)$ natije mishe $ \angle TCK\equal{}\angle TKC$ yeni hamoon chizi ke mikhastim...


ratio calculus derivative function logarithms algebra domain

Two differentiable real functions $ f(x)$ and $ g(x)$ satisfy
\[ \frac {f \prime (x)}{g \prime (x)} \equal{} e^{f(x) \minus{} g(x)}
\] for all $ x$, and $ f(0) \equal{} g(2003) \equal{} 1$. Find the largest constant $ c$ such that $ f(2003) > c$ for all such functions $ f$, $ g$.

take the natural log of both sides of the equation:

$ \ln f'(x) \minus{} \ln g'(x) \equal{} f(x) \minus{} g(x)$.

hope that helps......

Not really. The logarithm of a derivative is problematic.

[hide="Solution"]We can rearrange to $ f'(x)e^{ \minus{} f(x)} \equal{} g'(x)e^{ \minus{} g(x)}$, which is $ \minus{} \left(e^{ \minus{} f(x)}\right)^\prime \equal{} \minus{} \left(e^{ \minus{} g(x)}\right)^\prime$
The difference $ k \equal{} e^{ \minus{} f} \minus{} e^{ \minus{} g}$ is constant (assuming the domain is an interval). Since we don't know both of them at any point, we can't say much about this difference. From $ e^{ \minus{} f(0)} \equal{} e^{ \minus{} 1}$ and the fact that exponentials are nonnegative, we get $ k < e^{ \minus{} 1}$. From $ e^{ \minus{} g(N)} \equal{} e^{ \minus{} 1}$* we get that $ k > \minus{} e^{ \minus{} 1}$. Now $ f(N) \equal{} \minus{} \ln(e^{ \minus{} g(N)} \plus{} K)$ and $ e^{ \minus{} g(N)} \plus{} K$ is anything in $ (0,2e^{ \minus{} 1})$. That makes $ f(N)$ any element of $ (1 \minus{} \ln 2,\infty)$.

*I don't like problems with old years in them when the exact year doesn't matter. Update them.[/hide]

[quote="jmerry"]From $ e^{ \minus{} f(0)} \equal{} e^{ \minus{} 1}$ and the fact that exponentials are nonnegative, we get $ k < e^{ \minus{} 1}$. [/quote]

Sorry but I don't follow this line of the argument although I've been thinking about it for a while :P Could you explain further


hi,umm..I've tried to solve this problem several times but have not been successful,can ya guys help?

 Squares are constructed above the X-axis on the join of
 (2*n-1- 2,0),(2*n -2,0) for all natural no.s n.Is there a line which bisects every such square(ie.bisects the area of every such square)?if so,Find it's equation.

Hi
I cant understand :
Do you mean (2(n-1)-2 , 0) and (2n-2 , 0)
Or (2^(n-1)-2 , 0) and (2^n-2 , 0)
or sth else...
In the first case obviously the answer is "y=1" but...?
Good Luck

if a line goes through the midpoint of a square, it biscects the area, can you see why?


geometry
3D geometry
pyramid
Pascal\u0027s Triangle

So here's the big question that we need suggestions for in about 10 days or less.

What should we do during the opening ceremonies that would be crazy and fun?
Every team gets introduced and gets to walk out in front of everyone (briefly, for there are almost 100 teams).  It's seen as impressive and cool if you can do something interesting in those 15 seconds or so.

We're looking for suggestions for what we can do.  Singing, etc is out of the question because we don't have enough time.
According to David, the Japanese team members ran onto the stage and started sword-fighting at last year's IMO.

Any ideas?

[quote="lightrhee"][quote="Elyot"]
Singing, etc is out of the question because we don't have enough time.
[/quote]
Why not? The screwed-up version of 'O Canada' is less than 50 seconds. We can do this if we really want to(I doubt that Yufei would though).
I need to practice the French part. When I say 'mergez', it sounds like '[b]MAO[/b]je'.  :rotfl:[/quote]

Heh, you can't even type it right!  Mangez....you add that asian accent to everything!!!!!

Aww if you don't all sing that ruins it...too bad you didn't have enough time to do Canada is really big...

[quote="Sunny"][quote="Elyot"]There's no way we're gonna have enough time to do that... and it's MANGEZ!  The best pronounciation you could do without proper French nasal sounds is maonzhe.  (Zh as the s in pleasure)

I would love to get up there and spin some glowsticks or something.  Maybe we can buy some... I'm sure I could convince Felix and Dorette.[/quote]

Heh...just you though?  TEAM Canada...not Team Elyot...lol.  Still, extremely cool, but what about the other guys?  Give em something else to do....I wanna learn how to spin glowsticks![/quote]

Well I could have been teaching you at seminar... but no no no... you wanted to CLIMB TREES instead!!
A good site to learn basic moves is [url=http://www.homeofpoi.com]www.homeofpoi.com[/url].  Make yourself a pair of sock poi and try some of the basic moves.  Getting to weave and windmill gives you enough to impress people.

[quote="Elyot"]
Well I could have been teaching you at seminar... but no no no... you wanted to CLIMB TREES instead!!
[/quote]

You've seen the trees in Alberta!  Understand why I'm so thrilled to find a tree I can actually climb?!?!?!?!

[quote="Elyot"]There's no way we're gonna have enough time to do that... and it's MANGEZ!  The best pronounciation you could do without proper French nasal sounds is maonzhe.  (Zh as the s in pleasure)

I would love to get up there and spin some glowsticks or something.  Maybe we can buy some... I'm sure I could convince Felix and Dorette.[/quote]
This is sort of mean. You want to do cool things while we stand behind you and watch?
[hide="Canadian team"]:mad:  :)  :mad:  :mad:  :mad:  :mad: [/hide]

[hide="opening ceramony"]:mad:  :mad:  :mad:  :mad:  :mad:
[color=white]____ [/color]:bruce:  [/hide]

We need to do this as a team. :D

I've already told David this... but his post is damn funny.
:D

[quote="Sunny"]Hmm...sword fighting is hard to beat....maybe you can just pick David up and toss him around.  Or Lin, lol.[/quote]
Let's toss David around... he's reading over my shoulder right now and giggling like a little school girl.

Now's he's bursted into complete laughter.  We'll see how long we can keep this up for.
30 seconds...
reminds me of that time we were videotaping before the Piccolo attack and he couldn't stop laughing.

[quote="Elyot"]
Let's toss David around... he's reading over my shoulder right now and giggling like a little school girl.

Now's he's bursted into complete laughter.  We'll see how long we can keep this up for.
30 seconds...
reminds me of that time we were videotaping before the Piccolo attack and he couldn't stop laughing.[/quote]

Hahaha!  Hi David....don't laugh, your face will crack.

[quote="Elyot"]I've already told David this... but his post is - funny.
:D

[quote="Sunny"]Hmm...sword fighting is hard to beat....maybe you can just pick David up and toss him around.  Or Lin, lol.[/quote]
Let's toss David around... he's reading over my shoulder right now and giggling like a little school girl.

Now's he's bursted into complete laughter.  We'll see how long we can keep this up for.
30 seconds...
reminds me of that time we were videotaping before the Piccolo attack and he couldn't stop laughing.[/quote]
 :rotfl:
 :rotfl:
 :rotfl:
 :rotfl:
 :rotfl:
 :rotfl:
 :rotfl:
 :rotfl:

Oh. My stomach hurts. I shouldn't do this often.

[quote="d.lam.86"]Ahhh! Everyone's back and posting so quickly. It's taking me forever to read them. Anyway maybe you guys can fling tennis balls at each other with tennis racquets. LMAO! No—don't—I was just kidding!  :P Find something that everyone can learn and do in a reasonable time limit. GOOD LUCK. LMAO!  :D A friend of mine and I laugh at the most ridiculous things! Once we laughed during Jurassic Park—JURASSIC PARK! Just recently, we were laughing when Anakin turned over to the DARK SIDE and when he started burning.  :ninja:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:[/quote]

Anakin turns to the dark side?  WTF?  OMG!!!  You totally ruined it for me!!!  :'(

Sorry for ruining Star Wars III. I wasn't expecting people to NOT have watched it yet. I'm not a big fan so I didn't really like it and on that day, I got ripped off. I was in an unhappy mood anyway so that probably affected my perception of the movie quality. As to the Pascal's Triangle idea, it's brilliant but I doubt it's really safe. I expect everyone is well over 100 pounds so I doubt the idea will have any success. *hee* Once again, just my opinion which is probably not valued.  :agent:  I'm out of ideas.  :wallbash:  On another note, my last math question for a high school test that I did was about Pascal's Triangle. I didn't really know much about it so I just figured it out slightly with the example that the test provided. The next day, I found out my teacher had a complete book about Pascal's Triangle and it was quick interesting. I doubt my classmates were pleased. She insisted on giving homework about Pascal's Triangle.  :diablo:

[quote="Sunny"]Heh, David would giggle his entire way to the top of the pyramid.[/quote]
No doubt—LOL!  :rotfl: I wonder if someone can actually laugh like the emoticon.  :rotfl:

[quote="TheDreamer"][quote="Elyot"][quote="TheDreamer"]If you weren't being sarcastic, then haha, I haven't seen it either :(  :blush:

EDIT:

Maybe you guys could make a human pyramid depicting pascal's triangle

Hmm, prolly elyot Richard and Yufei on the bottom and david rhee on the top

___ :mad:
_ :mad:  :mad:
 :mad:  :)  :mad:[/quote]

That is an excellent idea... but it's something we'd need to practice for sure... David can't even open curtains properly!!  I don't know how we're going to be able to train him to get into a pyramid in < 20 seconds.  I think with enough training, we could probably pull that off.

It would be symmetrical too.[/quote]

David can't open curtains!?  Lol, is it because they were boobytrapped like at seminar?   :roll:[/quote]
Boobytrapped curtains... weird...  :ninja:

[quote="Elyot"][quote="Sunny"]Heh, no, waterboy and blinds boy were just incompetant.  Otherwise you'd see most of the video.[/quote]

Yeah... that kinda sucked...[/quote]
I did my best. He was nearly looked like waking up, so I got scared and couldn't finish it properly. :(

Why don't we pick up David and do something to HIM at opening ceremonies...?

[quote="Sunny"]Hahaha...I could so see it...David giggling the entire time as the other team members proceed to throw rubiks cubes at him...[/quote]
…solving all of them in record time.  ;)

[quote="Sunny"]Hahaha...I could so see it...David giggling the entire time as the other team members proceed to throw rubiks cubes at him...[/quote]
 :?

Sorry for this double post, but we only have one more day of training, and I really think we should do something at opening ceremony.
Yufei won't do anything stupid, things like singing and human pyramid.

So, what should we do?? Note again that we only have one more day of training :( .

[quote="TheDreamer"][quote="G-UNIT"]

Suggestion 2:

  Adapted from "How we do".

  Yes, yes singing is out of the question.  But I dont consider rapping the same as "singing".

  To begin, one of you must play the role of the boom-box (basically you must place your hand over your mouth and make loud, distorted sounds in a rhythimic flow.  Then the rest of you start "rapping":

                                                    This is how we do:

                We make a proof and solve a conjecture while we up in the club

                                                    This is how we do:

     Nobody solves math problems like the Canadian IMO math team so show us some love.

   Fressshhhh like Polya, impala Poincare

Crone hydraulics, fade away partitionssss

You don't want none,

amuteur better, run

when the beef is on, ill dis-prove that conjecture

  I would continue, but at this point ur 15 seconds or so would be up.

 [/quote]
Haha for our math club we tried making a rap to get more people into it, but the teacher thought it would corrupt the math club and bring in a bunch of ppl who thought it was a gang ^_^[/quote]

WEIRD × 2...  :ninja:

Aww.... come on Yufei.....spoil sport!

Guys, you gotta teach Yufei to have some fun every once in a while! Hehehe

[quote="lightrhee"][quote="G-UNIT"]I have three suggestions for what the Canadian IMO team should do at the opening ceremonies.

Suggestion 1:

  Bring 5 super-soakers filled with maple syrup.  When the anouncers call you guys up to the stage, you guys get all gung-ho and spray each other like mad.  Then when your super soakers runs out of maple syrup you guys roll around on the Canadian flag so it sticks to your syrup-covered body.  Finally, you guys put on aunthentic beaver coats ( made from beaver fur) and leave the stage in style.  I think this suggestion is the most patriotic and shows real Canadian pride.

Suggestion 2:

  Adapted from "How we do".

  Yes, yes singing is out of the question.  But I dont consider rapping the same as "singing".

  To begin, one of you must play the role of the boom-box (basically you must place your hand over your mouth and make loud, distorted sounds in a rhythimic flow.  Then the rest of you start "rapping":

                                                    This is how we do:

                We make a proof and solve a conjecture while we up in the club

                                                    This is how we do:

     Nobody solves math problems like the Canadian IMO math team so show us some love.

   Fressshhhh like Polya, impala Poincare

Crone hydraulics, fade away partitionssss

You don't want none,

amuteur better, run

when the beef is on, ill dis-prove that conjecture

  I would continue, but at this point ur 15 seconds or so would be up.

Suggestion 3:

  The only thing sensible and can actually be done on the stage within 15 second is simply, have the team jog around the stage waving the Canadian flag in the air.  And perhaps a few of you can execute some "pump-up-the-crowd" actions or something along those lines.  However, I also strongly recceomend suggestions 1 and 2,  :D  good luck!, and make us proud![/quote]
......

That was weird. :D[/quote]
Has the Olympiad started (since you're online right now)?

Why don't you ask Yufei what his ideas are?  I'm sure he's got something crazy up his sleeve :)

Heh...good luck to the IMO team once again.  You guys shoulda stayed in Calgary longer....oh well.

Yufei thinks they should just walk on stage like most of them are, lol.  But they have a team mascot which I tried to steal yesterday and today. Its this stuffed moose with a canadian flag tied around his neck....I want it!  Its really cute.


logarithms
inequalities
inequalities unsolved

Prove that \[ \log_ab\plus{}\log_bc\plus{}\log_ca\le \log_ba\plus{}\log_cb\plus{}\log_ac\] for all $ 1<a\le b\le c$.

[quote="April"]Prove that
\[ \log_ab \plus{} \log_bc \plus{} \log_ca\le \log_ba \plus{} \log_cb \plus{} \log_ac
\]
for all $ 1 < a\le b\le c$.[/quote]

$ \sum{log_ab}\le \sum{log_ac}$
$ \sum{\frac {lnb}{lna}}\le \sum{\frac {lnc}{lna}}$
$ \sum{(lnb)^2lnc}\le\sum{(lnc)^2lnb}$
$ (lna \minus{} lnb)(lnb \minus{} lnc)(lnc \minus{} lna)\ge 0$ last is true because $ 1 < a\le b\le c$

If we take $ log_{a}b$ = x and $ log_{b}c$ = y The inequality becomes x + y + 1/xy $ \le$ xy + 1/x + 1/y. And this is an obvious inequality since it equals to 0 $ \le$ (xy-1)(x-1)(y-1). (note that x and y are both greater than or equal to 1.

[quote="April"]Prove that
\[ \log_ab \plus{} \log_bc \plus{} \log_ca\le \log_ba \plus{} \log_cb \plus{} \log_ac\]
for all $ 1 < a\le b\le c$.[/quote]

$ \log_ba \plus{} \log_cb \plus{} \log_ac \minus{} (\log_ab \plus{} \log_bc \plus{} \log_ca)$

$ \equal{} \frac {\ln{\frac {c}{a}}\ln{\frac {b}{a}}\ln{\frac {c}{b}}}{\ln{a}\ln{b}\ln{c}}$

$ \ge 0$

Substitue $x=\ln a\wedge y=\ln b\wedge z=\ln c\implies 0<x\le y\le z$ Given inequality  is equivalent to $\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\ge \frac{x}{z}+\frac{y}{x}+\frac{z}{y}\iff (z-x)(z-y)(y-z)\ge 0$. Equality iff at least two of $x,y,z$ are equal.


modular arithmetic
LaTeX
number theory unsolved
number theory

Ok, my question is:

Show that if ab == 1 mod m, then
ord(m)a=ord(m)b

(Note that == means congruent) and ord(m)a means the order of a mod m

I know that if a^k==1 mod m, then the ord(m)a is the smallest integer k such that the congruence holds. For example,
ord(10)7=4 since 7^4==1 mod m.

I'm thinking, a^(ord(m)a)==1 mod m so maybe ab=a^ord(m)a and also:
b^(ord(m)b)==1 mod m so ab=b^ord(m)a

Am i on the right track? Thanks!

I think, it's better to do so. \[\forall k\in\mathbb{N}\qquad a^{k}b^{k}\equiv1 \pmod m,\] thus \[a^{k}\equiv1\qquad\Leftrightarrow\qquad b^{k}\equiv1.\] Don't you think so?

RIP, could you please tell me what the line that starts with the upside down A means? We haven't used that notation in my class....thanks!

The $\LaTeX$ code is \forall, and this is the meaning.

ok RIP I understand that
if ab == 1 mod m then (a^k)(b^k)==1 mod m

Then, in your next line it says:

a^k==1 <--> b^k==1

So, a^k and b^k are both congruent to 1 (mod something)...


[color=darkblue]Not use logarit, compare $ 2008^{2009}$ and $ 2009^{2008}$ ?[/color]

hello, it is known that for $ x\geq3$ is $ \left(1\plus{}\frac{1}{x}\right)^x<x$ .This is equivalent with $ (x\plus{}1)^x<x^{x\plus{}1}$. Let $ x\equal{}2008$ so we have $ 2009^{2008}<2008^{2009}$.
Sonnhard.

[color=darkblue]Thanh you! Are you use illation to prove clause: If $ x \ge 3 \Longrightarrow \left ( 1\plus{}\frac{1}{x} \right )^x<x$. So $ x\in N$.[/color]


geometry
geometric transformation
reflection
linear algebra
linear algebra unsolved

find all linear maps f from S to S where S is set of all n*n matrices with complex entries s.t.
     f(I)=I ,f(A^k)=f^k(A),for some integer k >1.

Hmmm, it is not even the second time I kindly ask you not to post problems from the very last issue of Mathematical Reflections, but you just seem to ignore my messages. This is not very fair, you know?


The ques. is just like that.
The villages are all located exactly the same with the figure. :?

If that is the case the best I can say is there in the attached document :lol: Hopefully, I haven't made any errors in calculations :oops:

And obviously, if you want to minimize the bridge length, just build a bridge across the river at one point, and roads.


geometry
rectangle
ratio
floor function
combinatorics open
combinatorics

I was working on the following problem today: "Prove that you can't place 100 points in a 13*18 rectangle, so that the distance between 2 points is at least 2"
(it's not hard but it's not my exact question: [hide](it can be improved to 96). I just "circled" each point with a circle of radius 1, and none of those should intersect, and we have 96 circles (area - $ 96\pi$) in a (13+2)*(18+2) (area- $ 300$) rectangle, with yields a contradiction.
[/hide])
My question is this - how good is this bound? What is the maximum number of points that can be placed in a X*Y rectangle, so that the distance between any 2 is at least d? (we can actually fix the d because it is sort of homogeneous). I think that in the mentioned example, 95 can't be obtained, and I hope I'm not wrong.

The largest fraction of the area of any rectangular subset of $ \mathbb{R}^2$ that can be covered by circles of any size is given by the so-called [i]hexagonal packing[/i], for which the ratio of covered area to total area is $ \frac {\pi}{2\sqrt {3}}$.

So for a rectangle with sides $ X,Y$, you can fit in no more than
\[ \left \lfloor \frac {2XY}{d^2\sqrt {3}}\right \rfloor
\]
circles of radius $ \frac {d}{2}$.

Thus, for the particular problem you are concerned with, you can fit in no more than
\[ \left \lfloor \frac {2\left(X \plus{} d\right)\left(Y \plus{} d\right)}{d^2\sqrt {3}}\right \rfloor
\]
points such that no two are at a distance of less than $ d$.  In your case this gives the upper bound $ 86$; in practice, it is likely quite a bit lower than that, since the above bounds become fairly tight only as the dimensions of the rectangle become large or as the radius of the circles becomes small.

For some more general discussion, try: http://mathworld.wolfram.com/CirclePacking.html


Is it still ok if I send in the entry form the day after the due date? (But my solutions were sent on October 10).


Let $x,y,z>0$ such that $xyz=x+y+z+2$.Prove that:

\[\sqrt{x}+\sqrt{y}+\sqrt{z}\geq 2(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{z}})\]

Put $ \frac{1}{\sqrt{yz}}= cosA $
 We need to prove : $ cosA+cosB+cosC \geq 2(cosAcosB+cosBcosC+cosCcosA) $
 It's enough to prove : $ cosA +cosB+cosC \geq \frac{2}{3} (cosA+cosB+cosC)^2 \iff \frac{3}{2} \geq cosA+cosB+cosC $ , which is true


Let $f: \mathbb R \to \mathbb R$ be a function such that
\[f(x^2 \plus{} x \plus{} 3) \plus{} 2f(x^2 \minus{} 3x \plus{} 5) \equal{} 6x^2 \minus{} 10x \plus{} 17\]
holds for all real $x$. Find $ f(85)$.

[quote="Rust"]$ f(x) \equal{} 2x \minus{} 3$.[/quote]
Are you sure?Note that $ f(x)$ is not a polynomial.
Let $ g(x)\equal{}f(x)\minus{}2x\plus{}\frac{9}{2}$
Then
$ g(x^2\plus{}x\plus{}3)\plus{}2g(x^2\minus{}3x\plus{}5)\equal{}0$
From this equation we can file some value of $ g(x)$ but can not find $ g(x)$ on R because $ h(x)\equal{}x^2\plus{}x\plus{}3$ is not inject in R.

[quote="Rust"]Let $ t(x) \equal{} f(x^2 \minus{} 3x \plus{} 5) \minus{} 2(x^2 \minus{} 3x \plus{} 5) \plus{} 3$, then $ f(x^2 \plus{} x \plus{} 3) \minus{} 2(x^2 \plus{} x \plus{} 3) \plus{} 3 \equal{} t(x \plus{} 2)$ and $ t(x \plus{} 2) \plus{} 2t(x) \equal{} 0,t(3 \minus{} x) \equal{} t(x)$.
Therefore $ f(n \plus{} 0.5) \equiv 0 \ \forall n\in Z$.
For prove $ t(x) \equiv 0$ we need continiosly f.[/quote]

The job is nearly finished :

You have  $ t(3\minus{}x)\equal{}t(x)$ and so $ t(x\plus{}2)\equal{}t(1\minus{}x)$
So $ t(x\plus{}2)\plus{}2t(x)\equal{}0$ becomes $ t(1\minus{}x)\equal{}\minus{}2t(x)$ and so $ t(x)\equal{}t(1\minus{}(1\minus{}x))\equal{}\minus{}2t(1\minus{}x)\equal{}4t(x)$ and so $ t(x)\equal{}0$ $ \forall x$

And so  $ f(x^2 \minus{} 3x \plus{} 5) \equal{} 2(x^2 \minus{} 3x \plus{} 5) \minus{} 3$
And so $ f(85)\equal{}167$

$ f(x)$ is not equalities $ 2x\minus{}3$ for all x.
Rust have a mistake with his solution.
$ x^2\minus{}3x\plus{}5$ not take all value on R.
We can file out an solution of this equation as follow :
$ f(x)\equal{}c,\forall x\in (\minus{}\infty,0)$ and $ f(x)\equal{}0$ on other interval.

[quote="TTsphn"]$ f(x)$ is not equalities $ 2x \minus{} 3$ for all x.
Rust have a mistake with his solution.
$ x^2 \minus{} 3x \plus{} 5$ not take all value on R.
We can file out an solution of this equation as follow :
$ f(x) \equal{} c,\forall x\in ( \minus{} \infty,0)$ and $ f(x) \equal{} 0$ on other interval.[/quote]

Rust gave a second more precise (and, according to me, correct) demo :

Let $ t(x) \equal{} f(x^2 \minus{} 3x \plus{} 5) \minus{} 2(x^2 \minus{} 3x \plus{} 5) \plus{} 3$
Then $ t(x \plus{} 2) \equal{} f(x^2 \plus{} x \plus{} 3) \minus{} 2(x^2 \plus{} x \plus{} 3) \plus{} 3$

So we have  $ f(x^2 \minus{} 3x \plus{} 5) \equal{} t(x) \plus{} 2(x^2 \minus{} 3x \plus{} 5) \minus{} 3$  and $ f(x^2 \plus{} x \plus{} 3) \equal{} t(x \plus{} 2) \plus{} 2(x^2 \plus{} x \plus{} 3) \minus{} 3$

And so $ 6x^2 \minus{} 10x \plus{} 17 \equal{} f(x^2 \plus{} x \plus{} 3) \plus{} 2f(x^2 \minus{} 3x \plus{} 5)$ $ \equal{} t(x \plus{} 2) \plus{} 2(x^2 \plus{} x \plus{} 3) \minus{} 3 \plus{} 2(t(x) \plus{} 2(x^2 \minus{} 3x \plus{} 5) \minus{} 3)$ $ \equal{} t(x \plus{} 2) \plus{} 2t(x) \plus{} 6x^2 \minus{} 10x \plus{} 17$

And so $ t(x \plus{} 2) \equal{} \minus{} 2t(x)$
We also have $ t(3 \minus{} x) \equal{} f((3 \minus{} x)^2 \minus{} 3(3 \minus{} x) \plus{} 5) \minus{} 2((3 \minus{} x)^2 \minus{} 3(3 \minus{} x) \plus{} 5) \plus{} 3$ $ \equal{} f(x^2 \minus{} 3x \plus{} 5) \minus{} 2(x^2 \minus{} 3x \plus{} 5) \plus{} 3$ $ \equal{} t(x)$

So $ t(3 \minus{} x) \equal{} t(x)$ and so $ t(x \plus{} 2) \equal{} t(1 \minus{} x)$
So $ t(x \plus{} 2) \equal{} \minus{} 2t(x)$ becomes $ t(1 \minus{} x) \equal{} \minus{} 2t(x)$ and so $ t(x) \equal{} t(1 \minus{} (1 \minus{} x)) \equal{} \minus{} 2t(1 \minus{} x) \equal{} 4t(x)$ and so $ t(x) \equal{} 0$ $ \forall x$

And so  $ f(x^2 \minus{} 3x \plus{} 5) \equal{} 2(x^2 \minus{} 3x \plus{} 5) \minus{} 3$
And so $ f(x) \equal{} 2x \minus{} 3$ $ \forall x\in[\frac {11}{4}, \plus{} \infty)$

And so $ f(85) \equal{} 167$

you can find find $ f(85$) using this method , but in general cases it isn't true , for example .
let$ f$be function defined from $ R$ to $ R$ such that $ (f(x))^2\equal{} f(2x)$. find $ f(0)$.
here there is two value $ 0$ and $ 1$ if $ f(x)\equal{}0$ for every $ x$ then $ f(0)\equal{}0$ but if $ f(x)\equal{}e^x$ then $ f(0)\equal{}1$.

[quote="FOURRIER"]Why don't we do it like this :

Lets look for a simple function $ f$ wich satisfies the conditions,

Let $ f(x) \equal{} ax \plus{} b$

We can find easily $ f(x) \equal{} 2x \minus{} 3$

and we deduce

Is this method true?

I think this is Rust's Method[/quote]

In fact, just by saying $ f(x)\equal{}2x\minus{}3$ is a solution, you can find [b][u]a[/u][/b] value for $ f(85)$.
The problem is that maybe different solutions exist leading to different values of $ f(85)$

So :
Either you consider that the problem says in an implicit way that there is a unique solution for $ f(85)$ and then the solution is trivial and you are right.
Either you consider that the problem does not say such a thing and you must show that there is a unique value $ 167$

And RUST used the first methoid in its first post
and nearly the second in its second post.

Let $ g(x)\equal{}x^2\minus{}3x\plus{}5\equal{}g(3\minus{}x)$, then $ x^2\plus{}x\plus{}3\equal{}g(x\plus{}2)$, therefore $ t(x)\equal{}f(g(x))\minus{}2g(x)\plus{}3$ satisfyed
$ t(x\plus{}2)\plus{}2t(x)\equal{}0,t(3\minus{}x)\equal{}t(x).$

[quote="FOURRIER"][quote="Rust"]Let $ t(x) \equal{} f(x^2 \minus{} 3x \plus{} 5) \minus{} 2(x^2 \minus{} 3x \plus{} 5) \plus{} 3$[/quote]

[quote="pco"]
Let $ t(x) \equal{} f(x^2 \minus{} 3x \plus{} 5) \minus{} 2(x^2 \minus{} 3x \plus{} 5) \plus{} 3$
[/quote]

How do you think of this?

Thanks  :)[/quote]

What do you mean ?

The phrase before, in my post (and that you carefully omitted) was :
"Rust gave a second more precise (and, according to me, correct) demo : "

I was just explaining completely Rust's demo (and my own end of Rust's demo)
So I used Rusts's phrases.

Is there any problem with this ?

And have you read all the posts ?

[quote="pco"]
What do you mean ?
[/quote]

I mean where the Idea of considering $ t(x) \equal{} f(x^2 \minus{} 3x \plus{} 5) \minus{} 2(x^2 \minus{} 3x \plus{} 5) \plus{} 3$ came from , and Rust answered me  but if you have somthn to add thank you  :lol: