$ (I) $ is the incircle of the triangle $ ABC $ . $ (O) $ is the circumcircle of the triangle $ ABC $ . $ (O_1) $ internally tangents $ (O) $ at $ D $ and tangents $ AB,AC $ . $ (O_2) $ internally tangents $ (O) $ at $ E $ and tangents $ AB,BC $ . $ (O_3) $ internally tangents $ (O) $ at $ F $ and tangents $ CA,CB $ . Prove that : $ AD , BE , CF $ are concurrent .


Given an integer N.
Which pairs of integers such that their sum is equal to N and the second number results from the first one by striking out one of its digits.

Hello 7erry,

[quote="7erry"][quote]2.3) For any n > 0 such that N >= 10^n, we have exactly two possibilities to write N as 10^n x + 2B with B < 10^n :
N = 10^n [N/10^n] + 2*(Mod(N,10^n)/2)
N = 10^n ([N/10^n]-1) + 2*((10^n + Mod(N,10^n))/2) [/quote]
How can you claim that there is not any possibility else? :)[/quote]

I don't understand your question.
If I want to write an even positive integer N as 10^n x + 2B with B < 10^n, n given in N*, I have only two cases :

1) B < 10^n /2 ==> 2B < 10^n ==> 2B = mod(N, 10^n) and x = [N/10^n]
2) 10^n/2 <= B < 10^n ==> 10^n <= 2B < 2*10^n ==> 0 <= 2B - 10^n < 10^n
So 10^n x + 2B = 10^n(x+1) + 2B - 10^n and I have just one other solution : 2B - 10^n = mod(N,10^n) and x = [N/10^n]-1

Is this the answer you're asking for ?

--
Patrick


I have the problem

Find all primes $ p$ and $ q$ such that $ p \plus{} q \equal{} (p \minus{} q)^3$.

The solution manual says $ p \minus{} q\equiv 2p\mod (p \plus{} q)$.
This is like step 1. How'd they get there

$ p\plus{}q\equiv0\pmod{p\plus{}q}$

$ \Rightarrow \minus{}q\equiv p\pmod{p\plus{}q}$

$ \Rightarrow p\minus{}q\equiv 2p\pmod{p\plus{}q}$


I recently came to know about MOEMS and would like to approach my child's elementary school to start a MOEMS groups. Any pointers on how to approach school?
Do people who become PICO ask for some re-embursement from school or is it done totally voluntarily?
Thanks
Gupta

Best way is to sit down with the principal and/or the head of the Parents Association. Bring with you an Informational Packet (available free from MOEMS at info@moems.org/866-781-2411). Most schools run and fund their teams, but in some cases, parents do it. The Parents Association might help. Reimbursement varies from school to school.

thankyou. I am going to contact the PTA president.

i have had this same problem...
thanks so much for going ahead and asking it! i almost forgot to  :lol:


Prove that any elementary row [column] operation of type 1 can be obtained by a succession of three elementulary row [column] operations of type 3 followed by one elementary row [column] operation of type 2.

Type 1: Switch any two rows/columns
Type 2: Multiply any row/column by a non-zero scalar
Type 3: Add a scalar multiple of one row/column to another

The single type 2 operation must be multiplication of a row by the scalar $ \minus{}1.$ How do we know this? Just track the effect of elementary row operations on determinants. Type 3 operations leave determinants unchanged. Type 1 operations multiply determinants by $ \minus{}1.$ Type 2 operations multiply determinants by the scalar in question.

To prove that this combination of 4 operations yields the elementary matrix in question, would it suffice to find 4 such operations that yield a type 1 matrix?  If so, how could I go about finding them?  If not, what would suffice?

Here's the process at work in a $ 2\times 2$ matrix. (1) Subtract the first row from the second; (2) add the second row to the first; (3) subtract the first row from the second; (4) multiply the second row by $ \minus{} 1.$

$ \begin{bmatrix}1 & 0 \\
0 & 1\end{bmatrix}\to \begin{bmatrix}1 & 0 \\
\minus{} 1 & 1\end{bmatrix}\to \begin{bmatrix}0 & 1 \\
\minus{} 1 & 1\end{bmatrix}\to \begin{bmatrix}0 & 1 \\
\minus{} 1 & 0\end{bmatrix}\to \begin{bmatrix}0 & 1 \\
1 & 0\end{bmatrix}.$

But the process is completely generic. In any matrix with two or more rows, pick two of those rows, and act as I did above on the first and second rows.

What we learn from this is that we don't need to have three types of elementary row operations; we can omit type 1 and  have the other two types suffice for all uses in proofs. (Logically sufficient, but inefficient.)

That's actually the same method I came up with and I generalized it for any $ m\times n$.  Thanks.


A typical ID number in a college class consists of a four-digit number, such as 0119. When an ID number is randomly selected, what is the probability that no two of its digits are the same? Express your answer as a decimal to the nearest thousandths.

We calculate the number of ID's in which no two digits are the same.  There are $ 10$ choices for the first digit, $ 9$ for the second, $ 8$ for the third, and $ 7$ for the fourth (for those familiar with this notation, we could just do $ _{10}P_4$).  Thus, there are $ 10 \times 9 \times 8 \times 7\equal{}5040$ ways to pick an ID with no two digits that are the same.  There are $ 10^4\equal{}10000$ total possible ID's, so the probability that no two digits are the same is $ \frac{5040}{10000}\equal{}\boxed{0.504}$.


Has anyone going to CA received any emails about going (besides the acceptance letter)?  I'm worried that I'm missing something...

yeah same, i haven't gotten anything.  i was sort of hoping for maybe course info, or confirmation that they received my info forms, but none so far.

I'm also worried by the lack of receipt of course choices. Hm...AND I'm going to be at another camp, meaning if AMSP mails the course stuff, my parents will have to somehow get them to me. Hrm.

I've also noticed a lack of information concerning where in UCSC to go to check in, get stuff, live for three weeks, etc. If that also comes in the mail, that would also pose...um...problems.


calculus integration quadratics symmetry

Find all integral solutions to: $ x^2\plus{}y^2\plus{}z^2\equal{}xyz.$

What is the method for tackling these kids of problems (because obviously you can complete the square or anything) and what is the answer to this specific question?

I got a 2.
x=y=z=3
x=y=z=0
for some reason I feel that there are no more solutions.

Is there any actual proof that there are no more solutions?

well it is not that hard if you think about it.
you know it is imposible to solve this equation if the values are different for $ x,y,z$ with one solution by using algebra so we suppose  $ x \equal{} y \equal{} z$

so we can turn the equation to $ x^2$+$ x^2$+$ x^2$=$ x^3$
well because there are three terms

so that would be $ 3x^2$=$ x^3$

then you divide both sidesby $ x^2$

well since this is a quadratic we have two solutions.now we factor it. $ x^2(3 \minus{} x)$

so one solution is $ 0$ and the other solution is $ 3$ because $ x \minus{} 3$=$ 0$

$ x \equal{} (3,0)$

another way you can get three is by dividing

$ \frac {3x^2}{x^2}$=$ \frac {x^3}{x^2}$

$ 3 \equal{} x$

so we know that $ 3$=$ x$

[quote="calculatorwiz"]well it is not that hard if you think about it.
you know it is impossible to solve this equation if the values are different for $ x,y,z$ with one solution by using algebra so we suppose  $ x \equal{} y \equal{} z$
[/quote]

But thats pretty intuitive isnt it? Is there like any proof that $ x\equal{}y\equal{}z$?

Also, calculatorwiz, by dividing by $ x^{2}$, you leave out solutions.
When you get to $ 3x^{2} \equal{} x^{3}$, you have to bring everything equal to 0 and then solve.
$ x^{3} \minus{} 3x^{2} \equal{} 0$
$ x^{2}(x \minus{} 3) \equal{} 0$
$ x \equal{} 0,3$

However, I'm also wondering how you can assume that $ x \equal{} y \equal{} z$. How do you find an answer if this is not true?

well you can asume $ x\equal{}y\equal{}z$

because there are no systems of equations given

you can not find all three values from one equation

[quote="calculatorwiz"]
you can not find all three values from one equation[/quote]

Thats a good point. However, say we were trying to find all integral solutions to $ 3x^{2} \plus{} y^{2} \minus{} 4y \minus{} 17 \equal{} 0$. Well we could turn that into $ 3x^{2} \plus{} (y \minus{} 2)^{2} \equal{} 21$, and from there its very easy to check all the cases and find all the solutions even if there is no system.

However, I do see your point. Since there is no constant, we cant solve this problem without assuming either $ x \equal{} y \equal{} z$ or $ x \equal{} y \neq z$.
since we already considered the former, I will do the latter.

$ x \equal{} y \neq z$. Then $ 2x^2 \plus{} z^2 \equal{} x^2z$.
$ z^2 \equal{} x^2(z \minus{} 2)$
$ \frac {z^2}{z \minus{} 2} \equal{} x^2$
we easily see that $ z \equal{} 6$, $ x \equal{} y \equal{} 3$ is a solution, as are its permutations because of symmetry-(3,3,6),(3,6,3),(6,3,3). So now I will kind of guess that the only solution to $ \frac {z^2}{z \minus{} 2} \equal{} x^2$ is (3,6).

So if anyone can prove that the only solution to $ \frac {z^2}{z \minus{} 2} \equal{} x^2$ is (3,6) we will be done, right?

If that is true, the only solutions are $ \boxed{(3,3,3), (0,0,0), (3,3,6), (3,6,3), (6,3,3)}$

Don't forget about $ (\minus{}3,\minus{}3,3), (\minus{}3,\minus{}3,6), (\minus{}3,3,\minus{}6)$, and all of the permutations of those as well...

[quote="calculatorwiz"]well you can asume $ x \equal{} y \equal{} z$

because there are no systems of equations given

you can not find all three values from one equation[/quote]
That is not true... since it is a Diophantine Equations it has many different values

well how do you find all the values did you just guess because you are not showing haw you gat the answers


solve in nonnegative integers the equation
$ 2^x\minus{}1\equal{}xy$

It 's similar to the problem
Find x such that $ 2^x \minus{} 1$ divided by x.
Let p is the smallest prime divisors of x.(1)
By little Fermat theorem we have:
      $ 2^{p\minus{}1} \minus{}1$ divided by p
and $ 2^x \minus{}1$ divided by p
because of (1),we have gcd(x,p-1)=1
so $ 2^{1} \minus{} 1$ divided by p.
Contraction.
So x don't have any prime divisors.
x = 1.
and y=1 ,too.
the root of the equation is (1,1)


geometry
geometric transformation
reflection
circumcircle
geometry proposed

Let $A',B',C'$ be the reflections of the vertices $A,B,C$ in the sides $BC,CA,AB$ respectively. Let $O$ be the circumcenter of $ABC$. Show that the circles $(AOA'),(BOB'),(COC')$ concur again in a point $P$, which is the inverse in the circumcircle of the isogonal conjugate of the nine-point center.

It shouldn't be difficult, but I found it cute :).

May be $A', B', C'$ --- reflections of point $O$? Then the solution:
Let $O_A$ --- the circumcenter of triangle $BOC$, $H_A$ --- projection the point $A$ on $BC$, $N$ --- the nine-point center, $N'$ --- isogonal conjugate of this and $M_C, M_B$ --- midpoints of segments $AB, AC$ respectively. Made the composition of inversion in point $A$ with radius is $\sqrt{{AB\cdot AC}\over 2}$ and reflection of the  $\angle BAC$ bisector. Then $M_C\rightarrow C, M_B\rightarrow B, O\rightarrow H_A$ (since $2Rh_a=AB\cdot AC$). Then line $AN\rightarrow AO_A$. Made the inversion with center $O$ and radius $R$. Then $O_A\rightarrow A'$, line $AO_A\rightarrow$ circle$(AOA')$, $N'\rightarrow P$.
Else this problem is wrong.

I assure you, the problem is not wrong. In fact, [url=http://mathworld.wolfram.com/MusselmansTheorem.html]here[/url]'s a link. $A',B',C'$ are, as stated, the reflections of $A,B,C$ in $BC,CA,AB$ respectively.

I just downloaded the article.  And it is by treegoner!

Sorry, the problem is true. Really, the points $A$, $O$, $A'$, reflection $O$ of $BC$ concircle, hence these facts are equivalent.

Dear Mathlinkers,
you will see on my website (http:/perso.orange/jl.ayme) article 10 Le point de Kostina... a synthetic proof of the Musselman's circle and of the isogonal of the nine point center.
Sincerely
Jean-Louis


please do not post the same question twice.

jamesstine, Someone have solved it for you!  [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=149331[/url]

[quote="jamesstine"]5a-12               3a-2
_________ - ___________
a^2-8a+15      a^2-8a+15

I made a mistake on the first denominator.[/quote]

Sorry we should have looked into your problem.

Your problem is $\frac{5a-12}{a^{2}-8a+15}-\frac{3a-2}{a^{2}-8a+15}$, isn't it? :)

You can correct by pushing Edit button.

Since there are common denominators it should just be:
\begin{eqnarray*}\frac{(5a-12)-(3a-2)}{a^{2}-8a+15}&=& \\ &=& \boxed{\frac{2a-10}{(a-5)(a-3)}}\end{eqnarray*}

Edit: (2a-1)=(2a-10) i can't type well =P

[quote="jamesstine"]I see how you got the denominator but I don't understand how you got (2a-1)[/quote]
Should be $2a-10$.

[quote="hunter34"]Since there are common denominators it should just be:
\begin{eqnarray*}\frac{(5a-12)-(3a-2)}{a^{2}-8a+15}&=& \\ &=& \boxed{\frac{2a-10}{(a-5)(a-3)}}\end{eqnarray*}
[/quote]

But $\frac{2a-10}{(a-5)(a-3)}= \frac{2(a-5)}{(a-5)(a-3)}= \frac{2}{a-3}$ so this is the simplest form.


integration
calculus
derivative
algebra
polynomial
symmetry
real analysis

I'm only going to work part of it, but enough to know that the formula is correct.

Assume $a=-h,\,b=h.$ The general case can be turned into this by translation. Assume that $f$ has as many continuous derivatives as we're going to use below in applying Taylor's theorem. The formula becomes:

$\int_{-h}^hf(x)\,dx=2hf(0)+\frac{h^2}3\left[f'\left(\frac h2\right)-f'\left(-\frac h2\right) \right] +\frac{h^5}{360}f^{[4]}(c)$

Replace $f(x)$ by its Taylor polynomial centered at zero. By the symmetry of the interval of integration, all of the odd terms of the Taylor polynomial drop out of the result. We get:

$\int_{-h}^hf(x)\,dx=2hf(0)+\frac{h^3}3f''(0)+\frac{h^5}{60}f^{[4]}(0)+O(h^7)$

Now look at the derivative:

$f'\left(\frac h2\right)=f'(0)+\frac h2f''(0)+\frac{h^2}8f'''(0)+\frac{h^3}{8\cdot 3!}f^{[4]} (0)+ch^4+O(h^5)$

$f'\left(-\frac h2\right)=f'(0)-\frac h2f''(0)+\frac{h^2}8f'''(0)-\frac{h^3}{8\cdot 3!} f^{[4]} (0)+ch^4+O(h^5)$

Subtract to get

$f'\left(\frac h2\right)-f'\left(-\frac h2\right)= hf''(0)+\frac{h^3}{24}f^{[4]}(0)+O(h^5),$ or

$\frac{h^2}3\left[f'\left(\frac h2\right)-f'\left(-\frac h2\right)\right]= \frac{h^3}3f''(0)+\frac{h^5}{72}f^{[4]}(0)+O(h^7)$

Subtracting what we have, we get

$\int_{-h}^hf(x)\,dx-\left[2hf(0)+\frac{h^2}3\left[f'\left(\frac h2\right)-f'\left(-\frac h2\right) \right]\right]$

$=\left(\frac1{60}-\frac1{72}\right)h^5f^{[4]}(0)+O(h^7)=\frac{h^5}{360}f^{[4]}(0) +O(h^7).$

Technically, this doesn't quite prove the mean-value form that Cezar Lupu wrote it in, but once we're this far, we know it's going to be true.

I do have doubts about the utility of a quadrature formula that involves evaluating a derivative.

[quote="Kent Merryfield"]The formula

$\int_{-h}^hf(x)\,dx\approx 2hf(0)+\frac{h^2}6\sqrt{\frac{10}3}\left[f'\left(\sqrt{\frac3{10}}\,h\right)- f'\left(-\sqrt{\frac3{10}}\,h\right) \right]$

should have even higher precision, assuming I did the algebra correctly.[/quote]

I like this formula, but I would like to see the remainder term "clean" and its derivative. ;)

The remainder term in the formula in #3 should be $Ch^7f^{[6]}(c)$ for some constant $C$ and some point $c\in(-h,h).$

I don't have the time or energy to figure out what the constant $C$ should be.


How do you install packages into TexShop? :help:
Also, where can you obtain packages? :help:

Please reply.

[quote="stevem"]TeXShop is just an editor so you don't install packages into it. You must first install LaTeX then you can add more packages to your LaTeX installation if you need them. You need to tell TeXShop where to find your latex files.

You install LaTeX on the Mac by using [url=http://www.tug.org/mactex/]MacTeX[/url]. There are full instructions on how to download and install there. The large version is an enormous 1.2GB download which means you will probably never have to look for extra packages, though there is a facility to do so built into the system.[/quote]
I already have TexShop installed, but I need to use the fancyhdr package and I don't know how.  I think I explained myself wrong.  I installed basictex instead of the full Mactex package if that makes a difference.  (BasicTex is a smaller version of MacTex.)

BasicTeX includes the TeX Live Manager which allows you to download and install packages. If you don't already have it, download [url=http://code.google.com/p/mactlmgr/]TeX Live Utility[/url] which is a "Mac OS X graphical interface for TeX Live Manager" that will help you to install packages without having to use the command line.

This application only works with Leopard (10.5.5).  I only have 10.4.11, so is there a similar application that works on my computer?  If not where can I find the code to run in Terminal to load packages?
 :help:

[url=http://www.tug.org/texlive/tlmgr.html]tlmgr - TeX Live package manager[/url]

I don't understand. Is this an application, or instructions for loading packages via command line?  If it's the latter, is this done in Terminal?  How?  If it's the former, how do you obtain it?

You need to type commands in Terminal. There are better instructions written especially for the Mac at [url=http://mactex-wiki.tug.org/wiki/index.php?title=TeX_Live_Manager]TeX Live Manager[/url].
There are 3 steps you need to follow there:
1. Update the manager [b]First: Update tlmgr[/b]
2. Tell it where to get the packages [b]Second: Set Your Server[/b]
You only need to do these 2 steps once. After that it is easy to install packages.
3. Install fancyhdr by typing into the Terminal [code]sudo tlmgr install fancyhdr[/code]


ok just found this nearly 16 year old thread lol
bumping this for no reason
also adventure10 upvoted the posts meaning that they saw it


Triangle ABC is acute-angled.Angle B is less than angle C.O is circumcentre.AO(extended) cuts BC at D.

The circumcentres of the triangles ABD,ACD are E and F respectively.

The sides BA, CA are extended beyond A to points G and H respectively  such that AG=AC, and AH=AB.

Prove that quadrilateral EFGH is a rectangle iff C-B =60 degrees

Is this problem from the shortlist of IMO2004?

You mean this problem? [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=205737#p205737]www.mathlinks.ro/Forum/viewtopic.php?p=205737#p205737[/url]


I was doing a math homework in LaTex, but when I type spanish with the acent mark like "factorización" the letter "ó" is omitted after compiled into pdf.
 How can I solve this problem? Do I need to include extra package?

Put \usepackage[latin1]{inputenc} in your preamble (ie after \documentclass and before \begin{document}). This tells LaTeX that you want to use special characters from Latin languages ie languages in Western Europe.


trigonometry
inequalities
function
geometry proposed
geometry

[u][b]The author of this posting is : darkmaster[/b][/u]
____________________________________________________________________

A triangle ABC has the property that $l_a$, $m_b$, $h_c$ are concurrent; $l_b$, $m_c$, $h_a$ are concurrent; $l_c$, $m_a$, $h_b$ are concurrent.
Prove that $\frac{\sin C}{\cos B} + \frac{\sin A}{\cos C} + \frac{\sin B}{\cos A} \geq 3\sqrt3$.

[u][b]The author of this posting is : Myth[/b][/u]
____________________________________________________________________

It is clear that $ABC$ is an acute triangle (since its altitudes lie inside $ABC$).
From Ceva theorem for $l_a$, $m_b$ and $h_c$ we obtain $\frac{\sin C}{\cos B} = \tan A$, and so on.
Therefore, LHS = $\tan A + \tan B + \tan C \geq 3\sqrt 3$ due to Jensen (the function $\tan x$ is convex on $\left[0,\pi/2\right)$).

[quote="darkmaster"]A triangle ABC has the property that $l_a$, $m_b$, $h_c$ are concurrent; $l_b$, $m_c$, $h_a$ are concurrent; $l_c$, $m_a$, $h_b$ are concurrent.
Prove that $\frac{\sin C}{\cos B} + \frac{\sin A}{\cos C} + \frac{\sin B}{\cos A} \geq 3\sqrt3$.[/quote]

I have a stronger assertion: Prove that $\frac{\sin C}{\cos B} + \frac{\sin A}{\cos C} + \frac{\sin B}{\cos A} = 3\sqrt3$.

In other words, prove that if we are given a triangle ABC such that $l_a$, $m_b$, $h_c$ are concurrent, that $l_b$, $m_c$, $h_a$ are concurrent, and that $l_c$, $m_a$, $h_b$ are concurrent, then this triangle ABC is equilateral.

BTW, Myth, I expected your answer to be "This is trivial :D "...

  Darij

[u][b]The author of this posting is : Myth[/b][/u]
____________________________________________________________________

[quote="darij grinberg"]In other words, prove that if we are given a triangle ABC such that $l_a$, $m_b$, $h_c$ are concurrent, that $l_b$, $m_c$, $h_a$ are concurrent, and that $l_c$, $m_a$, $h_b$ are concurrent, then this triangle ABC is equilateral.[/quote]

This is trivial.

[quote="darij grinberg"]BTW, Myth, I expected your answer to be "This is trivial :D "...[/quote]

I don't like such jokes.

[u][b]The author of this posting is : darkmaster[/b][/u]
____________________________________________________________________

[quote="darij grinberg"]In other words, prove that if we are given a triangle ABC such that $l_a$, $m_b$, $h_c$ are concurrent, that $l_b$, $m_c$, $h_a$ are concurrent, and that $l_c$, $m_a$, $h_b$ are concurrent, then this triangle ABC is equilateral.

BTW, Myth, I expected your answer to be "This is trivial :D "...

  Darij[/quote]

This "trivival" problem seems to be too touch for me. Can you give me a hint in proving this triangle ABC is equilateral. I have tried to use Ceva and the only thing I get is that " if $l_a$, $m_b$, $h_c$ are concurrent, $l_b$, $m_c$, $h_a$ are concurrent, then $l_c$, $m_a$, $h_b$ are also concurrent, and get the inequality"

[u][b]The author of this posting is : Myth[/b][/u]
____________________________________________________________________

Suppose contrary, and let $b$ be a medium side. Then $l_a$, $m_b$ and $h_c$ can't intersect in the same point.

[quote="darkmaster"]This "trivival" problem seems to be too touch for me. Can you give me a hint in proving this triangle ABC is equilateral. I have tried to use Ceva and the only thing I get is that " if $l_a$, $m_b$, $h_c$ are concurrent, $l_b$, $m_c$, $h_a$ are concurrent, then $l_c$, $m_a$, $h_b$ are also concurrent, and get the inequality"[/quote]

Well, it isn't trivial... but it isn't hard either.

At first, let's prove your assertion that

[b](1)[/b] If the lines $l_a$, $m_b$, $h_c$ are concurrent, and the lines $l_b$, $m_c$, $h_a$ are concurrent, then the lines $l_c$, $m_a$, $h_b$ are also concurrent.

In fact, by Ceva, the lines $l_a$, $m_b$, $h_c$ are concurrent if and only if $\frac{\sin C}{\sin A} = \frac{\cos B}{\cos A}$. Similarly, the lines $l_b$, $m_c$, $h_a$ are concurrent if and only if $\frac{\sin A}{\sin B} = \frac{\cos C}{\cos B}$, and the lines $l_c$, $m_a$, $h_b$ are concurrent if and only if $\frac{\sin B}{\sin C} = \frac{\cos A}{\cos C}$. Now, in order to prove [b](1)[/b], it is enough to show that the equations  $\frac{\sin C}{\sin A} = \frac{\cos B}{\cos A}$ and $\frac{\sin A}{\sin B} = \frac{\cos C}{\cos B}$ imply $\frac{\sin B}{\sin C} = \frac{\cos A}{\cos C}$. But this is clear, since $\frac{\sin B}{\sin C} = \left(\frac{\sin C}{\sin A} \cdot \frac{\sin A}{\sin B}\right)^{-1}$ and $\frac{\cos A}{\cos C} = \left(\frac{\cos B}{\cos A} \cdot \frac{\cos C}{\cos B}\right)^{-1}$. Thus, [b](1)[/b] is proven.

Now, we have to show that under the conditions of [b](1)[/b], the triangle ABC is equilateral. Well, since these conditions are cyclic (although not symmetric), we can assume WLOG that the angle A is the smallest among the angles A, B, C. This means $A\leq B$ and $A\leq C$. Now, as Myth showed, the triangle ABC must be acute-angled, and since the function sin is increasing and the function cos is decreasing on [0; 90], it follows that $\sin A\leq \sin C$, while $\cos A\geq \cos B$.

But we know that $\frac{\sin C}{\sin A} = \frac{\cos B}{\cos A}$. This immediately contradicts $\sin A\leq \sin C$ and $\cos A\geq \cos B$ except for the case when sin A = sin C and cos A = cos B. Thus, A = B = C, and triangle ABC is equilateral.

  Darij

[u][b]The author of this posting is : Myth[/b][/u]
____________________________________________________________________

Hmm... Actually, solution is really trivial and doesn't use any calculations or trigonometry.

You're right. It's even easier than I thought...

  Darij


Quick Question:

When you have the $ |3x\plus{}3|\ge0$. Then would it be that $ 3x\plus{}3\ge0$ and $ \minus{}(3x\plus{}3)\ge0$

or would it be $ 3x\plus{}3\ge0$ and $ \minus{}(3x\plus{}3)\le0$

..or something else entirely (i think its the second one i mentioned

Well, it's actually both...all absolute values of real numbers are nonnegative. :P

But if you have something like $ |3x \plus{} 3|\ge5$, then the two relevant inequalities are $ 3x \plus{} 3\ge5$ and $ \minus{} (3x \plus{} 3)\le5$ (think "it has to be outside the range $ ( \minus{} 5,5)$")

If you had $ |3x \plus{} 3|\le5$, then you would just get $ \minus{} 5\le3x \plus{} 3\le5$ (think "it has to be in the range $ [ \minus{} 5,5]$").

@Z-coli: Thanks for expanding. But you end up memorizing anyway. :P

Well, for things like this I don't bother memorizing.  You just have to understand why it works.
Let's say you have $ |x| \equal{} 5$.  You know that $ x \equal{} 5, \minus{} 5$.  Now, let's say you have $ |x| < 5$.
If $ x$ is positive, it has to be less than $ 5$, or if it's negative,
it has to be greater than $ \minus{} 5$ (If it is less than $ \minus{} 5$, it's absolute value will be greater
than $ 5$).  So, you know that either $ \minus{} 5 < x < 5$.  The same reasoning goes for this problem.


floor function
combinatorics proposed
combinatorics

A rectangular table $ 9$ rows $ \times$ $ 2008$ columns is fulfilled with numbers $ 1$, $ 2$, ...,$ 2008$ in a such way that each number appears exactly $ 9$ times in table and difference between any two numbers from same column is not greater than $ 3$. What is maximum value of minimum sum in column (with minimal sum)?

Could someone help me with this problem ? I don`t know how start solving !!!
(sorry for my bad English)

:wink:  I know that Dirichlet Principle hepls but How ???

We will denote by $ \sigma(c)$ the sum of the entries of column $ c$. We claim $ \max \min \sigma(c) \equal{} 24$, realized for example by the table with three columns equal to $ (1,1,1,3,3,3,4,4,4)$ (having $ \sigma(c) \equal{} 24$), one $ (2,2,2,2,2,5,5,5,5)$, one $ (2,2,2,2,5,5,5,5,5)$, and the rest $ (n,n,n,n,n,n,n,n,n)$, for $ 6\leq n\leq N\equal{}2008$.

The $ k$ columns contaiming an entry equal to $ 1$ can only have their entries from the set $ \{1,2,3,4\}$. Since there are only $ 9\cdot 4$ such values available, it means that $ 9k \leq 9\cdot 4$, hence $ k\leq 4$.

If $ k\equal{}1$, that column has $ \sigma(c_1) \equal{} 9\cdot 1 < 24$;

If $ k\equal{}2$, those columns have $ \sigma(c_1) \plus{} \sigma(c_2) \leq 9\cdot 1 \plus{} 9\cdot 4 \equal{} 45$, hence one of them has $ \sigma(c) \leq \lfloor 45/2 \rfloor \equal{} 22 < 24$;

If $ k\equal{}4$, those columns have $ \sigma(c_1) \plus{} \sigma(c_2) \plus{} \sigma(c_3) \plus{} \sigma(c_4) \equal{} 9\cdot 1 \plus{} 9\cdot 2 \plus{} 9\cdot 3 \plus{} 9\cdot 4 \equal{} 90$, hence one of them has $ \sigma(c) \leq \lfloor 90/4 \rfloor \equal{} 22 < 24$;

If $ k\equal{}3$, finally, those columns have $ \sigma(c_1) \plus{} \sigma(c_2) \plus{} \sigma(c_3) \leq 9\cdot 1 \plus{} 9\cdot 3 \plus{} 9\cdot 4 \equal{} 72$, hence one of them has $ \sigma(c) \leq \lfloor 72/3 \rfloor \equal{} 24$. A model has been presented with the claim above.


group theory
abstract algebra
superior algebra
superior algebra theorems

Prove this facts about supersolvable groups: The commutator subgroup of a supersolvable group is nilpotent.

I'm surprised no one answered this. It is a pretty neat result.

[hide="hint"]
Try to prove $ [G', G_k]\subset G_{k\plus{}1}$ where $ G\equal{}G_0\triangleright \cdots \triangleright G_N\equal{}1$ is the normal series with cyclic quotients, by examining the action of $ G$ on $ G_k/G_{k\plus{}1}$.
[/hide]


find the value of :

$\sqrt{1+1\cdot\sqrt{1+2\cdot \sqrt{1+3\cdot\sqrt{1+4\cdot\sqrt{1+5\cdot\sqrt{1+\cdots}}}}}}$

I'm pretty sure that's infinite. Don't know how to prove it though.

[quote="amirhtlusa"]i don't think so....
i have seen a similar problem before, pretty sure it converges...[/quote]

Yes, surely...

A first step:

[hide]Let $a_n = \sqrt{1 + \sqrt{1 + 2\sqrt{1 + 3\sqrt{\ldots \sqrt{1+n\sqrt{1}}}}}}$.  Suppose $a_n > 2$.  Square this out and subtract one; we get
$\sqrt{1 + 2\sqrt{1 + 3\sqrt{\ldots \sqrt{1+n\sqrt{1}}}}} > 3$.   Now, square this out, subtract one and divide by 2:
$\sqrt{1 + 3\sqrt{\ldots \sqrt{1+n\sqrt{1}}}} > 4$.

At the $i$th step, I claim we have as a consequence of our assumption that
$\sqrt{1 + i\sqrt{\ldots \sqrt{1+n\sqrt{1}}}} > i + 1$.  Then squaring, subtracting 1 and dividing gives us at the $i+1$st step that
$\sqrt{1 + (i+1)\sqrt{\ldots \sqrt{1+n\sqrt{1}}}} > i + 2$.
Thus, by induction the claim is true.  However, taking this to the $n-1$st step results in the claim that $\sqrt n > n$ (or something like that -- I don't think my indices work out exactly, but it's close enough), clearly absurd for integers.  Thus our sequence is certainly bounded above by 2.  As it is also certainly increasing, we know that it must have some real limit value less than or equal to 2.
[/hide]

AArg, i was trying to majorizate or minorizate noticing some inequalities of each radicals with $2<1+a_i$ since every radical $a_i$ is more than 1, and evalutating the approssimative expression of $S-\sqrt{S^2-1-S}$...but i arrived nowhere...so tryed with continueed function, because it seems me that could have some joining points with $e^{\frac{1}{e}}$


A printer can print 10 pages of text per minute or 4 pages of graphics per minute. How many minutes will it take to print 31 pages of text and 7 pages of graphics?

You first divide 31 by 10 so you'll need 3.1 minutes to print those pages.
Next you divide 7 by 4 so you'll need 1.75 minutes to print the graphics.

3.1+1.75=     4.85

I reached the same conclusion yet the review says the answer is $\frac{417}{20}$.


Let $\ ABC$ be an acute triangle, where we take $\ M$ on $\ (AB)$ and $\ N$ on $\ (AC)$ such that we have
 $\ <AMN$  equal $\ <ACB$.Prove that $\ AM*AN*MN\geq\frac{16S^2}{AM+AN+MN}$, where the notation is known.

S is the area of triangle ABC or of the triangle AMN? If it is the area of the triangle AMN why do we need triangle DEF? If not the problem seems wrong...can't you make AM,AN,MN arbitrary small?

What are you talking about? :ninja: OK, $\ S$ is the area of $\ AMN$, but where is triangle $\ DEF$?At this problem you can use the Law of Sin!


let a, b, c is real positive and $ a^{2}\plus{}b^{2}\plus{}c^{2}\equal{}9$

Prove that
$ 2(a\plus{}b\plus{}c) \minus{} abc \leqslant 10$.

[quote="aktyw19"]let a, b, c is real positive and $ a^{2} \plus{} b^{2} \plus{} c^{2} \equal{} 9$

Prove that
$ 2(a \plus{} b \plus{} c) \minus{} abc \leqslant 10$.[/quote]
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=52
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=122166


A square tray has nine houses. A piece on the tray can move for the houses as the figure.
Consider a piece initially in the centre of the tray and it's moved at random.
Find the probability that after 10 movements, the piece is back to the centre of the tray.

Notice that after each odd number of movements the piece can be at the four side square with equal probability then for the next movement the center one has $ \frac 12 \cdot (\frac 14 \plus{}\frac 14 \plus{}\frac 14 \plus{}\frac 14)\equal{}\frac 12$ probability and the vertices $ \frac 14 \cdot (1\minus{}\frac 12)\equal{}\frac 18$.

To sum up:
after $ 2n\minus{}1$ movements each side has probability $ \frac 14$ and the vertices and the center $ 0$
after $ 2n$ movements each side has probability $ 0$ and the vertices $ \frac 18$ and the center $ \boxed {\frac 12}$

Hi mszew,

The answer that i have is:

[hide="Answer"]$ \frac{1}{3}$ [/hide]

I completely fail to understand the movement described in the question.  From the center square, I can move to 6 other squares?  What about from the upper-right corner?  What about from the top-center?  What about from the left-center?

Sorry about that. I did a figure to make it easy to understand. Seems wasn't a good idea.

Tell me if it's not clear yet.

Top-center as the bottom-center has $ 4$ possible movements.

Thanks.

Okay.  There's no way that the answer is $ \frac {1}{3}$.  One way to approach this is recursively: let $ a_n$ be the number of ways to be back at the middle square after $ n$ moves, $ b_n$ be the number of ways to be to the left-center square after $ n$ moves (which is the same as the number of ways of being in the right-center after $ n$ moves), $ c_n$ be the number of ways of being in a given corner after $ n$ moves (all corners are symmetric) and $ d_n$ be the number of ways of being in the center-top (equiv. center-bottom) after $ n$ moves.  Then $ a_n \equal{} 2b_{n \minus{} 1} \plus{} 4c_{n \minus{} 1}$, $ b_n \equal{} a_{n \minus{} 1} \plus{} 2d_{n \minus{} 1}$, and so on.  It should take less than 5 minutes to construct a table of values out to $ n \equal{} 10$.  I get that the final answer is $ \frac {a_{10}}{a_{10} \plus{} 2b_{10} \plus{} 4c_{10} \plus{} 2d_{10}} \equal{} \frac {3345}{8077}$.

This situation is bipartite (for even $ n$, $ b_n \equal{} c_n \equal{} 0$, for odd $ n$, $ a_n \equal{} d_n \equal{} 0$ and this makes it relatively easy to show that in the limit we have $ \frac{a_n}{d_n} \to \sqrt{2}$ and $ \frac{b_n}{c_n} \to \sqrt{2}$.


Let  S is set of 669 distinct positive integers which don't exceed 2003. Prove that exist a;b;c  (not necessarily distinct) in S such that $a+b-c\vdots 2003$

I don't completely understand your question.  If you mean $a + b - c = 2003$, then the problem is invalid because the largest possible sum for the set $S = \{1, 2, 3... \}$ that contains the first 669 positive integers is $669 + 669 - 1 = 1337 < 2003$.

If you mean $(a + b - c) | 2003$, then the problem is still invalid because the set $S = \{2, 4, 6...\}$ containing any 669 of the 1001 even integers $1 < n < 2003$ will have an even value of $a + b - c$ no matter what numbers are chosen out of the set.

Please clarify your question.

I'm sure the problem is correct.$a+b-c \vdots 2003$ means  2003|a+b-c.

if $S=\{1;2;...\}$  we can choose $1;2;3$ because  $2003|1+2-3$


[quote="HelloWorld"]How does it get the value $ \lambda$[/quote]
Dear friend,
I dont know the way the author found it but it can be solved with the HOlder Inequality here, even with the more general
\[ (a^3\plus{}x)(b^3\plus{}y)(c^3\plus{}y) \ge k (a\plus{}b\plus{}c)^3\]
where $ a,b,c>0$ and $ x,y,z$ are the positive constants. :)
Let try with it!
Regards

[quote="B_bnguyen"]I think r is root of an equation. Reads this : http://www.sosmath.com/algebra/facto...11/fac111.html  :D  Good luck, my friend[/quote]
Dead Link!!!
I want to know how does it get that equation. :)

[quote="HelloWorld"]How does it get the value $ \lambda$[/quote]by the hammer :ninja:


i have 3 identical cylinder and 18 identical cubes, how many ways can i stack them(1 on top of each other) to make a tower?

permutation? combination? it has been too long since i did these.

If you have $n_{1}$ objects of type $1, n_{2}$ objects of type $2,\dots,n_{k}$ objects of type $k$, then you can order them in $\frac{(n_{1}+n_{2}+\dots+n_{k})!}{n_{1}!n_{2}!\dots n_{k}!}$ ways.

In the given problem, the answer is $\frac{21!}{18!3!}=\frac{21\cdot 20\cdot 19}{6}=1330$

[hide="solution"] you have 21 objects, therefore you have 21!/(18!*3!) ways to place cubes in the tower. each way to place 18 cubes defines the places of the cylinders. since the cylinders are indentical as are the cubes, I believe that should be the answer.[/hide]

Another way you can look at the problem, although the previous two solutions given are better and the way I would have done them, is to view them as cubes and cylinders in a bag that you are picking out of. Then, multiplying fractions, it is easy to see the same answer of:

1330

This should work. I tried it out and it worked for me.

But I would reccomend sticking to the original solutions provided by sapphyre571 and Farenhajt.


Prove the inequality

 \sum (1+a 2 ) 2 (1+b 2 ) 2 (a-c) 2 (b-c) 2  \geq (1+a 2 )(1+b 2)(1+c 2 )((a-b)(b-c)(c-a)) 2

where the initial sum is over cyclic permutation of a,b,c.

This is so monstruous and at the same time so nice.
   Of course, let a<>b<>c. It becomes sum [(1+a^2)(1+b^2)]/[(1+c^2)(a-b)^2]>=1. Now, by AM-GM we have [(1+a^2)(1+b^2)]/[(1+c^2)(A-B)^2]+[(1+b^2)(1+c^2)]/[(1+a^2)(b-c)^2]>=2(1+b^2)/|(b-a)(b-c)| and three other inequalities. Sum up and you reduced the problem to sum (1+b^2)/|(b-a)(b-c)|>=1. This is true, since |sum (1+b^2)/(b-a)(b-a)|=1.


limit logarithms function inequalities integration geometry rectangle

If $ x_n\equal{}\frac{1}{1\cdot n}\plus{}\frac{1}{2\cdot (n\minus{}1)}\plus{}\frac{1}{3\cdot (n\minus{}2)}\plus{}\cdots\plus{}\frac{1}{(n\minus{}2)\cdot 3}\plus{}\frac{1}{(n\minus{}1)\cdot 2}\plus{}\frac{1}{n\cdot 1}$
Find$ \displaystyle\lim_{n\to\infty}{x_n}$ :blush:

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

As a result, $ x_{n}\equal{}\sum_{k\equal{}1}^{n}\frac{1}{k((n\plus{}1)\minus{}k)}\equal{}\frac{1}{n\plus{}1}\sum_{k\equal{}1}^{n}\left[\frac{1}{k}\plus{}\frac{1}{(n\plus{}1)\minus{}k}\right]\equal{}\frac{2H(n)}{n\plus{}1}$.

Therefore, $ \lim_{n\rightarrow\infty}x_{n}\equal{}\lim_{n\rightarrow\infty}\frac{2H(n)}{n\plus{}1}\equal{}\lim_{n\rightarrow\infty}\frac{2\ln n}{n\plus{}1}\equal{}0$

(Note that in here $ H(n)$ refers to a harmonic function and $ \lim_{n\rightarrow\infty}\frac{H(n)}{\ln n}\equal{}1$)

[quote="Timestopper_STG"]$ \frac {1}{k} \plus{} \frac {1}{(n \plus{} 1) \minus{} k} \equal{} \frac {n \plus{} 1}{k((n \plus{} 1) \minus{} k)}$

As a result, $ x_{n} \equal{} \sum_{k \equal{} 1}^{n}\frac {1}{k((n \plus{} 1) \minus{} k)} \equal{} \frac {1}{n \plus{} 1}\sum_{k \equal{} 1}^{n}\left[\frac {1}{k} \plus{} \frac {1}{(n \plus{} 1) \minus{} k}\right] \equal{} \frac {2H(n)}{n \plus{} 1}$.

Therefore, $ \lim_{n\rightarrow\infty}x_{n} \equal{} \lim_{n\rightarrow\infty}\frac {2H(n)}{n \plus{} 1} \equal{} \lim_{n\rightarrow\infty}\frac {2\ln n}{n \plus{} 1} \equal{} 0$

(Note that in here $ H(n)$ refers to a harmonic function and $ \lim_{n\rightarrow\infty}\frac {H(n)}{\ln n} \equal{} 1$)[/quote]

Yes, the proof is nice, but the problem appears before the Harmonic Series in my book. I've got another proof.
$ \frac{2}{n\plus{}1}\displaystyle\sum_{i\equal{}1}^n{\displaystyle\frac{1}{i}}<\frac{2}{n}\displaystyle\sum_{i\equal{}1}^n{\displaystyle\frac{1}{i}}<2\sqrt{\displaystyle\frac{\displaystyle\sum_{i\equal{}1}^{n}{\frac{1}{i^2}}}{n}}$ < $ 2\sqrt{\frac{2}{n}}$
which uses the Cauchy Inequality or $ A<Q$

Why is  $ \lim_{n\rightarrow\infty}\frac{H(n)}{\ln n}\equal{}1$ ? . can anybody give me a proof ?

[quote="vijaymenon"]Why is  $ \lim_{n\rightarrow\infty}\frac {H(n)}{\ln n} \equal{} 1$ ? . can anybody give me a proof ?[/quote]

For large n, we have $ H_{n} \equal{} \ln n \plus{} \gamma \plus{} \frac{1}{2n} \plus{} O\left(\frac{1}{n^2}\right)$. Now divide through by   $ \ln n$ and take the limit as $ n \to \infty$

But what is $ \gamma$ and $ O\left(\frac{1}{n^{2}}\right)$ . i dont understand ?? please help me

akech, when someone asks to have something explained, that generally means using ideas or arguments that are [i]simpler[/i] than the statement to be explained, not more complicated.

Draw the region measured by $ \int_{1}^n \frac {1}{x} \, dx$.  Now draw two sets of $ n$ approximating rectangles for this integral, one using left endpoints and one using right endpoints.  Conclude that $ H_n \minus{} 1 < \ln n < H_{n \minus{} 1}$.  Then note that $ \ln n \to \infty$ and conclude.

$ \gamma : \equal{} \lim_{n \to \infty} H_n \minus{} \ln n$ is a constant (Euler's constant or the Euler-Mascheroni constant).

I tried to rig a power series/generating function style argument from $ (\ln(1\minus{}t))^2,$ but I haven't got the bugs worked out yet.

To show that the harmonic series is a lower order infinity than n simply fiddle around with the AM-GM inequality on n + H(n). Otherwise Stolz-Cesaro.


Does anybody know what the top score was for Indiana 2009 Mathcounts?  (for individuals....was it like 46/46 or something insane??)Indiana doesn't have an official website for this type of thing.  Also, can someone link me to the Calumet Chapter Mathcounts results(I can't find it, of course though I know the results.  I somehow managed first place despite the fact I don't even know the quadratic formula.  Probably lucky guesses.... :blush:I kinda wanted to know my score, but there was a computer bug and I never saw the results.  I probably got somewhere in the ballpark of 25-28 out of 46, which makes me feel a little guilty for people who fail chapter with near perfect scores)

Please, somebody answer before this message gets dropped down further!

Maybe you should post in [url=http://www.artofproblemsolving.com/Forum/index.php?f=243]Indiana forum[/url]?

I also copied it onto the Indiana forum.  Maybe somebody will answer me there..........


Hello again,

I have attempted a problem, though I do not know if it was done right or wrong, I don't know how you guys wrute all the math symbols in the forum, so I'm just going to scan the solution I wrote out:)

The Question is:

dN/dt = 5- 1/2 N                 N(2) = 3

Here's my solution!:

[url=http://img405.imageshack.us/my.php?image=imgcj5.jpg][img]http://img405.imageshack.us/img405/4904/imgcj5.th.jpg[/img][/url]
[Sorry I don't know how to resize!]

How is it?  Right/Wrong?  Thanks in advance for any feedback!

You could check your answer by plugging it into the equation...

I found two mistakes in the left column: the antiderivative of $ \frac {1}{5 \minus{} N/2}$ should have $ \minus{} 2$ in front; also, $ 5$ disappeared without trace near the bottom.

Hi,

Okay i tried to solve that equation.. is the -10 at the end that you are missing???

I got N= 3e^4-2t - 10  :maybe: I am not too sure.

Use parentheses! The answer is definitely not $ 3e^4\minus{}2t\minus{}10$, but that's probably not what you meant anyway. As mlok said, it's easy to check by the equation: calculate $ N'(t)\minus{}5\plus{}\frac12N(t)$ for your alleged $ N$. If it's not zero, it's not a solution. If it's zero and $ N(2)\equal{}3$, it is a solution.

[url=http://img81.imageshack.us/my.php?image=img2rf5.jpg][img]http://img81.imageshack.us/img81/1606/img2rf5.th.jpg[/img][/url]

Ok, I Re-added that 5 I somehow forgot to carry on, though I'm not sure how to check if it's right, I tried following what mlok said bt couldn't understand it completely, is the above solution correct?


group theory
abstract algebra
function
symmetry
superior algebra
superior algebra unsolved

OK, here is the theorem statement.

Let $ G, G'$ be groups, and let $ H$ be a subgroup of $ G\times G'$ such that the two projections $ p_1: H\rightarrow G$ and$ p_2: H\rightarrow G'$ are surjective. Let $ N$ be the kernel of $ p_2$ and $ N'$ the kernel of $ p_1$. One can identify $ N$ as a normal subgroup of $ G$, and $ N'$ as a normal subgroup of $ G'$. Then the image of $ H$ in $ G/N\times G'/N'$ is the graph of an isomorphism $ G/N\approx G'/N'$.

Now, the proof says that:

Let $ h\equal{}(g,g') \in H$. Consider the map $ H \rightarrow G/N \times G'/N'$ defined by $ h \mapsto (gN, g'N')$. The image of $ H$ under this map is $ \{(gN,g'N') | h \in H \}$. This relation is the graph of a well-defined function $ G/N \rightarrow G'/N'$ provided $ gN\equal{}N \Rightarrow g'N'\equal{}N'$, essentially an application of the vertical line test.

Since $ gN \equal{} N$ (more properly, $ (g,e')N \equal{} N$), we have $ (g,e') \in N \subset H$. Thus $ (e,g') \equal{} (g,g')(g^{\minus{}1},e') \in H$, whence $ (e,g') \in N'$, that is, $ g'N' \equal{} N'$. Note that by symmetry, it is immediately clear that $ g'N'\equal{}N' \Rightarrow gN\equal{}N$, i.e., this function also passes the horizontal line test, and is therefore one-to-one. The fact that the map is a homomorphism and is surjective also follows trivially.

I don't understand how the  $ gN\equal{}N \Rightarrow g'N'\equal{}N'$ implies that the function passes the vertical line test. This is probably something obvious, but i'm not seeing it. Thanks.

If f:A→B is a homomorphism, then its graph is Gr(f) = { (a,f(a)) in A × B }.  Since (a,f(a))*(b,f(b)) = (ab,f(a)f(b)) = ( ab, f(ab) ) in Gr(f), Gr(f) is a subgroup of A × B.

A subgroup H of A × B is a graph if and only if for every a in A, there is exactly one b with (a,b) in H (vertical line test).

Now suppose that H is not the graph of a function, but there is at least one b for every a with (a,b) in H.  Then there must be some a in A and b1, b2 in B such that both (a,b1) and (a,b2) are in H.  Since H is a subgroup, H also contains (a,b1)/(a,b2) = (1,b1/b2).  In other words, if H is not the graph of the function then there must be some non-identity b such that (1,b) is in H (or some a such that there is no b with (a,b) in H, but that part is taken care of in your situation).

Now (gN = N ⇒ g'N' = N') has the nice effect that (1,b) in im(H) implies b=1.  That is, if a=1, then b=1.  Here "a=1" is written gN=N, and "b=1" is written g'N' = N'.


[quote="cowboy"]Find all the positive integer k satisfied 1<K<10 such that the following system of equation:
kx^2+y^2=z^2 and x^2+ky^2=t^2 have the positive integer roof.[/quote]

Let me make it so it's easier to read: ;)

Find all positive integers $k$ that satisfy $1<k<10$ such that the following equations have a positive integer root:

$kx^2+y^2=z^2$ and $x^2+ky^2=t^2$.

Does it have to be $x\neq{y}\neq{z}\neq{t}$?


prove that in any acute triangle, there exists a point P  such that the projections of this point with respect to the the three sides of the triangle are the vertices of an equilateral triangle. i know of a related problem from the romanian mo in 92 by C. Cocea but this one is a little different  :P

This one is from the Tournament of Towns.
Consider the locus of the point st <BPC=60+A. This is a circular arc.
Now the proof should be trivial :)

It's one of the isodynamic points. There are two such points, but one of them is not inside the triangle, I guess.


integration
calculus
real analysis
real analysis unsolved

Find limits and(or) equivalents for the following sequences,functions(or to be more specific...study them):
1)$\sum_{k=0}^n \frac{1}{(n+k)^{\alpha}}$,where $\alpha>0$
2)$(\frac{1}{n}\sum_{k=1}^n a_k^{\frac{1}{n}})^n$,where $a_n$is a positive increasing sequence.
3)$\sum_{k=1}^n ln(1+\frac{(-1)^k}{k^a(ln(k))^b})$,where $a,b>0$
4)$f(x)=\sum_{n=1}^{\infty} (-1)^{n-1}ln(1+\frac{x}{n})$(Equivalent when $x$ goes to $\infty$)
5)$f(x)=1+\sum_{n=1}^{\infty} (1+\frac{1}{n})^{n^2}\frac{x^n}{n!}$(Equivalent when $x$ goes to $\infty$)
6)$\int_0^{\infty}(1+\frac{x}{n})^{-n}x^{-\frac{1}{n}}dx$
7)$\int_0^{\sqrt(n)}(1-\frac{t^2}{n})^n dt$

Killer, Killer, I know one thing: you"ll never like computing equivalents... ;)  ;)

Hey guys...I know you all "love" these kind of exercises.Why dont you give me a hand to solve them?

Killer, please post one exercise [b]per topic[/b], or else all your topics will be closed.
And oh, for easy problems use the [url=http://www.mathlinks.ro/Forum/index.php?f=296]Calculus Computations and Tutorials[/url] forums from now on! Topic closed.


geometry
3D geometry
algebra
polynomial
vector
geometric series

Given $ a_1\equal{}1, a_2\equal{}2, a_3\equal{}2, a_4\equal{}1$ and $ a_{k}\equal{}\frac{a_{k\minus{}3}\plus{}a_{k\minus{}4}}{2}$

Find the explicit formula.

Totally not pretty -- cubic formula and all that jazz.  Lots of cube roots of things involving the square root of 33, etc.  The limiting value works out to a nice rational number that you should be able to calculate by looking at, say, the 100th term ;).  Besides, you know how to approach these in general, don't you?  (Nice work on Mandelbrot rd 1, btw.  Tech did pretty well overall, too.)

We make the guess that it can be modeled as the sum of several geometric series.

The characteristic polynomial is
$ 2a_k \equal{} a_{k \minus{} 3} \plus{} a_{k \minus{} 4}$

$ 2a_{k \plus{} 4} \equal{} a_{k \plus{} 1} \plus{} a_k$
$ 2x^4 \equal{} x \plus{} 1$

JBL, doesn't that mean we need to use the QUARTIC formula to get roots?
EDIT: I noticed that 1 is a root, so you can factor that out and use the cubic. NVM above comment. You're right about the ugly cube roots of 33.
So if you take those roots, you can express the general form like
$ \alpha \cdot R_1 ^ n \plus{} \beta \cdot R_2 ^ n \plus{} \gamma \cdot R_3 ^ n \plus{} \mu \cdot R_4 ^ n$

You can use the given values and set up a system of four equations to solve for the constants, and then you should be set.

I'm not sure how well this would work on a contest, but I THINK it kills the problem off. (note this only works on homogenous linear recursions)

Where is the problem from?  I thought initially that it would become periodic, but after calculating a few terms that's clearly not the case.  JBL and DFT are correct about the form the solution should take, as ugly as it might be.

[quote="DiscreetFourierTransform"]We make the guess that it can be modeled as the sum of several geometric series.[/quote]

"Modeled" implies a degree of inaccuracy.  Inspection tells us that homogeneous linear recurrences must have geometric series in their solution sets, and moreover that their solution sets are vector spaces.  The result follows.


calculus
integration
inequalities
derivative
geometry
trapezoid
real analysis

It's an inequality, and the inequality's pointing in the wrong direction.

A simple estimate: Since $e^{x^{2}}$ is convex (second derivative $(2+4x^{2})e^{x^{2}}$), the two-interval trapezoid estimate $\frac14(1+2e^{\frac14}+e^{1})$ is greater than the integral. This estimate is approximately $1.5716$, which is less than $\frac35e\approx 1.6310$.

Since $\int_{0}^{1}e^{x^{2}}dx\le \int_{0}^{1}e^{x}dx=e-1$, the estimate would follow from $e-1\le 3e/5$, which is the same as $e\le 5/2$... oops.  :oops:

It is better to use the inequality $e^{y}\le 1+y+\frac{y^{2}}2$ for $-1\le y\le 0$ and write the integral as $\int_{0}^{1}e\cdot e^{x^{2}-1}\,dx$ ;)

Let's see. It gives
\[\int_{0}^{1}e^{x^{2}}\,dx\le 1+\frac{e-1}3=\frac{e+2}3\]
So, we need to prove that $e+2<\frac{9}{5}e$ or $e=1+1+\frac12+\dots>\frac{5}{2}$, which is true. So, indeed, it works perfectly well too :).


Given that $ XY \equal{} 21$ and $ 1\leq x\leq 2$, find the sum of the upper and lower bounds of $ Y$. Express your answer as a decimal.

Upperbound: 21/1=21
Lowerbound: 21/2=10.5

Sum=31.5


Ok pick a number 10-99

Got it. Good.

Now multiply that number by 2

add 12 to that number

now subtract 10

add 6

now divide that number by 2

Finally subtract that number by your original number

Your answer should be (spoilers)

4.

oh yeah...pick a positive integer less than or equal to one
multiply it by 3.12435845247540548
then divide by 247824973051830413\

lemme guess the answer is 1 :o

obviously this is not mind reading

sorry :D

[quote="236factorial"]Take a number. Subtract the number from the number itself. result should be 0.  :D[/quote]

WHOA THAT'S WHAT I GOT OMG YOU'RE A PSYCHIC

whats with the restriction on only numbers 10-99...this works with any number, right?

i guess it would take a long time to do it with 5462462 as your number.

[quote="jmadsen"]Pick a number. I predict that your number is...
[hide]not 84756729738572876589278.128735882873875209[/hide][/quote]
apparently that is exactly what i picked your mind reading skills are horrible

[quote="jmadsen"]How is that reading minds? You still don't know what number I picked. This is more like mind reading:

Pick a prime number less than 50 with two odd digits. The two digits must also be different, so for example 11 won't work.

I predict that your number is
[hide]37[/hide][/quote]

WRONG

i picked 17

[quote="jmadsen"]How is that reading minds? You still don't know what number I picked. This is more like mind reading:

Pick a prime number less than 50 with two odd digits. The two digits must also be different, so for example 11 won't work.

I predict that your number is
[hide]37[/hide][/quote]

I pick 19..not 37

eh, I picked 37

well I picked 43 at first and then I was like wait, no

Pick a number between 1 and 3 [b][i]inclusive[/i][/b].

Is it...
[hide] 2! [/hide]

 :D  :D  :D  :D

[quote="Hexaditidom"]Red screwdriver

wow  :D[/quote]

waoh we picked the same thing

[quote="math92"]i got 62 :?[/quote]You all can't count :D

Huh, I got cabbage for the vegetable one.

Orange Screwdriver?  :huh:

Guess I'm not normal

[quote="Derek"]How much is:

[hide="1 + 5?"] [hide="3 + 3?"] [hide="4 + 2?"]  [hide="Repeat the number 6 in your head as fast as possible
for 15 seconds."]  [hide="You were thinking about a CARROT, weren't you!?!
Really weird, right!?!?!"][/hide][/hide][/hide][/hide][/hide]

EDIT: added screwed up hide tags[/quote]

i couldn't come up w/ a vegetable

[quote="sirkuku"]Pick a number 10-99

multiply by 2

add 47

subtract 16

add 10

add 5

subtract 8

divide by 2

subtract by your original number
Your answer should be
(Spoilers)!!!!!!!!!

19[/quote]

I'm gonna remove the mystery around these things...  :)

My number is $x$

$x \Rightarrow 2x \Rightarrow 2x + 47 \Rightarrow 2x + 31 \Rightarrow 2x + 41 \Rightarrow 2x + 46 \Rightarrow 2x + 38 \Rightarrow x + 19 \Rightarrow 19$

[quote="Hamster1800"][quote="Hexaditidom"]Yeah, I heard the 37 problem a while ago. Real interesting, just like this one:

 PICK A NUMBER
1      2      3      4

[hide]Why 3?[/hide][/quote]

I pick 3 because it's between $e$ and $\pi$ my two favorite numbers :D.

I'm going to post a particularly exhaustive number trick.

Pick a number above 10.
square it
subtract 4
multiply by 2 less than your number
divide by 2 more than your number
subtract 1
multiply by 111,111
divide by 1 less than your number
add 999,999
divide by 1001
subtract 111 times your number
and you get:

[hide="spoiler"]

AN EVIL NUMBER BWAHAHAHAHAHA
[/hide]

figure that one out :P[/quote]
At least we don't have to square root it...

[quote="Hamster1800"][quote="Hexaditidom"]Yeah, I heard the 37 problem a while ago. Real interesting, just like this one:

 PICK A NUMBER
1      2      3      4

[hide]Why 3?[/hide][/quote]

I pick 3 because it's between $e$ and $\pi$ my two favorite numbers :D.

I'm going to post a particularly exhaustive number trick.

Pick a number above 10.
square it
subtract 4
multiply by 2 less than your number
divide by 2 more than your number
subtract 1
multiply by 111,111
divide by 1 less than your number
add 999,999
divide by 1001
subtract 111 times your number
and you get:

[hide="spoiler"]

AN EVIL NUMBER BWAHAHAHAHAHA
[/hide]

figure that one out :P[/quote]

Pick $x$.

$x \Rightarrow x^2 \Rightarrow x^2 - 4 \Rightarrow (x - 2)^2(x + 2) \Rightarrow (x-2)^2 \Rightarrow (x - 3)(x - 1) \Rightarrow 111111(x-3)(x-1) \Rightarrow 111111(x - 3) \Rightarrow 111111(x + 6) \Rightarrow 111x + 666 \Rightarrow 666$

Wow. It was actually thought out and not a random list of commands. That's good.  :)

[quote="chesspro"]I'm gonna remove the mystery around these things...  :)

My number is $x$

$x \Rightarrow 2x \Rightarrow 2x + 47 \Rightarrow 2x + 31 \Rightarrow 2x + 41 \Rightarrow 2x + 46 \Rightarrow 2x + 38 \Rightarrow x + 19 \Rightarrow 19$[/quote]
[quote="chesspro"]Pick $x$.

$x \Rightarrow x^2 \Rightarrow x^2 - 4 \Rightarrow (x - 2)^2(x + 2) \Rightarrow (x-2)^2 \Rightarrow (x - 3)(x - 1) \Rightarrow 111111(x-3)(x-1) \Rightarrow 111111(x - 3) \Rightarrow 111111(x + 6) \Rightarrow 111x + 666 \Rightarrow 666$

Wow. It was actually thought out and not a random list of commands. That's good.  :)[/quote]

yes...we actually can do algebra :o

[quote="MysticTerminator"]yes...we actually can do algebra :o[/quote] :help:  :surrender: you must be kidding! :)

This is a better one:
Pick a number of three digits lets say
abc whith a and c not the same digit
Then write the number
|abc-cba|=def
now add
def+fed
Which is your final number?
[hide] Isn´t it 1089?[/hide]

[quote="jmadsen"]Pick a number. I predict that your number is...
[hide]not 84756729738572876589278.128735882873875209[/hide][/quote]
and if mine was this would this causes problems to you???

 :rotfl: i'm obiuovsly joking :rotfl:

I know where you live.

[hide]In time.  Unless you're a [hide]fairy[/hide], in which case, I'm sorry.[/hide]


[quote="kunny"]For $ 1\leq |a|\leq 3,\ 1\leq |b|\leq 3,\ 1\leq |c|\leq 3$, find the maximum of the following expressions.

(1) $ \frac {\sqrt { \minus{} a^2 \plus{} 4|a| \minus{} 3}\sqrt { \minus{} b^2 \plus{} 4|b| \minus{} 3}\sqrt { \minus{} c^2 \plus{} 4|c| \minus{} 3}}{|abc|}$

(2) $ (\sqrt { \minus{} a^2 \plus{} 4|a| \minus{} 3} \plus{} \sqrt {9 \minus{} a^2})(\sqrt { \minus{} b^2 \plus{} 4|b| \minus{} 3} \plus{} \sqrt {9 \minus{} b^2})(\sqrt { \minus{} c^2 \plus{} 4|c| \minus{} 3} \plus{} \sqrt {9 \minus{} c^2})$[/quote]
For (1) $ \sqrt { \minus{} 1 \plus{} \frac {4}{|a|} \minus{} \frac {3}{a^2}}\le\frac {1}{\sqrt{3}}$ so $ max \equal{} \frac {1}{3\sqrt {3}}$

[quote="kunny"]You seem to have a typo.[/quote]
Am I right? (now)

[quote="kunny"]For $ 1\leq |a|\leq 3,\ 1\leq |b|\leq 3,\ 1\leq |c|\leq 3$, find the maximum of the following expressions.

(1) $ \frac {\sqrt { \minus{} a^2 \plus{} 4|a| \minus{} 3}\sqrt { \minus{} b^2 \plus{} 4|b| \minus{} 3}\sqrt { \minus{} c^2 \plus{} 4|c| \minus{} 3}}{|abc|}$

(2) $ (\sqrt { \minus{} a^2 \plus{} 4|a| \minus{} 3} \plus{} \sqrt {9 \minus{} a^2})(\sqrt { \minus{} b^2 \plus{} 4|b| \minus{} 3} \plus{} \sqrt {9 \minus{} b^2})(\sqrt { \minus{} c^2 \plus{} 4|c| \minus{} 3} \plus{} \sqrt {9 \minus{} c^2})$[/quote]
(2) if I didnt miss in calculations $ \sqrt { \minus{} a^2 \plus{} 4|a| \minus{} 3} \plus{} \sqrt {9 \minus{} a^2} \equal{} \sqrt {3 \minus{} |a|}(\sqrt {3 \plus{} |a|} \plus{} \sqrt {|a| \minus{} 1})\le \frac {11\sqrt {3}}{6}$ so $ max \equal{} \frac {11^3\sqrt {3}}{72}$

[quote="kunny"][quote="dima ukraine"]$ \sqrt { \minus{} a^2 \plus{} 4|a| \minus{} 3} \plus{} \sqrt {9 \minus{} a^2} \equal{} \sqrt {3 \minus{} |a|}(\sqrt {3 \plus{} a} \plus{} \sqrt {a \minus{} 1})$[/quote]

Why the equality hold?[/quote]
Hmm I mean it


How do you know how to a attack a certain problem when it is not too easy?

Mathematicians intuition, a sixth sense if you will, will develop as you learn more and more problem solving techniques.
You can guess how to roughly solve the problem using this.

Sometimes I have no idea how to approach a problem, but after playing around with it, I seem to find a little possibility.  Then I pursue it until I solve the problem.  Of course, if the problem is way out my league, then that enlightenment will never come.

-interesting_move

also, asking ppl who kno the question for hints also works!!  :D


Well this is ,a very ordinary problem ,hope it insprises you  :) :

Let $C$ be a circle and $P,Q$ two fixed points (Q on the circle and P outside( in the yard  ) :)  ).

and $L$ is  fixed line which passes through $P$ and doesnt intersect the circle.

$X,Y$ are two points on $L$, that varies such that $XP.YP=K$ (a constant ) .let $QX$ and $QY$ meet the circle in $R$ and $S$ .prove that $RS$ passes through a fixed point.

[u]Size:[/u] constant or variable (measure).
[u]Pozition:[/u] fixed or mobile (mathematical object - matter, thing).

[i]Examples:[/i] "fixed point" and not  "constant point"; "mobile ray" and not "variable ray"; a.s.o.

Well thank you very much levi for your lessons but i think the problem is clear enough to be understand . :)

Evidently, I understood all ! It is O.K. Sorry, for the my profesional deformation (I am a retired teacher)! I suppose that now you teach, but soon you will teach others.
[i]With friendship and congratulations[/i] for your nice share on this site, [i]Ph-An[/i] (alias Levi).

[u]Answer.[/u] [b]Let[/b] the fixed circle $w$, the fixed line $d$ such that $d\cap w =\emptyset$ and the fixed points $Q\in w$, $P\in d$. For the mobile points $\{X,Y\}\subset d$ for which $PX\cdot PY=k^2\ (k>0)$ I define the points: $\{Q,\underline {R}\}\subset QX\cap w$, $\{Q,\underline{S}\}\subset QY\cap w$; $\{A,B\}\subset d$, $P\in (A,B)$, $PA=PB=k$;$\{Q,\underline{U}\}\subset QA\cap w$, $\{Q,\underline {V}\}\subset QB\cap w$.[b]Then[/b] the line $RS$ passes through the fixed point $F$, where $F\in UU\cap VV$ (XX - the tangent in the point $X\in w$ to the circle $w$).

[u]A minute solution.[/u] I note the intersections $M\in RS\cap QA$, $N\in RS\cap QB$
and the middlepoint $L$ of the segment $[MN]$.
Then ($h.d.$- [b]h[/b]armonic [b]d[/b]ivision; $h.p.$-[b]h[/b]armonic [b]p[/b]encil)
$(X,A,Y,B)-h.d.\Longleftrightarrow Q(X,A,Y,B)-h.p.\Longleftrightarrow (R,M,S,N)-h.d.$
$\Longleftrightarrow LM^2=LN^2=LR\cdot LS$ and the circle with the diameter $[MN]$
is orthogonally to the circle $w\Longrightarrow L\in UU\cap VV\Longrightarrow L\equiv F$.

[u]A ("malicious and pedantic") remark.[/u] The circle $w$, the line $d$ and the points $P,Q,A,B,U,V$ are fixed and the points $X,Y,R,S$ are mobilly. The length of the ray of the circle $w$ and the distances $[AB]$, $[PQ]$ a.s.o. are constant, but the distances $[XY]$, $[RS]$ a.s.o. are  variably.

I like this problem. My dear Lomos Lupin, are you an other solution ? I offer soon some problems from this family (with harmonical pencils and harmonical divisions).


geometry
circumcircle
parallelogram
rectangle
rhombus
symmetry
geometric transformation

Let $ABC$ be an acute-angled triangle. The tangents to its circumcircle at $A$, $B$, $C$ form a triangle $PQR$ with $C \in PQ$ and $B \in PR$. Let $C_1$ be the foot of the altitude from $C$ in $ABC$. Prove that $CC_1$ bisects $\angle QC1P$.

[color=blue][b]Lemma 1:[/b][/color] A tangential quadrilateral is also cyclic (i.e., bicentric) iff its contact quadrilateral has perpendicular diagonals.

Let $ABCD$ be the contact quadrilateral of a tangential quadrilateral $PQRS$  with the incircle $(I)$, $A \in SP, B \in PQ, \C \in QR, D \in RS$. Assume that a tangential quadrilateral is also cyclic with the circumcircle $(O)$. Invert it in its incircle $(I)$. The vertices $P, Q, R, S$ are carried into to the midpoints $P', Q', R', S'$ of of the chords $AB, BC, CD, DA$, i.e., they form the Varignon parallelogram $P'Q'R'S'$ of the contact quadrilateral $ABCD$ with the sides parallel to the diagonals $AC, BD$: $P'Q' \parallel S'R' \parallel AC$, $Q'R' \parallel S'P' \parallel BD$. At the same time, the parallelogram $P'Q'R'S'$ is inscribed in the inverted circumcircle $(O')$, hence, it is a rectangle. Thus $P'Q' \perp Q'R'$, which implies $AC \perp BD$. Conversely, if the tangential quadrilateral $PQRS$ is not cyclic, the inverted vertices still form a parallelogram $P'Q'R'S'$, which cannot lie on a circle (because the original vertices $P, Q, R, S$ do not), hence, it cannot be a rectangle. Consequently its sides $P'Q', Q'R'$ are not perpendicular, which implies that the diagonals $AC \parallel P'Q'$, $BD \parallel Q'R'$ of the contact quadrilateral $ABCD$ are not perpendicular either.

[color=blue][b]Lemma 2:[/b][/color] The diagonals of a tangential quadrilateral have the same diagonal intersection as the diagonals of its contact quadrilateral. The opposite sides of the contact quadrilateral intersect on the diagonals of the tangential quadrilateral.

Let $ABCD$ be the contact quadrilateral of a tangential quadrilateral $PQRS$ with the incircle $(I)$, $A \in SP, B \in PQ, \C \in QR, D \in RS$, and let the diagonals $AC, BD$ meet at a point $X$. Consider the cyclic contact quadrilateral as a degenerate cyclic hexagon $ABBCCD$ and denote $BB \equiv PQ, CC \equiv QR$ the tangents of its circumcircle $(I)$ at the points $B, C$. According to Pascal's theorem, the intersections $X \equiv AC \cap BD$, $K \equiv AB \cap CD$, $Q \equiv BB \cap CC$ are collinear. Likewise, consider the cyclic contact quadrilateral as a degenerate cyclic hexagon $AABCDD$ and denote $AA \equiv SP, DD \equiv RS$ the tangents of its circumcircle $(I)$ at the points $A, D$. According to Pascal's theorem, the intersections $X \equiv AC \cap BD$, $K \equiv AB \cap CD$, $S \equiv AA \cap DD$ are collinear. Consequently, the 4 points $X, K, Q, S$ are all collinear, i.e., the diagonal $QS$ of the tangential quadrilateral PQRS passes through the point $X$ and the opposite sides $AB, CD$ meet on this diagonal. In the same way, we can show that the other diagonal $PR$ also passes through the point $X$ and that the opposite sides $BC, DA$ meet on this diagonal. Thus the 2 diagonals intersect at the point $X$.

[color=blue][b]Lemma 3:[/b][/color] Let the opposite sides $AB, CD$ and $BC, DA$ of any quadrilateral ABCD with perpendicular diagonals (not necessarily cyclic) meet at points $K, L$, respectively. Then the diagonals $AC, BD$ intersecting at a point $X$ bisect the angles formed by the lines $KX, LX$.

Let $E, G$ be the intersections of the line $LX$ with the opposite sides $AB, CD$ and $F, H$ the intersections of the line $KX$ with the opposite sides $BC, DA$, forming another quadrilateral $EFGH$. In a polarity transformation with respect to an arbitrary circle $(X)$ centered at their common diagonal intersection $X$, the quadrilaterals $ABCD, EFGH$ are transformed into parallelograms $A'B'C'D', E'F'G'H'$. Since the diagonals $AC, BD$ are perpendicular to each other, $A'B'C'D'$ is a rectangle. The quadrilateral $EFGH$ is inscribed in the quadrilateral $ABCD$ and consequently, the rectangle $A'B'C'D'$ is inscribed in the parallelogram $E'F'G'H'$. Since the points $E, G, L$ and $F, H, K$ are collinear with the center $X$ of the polar circle, their polars $e = E'F', g = G'H', l = A'C'$ and $f = F'G', h = H'E', k = B'D'$ are parallel. The rectangle diagonals $l = A'C', k = B'D'$ are thus the middle lines of the parallelogram $E'F'G'H'$. Conversely, the parallelogram diagonals $n = E'G', m = F'H'$ are the middle lines of the rectangle $A'B'C'D'$. Consequently, the parallelogram $E'F'G'H'$ has perpendicular diagonals and it is a rhombus. The opposite sides $EF, GH$ and $FG, HE$ meet at the poles $M, N$ of the rhombus diagonals $m = F'H', n = E'G'$. Since these rhombus diagonals are parallel to the rectangle opposite sides $a = A'B', c = C'D'$ and $b = B'C', d = D'A'$, the poles $M, A, C$ and $N, B, D$ of these parallel lines are collinear with the center $X$ of the polar circle. Hence, the diagonals $AC, BD$ pass through the points $M, N$. The rectangle diagonals $l = A'C', k = B'D'$ are perpendicular to the lines $LX, KX$, identical with the diagonal lines $EG, FH$, connecting their poles $L, K$ with the center of the polar circle. Similarly, the rhombus diagonals $n = E'G', m = F'H'$ are perpendicular to the lines $NX, MX$, identical with the diagonal lines $BD, AC$, connecting their poles $N, M$ with the center of the polar circle. Because of the symmetry of the transformed configuration, the rhombus diagonals are bisectors of the angles between the rectangle diagonals. But this also means that the diagonals of the quadrilateral $ABCD$ are bisectors of the angles between the diagonals of the quadrilateral $EFGH$.

[color=red][b]Problem:[/b][/color] Let the altitude $CC_1$ of the acute angle triangle $\triangle ABC$ meet its circumcircle $(O)$ at another point $D$ different from the vertex $C$. Let the tangent to the circumcircle at the point $D$ meet the sides $QR, RP$ of the tangential triangle $\triangle PQR$ at points $S, T$, respectively. The quadrilateral $PQST$ is tangential and $ADBC$ is its contact cyclic quadrilateral with perpendicular diagonals $CC_1 \equiv CD \perp AB$ intersecting at the point $C_1$. This means that the quadrilateral $PQST$ is bicentric (see lemma 1), but we do not need this fact at all. Since the quadrilateral $PQST$ is tangential, its diagonals $PS \equiv PC_1, QT \equiv QC_1$ have the same intersection $C_1$ as the diagonals $AB, CD$ of its contact quadrilateral $ADBC$. Moreover, the opposite sides $BC, AD$ and $CA, DB$ intersect at points $K, L$ on the diagonals $QT, PS$ of the tangential quadrilateral $PQST$ (see lemma 2). Since the quadrilateral $ADBC$ has perpendicular diagonals, they bisect the angles formed by the lines $KC_1 \equiv QC_1, LC_1 \equiv PC_1$ (see lemma 3), i.e., $CC_1 \equiv CD$ bisects the angle $\angle QC_1P$ .

[b]Yetti, you are a great !.[/b] I always admired your solutions. You are unique on this site.
Thank you for your replies, for your effort in a clear and distinct get up.

Here is my solution. It isn't necessary as $ABC$ - acute triangle.

I note $T\in AB\cap CC,\ S\in AB\cap CR$. Because $CC_1\perp AB$, then the ray
$[CC_1$ bisects the angle $\angle PC_1Q$ iff the division $(TQCP)$ is harmonically.
It is wellknown that the ray $[CS$ is the simedian from the vertex $C$ of the triangle
 $ABC$ and $\frac{TA}{TB}=\frac{SA}{SB}=\left( \frac ba \right)^2$. Thus, $R\ (T,A,S,B)$ - harmonic pencil $\Longrightarrow$
$R\ (T,Q,C,P)$ - harmonic pencil $\Longrightarrow (T,Q,C,P)$ - harmonic division.

[b]Remark.[/b] Here is a wellknown property of the [i]harmonic division[/i]:
Let $A,B,C,D\in d$ be four points which belong to the same line $d$
(in the mentioned order) and the point $P\not\in d$. [u]Then if two from the following[/u]
[u]affirmations are truly, results that and the other affirmation is truly[/u]:

$\blacksquare \ 1.\ (A,B,C,D)$ - harmonic division $\left(\ \frac{\overline {BA}}{\overline {BC}}=-\frac{\overline {DA}}{\overline {DC}}\ \right);$

$\blacksquare \ 2.\ PB\perp PD;$

$\blacksquare \ 3.$ The ray $[PB$ bisects the angle $\angle APC$.

[b]A similar problem.[/b] Let $ABC$ be a triangle, the its circumcircle $w=C(O,R)$, the reflection $A'$ of the point $A$ w.r.t. the center $O$, the intersection $T\in BC\cap A'A'$ and the proiection $P$ of the point $A'$ on the line $OT$. Prove that the ray $[PA'$ bisects the angle $\angle BPC$.
[u]Indication.[/u] $I\in BC\cap PA'\Longrightarrow (B,I,C,T)$ is a harmonic division.

Hi

You have written fantastic solutions...

I have another solution:

Let Call C1 as H.

Let M be the foot of the altitude from H on BR
(I mean HM  L  BR)
and let N be the foot of the altitude from H on AR

As shown in the picture:

tanH2 = SinH2/SinH1 = a/b * x/h

= SinQ2/SinQ1 * x/h

tanH3 = inP2/SinP1 * y/h

tanH2/tanH3 = SinQ2/SinP2 * SinP1/SinQ1 * x/y

= PH*SinP1 / (QH*sinQ1) * x/y = HM/HN * x/y

But we know

BMH ~ ANH    so :

HM/y = HN/x

and it means  tan H2 = tan H3

So : H2 = H3

Well, the circumcircle of triangle ABC is the incircle of triangle PQR (since it touches the sides QR, RP and PQ of this triangle PQR); thus, the points A, B, C are the points of tangency of the incircle of triangle PQR with its sides QR, RP and PQ, so that the lines PA, QB and RC concur at one point, namely at the Gergonne point of the triangle PQR. Hence, if we denote this Gergonne point by K, then the points A, B, C are the points of intersection of the lines PK, QK and RK with the lines QR, RP and PQ. Now, the problem follows from http://www.mathlinks.ro/Forum/viewtopic.php?t=6088 post #2 Theorem 2, applied to the triangle PQR and the point K.

  darij

Sorry I coudnt send the picture

Let:

x=Ah   y=BH

U = (AC,QH)   a=CU   b=AU

V = (BC,PH)   c=CV   d=BV

h=AH

H1 to H4 = AHQ, QHC, CHP, PHB

Q1,Q2 = AQH, HQP

P_1,P2 = BPH, HPQ

Thanks

Sorry I couldnt send the picture

Let:

x=Ah   y=BH

U = (AC,QH)   a=CU   b=AU

V = (BC,PH)   c=CV   d=BV

h=AH

H1 to H4 = AHQ, QHC, CHP, PHB

Q1,Q2 = AQH, HQP

P_1,P2 = BPH, HPQ

Thanks

Sorry I couldnt send the picture

Let:

x=Ah   y=BH

U = (AC,QH)   a=CU   b=AU

V = (BC,PH)   c=CV   d=BV

h=AH

H1 to H4 = AHQ, QHC, CHP, PHB

Q1,Q2 = AQH, HQP

P1,P2 = BPH, HPQ

Thanks

If $ CC_1$, $ BN$ ,$ AM$ concur, then the problem will be true by Blanchett's theorem. They concur by Ceva iff $ \frac{CN}{NA} \cdot \frac{AC_1}{C_1B} \cdot \frac{BM}{MC}\equal{}1$. But $ \frac{CN}{NA}\equal{}\frac{[CQC_1]}{[AQC_1]}\equal{}\frac{CQ \cdot CC_1 \cdot sin QCC_1}{QA \cdot AC_1 \cdot sin QAC_1}$. We find the product we want is equal to $ \frac{CQ}{CP} \cdot \frac{BP}{AQ}$, and this is $ 1$ since $ CQ\equal{}AQ$ and $ BP\equal{}CP$, so we are done.

I really doubt that this belongs in the pre-Olympiad section.

As a comment, we can prove yetti's first and third lemmas in a much simpler fashion. Using his notation,

Lemma 1: Observe that $ \angle ASD\equal{}180\minus{}\text{arc}\, AD$, so $ \angle ASD\plus{}\angle BQC\equal{}360\minus{}\text{arc}\, AD\minus{}\text{arc}\, BC\equal{}360\minus{}2\angle (AC,BC)$. The equivalence follows easily.

Lemma 3: Let $ U\equal{}AC\cap LK$ and $ V\equal{}BC\cap LK$. Then looking at $ \triangle ALK$, concurrent cevians $ AU$, $ LB$, and $ KD$, and the collinear points $ B$, $ D$, and $ V$, we get that $ (K,L;U,V)\equal{}\minus{}1$. Then using the well-known lemma that Virgil mentions finishes the lemma.

sorry for spam but I think those kinds problem use complex number is more beutiful than a solution with many lemma or some strange geometry (at me, that is harmonic for example, sorry for my bad knowledge about geometry  :( ). I admire darij so much because in his solution often use many lemma  :)
 if we call the orgin of the circle is O. we have: $ q = \frac {2ac}{a + c}$; $ p = \frac {2bc}{b + c}$. $ r = \frac{2ab}{a + b}$ ${ c_1= \frac {1}{2}(a + b + c - ab/c}$.
so, the proof is complete. (please notice that $ \angle BC_1C$ congruent with $ \angle QC_1C(mod \pi)$  :wink:

[hide = 2-line Projective Solution]

First note that $AP$, $BQ$, and $CR$ concur at the Gergonne Point of $\triangle PQR$. Then, let $X = AB \cap PQ$ -> a well-known lemma implies that $(X , C ; P , Q) = -1$, and since $\angle XC_1C = 90$ (by definition), we have $CC_1$ bisects $\angle AC_1P$ by another well-known lemma.

[/hide]

Dear Mathlinkers,

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=503523

Sincerely
Jean-Louis


geometry
parallelogram
conics
parabola
search
rotation
geometry unsolved

Let $ ABCD$ be a parallelogram with no angle equal to $ 60^{\circ}$. Find all pairs of points $ E, F$ that lie in the plane of $ ABCD$ and are such that triangles $ AEB$ and $ BFC$ are isosceles, with bases $ AB,BC$, and triangle $ DEF$ is equilateral.

It will something like the intersection points of two parabolas, but I'm not sure when is easy to solve for that fourth degree equation...

May be there is another aproach...

 :)

It is equivalent to construct an equilateral triangle $ \bigtriangleup DEF,$ whose the vertices $ E,$ $ F,$ lie on two constant lines, here the midperpendiculars of the side-segments $ AB,$ $ BC$ respectively, of the given parallelogram $ ABCD.$

In my drawing, the vertex $ D,$ lies inwardly to the angle $ \angle MPN,$ where $ M,$ $ N,$ are the midpoints of $ AB,$ $ BC$ and $ P$ is the intersection point, of their midperpendiculars. In this case, I remember that we can construct, by an elementary way, two triangles as the problem states.

I will search my notes and if I am not mistaken, I will post here later more details.

Kostas Vittas.

rotate 60 degrees about D and intersect the image of one perpendicular bisctor with the original version of hte other

[b][size=100][color=DarkBlue]GENERAL PROBLEM. – An angle $ \angle XOY$ is given and let $ P$ be, an arbitrary point in the plane. Construct an isosceles triangle $ \bigtriangleup ABC$ with $ PA \equal{} PB$ and $ \angle APB \equal{} \angle \omega,$ whose the vertices $ A,$ $ B,$ lie on the lines $ OX,$ $ OY,$ respectively[/color][/size][/b].

For the solution, I post here only the schema and I think it is enough.

Kostas Vittas.

[quote="easternlatincup"]Well, vittasko, is this really the general problem of the original one?[/quote]
I think yes. We can consider the lines $ OX,$ $ OY,$ as the midperpendiculars of the side-segments $ AB,$ $ BC$ and the point $ P,$ as the vertex $ D,$ of the parallelogram $ ABCD,$ in the original problem and $ \angle \omega \equal{} 60^{o}.$

$ ($ then the labels $ A,$ $ B,$ of the general problem, are corresponded with the $ E,$ $ F$ ones, in your problem $ ).$

Never mind, if the vertex $ D$ of $ ABCD,$ lies inwardly to the angle formed by the midperpendiculars of $ AB,$ $ BC.$

Kostas Vittas.

[quote="vittasko"]...Never mind, if the vertex $ D$ of $ ABCD,$ lies inwardly to the angle formed by the midperpendiculars of $ AB,$ $ BC.$[/quote]
In the problem [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=163344]Find a locus[/url], it has been mentioned a construction by the same way, when the point $ P,$ lies inwardly to the angle $ \angle XOY.$

An interesting remark about this simple construction, based on the [b][size=100]Morley’s theorem[/size][/b], is that we can construct infinity geometrically trisectable angles.

Kostas Vittas.

[quote="easternlatincup"]I wonder, Kostas, how you know the two triangles you construct are all there are.[/quote]
I don't know how to construct other equilateral triangles, as your problem states.

So, I didn't search completely, if the two triangles constructed by the way I mentioned, are all there are.

I also wonder if may be constructed more than two equilateral triangles, as your problem states.

Kostas Vittas.

Yes, there are two equilateral triangles, but how do you prove there are no more than these two?


1. How does the mass percentage of carbon in a polyamide, for example (-HNCH2CO^-)x, compare with the mass percentage in a typical monomer, for example H2NCH2CO2H, from which it is formed?
*The numbers are subscripts.
A.The percentage of carbon in the polymer is greater.
B. The percentage of the carbon in the polymer is less.
C. The percentages are equal.
D. There is not enough information to make this comparison.
I really have no clue about this problem.  Please help.

2. Which one of the following compounds would be expected to show optical activity?
A.CH3CH=CHCOOH
B.HOOCCH2CH(CH3)CH2COOH
C.CH3CH(OH)COOH
D. HOCH2C(CH3)2CH2OH
*The numbers are subscripts.
I'm really confused about optical activity. I'd appreciate any help.[/code]

I don't know about the first one.

[quote="RHSLilSweetie07"]2. Which one of the following compounds would be expected to show optical activity?
A.CH3CH=CHCOOH
B.HOOCCH2CH(CH3)CH2COOH
C.CH3CH(OH)COOH
D. HOCH2C(CH3)2CH2OH
*The numbers are subscripts.[/quote]

I think an organic substance has optical isomers when a carbon is attached to groups that are not the same as any other.  I think C would be the correct answer.

For number 1, note that when you form the polymer, you remove two hydrogen atoms, and leave all the other atoms unchanged (except for the end units, where only one hydrogen is removed).  See if you can figure out the answer given that hint.

For 2, I would say it is a bad question.  Optical activity means that you have a mixture with an excess of one enantiomer.  One way of thinking about an object having an enantiomer is if you hold it up to a mirror, you do not get the same image back--for example, your hand.  Zakary is right in that a compound with a carbon attached to four different groups usually has an optical isomer (a synonym for enantiomer).  This isn't actually true in general, because there could be a carbon somewhere else in the molecule that is the other one's mirror image.

In short, optical activity is a property of a mixture, while existence of an enantiomer is a property of a compound.  The answer that is being looked for is indeed C, however.

Huh, I didn't realize that, gregx007.  Thanks!

how do you find the number of isomers of a hydrocarbon?

Try every appropriate combination of the atoms involved in the structure?  How difficult that would be depends on the complexity of the structure...

Yep, there's no easy formula.  Doing it systematically (least branched to most branched) and if you have time, writing out the IUPAC name to make sure you didn't accidentally duplicate one are usually the best things you can do to help yourself.


quadratics
calculus
integration
modular arithmetic
number theory

Try looking up [url=http://en.wikipedia.org/wiki/Quadratic_residue]Quadratic Residues[/url].

Uhhh.. does that mean that you can take square roots of quadratic residues mod p?  :?

I don't get it.. can someone explain, please?  :?

Yes, because one of the six rules of mods is that you can raise both sides to any power, and fractions count as powers, therefore you can take roots.

[quote="chess64"]The rule is [b]integral[/b] powers.

If we have $36 \equiv 25 \mod {11}$, then raising both sides to $\frac{1}{2}$ power, we have $6 \equiv 5 \mod {11}$, which is certainly not true.  ;)[/quote]

ahh...I see, yes, I thought that the rule applies to $all$ powers.

[quote="chess64"]The rule is [b]integral[/b] powers.

If we have $36 \equiv 25 \mod {11}$, then raising both sides to $\frac{1}{2}$ power, we have $6 \equiv 5 \mod {11}$, which is certainly not true.  ;)[/quote]

Actually  if A^2 = B^2 then A can be either B or -B, here too,  $6 \equiv -5 \bmod 11 $

[quote="chess64"]Can you raise a number to a fractional power in mods, i.e. take the square root? I don't think you can, but can anyone prove it?[/quote]Well I can't answer that question, but I just learned from mathworld that you can have rational numbers in congruences. For example, $2\cdot4\equiv1\pmod{7}\Longrightarrow2\equiv\frac{1}{4}\pmod{7}$. I didn't know that. If you want to find out about rational roots in congruences, maybe you should take an algebraic number theory course.  :D (I actually have no idea what algebraic number theory is, but it could have to do with that.)


college
inequalities proposed
inequalities

Let$a_1,a_2,...,a_{2000}$ be real numbers in the interval$[0,1]$.

Find the Maximum possible value of

$\sum_{1\leq i<j\leq 2000}(j-i)|a_j-a_i|$

The monthly contest of the Berkeley Math Circle.


If $x,y,z$ be positive real numbers, then

$\sum\frac{\sqrt{yz+4zx+4xy}}{3(y+z)}\leq\sum\sqrt{\frac{y+z}{8x}}$,

with equality if and only if $x=y=z$.

Hint.

Prove that
\[\left(\frac{\sqrt{z}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\right)\cdot \left(\sum_{cyc}\sqrt{\frac{\sum_{cyc}xy}{9(y+z)^{2}}+\frac{x}{3(y+z)}}\right)\le \frac12 \left(\sqrt{\frac{z+x}{8y}}+\sqrt{\frac{z+y}{8x}}\right) \]
EDIT: Actually it might be tougher than original, I'm not sure... :(
Sorry, its only correct under a certain range :( disregard (unless it can somehow be salvaged...)

For now I just have (bad) proof of the much weaker than your inequality:
\[\sum\sqrt{\frac{y+z}{8x}}\geq \frac32 \]
Let $x\leq y\leq z$. Then we have:
\[y+z-2x\geq x+z-2y\geq x+y-2z \]
Also,
\[\frac{1}{\sqrt{8x}}\geq \frac{1}{\sqrt{8y}}\geq \frac{1}{\sqrt{8z}}\]
Finally,
\[\frac{1}{\sqrt{y+z}+\sqrt{2x}}\geq \frac{1}{\sqrt{x+z}+\sqrt{2y}}\geq \frac{1}{\sqrt{y+x}+\sqrt{2z}}\]
since each is equivalent to a form of:
\[\sqrt{2}(\sqrt{z+x}+\sqrt{y+z})\geq \sqrt{x}+\sqrt{y}\]
which is superweak and clear.

Thus, the sequences $\left\{y+z-2x\right\}$ and $\left\{\frac{1}{\sqrt{8x}(\sqrt{y+z}+\sqrt{2x})}\right\}$ are similarly sorted. It follows that:
\[3\sum_{cyc}\frac{y+z-2x}{\sqrt{8x}(\sqrt{y+z}+\sqrt{2x})}\geq \left(\sum_{cyc}x+y-2z\right)\cdot\left(\sum_{cyc}\frac{1}{\sqrt{8x}(\sqrt{y+z}+\sqrt{2x})}\right)=0 \]
Thus,
\[\sum_{cyc}\frac{\sqrt{y+z}-\sqrt{2x}}{\sqrt{8x}}\ge 0 \iff \sum_{cyc}\sqrt{\frac{y+z}{8x}}\geq \frac32 \]
I'm sure there is a nicer proof of this, any ideas (I know its a little off topic but I want to work up to this problem...)

Actually, that was not mine. I just had heard that one from Hojoo Lee.  :blush:


blogs
geometry
3D geometry
sphere
AMC
AIME
ratio

Posting game
i post a problem, and u answer as fast as possible
the fastest poster gets a point
the goal is 10 points

Problem 1:
What is 1+2+3?

No, ans is 6

You wrote: ((((6!)!)!)!), honestly a pretty huge number

jk

Can I get a consolation half-point?

\begin{eqnarray*}12 + 12 + 5c3 & = & 24 + 10 \\
& = & 34\end{eqnarray*}

Shouldn't this be in like Classroom math or something?

Suggestion:

Problem 3: If a=3, b=4, c=5, what is:

$ (a^a + b^b + c^c)^{a + b + c}$

dojo this isnt a marathon, it's basically a who posts first game.  and only program posts probs.

Dojo:1
AIME15:1

Problem 3:
Go to my blog, and answer problem 8.

(Hint: blog under)
|
|
|
V

i think i did to...
use numbers [b]-20[/b] to [b]20[/b]
stevenmeow:1
AIME15:1
Dojo:1

Problem 4:
A runner runs from point A to point B at 40 mph. He runs back to A at a pace of 50 mph. What is the average speed? (Hint: You can get tricked even if you exploit [b]one[/b] mistake)

Nope. Hint: I never said that I traveled the same route from A to B and B to A.
Answer: Cannot be determined
Fine, 1 point for james4l :wink:
Dojo:1
AIME15:1
stevenmeow:1
james4l:1

Problem 5:
Two concentric circles have radii 5 cm and 8 cm. What is the area occupied by the 8cm circle but not the 5cm circle?
 :?:

stevenmeow:
Dojo:1
AIME15:1
james4l:1

Problem 6:
Find the space diagonal for a cube with surface area 54.
Express your answer as [b]sqrt x[/b]

isnt it sqrt 27?

side=3 diag = 3 root 3

Yes.
stevenmeow:3
Dojo:1
AIME15:1
james4l:1

A sphere is inscribed in a cube of diagonal $ \sqrt27$. Find the volume of the sphere, in the form [b]x$ \pi$[/b].

Yea, but your post says "fastest poster" not "fastest correct poster".
What's a space diagonal?
[b]$ \frac{9}{2}\pi$[/b]

ok fine... but the diagonal is infact $ 7\sqrt2$ if it is written like that so the side is $ \frac{7\sqrt6}{3}$ and the radius is $ \frac{7\sqrt6}{6}$ meaning the volume of the sphere is $ \frac{343\pi\sqrt6}{27}$

stevenmeow:3
Dojo:1
AIME15:1
james4l:1
aleph0:1
Yup.

#8
Find x in |2x+5|=|2x-4|.

stevenmeow:3
Dojo:1
AIME15:1
james4l:1
aleph0:1
dragon96:1
#9
AIME_is_hard and program4 have weights in a ratio of 7:4. Our total weight is 3300 tons. Find program4's weight in pounds.

stevenmeow:3
Dojo:1
AIME15:1
james4l:1
aleph0:1
dragon96:1
ernie:1
#TEN
Find x. Specify which column and row x is in.

asdjfoiasdjfae
asdfaesrasefd
sjdfhinioziocjio
dfsakdfkasjief
dskfasigioaejs
sdfswoiejffdff
fasdkfjaieffsdf
asdfsefrfsdfbgp

ugg there is no x, unless you count the ones in your text.

2 x's exist.

X1- row 6, column 9
X2- row 6, column 37

stevenmeow:3
Dojo:1
AIME15:1
james4l:1
aleph0:1
dragon96:1
ernie:2
new rule:answers must be [color=cyan][size=200][u][i][b]correct[/b][/i][/u][/size][/color]
#11
worth 1 point :wink:
Simplify (13364+4534)*456+42456

stevenmeow:3
Dojo:1
AIME15:1
james4l:2
aleph0:1
dragon96:1
ernie:2
#12
Find the derivative of $ 5x^3\plus{}4x^2\plus{}3x\minus{}215648651235456.$

stevenmeow:3
Dojo:1
AIME15:1
james4l:2
aleph0:1
dragon96:1
ernie:2
cf249:1
#lucky 13:
Find f(x) if f(x^2)=x^4.

Facts:
MATHCOUNTS state tryout is on 3/13/08 friday the 13th, also 1 day before pi day

stevenmeow:3
Dojo:1
AIME15:2
james4l:2
aleph0:1
dragon96:1
ernie:2
cf249:1
Mine is similar
#14
What is f(5) if f(n+1)=n^3-6n^2-7i if i=sqrt(-1)?

stevenmeow:3
Dojo:1
AIME15:2
james4l:3
aleph0:1
dragon96:1
ernie:2
cf249:1
#15
AIME15 is hiking at 30 mph. program4 is hiking the same route one hour later at 60 mph. program4 started hiking at 2:00PM. When did program4 catch up with AIME15?
Include AM or PM in your answer.

stevenmeow:3
Dojo:1
AIME15:2
james4l:3
aleph0:2
dragon96:1
ernie:2
cf249:1
i cant come on too often
#16
Find the value of the expression
f(x)+x=x^2
if f(x)=x^2+f(0)+13x-f(x-x)
and x=1

stevenmeow:3
Dojo:1
AIME15:2
james4l:3
aleph0:3 :wink:
dragon96:1
ernie:2
cf249:1
#17
I travel at 50mph. How long, in seconds, does it take for me to travel 29 1/3 miles? Answer in seconds as a mixed number.

[quote="aleph0"][quote="program4"]
#16
Find the value of the expression
f(x)+x=x^2
if f(x)=x^2+f(0)+13x-f(x-x)
and x=1[/quote]
"f(x)+x=x^2" is an equation, not an expression. Therefore this question is not possible.[/quote]
a=b is an expression that has value 1 or 0, depending on whether it is true or not
lets try some examples
1=1 has value 1
1=2 has value 0
$ \sqrt [6]{9 \plus{} 4\sqrt5} \equal{} \frac {1 \plus{} \sqrt5}{2}$ has value 1
the given eqpression is false, so it has value 0

meanwhile, for #17, i have 2112

For program's problem

D=29 + 1/2
and
R=50

So; D=RT, or D/R = T

(59/2)/50 = T

59/100 = T
but 59/100 is just the number of hours, so we convert it to seconds by multiplying it by 3600. The 100's cancel out to give us;

59(36) = 2124

So the answer is[b][size=200] 2124!!![/size][/b]

^THIS POST WAS MADE 58 MINUTES AFTER MY BIRTH.


Give quadrilateral $ ABCD$ with $ \angle CAD \equal{} 45^\circ$, $ \angle ACD \equal{} 30^\circ$, $ \angle BAC \equal{} \angle BCA \equal{} 15^\circ$, find the value of $ \angle DBC$.

[size=117][color=darkred][b][u]Nice and easy problem ![/b][/u][/color][/size]

[quote="April"]$ \{\begin{array}{c} \mathrm {quadrilateral\ } ABCD \\
\ = = = = = = = = = = = = = = = = = \\
\ m(\widehat {CAD}) = 45^{\circ}\ ,\ m(\widehat {ACD}) = 30^{\circ} \\
\ m(\widehat {BAC}) = m(\widehat {BCA}) = 15^\circ\end{array}$ $ \implies$ $ m(\widehat {DBC}) = \ ?$[/quote]
[color=darkblue][b][u]Method I[/u] (synthetical).[/b] Construct $ \{\begin{array}{c} E\in (CD) \\
\ EA = EC\end{array}$.

Therefore, $ BE\perp AC$ and $ ABED$ is cyclically $ \implies$

$ ABD$ is equilateral $ \implies$ $ BC = BD$ $ \implies$ $ \boxed {\ BD\perp BC\ }$.

[b][u]Method II[/u] (trigonometric).[/b] Denote $ m(\widehat {DBC}) = \underline {x\in (0,150)}$. Therefore,

$ \|\begin{array}{c}
\sin (30 + x)\sin 45\sin (45 + x)\sin 15 = \sin x\sin 15\sin (x - 30)\sin 30\\\
2\sin (30 + x)\sin 45\cos (45 - x) = \sin x\sin (x - 30)\\\
\sin (30 + x)(\cos x + \sin x) = \sin x\sin (x - 30)\\\
\sin x[\sin (30 + x) - \sin (x - 30)] + \cos x\sin (30 + x) = 0\\\
2\sin x\sin 30\cos x + \cos x\sin (30 + x) = 0\\\
\cos x[\sin x + \sin (30 + x)] = 0\\\
\cos x = 0\ \ \vee\ \ \sin x + \sin (30 + x) = 0\\\
x = 90\ \ \vee\ \ x = 165\\\
\boxed {\ x = 90\ }\ .\end{array}\|$[/color]


See... this was  a GPML contest problem and I figured out how to do it,, but there is no way for me to check the anwer.
Does anybody else have answer to this question?

For a family of functions F(x,y,z), the inputs x,y and z each equal 0 or 1.  Also, for each input (x,y,z), F(x,y,z) equals 0 or 1.  How may distinct such functions F(x,y,z) exist?

A) 16
B) 28
C) 64
D) 256
E) None of the above

Another question about vectors!

Let the set of all roots of the equation x^12 = 1 be graphed as vectors OP(I couldn't type it but there was a little arrow on the top, and a small i in the down right hand corner) on the complex plane with origin O.  What is the smallest angle between two of these vectors?

A)10 degrees
B) 15 degrees
C) 20 degrees
D) 25 degrees
E) None of the above

Thanks a lot!

Well, there are $2^3=8$ distinct inputs (i.e. numbers in the domain).  Each of these can be either 0 or 1.  So there are $2^8=256$ total functions.


modular arithmetic
function
number theory
least common multiple
algebra
polynomial
geometry

Prove that $ n^9 \minus{} n^3$ is divisible by 9 for all integers $ n$.

Casework:
(1) $ n\equiv 0 \pmod 3 \implies n\equal{}3a$ so $ n^9\minus{}n^3 \equal{} 3^9 a^9 \minus{} 3^3 a^3$ which is definitely divisible by $ 9$.
(2) $ n \equiv 1\pmod 3 \implies n\equal{}3a\plus{}1$ so $ n^9\minus{}n^3 \equal{} (3a\plus{}1)^9 \minus{} (3a\plus{}1)^3$
By binomial theorem, the coefficients of $ a^9, a^8, a^7 \ldots a^2$ are all divisible by $ 9$ since the $ 3$ is substituted into the formula along with the $ a$.  The constant is $ 0$, so we only need to compute the coefficient of $ a$: $ 9*3*1^8 \minus{} 3*3*1^2 \equal{} 18$ which is divisible by $ 9$.  So $ 9|n^9\minus{}n^3$ in the case $ n \equiv 1 \pmod 3$
(3) $ n \equiv 2 \pmod 3 \implies n\equal{}3a\plus{}2$ so $ n^9\minus{}n^3 \equal{}(3a\plus{}2)^9\minus{}(3a\plus{}2)^3$.  Every coefficient in this expansion is guaranteed to be divisible by $ 9$ except the coefficient of $ a$ and the constant.  But the constant is just $ 0$, and the coefficient of $ a$ is $ 2^8 (9*3\minus{}3*3) \equal{} 2^9*9$ which is divisible by $ 9$.  So $ 9|n^9\minus{}n^3$ in the case $ n \equiv 2 \pmod 3$

Thus $ 9|n^9\minus{}n^3$ for all $ n \in \mathbb{Z}$

Simpler proof:

By euler's totient theorem, since $ \phi(9) \equal{} 6$, $ n^7 \equiv n \pmod 9$.  Multiply both sides by $ n^2$ and rearrange to get that $ n^9\minus{}n^3 \equiv 0 \pmod 9$ as desired.

That's right. Euler's Totient Theorem is the background of the problem.
Without the knowledge, we can solve the problem.

Just factorize $ n^9\minus{}n^3$.  :lol:

Extension: what's the largest integer $ m$ such that $ m$ divides $ n^9 \minus{} n^3$ for all positive integers $ n$?

First of all, note that $ 2^9 \minus{} 2^3 \equal{} 504$, so $ m|504$.  Thus we know $ m \le 504$.  Since $ 504 \equal{} 2^3*3^2*7$, from the carmichael function we have $ \lambda(504) \equal{} lcm(\lambda(8), \lambda(9), \lambda(7)) \equal{} lcm(2,6,6) \equal{} 6$.  Thus we have $ n^6 \equiv 1 \pmod{504} \implies n^9 \minus{} n^3 \equiv 0 \pmod{504}$ if $ gcd(n,504)\equal{}1$.

Question: how do we prove that this holds when $ gcd(n,504) \neq 1$?  As far as I can tell, carmichael's theorem only holds when $ n$ and $ 504$ are relatively prime.

On that note, I should correct myself from earlier:
[quote="facis"]
By euler's totient theorem, since $ \phi(9) \equal{} 6$, $ n^6 \minus{} 1 \equiv 0 \pmod 9$ if $ gcd(n,9) \equal{} 1$. Then we can multiply both sides of $ n^6 \minus{} 1\equiv 0 \pmod 9$ by $ n^3$ to get $ n^9 \minus{} n^3 \equiv 0 \pmod 9$

In the case $ gcd(n,9) \neq 1$, $ 3|n$ so $ n^3 \equiv 0 \pmod 9$.   Thus we can multiply both sides of $ n^3 \equiv 0 \pmod 9$ by $ n^6 \minus{} 1$ to get $ n^9 \minus{} n^3 \equiv 0 \pmod 9$.

So the desired result holds for all integers.
[/quote]

[quote="kunny"]Just factorize $ n^9 \minus{} n^3$.  :lol:[/quote]

Excuse my lack of knowledge in number theory, but how would using the factorization $ n^9\minus{}n^3 \equal{} n^3(n\plus{}1)(n\minus{}1)(n^2\minus{}n\plus{}1)(n^2\plus{}n\plus{}1)$ help to prove that the expression is always divisible by 3 for all integer $ n$?

It might be more obvious if you factored $ n^9\minus{}n^3\equal{}n^3(n^3\plus{}1)(n^3\minus{}1)$.  :wink:

Yeah, thanks; I always neglect the easy ways!

hmm. I knew that the factorization could be used to show that $ 3|n^9 \minus{} n^3$, but I was a little skepitical about using that to show $ 9$.  But after a little number crunching (and using the fact that for Polynomial $ f(x)$, $ x,k,p \in \mathbb{Z}$, $ f(x \minus{} kp) \equiv f(x) \pmod{p}$ so we only need to cube $ 0,1,2\ldots p \minus{} 1$ to find all cubic residue classes in mod $ p$):

$ 0^3 \equiv 3^3 \equiv 6^3 \equiv 0 \pmod{9}$
$ 1^3 \equiv 4^3 \equiv 7^3 \equiv 1 \pmod{9}$
$ 2^3 \equiv 5^3 \equiv 8^3 \equiv 8 \pmod{9}$

So we have $ 3$ cases:
$ n^3 \equiv 0\pmod{9}$ in which case $ 9|n^3$
$ n^3 \equiv 1\pmod{9}$ in which case $ 9|n^3 \minus{} 1$
$ n^3 \equiv \minus{} 1\pmod{9}$ in which case $ 9|n^3 \plus{} 1$

Putting these cases together, $ 9|n^9 \minus{} n^3$ for any integer $ n$.

Now lets tackle $ 504|n^9 \minus{} n^3$.  Since we have already shown $ 9|n^9 \minus{} n^3$, we only need to show $ 2^3|n^9 \minus{} n^3$ and $ 7|n^9 \minus{} n^3$

$ 8|n^9 \minus{} n^3$:

If $ n^3$ is odd, then both $ n^3 \plus{} 1$ and $ n^3 \minus{} 1$ are even, and since they are consecutive evens, one is a multiple of $ 4$ and thus $ 8|(n^3 \plus{} 1)(n^3 \minus{} 1)$.

If $ n^3$ is even, then $ n$ must be even and thus $ 8|n^3$.

In either case, $ 8|n^9 \minus{} n^3$.

$ 7|n^9 \minus{} n^3$:

$ 0^3 \equiv 0 \pmod{7}$
$ 1^3 \equiv 2^3 \equiv 4^3 \equiv 1 \pmod{7}$
$ 3^3 \equiv 5^3 \equiv 6^3 \equiv 6 \pmod{7}$

So we have 3 cases:
$ n^3 \equiv 0 \pmod{7}$, in which case $ 7|n^3$
$ n^3 \equiv 1 \pmod{7}$ in which case $ 7|n^3 \minus{} 1$
$ n^3 \equiv \minus{} 1 \pmod{7}$ in which case $ 7|n^3 \plus{} 1$
Thus $ 7|n^9 \minus{} n^3$ in any circumstance.

Putting all of the above together, $ 504|n^9 \minus{} n^3$ for all $ n$.


where is the homogeneous center of cicumcenter of a triangle and the inscribed circle of it.?
is it a good point or not? :D

- The external center of similitude of the circumcircle and the incircle of a triangle ABC is the isogonal conjugate of the Nagel point of triangle ABC (this was proven in http://www.mathlinks.ro/Forum/viewtopic.php?t=23668 ). It is $X_{56}$ in [url=http://faculty.evansville.edu/ck6]Clark Kimberling's Encyclopedia of Triangle Centers[/url].

- The internal center of similitude of the circumcircle and the incircle of a triangle ABC is the isogonal conjugate of the Gergonne point of triangle ABC (for the proof, see the link at the end of http://www.mathlinks.ro/Forum/viewtopic.php?t=23668 post #3). It is $X_{55}$ in [url=http://faculty.evansville.edu/ck6]Clark Kimberling's Encyclopedia of Triangle Centers[/url].

These two centers of similitude have some other, less simple properties. Try searching for "center similitude circumcircle incircle" on ML.

  darij

thanks a lot darij,you realy helped me.you know,i think you are one of the best in geometry,you are so cool in it. :D


Find the smallest integers that make $ x^{2} \minus{} 1666y^{2} \equal{} 1$.

hello, $ x\equal{}2449,y\equal{}60$ fulfills our equation.
Sonnhard.

could you offer an explanation with your answer

There is also the obvious solution:
$ x\equal{}1$ and $ y\equal{}0$


geometry
3D geometry
algebra
factorization
sum of cubes

[hide]Rewrite as $a^{3}+b^{3}=(a+b)^{3}=a^{3}+3ab(a+b)+b^{3}\implies 3ab(a+b)=0$

Thus, the solutions to the equation are:
$16x^{2}-9=0 \implies x= \pm \frac{3}{4}$
$9x^{2}-16=0 \implies x= \pm \frac{4}{3}$
$25x^{2}-25=0 \implies x= \pm 1$
[/hide]

Oops... I rewrote $a^{3}+b^{3}$ as sum of cubes and canceled off the $a+b$, and completely forgot about the $(a+b)^{2}$ on the other side :blush: .


Arqady, does work your solution also for the inequality

$6(a-b)^{2}(b-c)^{2}(c-a)^{2}\geq \sum_{cyc}(a^{2}-b^{2})(a^{2}+2b^{2})c^{2}?$

Both inequalities become equality for $a=1, b=2, c=4$.


Prove that if n is an even natural number, then the number 13^n+6 is divisible by 7.

[hide="Solution"]
Let $n=2k \ \forall \ \mathbb{Z}^{+}$. We have that
\[13^{2}\equiv 1 \pmod{7}\Rightarrow 13^{2k}\equiv 1 \pmod{7}\]
Therefore,
\[13^{2k}+6 \equiv 7 \equiv 0 \pmod{7}\]
[/hide]

$13 \equiv 6 \equiv-1 \pmod{7}$, hence the equation is equivalent to $(-1)^{N}+(-1) \equiv 0 \pmod{7}$

i agree, this problem was pretty straightforward if your were familiar with modular arithmetic.

[quote="mdk"]when doing this problem, I thought it was way too easy for an olympiad problem.[/quote]

It probably was not a national olympiad. Sometimes, olympiad can just mean: math competition.