I have a question...
this was a MC problem i had seen...

Jane and Emily are going to run a marathon.  Jane arrives 16 minutes before Emily does and starts running.  If Jane runs at 9 minutes a mile, and Emily runs at 8 1/4 minutes a mile, how many minutes will it take Emily to catch up to Jane?

This is what the answer said...  explain how...
[hide]176[/hide]

note that it says MINUTES A MILE, not miles a minute...

Well, convert to mpm (miles per minute)!
[hide]9 minutes per mile = 1/9 mpm
33/4 minutes per mile = 4/33 mpm
Jane has 16 minutes to run already, = 16/9 miles ahead
Each minute Emily catches up by 4/33 - 1/9 miles = 12/99 - 11/99 = 1/99 of a mile
(16/9)/(1/99) = 16*11 = 176[/hide]

Yeah... that one confused me until I realized I had to convert.  :rotfl:

much simpler way:

[hide]
 Since Jane runs 9 minutes a mile, and Emily runs 8 1/4 minutes a mile, for every mile Emily does, Jane gains 3/4 minutes on her. Since she is 16 minutes ahead, 16/(3/4) = 64/3. So Emily has to do 64/3 miles before Jane catches up. Since she does 8 1/4 minutes per mile, (8 1/4)(64/3) = 176.
[/hide]


What age do you act?  in relation to people your age around you that is

[quote="G-UNIT"]The real question you should be asking (and then later PMing me with the answer) is:

  What is your credit card number?[/quote]

8675 3091 6182 7185

umm i really dont think credit card number is important without the credit card company  :D

nonsense just try all the companies the hard part is picking the number

strangely enough I have exactly the same number as joml now isn't that weird

Can someone tell me all of the credit card sites please...you know..just for reference?

Wtf?  How can someone act more than 120 years old?

And how does someone edit their post 47 times?

[quote="catcurio"][quote="h_s_potter2002"]Wtf?  How can someone act more than 120 years old?
[/quote]
It says "percent of current age," so 120%.

[quote="h_s_potter2002"]And how does someone edit their post 47 times?[/quote]
If you mean la mafioso, I think that's part of his signature. Notice it says that it was edited on 12/7/05. ;)[/quote]

Oh  :blush:

Hmm, where's the option for $-\exp(\pi^2) \%$ ???

It appears when you PM an "auenthentic" credit card number.

Lockem thisem topicem. This hasn't been posted on for nigh on to three years now.

wtf...how do u randomly activate a 3 year old topic..lets lock this or let it die plz.


Source: 1987 AHSME #12 [from anniversary edition]

In an office, at various times during the day the boss gives the secretary a letter to type, each time putting the letter on top of the secretary's inbox.  When there is time, the secretary takes the top letter off the pile and types it.  If there are five letters in all, and the boss delivers then in the order 1 2 3 4 5, which of the following could NOT be the order in which the secretary types them?

(A) 12345
(B) 24351
(C) 32415
(D) 45231
(E) 54321

Give your answer and explain your reasoning.

Here goes. [hide] A obviously works, if the letters are typed as they are delivered. E also is very obvious - all the letters were delivered, and then typed. B works as well - 1 and 2 are delivered, 2 is typed, 3 and 4 are delivered, 4 then 3 are typed, 5 is delivered and typed, 1 is typed. C works - 1,2,3 are delivered, 3,2 are typed, 4 is delivered and typed, 1 is typed, 5 is delivered and typed. We're left with D, which doesn't work. 1,2,3, and 4 must be delivered in that order, 4 must be typed, 5 delivered and typed, and then we're left with the problem that 3 is on top of 2, so has to be typed before 2....[/hide]

I wonder -- is there a nice way to classify all those combinations which are or are not possible?  Or calculating how many of them there are?  (A pattern search is probably in order here.)

[quote="paladin8"]the multiple choice format gives it away... an interesting question however.[/quote]

This is one of those multiple-choice-required questions.  That is, there is more than one order that cannot be achieved, so if you didn't know the options, you wouldn't know which answer is desired.

Well, I believe it has to do with trying to get at a paper that's already far down in the pile.  For instance, with 45231, we start:

The first letter typed up is 4.  So far so good.  Then 5.  Clearly, all of the letters have been put in the pile now.  Then 2.  Bad.  3 will always be above 2, so we can't skip 3.

With 34152, we run into another problem:

First letter is 3.  So far so good.  Then 4.  Fine.  Then 1.  No good.  We jumped downward past 2, and since 2 is always above 1, we cannot do this.

With 12534, we run into the same problem again:

First letter is 1.  That's fine.  Boss puts paper in, secretary takes it out.  Then 2.  Once again, boss puts paper in, secretary takes it out.  Then 5.  Uhh...okay...so the boss puts 3 4 5 in, and the secretary has just taken the top one (5) out.  Then 3.  Bad.  We can't get at 3 until we get at 4, because 4 is always above 3.

For that matter, the sequence 31254 wouldn't work, since you're jumping from 3 to 1 and skipping 2.

If the only way to not get something to work is to have a downward jump, then it wouldn't be terribly hard to calculate the number of possibilities.

Don't tell me it's doable -- I know it's doable.  There are only 120 possibilities, so worst-case scenario is to list out all 120 and go checking each one individually.  My question is, can you do it?  (Preferably without listing all of them :) )

We can set up some general rules:

Going from left to right, the numbers can increase by any amount, but can only decrease by one (unless the number one less than the current has already been taken, in which case it can decrease to the next available number)

So many combinations are ruled out by an intelligent person making a list right away - like all those starting with
51
52
53
41
42
31
And, of course, that can be easily extended to include the first three digits. But just that list alone rules out many possibilities (36?)

So, yeah, I'd say that they could be listed without too much work.


search
inequalities
function
inequalities unsolved

Let $ a,b,c$ reals, such that a,b,c>=0, but not all 0; prove:
$ \frac{bc}{b^{2}\plus{}c^{2}\plus{}3a^{2}}\plus{}\frac{ca}{c^{2}\plus{}a^{2}\plus{}3b^{2}}\plus{}\frac{ab}{b^{2}\plus{}a^{2}\plus{}3c^{2}}\leq\frac{3}{5}$

good, I found another solution:
If the inequality is true for some $ a, b, c$, then for $ ka,kb,kc$  for $ k > 0$ also is true;
so let: $ a^{2}\plus{}b^{2}\plus{}c^{2}\equal{} 1$, so the LHS is: $ abc(\sum_{cyc}{\frac{1}{a(1\plus{}2a^{2})}})$
consider the function $ f(x) \equal{}\frac{1}{x(1\plus{}2x^{2})}$, and $ g(x) \equal{} x\plus{}2x^{3}$
the second derivate of this function is $ 12x > 0$, so for $ x > 0$, g is convex then f is concave in x>0
Then for Jensen´s inequality: $ v \equal{}\frac{a\plus{}b\plus{}c}{3}$
$ abc(\sum_{cyc}{\frac{1}{a(1\plus{}2a^{2})}})\leq\frac{3abc}{v(1\plus{}2v^{2})}\leq\frac{3}{5}$
if and only if: $ 135abc\leq (a\plus{}b\plus{}c)(11a^{2}\plus{}11b^{2}\plus{}11c^{2}\plus{}4ca\plus{}4ab\plus{}4bc)$, this is true for AM-GM inequality

[quote="cuenca"]
consider the function $ f(x) \equal{}\frac{1}{x(1\plus{}2x^{2})}$, and $ g(x) \equal{} x\plus{}2x^{3}$
the second derivate of this function is $ 12x > 0$, so for $ x > 0$, g is convex then f is concave in x>0
[/quote]
$ f''(x)\equal{}\frac{2(1\plus{}6x^{2}\plus{}24x^{4})}{x^{3}(1\plus{}2x^{2})^{3}}>0$ for $ x>0.$
Thus, $ f$ is convex function. :wink:

[quote="cuenca"]
consider the function $ f(x) \equal{}\frac{1}{x(1\plus{}2x^{2})}$, and $ g(x) \equal{} x\plus{}2x^{3}$
the second derivate of this function is $ 12x > 0$, so for $ x > 0$, g is convex then f is concave in x>0
[/quote]
How did you know that f is concave? :)

$ \sum_{cyc}\frac{ab}{a^{2}\plus{}b^{2}\plus{}3c^{2}}\leq\sum{cyc}\frac{ab}{2ab\plus{}3c^{2}}\equal{}\sum_{cyc}\frac{1}{2\plus{}\frac{3c^{2}}{ab}}.$
Let $ \frac{a^{2}}{bc}\equal{}x,\frac{b^{2}}{ac}\equal{}y,\frac{c^{2}}{ab}\equal{}z$,with $ xyz\equal{}1$.
We must prove that :
$ \sum_{cyc}\frac{1}{2\plus{}3x}\leq\frac{3}{5}$,with $ xyz\equal{}1$
Can anybody prove it???(or maybe it is wrong)

I'm not sure,but try $ x\equal{}1,y\equal{}2,z\equal{}\frac{1}{2}$.

Your inequality is wrong.
If $ x\equal{}1,y\equal{}2,z\equal{}\frac{1}{2}$.then
$ \sum\frac{1}{2\plus{}3x}\equal{}\frac{171}{280}\geq\frac{3}{5}$. :rotfl:


define a sequence such that $ a1\equal{}2 ,a2\equal{}3and an\equal{}3an\minus{}2 \plus{}2an\minus{}1\equal{}an?$

[quote="vaibhav2903"]define a sequence such that $ a1 = 2 ,a2 = 3and an = 3an - 2 + 2an - 1 = an?$[/quote]

If the question is "what is the general term of the sequence defined as $ a_1 = 2$ ,$ a_2 = 3$ and $ a_n = 3a_{n - 2} + 2a_{n - 1}$" , then :

Characteristic equation is $ x^2 - 2x - 3 = 0$ whose roots are $ - 1$ and $ 3$ and so $ a_n = \alpha( - 1)^n + \beta 3^n$ with :
$ a_1 = 2 = 3\beta - \alpha$
$ a_2 = 3 = 9\beta + \alpha^$

And so $ \alpha = - \frac 34$ and $ \beta = \frac 5{12}$

Hence the result : $ \boxed{a_n = \frac {3( - 1)^{n - 1} + 5\cdot 3^{n - 1}}4}$

[quote="vaibhav2903"]this is wrong[/quote]

What is wrong ?

Is my understanding of the problem "what is the general term of the sequence defined as $ a_1 \equal{} 2$ ,$ a_2 \equal{} 3$ and $ a_n \equal{} 3a_{n \minus{} 2} \plus{} 2a_{n \minus{} 1}$" wrong ?
In such a case, please make your question more clear and explain what I misunderstood in your problem statement.

Is my understanding of the problem true but my result wrong ?
In such a case, explain why I am wrong.
I checked my result and it seems quite OK for me :
$ a_n \equal{} \frac {3( \minus{} 1)^{n \minus{} 1} \plus{} 5\cdot 3^{n \minus{} 1}}4$

$ a_1\equal{}\frac{3\plus{}5}4\equal{}2$

$ a_2\equal{}\frac{\minus{}3\plus{}15}4\equal{}3$

And obviously $ a_n \equal{} 3a_{n \minus{} 2} \plus{} 2a_{n \minus{} 1}$ $ \forall n>2$

vaibhav2903 , could you answer my question, please ?

[quote="vaibhav2903"]i typed it wrong .it was actually,$ a_{n} \equal{} 2a_{n \minus{} 2} \plus{} 3a_{n \minus{} 1}$and $ a_{1} \equal{} 1$.sorry for that.and your solution is correct.i tried that one.[/quote]

So I think that the question is "what is the general term of the sequence defined as $ a_1 \equal{} 2$ ,$ a_2 \equal{} 3$ and $ a_n \equal{} 2a_{n \minus{} 2} \plus{} 3a_{n \minus{} 1}$"  (I added the $ a_2$ value) then :

Characteristic equation is $ x^2 \minus{} 3x \minus{} 2 \equal{} 0$ whose roots are $ r_1\equal{}\frac{3\minus{}\sqrt{17}}2$ and $ r_2\equal{}\frac{3\plus{}\sqrt{17}}2$ and so $ a_n \equal{} \alpha r_1^n \plus{} \beta r_2^n$ with :
$ a_1 \equal{} 2 \equal{} \alpha r_1\plus{}\beta r_2$
$ a_2 \equal{} 3 \equal{} \alpha r_1^2\plus{}\beta r_2^2$

And so $ \alpha \equal{} \frac 1{r_1}$ and $ \beta \equal{}\frac 1 {r_2}$

And so $ \boxed{a_n\equal{}\left(\frac{3\minus{}\sqrt{17}}2\right)^{n\minus{}1}\plus{}\left(\frac{3\plus{}\sqrt{17}}2\right)^{n\minus{}1}}$

[quote="vaibhav2903"]its $ a_{1} \equal{} 1$ not $ a_{1} \equal{} 2$[/quote]

Ok, I think you are able to modify my solution in order to match $ a_1\equal{}1$.

[url http://www.artofproblemsolving.com/Forum/viewtopic.php?t=331522]

[url http://www.artofproblemsolving.com/Forum/viewtopic.php?t=331524]

A must read for you: http://www.mathlinks.ro/viewtopic.php?t=135914 .


In parallelogram ABCD, M and N are the midpoints of sides AB and CD, respectively. If AB=24 and the area of parallelogram ABCD is 120, what is the area of the region shared by $\triangle ANB$ and $\triangle CMD$?

i think its [hide] 20 but i am not sure, so i will keep my reasoning until i am sure[/hide]

I got
[hide]30[/hide]
but I am unsure too.

And I will keep my solution until someone tells me it's right (probably not).

Hmm.. I got [hide]30[/hide] too.

Here's a picture:

[hide]30, from my own drawing (it's one fourth of 120)[/hide]


Hi people

Just wondering

Time ago I'd like to compose a PDF made in LaTeX, but already know that I have to download LaTeX and install it.

The point is I don't know what download, how to install LaTeX neither how to make PDF in LaTeX.

I'd really appreciate your help.

I hope you can tell me what I have to do step by step.

You'll find everything you need in the AoPS guide starting at http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php


The $n$th triangular number, $t_{n}$, is defined by $t_{n}=\frac{n(n+1)}{2}$, $n\geq1$ (the sum of the first $n$ positive integers). There are two pairs of triangular numbers such that their sum and positive difference are triangular numbers. Find them.

[hide]$(21,15)$ is one pair, since $21+15=36$ and $21-15=6$. The other pair seems to be a little more difficult to find though...[/hide]


Gauss
geometry
power of a point
radical axis
geometry unsolved

Let ABCD be a convex quadrilateral and let E and F be the points of intersections of the lines AB, CD and AD, BC, respectively.
Prove that the midpoints of the segments AC, BD, and EF are collinear.

Thats the gauss line of the complete quadrialateral.....

[u][i]proof[/i][/u]
First of all we have to use a theorem that tells us that the circles with diameters the diagonals of the qomplete quadrilateral form a bundle....

Let $ C_1,C_2,C_3$ be the circles with diameter $ BD,FE,AC$.

These 3 circles form a bundle.

Let $ l$ be the radical axis of these tree circles,and $ K,L,N$ the midpoints of $ AC,BD,EF$.

From well known thereom we now that $ KL \bot l$ and $ KN \bot l$ (note that $ K,L,N$ are the centers of $ C_1,C_2,C_3$).

But from a point there is only one perpendicular line to a given line.
So $ K,L,N$ are collinear....

Dimitris

Ask Jean-Louis, he will provide you a link where to find 30 various proofs of this theorem.

Best regards,
sunken rock

Let $ M,N,L$ be the midponts of $ AC,BD,EF$ and $ P,Q,R$ the midpoints of $ BC,CE,EB.$ It's is clear that $ L,M,N$ lie on the sidelines $ RQ,QP,PR$ of the medial triangle $\triangle PQR$ of $ \triangle BCE,$ hence:

$ \frac {\overline{LD}}{\overline{LR}} \equal{} \frac {\overline{FC}}{\overline{FB}} \ \ , \ \ \frac {\overline{RN}}{\overline{NP}} \equal{} \frac {\overline{ED}}{\overline{DC}} \ \ ,  \ \ \frac {\overline{PM}}{\overline{MD}} \equal{} \frac {\overline{BA}}{\overline{AE}} \ \Longrightarrow$

$ \frac {\overline{LD}}{\overline{LR}} \cdot \frac {\overline{RN}}{\overline{NP}} \cdot \frac {\overline{PM}}{\overline{MD}} \equal{} \frac {\overline{FC}}{\overline{FB}} \cdot \frac {\overline{BA}}{\overline{AE}} \cdot \frac {\overline{ED}}{\overline{DC}} \equal{} 1$

By the converse of Menealus' theorem, the points $ M,N,L$ are collinear.

For other (probably) new solution, please, see http://www.mathlinks.ro/viewtopic.php?p=1114247#1114247.

Best regards,
sunken rock


function
real analysis
real analysis solved

The problem is to constract a real function such that for any interval $(a,b)$ and every real $c$, there exists $x\in (a,b)$ such that $f(x)=c$.

I have found such a constraction using the fact that Cantor's set in $[0,1]$ has the power of the continuoum.
I'm intrested in seeing more solutions.

I'm pretty sure this has come up before on the forum, and constructions were sketched.

Let $(a_{\alpha})_{\alpha\in \mathbb R}$ be a system of reals with the property that there are no two of them whose difference is rational (i.e. a system of representatives for the relation $x\equiv y\iff x-y\in\mathbb Q$). Now make $f(x)=\alpha$ iff there is a rational $r$ s.t. $x=\alpha+r$.

Dear Grobber,
Do you remember my silly post about Continuoum Hypothesis ?
To tell you the truth, this silly question come to me while i was trying to solve this problem.
The constraction you are proposing is exactly my first constraction. But i turned it down because i just don't know whether or not this system of reals has the power of the continuoum.

grobber's construction is off in a detail; we need a 1-1 correspondence between $\mathbb{R}/\mathbb{Q}$ and $\mathbb{R}$ to make it work. On the other hand, showing that these two sets have the same cardinality is easy. Just use the well-known fact that $\mathrm{card}(A\times B)=\max(\mathrm{card } A,\mathrm{card }B)$ if $A$ and $B$ are infinite.

perhaps a more explicit construction:
$x=a.x_1x_2..x_k..$, the decimal expansion of $x$.
Let $X_n$ be the number of $1$ among $x_1,x_2,..,x_n$.
Take $g(x)=limsup(\frac{X_n}{n})$. Then $g$ maps every interval onto $[0;1]$.
So it is easy to conclude.

jmerry, i don't understand what you are saying.
Grobber's system of reals doesn't miss only the rationals but uncountably many irrationals too.
What am i missing?

alekk, very clever construction.  :)

Here's what jmerry is saying (I did not say it only because I thought I didn't have to :)):

Let $M$ be that set, and let $|M|$ be its cardinal. Every real can be written in a unique way as $m+r,\ m\in M,r\in\mathbb Q$. This means that $\mathbb R$ is in bijective correspondence with $M\times \mathbb Q$, so $c=|M|\cdot \aleph_0$, which equals $|M|$, from what jmerry said. This is precisely what you want: $|M|=c$.

I missunderstood jmerry's words.
It's clear now, Thanks  :)

(So, my first constraction finally works  :P  :P )


prove: $\not\exists k\in\mathbb{Z}$ s.t.
$(n+1)^4-2n(n-1)=k^2$ for some $n\in\mathbb{Z}$

Lets write $m=n+1$.
Then the desired number is $m^4-2(m-1)(m-2) = m^4-2m^2+6m-4$.
But $(m^2-1)^2 = m^4-2m^2+1$. So for $m\geq 0$, the desired number lies between consecutive squares, so is not a square. The same holds for $<0$, too, just with inverse direction.


Mr. UltraGenius creates a spammosphere, where everyone is supposed to spam. Mr. Supergenius spams 20000 times per second using a computer program. He spams 5 more every second later, for example in the first second he spams 20000 times, the second 20005, the third 20010, and so on. How many spam messages whould he have spammed after 2 days, 2 hours, 2 minutes, and 2 seconds?

anirudh, do you have the correct answer for this? Somehow, I doubt it.

Are we allowed to use calculators?

[quote="i_like_pie"][hide]He spams for a total of 180122 seconds.

$\sum_{n=0}^{180121}20000+5n$[/hide][/quote]

1. That's probably above MathCounts level

2. You didn't answer the question

3. Please show how you got the number of seconds for other people

[hide]$172800+7200+120+2=180122.$ He spams a total of 180122 seconds. This can be expressed as $\sum_{n=0}^{180121}20000+5n.$ Notice that the maximum value of $n$ is one less than the total amount of seconds he spammed. This is because $n$ starts at $0,$ not $1.$ The answer is $84,711,806,905.$[/hide]

[quote="anirudh"]Actually, it is very easy to solve it if you make it into a Summation. Then you can input it into a calculator or computer program and get the answer. Some websites can do it too if you don't have a TI-89.[/quote]
#1: Most graphing calculators can calculate summations (except I forgot how :D)

#2: You don't need a summation. You could input, on a regular calculator, what alex__ib did.

IN a TI-89, you can use the sum to do $\sum \text{(Variable on side, Variable such as }n\text{, The bottom number start, The end Number.}$

This is assuming everyone has a TI-89?  Especially when MC nats gives TI-84+Silvers?

You could simply make the numbers smaller so we could handcalculate the numbers.

Summations should NOT be involved in any way in MathCounts.

[quote="alex__ib"]You are insane if you put us to calculate that  :P .

[hide]our wanted result is
$20000\times180122+5(1+2+3+4+5+.................+180121)=3602440000+5\times\frac{180121\times180122}{2}=8471806905$[/hide]

DUDE YOU ARE OUT OF YOUR MIND!!!!!!!  :rotfl:[/quote]
You forgot a 1 in the millions place. :wink:

The TI-84 can do it. It's under list, i think, but I'm not sure exactly how to do it.

@Ignite: silly comment. Summations appear all the time in Mathcounts. Ever see this? $1+2+...+n = \frac{n(n+1)}{2}$. Well it's no different if we write it as $\sum_{i=1}^{n}i$. And that's all this problem really is.

See, anirudh: This is the type of topic that I'm referring to when I say spam.  Your topics really aren't about math AT ALL.  Maybe you think you look smart posting all of this about the rules of Mathcounts, but really, all it does is piss people off and make you look [i]stupider[/i]--you're not actually doing the math at all--when you try to act "smart". :mad:

[quote="anirudh"]You are allowed a TI-89 in MC.

[quote][b]Calculators that have the ability to enter letters of the alphabet but do not have a keypad in a standard typewriter arrangement are acceptable.[/b][/quote]

Such is the case of the 89Tit

[quote][b]Where calculators are permitted, students may use any calculator that does not contain a QWERTY (i.e., typewriter‑like) keypad.[/b][/quote]

The 89Tit does not fit this category either.

[hide="Image Here"][img]http://www.ticalc.org/images/calcs/89-ti-big.gif[/img][/hide]

Thus, the TI-89 Passes both categories, and therefore is allowed.  :wink:[/quote]

Again, this is assuming everyone has the, oh, 200 dollars or so to spend on a TI-89 when we could have solved your problem with more elegant methods?

i think TI-89s are not allowed for another reason....

like for special programming ways or soemthing...

[quote="i_like_pie"][quote="ragnarok23"]I suppose the hackers have done this too, anirudh?[/quote]
Owned. :rotfl:

I think this topic should be locked. The problem is solved, and now people are spamming.[/quote]
Yes, it really should be.  :wink:  Locked.


I'm taking finite math over the summer and since class doesn't start until June 5th I want to get a head start on it.

The description of the class says that we will be going over these topics- Logic, sets, partitions, counting, probability rules and Bayes’ Theorem, vectors, matrices, and linear programming are discussed. Applications to the behavioral sciences and to business are included.

I have a hard time with math so a site that explains everything thoroughly with lots of examples would be great. Thanks in advance.


[quote="rem"]
I'll start. My school is really boring, we don't get any homework(at least me), and so I do nothing in school.[/quote]

Me too. I rarely get homework(we get some optional :huuh:  homework, but not much real homework :D ). I had more piano practice this year than homework(I used to practice monthly, however the last time I touched piano was in the winter camp, which was in January) :rotfl:

_____________________________________________
EDITED: found out that homework is not countable, so no plural

Homework...yeah.

I like homework if you can gain from it.  With classes that I'm taking now, you either do your homework, or you fail.  (Ok that's a bit of an exaggeration; I tend to minimize homework time by intentionally missing out on what I feel isn't necessary).  What I like about my current school, is that homework is never mandatory.

I think Ontario Secondary School pedagogy is questionable.  When I went to A.B. Lucas in London, teachers tended to stuff you with "busy work" or "homework" in 10th, 11th, and even 12th grade classes!  Actual learning was minimal, and grades were heavily emphasized on work that quite frankly, I felt was a waste of time.  I didn't see a point to much of what was taught in Ontario Secondary Schools and consequently, I was very dissatisfied with the quality of education.

[quote="Sunny"]I'M DONE I'M DONE I'M DONE IB!!!!!!!!!

My last IB test was today.

Soo...yeah, I go to a small town school, in the middle of no where.  And it has IB.  Actually, its the only non-city IB school in the world.  Nothing special about it.[/quote]Congratulations, n that's kinda amazing about the only non-city IB school in the world.  Also, look at your post count.  That's nice eh?

Anyway, school.. uhh, well, for the homework thing - it really depends.  When I was a kid in elementary school, I was lucky in that math made a lot of sense to me, so it wasn't very difficult for me to do well.  However, because I wasn't challenged, I got so sick of doing the homework, and I stopped completely.  Going on to high school, I found that I was very slow and could never finish tests, even though I knew everything very well.  So for me, I find that homework helps.  In calculus this year, if I hadn't done my homework, I don't think I would have done nearly as well.

So, it really depends on the person.  And sometimes, you may think it's pointless, but it may hinder you in the future if you don't do it now.  Sometimes, the stuff they teach is just too basic and you have already mastered the material and understand it on very many levels - in this case, you are in the wrong class, probably.

I agree with 'the master', while its NOT hard to wrap your head around highschool concepts, you still need muscle memory for basic operations, or practice looking for other cases etc.

same principle of 'learning through problem solving', we solve hard problems, to make the easy problems rediculousely easy, becuase when you 'understand something', you'll never forget it.

Its true- doing homework helps!! In grade 10, I didn't use to do any homework, in fact, I never took my math textbook home and I use to get perfect on all the math tests, so I used to think that I am too good at math. I realised that I am not good at math when I started to write math cotests in which I got good score but not perfect scores. Now, I do extra questions that are not assigned for homework to improve my math skills.

[quote="rem"]Well, we have three university threads so why not create a schools thread, taking into account the fact that most aopsers are middle or high school students.
Here we can discuss how boring school is, personal school experiences, what cool courses are offered at schools, chat with ppl from same school, etc
Please don't discuss contests here.
I'll start. My school is really boring, we don't get any homework(at least me), and so I do nothing in school.
Does anyone here know the William Lyon Mackenzie school in Toronto? I'll be moving in the area around that school so I'll have to go there, so I want to know more about this school.[/quote]
wait...
alex, you're moving to TORONTO?!

My school is more of Art school. Students doing pretty good on math but school is not interested in it. Teachers are just like take some competitions and then they don't announce the results.
They even don't know what my mark is for COMC and other competitions for this year and last year.
Every year, Yrok Region has Mathematics Award Night, and at there ppl who did good job in COMC, CMC, and so on get some awards. But last year, my math teachers didn't realize that I got provincial medal so it took me 5 months to get my certificate. Again this year, I got 2 awards and they forgot to give me invitation letter. So, guess what. I don't know where is my certificates now,,,-_-. And again, my math teachers didn't realized that I got provincial medal in COMC this year. So now, I just gave up to get some info from my math teachers LOL.

Actually my school is one of the schools that gives tons of hwks. But, it does not affect mark that much. Just teachers hate you, if u don't do hwk.LOL and I was the one who did none of the hwks for Caculus last semester..LOL... I was kinda angry at all my math teachers. For my mid-term, my cal. teacher gave me N(="needs inprovement") for my homework section, but I still didn't do any hwk and lastly he gave me E(="Excellent").

School itself is fun tho. Students r good&funny, Classes r pretty interesting, and my school have lots of events. I think my school will be the one which has some sort events at least twice every month.

BTW, Alex r u moving to toronto?

[quote="rem"][quote="lightrhee"]we get some optional :huuh:  homework [/quote]
OMG this is hilarious :rotfl: . The first time in my life I've heard something this funny about school. Here's a fun fact about homework in my school: if you take Math gr 9 or 10, you have to do homework, teachers check it, and the homework mark is worth a lot. (about at least 30% :o ). In comparison, for Math in senior grades, teachers don't check homework becasue they think "senior students are responsible" :rotfl: *cough cough not me -- i am not a senior student*. The funniest part is, it's much more important to do homework in senior grades than in gr 9 and 10 espescially in Math. Life is funny and stupid :)[/quote]

LoL, well I think that might actually be smart because by making homework worth nothing, the teacher gives you no insentive to do the homework.  However, most students need to, or should do their homework in math so that they do well on the tests.  I guess that isn't a problem for most people here, but it's kind of a way of building independence.

Personally for homework... Yeah... we get... some... but it's all busy work and repetitive tasks.  :mad: My math mark is comprised 85% of tests, 10% of "assignments" which are actually just pretests, a test before each test except we can take it home and it's essentially the same format as the test, same length and difficulty, just that it's for review, and 5% on our manners in class.  Then that is all comprised as a mark out of 100, and ends up being 70% of our final mark with the other 30% being our final... in other classes... we don't learn anything about what we're doing, we just get assignments relating to it.  Such as for english, our teach doesn't go over how to do an essay, he just tells us to do one.  History, we copy maps and boring stuff... history is essentially just memorizing stuff which is borrrinnng =/

Oh, and chemistry is easy... it's essentially just simple math made into a new class and given a different name :D

[quote="nneonneo"]Walter Murray isn't exactly known for math. We have no IB, no AP or anything like that :(

However, we DO have a wonderful math teacher who is willing to try new things :D

We didn't even write the COMC, Euclid, or AMC before my grade 10 year, which is pretty unfortunate (mind you, we did very well on Euclid as a school, placing 7th nationally!)

Still...I would love to be from a more math-oriented school...oh well, you get what you get.

I'm finishing grade 11 (first final was today) and I'm in the Advanced Program--it's not AP, but it's purportedly for students who are gifted in learning.[/quote]

No man, walter murray is world wide famous :P

I'm guessing it's you Bobby :P this is Chris ;)

And yes, Mrs Christ is amazing!

And yes, our advanced program is a joke!

And yes, we have to write more contests  :P

Mrs. Christ? Are you serious?

I go to Lord Byng Secondary in Vancouver. We have two really good senior teachers, so we do well on the provincial exams, and also on the contests. Unfortunately this year we didn't do much enrichment in Math 12 Enriched. Part of the problem is that over the past 10 years the curriculum in BC has been degrading. The Ministry keeps removing things from the curriculum, first geometry and calculus, and now conics. They also used to have a scholarship exam which was problem-solving oriented, which they removed. They tried a Geometry 12 course a few years ago but it failed, so there is basically only one math 12 course. There is an Applications of Mathematics 12, which is basically an introduction to linear algebra, but it's not taken seriously.

[quote="otaku"]
wait...
alex, you're moving to TORONTO?![/quote]
Maybe. Maybe in summer maybe in autumn. Maybe.

man, that's awesome
you shoulda told me
anyway, give me a shout if you decide to come


limit
logarithms
calculus
integration
factorial
function
calculus computations

How do you find $ \lim_{x \to \infty}(ln(x!) \minus{} (x\plus{}.5)(lnx\minus{}1))$?

$ \ln(n!)\equal{}\sum_{k\equal{}1}^n \ln k$. Estimate this sum with an integral; that gets you the $ n\ln n$ dominant term, and more depending on how carefully you do the approximation. This has certainly appeared before.

You can use $ x! \approx \sqrt { 2 \pi x} \left( \frac {x}{e} \right)^x$, which is called Stirling's approximation. $ (*)$

Now, the problem becomes an exercise in elementary algebraic manipulations

$ (*)$ It can be somewhat intuitively determined that the growth rate of the factorial is proportional to that of $ \sqrt {x} \left( \frac {x}{e} \right)^x$; if the integral divided by the interval size gives the arithmetic mean, then we can make an analogy that the exponential of the arithmetic mean of the logarithm of a function gives the continuous geometric mean. We can use such a method to approximate the discrete product of a factorial by an integral, which works out to be proportional to Stirling's approximation.

I guess the question then becomes, how do you prove Stirling's approximation

Look here : [url]http://www.sosmath.com/calculus/sequence/stirling/stirling.html[/url]  :)

Well, my post was how you prove Stirling- at least to within $ O(1)$ in the logarithm.


Are you smart or an average student in the main subjects?

First of all, IQ measurements are test-dependent.  One test might put you at 120, another at 150.  And second, where's the option for 130+? :lol:

CUMULATIVE RESULTS!!!
Also, I put 130+, and it disappeared.

This is not appropriate discussion for the NC forum.

Locked.


[hide]
RRT gives $x=7$, then quadratic formula gives $x=4+3i$ and $x=4-3i$.[/hide]


he creado un foro de la olimpiada de yucatan en donde compartiremos material, examenes, soluciones de las diversas olimpiadas al igual que algunos resultados de esta delegacion, apenas esta comenzando y por lo que no hay muchas cosas, pero poco a poco se iran agregando.

el link es: http://ommyucatan.mundoforo.com


The number $\sqrt{17+12\sqrt{2}}$ is an example of a number expressed in "embedded radical form" because of the $\sqrt{2}$ under the radical. What is the equivalent value of this number expressed in simplest radical form $a+b\sqrt{3}$ where $a$ and $b$ are integers and $c$ is a positive integer containing no perfect square factors greater than 1?

ok, number one i think u typed the problem wrong... but if it was a+b(squareroot2), instead of 3, it would be 5..

EDIT: and i read the quetsion wrong, i thought it asked for a+b, which is 5

Yeah, he/she typed it wrong.

3+2sqrt(2) is the answer.

No that cant be.. if the 3 was a 3 there would be no solution, you can get a sqr2 from sqr3..

[quote="#H34N1"][img]7255[/img]

Yes. the 3 is supposed to be a 3[/quote]
Well, according to that, the answer should be $3+2\sqrt{2}$.  To figure this out, I guessed and checked.  By the way, I think you meant that the 3 was supposed to be a c.


In rectangle ABCD, points P and Q lie on $ \overline{AB}$ and $ \overline{DC}$ respectively. $ \angle PMQ$ is a right angle, M is the midpoint of $ \overline{BC}$ and $ PB \equal{} \frac43BC$. What is the ratio $ PM: MQ$? Express your answer as a common fraction.
[asy]pair A,B,C,D,P,M,Q;
C = (0,0); B = (0,48); P = (-64,48); M = B/2;
Q = (-9,0); A = (-80,48); D = (-80,0);
draw(Q--M--P--B--C--D--A--P);
label("A",A,W);label("B",B,E);label("C",C,E);
label("D",D,W);label("M",M,E);label("Q",Q,S);
label("P",P,N);[/asy]

[quote="5849206328x"]Let $ BM\equal{}BC\equal{}x$.  Then, $ PB\equal{}\frac{8x}{3}$.

Triangle $ PBM$ is similar to triangle $ MCQ$, so \[ \frac{PM}{MQ}\equal{}\frac{\frac{8x}{3}}{x} \Rightarrow \frac{PM}{PQ}\equal{}\boxed{\frac{8}{3}}\][/quote]

BM does not equal BC.

Actually, BM=BC because in the problem, it states that M is the midpoint of BC.


lol same
in reality...
at home i wake up between 8 and 9
at school, if i have work to finish up, around 6
if i dont, 7:55 for an 8 o clock class... -.-

it depends on if there is school or not. without school, i sleep till like 11 or 12. on a school day i wake up at 6:29 - just in time to get ready and make it school before i am tardy.

6:00 or earlier most of the time. I like getting up early and going outside, it's really nice. The problem is I also like staying up late.

The time I wake up at varies.  Some days I get up with my alarm others I sleep through it. Also, my clock is set like 40 minutes fast (I'm too lazy to fix it :D ).  On average I wake up at about 5:30 or so.

[quote="sky9073"]I was involuntarily woken up today at 2:00 pm. Could've gone on sleeping for several hours..[/quote]

Sleeping during class, eh?  (or is that just a typo)

Okay. How about we try to keep this clean. If you have to -------- out the word, it's probably not appropriate so don't use it. :-)

i get up at around 6:10 in order to get to school on time. but over the weekends, i get up at around 11.

[quote="sky9073"]Spring break. Absolutely wonderful. :) I slept for 12 hours straight.[/quote]

heehee, sleep at 2 wake up at 2, ladeda.

spring break i wake up at about 10:30. summer i wake up 12-2

i wake up at seven , i eat something(if i remember to) and then i take my bike and if there's math at school i go to school if that's not a mate school i go school if i remember the way :D

well...i'm [i]supposed[/i] to get up around 6:30 on school days... ;) i usually don't get up til 7 or shortly after tho. on weekends i get up 'round 10 or so on saturday and 7:30 on sundays for church. then summers i sleep til 9 or 10. kinda depends whether i'm in canada or not.

In my winter vacation, i slept at 2 every night and got up at 12, i really miss the break!

Spring break or not, my dog wakes me up before 6:00 AM.

i wake up at around 6:30, but during spring break, i wake up at 9 (can't sleep till after 9 for some reason...not that i haven't tried...)

On a school day I wake up around 6:30, but on a weekend I usually sleep till 2.

School days 5:30, weekends 8:00-9:00, during long breaks 10:00-11:00.

About 6:30 on school days, 10 o' clock on any other day

Around 4 to 5:30, before my parents wake up so I can play Warcraft and Maplestory in the morning. I stay up to around 11 PM to 2 AM, thanks to caffeine. Not too good for my health, but... :P


[quote="archimedes1"]Given triangle $ \triangle ABC$ with $ AB=AC=7$ and $ BC=2$:
Let $ I$ be some point in the interior of the triangle such that lines $ IB$ and $ IC$ bisect angles $ B$ and $ C$, respectively.  Now let $ I$ be the midpoint of side $ DE$ of equilateral triangle $ \triangle ADE$.  Find the side length of the equilateral triangle.[/quote]

[hide]
[asy]pair B,A,C,I; B=(0,0); A=7*dir(81.7867892983); C=2*dir(0); I=incenter(A,B,C); draw(A--B); draw(B--C); draw(C--A); dot(I); label("I",I,NE); label("A",A,N); label("B",B,SW); label("C",C,SE); draw(incircle(A,B,C)); draw(I--(dir(81.7867892983))); draw(A--I); draw(B--I); draw(C--I); draw(I--midpoint(B--C));[/asy]

Note that I is the incenter of the triangle.
The height of the isosceles triangle is $ 4\sqrt{3}$. The inradius works out to be:
$ r = [ABC]/s = \sqrt{3}/2$.

Note that AI is just the difference between the altitude and the inradius, so

$ AI = 4\sqrt{3}-\sqrt{3}/2= 7\sqrt{3}/2$

Hence, since this is the altitude of the equilateral triangle, the side length is

$ 2/\sqrt{3}\cdot 7\sqrt{3}/2= \boxed{7}$[/hide]

EDIT: Now we have a picture. :) I'm not sure if my answer was right, though. It seems too complicated compared to the other problem.

EDIT EDIT: Didn't see your comment. :blush:  But now it's more ugly...

EDIT EDIT EDIT: I hope it's right now...

I edited a lot of times since my latex is messed up.

Last edit (hopefully): I thought that $ BC=1$... :wallbash_red:

It may well be too easy for you; why not let others have a chance? :)
By the way, you made a slight computation error.

[quote="vishalarul"][quote="vikas"]I get $ 7$.[/quote]

Can you provide a solution? My answer is awfully close to yours...[/quote]
well$ r=\frac{\sqrt{3}}{2}$....

[quote="archimedes1"]I guess I could have made the problem nicer (whoa! much nicer) by making the equilateral triangle $ IFG$.  Where F and G are on BC...[/quote]

[hide="In that case,"]$ r=\frac{\sqrt{3}}{2}$

(see the other solution on how to derive this).

This is the altitude of $ IFG$, so the side length is
$ 2/\sqrt{3}\cdot r = \boxed{1}$.[/hide]

I'd say that this problem is easier than Triangle Geometry #1.


geometry
incenter
search
geometry proposed

Let $ (I)$ be the incircle of $ \triangle ABC$ and $ D,E,F$ are their points of tangency. $ P$ is an arbitrary point inside $ \triangle ABC$. $ AP,BP,CP$ cut $ BC,CA,AB$ at $ A',B',C'$. Let $ A''$ be the intersection of $ B'C'$ and the tangent of $ (I)$ which passes through $ A'$ and different from $ BC$. $ B''$ and $ C''$ are defined similarly. Prove that $ A'',B'',C''$ are collinear.

I think that the problem is also true, in its more general configuration of the triangle $ \bigtriangleup A'B'C',$ as the [b][size=100]Cevian triangle[/size][/b] of an arbitrary point $ P$ inwardly to it, instead of the incenter $ I.$

If I am not mistaken, we can get ( applying the [b][size=100]Brianchon theorem[/size][/b] and the [b][size=100]Desarques theorem[/size][/b] ) the result as the proposed problem states, from the discussion about the generalization of a familiar problem posted as [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=140191]Intersting and hard[/url], by [b][size=100]cvix[/size][/b].

So, I will search my archive and if I am not wrong, I will post here later more details.

Kostas Vittas.

I'm terribly sorry Mr Kostas Vittas, the problem was mis-typed. :blush:  It is exactly what you say, $ A',B',C'$ are the intersection of $ AP,BP,CP$ with $ BC,CA,AB$. I have edited the problem. :blush:


The product of two positive fractions is 1/9. The larger fraction divided by the smaller fraction is 4. What is the sum of the two fractions? Express your answer as a common fraction.

call the larger # x and the smaller one y

$ \frac {x}{y} \equal{} 4$
and
$ xy \equal{} \frac {1}{9}$

rearrange the first equation to get

$ x \equal{} 4y$
plug in to 2nd equation
$ 4y^2 \equal{} \frac {1}{9}$
$ y^2 \equal{} \frac {1}{36}$
$ y \equal{} \frac {1}{6}$
$ x \equal{} \frac {4}{6} \equal{} \frac {2}{3}$
$ \frac {2}{3} \plus{} \frac {1}{6} \equal{} \frac {5}{6}$


function
quadratics
number theory proposed
number theory

Let $ N$ be an integer whose base $ b$ representation is $ 777$. Find the smallest natural number $ b$ for which $ N$ is the fourth power of an integer.

[hide]Rewrite N as $ 7(b^2\plus{}b\plus{}1)$  In order for N to be a fourth power, $ b^2\plus{}b\plus{}1$ must be of the form $ 343\cdot x^4$ for some integer $ x$.  Since $ b^2\plus{}b\plus{}1$ is an increasing function when $ b$ is positive, we want $ x$ to be as small as possible.  We start by trying $ x\equal{}1$ and by some miracle, we end up with a factorable quadratic, with solutions $ b\equal{}\minus{}19$ and $ b\equal{}18$.  Rejecting the negative solution, we get $ b\equal{}18$.[/hide]


calculus
derivative
calculus computations

I can't solve this differentiation problem, can anyone help me?

f(x) = e^(x) * log(base(e))(21+5x)

should i use product rule ??? I am quite confused because exponential and log are in one equation ...

Please help me

$ f(x) \equal{} e^x \ln(5x\plus{}21)$

Yes, you should use the product rule. Exp and Log don't seem to have much influence on the method of solving and you can proceed as usual, (it seems like a totally random expression to me...)


function
number theory unsolved
number theory

For every natural number $ m$ and every integer $ a$ the number $ \sum_{d|m}\sigma(d).a^{\frac{m}{d}}$ is divisible by $ m$ where $ \sigma(d)$ is Mobiis function defined by conditions:
1) $ \sigma(1)\equal{}1$
2) $ \sigma (n)\equal{}0$ if the natural number $ n$ is divisble by the square of natural number >1
3) $ \sigma (n)\equal{}(\minus{}1)^{k}$ if the natural number $ n$ is the product of $ k$ different prime factors

Möbius function is usually denoted by $ \mu(\cdot)$.

Let $ N(m) \equal{} \sum_{d|m}\mu(d) a^{\frac{m}{d}}$. We need to show that $ m$ divides $ N(m)$.

It is easy to see that for prime $ p$ and positive integer $ k$, we have
\[ N(p_k)\equal{}a^{p^k}\minus{}a^{p^{k\minus{}1}}\equal{}a^{p^{k\minus{}1}}(a^{p^{k\minus{}1}(p\minus{}1)}\minus{}1)\]
which is divisible by $ p^k$ by Euler's Totient Theorem.

To finish the proof, note that $ N(m)$ is a multiplicative function (as $ \mu(\cdot)$ is such), i.e., for $ m\equal{}p_1^{k_1}p_2^{k_2}\dots p_\ell^{k_\ell}$ where $ p_i$ are distinct primes, $ N(m)\equal{}N(p_1^{k_1}) N(p_2^{k_2})\dots N(p_\ell^{k_\ell})$ that is divisible by $ m$.


[quote="amir2"]Let $a, b, c$ be real numbers. Prove that there exists a triangle with side
lengths $a, b, c$ if and only if
$2(a^4 + b^4 + c^4) > (a^2 + b^2 + c^2)^2.$[/quote]
You have reversed the sign. The inequality should be
$(a^2+b^2+c^2)^2>2(a^4+b^4+c^4)$
$\Leftrightarrow (a+b+c)(a+b-c)(b+c-a)(c+a-b)>0$
$\Leftrightarrow a+b>c, b+c>a, c+a>b$


You can just search for programs online...

phelpedo: I know I can, for some reason it's screwing up.

jorian

I do, but I guess its a little too late.

nice one alex! :lol:

who are the 2 others in ur grade that have ti89's?

i'm the only one that has one greater than an 84 in 8th grade (89 TITANIUM!!!)

jorian

Aside from me, the only person you know who has one in my grade is Kyle. A couple other people have them, but it is mostly people who don't really know how to get the most out of it. They only got it because it is the most expensive. Also http://www.ticalc.org has the best programs and games. (Although it is always good to just make a few simple ones yourself).

All these fancy calculators will be useless on the AMC next year eh


function
LaTeX
real analysis
real analysis unsolved

let (Yn) and (Xn) two sequences and (Xn) is limeted ( there is M so that |xn|<=M for every n ) , f a continuous function so that for every x , |f(x)|<=c(1+|x|) where c>o , we know that Yn=Yn-1+Xn-1f(Yn-1) and f(x)=0 have a unique solution, show that (Yn) converge

A) Learn basic $ \text{\LaTeX}$, please.

B) Take $ f(x)=x$, $ X_n=1$, $ Y_n=2^n$ for a counterexample :roll:


geometry
circumcircle
cyclic quadrilateral
power of a point
radical axis
geometry proposed

In $ \triangle ABC$, let $ O$ be the circumcenter. A line perpendicular to $ AO$ cuts $ AB,AC$ at $ M,N$ respectively. $ MC,NB$ intersect at $ K$. Let $ H$ be the orthocenter of $ \triangle ABC$, and $ P$ the orthocenter of $ \triangle AMN$. Show that $ H,P,K$ are collinar.

It is easy to see $ B,C,N,M$ are cyclic. Let center of circle $ BCNM$ be $ O$.  Since $ BH$ is perpendicular to $ NC$ and $ O$ is circumcenter of triangle $ BNC$, $ BH$ and $ BO$ are isogonal line wrt $ \angle KBC$. Similarly, $ CH,CO$ are isogonal line wrt $ \angle KCB$. Hence $ H,O$ are isogonal points wrt $ \triangle KBC$. Hence $ H$ lie on isogonal line of $ KO$ wrt $ \angle BKC$. Similarly, $ P$ lie on isogonal line of $ KO$ wrt $ \angle NKM$ ($ P,O$ are isogonal points wrt $ \triangle KNM$). Hence $ H,P,K$ are collinear.

Nice solution!  :first:

Anyway, I already used point $ O$. Please use another point name to avoid confusion.

[b][size=100][color=DarkBlue]EQUIVALENT PROBLEM. – In every cyclic quadrilateral, the orthocenters of the triangles formed by its sidelines taken per three of them, lie on a line pass through the point of intersection of its diagonals[/color][/size][/b].

Let $ ABCD$ be a cyclic quadrilateral with circumcenter $ O$ and we denote as $ E,$ the point of intersection of its sidelines $ AB,$ $ CD.$

It is well known that the lines $ AD,$ $ BC,$ are antiparallel each other, with respect to the lines $ AB,$ $ CD$ and so, we have that $ O_{1}E\perp BC,$ where $ O_{1}$ is the circumcenter of the triangle $ \bigtriangleup ADE.$

Hence we have the configuration as the [b][size=100]massnet’s[/size][/b] problem states.

$ \bullet$ From cyclic quadrilateral $ ABCD,$ we have $ (KA)\cdot (KC) = (KB)\cdot (KD)$ and so, the point $ K\equiv AC\cap BD,$ lies on the common cord $ MN,$ of the circles $ (O_{2}),$ $ (O_{3}),$ with diameters the segments $ AC,$ $ BD$ respectively, as their radical axis.

From cyclic quadrilateral $ ADA'D'$ we have $ (HA)\cdot (HA') = (HD)\cdot (HD')$ and so, the point $ H\equiv AA'\cap DD',$ lies on the line segment $ MN,$ because of the segments $ AA',$ $ DD',$ are the cords of the circles $ (O_{2}),$ $ (O_{3}),$ respectively.

Similarly the point $ P\equiv BB'\cap CC',$ lies also on the line segment $ MN.$

Hence, the points $ H,$ $ P,$ $ K,$ are collinear and the proof is completed.

Kostas Vittas.

Another problem involve orthocenters collinear:    Let $ABCD$ be a quadrilateral  and a point $P$ in this plane. Through $P$ we draw two lines, one parallel to $BC$ and meets $AC$ at $M$, one parallel to $AD$ and meets $BD$ at $N$. Let $H_1, H_2, H_3$ are the orthocenters of triangles $PBC, PAD$ and $PMN$. Prove that $H_1, H_2, H_3$ are collinear if and only if $A, P, B$ are collinear. Let solve :) !

My solution:

Let $ B', M' $ be the projection of $ B, M $ on $ AC $ , resp .
Let $ C', N' $ be the projection of $ C, N  $ on $ AB $ , resp .

Since $ MN $ and $ BC $ are anti-parallel WRT $ \angle BAC $ ,
so $ B, C, M, N $ are concyclic.

Since $ NP \cdot PN'=MP \cdot PM' , BH \cdot HB'=CH \cdot HC' , BK \cdot KN=CK \cdot KM $ ,
so $ P, K, H $ all lie on the radical axis of $ \{ \odot (BN) , \odot (CM) \} $ .

Q.E.D

Dear Mathlinkers,
another point of view…

1. B', C'	the feet of the B, C-altitudes of ABC
2. Y, Z	the points of intersection of BB' and CN, CC' and BN
3. According to Brianchon hexagon,	YZ // MN
4. the triangles HYZ and PNM being homothetic… we are done…

Sincerely
Jean-Louis

My solution: note that quadrilateral $ MNCB $ is cyclic so orthocenter of $\triangle AMN , H $ and $ K $ lie on the steiner line DONE.

Dear Mathlinkers,

http://jl.ayme.pagesperso-orange.fr/Docs/Orthique%20encyclopedie%207.pdf   p. 23...

Sincerely
Jean-Louis


Let $A,B \in M_{2}(\mathbb{R})$  such that their eigenvalues are 1and 3, 2 and 4 respectively.
Can it be that eigenvalues of A+B are 4 and 5; 1 and 9?
What if we change $\mathbb{R}$ to $\mathbb{C}$?
I have    solved the easiest part 4and 5, but  cant do anything with 1,9. Please help me

Hint:
Consider the matrices
\[A = \begin{pmatrix}1 & a\\ 0 & 3 \end{pmatrix}\qquad B = \begin{pmatrix}2 & 0\\ b & 4 \end{pmatrix}.\]

What are the eigenvalues of $A$ and $B$? What are the eigenvalues of $A+B$ (as a function of $a$ and $b$)?


abstract algebra
group theory
superior algebra
superior algebra unsolved

If G is a group soluble and simple.Then G is abelian?
HELP ME ....PLEASE   :(
Thanks
sastra

Soluble requires the existence of a normal subgroup with nontrivial cyclic factor group. Simple requires the existence of only one normal subgroup with nontrivial factor group.
Thus you have a cyclic group, clearly abelian.


A sum of money is invested in the Mullionaire's Club Bank at 10% compounded continuously. Determine after how many years there will be a 150% increase in the original investment.
A) $ (0.1)ln1.5$ years
B) $ 10ln1.5$ years
C) $ ln15$ years
D) $ 1.5$ years
E) None of the Above

[hide="??"]Compounded Continuously is $ Pe^{rt}$. A 150% increase will be 2.5P. Setting those two equal we now have an equation of $ e^{0.1t} \equal{} 2.5$. Solving for t, I get $ 10In2.5$ years. However the answer key told me it was B, and now I'm confused as to take the wording of 150% increase as just 1.5, or actually increase it by 150% to get 2.5[/hide]

hmm....

Well i agree with you. A 150% increase must be 100% + 150% = 250%
If the problem stated asked how much time will it take for the initial sum of money to grow to 150% its original, then their answer would be correct.

That's worded a little weirdly, it should say how long should it take for to achieve 150% of the original balance.

So
$ Pe^{.1t} \equal{} 1.5P$
$ .1t \equal{} \ln 1.5$
$ t \equal{} 10 \ln 1.5 \rightarrow B$


It seems that GIF attachments don't get displayed within the post, unlike JPG attachments, though it wasn't so at some earlier point. What's the problem there?

Also, JPGs get automatically shrinked/stretched to 500 pixels, which often make them ugly. Can that somehow be changed (no automatic resize for small enough pictures, for example)?

Um, since when do JPEGs get shrunk?

http://www.artofproblemsolving.com/Forum/viewtopic.php?mode=attach&id=10792

-703*437 JPEG, original size.

Or do you mean the thumbnails?

Also, GIFs do get displayed within the post: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=933261#933261

Here are two attachments - I get them both as "Download", no preview.

It seems to be a bug. Thumbnail is generated however it is not displayed. Will look into it soon.

Ok, this seems to be fixed at this time. Most likely a permission issue on the thumbnails directory.


[quote="Omid Hatami"]$I$ is incenter of triangle $ABC$. Incircle of $ABC$ touches $AB,AC$ at $X,Y$. $XI$ intersects incircle at $M$. Let $CM\cap AB=X'$. $L$ is a point on the segment $X'C$ that $X'L=CM$. Prove that $A,L,\boxed{Y}$ are collinear iff $AB=AC$.[/quote]Maybe you made a bit of mistake : in the my opinion, I think ... $AB=AC\Longleftrightarrow$ the points $A,L,I$ are collinearly !

[quote=Virgil Nicula][quote="Omid Hatami"][color=darkred]$I$ is incenter of triangle $ABC$. Incircle of $ABC$ touches $AB,AC$ at $X,Y$. $XI$ intersects incircle at $M$.
Let $CM\cap AB=X'$. $L$ is a point on the segment $X'C$ that $X'L=CM$. Prove that $A,L,I$ are collinear iff $AB=AC$.[/color][/quote]
[color=darkblue][b]Proof.[/b] From a well-known property, $AX'=p-b\ .$ Therefore, $\boxed{\ L\in AI\ }\Longleftrightarrow$ $\frac{AX'}{AX}\cdot \frac{IX}{IM}\cdot\frac{LM}{LX'}=1$ $\Longleftrightarrow$ $\frac{LM}{LX'}=\frac{p-a}{p-b}$ $\Longleftrightarrow$
$\frac{MX'}{LX'}=\frac{c}{p-b}$ $\Longleftrightarrow$ $\frac{MX'}{MC}=\frac{c}{p-b}$ $\Longleftrightarrow$ $\frac{MX'}{CX'}=\frac{c}{p-b+c}$ $\Longleftrightarrow$ $\frac{2r}{h_{c}}=\frac{c}{p-b+c}$ $\Longleftrightarrow$ $2r(p-b+c)=2pr$ $\Longleftrightarrow$ $\boxed{\ b=c\ }\ .$[/color][/quote]

DEAR YOU,
why  $AX'=p-b\ .$ and  $\frac{2r}{h_{c}}=\frac{c}{p-b+c}$ ?

See http://www.mccme.ru/~akopyan/papers/EnGeoFigures.pdf 2.8). Is well-know that $AX' = XB$.Since $X'L = MC$, $L$ is the Nagel point. So $A,L$ and $I$ are collinear if $AB = AC$ because $AI$ would altitude of $ABC$.


A right square pyramid's base has edges measuring 20 feet each. If the pyramid's height is 27 feet, what is its volume, in cubic feet?

$ \text{V}\equal{}\frac{1}{3}[B]h$
[B]=area of base

$ \text{V}\equal{}\frac{1}{3}\cdot27\cdot20^2$
$ \text{V}\equal{}9\cdot400$
$ \text{V}\equal{}\boxed{3600}$


The sum of the first n integers is $ \frac{n(n\plus{}1)}{2}$, so in this case we have $ \frac{100(101)}{2}$, which simplifies to [u]5050[/u]

no man,
that is the easiest way

just remember that consecutive numbers sum for 1 2 3 4 5 6 .....

is (n(n+1))/2    ------ its a formula, and n represents the number you are goin to
so in your case 100

100 x 101
-----------    = 5050
      2

[quote="shaun ong"]Er can u say it in a more elimentery way? I dun understand what are u doing[/quote]

The elementary way, I would say, would be to get your calculator and add them up.  Of course, no one has time for this.  That's why there is a formulas for the sum of the first positive n integers: $ \frac{n(n\plus{}1)}{2}$.  You plug in the sum of the first whatever numbers as n, so:

$ \frac{100(101)}{2}\equal{}50(101)\equal{}5050$

There is a neat story of the famous mathematician Gauss, who was given this problem in elementary school as busiwork for punishment. The teacher's idea was to have some peace and quiet while everyone worked out the arithmetic but Gauss solved it within seconds, astonishing the teacher, who deemed he cheated somehow.

And the picture proof is shown on small area beneath the main menu for the sum of the first n numbers. Bascially, it has the numbers shown as dots and lined up in a right triangle formation. Then, they flip the array of dots so that one side is n and the other is n+1. The area of needed area is n(n+1)/2. If you don't know what I mean keep refreshing the page you're at until you get it.

[quote="minicon"]Yeah, Gauss was awesome.
Is there something like that for the product of the first n integers?[/quote]

The product of the first n integers... huh, it's not that bad on a calculator, just do n! :wink:

but for a formula, I have no clue

If you want a general formula for factorial you have,

$ n! \equal{} n(n\minus{}1)(n\minus{}2) \cdots  1$.

Duh... Factorials.  Well I guess we all have our stupid moments...

Ok I get wat u mean about that formula but what does this formula do orginaly?

I'll try to make it simpler.
$ 1\plus{}2\plus{}3\plus{}...\plus{} n \equal{} \frac {n(n \plus{} 1)}{2}$
Edit: omg, this topic is haunted.  I keep on making the stupidest mistakes in the world :wallbash_red:

um, isn't it supposed to be \[ 1\plus{}2\plus{}3\plus{}...\plus{}n\equal{}\frac{n(n\plus{}1)}{2}\]

Yeah Thanks for the help , I better memorize this useful formula  :)

So, the sum of the first n positive integers is $ \frac{n(n\plus{}1)}{2}$, the sum of the first n positive odds is $ n^2$ and the sum of the first n positive evens is $ n(n\plus{}1)$, all can be verified with induction.

[quote="wz1118"]Since 1+99=100, 2+98=100, ...
There are 49 0f those.
All of them added together, it become 49 hundred or 4900
Then add 100
4900+100=5000
Then add the remainder 50
5000+50=5050

[b]Solution 5050 %)%)[/b][/quote]

If the legendary story is true, thats exactly how gauss did it. His teacher gave them that problem thinking it to be a toughie so that he could go to sleep. ;)

In the above question. Group them as:
1+100
2+99.............

you got 50 Pairs each of which sum to 101. So the answer is 5050. Thats how we get the general formula mentioned above.

why is the topic revived, anyway thanks to the person giving me this formula, i have no probs in summing up consecutive numbers


On her science test, Nancy answered 5/6 of the questions correctly. If Nancy answered 18 of the 1st 27 questions correctly, then the total number of questions must be at least ?

I did this one the other day. I'm not sure exactly what I did. I can't make sense of it now. I'm just copying what I have. I think that my answer is correct though.

5/6 = 10/12 = 15/18 = 20/24 = 25/30

18/27 were correct

1st correct = 2/3 correct

18/27 = 2/3

18/27 = x/50

5/6 = n/36

30/36

There would have to be at least 36 question.

[quote="pi^2/6"]30 correct/36 total doesn't make sense because if she got 18 out of the first 27, then she missed at least 9.

so the numerator is at least 9 greater then the denominator, so I think's it's 45 right/54 total.[/quote]

I missed that. Thank you!


What is the number of square meters in the area of a square
if the length of a diagonal is 18 meters?

[asy]size(50);
draw((0,0)--(1,0)--(1,1)--(0,1)--cycle); draw((0,1)--(1,0)); label(rotate(-45)*"18 m",(.5,.5),NE,fontsize(8pt));[/asy]

I'm not sure, but I think you use

$ a^2\plus{}b^2\equal{}c^2$

By Pythagorean Theorem, a diagonal of a square is $ \sqrt {2}$ times larger than the side length. Thus, the side length is $ \frac {18}{\sqrt2}$, or $ 9\sqrt {2}$ Square that to get $ \boxed{162}$

Edit: thanks for the correction. Your method is probably better than this one anyways.

No... The area. The area of any rhombus is half the product of the lengths of the diagonals, and since is a square is a rhombus, the answer is 162.

both ways work
another way is to use the pythagoream theory a little differently

since it's a square, the lengths of the legs will be equal, so

$ a^2 \plus{} b^2 \equal{} c^2$
since $ a$ and $ b$ are the same, we get
$ 2a^2 \equal{} c^2$
we plug in the c value which is given as 18
$ 2a^2 \equal{} 18^2$
$ 2a^2 \equal{} 324$
divide by 2 on each side
$ a^2 \equal{} 162$

since the area of a square is $ s^2$, then $ a^2$ would be our answer, so the area is $ \boxed{162}$

[quote="lifeisacircle"]By Pythagorean Theorem, a diagonal of a square is $ \sqrt {2}$ times larger than the side length. [/quote]

Just a note about this:

In case you are wondering, the reason for this is that the right triangle created when you draw a diagonal in a square is an isosceles triangle. In all isosceles right triangles, the hypotenuse(or in this case the diagonal) is exactly $ \sqrt{2} \times s$, where $ s$ is the length of one of the legs.

[asy]
size(50);
draw((0,0)--(1,0)--(1,1)--(0,1)--cycle);
draw((0,1)--(1,0)); label(rotate(-45)*"18 m",(.5,.5),NE,fontsize(8pt));
draw((1,0)--(2,1)--(1,2)--(0,1)--cycle);
[/asy]

Here is a diagram to show that $\frac{1}{4}$ of the big square = $\frac{1}{2}$ of the small square. Therefore, the area of the big square is $18^{2}$, and the area of the small square is half of that, which is $\frac{18^{2}}{2}=\frac{18\cdot18}{2}=18\cdot9=\boxed{162}$.


A function $ f: R^3\rightarrow R$ for all reals $ a,b,c,d,e$ satisfies a condition:

\[ f(a,b,c)\plus{}f(b,c,d)\plus{}f(c,d,e)\plus{}f(d,e,a)\plus{}f(e,a,b)\equal{}a\plus{}b\plus{}c\plus{}d\plus{}e\]

Show that for all reals $ x_1,x_2,\ldots,x_n$ ($ n\geq 5$) equality holds:

\[ f(x_1,x_2,x_3)\plus{}f(x_2,x_3,x_4)\plus{}\ldots \plus{}f(x_{n\minus{}1},x_n,x_1)\plus{}f(x_n,x_1,x_2)\equal{}x_1\plus{}x_2\plus{}\ldots\plus{}x_n\]

For a given $ n$, add the equations for $ (a,b,c,d,e) \equal{} (x_1, x_2, x_3, x_4, x_5)$, $ (a,b,c,d,e) \equal{} (x_1, x_2, x_3, x_4, x_6)$, $ \ldots$, $ (a,b,c,d,e) \equal{} (x_{n \minus{} 1}, x_n, x_1, x_2, x_3)$, $ (a,b,c,d,e) \equal{} (x_n, x_1, x_2, x_3, x_4)$. Divide both sides by $ 5$ to get the desired equality.

Seems a bit easy. :maybe:

EDIT: Yeah, you're right, I misread the problem. I'll rethink this and maybe post another reply.

how to get rid of $ f(x_4, x_5, x_1)$ that appear in the equation for $ (a,b,c,d,e)\equal{}(x_1,x_2,x_3,x_4,x_5)$?

i think you misunderstood the problem :huh:

It would suffice inductively if we could prove the identity

$ f(x_{n \minus{} 2}, x_{n \minus{} 1}, x_n) \plus{} f(x_{n \minus{} 1}, x_n, x_1) \plus{} f(x_n, x_1, x_2) \minus{} f(x_{n \minus{} 2}, x_{n \minus{} 1}, x_1) \minus{} f(x_{n \minus{} 1}, x_1, x_2) \equal{} x_n$.

In other words, letting $ x_{n \minus{} 2} \equal{} a, x_{n \minus{} 1} \equal{} b, x_n \equal{} c, x_1 \equal{} d, x_2 \equal{} e$,

$ f(a, b, c) \plus{} f(b, c, d) \plus{} f(c, d, e) \equal{} c \plus{} f(a, b, d) \plus{} f(b, d, e)$.

Substituting the original condition gives

$ a \plus{} b \plus{} d \plus{} e \equal{} f(d, e, a) \plus{} f(e, a, b) \plus{} f(a, b, d) \plus{} f(b, d, e)$,

so it suffices to prove the desired statement for $ n \equal{} 4$ to prove it for all $ n$.  I'll keep working.

For $ n>5$ sum up all five term subsums of $ f(x_1,x_2,x_3) ... f(x_n,x_1,x_2)$.
Getting $ n(f(x_1,x_2,x_3)\plus{} ... \plus{}f(x_n,x_1,x_2)) \equal{} n(x_1\plus{}x_2\plus{}...x_n)$.

Okay, I think I got this, finally.

[hide="Solution"]Step 1: $ f(0,0,0) \equal{} 0$.
Put $ (a,b,c,d,e) \equal{} (0,0,0,0,0)$ into the given equation.

Step 2: $ f(a,a,a) \equal{} a$ for all reals $ a$.
Put $ (a,b,c,d,e) \equal{} (a,a,a,a,a)$ into the given equation.

Step 3: $ f(a,0,0) \plus{} f(0,a,0) \plus{} f(0,0,a) \equal{} a$ for all reals $ a$.
Put $ (a,0,0,0,0)$ into the given equation and use step 1.

Step 4: $ f(a,0,b) \equal{} f(a,0,0) \plus{} f(0,0,b)$ for all reals $ a,b$.
Put $ (a,0,b,0,0)$ into the given equation and use step 3.

Step 5: $ f(a,b,0) \plus{} f(b,0,a) \plus{} f(0,a,b) \equal{} a \plus{} b$ for all reals $ a,b$.
Put $ (a,b,0,0,0)$ into the given equation and use step 1 and step 4.

Step 6: $ f(a,b,a) \plus{} f(b,a,b) \equal{} a \plus{} b$ for all reals $ a,b$.
Put $ (a,b,a,b,0)$ into the given equation and use step 5.

Step 7: $ f(a,b,c) \plus{} f(c,a,b) \plus{} f(b,c,a) \equal{} a \plus{} b \plus{} c$ for all reals $ a,b,c$.
Put $ (a,b,c,a,b)$ into the given equation and use step 6.

Step 8: $ f(b,a,a) \plus{} f(a,a,c) \equal{} f(b,a,c) \plus{} a$ for all reals $ a,b,c$.
Put $ (b,a,a,a,c)$ into the given equation and use step 2 and step 7.

Step 9: $ f(a,b,c) \plus{} f(b,c,d) \plus{} f(c,d,a) \plus{} f(d,a,b) \equal{} a \plus{} b \plus{} c \plus{} d$ for all reals $ a,b,c,d$.
Put $ (a,a,b,c,d)$ into the given equation and use step 2 and step 8.

Then induct on $ n$ and use t0r's hint.[/hide]

Definition of cancer.

We claim that the result holds for $n=4$. Plugging in $(a,a,a,a,a)$ gives $f(a,a,a)=a$. Taking $(a,a,b,a,a)$ and invoking $f(a,a,a)=a$ gives
\begin{align} f(a,a,b)+f(a,b,a)+f(b,a,a)=2a+b \end{align}Using $(1)$, the pair $(b,a,a,b,a)$ gives
\begin{align*} 2b+3a=[f(b,a,a)+f(a,a,b)+f(a,b,a)]+f(a,b,a)+f(b,a,b)=2a+b+f(a,b,a)+f(b,a,b), \end{align*}or \begin{align*}\tag{2} f(a,b,a)+f(b,a,b)=a+b.\end{align*}Applying the pair $(d,a,b,a,b)$ and using $(2)$ admits $$2a+2b+d=f(d,a,b)+[f(a,b,a)+f(b,a,b)] + f(a,b,d)+f(b,d,a)=a+b+f(d,a,b)+f(a,b,d)+f(b,d,a),$$which implies $f(d,a,b)+f(a,b,d)+f(b,d,a)=a+b+d$, i.e. the result holds for $n=3$.

Lemma: $f(d,a,a)+f(a,a,b)=a+f(d,a,b)$.
Proof: Let us add $f(a,a,d)+f(a,d,a)+f(b,a,a)+f(a,b,a)$ to both sides. Grouping the LHS and using the result for $n=3$, $$[f(d,a,a)+f(a,a,d)+f(a,d,a)]+[f(a,a,b)+f(b,a,a)+f(a,b,a)]=4a+b+d,$$Simplifying the RHS using the pair $(d,a,b,a,a)$ along with the problem condition, $$a+f(d,a,b)+f(a,b,a)+f(b,a,a)+f(a,a,d)+f(a,d,a)=4a+b+d$$as well. Since all our steps are reversible the lemma is established. $\square$

Plugging in the pair $(a,b,c,d,a)$, we obtain $$2a+b+c+d=f(a,b,c)+f(b,c,d)+f(c,d,a)+f(d,a,a)+f(a,a,b).$$Using the lemma, $a+b+c+d=f(a,b,c)+f(b,c,d)+f(c,d,a)+f(d,a,b)$, which means the result holds for $n=4$.

Now we shall prove the statement via induction on $n$. The base case for $n=5$ is already done. Assume that \[ f(x_1, x_2, x_3) + f(x_2, x_3, x_4) + \cdots + f(x_{n-2},x_{n-1},x_n)+f(x_{n-1}, x_n, x_1) + f(x_n, x_1, x_2) = x_1 + x_2 + \cdots + x_n\]is valid. We wish to show \[ f(x_1, x_2, x_3) + f(x_2, x_3, x_4) + \cdots + f(x_{n-2},x_{n-1},x_n)+f(x_{n-1}, x_n, x_{n+1}) + f(x_n, x_{n+1}, x_1)+f(x_{n+1},x_1,x_2) = x_1 + x_2 + \cdots + x_{n+1}\]Note $f(x_1, x_2, x_3) + f(x_2, x_3, x_4) + \cdots + f(x_{n-2},x_{n-1},x_n)=x_1+x_2+\cdots+x_n-f(x_{n-1}, x_n, x_1)- f(x_n, x_1, x_2)$, so it suffices to prove $$f(x_{n-1}, x_n, x_{n+1}) + f(x_n, x_{n+1}, x_1)+f(x_{n+1},x_1,x_2)=x_{n+1}+f(x_{n-1}, x_n, x_1)+ f(x_n, x_1, x_2)$$Adding $f(x_1,x_2,x_{n-1})+f(x_2,x_{n-1},x_n)$ to both sides and using the given relation, we must show $$x_{n-1}+x_n+x_{n+1}+x_1+x_2=x_{n+1}+f(x_{n-1}, x_n, x_1)+ f(x_n, x_1, x_2)+f(x_1,x_2,x_{n-1})+f(x_2,x_{n-1},x_n),$$which easily follows from the fact the relation holds for $n=4$. All of our steps were reversible so the original equation must have been true.

I think that this problem is not that terrible, as it can be reasonably solved in half an hour if we work backwards from the inductive hypothesis. In other words, the inductive step reduces to an equality in five variables, and so we can use the given condition to continually reach equivalent statements with less variables until we can eventually just prove them.

We will use $abc$ as shorthand for $f(a, b, c)$. We will use $P(a, b, c, d, e)$ to denote the assertion in the problem.

[b] Lemma 1. [/b] $ccc = c$ for any $c \in \mathbb{R}$.

[b] Proof. [/b] Trivial by $P(c, c, c, c, c).$

$\blacksquare$

[b] Lemma 2. [/b] $cec + ece = e+c$ for any $e, c \in \mathbb{R}$.

[b] Proof. [/b] By $P(e, e, c, c, c)$ and Lemma $1$ we obtain:

$$(cce + cee + eec + ecc) = 2e + 2c.$$

By $P(e, c, c, c, c) + P(c, e, e, e, e)$ and Lemma $1$ we obtain:

$$(cce + cee + eec + ecc) + cec + ece = 3e + 3c.$$

Subtracting the above two equations finishes.

$\blacksquare$

[b] Lemma 3. [/b] $bce + c = bcc + cce$ for any $b, c, e \in \mathbb{R}.$

[b] Proof. [/b] By $P(b, c, e, c, e) - P(b, c, c, c, e)$ we know that:

$$bcc + ccc + cce - c = bce + (cec + ece) - e.$$

By Lemmas $1, 2$, we get:

$$bcc + c + cce - c = bce + (e + c) - e \Rightarrow bcc + cce = bce + c,$$

as desired.

$\blacksquare$

[b] Lemma 4. [/b] $bce + cef + efb + fbc = b+c+e+f$ for any $b, c, e, f \in \mathbb{R}.$

[b] Proof. [/b] By $P(c, e, f, b, c)$ we know:

$$cef + efb + fbc = c+e+f+b+c - (bcc + cce).$$

By Lemma $3$, we can rewrite the RHS as:

$$c+e+f+b+c - (bce + c) = c+e+f+b - bce,$$

and so therefore we have

$$cef + efb + fbc = c+e+f+b - bce \Leftrightarrow bce + cef + efb + fbc = c+e+f+b,$$

as desired.

$\blacksquare$

[b] Lemma 5. [/b] For any $b, c, e, f \in \mathbb{R}$, function $g: \mathbb{R} \rightarrow \mathbb{R}$ given by $g(x) = bcx + cxe + xef - bce - cef$ for each $x \in \mathbb{R}$ is identical to the identity function (i.e. $g(x) \equiv x$).

[b] Proof. [/b] First of all, we know from $P(b, c, x, e, f) - P(b, c, 0, e, f)$ that:

$$g(x) - g(0) = x.$$

Therefore, it suffices only to show that $g(0) = 0$, i.e. that:

$$bc0 + c0e + 0ef = bce + cef.$$

From $P(b, c, 0, e, f)$ and Lemma $4$ we have:

$$bc0 + c0e + 0ef = b+c+e+f - efb - fbc = bce + cef,$$

as desired.

$\blacksquare$

We are now ready to solve the problem. We will prove it by induction on $n.$ If $n = 5$, we are done by the condition of the problem. Suppose that it holds when $n = k$, for some $k \ge 5.$ When $n = k +1$, we know from Lemma $5$ that $f(x_1, x_2, x_3) + f(x_2, x_3, x_4) + f(x_3, x_4, x_5) - f(x_1, x_2, x_4) - f(x_2, x_4, x_5) = x_3.$ Therefore, adding this to the equality obtained when we apply the inductive hypothesis on $(x_1, x_2, x_4, x_5, x_6, \cdots, x_{k+1})$, the induction is complete.

$\square$

The error in this solution is that the problem was not interpreted correctly, here for example

[quote=Dark Wind]f(a,0,0) + f(0,0,0) + f(0,0,0) + f(0,0,0) + f(0,0,0) = a + 0 + 0 + 0 + 0 <=>
f(a,0,0) = a[/quote]

the correct equality (assuming $(a,b,c,d,e)=(a,0,0,0,0)$) would be $f(a,0,0)+f(0,0,0)+f(0,0,0)+f(0,0,a)+f(0,a,0)=a+0+0+0+0=a$


a chessboard has the form of a cross, obtained from a 4 *4 chessboard by deleting the corner squares. can a knight travel around this board, pass through each square exactly once , and end on the same square he starts on?

Sorry for my lack of graphic-making skills.

[hide]
Label the squares thusly:

[code]
_12_
8AB3
7CD4
_65_
[/code]

We are essentially trying to find a closed Hamiltonian circuit on the graph with the squares as vertices and all possible knight moves as the edges. (side note GAH annoying automatic browser spell check marking up my entire post with no math vocab whatsoever!!!!)

Because the tour is closed, the knight will enter each square from one square and exit from another square. Note that from each of the squares $ A,B,C,D$ there are only have two possible knight moves. Therefore these moves/edges must be part of the tour. So far we have some segments of the tour:
[code]4-A-5
6-B-7
2-C-3
1-D-8[/code]
Square 8 only has two possible squares from which the knight could have arrived (or to which the knight could go, which ever way you want to look at it), namely 2 and 6. WLOG we choose 2. It is easy to come up with the tour
1-D-8-2-C-3-5-A-4-6-B-7-1
after that.
[/hide]


I got a few questions correct (in a row) and my points (on the stats page, where it displays your points, karma, etc.) went DOWN????? Plz help!!! I have no clue why  :(  :mad:

is the lower number the better?

I had

POINTS: 255 of 1876

now it is

POINTS: 203 of 1876

Is that good? Lower, better?

The points/rank/percent/stuff are not updated real time.

That's your rank. You went up. That's good.


Can someone please post AMC10A 2008 problem #2, #3, #6, #7, #10, #11 & #12?

Thanks,

I'm trying to get a copy of it to post it, but it might be a couple days as posting a contest takes a lot of work.


Two equal parallel chords are drawn 8 inches apart in a circle of radius 8 inches.  The area of that part of the circle that lies between the chords is:
\[ \text{(A)}\ 21\frac{1}{3} \pi - 32\sqrt{3} \qquad \text{(B)}\ 32\sqrt{3} + 21\frac{1}{3} \pi \qquad \text{(C)}\ 32\sqrt{3} + 42\frac{2}{3} \pi \]
\[ \qquad \text{(D)}\ 16\sqrt{3} + 42\frac{2}{3} \pi \qquad \text{(E)}\ 42 \frac{2}{3} \pi \]

[hide="Answer"]
$(B)$

The length of each chord is $8\sqrt{3}$.  The total area between the chords can be calculated by splitting the area into two triangles and two "slices" of the circle.

$\displaystyle Triangles = 4\times8\sqrt{3} = 32\sqrt{3}$
$\displaystyle Slices = 2\times\frac{1}{6}\times64\pi = \frac{64\pi}{3}$
$\displaystyle Area = 32\sqrt{3} + \frac{64\pi}{3} = 32\sqrt{3} + 21\frac{\pi}{3}$
[/hide]


geometry
ratio
combinatorial geometry
combinatorics proposed
combinatorics

There are $ n$  straight lines on the plane, which divide it on areas. They are painted in chessboard order: the areas, which have the the same sides, should be painted in different colors. What is the maximal difference between number of black and white areas?

[quote="seemmeriast"][quote="srikanth"]I donot understand exactly. Consider the diagram (how.jpg). As $ n$ increases we can see that the ratio of area of  $ ABCD$ to $ EFGH$ tends to 0. For a even $ n$ we can place the horizontal lines far apart to decrease the ratio. Do you intend to include a additional condition like keeping the intersection points of the lines within a convex hull of points of finite radius or circumscribing the intersection points within a circle of finite radius ?[/quote]
I think that means [b]"the difference between the number of areas painting black and painting white"[/b], not the ratio of area[/quote]
You are right

Here's picture for 7 lines. The difference is 7.
Who has other variants?


Hi guys,

  Last week  i faced to the problem:

Problem:
               Find all prime number p and positive integer number n such that
  15+p^n is a square

Very glad to see solutions and comments

Hi pluricomplex,

Nice of you to join us. Please read carefully each subforum's destination before posting. The solved problems forum is the place where the problems are moved after they have been solved / discussed on the forum first :) ;)

hei, maybe you could post the solution or send it to me by private msg ?
it seems like a very difficult problem for me.

I'm sure that this is a very hard (if not unsolvable) problem. So, if anyone has found something interesting about it, please post that. Thanks.

Dear friends,
Thanks for care in my problem, but indeed i still do not finish it.But in the case p=4k+3 ,we can solve it easy!Try this case and this is only i can tell about the problem!
This come from when i did the problem: Find all n such that 15+ 3^n is a square!I think we can prove for the general case but....... :(

The case p is 3 modulo 4 is simple. Suppose n>1 such that p^n+15=x^2. First, if n is odd, then p^n+15 is 2 modulo 4, false. Thus, we can assume n=2k. Then (x-p^k)(x+p^k)=15. First case: x-p^k=1 and x+p^k=15. Then p^k=7 and we are done. Second case: x-p^k=3 and x+p^k=5. Then p^k=4. As you can see, if I'm not wrong, the case p mod 4=3 is almost trivial.


Find the sum:

$\sum_{k=1}^{m}{\frac{1}{sin(kn)sin((k+1)n)}}$

where $m$ is an odd positive integer and $n\geq1$ is also an integer, angles are considerer in radians.

(
Do something like:
$\sum_{h=0}^{n}h^{2}=\frac{(n)(n+1)(2n+1)}{6}$
)
(in spanish we use to write sen insted of sin jiji)
 :)  :lol:  :!:

What is $\text{sen}$?  Is $\text{sen}\equiv\sin$ true?


probability
ratio
geometry
geometric transformation
articles

I have no idea where to put this since it's as much high school stats as it is bio, but I need more help with the calculations than anything else. I have a SNP with 90 alleles [A/G] with frequencies of 16 AA, 35 AG, and 39 GG. I need to know how to calculate the p-value and determine whether this fits the Hardy-Weinberg equilibrium. I've searched the internet but haven't found anything resembling an equation I can use since they are mostly poorly explained. Any help would be appreciated.

If the population is in Hardy-Weinberg equilibrium then $ \frac {16}{90} \equal{} q^2, \frac {35}{90} \equal{} 2pq, \frac {39}{90} \equal{} p^2$.  What you have to determine is whether this system is consistent, so solve for either $ p$ or $ q$ and find out if you can find a value of $ p$ that actually produces the above frequences.

[hide="Hardy-Weinberg Principle"]
For a gene pool in equilibrium, if the frequency of dominant allele $ A$ is $ p$ and recessive allele $ a$ is $ q$, then
\begin{align*} AA & = p^2 \\
Aa & = 2pq \\
aa & = q^2 \end{align*}
[/hide]

[hide]
(1) $ p = \sqrt {\frac {39}{90}} \approx .58$.
(2) $ Aa = 2\sqrt {AA\cdot aa} = 2\sqrt {16\cdot 39} \approx 50 \neq 35$. Therefore, it's not in equilibrium.
[/hide]

Thanks, I saw those equations on the internet but couldn't make head or tail of it. =)

One more question:
If one of the combinations is missing, [A/G] has 0 freq. for example, would that mean it's automatically not going to fit the principle? Since it would be impossible for the other equations to work out to a 0.

Edit: I'm asking because I have a couple problems with only 2 out of 3 combinations and they ask me to find whether or not it's with the equilibrium and I don't think they can [i]all[/i] be no.

Use the fact that the sum of the proportions must add up to one.  (That is, if we let $ P(x)$ represent the proportion of the population with genotype $ x$ then we have $ P(AA)\plus{}P(Aa)\plus{}P(aa) \equal{} 1$).  Probably the proportion not given is not zero.

The porportion not given is zero, I'm sure since the other 2 adds up to the given total.

[quote="SMCount"]I've searched the internet but haven't found anything resembling an equation I can use since they are mostly poorly explained. Any help would be appreciated.[/quote]

The Hardy-Weinberg principle takes two alleles [b]frequencies[/b], $ p$ and $ q$, and modifies them to estimate genotype/phenotype frequencies in a population. This is only a simulation, and the population must satisfy certain aspects. The population must be [b]infinite, or very large,[/b] to minimize large and abnormal changes in allele frequences. The population must also [b]randomly mate[/b], so sexual selection does not occur. This gives each allele an equal footing it's ability to be passed down. Also, the population must not experience any [b]mutations[/b], or this will change the allele proportions. Similarly, the population must not be evolving, and natual selection [b]does not[/b] occur. You can see for youself how these factors affect the allele population by completing the [url=http://www.phschool.com/science/biology_place/labbench/lab8/intro.html]AP Biology Lab 8 - Population Genetics[/url].

To derive the Hardy-Weinberg prinicple, we take the two allele frequencies, $ p$ and $ q$. Because these frequencies are parts of a whole, we can immediately set up the equation -
\[ p \plus{} q \equal{} 1
\]
Where the frequency of $ p$ and the frequency of $ q$ must add up to $ 1$. This equation is helpful in determining the frequency of the other allele when one is present.

Now, by squaring, we get
\[ (p \plus{} q) \equal{} 1
\]
\[ p^2 \plus{} 2pq \plus{} q^2 \equal{} 1
\]
The last equation gives the genotype frequency of homozygous dominant ($ p^2$), heterozygous ($ 2pq$), and homozygous recessive ($ q^2$).

Right, the science part I have down, it's the math part that I have trouble with. I understand how to calculate for a given population assuming that it's under hardy-weinburg expectations, however I don't understand how to calculate whether or not a given certain number of different alleles fit the principle or not. (Well I kinda do now) The puzzling thing is when there is when one of the three combinations = 0 since logically that would automatically say HWE is false but there are too many problems for me to think that all of them are false.

wait.. this sounds like a chi-squared situation.  To do a chi-squared test:

1) calculate expected counts by Hardy-Weinberg principle.  (All must be greater than 5 or the test won't be valid)
2) calculate the chi-square statistic: $ \chi^2 \equal{} \sum \frac {(observed \minus{} expected)^2}{expected}$
3) find the p-value.  With a TI-84, you can go to 2nd->VARS (DISTR), and choose #8, x2cdf.  Then compute $ 1 \minus{} \chi^2cdf(0,\chi^2, 2)$, where $ \chi^2$ is the chi-squared statistic computed in part 2. (p-value is the right tail probability)

The p-value obtained is the probability that, if the hardy-weinberg principles is an accurate model, that a sample such as the one observed would happen by chance.  The test never gives a definitive answer; usually, however, if we have a p-value of .05 or less we would reject the null-hypothesis (that the observed sample follows hardy-weinberg) and conclude that it probably does not.

Somehow I doubt you are supposed to be running statistical tests like these for science homework, but... maybe.  Anyway, I hope this helps.

Chi-squared?  It's nothing like that.

Like I said, all you have to do is figure out whether the system I gave you is a consistent one.  To do so, solve for $ q \equal{} \sqrt{ \frac{16}{90} }$ (or whatever the relevant allele ratio is) and use this number to find $ p \equal{} 1 \minus{} q$.  Now if it is [i]also[/i] true that $ 2pq \equal{} \frac{35}{90}$, then the system is consistent.  Otherwise, it's not.

t0rajir0u, perhaps we are answering different questions.  However, I doubt that a small population will fit the principle exactly, so I think a chi-squared test is appropriate to account for variations due to chance.  Perhaps, however, we may want to use a larger alpha level because we really want to see if it fits the model, not if the model could be accurate (which is usually what is tested).

As for testing whether it is an exact fit, all we need to do is solve for both $ p$ and $ q$ from what we know about the second generation, and then check to see if $ 2pq$ gives the proportion of heterozygous individiuals in the 2nd generation.

Ninety alleles is not a large population, and by that you are correct. However, I do believe that that ninety alleles is enough to prevent genetic drift. You can add a chi-square analysis as a supplement to the answer, but I do not believe it is necessary.

Thanks to all of your help I found the p-value for the practice set of 90 alleles. The real task is finding p-value for 2000markers which I did through a helpful program, however I ran into trouble again. For my presentation I need to present the 'adjusted p-value for multiple tests'; I have no idea what that means. The 'book' (more of a manual / dictionary thing that I use for 'research') defined it as 'the cutoff number used divided by the total number of tests gives the Ap (adjusted p-value) which then modifies raw p-values to give more stringent results'.

I took everything under the normal 5% [.05] for raw p-value. According to my interpretation of the book it says to do $ .05/2000$ for the adjusted p-value, but wouldn't that mean all the p-values are the same? I'm completely clueless here  :?:

Perhaps look here:[url]http://www.ats.ucla.edu/STAT/sas/library/multtest.htm[/url].  In short, (I think) the idea is if you do multiple one or two-sided t (or z) tests, then you have to consider that it becomes more probable that one of the tests leads to a type I error (that you reject the null hypothesis, here the Hardy-Weinberg principle, because the p-value obtained is less than the alpha-level you set at the beginning).  For instance, if you conduct 20 tests at the .05 level, you would expect that 1 of the tests produces a type I error.  Thus you need to change your ALPHA LEVEL to make this less likely (NOT your p-values).

The actual equation:  if we want an overall alpha level of $ A$, and we have an alpha level of $ a$ for each individual test of $ n$ tests then we need $ A \equal{} 1 \minus{} (1 \minus{} a)^n \implies a \equal{} 1 \minus{} (1 \minus{} A)^\frac {1}{n}$.  I believe the method that your book is giving you ($ a \equal{} A/n$) is slightly more conservative than this calculation.

Translation for those without much Stat knowledge:
The idea is that we are making decisions based on the p-values (the probabilities that, if our assumptions (null hypothesis) about the population are true, we would obtain a sample as far or further from predicted as the one we obtained).  The alpha level is the threshold we set for the p-value; if the p-value is less than the alpha level, we reject the null hypothesis and say that the sample is significant evidence against the null at the alpha-level; if the p-value is greater than the alpha-level, then we accept the null hypothesis.  A type 1 error is the error of rejecting the null hypothesis when it is actually true.  We are trying to keep down the chance of committing ANY type 1 errors.  The more tests we do with one set of data, the larger chance of having a type 1 error since we have more chances to produce type errors, so we must shrink our alpha levels accordingly to shrink the chance of a type 1 error. (probability of a type 1 error is the alpha level when we only do one test; with multiple tests, $ P(any type 1 errors) \equal{} 1 \minus{} (1 \minus{} a)^n$ where $ n$ is the number of tests and $ a$ is the alpha level of each individual test)

perhaps this explains it better: [url]http://www.pubmedcentral.nih.gov/articlerender.fcgi?artid=117123[/url]

I actually also read the ucla article but not the nih one, thanks for the explanation. So, and please correct me if I misunderstood, I modify my alpha level for the multiple tests by using $ a \equal{} 1 \minus{} (1 \minus{} A)^\frac {1}{n}$ and further reject any p-values greater than the new alpha level from my original set of p-values?

You do not accept/reject p-values.  You accept/reject hypotheses based on the p-values.  The p-value of a single test is the probability that the null-hypothesis of that test (some claim about the distribution of the population) is true.  So if a p-value is greater than the new alpha level then you accept the null, and if it is less than the new alpha level then you reject the null of that specific test.  I think the point of the multiple comparisons is, unlike a chi-squared test, we see exactly which part of the overall claimed distribution is incorrect.

Yes, you can use $ a\equal{}1\minus{}(1\minus{}A)^{1/n}$ where you probably want $ A\equal{}.05$ so all the tests together have a 5% chance of giving you the wrong conclusion.  The method your book suggests is $ a \approx A/n < 1\minus{}(1\minus{}A)^{1/n}$, but there is not a large numerical difference between the two methods (about 3% error at larger $ n$'s) though the approximation is conservative.


algebra
polynomial
number theory proposed
number theory

Show that if $p$ is prime, so $p^{p}-1$ has a prime factor that is congruent to 1 modulo $p$.

Let $q|p^{p}-1$ be a prime.
Then the order of $p \mod q$ is either $1$ or $p$ (it is a divisor of $p$).
If the order is $1$, we would get $q|p-1$, not dividing $\frac{p^{p}-1}{p-1}=p^{p-1}+...+p^{2}+p+1 \equiv p \mod q$.
Thus let $q$ be a divisor of $p^{p-1}+...+p^{2}+p+1$, then the order of $p \mod q$ is $p$, thus $p|q-1$.

Or just pwn it with the main theorem on cyclotomic polynomials $\Phi_{n}(x)$:

Let $n \in \mathbb{N}$. If there is an $a \in K$ (a field) with $\Phi_{n}(a)=0$, then $n_{K}=0_{K}$ or $\mathrm{ord}_{K}(a)=n$.


Let g(x) be defined by g(0) = 1 and g(x) = ln(1+x2)/x2, x  :!=: 0.
(It is easy to show that lim(x:downto:0) = 1 = g(0), so g(x) is continuos for all x)

Using ln(1+x) =  :Sigma: (-1)^n*x^(n+1)/(n+1) for -1<x <= 1, I showed that
ln(1+x2)/x2  =  :Sigma: (-1)^n*x^2n/(n+1) for 0<Abs(x)<1.
For x = 0 this series coverge to ln(1) = 0  :!=: 1 = g(0).
The question is then:
Is :Sigma: (-1)^nx^2n/(n+1) the Mac-Laurin series for g?

[quote="Antilopen"]
ln(1+x2)/x2  =  :Sigma: (-1)^n*x^2n/(n+1) for 0<Abs(x)<1.
For x = 0 this series coverge to ln(1) = 0  :!=: 1 = g(0).
[/quote]
Here is your mistake: for $x=0$ the first term of your series is $(-1)^0 x^0/(2\cdot 0+1)=1$ and all other terms are $0$, so the series actually converges to $1=g(0)$.
The answer to your question is "Yes, it is the Taylor-Mac-Laurin series for $g$".


function
complex analysis
complex analysis unsolved

Prove that:
  Power series of an analytic function f doesn't converge at any point on its convergent circle if on which there is at least one pole of function f.

 :roll:

If $f(z)=\sum_{n=0}^{\infty}c_{n}z^{n}$ in the open unit disk and $c_{n}\to 0$, then $f$ does not have a pole on the unit circle. Indeed, $|f(z)|(1-|z|)\le \sum_{n=0}^{\infty}|c_{n}z^{n}|(1-|z|)\le (1-|z|)\sum_{n=0}^{N}|c_{n}|+\max_{n>N}|c_{n}|$ for any positive integer $N$, which implies $|f(z)|(1-|z|)\to 0$ as $|z|\to 1$.

To mlok: I think you misunderstood me. What i want to show is:

if the funtion f has at least one pole on the unit circle, then the power series of function f doesn't converge at any point on the boundary of this unit circle.

In addition, |f(z)|.(1-|z|)--> 0 doesn't lead to that f doesn't have any pole on unit circle.

@hoctoandenbaogio: Mlok's solution is correct. What he showed is that if the series converges at at least one point on the unit circle, then there are no poles on the circle, which is completely equivalent to the statement of the problem. As to your second remark, it is just false. If $f$ has a pole at $\zeta$ with $|\zeta|=1$, then, near that pole, we have $|f(r\zeta)|(1-r)\ge c>0$, so $|f(z)|(1-|z|)$ cannot tend to $0$ when $|z|\to 1$.

[u]To fedja[/u]: You said that If $f$ has a pole at $\zeta$ with $|\zeta|=1$, then, near that pole, we have $|f(r\zeta)|(1-r)\ge c>0$ ????
But why?

Write down $f$ as a Laurent expansion around $\zeta$ and use this to evaluate $|f(r\zeta)|(1-r)$


AMC
AIME
trigonometry
probability
AIME I
AMC 10
number theory

I am hoping to make the AIME (I think i will with an average AMC 10 score of 120-135) but i am not well informed of the AIME. I know about the time, # of questions, no calculators, essentially the rules. What I don't know about are the types of questions on the test. What kinds of concepts are on the AIME? log, modular arithmetic, trig, geo, calc??? Please tell me soon!!! Thank You.

You can find a lot of old AIME's here:

http://www.kalva.demon.co.uk/aime.html

I personally think that the American Invitational Math Examination has great problems (much nicer than amc 12 problems, even though their much harder).  From my experience with doing kalvademon's tests, i would say that trigonometry is essential (identities are important) but again, they understand that you dont have a calculator, so the trigonometry is limited to easy angles that are easy to work with.  and that you should be pretty good with combinatorics.  My main focus right now should be on sequences and series (because im abnormally bad at these).  There is a lot of sequences and series problems on the Invitational test.  Factors, divisibility, lcm's, gcf's, abc's whatever are important topics in number theory that you should know.
Another thing:
There's a long way to solve an Invitational problem, and there's a short way.  With the long way, you could be on a single problem for hours at a time (especially without a calculator) and then there's a huge probability of error.  But, if you actually see patterns or recognize that two triangles are similar, the time it takes to solve the problem becomes a lot shorter.  Why are demon's solutions to the invitational round so short?

[quote="tongos"]  Why are demon's solutions to the invitational round so short?[/quote]

They are solution sketches.  I find that to understand them at all you almost have to have solved the problem yourself.

Additionally, them having short, hard to understand solutions encourages people to buy the real AIME's which come with the official solutions which are much nicer.


How would I do this?

let x, y, and z be positive real numbers such that x+y+z=1.  Show that 1/x + 1/y + 1/z is > or = 36

Unless I'm missing something, this isn't true.

Take $x=y=z=1/3$. Then $x+y+z=1$, but $1/x+1/y+1/z = 9$.

You can prove this is the minimum by using Cauchy, I believe.

He must write 9 and not 36 and then to use twice AM - GM or to rewrite the inequality as

   $(x+y+z)(\frac{1}x + \frac{1}y + \frac{1}z ) \geq 9$ and expandb But  $a/b + b/a \geq 2$  .

 Babis

[quote="pkothari13"]i think you have to use that other form of the inequality:

$\displaystyle\frac{1^2}{x} + \displaystyle\frac{2^2}{y} + \displaystyle\frac{3^2}{z} \geq \displaystyle\frac{(1+2+3)^2}{x+y+z} = \displaystyle\frac{36}{1}.$

is this right?[/quote]

Nice solution using Cauchy-Schward the second form  :clap2: .


Who's doing the Team Scramble this year?  I know my school is...

I'm fairly sure that the schools that do one NA usually follow up with the next one....anyway, we are doing it.  Sad group of kids, though. :( (compared to previous years, that is)

So how did everyone do?

My school has quite a big math team, yet no one in my school wants to work hard on math, and thus no one from my school has yet made USAMO, and thus we do not do that well in worthwhile competitions, so I don't think we did that great (I must ask my teacher tomorrow

EDIT: our school only got 58. and we had at least 25 or 30 people, all but 2 of whom were juniors and seniors...I am glad I contributed 8 of those points.


If $ F(a, b, c, d) \equal{} a^b \plus{} c \times d$, what is the value of $ b$ such that $ F(6, b, 4, 3) \equal{} 48$?

$ 6^b\plus{}4\times3\equal{}48 \iff 6^b\equal{}36 \iff b\equal{}2$


geometry
parallelogram
ratio
geometric transformation
reflection
geometry solved

Menelaus works fine:

Let $X,Y$ be the intersection pts between the tangent in $C$ to the circle and $AB,AD$ respectively. We need to show that $\mathcal E_{1}=\frac{XM}{MA}\cdot\frac{AN}{NY}=\frac{XB}{BA}\cdot\frac{AD}{DY}=\mathcal E_{2}$.

First of all, notice that $M,N$ are the projections of $C$ on $AB,AD$ respectively. We thus have $\frac{XM}{MA}=\frac{XM}{MC}\cdot\frac{MC}{MA}=\left(\frac{XC}{CA}\right)^{2}$. In the same manner we get $\frac{AN}{NY}=\left(\frac{CA}{CY}\right)^{2}$. From these two we get $\mathcal E_{1}=\frac{XC^{2}}{CY^{2}}\ (*)$.

We also have $\mathcal E_{2}=\frac{XB\cdot BC}{CD\cdot DY}=\frac{S_{XBC}}{S_{CDY}}$, and since $XBC,CDY$ are similar, that ratio is equal to $\frac{XC^{2}}{CY^{2}}$. From $(*)$ we see that $\mathcal E_{1}=\mathcal E_{2}$, Q.E.D.

Another solution:

Let the tangent to the circle k at C meet the line MN at a point T. We want to show that the point T lies on the line BD.

Let O be the center of the circle k. We know that the points A and C are diametrically opposite to each other on the circle k. Let U be the point diametrically opposite to the point N on the circle k. Then, the segments AC and NU are diameters of the circle k, and hence, the center O of the circle k is the midpoint of these two segments. This means that the points C and U are the reflections of the points A and N in the point O. Consequently, CU || AN. This, together with BC || AN, yields that the points C, U and B are collinear. Now, applying the Pascal theorem to the degenerate cyclic hexagon NMACCU, we see that the points NM $\cap$ CC, MA $\cap$ CU and AC $\cap$ UN are collinear. Now, NM $\cap$ CC = T, MA $\cap$ CU = B, and AC $\cap$ UN = O (the latter because the segments AC and UN are diameters of the circle k and thus intersect at its center O). Hence, the points T, B and O are collinear. Similarly, the points T, D and O are collinear. Thus, the point T lies on the line BD. Proof complete.

See also http://www.mathlinks.ro/Forum/viewtopic.php?t=56624 .

  Darij

[quote="yamilimot"]Given a non-rectangular parallelogram ABCD, a circle k with diameter AC
intersects lines AB and AD in points M and N (other than A). Prove that
lines BD, MN and the tangent at C to k are concurrent (or all parallel).[/quote]

If K=CB^MN and I=AB^tC (where tC is the tangent to k at C) then, by simple angle chasing, one can see that <NMA = <KCI, i.e. CIMK is cyclic, from which follows that <CKI = <CMI = 90�. This directly leads to the fact that DNC and BKI are homotetics, which is equivalent to what is requested.