What is the value of one minus nine over ten, OVER, one minus 99 over 100?

What is 1.2727272727... expressed as a fraction?

If a shpere's hieght is doubled its radius is halved, what is the ratio of the old volume to the new volume?

A bean of paper wieghs 0.35 cubic grams. If a ream of paper has a box which measures  20 15 10, how much does it wieght?

[quote="isabella2296"][hide="Number 1"]

This is:

$ \frac {1 \minus{} \tfrac{9}{10}}{1 \minus{} \tfrac{99}{100}} \equal{} \frac {\tfrac{1}{10}}{\tfrac{1}{100}} \equal{} \frac {1}{10} \times 100 \equal{} \boxed{10}$

[/hide]

[hide="Number 2"]

Ignore the 1 for now. We have

$ 0.2727272727... \equal{} x$

$ 27.2727272727... \equal{} 100x$

$ 27 \equal{} 99x$

$ \frac {27}{99} \equal{} x$

Now tack the one on to get $ 1 \frac {27}{99}$.

[/hide][/quote]

thanks! can you do the volume one too please? :]

You're welcome.

I'm not sure what PiLuvah means by "height" though. Diameter?

If you want to express any periodic number into a fraction follow this way:

Let be  x.rrrrrrrrr......
with x-integer and rrrrrr.....-period

x.rrrrrr......=x+0.rrrrrrr       we can express 0.rrrrrrr  as  r/9  (why? Because for any intger 0>r<9 we have $ \frac{r}{9}\equal{}0.rrrr...$) x.rrrrr....=x+0.rrrr....=x+r/9=$ \frac{9x\plus{}r}{9}$

[quote="Ttocs45"]Also, if it is 0.ababababab..., then the fraction is ab/99.
If it is 0.abcabcabcabcabc..., then the fraction is abc/999.
If it is 0.abcdabcdabcdabcdabcd..., then the fraction is abcd/9999.[/quote]

OH wait im sorry i meant the shape with the volume= 1/3 pi r² h

Are you talking about a cone?
If you are, than it would be 1/2 of the old volume, because the length and width are both halved and the height is doubled.

Yay post 42!

If you have a cone with radius $ 1$ and height $ 1$, the volume will be $ \frac{\pi}{3}$.

Halving the radius and doubling the height causes the area to be $ \frac{\pi \frac{1}{2}^2 \times 2}{3}$, or $ \frac{\pi}{6}$.

Thus, the ratio is clearly $ \boxed{\frac12}$.

And for your fractional one, the simplified answer is $ \boxed{\frac{14}{11}}$.

[quote="ernie"]If you have a cone with radius $ 1$ and height $ 1$, the volume will be $ \frac {\pi}{3}$.

Halving the radius and doubling the height causes the area to be $ \frac {\pi \frac {1}{2}^2 \times 2}{3}$, or $ \frac {\pi}{6}$.

Thus, the ratio is clearly $ \boxed{\frac12}$.

And for your fractional one, the simplified answer is $ \boxed{\frac {14}{11}}$.[/quote]

I think by 1.27272727 and so fourth, he meant 1.27 bar, and 14/11= 1.272727273.

[quote="mewto55555"]No it doesnt. Maybe it rounds on a calc, but x/11 (where x<11) is .9x repeating[/quote]

Ohhh okay. :o

To avoid confusion in the future, use the code

[code]0.\overline{9x}[/code]

to produce

$ 0.\overline{9x}$.


I would like to know of any sources that show a relation between Pathagorean Triplets, Square Triangular numbers and Pell-like sequences.
My main concern is to find anything relating to how Pell-like sequences lead to solutions of the paired equations:
$a^{2} + b^{2} = c^{2}$
 and
 $\mid a - b \mid = C \mid {a,b,c,C} \in I^{+}$

Thanks, I do not have access to searching the math publications. I am confident that this has been published somewhere and would like to know where.

Why reposting it¿ http://www.mathlinks.ro/Forum/viewtopic.php?t=81107


$ a$ ,$ b$ and $ c$ $ \in$ $ R$  such that $ a>0$

let $ P(x)\equal{} ax^{2}\plus{}bx\plus{}c$

we suppose $ M\equal{}b^{2}\minus{}4ac$ and $ b>\equal{} 1/(8a)$ .
prove that $ P(M) >\equal{} 0$.

If $ M\leqslant 0$, then $ P$ is always non-negative and there's nothing much to prove. If $ M > 0$, then observe

$ f(M) \equal{} 4a^2M^2 \plus{} (4ab \minus{} 1)M \plus{} b^2$

Since its discriminant is $ (4ab \minus{} 1)^2 \minus{} 16a^2b^2 \equal{} 1 \minus{} 8ab\leqslant 0$, we have

$ 4a^2M^2 \plus{} (4ab \minus{} 1)M \plus{} b^2\geqslant 0$

$ (2aM \plus{} b)^2\geqslant M$

$ 2aM \plus{} b\leqslant \minus{} \sqrt {M}\lor 2aM \plus{} b\geqslant\sqrt {M}$

$ M\leqslant { \minus{} b \minus{} \sqrt {M}\over 2a}\lor M\geqslant{ \minus{} b \plus{} \sqrt {M}\over 2a}$

Therefore $ M$ lies outside the interval of negativity of $ P(x)$, which means $ P(M)\geqslant 0$.


Firstly, which is the odd one out

28p , 36p , 48p , 67p , 85p

the 'p' is for pence which is UK money, I'm from the UK  :)

what is the next two numbers in this sequence

3 , 5 , 8 , 13 , 20 , 31 , __ ? , __ ?

[quote="HenrikLarsson"]

what is the next two numbers in this sequence

3 , 5 , 8 , 13 , 20 , 31 , __ ? , __ ?[/quote]

He he, I got this one in like 4 seconds  :D

Look at the difference,

2,3,5,7,11

Primes number!!

So you need to add 13 to the next one, and then 17 to the next one.

The one and only correct sequence is $3 ,\ 5 ,\ 8 ,\ 13 ,\ 20 , \31 ,\ 23,\ 42,\ 105,\ 481,\ 1337,\ 65537,\ \pi$

Oh, and this problem isn't math -> completely wrong forum ;)


vector
topology
real analysis
real analysis unsolved

Let $ X$ be a topological vector space that is of second Baire category with respect to itself. Show that if a closed convex subset of $ X$ has an internal point, then it has an interior point.

[hide="Hint"]Given an internal point $ y$ of the closed convex set $ C$ consider the family of (closed) subsets \[ X_n\equal{}\left\{x\in X| \forall t\in[0,1/n]: y\plus{}tx\in C\right\}\]for $ n>0$.[/hide]

The sets $ X_{n}$ are closed because of the continuity of addition and scalar multiplication?

Then $ X\equal{}\bigcup X_{n}$ since $ y$ is an internal point so some $ X_{n}$ has an interior point and it would be an interior point of $ X$?

[quote="chungyc"]The sets $ X_{n}$ are closed because of the continuity of addition and scalar multiplication?[/quote]... and because $ C$ is closed.
[quote="chungyc"]... it would be an interior point of $ X$?[/quote]No.

And a final question: Where do you really need the assumption that $ C$ is convex?

Some $ X_{n}$ has an interior point $ x$ so $ x \in O \subset X_{n}$ for some open set $ O$.
Now $ y\plus{}\frac{1}{n}O$ is open and $ y\plus{}\frac{1}{n}x \in y\plus{}\frac{1}{n}O \subset C$ so $ y\plus{}\frac{1}{n}x$ is an interior point of $ C$?

Correct. The claim was that $ C$ has non-void interior. This is shown now.

What about the final question?  :wink:

Nor can I.  :P But you it is sufficient in order to show that the internal point is contained in the interior of $ C$.


algorithm
percent
algebra proposed
algebra

A)  Show the steps in the square root algorithm for x^2 + 2x + 3
Extract the root out to and including the fourth term.  Then plug
in x = 10 into those four terms to come up with a value.

B)  Plug x = 10 into x^2 + 2x + 3 first.  Then take the square root
of that and compare the result to the final value you got in A).

General example:  Explanation follows it.

..........................................a ..    +..    b
....................................._________________
..................................\/.....a^2 + 2ab + b^2
.......................................- a^2
...........................b...........------
..................2a + ___.....................2ab + b^2
..............................................- ( 2ab + b^2)
.................................................------------
................................................................0

The square root of the 1st term here, a, is written atop the radical, is squared and written under a^2 and subtracted.  Then the next pair is brought down, 2ab + b^2.  Off to the left you're going to form a divisor.  You take everything you have  at the top, which happens to be the variable a, double it, and add a space.  You have to determine what number, variable(s), or combination of them go in the space (underscore).  Double the a to get 2a.  To find what goes in the space, divide the first term of the pair you brought down, the 2ab, by the 2a to get a quotient of b.  It is this b that simultaneously goes in the space and as the next term as part of the answer at the top, as in a + b.  Multiply the b back through by 2a + b created on the left and write them under the pair brought down.  Subtract and in this particular case the remainder is zero so the process stops.

Well especially for my good old troll: the problem is very EASY.

The only problem in this thread is to understand what your multiple accounts speak of...

ZetaX,

TROLL, or whatever that is you CHOOSE to call yourself as you hide behind your
anonymity,

you brought up "EASY" did you not?  I am against ANYONE advertising problems as such.
It doesn't make sense and is nonprofessional.  Many here cry "sour grapes" when they can't get problems.  Many problems' solutions are posted with too few steps as if we're supposed to read the poster's mind.

I believe, when appropriate in giving examples to make it clearer.  And I never resort to
insulting language such as "simple," "obvious, " "it is clear," "easy," "naturally follows,"
or similar words and phrases because I know they are so subjective.  I am mystified that
mathematicians would use them too.

Zetax, if you did not come on this thread to add to this challenge problem, then please
** take your trolling somewhere else.  **


x,y and z are in Z-{1}
such that:
(x+y=<xy)and(y+z=<xz)

prove that :x+z=<xz

$ (x\minus{}1)(y\minus{}1)\geqslant 1\land (y\minus{}1)(z\minus{}1)\geqslant 1$

1. $ y\geqslant 2\implies x\minus{}1\geqslant 1\land z\minus{}1\geqslant 1\implies (x\minus{}1)(z\minus{}1)\geqslant 1$

2. $ y\leqslant 0\implies x\minus{}1\leqslant \minus{}1\land z\minus{}1\leqslant \minus{}1\implies (x\minus{}1)(z\minus{}1)\geqslant 1$

Hm, someone rated my previous post as spam (1 diamond). I really can't see any mistake in the proof - can someone help?

you have  a mistake because  we  have y+z=<xz not  y+z=<yz

That's an obvious error in the statement, since it doesn't hold. Take for example $ x\equal{}\minus{}1, y\equal{}\minus{}5, z\equal{}2$. Then $ (\minus{}1)\plus{}(\minus{}5)\leqslant (\minus{}1)(\minus{}5)$ and $ (\minus{}5)\plus{}(2)\leqslant (\minus{}1)(2)$, but $ (\minus{}1)\plus{}(2)\not\leqslant (\minus{}1)(2)$

but we have  x , y and z in Z-{1}
it doesn 't have exeption :roll:

What are you talking about? $ \mathbb{Z}\setminus\{1\}$ means "any integer except $ 1$", and neither of my numbers is equal to $ 1$.

Perhaps you're just too tired... Or a spammer.


algebra
system of equations
algebra unsolved

Solve the system of equations:
$\{\begin{array}{l}(1+\frac{12}{3x+y}).\sqrt{x}=2\\(1-\frac{12}{3x+y}).\sqrt{y}=6\end{array}$

[quote="Rust"]
From (1) and (2) $a+b=\frac{1}{2}$.[/quote]
Are you sure? :P

[quote="cuopbientcd"]Let $z=\sqrt{y}+i\sqrt{3x}$ then I can solve $z$.That ia a solution.[/quote]
Can you post your solution?

a solution uses complex number!!!i want to read it,can you post your solution cuopbientcd??

$ \{\begin{array}{l}(1+\frac{12}{3x+y}).\sqrt{x}=2\\(1-\frac{12}{3x+y}).\sqrt{y}=6\end{array} $

LET $f(x,y)=(1+\frac{12}{3x+y})\sqrt{x}-2$ then $f(-x,-y)=i(1-\frac{12}{3x+y})\sqrt{x}-2$, or,$f(-x,-y)=6i*\sqrt{\frac{x}{y}}-2$ ,or,$f(x,y)=6i*\sqrt{\frac{x}{y}}-2=0$,then ,$i*\sqrt{\frac{x}{y}}=1/3\Rightarrow{9x+y=0}$
putting $9x+y=0$ in the given two eq. we get $x^2-8x+4=0,x^2+8x+4=0$ , comparing we get $x=0,y=0$ ,which does not satisfy the given two eq.

I don't know what is the solution really, but I think the given eq. has no solution.

[hide="My hint"][quote="hien"]Solve the system of equations:
$\{\begin{array}{c}{l}(1+\frac{12}{3x+y}).\sqrt{x}=2(1)\(1-\frac{12}{3x+y}).\sqrt{y}=6(2)\end{array}$[/quote]
we observe that $x>0$ and $y>0$. Hence, dividing $(1)$ by $\sqrt{x}$, dividing $(2)$ by $\sqrt{y}$, we have
$\{\begin{array}{c}{1}(1+\frac{12}{3x+y})=\frac{2}{\sqrt{x}}(3)\(1-\frac{12}{3x+y})=\frac{6}{\sqrt{y}}(4)\end{array}$
Plus $(3)$ and $(4)$, and subtract $(3)$ and $(4)$, we have a new system equation
$\{\begin{array}{c}{l}2=\frac{2}{\sqrt{x}}+\frac{6}{\sqrt{y}}(*)\\frac{24}{3x+y}=\frac{2}{\sqrt{x}}-\frac{6}{sqrt{y}}(**)\end{array}$

Multiplying $(*)$ and $(**)$, we have:
$\frac{48}{3x+y}=\frac{4}{x}-\frac{36}{y}$
This is equivalent to $27x^2-y^2+18xy=0$[/hide]

p/s: this kind of system equation first appeared in VMO in the 1990s

[quote="npqhuy"]we observe that $x>0$ and $y>0$[/quote]
the solution may be complex, then $x,y$ may be less than $0$.
So you hint gives $ 27x^2-y^2+18xy=0$ ,my solution gives there is no solution for $9x+y=0$ ,then only we get $y=3x$
Hence the solution is $(4+2\sqrt{3},12+6\sqrt{3})$

Hmmm. So everybody gets $\frac {1} {\sqrt{x}} + \frac {3} {\sqrt{y}} = 1 $, and with $1 + \frac {12} {3x+ y} = \frac {2} {\sqrt{x}}$ we can get $\sqrt{y}$ from the first and $y$ from the second. Equaling these, and denoting $t=\sqrt{x}$ I got (with possibly miscalculations) $3t^4 -12t^3 + 42t^2 - 48t + 12 = 0$, which has two positive real roots. The path is clear, anyways, so unless the solution has a hidden nice trigonometric form, what's the hype about it ...

[quote="npqhuy"][quote="MANMAID"]
the solution may be complex, then $x,y$ may be less than $0$.
So you hint gives $ 27x^2-y^2+18xy=0$ ,my solution gives there is no solution for $9x+y=0$ ,then only we get $y=3x$
Hence the solution is $(4+2\sqrt{3},12+6\sqrt{3})$[/quote]
First of all, this system equation is from a VMO quiz, and we just need real roots, not the complex ones. Secondly, I wrote my hint according to my memory, as I had solve this before. Therefore, maybe I have something wrong with my calculations. Please forgive me.
by the way, have you checked your solution?[/quote]

Thank you for this imformation, now I will not check any complex number solution :lol: in VMO problems.


What is the value of $ x$ in $ 222^x \minus{} 111^x \equal{} 111^x \cdot 7$?

Re-write the expression as $ 222^{x}\equal{}2^{x}(111^{x})\equal{}8\cdot 111^{x}$.  Since $ 111^{x}\neq 0\forall x$, we can divide it out to get $ 2^{x}\equal{}8$, so $ x\equal{}3$.

hello, we have $ 222^x\minus{}111^x\equal{}111^x\cdot7 \Leftrightarrow 111^x(2^x\minus{}1)\equal{}111^x\cdot7$, from here we get that it must be $ 2^x\equal{}8$, we have $ x\equal{}3$.
Sonnhard.


you are buying 15 doughnuts for your MATHCOUNTS team and coach.  If the doughnuts are packaged singly or in sets of 3 or 4, in how many different ways can you select your order?

[hide]16[/hide]

umm... i got 15.

I don't know a good way, and i think there is, but i just can't think of it...

anyways... i made a chart

4's     3's     1's

3         1         0
3         0         3

2         2         1
2         1         4
2         0         7

1         3         2
1         2         5
1         1         8
1         0         11

0         0         15
0         1         12
0         2         9
0         3         6
0         4         3
0         5         0

that's 15, what did u have that i didn't have?

yes, but what do you mean by your last sentence thing

perhaps he has a penchant for putting random numbers in spoiler. it is not your place to question the great neal.

oh wow, thanks for the late warning, and it wasn't #16 or anything. sorry neal.


[color=olive]Find all primes of the form 2^(2^n)+5, where n is a nonnegative integer[/color]

I've calculated that the only prime number is 7, that is n = 0, is it right?  If so, how can prove that n>0 are not primes?

Thanks very much!!

I have learnt about mod yet, so is there any other way to do this problem?

$2^{2^{n}}+5=(3-1)^{2^{n}}+3+2$, and when you develop using Newton's binomial formula, if $n \geq 1$, you notice that your expression is divisible by $3$.

could you explain that a little more, because i dont what newtons binomial theorem is.

http://mathworld.wolfram.com/BinomialTheorem.html

It's a little easier than that.  For $n > 0$ we have $2^{n}$ is even.  Therefore,

$2^{\text{even number}}+5 = \left( 2^{ \text{even number}}-1 \right)+6$

Since $2 | \text{even number}$ it follows that $2^{2}-1 | 2^{ \text{even number}}-1$ (easy lemma).  Thus $3$ divides both terms.


in a regular game, doubles are needed on two cubical dice to get out of the pit. in a revised game, doubles or triples are needed on three cubical dice to get out. how many times as likely is it for a player to get out of the pit on one toss under the new rules compared to the old rules?

can anyone explain it to me please because i dont understand how to do it.
thanks

Solutions in spoiler:

[hide]In the original situation you have a 1/6 chance of getting doubles.

In the second situation, there are two separate cases: getting doubles and getting triples.

For doubles: Let's assume that the first and second dice are equal and the third one doesn't match. The probability of that is 1/6*5/6= 5/36. (Whatever the first dice is, the second dice has a 1/6 chance of matching and the third dice has a 5/6 chance of not matching.) We have to multiply 5/36 by 3, because there are three situations: the first and second match, the first and third match, and the second and third match. 5/36*3= 15/36.

For triples: Whatever the first one is, the second and third ones each have a 1/6 chance of matching. 1/6*1/6 = 1/36.

The probability of your getting out of the pit is 16/36, which is 8/3 times larger than the original 1/6.

Alternately, to find the probability of getting out of the pit, you could try to find the probability of staying in. For this to happen, all three of the dice have to come out to different numbers. Whatever you roll on the first dice, the second one has a 5/6 chance of not matching, and the third one has a 4/6 chance of not matching the number rolled on either the first or second dice. The probability comes out to 4/6*5/6= 20/36. 1-20/36= 16/36, which is the probability that you escape the pit. [/hide]

I disagree, try to count the total possible outcome then count the total possible ways in which you could get a double, then triple. I got a different answer that way...

I can not quite figure out what's wrong with your logic...intriguing...

the answer is that its 8/3 times as likely to be able to come out of the pit the new way.

Solution by counting:

[hide]For the original toss: 6 sets of doubles work (1,1) (2,2) (3,3) (4,4) (5,5) (6,6) out of 36 possible sets of doubles (6*6). The probability of getting out of the pit is 1/6.

For the toss with three dice: six sets of triples. There are 90 sets of doubles. Explanation: Let's say the person rolls a 1,1,2 in some order. There are three ways for the person to do this: 1,1,2; 1,2,1; 2,1,1. Multiply 3 by 5; the number that does not match the 1s can be 2,3,4,5 or 6. 15*6 (since there are 6 possible numbers that can be doubles) = 90.  90+ the 6 sets of triplets = 96. 96/216= 16/36, which is 8/3 times greater than the original probability. [/hide]

Perhaps you're overcounting by repeating the triplet sets in your count of doubles?

Whoops :oops:
You'r completely right, I counted 18 extra cases...


Yes, it's an other way to say geometric mean.

[hide]

log_512(7) = log(7) / 9* log(2), ...

1/3 log2(7) [/hide]


Inscribed circle of triangle ABC touches the sides AB, BC, AC at points M, N, K respectively. The lines MN and AC intersect at point E. I is the centre of the inscribed circle. Prove that BK and IE are perpendicular to each other.

can you explain what 'polar and WRT incircle is' please it's very important


Two angles of an isosceles triangle measure $ 70^\circ$ and $ x^\circ$. What is the sum of the three possible values of $ x$?

$ \textbf{(A)}\   95    \qquad
\textbf{(B)}\    125   \qquad
\textbf{(C)}\    140   \qquad
\textbf{(D)}\    165   \qquad
\textbf{(E)}\     180$

The possible values of x are 70 (70-70-40), 40 (70-70-40), and 55 (70-55-55). Therefore, the sum of all possible values of x is 165.

May I please ask what a 70-70-70 triangle is?  :wink:

[quote="PhireKaLk6781"]May I please ask what a 70-70-70 triangle is? :Wink:[/quote]

It can be drawn on a sphere.

Anyway, I do not see where the solution poster mentioned a 70-70-70 triangle...

I believe in this question .....
Three probabilities are determined.  students can be confused in the first case and the third case because it just look similar when people just look with the angle that measure
one is (70,70=x,40) and (70, 55=x, 55) and (70, 70, 40=x) --> this thing means x can be a one of isosceles triangle's two same angles or  70 must be angle measured as other things  or the angle which two same angle exist and other angle that left


Is there anywhere I can find National mathcounts stuff online? sprints would be especially good.

http://freewebs.com/tzpublicsite
Also use the search function at the top, question has been asked many times (including by me :P) so you'll probably find more resources in old topics.


I was talking to a first-year student in University of Waterloo.  He commented that the "algebra" I am learning right now in high-school is not algebra.  it is merely "arithmetic".  He considers Galois theory to be real algebra.  Considering he isnt really a fan of math (he is an engineering student), this bothers me a little (actually it bothered me VERY MUCH).  What do you guys think?

Anyone who is trying to bug you by setting up his own personal definition of "algebra" is just engaging in logomachy and is a sophist (said tokenadult, just to throw some other words around). The kind of thing to which the term "algebra" was first applied, the content of the book by Muhammad ibn Musa al-Khwarizmi, is the kind of thing east Asian kids study in junior high and American kids study in senior high: good-old school algebra. There shouldn't be any quibbling about that.

An undergraduate mathematics major will study what is variously called "modern algebra" or "abstact algebra" and will learn about many generalizations of algebraic structures. I suppose the engineering student you met has heard about such ideas. I have seen Usenet posts by mathematicians who say that kids ought to learn abstract algebra BEFORE they learn conventional high school algebra, but I'm not following that advice. Joseph Gallian's book about abstract algebra, which I can get from my friendly public library, is a good introduction to what abstract algebra is all about. It's indispensable for tying together certain aspects of higher mathematics, and is very cool for those who like it, but its elegance is not a reason to disparage secondary school algebra.

I guess this just proves that advanced mathematical education doesn't keep a person from being a twerp.

[quote="tokenadult"]Anyone who is trying to bug you by setting up his own personal definition of "algebra" .[/quote]

Notice that this was a Canadian. I do think that people from other countries use the word "arithmetic" to cover higher level courses than Americans associate with the word.

i agree with that waterloo student, to a certain extent, however, one may wonder what is the difference between arithmetic and algebra, i believe that everything in high school is pre-algebra because simply most high schools do not pick their students; therefore, the textbooks and the topics in which the curriculum wishes to cover is really not that hard, since they students r not picked, but anyone who lives in the area can attend. In another words, the materials which they teach in high school is intended so that most, if not all of the students could understand which some or very little practice. since the students in which r attending universities have been picked, therefore the topics and pace in which they cover in universities r according to the schools, which is more difficult in high school. i will be attending University of Waterloo next year either in math or engineering. although i think arithmetic is part of algebra, but then that's just what i think..


hi phelphedo, worthawholebean, Ignite168, and Dan Tobias who doesn't go on this site (i go to his school, smith middle school)

-----------------------------------------------------------

how annoying was dan at nats, mtms said he was fooling around at 1am (did he have too much sugar?)

-----------------------

-jorian

Phelpedo lives in Massachusettes, dunno what you're talking about.

I certaintly didn't make it to ct nat 06 either so I have no idea what you're talking about.

At 1am we were up with PA(even though we had to leave at 5am to catch our plane. We missed it.)

Perfect definitely wasn't on the CT team.

Ignite definitely was on the CT team...

According to popular rumor, Dan Tobias was quite annoying. Some coach (Ravi B maybe?) even suggested that he be banned from this year's comp for holding up the elevators.

NATIONALS WAS NOT AN ACCURATE ASSESSMENT OF MY ABILITIES.

In reality, I'm actually much worse.

Nationals was not an accurate assessment of MY abilities. I wouldn't have gotten a 13 on sprint if you guys hadn't kept us up 'til 1 in the morning calling us.

Then again, Dan was my roommate. So you probably had nothing to do with it.

Pffft.

no offense guys, but this is pretty funny knowing dan

and just so you know, there is a very low chance he'll make it this year, he got in kind of luckily last

worthawholabean...could u pm me exactly what he did (im going to put it against him, don't worry)
-jorian

p.s. how do u get a 13 on sprint?

[quote="Ignite168"][quote="junggi"][quote="Ignite168"]
In reality, I'm actually much worse.[/quote]
yes, I've noticed.[/quote]

AoPS = Reality.

Nothing you do on AoPS is applicable to real life.[/quote]
What are you saying? After you log off AoPS you suddenly become a lot smarter?

[quote="jhredsox"]no offense guys, but this is pretty funny knowing dan

and just so you know, there is a very low chance he'll make it this year, he got in kind of luckily last
[/quote]

Ah, I see. And you have a much higher  chance than him, right  :wink:

[quote="jhredsox"]no, i never said i would get to nats, im just saying that knowing dan, its amazingly hard to believe he got 2nd at nats when we have smarter kids (not me - for now) that always do better than him on practices

oh well, chances are he won't make it this year
-jorian[/quote]

Dan got 2nd at Nats?

o sry i meant states :maybe:

....when dan called, was it funny at first and then it was ridiculous, or was it ridiculous from the start?
-jorian

[quote="kyyuanmathcount"]Yeah... I'm a michigander, but anyways, PenguinIntegral, isn't he Patrick Hulin?

I thought they caught andrew ardito doing that elevator thing.  On the subject of accurate measurings of our abilities, what would you say about 22 sprint and 3 target?  If I had gotten just 4, I would have made top 57, 7 or 8 would have made countdown.[/quote]

How about 16 sprint and 4 target? -_-

BTW, I SAW tobias doing the elevators. I entered, he pushed all the buttons, and ran out.

Guess what, jorian...

[quote]
....when dan called, was it funny at first and then it was ridiculous, or was it ridiculous from the start? [/quote]

That's a stupid question. No more stupid questions please, kthxbai.

its not really a ridiculous question
-jorian

I don't even understand the question.

It looks like you mean "when dan [was] called...", in which case not only does it make no sense but it appears to be a not-so-subtle flame...I must also confess to being rather confounded by your use of "ridiculous."

well i mean, was it funny at first, and he did it too much, or was it never funny

and phelpedo, did u get to nats from ct ever?
-jorian

[quote="Phelpedo"]I suppose this is some inside joke...and yes I made it in '05.[/quote]

Don't think it's a joke, more like a question.

............so are u gonna answer it?
-jorian

u have time every day, u stay up to 3 am
-jorian

I have never been to nationals. Say what school grade and town are you guys in. I go to Fairfield Woods Middle school, I am obssessed with math I am in 7th grade and I live in noneother than Fairfield.

please don't spam this topic as there is already a topic for this. :wink:
-jorian


[quote="xoop"][Removed][/quote]
yep this is one big problem  :D

I'm pretty sure this problem is formally undecidable...


integration
calculus
logarithms
calculus computations

$ \int_{0}^{1}x^9lnxdx$
It occurred to me that rather than repeated integration by parts, it'd be much easier if you added in another variable and integrated with respect to it. Is this allowable?
$ \int_{0}^{1}x^nlnxdx$

$ \equal{} \int_{0}^{1}\frac {dx^n}{dn}dx$

$ \equal{} \frac {d}{dn}\int_{0}^{1}x^ndx$

$ \equal{} \frac {d}{dn}\frac {1}{n \plus{} 1}$

$ \equal{} \frac { \minus{} 1}{(n \plus{} 1)^2}$

$ \equal{} \minus{}.01$

Of course. Leibniz's integral rule ensures that such manipulation is valid.

$ \int_{0}^{1} x^9 \ln x \, dx \equal{} \left. \int_{0}^{1} \frac{\partial x^t}{\partial t} \, dx \right|_{t\equal{}9} \equal{} \left. \frac{d}{dt} \int_{0}^{1} x^t \, dx \right|_{t\equal{}9} \equal{} \left. \frac{d}{dt}\frac{1}{1\plus{}t} \right|_{t\equal{}9}$

I'll also point out that the integration by parts you were avoiding isn't "repeated." Integrate the power of $ x,$ differentiate the logarithm, and finish the problem after one integration by parts.

$ \int_0^1x^p\ln x\,dx\equal{}\minus{}\frac{1}{(p\plus{}1)^2}$ for any $ p>\minus{}1.$


Two cards are drawn without replacement from a standard deck of cards. If it is known that neither card is a spade, what is the probability that both cards are diamonds?

[hide]Since we know neither card is a spade, we can simply remove the spades from consideration and say we have a deck of 39 cards, 13 of which are diamonds.  So the probability that both are diamonds is $\displaystyle\frac{13}{39}*\displaystyle\frac{12}{38}=\displaystyle\frac{2}{19}$[/hide]

Thanks, that's what I got, but I wasn't sure if we could just "remove" the 13 cards.


function
LaTeX
geometry
circumcircle
inequalities
geometry proposed

Let ABC be a triangle. AB = c, BC = a, AC = b.
 M is a point and MA = x, MB = y, MC = z.
  Denote f(M) = ( \sum a^4*x^4)/( \sum b^2*c^2*y^2*z^2).
  Find the locus of M such that f(M) is an integer.

It was probably the strange "integer" condition that kept most MathLinkers from solving this problem... it's a nice problem anyway.

I will write your function f(M) in LaTeX notation:

$\displaystyle f\left(M\right) = \frac{a^4x^4+b^4y^4+c^4z^4}{b^2c^2y^2z^2+c^2a^2z^2x^2+a^2b^2x^2y^2}$.

You ask for the locus of all points M such that f(M) is an integer. In fact, this locus is the union of the circumcircle of triangle ABC with the set $\left\{ I_1;\;I_2\right\}$, where $I_1$ and $I_2$ are the two isodynamic points of triangle ABC.

[i]Proof.[/i] I will denote X = ax, Y = by, Z = cy. Note that the numbers x, y, z are nonnegative (as distances between points!), and hence the numbers X, Y, Z are nonnegative, too. Since X = ax, Y = by, Z = cy, we can rewrite our function f(M) as

$\displaystyle f\left(M\right) = \frac{X^4+Y^4+Z^4}{Y^2Z^2+Z^2X^2+X^2Y^2}$.

The number f(M) is an integer if and only if one of the following cases occurs:

[b]Case 1.[/b] The number f(M) is a negative integer.
[b]Case 2.[/b] The number f(M) = 0.
[b]Case 3.[/b] The number f(M) = 1.
[b]Case 4.[/b] The number f(M) is an integer $\geq$ 2.

Now, case 1 is clearly impossible, since

$\displaystyle f\left(M\right) = \frac{X^4+Y^4+Z^4}{Y^2Z^2+Z^2X^2+X^2Y^2}$

is nonnegative per definitionem.

Case 2 corresponds to f(M) = 0, what, because of

$\displaystyle f\left(M\right) = \frac{X^4+Y^4+Z^4}{Y^2Z^2+Z^2X^2+X^2Y^2}$,

must yield $X^4+Y^4+Z^4=0$. Since $X^4$, $Y^4$, $Z^4$ are all nonnegative, it must follow that X = Y = Z = 0, so that x = y = z = 0, what is impossible, since there is no point M having the distance 0 from every vertex of the triangle ABC.

Case 3 gives

$\displaystyle 1 = f\left(M\right) = \frac{X^4+Y^4+Z^4}{Y^2Z^2+Z^2X^2+X^2Y^2}$,

so

$X^4+Y^4+Z^4=Y^2Z^2+Z^2X^2+X^2Y^2$,

or, equivalently,

$X^4+Y^4+Z^4-Y^2Z^2-Z^2X^2-X^2Y^2=0$.

But

$\displaystyle X^4+Y^4+Z^4-Y^2Z^2-Z^2X^2-X^2Y^2=\frac12 \cdot \left(\left(Y^2-Z^2\right)^2+\left(Z^2-X^2\right)^2+\left(X^2-Y^2\right)^2 \right)$.

Hence, we must have

$\left(\left(Y^2-Z^2\right)^2+\left(Z^2-X^2\right)^2+\left(X^2-Y^2\right)^2 \right)= 0$.

But since $\left(Y^2-Z^2\right)^2$, $\left(Z^2-X^2\right)^2$, $\left(X^2-Y^2\right)^2$ are all nonnegative, this must yield $Y^2-Z^2=Z^2-X^2=X^2-Y^2=0$, so $X^2=Y^2=Z^2$. Since X, Y, Z are nonnegative, this entails X = Y = Z, or, in other words, ax = by = cz, so that $a \cdot MA = b \cdot MB = c \cdot MC$. In this case, the point M must be one of the two isodynamic points $I_1$ and $I_2$ of triangle ABC.

Now consider the case 4. We have

$\displaystyle \frac{X^4+Y^4+Z^4}{Y^2Z^2+Z^2X^2+X^2Y^2} = f\left(M\right) \geq 2$,

so that

$X^4+Y^4+Z^4 \geq 2\cdot \left(Y^2Z^2+Z^2X^2+X^2Y^2\right)$,

and

$2\cdot \left(Y^2Z^2+Z^2X^2+X^2Y^2\right) - \left(X^4+Y^4+Z^4\right) \leq 0$.

But

$2\cdot \left(Y^2Z^2+Z^2X^2+X^2Y^2\right) - \left(X^4+Y^4+Z^4\right) = \left(X+Y+Z\right) \left(Y+Z-X\right) \left(Z+X-Y\right) \left(X+Y-Z\right)$.

Thus,

$\left(X+Y+Z\right) \left(Y+Z-X\right) \left(Z+X-Y\right) \left(X+Y-Z\right) \leq 0$.

Now, WLOG assume that $X \leq Y \leq Z$.

If the point M coincides with one of the vertices of triangle ABC, then it obviously lies on the circumcircle of triangle ABC. Now we will only consider the case when the point M doesn't coincide with any of the vertices of triangle ABC. In this case, the numbers x, y, z are all nonzero. Since we also know that these numbers x, y, z are nonnegative, it follows that the numbers x, y, z are strictly positive. Hence, the numbers X, Y, Z must be strictly positive, too. Thus, X + Y + Z > 0. Also, Y + Z - X > 0 (since $Y \geq X$ and Z > 0). Also, Z + X - Y > 0 (since $Z \geq Y$ and X > 0). Hence, the inequality

$\left(X+Y+Z\right) \left(Y+Z-X\right) \left(Z+X-Y\right) \left(X+Y-Z\right) \leq 0$

can hold only if $X+Y-Z \leq 0$. This means that we must have $X+Y-Z \leq 0$, i. e. we have $ax+by-cz \leq 0$, what can be also written in the form $ax+by \leq cz$, so $BC \cdot MA + CA \cdot MB \leq AB \cdot MC$. But after the Ptolemy inequality for the quadrilateral ACBM, we have $BC \cdot MA + CA \cdot MB \geq AB \cdot MC$, with equality only if the point M lies on the circumcircle of triangle ABC, and, more exactly, on the arc AB of this circumcircle. Now, the inequalities $BC \cdot MA + CA \cdot MB \leq AB \cdot MC$ and $BC \cdot MA + CA \cdot MB \geq AB \cdot MC$ combined yield $BC \cdot MA + CA \cdot MB = AB \cdot MC$, so we actually have the equality case, and the point M lies on the circumcircle of triangle ABC, and, exactly, on the arc BC. (Note that we had assumed WLOG that we have $X \leq Y \leq Z$. With other assumptions, the point M could lie on the arc CA or on the arc AB, but in any case, the point M lies on the circumcircle of triangle ABC.)

Hence, a point M for which the value f(M) is an integer must either be one of the two isodynamic points $I_1$ and $I_2$ of triangle ABC, or lie on the circumcircle of triangle ABC. By reversing the proof above, we can easily find that conversely, if a point M either is one of the two isodynamic points $I_1$ and $I_2$ of triangle ABC, or lies on the circumcircle of triangle ABC, then f(M) is an integer. Hence, the locus of all points M such that f(M) is an integer is the union of the circumcircle of triangle ABC with the set $\left\{ I_1;\;I_2\right\}$. And we are done.

  Darij


function
calculus
derivative
real analysis
real analysis unsolved

Hi! I need some help with the following problem:

Let $ x \in \mathbb{R}^n, \, t \in \mathbb{R}, \, f: \mathbb{R}^n \to \mathbb{R} $ with

$ f(tx) = tf(x) $, where $ f $ is differentiable for all $ x \in \mathbb{R}^n \backslash \{0\} $.

I could already show that in this case we have: $ f(x) = \langle \mathrm{grad} f(x), x \rangle $, where $ \mathrm{grad} $ is the gradient and $ \langle \cdot, \cdot \rangle $ is the standard scalar product in $ \mathbb{R}^n $.

In what I didn't succeed yet, is to show that it follows from these properties, that all directional derivatives of $ f $ in $ 0 \in \mathbb{R}^n $ exist.

I suppose this should be no hard problem for all of you, but could you please help me?

If $t$ is a constant fixed in advance, the relation you are looking for is false. If $f(tx)=tf(x)$ for all pairs $x\in V,t\in\mathbb{R}$, then the restriction of $f$ to a particular line through the origin is linear, and of course the directional derivative at the origin is the value at distance 1 on the line. This does not require $f$ to be differentiable anywhere.


Let S(x) denote the sum of digits for a positive integer x. Do there exist positive integers a,b,c such that S(a+b)<5, S(a+c)<5, S(b+c)<5 and S(a+b+c)>50?


If you roll a pair of dice, what is the probability expressed as a common fraction that both numbers are 2's or that the sum is less than 6?

The probability of getting both 2's is $ \frac{1}{6}\cdot\frac{1}{6}\equal{}\frac{1}{36}$. The probability of rolling less than 6 is $ \frac{1\plus{}2\plus{}3\plus{}4}{36}\equal{}\frac{10}{36}$. Thus, our answer is $ \frac{1}{36}\plus{}\frac{10}{36}\equal{}\boxed{\frac{11}{36}}$.

But if you have both 2's, the sum would be less than 6 anyway.

10/36
= 5/18

Swimming is cool  :D


Need help with the system

sin(x) + 2 sin(x + y + z) = 0

sin(y) + 3 sin(x + y + z) = 0

sin(z) + 4 sin(x + y + z) = 0

I used and expanded trig. identities. The cosines intefere.
And, I have sin(z) = 2 sin(x) and sin(y) = 3/2 sin(x)

But it's very messy. Need help, pls.

sin (x+y+z)
= sin [(x+y)+z]
= sin (x+y) cos z + cos (x+y) sin z
= (sin x cos y + cos x sin y) cos z + (cos x cos y - sin x sin y) sin z
= sin x cos y cos z + cos x sin y cos z + cos x cos y sin z - sin x sin y sin z

but, after that it gets messy.

Did you propose this yourself? Where is the problem from? It seems unsolvable with elementary methods...

I recently joined a math club, and this problem was dumped on me.

I finally found the solution, so here it is:

So, we have the sytem

sin(x) + 2 sin(x + y + z) = 0

sin(y) + 3 sin(x + y + z) = 0

sin(z) + 4 sin(x + y + z) = 0

Notice that we have in common sin(x+y+z) in all 3 equations.

2*sin(x+y+z) = -sin x
3*sin(x+y+z) = -sin y
4*sin(x+y+z) = -sin z

sin(x+y+z) = -1/2 sin x = -3/4 sin y = -1/4 sin z

Let's multiply by (-12), and we get

6 sin x = 4 sin y = 3 sin z

x = 0, y = 0, z = 0 (trivial solution)
x = pi, y = 0, z = 0
x = 0, y = pi, z = 0
x = 0, y = 0, z = pi
x = pi, y = pi, z = 0
x = pi, y = 0, z = pi
x = 0, y = pi, z = pi
x = pi, y = pi, z = pi.

[quote="Jean-Louis"]
6 sin x = 4 sin y = 3 sin z

x = 0, y = 0, z = 0 (trivial solution)
x = pi, y = 0, z = 0
x = 0, y = pi, z = 0
x = 0, y = 0, z = pi
x = pi, y = pi, z = 0
x = pi, y = 0, z = pi
x = 0, y = pi, z = pi
x = pi, y = pi, z = pi.[/quote]

Umm, those are all trivial solutions. (At all of the solutions, we just have $0=0=0$.) Can you find the non-trivial solutions? Do they exist?


$ \frac {1}{x_{1}} \plus{} \frac {1}{x_{2}} \plus{} \frac {1}{x_{3}} \plus{} ....\plus{}\frac {1}{x_{n}} \plus{} \frac {1}{x_{1}x_{2}x_{3}...x_{n}} \equal{} 1$

Find the solution, $ x_{1},x_{2},....,x_{n}$ are positive integers.

Working hard with this problem,  I finally got that I can't solve this.

$ x_1 \equal{} 2,x_{n \plus{} 1} \equal{} x_1x_2\cdots x_n \plus{} 1$

Not sure if there are others.

A related problem is Znam's problem:
[url]http://en.wikipedia.org/wiki/Zn%C3%A1m%27s_problem[/url]


Find all positive integers $ (m,n)$, such that

$ 2m^2 \plus{} 3nm \minus{} 2n^2 \equal{} 71$

Stop! Why do you post the problem in 2 places?

[url]http://www.mathlinks.ro/viewtopic.php?p=1530185#1530185[/url]

Should be locked!

Japanese Communities Moderator

kunny

sorry!!! i thought this was the original place i posted it and obviously I couldnt find my post...


calculus
integration
limit
trigonometry
calculus computations

Find $ \lim_{n\to\infty} \int_0^1 x^2|\sin n\pi x|\ dx\ (n\equal{}1,\ 2,\cdots)$.

Using the theorem discussed here: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=296&t=417146&p=2351074#p2351074

$\lim_{n\to\infty}\int_{0}^{1}x^{2}|\sin n\pi x|\,dx$ $= \left(\frac{1}{2}\int_{0}^{2}|\sin(\pi x)|\,dx\right)\left(\int_0^1 x^2\,dx\right)$ $= \left(\frac{1}{2}\cdot\frac{4}{\pi}\right)\left(\frac{1}{3}\right)$ $= \frac{2}{3\pi}$

@AYMANE: The correct expression should be $\dfrac{(2n^2+1)\pi^2-12}{3n^2\pi^3}$.


[geogebra]09b5480cf12978c2fd55162748161cd8c322adc9[/geogebra]

That is, ofcourse, the construction of the pentagon.

But obvously it is wrong?

I believe this was the correct construction for two adjacent vertices of a regular pentagOn:

 [geogebra]5d671ba6872cccea2e2620a5ebce15e7f04850f7[/geogebra]

The line in there is  not the median: it's the angle bisector.

[quote="cognos599"][hide="Another technique"]Draw 5 points, no 3 of which are collinear and connect them, or did you mean a REGULAR pentagon[/hide][/quote]

You give no way to construct them so that they are assuredly noncollinear.

*invade* there is an easy method on AoPS wiki entry: pentagon


Let $\triangle ABC$ be a triangle and denote $a=BC, b=CA, c=AB$. The internal angle bisectors intersect the side $BC, CA, AB$ respectively at $L,M,N$; the external angle bisectors intersect the sidelines $BC, CA, AB$ at $L_1, M_1, N_1$.  If $a > b > c$ show that:

\[ \frac{1}{{LL_1 }} - \frac{1}{{MM_1 }} + \frac{1}{{NN_1 }} = 0   \]

Just use the substitution $BL=\frac{ac}{b+c}$ and $BL_{1} = \frac{ac}{b-c}$ and so on. Then \[\frac{1}{LL_{1}} + \frac{1}{NN_{1}} - \frac{1}{MM_{1}} = \frac{b^2-c^2}{2abc} + \frac{a^2-b^2}{2abc} - \frac{a^2-c^2}{2abc} = 0.\]


Prove that $x^{x+1}=(x+1)^{x}$ has only one root

[hide="Solution"] The problem is equivalent to

$\frac{\ln (x+1)}{x+1}= \frac{\ln x}{x}$

But the function $f(x) = \frac{\ln x}{x}$ has only one maxima and no minima (this is not hard to prove or to see graphically).  In essence, we need a horizontal line of length $1$ such that both endpoints are on the graph, and it is not hard to deduce that only one such placement is possible. [/hide]


There have been threads about this before, but they age off this board over time. My avatar is a photograph of Isreal M. Gelfand, a Russian mathematician (now living in the United States) who is one of my favorite authors of mathematics textbooks.

Here are two quotations from Gelfand:

"I would like to make one comment here. Some of my American colleagues have explained to me that American students are not really accustomed to thinking and working hard, and for this reason we must make the material as attractive as possible. Permit me to not completely agree with this opinion. From my long experience with young students all over the world, I know that they are curious and inquisitive and I believe that if they have some clear material presented in a simple form, they will prefer this to all artificial means of attracting their attention--much as one buys books for their content and not for their dazzling jacket designs that engage only for the moment. The most important thing a student can get from the study of mathematics is the attainment of a higher intellectual level."

"Students have no shortcomings, they have only peculiarities. The job of a teacher is to turn these peculiarities into advantages."

[quote="theone853"]one has actually yet to age off
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=14123
we should really make like a sticky or something of it because it comes up so often along with age, names, locations, and sigs[/quote]

Good idea. Done. We can keep that there and not deal with these threads every couple of days.


trigonometry
geometry
function
angle bisector
geometry solved

$AD$ is the altitude on side $BC$ of triangle $ABC$. If $BC+AD-AB-AC = 0$, find the range of $\angle BAC$.

[i]Alternative formulation.[/i] Let $AD$ be the altitude of triangle $ABC$ to the side $BC$. If $BC+AD=AB+AC$, then find the range of $\angle{A}$.

well, sin(A) = sin(B) + sin(C) - sin(B)sin(C), so 0< A < 180 is all fine, right?

I write your relation such: ( 1 - sinB ) ( 1 - sinC ) = ( 1 - sinA ) and is it easy a.s,o ???

My solution uses another approach:

[quote="orl"]$AD$ is the altitude on side $BC$ of triangle $ABC$. If $BC + AD - AB - AC = 0$, find the range of $\angle BAC$.[/quote]

First we show:

[color=blue][b]Lemma 1.[/b] Let D be the foot of the altitude to the side BC of a triangle ABC. Then, BC + AD - AB - AC = 0 holds if and only if $\tan\frac{B}{2}+\tan\frac{C}{2}=1$.[/color]

[i]Proof of Lemma 1.[/i] Let $s=\frac{a+b+c}{2}$ be the semiperimeter, $\Delta$ the area and $r_a$ the A-exradius of triangle ABC. Then, on one hand, $\Delta=\frac12\cdot a\cdot AD$, since AD is the altitude to the side BC of triangle ABC, but on the other hand, $\Delta=\left(s-a\right)r_a$ by a well-known formula. Hence, $\frac12\cdot a\cdot AD=\left(s-a\right)r_a$; equivalently, $AD=\frac{2\left(s-a\right)r_a}{a}$. On the other hand, b + c - a = (a + b + c) - 2a = 2s - 2a = 2 (s - a). Thus,

$BC+AD-AB-AC=BC+\frac{2\left(s-a\right)r_a}{a}-AB-AC=a+\frac{2\left(s-a\right)r_a}{a}-c-b$
$=\frac{2\left(s-a\right)r_a}{a}-\left(b+c-a\right)=\frac{2\left(s-a\right)r_a}{a}-2\left(s-a\right)$
$=2\left(s-a\right)\frac{r_a-a}{a}$.

Since s - a > 0, we thus have BC + AD - AB - AC = 0 if and only if $r_a-a=0$, what is equivalent to $a=r_a$.

Now, let the A-excircle of triangle ABC have the center $I_a$ and touch the side BC at a point X. Then, $I_aX=r_a$, and $I_aX\perp BC$. But, being the A-excenter of triangle ABC, the point $I_a$ lies on the external angle bisector of the angle ABC, and thus $\measuredangle I_aBX=90^{\circ}-\frac{B}{2}$. Hence, in the right-angled triangle $BXI_a$, we have $BX=I_aX\cdot\cot\measuredangle I_aBX=r_a\cdot\cot\left(90^{\circ}-\frac{B}{2}\right)=r_a\cdot\tan\frac{B}{2}$. Similarly, $CX=r_a\cdot\tan\frac{C}{2}$. Thus, $a=BC=BX+CX=r_a\cdot\tan\frac{B}{2}+r_a\cdot\tan\frac{C}{2}=r_a\cdot\left(\tan\frac{B}{2}+\tan\frac{C}{2}\right)$. Hence, $a=r_a$ is equivalent to $\tan\frac{B}{2}+\tan\frac{C}{2}=1$. But we saw before that BC + AD - AB - AC = 0 is equivalent to $a=r_a$. Hence, BC + AD - AB - AC = 0 is equivalent to $\tan\frac{B}{2}+\tan\frac{C}{2}=1$, and Lemma 1 is proven.

Another lemma now:

[color=blue][b]Lemma 2.[/b] For a given angle x such that 0° < x < 90°, the existance of two angles y and z satisfying the conditions 0° < y < 90°, 0° < z < 90°, y + z = x and $\tan y+\tan z=1$ is equivalent to $45^{\circ}<x\leq 2\arctan\frac12$.[/color]

[i]Note.[/i] $2\arctan\frac12\approx 53,13^{\circ}$.

[i]Proof of Lemma 2.[/i] First, we show that if there exist two angles y and z satisfying the conditions 0° < y < 90°, 0° < z < 90°, y + z = x and $\tan y+\tan z=1$, then $45^{\circ}<x\leq 2\arctan\frac12$.

In fact, the function f(t) = tan t is convex on the interval [0°; 90°[; hence, $\tan y+\tan z\geq 2\tan\frac{y+z}{2}$. In other words, $1\geq 2\tan\frac{x}{2}$. Hence, $\frac12\geq\tan\frac{x}{2}$, so that $\arctan\frac12\geq\frac{x}{2}$ and $2\arctan\frac12\geq x$. On the other hand, by the tangens addition formula, $\tan\left(y+z\right)=\frac{\tan y+\tan z}{1-\tan y\tan z}$. In other words, $\tan x=\frac{1}{1-\tan y\tan z}$. Since tan y and tan z are positive (as 0° < y < 90° and 0° < z < 90°), we have 1 - tan y tan z < 1; but tan x is also positive (since 0° < x < 90°), and thus, tan x > 1, and thus x > 45°. Combining this with $2\arctan\frac12\geq x$, we obtain $45^{\circ}<x\leq 2\arctan\frac12$.

Remains to prove the converse: If $45^{\circ}<x\leq 2\arctan\frac12$, then there exist two angles y and z satisfying the conditions 0° < y < 90°, 0° < z < 90°, y + z = x and $\tan y+\tan z=1$.

In fact, first define the angles y and z as follows: y = 0°, z = x. This will be called the "starting position". Of course, y + z = x is satisfied in this position, but $\tan y+\tan z=1$ does not hold.

Now, let's start continuously increasing y and decreasing z at the same speed, so that the sum y + z = x remains invariant. Then, the angles y and z come closer and closer to each other, and at the end both of them come together: $y=z=\frac{x}{2}$. This will be called the "ending position".

Now look at what happens with the value of tan y + tan z while we increase y and decrease z. Since the function f(t) = tan t is continous on the interval [0°; 90°[, this value of tan y + tan z behaves continuously. In the "starting position", the value of tan y + tan z equals tan 0° + tan x = tan x; this is > 1, since x > 45°. In the "ending position", the value of tan y + tan z equals

$\tan\frac{x}{2}+\tan\frac{x}{2}=2\tan\frac{x}{2}\leq2\tan\arctan\frac12$           (since $x\leq 2\arctan\frac12$ yields $\frac{x}{2}\leq\arctan\frac12$)
$=2\cdot\frac12=1$.

Hence, by the intermediate value theorem, somewhere between the "starting position" and the "ending position", the expression tan y + tan z must take the value 1. Thus, we have proved the existence of two angles y and z satisfying the conditions 0° < y < 90°, 0° < z < 90°, y + z = x and $\tan y+\tan z=1$. Hence, Lemma 2 is proven.

Now, the problem becomes immediate: After Lemma 1, the condition BC + AD - AB - AC = 0 is equivalent to $\tan\frac{B}{2}+\tan\frac{C}{2}=1$. Hence, we are looking for the range of angles A such that 0° < A < 180° and such that there exist angles B and C with 0° < B < 180°, 0° < C < 180° and A + B + C = 180° (these are the conditions for the existence of a triangle with angles A, B, C) which satisfy $\tan\frac{B}{2}+\tan\frac{C}{2}=1$. Set $x=90^{\circ}-\frac{A}{2}$, $y=\frac{B}{2}$, $z=\frac{C}{2}$; then, the conditions 0° < A < 180°, 0° < B < 180°, 0° < C < 180° rewrite as 0° < x < 90°, 0° < y < 90°, 0° < z < 90°, the condition A + B + C = 180° rewrites as y + z = x (since $y+z-x=\frac{B}{2}+\frac{C}{2}-\left(90^{\circ}-\frac{A}{2}\right)=\frac{A+B+C}{2}-90^{\circ}=\frac{A+B+C-180^{\circ}}{2}$), and the condition $\tan\frac{B}{2}+\tan\frac{C}{2}=1$ rewrites as $\tan y+\tan z=1$. Now, by Lemma 2, the existence of two angles y and z satisfying these conditions is equivalent to $45^{\circ}<x\leq 2\arctan\frac12$. Since $x=90^{\circ}-\frac{A}{2}$, this rewrites as $45^{\circ}<90^{\circ}-\frac{A}{2}\leq 2\arctan\frac12$. Upon subtraction from 90°, this becomes $45^{\circ}>\frac{A}{2}\geq 90^{\circ}-2\arctan\frac12$, and multiplication with 2 transforms this into $90^{\circ}>A\geq 2\left(90^{\circ}-2\arctan\frac12\right)$. This rewrites as $90^{\circ}>A\geq 180^{\circ}-4\arctan\frac12$. Thus, the range of our angle A is $\left[180^{\circ}-4\arctan\frac12;\;90^{\circ}\right[$.

Note that $180^{\circ}-4\arctan\frac12\approx 73,74^{\circ}$.

  Darij

I think you thought too complex( although you solution is not very special)
My solution is bases on the problem:
find all A in (0,pi) that the equation f(x)=sinx+sin(x+A)-sinx sin(x+A)- sinA has solution x in (0,pi-A).
simply by using derivative.

If $BC = a$, $CA = b$ and $AB = c$ then either $a \le \min\{b,c\}$ or $a \ge \max\{b,c\}$. Does this lead anywhere?


Find all real x>o such that x-[x], [x], x (where [x] denotes the greatest integer not greater than x) form a geometric progression.

[quote="Rust"]$ x \equal{} \frac {1 \plus{} \sqrt 5}{2}.$[/quote]

Working, please?

We have:

$ \frac{[x]}{x\minus{}[x]} \equal{} \frac{x}{[x]} \equal{} \frac{[x]\plus{}x}{x\minus{}[x]\plus{}[x]} \equal{} \frac{x\plus{}[x]}{x} \equal{} 1 \plus{} \frac{[x]}{x}$.

Solve for $ \frac{x}{[x]}$ and deduce the conclusion.


Respond on the poll, and post your thoughts here.  I personally liked #4 the most - it took me a while to work out, and the answer I came up with was very satisfying.

[quote="SplashD"][quote="bpms"]Was number 4 $k^{2}+k$?[/quote]

it works but $k^{2}+k-1$ is the smallest[/quote]
NOOOO!!!!
I had that, but I changed it. Oh well

[quote="me@home"]If you say that the "basic" solution to 4 is satisfying, then I don't know what you're going to say when you find the generalization to $c$ colors... it is amazing (and the problem is trivial in comparison)[/quote]
What did you get for the generalization? I didn't prove it, I just stated it, so I don't know if it's correct.

I got $k^{c}+k^{c-1}-1$, which probably isn't correct.

Did you guys get:

[hide]
2023, 2083 for #1
114 degrees and 2 for #2
$\frac{4}{3}n^{3}+2n$ for #3 (I think that's what I got, don't remember)
3/2 for #5[/hide]

grr for #4 I only had time to do $k=2$, which I got right, thank god, so maybe 5 5 5 1 5, I think I got either 2 or 5 commended if I got the correct answer.

[quote="13375P34K43V312"]Did you guys get:

[hide]
2023, 2083 for #1
114 degrees and 2 for #2
$\frac{4}{3}n^{3}+2n$ for #3 (I think that's what I got, don't remember)
3/2 for #5[/hide][/quote]
Yes, you can read the other topic with solutions.

I liked number 5 best, because I am pretty bad at geometry, and I was proud that I got it right. I found number 4 to be the easiest.

so close!!!
$\frac{k^{c+1}-2k^{c}+1}{k-1}=f(k-1,c)$
if I remember correctly

Will the graders take off points if a generalization is incorrect?

I was wondering, why do people generalize? to get a commended? or just to further explore the problem?

Xantos C. Guin, didn't you have a post here? Because I quoted it and got a message like "you need special permission to post in this forum" and I pushed "BACK" and it said "this topic does not exist."

And yeah, 13375P34K43V312 probably meant $\frac{4}{3}n^{3}+\frac{2}{3}n$ because plugging in $n=1$, we quickly see that it's impossible.

Although I liked #4 because it was very difficult and was very satisfying to eventually solve, I just thought that #5 was a really cool geo problem  :cool:

Yes! I got #1,2,3 definitely right!!

Well, what can I say, I got 15 for the Round 1...   :oops:

I have a brain of a llama... :ninja:

4 took me the longest so the joy i got from finishing that problem is the greatest.

I'm not great with geometry but the solution I found for number 5 impressed me, especially since it was all geometry and no trig-bashing.  4 still gets my props, it took me a while playing around with it to understand the question.

Number 4 was far and away my favorite of this one ... it might be one of my favorites of all time, along with the one triangle with the altitudes in the circle with the ratios from last year (wow that was vague)

I liked how much kind of thinking it took.  It was pretty easy to get that it had to be at least $k^{2}+k-1$, but it took a little bit of thinking/casework to show that it could never, ever be less than that.

#5 was second.

JB

I liked 5 the best, followed by 4, then 3, then 2, then 1.

4 really annoyed me...  I spent forever on it, and eventually I got the answer, but I had lots of trouble justifying it.


Find all polynomials $ f(x)$ with real coeficients such that for all positive integer $ n$ holds $ [f(\frac{1}{2})] \plus{} [f(1 \plus{} \frac{1}{2})] \plus{}...\plus{} [f(n \plus{} \frac12)] \equal{} f(1) \plus{} f(2) \plus{}...\plus{} f(n)$;
where $ [a]$ denotes the integer part of a real number $ a$.

If $ deg f(x) >1$, then $ |f'(x)|\to \infty$, therefore for big k  $ |f(k\plus{}\frac 12 )\minus{}f(k)|\ge|f'(k)\frac 12 >1$. and don't $ \sum_{r\equal{}0}^n [f(k\plus{}\frac 12 )] \equal{}\sum_{k\equal{}0}^n f(k).$
If deg f =0, then $ f(x)\equal{}c, \{c\}\equal{}$ or c is integer.
If $ deg f(x)\equal{}1$, then $ f(x)\equal{}x\plus{}n,n\in Z$.


Prove that $\frac{e}{(1+\frac{1}{n})^{n}}<1+\frac{1}{2n}$ for all integers $n\geq 1$.

This is related to Morrow's inequality (right spelling?);$\frac{e}{2n+2}<e-\left(1+\frac{1}{n}\right)^{n}<\frac{e}{2n+1}.$ I think.
I have posted it before, but I can't find it. :(


number theory
prime factorization
number theory unsolved

$ x,y,z$ are natural numbers such that $ gcd(x,y,z)\equal{}1$ and $ \frac{1}{x}\plus{}\frac{1}{y}\equal{}\frac{1}{z}$
Proof that $ x\plus{}y$ is a perfect square.

[hide="My Solution"]$ \frac {1}{x} \plus{} \frac {1}{y} \equal{} \frac {1}{z}$
$ \implies z(x \plus{} y) \equal{} xy$
$ \implies (x \plus{} y)|xy$
Suppose $ p$ is such a prime that $ p^{2k \minus{} 1}|(x \plus{} y)$ [$ k$ is a natural number]
$ \implies p^{2k \minus{} 1}|xy$
By the pegion-hole-principle: $ p^k|$(one of $ x,y$)
WLOG $ p^k|x$
We also have $ p^k|(x \plus{} y)$
Hence $ p^k|(x \plus{} y) \minus{} x\implies p^k|y$
So $ p^{2k}|xy$
Since $ (x,y,z) \equal{} 1$, $ p\not |z$
But $ z(x \plus{} y) \equal{} xy$
Hence $ p^{2k}|(x \plus{} y)$
So for any prime $ p$, if $ p^{2k \minus{} 1}|(x \plus{} y)$, then $ p^{2k}|(x \plus{} y)$
So all primes are raised to even powers in the prime factorization of $ (x \plus{} y)$
Hence $ (x \plus{} y)$ is a perfect square.[/hide]

[quote="Przemo"]$ x,y,z$ are natural numbers such that $ gcd(x,y,z) \equal{} 1$ and $ \frac {1}{x} \plus{} \frac {1}{y} \equal{} \frac {1}{z}$
Proof that $ x \plus{} y$ is a perfect square.[/quote]
[hide="Solution"]
Let $ z\equal{}ka$ and $ x\equal{}kb$ where $ k\equal{}\gcd (x,z)$.  Since if $ \gcd (y,k)>1$, then $ \gcd (y,z, x)>1$, we have that $ \gcd (y,k)\equal{}1$, so if $ \gcd (z, y)\equal{}l$, then $ l|a$.  Let $ ld\equal{}a$ and $ y\equal{}lc$.  Upon expanding, we get that $ z(x\plus{}y)\equal{}xy$, so $ kld(kb\plus{}lc)\equal{}lkcb$, meaning that $ d(kb\plus{}lc)\equal{}bc$.  Now, we have that $ \gcd (d, c)\equal{}1$ and $ \gcd (d, b)\equal{}1$, so $ \gcd (d, bc)\equal{}1$.  Yet, $ d|bc$, so $ d\equal{}1$.  This leaves us with $ (b\minus{}l)(c\minus{}k)\equal{}lk$.  Assume that $ \gcd (c\minus{}k, k)>1$.  Then, $ \gcd (c, k)>1$, meaning that $ \gcd (c, \gcd (x,z))>1$, so $ \gcd (x,y,z)>1$, which is a contradiction.  This means that $ c\minus{}k$ and $ k$ are coprime, and analogously are $ b\minus{}l$ and $ l$.  This means that $ b\minus{}l\equal{}k$ and $ c\minus{}k\equal{}l$, so $ b\equal{}c\equal{}k\plus{}l$.  We conclude that $ x\plus{}y\equal{}kb\plus{}lc\equal{}(k\plus{}l)^2$, which is a perfect square. [/hide]

[quote="The QuattoMaster 6000"] Now, we have that $ \gcd (d, c) \equal{} 1$[/quote]
Why is this?
BTW, I know I'm posting on the same page as you a lot, that's because I think your proofs are very easy to read :)

Sorry for not clearing that up in my solution:

Say that $ d$ and $ c$ share a factor.  Then, $ \gcd(a, y) > l$ and since $ a | z$, we would have that $ \gcd (z, y) > l$, which is a contradiction.


IT REALLY IS DANG HARD! I NEED SOME HELP PLEASE :(  :rotfl:  :( !

I agree that some of the counting problems are difficult. (which is probably why my Alcumus score keeps on going down because I get these problems wrong...)
What types of problems do you need help on? If you have trouble on a specific problem, you can post the problem [url=http://www.artofproblemsolving.com/Forum/index.php?f=540]in this part of the forum.[/url]

yes it would be helpful to know exactly what subjects you require help on before people can provide help.
Have you checked the video lessons for your subjects?

Mostly, when you first start off, it may seem hard at first. Don't worry, and if you have any problems, post them up on the Alcumus Problem Discussion board, although its best not to just get your Alcumus score up by putting specific problems on before you've even tried them.

Video lessons are another thing to use; if you get stuck on some problems, it might be helpful to watch the problems. But, also, some problems you might not get help with from the video lessons; they're taken from the MATHCOUNTS/AMC tests, and are rather harder than others.

If you get stuck on a SPECIFIC topic and watched all of the videos, then maybe you should ask for help on the message boards concerning this subject, or thoroughly go over the solutions.

Hoped this helps!

Yeah, i'm getting better at it, i actually passed a topic. I just need a lot of help on casework counting. :rotfl:  :rotfl:


algebra
polynomial
quadratics
calculus
integration
Rational Root Theorem
complex numbers

[quote="ch1n353ch3s54a1l"][quote="jhcreinhardt"]Find the monic polynomial with integer coefficients of least degree such that $p(\sqrt{2}+\sqrt{5})=0$[/quote]

Uh...what? Sorry, I don't see what the problem's asking... what's p...??...[/quote]
the polynomial is p(x)

[hide]$x = \sqrt2+\sqrt5$
$x-\sqrt2 = \sqrt5$
$x^{2}-2x\sqrt2+2 = 5$
$x^{2}-3 = 2x\sqrt2$
$\boxed{x^{4}-14x^{2}+9 = 0}$[/hide]

there was a problem like this posted recently. I used same method as farenhajt.

[quote="breez"]What's the logic behind setting x equal to the radicals?[/quote]
However you manipulate the equation, $x=\sqrt2+\sqrt5$ will always be true. So we are trying to manipulate the equation into the form $f(x)=0$, where $f(x)$ is a polynomial. $f(\sqrt2+\sqrt5)=0$, so $\sqrt2+\sqrt5$ is a root of the polynomial $f(x)$.

[hide]This was like something I posted.  Woot.

$x=\sqrt{2}+\sqrt{5}$

So $x^{2}=7+2\sqrt{10}$

So $x^{2}-7=2\sqrt{10}$

$x^{4}-14x^{2}+49=40$

So the monic polynomial is

$x^{4}-14x^{2}+9=0$[/hide]

[quote="jclarkemathL314159"]What's a monic polynomial?[/quote]

leading coeff is 1

The question actually asks for the monic polynomial with integer coefficients of least degree, so I'm not sure if what you have is actually enough. There's one more easily overlooked detail that proves there is not a monic polynomial with integers coefficients and a lesser degree.

A polynomial with integer coefficients has its irrational roots come in conjugate pairs.  The given value's conjugate won't give a quadratic with integer coefficient, and a cubic polynomial couldn't have another because the third root would have to be rational (rrt).  So lowest degree here is quartic.

[quote="jhcreinhardt"] a cubic polynomial couldn't have another because the third root would have to be rational (rrt).[/quote]

Care to justify this?  RRT = Rational Root Theorem says nothing about forcing polynomials to have rational roots, it just says what their rational roots have to look like.

[quote="jhcreinhardt"]what i was trying to say is that a cubic wouldnt have 3 irrational roots if it had integer coefficients.  it would have the same two as a quadratic plus a rational one.[/quote]

Really?  So which is the rational root of $x^{3}-3x+1 = 0$?

:D

Let's clear some things up.

[b]Rational Root Theorem:[/b]  The rational roots of a polynomial with integer coefficients $a_{0}+a_{1}x+...+a_{n}x^{n}$ are of the form $\frac{p}{q}$ where $(p, q) = 1, p | a_{0}, q | a_{n}$.

Note that this theorem says nothing about the number of such roots.  The lemma to which I think you refer is this:

[b]Lemma:[/b]  A polynomial with [i]real[/i] coefficients has an even number of [i]imaginary[/i] roots.  Equivalently, if $z \in \mathbb{C}$ is a root of some $P(x) \in \mathbb{R}[x]$, then it follows $\bar{z}$ is also a root.

Note that this is real/imaginary, not rational/irrational.  (By an imaginary number I refer to a member of the set $\mathbb{C}-\mathbb{R}$; the complex numbers with nonzero imaginary part.)

The original problem and its implications were discussed in my Galois theory class; the stronger statement is that $\mathbb{Q}[ \sqrt{2}][ \sqrt{5}]$ and $\mathbb{Q}[ \sqrt{2}+\sqrt{5}]$ are isomorphic.

[quote="jhcreinhardt"]perhaps i dont quite understand descartes rule of signs.  when a coefficient is zero, this counts as a sign change?  i didnt think it did.  in the example you gave, $f(-x)=-x^{3}+3x+1$  does that count as one or two sign changes?[/quote]
one i believe.

It's sign changes of nonzero coefficients. There are can be 3 real roots or 1 real root and 2 imaginary roots. But I don't think descarte's rule tells whether the 3 must be rational/irrational or come in pairs of irrational.

However, if an integral polynomial has 3 roots, I would imagine that the irrational roots come in pairs, as do the complex roots.

[quote="breez"]I would imagine that the irrational roots come in pairs[/quote]

Exactly what jhcreinhardt believed, and, as JBL and I pointed out, exactly wrong.

In his example, $x^{3}-3x+1$ has three irrational (real) roots.

the first time i mentioned rrt i was meant descartes.  i made the mistake of thinking that descartes rule of signs applied to irrational and rationals.  That's why the basis of my logic failed...completely.  Thanks for pointing this out...


if A*B means A*B/2 then what does (3*5)*8 equal?


geometry
trigonometry
perimeter
AMC
AMC 12
rectangle

1)Consider a triangle which have three circles tangent between them, and tangent with the sides of that triangle. If the three circunferences have a radius of 3, then which is the perimeter of that triangle??

2)If $ sinx \plus{}cosx \equal{}\frac{1}{5}$, and $ 0<x<\pi$, then which is the value of $ tanx$???

2).

$ \sqrt{2} \sin(x\plus{}\frac{\pi}{4}) \equal{} \frac{1}{5}$

$ \sin(x\plus{}\frac{\pi}{4}) \equal{} \frac{1}{5\sqrt{2}}$

$ x\plus{}\frac{\pi}{4} \equal{} 0.142, \pi\minus{}0.142, ...$

Therefore $ 0 < x < \pi \implies x \equal{} \pi \minus{} 0.142$, that is $ x\equal{}\frac{\pi}{4}$ is in the second quadrant, where $ \tan$ is negative.
Then, using $ \cos^{2} \equal{} 1\minus{}\sin^{2}, \sec \equal{} \frac{1}{\cos}, 1\plus{}\tan^{2} \equal{} \sec^{2}$, we get

$ \tan (x \plus{} \frac{\pi}{4}) \equal{} \frac{\minus{}1}{7}$

Therefore, $ \frac{\tan x \plus{} 1}{1\minus{} \tan x} \equal{} \frac{\minus{}1}{7}$

$ 7t \plus{} 7\equal{}\minus{}1\plus{}t$, where $ t\equal{}\tan x$,

$ \tan x \equal{} \frac{\minus{}4}{3}$.

For 1, do you mean the three circles are tangent inside or outside the triangle?  Plus, does "the three circumferences have radius of 3" mean that all three [u]circles[/u] have radius 3? For example,
[geogebra]279579cfec2062e86ae37983ea039db643d611c0[/geogebra]

[quote="Chalao"]Is a right triangle which have the three circles inside it. The three circles touch each other, and each one touches two sides of the triangle.[/quote]

Chalao could you provide a sketch of the proble cuz I do not see how that triangle can be right-angled, my figure is similar to bbgun34's , the triangle is equilateral if the three circles are equal.

[img]http://i204.photobucket.com/albums/bb126/agerador/triangle.jpg[/img]

Something like this, supposing the radius is 3 fot the three circles (i know my drawing is not the best :wink:, but the three circles suppose to be equal)

Please someone help me solving the triangle problem :lol:

Then that should definitely have been "equilateral triangle", not "right triangle", unless we're working with spherical geometry.

[hide="Alternate solution to #2:"]
Let $ a \equal{} \sin {x}$ and $ b \equal{} \cos {x}$.
Then $ a\plus{}b \equal{} \frac{1}{5}$ and $ a^2 \plus{} b^2 \equal{} 1$. We wish to find $ \frac{a}{b}$.
$ b \equal{} \frac{1}{5} \minus{} a$, so $ a^2 \plus{} \left(\frac{1}{5} \minus{} a\right)^2 \equal{} 1$
$ \Rightarrow 2a^2 \minus{} \frac{2}{5}a \minus{} \frac{24}{25} \equal{} 0$
$ \Rightarrow 25a^2 \minus{} 5a \minus{}12 \equal{} 0$
$ \Rightarrow a \equal{} \frac{5 \plus{} \sqrt{25\minus{}(4)(25)(\minus{}12)}}{50} \equal{} \frac{4}{5}$ (choosing the plus sign because $ 0 < x < \pi$: making $ \sin {x}$ positive)
$ \Rightarrow b \equal{} \frac{1}{5} \minus{} a \equal{} \minus{}\frac{3}{5}$

So $ \frac{a}{b} \equal{} \minus{}\frac{4}{3}$[/hide]

Very easy and clean solution gauss202!! What about the triangle problem!?

Options for the triangle problem:

(a)$ 36\plus{}9 \sqrt{3}$   (b) $ 36\plus{}6\sqrt{3}$   (c) $ 18\plus{}18\sqrt{3}$

I still think that there must be a problem with the problem.

Suppose it isn't necessary a right triangle, then..... is there a way to get any of the posible solutions listed above?

hmmm... Where did you get #1 from? An AMC12? Can you please post a link to it?(This might solve some of the doubts about the problem)

So just look at the area near on vertex. Let the center of the circle be O, the vertex be A, and the points of tangency be B and C. BO=CO=3

$ m\angle ABO\equal{}m\angle ACO\equal{}90^{\circ}$

$ m\angle A\equal{}60^{\circ}$

$ m\angle BOC\equal{}120^{\circ}$

Draw in AC, creating two 30-60-90 triangles.

$ AB\equal{}AC\equal{}3\sqrt{3}$

There are 6 such segments on the triangle.

Look at the rectangle formed by connecting the centers of 2 of the circles to the points of tangency on their shared side.

Let the centers be $ O_1$ and $ O_2$ and the points of tangency be $ D$ and $ E$

$ DE\equal{}O_1O_2\equal{}3\plus{}3\equal{}6$

There are 3 such segments on the perimeter.

So the perimeter is $ 6(3\sqrt{3})\plus{}3(6)\equal{}18\sqrt{3}\plus{}18$ which is C.


function
real analysis
real analysis theorems

Say that you have $f: \mathbb{R}\rightarrow\mathbb{R}$ and $g: \mathbb{R}\rightarrow\mathbb{R}$.  What is the best way to find, say, the minimum distance between the curves?  Normally, I would take two points $P=\left(x,f\left(x\right)\right)$ and $Q=\left(y,g\left(y\right)\right)$.  Minimizing the square of the distance would produce the desired result, so I could say

\[ d^{2}\left(P,Q\right)=\varphi\left(x,y\right)=\left(f\left(x\right)-g\left(y\right)\right)^{2}+\left(x-y\right)^{2}.  \]

What is the best way to minimize this function?  I mean, does it make sense to solve $\frac{\partial\varphi}{\partial x}=\frac{\partial\varphi}{\partial y}=0$?


probability
Euler
number theory
relatively prime

What is the probability that a fraction $ \frac{a}{b}$ whose numerator and denominator are chosen at random among all positive integers turns out to be irreducible?

[hide="Solution"] There is a $ \frac{1}{p}$ probability of a given number $ n$ being divisible by $ p$.  The probability that two numbers $ a, b$ will not both be divisible by $ p$ is therefore $ 1-\frac{1}{p^{2}}$.

The probability that two numbers $ a, b$ are relatively prime is the probability that they won't both be divisible by any prime $ p$, which approaches

$ P = \prod_{p \in \mathbb{P}}\left( 1-\frac{1}{p^{2}}\right)$

But the Euler product identity tells us that

$ \frac{1}{P}= \zeta(2) = \sum_{k=1}^{\infty}\frac{1}{k^{2}}= \frac{\pi^{2}}{6}$

So $ P = \boxed{ \frac{6}{\pi^{2}}}$. [/hide]


Solve for $ x$
$ \parallel{}|...\parallel{}|x\minus{}1|\minus{}2|\minus{}3|\minus{}...|\minus{}n\plus{}1|\minus{}n|\equal{}0$

$ x\equal{}\pm1\pm2\pm3...\pm n$.
Let $ N\equal{}\frac{n(n\plus{}1)}{2}$. Then $ x\equal{}2k\minus{}N, \ k\equal{}0,1,...N$.


Let $ ABC$ is triangle, let $ BC\equal{}a, CA\equal{}b, AB\equal{}c$. Find place of point $ M$ if: $ \boxed {a.MA^2\plus{}b.MB^2\plus{}c.MC^2\equal{}abc}$.

[quote="thanhnam2902"]Let $ ABC$ is triangle, let $ BC \equal{} a, CA \equal{} b, AB \equal{} c$. Find place of point $ M$ if: $ \boxed {a.MA^2 \plus{} b.MB^2 \plus{} c.MC^2 \equal{} abc}$.[/quote]
We have $ a.IA^2 \plus{} b.IB^2 \plus{} c.IC^2 \equal{} abc$  :D
$ a.MA^2 \plus{} b.MB^2 \plus{} c.MC^2 \equal{} abc$
 $ \Leftrightarrow (a \plus{} b \plus{} c)IM^2 \plus{} 2.\vec{MI}(a.\vec{IA} \plus{} b.\vec{IB} \plus{} c.\vec{IC}) \plus{} a.IA^2 \plus{} b.IB^2 \plus{} c.IC^2 \equal{} abc$
 $ \Leftrightarrow (a \plus{} b \plus{} c).IM^2 \equal{} 0$
 $ \Leftrightarrow I \equiv M$

Thank. It's true.  $ \boxed {a.MA^2 \plus{} b.MB^2 \plus{} c.MC^2 \equal{} abc \Longleftrightarrow I\equiv M}$

Generally, exists the identity (in $ X$) $ \boxed {\ \frac {a\cdot XA^2\plus{}b\cdot XB^2\plus{}c\cdot XC^2}{a\plus{}b\plus{}c}\equal{}XI^2\plus{}2Rr\ }$ .

For $ X: \equal{}I$ obtain the required (particular) relation because $ 4Rpr\equal{}abc$ .


Personally, I like the design of this website... very chic indeed. However, my eyes start to read all the text on the forums as a blur after a while. Is it possible to have an option so that we may change the text size bigger?

I would appreciate that,
IW.OAV

This fall we'll look into offering some alternate layouts.  It's quite a bit of work maintaining one, but we may be able to find a way to offer a couple of others, or at least to make a few things customizable.

It's easier to read, but to me it looks stretched a bit out of proportion. It's probably fine, but I'm rather picky about what fonts I like.

[quote="s0mp"]IW.OAV, if ur mouse has a wheel control+wheel will change the font size[/quote]

I can get the fonts on some of the artofproblemsolving.com pages to resize (using the toolbar font resizing command on Windoze IE) but I can't get the forum messages to resize at all. I am 40-something years old, using eyeglasses for presbyopia, and it would be a BIG HELP to have forum text that really, truly resized according to user-friendly Web standards. I imagine some obnoxious lawyer could even say the site is not sufficiently friendly to the visually impaired if the forum text doesn't resize in a straightforward manner.

For what it's worth, there's a good reason why most sites that try to make money put black text on a white background--for most users that really is the most readable, and thus the most conducive to online sales. Amazon.com, as usual, provides a good example of readable color schemes and resizable text.

This is just friendly advice from a new member who has already recommended this site to friends online and who wants to see this site thrive and succeed.

Can you get the font size to increase on other phpbb boards (like the one you cited in another thread)?  If so, I'll look into how we can get the resizing to work here.  If not, then a it's a phpbb problem that I'm not likely to be able to fix.

We expect to experiment with other color schemes when we go to a dedicated server either this fall or next spring.

The reason why the shift+mouse wheel (or just mouse wheel trick in general) is that this forum is different than just HTML. It is kind of like a different program. (Technically it's not, but it's much easier to understand that way) The words on the left side of the page are images, and those also doesn't resize by the mouse wheel. The only thing the mouse wheel changes is the font size of basic HTML text, and it only makes it large or small to a certain extent.
If you really want big text, you can copy the text onto Microsoft Word, Notepad, Simple Text, or other text editors and enlarge it's font size.

I spent an hour or so tonight messing with the colors - the light on dark is pretty deeply entrenched in this template; it will have to wait until we switch servers.

I can't get the font size to increase on a couple other phpbb boards, so I suspect this is a problem that I'm not going to beat easily, for the reason Tare mentions.  Thus, when we go to the new server I'll create a 'large text' style as well (which will be less maintained than the regular text, but it should be ok).

So there would be two forums? Or do we each customize our settings?

There will be one forum.  There may be a second style available, but only if it's very, very simple for me to maintain.

I know I'm asking too many questions, but what exactly do you "maintain"? And is the maintaining too hard for moderators to help out?

Hmm, that's weird

How come the font changes aren't registering on my screen???

The font change haven't been done yet. It happens when AoPS move to a dedicated server.

[quote="rrusczyk"]I mean making code changes when there's reason to (fundamental changes to the site like site structure changes or database alterations, or little tweaking changes to colors/sizes/etc).  This really isn't something the moderators could help with.  Sorry.[/quote]

Thanks for making the quick fix of having a forum style with white background and black text.  :D  That helps. I used View | Source to find the URL for the forum style sheet, and then looking at the style showed at once what the problem is: hard-coding font sizes. That is ALWAYS a bad usability idea

http://www.useit.com/

and since I am a fan of this site, and realize you are busy with a lot of transitions coming up, I will simply make the friendly suggestion that when you can have all your own forum code on a dedicated server, make SURE that none of the style sheets specify any font sizes (other than, possibly, relative sizes compared to body text default size). Users will see the pages in their own default size, and any user who chooses to will be able to resize at will. These days, pages that are hard-coded to a too-small font size are some of the main annoyances on the Web, out-annoying even the old deprecated <blink>BLINK</blink> from Netscape.

Thanks for what has been fixed already; best wishes on fixing more in the next iteration of rewriting the forum style sheet.

Just as a comment, what tokenadult said about the font sizes is very true - don't hard code that.  That having been said, Mozilla (web browser available for free [url=http://www.mozilla.org/]here[/url]) can change font sizes even when they are hard coded.  Ctrl-+ and Ctrl-- do the trick.  And yes, I just tested it on the board and it works.

is the font size so big that it overlaps, or is it just my computer?

Netscape is also based on Mozilla so the commands confuted gave also work on Netscape.

On the Forum page, near the top right, you can also specify the font size you prefer: small, medium, or large.


calculus
integration
trigonometry
calculus computations

In [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=145922]this topic[/url], the integral

$\int_{0}^{\infty}\frac{e^{-x}\sin x}x\,dx$

was the topic of conversation. Both jmerry and I argued that it is convergent.

Compute its exact value.

Let $I(\alpha)=\int_{0}^{\infty}e^{-\alpha x}\frac{\sin x}{x}dx$.

Then $I'(\alpha)=-\frac{1}{1+\alpha^{2}}$. Hence,

$I(\alpha)=-\arctan \alpha+C$. If $\alpha=0$, we get that $C=\frac{\pi}{2}$.
Therefore we get that

$I(\alpha)=\frac{\pi}{2}-\arctan \alpha=\arctan \frac{1}{\alpha}$.

I hope that there are no mistakes.

Meaning $\frac{\pi}4$ in the case $\alpha=1.$ That was my solution as well.

Slightly different method (but essencially identical to didilica's solution):

\begin{eqnarray*}\int_{0}^{\infty}\sin x\frac{e^{-x}}{x}\, dx & = & \int_{0}^{\infty}\sin x \left( \int_{1}^{\infty}e^{-sx}\, ds \right) \, dx\\ & = & \int_{1}^{\infty}\int_{0}^{\infty}\sin x e^{-sx}\, dx ds \\ & = & \int_{1}^{\infty}\frac{1}{1+s^{2}}ds \\ & = & \frac{\pi}{4}\end{eqnarray*}

Haha, it's equal to sos' solution:

$ \frac{{\sin x}}{x} = \int_0^1 {\cos (ux)\,du} .$ Now construct a double integral as follows and reverse the integration order,

$ \int_0^\infty  {\frac{{\sin x}}
{{xe^x }}\,dx}  = \int_0^\infty  {\int_0^1 {e^{ - x} \cos (ux)\,du} \,dx}  = \int_0^1 {\int_0^\infty  {e^{ - x} \cos (ux)\,dx} \,du} .$

For the inner integral,

\begin{eqnarray*}\int_0^\infty  {e^{ - x} \cos (ux)\,dx}  &=& \operatorname{Re} \int_0^\infty  {e^{ - (1 - iu)x} \,dx}\\&=& \operatorname{Re} \frac{1}{{1 - iu}}\\&=& \frac{1}{{1 + u^2 }}.\end{eqnarray*}

And finally $ \int_0^\infty  {\frac{{\sin x}}{{xe^x }}\,dx}  = \int_0^1 {\frac{1}{{1 + u^2 }}\,du}  = \frac{\pi }{4}.$


function
calculus
integration
derivative
calculus computations

$ y\frac{dy}{dx}$=$ \frac{1}{2}\frac{d(y^2)}{dx}$

Now, I can prove it starting with the RHS and using the chain rule, but can someone do it the other way for me?

What do you mean by "the other way"? It helps to ask a clear question- in this case, I think the one that makes sense is this:
Given a function $ y(x)$, solve the differential equation $ z'(x)\equal{}y(x)y'(x)$.
That's an integration problem, and we have an obvious substitution (which is of course just the chain rule written backwards), giving solutions $ z(x)\equal{}\frac12(y(x))^2\plus{}C$ for constants $ C$.

Not quite what I meant, sorry. Notice that on the RHS, the $ y^2$ is inside the differentiation operator. What I was hoping for was for someone to start from

$ y\frac {dy}{dx}$ and get to $ \frac {1}{2}\frac {d(y^2)}{dx}$

Actually I think I got it:

$ \frac {d}{dx}\left( \int(y*y')\right) \equal{} \frac {d}{dx}\left(\frac {1}{2}y^2\right) \equal{} \frac {1}{2}\frac {d(y^2)}{dx}$

Nothing hard, it was just bugging me because the textbook keeps glossing over intermediate steps.

Thanks


We're trying to find sites that would be good to have provide a link to AoPS.  Mainly, we're looking for sites that might link to AoPS which have traffic from the type of people that would enjoy being part of the AoPS community.  What sites would you recommend we asked to be linked from?  Also, if you know people involved with any of these sites, please ask them to link to us!

Well, I have a link to the site in my personal webpage  :D .

But now seriusly, just a bunch of ideas: There are some math or math-related summer camps that have their own webpages, may be they would be interested to exchange links. Also, potentially many of the Pi Mu Epsilon local pages could be a good place to try. Local Math Competitions and National Math Olympiads would do nice also. There are some small online math related competition (PQRST is one of them) that could be a good place to link to the site.

Best,

I have a link to AoPS on the webpage I made for my web design class. It's on our school's website. Doesn't help much but... :lol:

And does the official MathCounts website have a link to here? If not, it should.

No, MATHCOUNTS hasn't put us in their official link list (we were in the old one) - but it wouldn't hurt to suggest it to them :)


[quote="highlander"]Solve $ 2^x \plus{} 6^x \equal{} 3^x \plus{} 5^x$.[/quote]
Let $ f(a) \equal{} a^x$. Then we have $ f(2)\plus{}f(6) \equal{} f(3)\plus{}f(5)$.
$ f'(a) \equal{} xa^{x\minus{}1}, f''(x) \equal{} x(x\minus{}1)a^{x\minus{}1}$. If $ x \neq 0,1$ then $ f$ is strictly convex or concave on $ [2;6]$ and by Karamata we find that $ f(2)\plus{}f(6) > f(3) \plus{} f(5)$ (convex) or $ f(2)\plus{}f(6) < f(3) \plus{} f(5)$ (concave) since $ (2,6) \succ (3,5)$. So there is just two solutions $ x\equal{}0,1$ which is seen to fit after insertion.


AMC
USA(J)MO
USAMO
AMC 12
AIME
trigonometry
AMC 12 A

To figure out a potential floor, can you all post your index along with your AIME score?

I wish it's below 227.5. think it's probably 224 or 223 where someone got a 144 and 8.

AMC 12 A score: 150 :)
AIME: 10 ;)
index: 250

guess for cutoff: 220~ish

AMC score: 115.5
AIME: 10

So... my index is 215.5

This might be enough to squeak in...

AMC Score: 137
AIME Score: 11

Index: [b]247[/b]

I hope that this is about enough to get in.

Grade: 11
AMC 12A: 123
AIME I: 8
Index: 203

Oh well.

[quote="zanttrang"][quote="pkerichang"]206.5...don't think I'll make it
AIME: 8[/quote]
Are you my clone or something Eric?  :P I got EXACTLY the same AIME score and index.  I definitely don't think that we're going to make USAMO, even with the expansion.  Oh well, there's always one more year to study and prepare (and now I can focus on doing that last round of USAMTS too).[/quote]

Hey, that's cool!!! you're Erik, I'm Eric, we both got 75 on USAMTS and live in CA and now we have the same index...Well, let's get a perfect score for USAMTS together!!! :lol:

Darn it, now I feel stupid.

Grade: 10
AMC 12-A: 115
AIME: 1
Index: 125

Not even close.  Should have been a 3 or 4 on the AIME though, I just way overthought problems like 2 and 4...

What tips do you have for someone like me who wants to get a lot better at these.  I know I can do better than this, but... anyway, what kind of prepartation do you recommend?  I hope to spend at least an hour a day working on stuff over the summer, mainly just 'cus I had a lot of fun practicing for the AIME the last few days...

Darn... last year my index was 257... this year 224... oh wait! 234! phew... i thought it was 224. sigh of relief. I should make it with that i think.

Seems like cutoff will be around 215...other thoughts?

[quote="ThAzN1"][quote="linus9891"]AMC 12A: 144.0
AIME: 11
USAMO INDEX: 254.0[/quote]

what a noob, LINUS :P (jk)

[quote="probability1.01"]AMC 12: 144
AIME: 14
Index: 284

I hope I can squeak by...[/quote]

Aww you beat me  :)

132/15 => 282 Hmm... it's gonna be a tough call[/quote]

Tony.. man.. that hurts my feelings. :D. Congratulations Zhai for OWNING Tony!

[quote="Halluci"]AMC 12A: 139.5
AIME: 10 (meh... lots of stupid mistakes)
Index: 239.5
[/quote]

AMC 12A: 139.5
AIME I: 10 (meh...here's a story:
Number 12, I told myself, hey, $\cos 3x = \cos 4x \cos x + \sin 4x \sin x = A + B$ and $\cos 5x = \cos 4x \cos x - \sin 4x \sin x = A - B$. After manipulating the given equation I have $A^3 = AB^2$. So instead of factoring into $0 = A(A + B)(A - B) = \cos 4x \cos x \cos 3x \cos 5x$ from the get go, I set up the three equations:
$\cos 4x \cos x = 0$
$\cos 4x \cos x = \sin 4x \sin x$
and $\cos 4x \cos x = -\sin 4x \sin x$
where I then proceeded to used double angle formulas and many other ugly things to get to an equation in $\cos x$ but from there my values were ugly and I had to pull out the golden triangle to find all of the crazy angles that produce this cosine. After [b]45 MINUTES[/b] of this pain, I summed my answers and got a number more than 999!!!! I had 2 more than the 6 possible solutions and then when I went back I saw I could've been finished much earlier. Anger, to say the least.

probability1.01 and ThAzN.... yall are beasts.

i took the AIME II last year d343seven, and it was much a little bit harder than this year's AIME I, in my opinion.

I'm not a junior or senior, but...

AMC12: 130
AIME: 8 (really mad about that, I should have had 11..... I'm a victim of stupid mistakes, no matter how hard I try :( )
Index: 210

All of you have really indexes... :(

Do I have any chance with a 213?

I'm expecting 9 on AIME II since I got a 11 on AIME I when I took it for practice, so with 123 AMC that gives me 213.

It seems VERY likely that the index will be between 210 and 219, I'd say on the lower end of that. If you say $\ge 220$, I have a feeling you didn't read the announcement about USAMO expansion.

Anyway I'm a 9th grader:

122 on AMC 12

I took AIME I for practice and got tired and stopped after 1 hour and 40 minutes, having answered 8 questions. I got 2 of them wrong (not stupid mistake but misread problem on both of them!), so 6.

Michiganders are predicting 230's and we definently read the announcement. Maybe we're just pessimistic.

[quote]It seems VERY likely that the index will be between 210 and 219, I'd say on the lower end of that. [/quote]

I'm curious as to why this is so. It certainly seems to be the consensus here - that the floor will be 7 or 8 and the index between 210 and 220. I haven't really understood why.

Anyone willing to explain?

My opinion here (and it's based only on one state, so it's not statistically accurate) is that the index will be quite high - around the same as last year, in the 230+ range and the floor will be a 9. Besides the expansion, any reason for that not to happen?

[quote="tetrahedr0n"]I'm curious as to why this is so. It certainly seems to be the consensus here - that the floor will be 7 or 8 and the index between 210 and 220. I haven't really understood why.
[/quote]

Because of the expansion, statistics from only one or two states will be skewed. You can use the index from last year to follow our reasoning, and because AIME I has far more takers than AIME II, ignore the later for now.

Taking the 233 index, we agree that the expansion pulls us down somewhere between 7 - 13 (some say 5 - 10, some say 10 -15...w/e) points. That leaves you at 226 - 220 already. Then factor in the more difficult AIME - more difficult for the 6 - 10 scorers at least - for another 5 - 10 points. You basically end up between 210 and 221, which is a rather...large...range. Personally, I'm thinking it'll be in the middle of that and placing my bets around the 215-217 range.

As for the index, you basically subtract 140 from that number right there and divide by 10? 7.5 - 7.7. However, the floor will PROBABLY be set by someone with 144/146.5/150 so that gives you about 7. 6 seems irrational from the data we have. 8 would be the case if (and this is possible) the 139.5-150 scorers all got above 7 on the AIME, in which case the floor would be set by someone in the 130 - 138 range.

7 is more likely, but 8's a possibility. And yes, I realize my "evidence" and "data" and both very crude, but using that in combination with common sense you can see where that takes us...

[quote="beta"]I have a feeling that Michiganders are too pessimistic about their improvements.[/quote]

Must be from getting their aases kicked by ohio state so many times.  ;)

[quote="yif man12"]Must be from getting their aases kicked by ohio state so many times. [/quote]

Them's fightin' words ...  :D


Given that $ 2^{37}\minus{}1$ is composite and has a prime factor $ 200<p<300$, find $ p$.

We know that $ 2^{p \minus{} 1} \equiv 1 \bmod p$, so $ 37$ must divide $ p \minus{} 1$, which gives the possibilities $ p \equal{} 297, 223$.  The former is not prime, so the only possibility is that $ p \equal{} \boxed{223}$.


i'm doing a practice test this weekend for a test on monday.... i have done these 2 problems 3 times and cannot get it right.. he gave us the answer key to the practice test...
1st question is "SOLVE THE FORMULA FOR THE SPECIFIED VARIABLE"

A=1/2h(b1+b2)    SOLVE FOR b1  (the 1/2 is a fraction)

2nd question, is to say problem is conditional, contradiction, or identity... i know its conditional by process of elimination... but i cant figure out how to get the answer right...

3X+5(X+1)+2=7-2X

i realize these are probably simple to everyone... but, i gotta start somewhere...

thanks for the help...
tony


quadratics
algebra
polynomial
Vieta
quadratic formula

If $ x_1$ and $ x_2$ are the roots of the equation $ x^2\minus{}3x\plus{}1\equal{}0$ then the sum $ x_1\plus{}x_2$ is:
$ (A)$ $ 1$
$ (B)$ $ 0$
$ (C)$ $ 3$
$ (D)$ $ 2$
$ (E)$ $ 4$

[hide="Answer"]Quadratic formula gives $ x\equal{}\frac {3\pm \sqrt{5}}{2}$

their sum is $ \frac{3\plus{}\sqrt{5}}{2} \plus{} \frac{3\minus{}\sqrt{5}}{2} \equal{} \frac{6}{2}\equal{}3$[/hide]

Use vieta's formula:

For a quadratic equation in the form of ax^2+bx+c=0,
the sum of the roots is -b/a and the product of the roots is c/a.

In this case the equation is x^2-3x+1.
The sum of the roots is -(-3)/1 =3/1=[b]3[/b]
(In this case the coefficient of x^2 is obviously 1)

Answer is C.

$ x1 \plus{} x2 \equal{} \minus{} b/a$

$ \minus{} ( \minus{} 3)/1 \equal{} 3$


See http://www.mathlinks.ro/Forum/viewtopic.php?t=32539 , http://www.mathlinks.ro/Forum/viewtopic.php?t=64497 , http://www.mathlinks.ro/Forum/viewtopic.php?t=1339 .

  darij


Let $ h$ be a positive integer. The sequence $ a_n$ is defined by $ a_0\equal{}1$ and

$ a_{i\plus{}1}\equal{}\frac{a_i}{2}$ if $ a_i$ is even, and $ a_{i\plus{}1}\equal{}a_i\plus{}h$, if $ a_i$ is odd.

For which $ h$ is there $ n>0$ such that $ a_n\equal{}1$?

If $ h$ is even, then never $ a_n\equal{}1$.
Let $ h$ is odd. We can define $ b_0\equal{}1, b_{n\plus{}1}\equal{}b_n/2$ if $ b_n$ even, else $ b_{n\plus{}1}\equal{}(b_n\plus{}h)/2$. All $ b_i$ are some $ a_j$. It equavalent for $ b_n\equal{}a_j$: $ b_{n\plus{}1}\equal{}a_{j\plus{}1}$ if $ a_j$ even and $ b_{n\plus{}1}\equal{}a_{j\plus{}2}$ else.
Obviosly $ b_n\equal{}2^{\minus{}n}\mod h$, therefore $ b_{T_h}\equal{}1$, were $ T_h$ minimal k, suth that $ 2^k\equal{}1\mod h$.

[quote="Rust"]
Obviosly $ b_n \equal{} 2^{ \minus{} n}\mod h$, therefore $ b_{T_h} \equal{} 1$, were $ T_h$ minimal k, suth that $ 2^k \equal{} 1\mod h$.[/quote]

Your writing doesn't seem clear. Please write clearer.


projective geometry
combinatorics unsolved
combinatorics

A group of students have created message lists on the internet. Information regarding the lists are as follows:

• For any two students there is exactly one list which includes both of these students.
• For any two lists, there is exactly one student common in both of these lists.
• The number of students, the number of lists and the number of students in any list are all greater than two.

How many students and lists are there?
Give your answer for the minimum number of lists.

These are the axioms of a projective plane. It's easy to show that the number of students is the same on each list (so call it $n+1$), and each student belongs to $n+1$ lists. There are $n^2+n+1$ students (points) totally, and $n^2+n+1$ lists (lines). The smallest configuration occurs for $n=2$, and it's the [url=http://www.maths.monash.edu.au/~bpolster/fano.html]Fano plane[/url].


inequalities
calculus
inequalities proposed

Given $ a, b, c > 0$ and $ abc\equal{}1$. Prove that: $ \sqrt{a^2\plus{}1}\plus{}\sqrt{b^2\plus{}1}\plus{}\sqrt{c^2\plus{}1} \leq\ \sqrt{2}(a\plus{}b\plus{}c)$
 :)

Case 1: $ a\plus{}b\plus{}c \geq ab\plus{}bc\plus{}ca$ then

$ \sqrt{a^{2}\plus{}1}\plus{}\sqrt{b^{2}\plus{}1}\plus{}\sqrt{c^{2}\plus{}1}\equal{}\sqrt{a^{2}\plus{}abc}\plus{}\sqrt{b^{2}\plus{}abc}\plus{}\sqrt{c^{2}\plus{}abc}\\ \leq (a\plus{}b\plus{}c)(a\plus{}b\plus{}c\plus{}ab\plus{}bc\plus{}ca) \leq 2(a\plus{}b\plus{}c)^2$
Or $ a\plus{}b\plus{}c \geq ab\plus{}bc \plus{}ca$

Case 2: $ ab\plus{}bc\plus{}ca \geq a\plus{}b\plus{}c$
I wait for a solution

[quote="nguoivn"]Given $ a, b, c > 0$ and $ abc \equal{} 1$. Prove that: $ \sqrt {a^2 \plus{} 1} \plus{} \sqrt {b^2 \plus{} 1} \plus{} \sqrt {c^2 \plus{} 1} \leq\ \sqrt {2}(a \plus{} b \plus{} c)$
 :)[/quote]
$ a\equal{}x^{2},b\equal{}y^{2},c\equal{}z^{2}$

$ \Leftrightarrow \sum \frac{x^{4}\minus{}1}{\sqrt{2}x^{2}\plus{}\sqrt{x^{4}\plus{}1}}\geq 0
\Leftrightarrow \sum (x\minus{}1)\frac{(x\plus{}1)(x^{2}\plus{}1)}{\sqrt{2}x^{2}\plus{}\sqrt{x^{4}\plus{}1}}\geq 0$
$ f(x)\equal{}\frac{(x\plus{}1)(x^{2}\plus{}1)}{\sqrt{2}x^{2}\plus{}\sqrt{x^{4}\plus{}1}}$
f'(x)> 0
suppose $ x\geq y\geq z$ and apply ineq Chebyshev :
$ \sum (x\minus{}1)\frac{(x\plus{}1)(x^{2}\plus{}1)}{\sqrt{2}x^{2}\plus{}\sqrt{x^{4}\plus{}1}}\geq \frac{1}{3}(x\plus{}y\plus{}z\minus{}3) \sum \frac{(x\plus{}1)(x^{2}\plus{}1)}{\sqrt{2}x^{2}\plus{}\sqrt{x^{4}\plus{}1}}$
xyz=1 => x+y+z $ \geq$ 3
done !

[quote="iqtimo"][quote="nguoivn"]Given $ a, b, c > 0$ and $ abc \equal{} 1$. Prove that: $ \sqrt {a^2 \plus{} 1} \plus{} \sqrt {b^2 \plus{} 1} \plus{} \sqrt {c^2 \plus{} 1} \leq\ \sqrt {2}(a \plus{} b \plus{} c)$
 :)[/quote]
$ a \equal{} x^{2},b \equal{} y^{2},c \equal{} z^{2}$

$ f(x) \equal{} \frac {(x \plus{} 1)(x^{2} \plus{} 1)}{\sqrt {2}x^{2} \plus{} \sqrt {x^{4} \plus{} 1}}$
f'(x)> 0
[/quote]
Are you sure?  :wink:
Indeed, I don't like to calculus  :oops: . We can solve this old result by only CS and Am-Gm  :)

Wth the stronger, I haven't found any nice proof for it but can_hang found a very nice proof  :oops:
Given $ a, b, c > 0$ and $ abc \equal{} 1$. Prove that:
$ \sqrt {8a^2 \plus{} 1} \plus{} \sqrt {8b^2 \plus{} 1} \plus{} \sqrt {8c^2 \plus{} 1} \leq\ 3(a \plus{} b \plus{} c)$

I found this problem more than interesting!
Let's us consider the following estimation
$ \sqrt {\frac {x^2 \plus{} 1}{2}} \le x \plus{} 1 \minus{} \sqrt {x}$
We are done!
[hide=" @"]Anh Quoc Anh co may bai hay ma cu giau anh em mai  :D  [/hide]

I am pretty sure that his solution is the same as me!  :D
The second problem is quite tough. I use mixing variables but it gives me an ugly solution.

[quote="great math"]I use mixing variables but it gives me an ugly solution.[/quote]
It's ugly but very effective.  :wink:

[quote="nguoivn"]Given $ x, y, z > 0$ and $ xyz \equal{} 1$. Prove that: $ \sqrt {x^2 \plus{} 1} \plus{} \sqrt {y^2 \plus{} 1} \plus{} \sqrt {z^2 \plus{} 1} \leq\ \sqrt {2}(x \plus{} y \plus{} z)$
 :)[/quote]
[quote="Agr_94_Math"]Can Mr. Nguoivn post his proof using CS and AM GM?[/quote]
The first, let $ x \equal{} \frac {a^2}{bc}$,...
We need to prove: $ \sum\ a\sqrt {a^4 \plus{} b^2c^2} \leq\ \sqrt {2}(a^3 \plus{} b^3 \plus{} c^3)$
Using CS, wee only need to prove: $ 2(a^3 \plus{} b^3 \plus{} c^3)^2 \geq\ (a \plus{} b \plus{} c)(\sum\ a^5 \plus{} abc\sum\ ab)$
By AM-Gm easily to see that $ 3\sum\ a^3b^3 \geq\ abc(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)$
So, we only need to prove: $ \sum\ a^6 \plus{} \sum\ a^3b^3 \geq\ \sum\ ab(a^4 \plus{} b^4)$
<=> $ \sum\ [(a^2 \plus{} ab \plus{} b^2)^2 \minus{} 2ab(a \plus{} b)^2](a \minus{} b)^2 \geq\ 0$
By Am-Gm, easily to see that $ Sa, Sb, Sc > 0$. We have done  :)


calculus
function
analytic geometry
graphing lines
slope
rate problems

How many hours does it take a train traveling at an average rate of $ 40$ mph between stops to travel $ a$ miles it makes $ n$ stops of $ m$ minutes each?

$ \textbf{(A)}\ \frac{3a\plus{}2mn}{120} \qquad
\textbf{(B)}\ 3a\plus{}2mn \qquad
\textbf{(C)}\ \frac{3a\plus{}2mn}{12} \qquad
\textbf{(D)}\ \frac{a\plus{}mn}{40} \qquad
\textbf{(E)}\ \frac{a\plus{}40mn}{40}$

[quote="ckck"][quote="AIME15"]How many hours does it take a train traveling at an average rate of $ 40$ mph between stops to travel $ a$ miles it makes $ n$ stops of $ m$ minutes each?

$ \textbf{(A)}\ \frac {3a \plus{} 2mn}{120} \qquad \textbf{(B)}\ 3a \plus{} 2mn \qquad \textbf{(C)}\ \frac {3a \plus{} 2mn}{12} \qquad \textbf{(D)}\ \frac {a \plus{} mn}{40} \qquad \textbf{(E)}\ \frac {a \plus{} 40mn}{40}$[/quote]
[hide]Using $ D\equal{}rt$, we have $ a\equal{}40t$ so $ t\equal{}\frac{a}{40}$. But $ n$ stops of $ m$ minutes each results in added time of $ mn$. So $ \frac{a}{40}\plus{}mn\equal{}\frac{a\plus{}40mn}{40}$ E[/hide][/quote]
mn is the time in minutes. You need to divide it by 60

[hide="How to cheat"]If you have the choices, plug in some simple values to eliminate options.

$a=40, m=0, n=0$ should give an answer of $1$ hour.  Only $A, D, E$ give this answer.

$a=20, m=1, n=30$ should also give an answer of $1$ hour - you spend half an hour going $20$ miles, and stop for the other half hour.  Only $\boxed{A}$ gives this answer.[/hide]

[hide="I can't believe it's not multivariable calculus"]Holding all else constant, increasing $a$ by $40$ should increase our function by $1$.  You can think of this as the "slope" of the function with respect to $a$, and the slope is $\frac{1}{40}$.  Thus, the function should have $\frac{a}{40}$ in it.

Holding all else constant, increasing $m$ by $60$ should increase our function by $n$, the number of stops.  Thus, the "slope" of the function with resepect to $m$ should be $\frac{n}{60}$.  Thus, the function should have a $m\frac{n}{60}$ in it.

Summing those two gives answer $\boxed{A}$ when you get a common denominator.

Don't get too wrapped up in what I mean by "slope".  The graph is sort of a 4-dimensional graph because there are four variables.  There are three separate "slopes", if you consider time to be the independent variable and $a, m, n$ to be the three dependent variables.  (There are even more slopes if you allow time to vary.)  It will all become very clear in multivariable calculus.[/hide]

Sorry for reviving but here is the [hide="Easier Way"]The time for the trip part is distance over speed which equals $\dfrac{a}{40}$

The time for the stops are the number of stops times time of each stop in hours which is $\dfrac{mn}{60}$

So total time is $\dfrac{a}{40} + \dfrac{mn}{60}$ which equals $\frac{3a+2mn}{120}$ so the answer is $ \boxed{A} $[/hide]


abstract algebra
induction
superior algebra
superior algebra theorems

I am trying to answer exercises on my book, but there are no solutions for the proof part. I hope you can help me. This is just a practice for the incoming exam.

1.) Prove that the order of $S_{n}$ is $n!$.
2.) Show that S1 and S2 are cyclic groups.
3.) Show that for $n \geq  4$, A_n is non abelian and noncyclic.

Thanks

[quote="guile"]

1.) Prove that the order of $S_{n}$ is $n!$.[/quote]
Induction on $n$.
[quote]
2.) Show that S1 and S2 are cyclic groups.
[/quote]

Check direct!

[quote]
3.) Show that for $n \geq 4$, A_n is non abelian and noncyclic.
[/quote]
We only need prove $A_{4}$ nonabelian.

P.S: if non abelian then  noncyclic.

Look at the elements of $A_{4}$ and make an educated guess...
$(123) (134) = (234)$
$(134) (123) = (124)$

It's not very hard, you know.

Yeah I know that, but I was trying to ask is the general proof. That is just an single example which is $A_{4}$. What about the general term $A_{n}$ given the abovementioned condition?

That example works in $A_{n}$ for $n \geq 4$  :dry: