Prove that the length of the angle bisector from A in triangle ABC is
(1/(a+b)) X ((ab)(a^2+b^2-c^2))^(1/2)

Let $ ABC$ be a triangle with bisector $ AD$;$ BC \equal{} a;AC \equal{} b;AB \equal{} c;AD \equal{} l_a$,we  have:
$ l_a \equal{} \frac {2bc}{c \plus{} b}.cos(\frac {A}{2})$
[i][b]Proof[/b][/i]
We have:
$ \frac {BD}{DC} \equal{} \frac {c}{b}$
so $ BD \equal{} \frac {ac}{b \plus{} c}$
In triangle $ ABD$,we have
$ \frac {AD}{sinB} \equal{} \frac {BD}{sin(\frac {A}{2})}$
=>$ AD \equal{} \frac {ac}{b \plus{} c}.\frac {sinB}{sin(\frac {A}{2})}$
=>$ l_a\equal{}\frac{2ac}{b\plus{}c}.cos(\frac{A}{2}).\frac{sinB}{sinA}$
=>$ l_a \equal{} \frac {2bc}{c \plus{} b}.cos(\frac {A}{2})$

[hide="Hint"]Extend $ AD$ until it hits the circumcircle of $ \triangle ABC$.[/hide]

I feel stewart's theorem may be useful here. But i'm probably wrong...

[quote="Bugi"][hide="Another way"]

Stewart's theorem [/hide][/quote]

Eh? :P

OK, full solution

[hide="Solution"]

AD=l - bisector
BD=m, CD=n

$ m\equal{}\frac{ac}{b\plus{}c},n\equal{}\frac{ab}{b\plus{}c}$

Recall Stewart's theorem: $ b^2m\plus{}c^2n\equal{}a(l^2\plus{}mn)$, rearrange and express $ l$. Satisfied?  :P

[/hide]

Oops, it was directed at Poincare, not you. I didn't realize the ambiguity until now. :blush:

Lol...That was funny :D

This was covered in volume 2 i think...


topology
superior algebra
superior algebra unsolved

Either you are equal to sepidetaghizade or you should ask him/her to build a group on this...
http://www.mathlinks.ro/Forum/viewtopic.php?t=119096


Let $ n\geq2$ be a positive integer and for real numbers $ a,b$ we have:

$ a^{n}=a+1,\;\; b^{2n}=b+3a$

which one is biGger: $ a$ or $ b$?

This Problem purposed in American Mathematical Olympiads and the answer:

tha was really nice!Beautiful solution Navid :lol:


Find all permutations $ (\mathop x\nolimits_1,\mathop x\nolimits_2,\mathop x\nolimits_3,\mathop x\nolimits_4,\mathop x\nolimits_5,\mathop x\nolimits_6,\mathop x\nolimits_7,\mathop x\nolimits_8,\mathop x\nolimits_9)$ of $ (1,2,3,4,5,6,7,8,9)$ such that for any $ i>j,\left| {\left. {\mathop x\nolimits_i \minus{} \mathop x\nolimits_j} \right|} \right.\ne i\minus{}j$

by using the help of mathematica

lista[n_] := lista[n] = Flatten[Table[{i, j}, {
    i, 1, n - 1}, {j, i + 1, n}], 1]
perm[n_] := perm[n] = Permutations[Range[n]]
chequeo[per_, {j_, i_}] := Abs[per[[i]] - per[[j]]] == i - j
cheq[per_, lista_] := {Or @@ ((chequeo[per, #]) & /@ lista ), per}
f[n_] := Table[cheq[Part[perm[n], a], lista[n]], {a, 1, n!}]
Cases[ f[9],  {False, b_}] // ColumnForm

There are $ 352$ cases...however it is very interesting that there only $ 4$ permutations of $ \{1,2,3,4,5,6\}$ with that property.

please show me one permutation please mszew

thank you

[quote="Fermat's Little Turtle"]How would one go about solving this for smaller sets? Such as the 6 element set.[/quote]  He showed precisely the computer code he used to do it  :wink:

Equivalently, the question asks for the number of ways of placing 9 nonattacking queens on a 9 by 9 chessboard.  The corresponding sequence is [url=http://www.research.att.com/~njas/sequences/A000170]A000170[/url] in the [url=http://www.research.att.com/~njas/sequences]OEIS[/url].

One example of such a permutation is 136824975.

Another problem

Find all permutations $ (\mathop x\nolimits_1,\mathop x\nolimits_2,\mathop x\nolimits_3,\mathop x\nolimits_4,\mathop x\nolimits_5,\mathop x\nolimits_6,\mathop x\nolimits_7,\mathop x\nolimits_8,\mathop x\nolimits_9)$ of $ (1,2,3,4,5,6,7,8,9)$  such that for any $ i > j,\left| {\left. {\mathop x\nolimits_i \minus{} \mathop x\nolimits_j} \right|} \right.\ne i \minus{} j$ or $ 9 \minus{} (i \minus{} j)$

the answer is no such the permutation .

Can one show me any solutions please ?

Change the code for the aditional condition and there no such permutations, however it remains interesting that the same $ 4$ six number permutation fullfils that aditional criteria.

There are 2,10,4,4,0,0 for the 5,6,7,8 and 9 permutations.


geometry
geometric transformation
dilation

Hey guys,
Does anyone know in what way i should organize my CV / resume from high-school?
Can you give me some templates/examples?
Thanks in advance,
:)

Here are some things to include.

After reading them, note that you should not go over 2 pages -- if you do, you are almost certainly diluting the good stuff. In the resume, it is [b]your[/b] job (not admissions') to determine the highlights of your experiences and accomplishments. Also, make your margins narrower if you must, and use size-10 font for the body.

[b]Header:[/b]
- Name
- Date of Birth
- SSN (for identification purposes -- don't leave temporary copies with your SSN lying around)
> Header should be grey or dark grey, but not black.

[b]Top:[/b]
- Title (Resume, Curriculum Vita/Vitae/Vitæ)
- Name
- website url (if you have one that you want to post)
- optionally, contact information (if you are attaching your resume on Common App, as additional information, it is not very useful to post all your contact info. Otherwise, it can make it look more professional).

[b]Body:[/b] I would suggest the following sections (if applicable), in the order listed:
I. Education
- List your high schools and the dates.
- This is important if you have more than one school; that way, admissions officers won't superficially look and note that you weren't in any particular school organization for 4 years, and will (hopefully) take more time to absorb the things they see.
- If you only went to one high school, but have a highly competative GPA and/or class rank, you can still put an education section, include the one high school, and your rank and GPA. Coherence is a virtue.

II. Employment / Internships / Employment and Internships / Work Experience (if this section is substantial)
- Pick the title that is most relevant.
- Order in reverse chronological order, with the most recent jobs first
- For each, include the title of your position (if applicable), a short description of responsibilities, achievements (NOT how much money you made), contributions, skills learned, or anything else that adds dimension to your resume
IMPORTANT CAVEAT - If you started your own business for fun, don't be like, "Blah blah blah, Inc: Founder and CEO" "I oversee all operations and make executive business decisions", unless there is substance behind the words, because that sounds stupid and childish. Don't say you were in charge of client relations or something if you had 3 clients. Don't say you were in charge of advertising if "advertising" means posting it on eBay.

III. Community Service
- Include only if you have a significant amount of this. Don't put too much fluff here. If you have only one or two things here, place them under Activities instead.
- Organize in reverse chronological order. Include what you did/achieved/gained/contributed and approximate time investment.

IV. Activities
- This should be organized in the most relevant way
- If you have SIGNIFICANT athletic experience, you can create a section "Athletics" preceeding this section.
- If you want to focus on a few major ones, than write down the names of the major clubs first, along with a description of what you did/accomplished/gained/contributed, much like for the jobs, as well as the dates. Then, list less significant ones without detail.
- Alternately, list all your activities temporally, and include .5 - 2 lines briefly describing the above for each club.
IMPORTANT NOTE: Don't put down stupid things. Some people think that adding a minor activity will either help them or just be overlooked, and cannot hurt. This is wrong. If something contributes to your resume, put it in. If it does not, then it is just diluting the rest of your resume!

V. Honors and Awards
- Important awards/distinctions you have achieved. Be very discriminating in what you put!
IMPORTANT NOTE: Same as for Activities, but even more important because there is little to describe for awards, usually. It is better to list 2 or 3 important awards alone, than to list 2 or 3 important awards scatterd among 7 unimportant awards.

VI. Other Skills / {any skills you want to include}:
- Use own discretion on whether or not to include it
- Language proficiencies
- Computer skills (mostly for programming languages. Software familiarity is probably irrelevant for applying to college, though very important for applying for jobs)
NOTE: Do NOT include sports, art, or hobbies here.

VII. Hobbies
- Hobbies go here.
- Use own discretion on whether or not to include it
- Things that you chose not to put into activities because they would dilute your good stuff might go here.
EXAMPLE: Someone who is heavily into math and science and has a random snowboarding club experience might rather put down snowboarding as a hobby instead of putting it in with their list of activities, which can dilute the list and/or look like fluff.
- Include (if it helps you) how long you've been involved

High School Resume Template Tips
· If you have work experience, then highlight your responsibilities and achievements relevant to the job you are applying for. For example if you are going for a customer service job, highlight your customer service skills.
· If you do not have any work experience simply remove this heading from your resume or CV and focus more on Relevant Skills. List your attributes and capabilities in order of what you know the employer wants.
· Limit your resume or CV to maximum 2 sides of paper.
· Finally proofread your resume.


If $ a \plus{} b \plus{} c \equal{} 2$, and $ a \times b \plus{} 2 \times c \equal{} 0$

then $ a^3 \plus{} b^3 \plus{} c^3 \equal{} ?$

hello, by solving the system $ a\plus{}b\plus{}c\equal{}2,ab\plus{}2c\equal{}0$ we get $ a\equal{}\minus{}c,b\equal{}2,c\equal{}c$ or $ a\equal{}2,b\equal{}\minus{}c,c\equal{}c$ so we have
$ a^3\plus{}b^3\plus{}c^3\equal{}\minus{}c^3\plus{}8\plus{}c^3\equal{}8$.
Sonnhard.

$ \left\{\begin{array}{l}
a\plus{}b\plus{}c \equal{} 2 \\
a b \plus{} 2 c \equal{} 0
\end{array}\right.
\Rightarrow
\left\{\begin{array}{l}
(a, b, c) \equal{} (\minus{}c, 2, c) \\
\text{or} \\
(a, b, c) \equal{} (2, \minus{}c, c)
\end{array}\right.$

So, $ a^3\plus{}b^3\plus{}c^3 \equal{} 8$

[hide]

I basically solved the system of equations to get:
b = 2
a = -c
c = c

From there you do:

$ \minus{}c \plus{} 2 \plus{} c \equal{} 2$
and
$ c \equal{} 0$

so:

$ \minus{}0^3 \plus{} 8 \plus{} 0^3 \equal{} \boxed{8}$

[/hide]

Substitute for c in the second equation, you get
$ ab \minus{} 2a \minus{} 2b \plus{} 4 \equal{} 0$
$ (a \minus{} 2)(b \minus{} 2) \equal{} 0$
$ a \equal{} 2$ or $ b \equal{} 2$
This is how you solve the system.


Add seven different fractions with numerator 1 to get 1.

[hide]1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/48 + 1/96 = 1[/hide]
There are actually many possibilities...:D

Lightspeed's solution works. Math92's was very creative.  :lol:

Do they all have to be different? I'd enjoy doing this:
[hide]$\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{12}+\frac{1}{12}$[/hide]

[quote="math92"]oh, it doesn't work because all the fractions have to be different.  Well, lightspeed's solution works.[/quote]

Otherwise, math92's would work

If math92's solution was a valid answer to this problem, I wouldn't have posted it in the mathcounts forum.

Yeah it would be a little too easy then.

So, lightspeed's solution is one of the answers.

[hide]1/2+1/3+1/12+1/24+1/48+1/72+1/144[/hide]
your right, there ARE a lot of solutions. In fact, there's an infinite # of them.


geometry
function
integration
real analysis
real analysis theorems

Prove that the area under the curve from a to b is equal to the antiderivative of b minus the antiderivative of a please.

You need to define what you call area and you need to give hypotheses on $f$.

for what type of function?

i think it'd be too hard to prove it for all functions.

$F(b) - F(a) = \int^b_a f(x) dx$

where $f(x) = \frac{F(x+dx)-F(x)}{dx}$

that's the best i can do in terms of a proof, but what that means is f(x) how much F(x) changes.  so F(b) minus F(a) is the cumulitive change from a to of f(x), which can be reguarded as an area.


inequalities
floor function
function
integration
logarithms
ceiling function
calculus

For a natural number $ n$ and $ n>1$ show that $ \frac{1}{n}\plus{}\frac{1}{n\plus{}1}\plus{}\frac{1}{n\plus{}2}\plus{}....\plus{}\frac{1}{n^{2}}>1$.

I have solved thsi but I don't like my proof. So can anyone out her give me a nice and ingenious one?

I tried but what I got was very clumsy! Any other approach?

[hide="Solution"]By $ \text{AM} > \text{HM}/\text{Cauchy Schwarz}$
$ \left(\sum_{i\equal{}n}^{n^2} i\right)\left(\sum_{i\equal{}n}^{n^2} \frac{1}{i}\right)>\left(\sum_{i\equal{}n}^{n^2} 1\right)^2$
(Equality doesn't occur since all terms are unequal.)
So we need to prove that
$ \left(\sum_{i\equal{}n}^{n^2} 1\right)^2\geq \left(\sum_{i\equal{}n}^{n^2} i\right) \iff (n^2\minus{}n\plus{}1)^2\geq \left(\sum_{i\equal{}1}^{n^2} i\minus{}\sum_{i\equal{}1}^{n\minus{}1} i\right)$
$ \iff (n^2\minus{}n\plus{}1)^2\geq \left(\frac{n^2(n^2\plus{}1)}{2}\minus{}\frac{(n\minus{}1)n}{2}\right)$
$ \iff 2(n^2\minus{}n\plus{}1)^2\geq (n^4\plus{}n^2\minus{}n^2\plus{}n)$
$ \iff 2(n^2\minus{}n\plus{}1)^2\geq n(n\plus{}1)(n^2\minus{}n\plus{}1)$
$ \iff 2(n^2\minus{}n\plus{}1)\geq n(n\plus{}1)$
(Note that $ n^2\minus{}n\plus{}1>0$)
$ \iff n^2\minus{}3n\plus{}2\geq 0\iff (n\minus{}1)(n\minus{}2)\geq 0$
This is obvious for $ n>1$[/hide]

We have from Cauchy-Schwarz
\begin{align*}\sum_{i=n}^{n^2}\frac 1i&=\frac 12\sum_{k=0}^{n^2-n}\left(\frac{1}{n+k}+\frac{1}{n^2-k}\right)\\
&\ge\frac 12\sum_{k=0}^{n^2-n}\frac{4}{n^2+n}\\
&=\frac{2(n^2-n+1)}{n^2+n}\ge 1\end{align*}
where the last inequality follows from $ 2n^2-2n+2\ge n^2+n\iff (n-1)(n-2)\ge 0$, which is true. And since equality does not hold in the Cauchy-Schwarz for $ n=2$, equality never holds.

I don't understand what u said. Are u keeping the ineq intact and Substituting $ n^2$ with a smaller $ k$? :maybe:

Yes, I am and I want to know which is the smallest such $ k$?

[quote="manjil"]Can we relplace $ n^{2}$ by a $ k$ and which is the smallest value of $ k$ such that the inequality holds.[/quote]
[hide="Conjecture"]$ \lfloor en \minus{} 1\rfloor$
Where $ e$ is the exponential constant[/hide]

That is very interesting indeed!
When we take $ n\equal{}2$ it shows that $ \frac{1}{2}\plus{}\frac{1}{3}\plus{}\frac{1}{4}>1$ i.e., the $ k\equal{}n\plus{}2$.

Again when we take $ n\equal{}3$ it shows that $ \frac{1}{3}\plus{}\frac{1}{4}\plus{}\frac{1}{5}\plus{}\frac{1}{6}\plus{}\frac{1}{7}>1$ i.e. the $ k\equal{}n\plus{}4$ now!

When we take $ n\equal{}4$, it shows that $ \frac{1}{4}\plus{}\frac{1}{5}\plus{}\frac{1}{6}\plus{}\frac{1}{7}\plus{}\frac{1}{8}\plus{}\frac{1}{9}\plus{}\frac{1}{10}>1$
So lastly our $ k\equal{}n\plus{}6$!

I think the generalisation follows now, doesn't it! :lol:
For any $ m>1$ the desired $ k$ is expected to be $ k\equal{}m\plus{}2(m\minus{}1)$, I hope! :)

But I haven't proved it yet. :(

[quote="Sunkern_sunflora"]I think the generalisation follows now, doesn't it! :lol: [/quote]
Unfortunately it doesn't :!:
[quote="Sunkern_sunflora"]For any $ m > 1$ the desired $ k$ is expected to be $ k \equal{} m \plus{} 2(m \minus{} 1)$, I hope! :) [/quote]
$ \sum _{i \equal{} 6}^{15}\frac {1}{i}\approx 1.034896 > 1$
So for $ n\equal{}6$, $ 15$ is the minimum $ k$
But your formula says that the minimum is $ 16$

$ k\leq(3n\plus{}1)$. Now guys you can do better!

[quote="manjil"]$ k\leq(3n \plus{} 1)$. Now guys you can do better![/quote]
My formula is true even at $ 10000$. And I think it is always true. We just need to prove it.

[hide="Narrow range of k proved"]Since $ \frac{1}{x}$ is a decreasing function
$ \int_{m}^{m\plus{}1} \frac{1}{x} dx\leq \frac{1}{m}\leq \int_{m\minus{}1}^{m} \frac{1}{x} dx$ for $ m>1$
$ \int_{a}^{b\plus{}1}\frac{1}{x} dx\leq \sum _{i\equal{}a}^{b} \frac{1}{i}\leq \int_{a\minus{}1}^{b}\frac{1}{x} dx$
for integers $ b>a>1$
Suppose $ k$ is the required minimum.
We need to have
$ \sum _{i\equal{}n}^{k} \frac{1}{i}\geq 1$
[b]and[/b]
$ \sum _{i\equal{}n}^{k\minus{}1} \frac{1}{i}\leq 1$
We know that
$ \sum _{i\equal{}n}^{k} \frac{1}{i}\leq \int_{n\minus{}1}^{k}\frac{1}{x} dx\equal{}\ln k\minus{}\ln (n\minus{}1)\equal{}\ln (\frac{k}{n\minus{}1})$
and
$ \sum _{i\equal{}n}^{k\minus{}1} \frac{1}{i}\geq \int_{n}^{k}\frac{1}{x} dx\equal{}\ln k\minus{}\ln n\equal{}\ln (\frac{k}{n})$
So
$ \ln (\frac{k}{n\minus{}1})\geq 1\implies \frac{k}{n\minus{}1}\geq e\implies k\geq en\minus{}e$
[b]and[/b]
$ \ln (\frac{k}{n})\leq 1\implies \frac{k}{n}\leq e\implies k\leq en$
So $ en\geq k\geq en\minus{}e$
This narrows down our value of $ k$ [b]very much[/b].[/hide]

Brilliant Akashnil! I thought it would follow generalisation, unfortunately it doesn't! :(
Anyway, your proof really narrows down $ k$ a lot. I think combining your proof and Manjil's conjecture, the problem gets solved as well.
 :)

[quote="Sunkern_sunflora"]I think combining your proof and Manjil's conjecture, the problem gets solved as well.
 :)[/quote]
It was my conjecture. However my proof allows $ 2$ or $ 3$ values for $ k$ for all $ n$. Hopefully the interval doesn't get larger as $ n$ gets larger. However I still couldn't pinpoint it :(
I have also proved that a stronger range $ en > k > en \minus{} \frac {e \plus{} 1}{2}$ is possible using the convexity of $ \frac{1}{x}$. But still there are $ 2$ possibility for $ k$ for some $ n$.
However we have got a very nice and strong inequality $ \sum_{i \equal{} n}^{\lceil en \rceil} \frac {1}{i} > 1$
Lets wait for someone like [t0rajir0u] to solve the problem of $ \min k$ completely.

I thought I was a duffer that I could not solve the $ mink$ but it turns out that its really much more complex than I thought!

Anyway I was trying to work out on a proof of the inequality without the use of any standard inequality and I have got something which I will post soon enough!

About the $ min k$ problem. I have not been able to solve it completely and each time I think I have done it something occurs. Anyway I sincerely hope someone will be able to solve that!

Or else we can see if we can get some lower bounds for the left hand side expression taking that as a function of $ n$ and finding the minimum $ k$ which satisfies it that is find $ f(k)$!

[quote="manjil"]Maybe we can solve the problem for minimum $ k$ by taking the rleft hand side as an integral of some function? It is just an idea, can anyone see if it will work?[/quote]
Have you read post #15 in the thread?


Hi,

On most math books and articles theres always a label on the extreme right of every equations, like (1), [17]...

How can you do that in \LaTeX\ ?

Thx

Use [code]\begin{equation} \int_{-\infty}^\infty e^{-x^2}\; dx = \sqrt{\pi} \end{equation}[/code] to get automatic equation numbers. If you want to number them on your own, do it with [code]\begin{equation*} \Gamma(s)=\int_0^\infty e^{-t} t^{s-1} \; dt \tag{1}\end{equation*}.[/code]

Thank you !

It's quite simple...

However...

[quote]\begin{equation*} \Gamma(s)=\int_0^\infty e^{-t} t^{s-1} \; dt \tag{1}\end{equation*}.[/quote]

That does'nt work, really don't know why...

Sorry I wasn't thinking. You still use equation rather than equation* even when you want to choose the equation numbers. But then you also use tag. If you don't want the parentheses around the equation number, you can use \tag*.


Polygon $ ABCDEF$ is a regular hexagon. What is the measure in
degrees of angle $ ABF$?

This problem had an intended diagram, but it's missing. Unless someone finds the picture, it's impossible to solve.

Actually you don't need the picture, since its a regular hexagon.

I know the answer is 30, don't know how to explain it.

Hint: The vertices are labeled in order (A then B then C, etc.)

Yeah, like ernie said, no diagram necessary.

The basic fact from geometry:

For a regular polygon of n sides, using degrees:
$ \text{"sum of interior angles"}\equal{}(n\minus{}2)\times 180$

F.E.  triangle gets you 180, square gets you 360, hexagon 720 etc.

Then each of the 6 interior angles of a regular hexagon is 120.  the angle ABF is part of the isocoles triangle ABF. Of course the angle FAB is 120 degrees (because it is an interior angle), and since it is isocoles, the other two angles would have to be 30.  (ABF is one of those angles :) )

Diagram:

[asy]pair A=(0,0),
B=A+4*dir(0),
C=B+4*dir(60),
D=C+4*dir(120),
E=D+4*dir(180),
F=E+4*dir(240);
label("A", E, N);
label("B", F, W);
label("C", A, S);
label("D", B, S);
label("E", C, W);
label("F", D, N);
draw (A--B--C--D--E--F--cycle);
draw(F--D);[/asy]

Triangle ABF is isosceles, and the sum of <ABF and <AFB is 60 degrees, so <ABF is half that, or 30 degrees.


Let $n$ be a positive integer number. Prove that: $\frac{1.3.5..(2n-1)}{2.4.6...2n}<\frac{1}{\sqrt{3n+1}}$

What exactly do you mean?
If you're talking about a sequence of numbers, that's wrong.
$135/246<\frac{1}{\sqrt{7}}$ is not true.
If you're talking about multiplication, that's also wrong since
$1/2=1/\sqrt{4}$.
It does however work for all integer n greater than 1.
[hide]Base: n=2 where $3/8<1/\sqrt{7}$
Now assume it true for some n.  Let's show it true for n+1.
$\frac{2n+1}{2n+2}*\frac{1*3*5*7*9*...*(2n-1)}{2*4*6*8*...*(2n)}<1/\sqrt{3n+1}*\frac{\sqrt{3n+1}}{\sqrt{3n+4}}$.  This is equivalent to showing $2n+1/2n+2<\frac{\sqrt{3n+1}}{\sqrt{3n+4}}$.  Squaring and then cross multiplying, one gets $12n^{3}+28n^{2}+19n+4<12m^{3}+28n^{2}+20n+4$ which is true since n is positive.[/hide]

Yes. this is my error for typing. This inequalitu olny true for all n is a positive integer number that is larger or equal to 2


In circle with center $ O$, the measure of $ \angle RIP$ is $ 36^\circ$ and $ OR\equal{}10$ cm. Find the number of centimeters in the length of arc $ RP$. Express your answer in terms of $ \pi$.
[asy]draw((1,0)..(0,1)..(-1,0)..(0,-1)..cycle);
draw((0,0)--(4/5,3/5)--(-1,0)--(4/5,-3/5)--cycle);
label("$O$",(0,0),W); label("$I$",(-1,0),W); label("$R$",(4/5,3/5),ENE); label("$P$",(4/5,-3/5),ESE);[/asy]

If $ m\angle RIP \equal{} 36^\circ$, then $ m\angle ROP \equal{} 72^\circ$, so $ RP$ makes up $ \frac {72}{360} \equal{} \frac {1}{5}$ of the circumference.  The circumference is $ 2(10)\pi \equal{} 20\pi$, so $ RP \equal{} \boxed{4\pi}$

In general, as long as $ I$ is not on minor arc $ RP$, $ m\angle RIP\equal{}\frac{m\angle ROP}{2}$


MIT college Stanford Harvard Yale Princeton

Hello guys,
I know that there isn't any particular formula that will help you to get into Ivy League, but I want to ask you to share your comments about my CV, extracurriculars, etc.
Just say what do you think and if I am going to be able to get into any of the Ivys.
Also have in mind that I am an international student !
Member of the Model United Nations club.
Member of the Business Club.
Member of the Math Club for more than 4 years.
A few competitions won in mathematics.
Numberous websites created.
Excellent knowledge of PHP/HTML/Java.
Average GPA - 95 %.
Member of Honor Roll.
Founder of a club, which aimed to help people understand better the material studied in math classes and its president.
Member of the snowboard team.
Member of the basketball team.
Writing for the school magazine + his editor for 1 year.
Spoken Languages: Russian, English, Spanish - excellent and German, Latin - basic
Outside school: Swimming.
Thank you in advance,
:)

Your scores are pretty important so make sure those are high.  Your EC's are fine but honestly, I don't see anything that would greatly distinguish you from the pack.

Here is your competition: http://talk.collegeconfidential.com/what-my-chances/

Not all of these people are applying to Ivies, obviously, but any are.  Just so you (and everyone else) can see the competition.

What do terms like Match/high match/slight reach mean? Match - accept?

They're basically measures of chances of being accepted.  Match means that one should have a fairly good chance of getting in.  Reaches (typically Ivies, MIT, Stanford, and other top schools) would mean you're more likely to not get in.  Slight reach would be you certainly could get in, maybe 50/50 or 40/60.  High reach would be unlikely.

More information: Perfect score on SATs.
+ Participated in different conferations.
Vice-President of the business club.
A few literature competitions won (about 4).
Many publications in the newspaper and the literature magazine.

Look, we can't put a percentage on your chances. Your list of EC's, test scores, and grades suggests that any ivy league admissions staff person would be very comfortable with your credentials. Stop worrying about those and move on. Just think about your essays and about finding the right people to write effective letters of recommendation.

Through your entire application package, try to develop a comprehensive image of your strengths, interests goals, and personality. At the focus of this image should be a [i]clear, coherent impression of who you are[/i]. The effectiveness of your application will depend on how well you do this, so keep this in mind when you write your essays and choose people to write your letters.

I think my recommendations are really good and I also have actually stated who I am.
So I think applying to those universities:
Harvard
Cornell
Brown
Cambridge
Oxford
Yale
Princeton
and I doubt which universities to choose to be safety ones.
Any ideas/comments? :roll:

None of those can be considered "safety" schools.  Those are all reaches, simply because top-tier schools are reaches for everyone.

I know.
What I am saying is I am wondering which universities to be my safeties (not from those on the list) but ones that you suggest :)

Where do you live?  Where do you want to study?  Usually, public universities in your state are considered safties (there are exceptions, such as UC Berekely, UCLA, UVA,... schools that difficult to get into anyway).

Some of my safties were SUNY Binghamton and SUNY Buffalo.  NYU would have been another one had I actually decided to apply, but we decided it wasn't worth the money.

I am an international student from Europe

According to me your ec's are fine.
just one thing if you are applying for Fin Aid then your chances at the ivy league colleges are very low(since you are an international student).
But don't worry mate,we are in the same boat.

However,if you are not not applying for Fin Aid then your chances are fair for the ivy league colleges but i think MIT or CALTECH are still very tough.

It is very unpredictable(especially for me........) to chance anyone who is applying at HYPM or others.............

Still best of luck to you :)

By the way, you can't apply to both Oxford and Cambridge.

In terms of safeties, there are lots of pretty good state schools that you can look at. These are also cheaper. But you need to have an idea of:

1. What you want to study
2. Where you want to study

Why can't I apply to both Cambridge and Oxford?
Having in mind that I am undergraduate, I am not pretty sure yet, but maybe I will study in a business school or law :)
Can you name some good safety universities?

I believe Oxford and Cambridge tell people not to apply to both because they receive such a high volume of applicants. Look it up to be sure.


limit
trigonometry
calculus
calculus computations

Find the following limit.

\[ \lim_{x\rightarrow 0} \frac{\sqrt{\cos 5x}\minus{}\sqrt{\cos 3x}}{x^2}\]

Please find the limit without using l'Hopital rule for my students.

[quote="kunny"]Find the following limit.
\[ \lim_{x\rightarrow 0} \frac {\sqrt {\cos 5x} \minus{} \sqrt {\cos 3x}}{x^2}\]
Please find the limit without using l'Hopital rule for my students.[/quote]

$ \begin{gathered} \mathop {\lim}\limits_{x \to 0} \frac {\sqrt {\cos 5x} \minus{} \sqrt {\cos 3x}}{x^2} \equal{} \mathop {\lim}\limits_{x \to 0} \frac {{\cos 5x \minus{} \cos 3x}}{{{x^2}\left( {\sqrt {\cos 5x} \plus{} \sqrt {\cos 3x} } \right)}} \equal{} \hfill \\
\equal{} \minus{} 2\mathop {\lim }\limits_{x \to 0} \frac {{\sin x\sin 4x}}{{{x^2}\left( {\sqrt {\cos 5x} \plus{} \sqrt {\cos 3x} } \right)}} \equal{} \minus{} 8\mathop {\lim }\limits_{x \to 0} \frac {{\dfrac {{\sin x}}{x}\dfrac {{\sin 4x}}{{4x}}}}{{\sqrt {\cos 5x} \plus{} \sqrt {\cos 3x} }} \equal{} \hfill \\
\equal{} \minus{} 8 \cdot \frac {{\mathop {\lim }\limits_{x \to 0} \dfrac {{\sin x}}{x} \cdot \mathop {\lim }\limits_{x \to 0} \dfrac {{\sin 4x}}{{4x}}}}{{\mathop {\lim }\limits_{x \to 0} \left( {\sqrt {\cos 5x} \plus{} \sqrt {\cos 3x} } \right)}} \equal{} \minus{} 8 \cdot \frac {{1 \cdot 1}}{{1 \plus{} 1}} \equal{} \minus{} \frac {8}{2} \equal{} \minus{} 4. \hfill \\
\end{gathered}$

If I asked the same question in my own class, how would I prefer that my students see it? For that matter, how would I see it myself? The answer is power series, and power series are the reason I can work this problem without writing anything down. (Not looking at DeMath's post - OK, the answer is $ \minus{}4,$ right?)

$ \sqrt{\cos 5x}\equal{}\sqrt{1\minus{}\frac{5^2x^2}2\plus{}\cdots}\equal{}1\minus{}\frac{25x^2}{4}\plus{}\cdots$

Similarly, $ \sqrt{\cos 3x}\equal{}1\minus{}\frac{9x^2}4\plus{}\cdots.$

Subtract those. The $ 1$'s cancel, and $ \minus{}\frac{25x^2}4\minus{}\left(\minus{}\frac{9x^2}4\right)\equal{}\minus{}\frac{16x^2}4\equal{}\minus{}4x^2.$ Divide by $ x^2$ and there's our answer.


USAMTS
analytic geometry
calculus
geometry

What's your favorite problem from this round?

I'd have to say it was #5, with #4 a close second.
1,2 were too easy; 3 was easy but hard to write out well.

I used a neat coordinate geometry/differential calculus approach and ended up with 24.

It was fun.

 :D

I liked number 4. I discovered on the day it was due that my answer was completely wrong, so I spent the rest of the day working out the right answer with a wierd coordinate geometry approach.


[quote="beans"][quote="pardesi"]Why is no one here posting Problems from Physics??
We have Had maths Problems Here...Chemistry in the Chemistry Forum and none in the Physics Section Obviously .
if people do agree i will post one or anyone can start... :D[/quote] Don't tell anyone, but not many people like Physics! :P  :wink:[/quote]

I dont think so beans.. Perhaps since this forum is math oriented it is so.. But defn we have equal no of physicists than mathematicians if not more :D

actually there are two kinds of people who think physics is easy and donn't think much about it...and the other who think it's easy and think much about it...those of the first kind think it's easy..and those of second think it's interesting


inequalities
function
logarithms
advanced fields
advanced fields unsolved

[b] Let  $e$ be Napier's constant and define the   function $f: {\mathbb N} \times ( 0,\infty) \to (0,\infty)$ by  $f(n,p)= \left( \frac{1+\left(1+\frac{1}{n}\right)^p }{2}\right)^{1/p}\; .$
To find the best inequalities of form  (i.e. the best  $(p,q) \in (0,\infty)\times (0,\infty)$)
\[ \begin{array}{|c|} \hline \\ \displaystyle f(n,p) <\displaystyle \frac{1}{e}\left(1+\frac{1}{n}\right)^n < \displaystyle f(n,q) \\ \\ \hline \end{array}\; ,\; \; \; \; \left(\forall n\in {\mathbb N}^* \right).  \]
[color=green]  Prove that above inequalities are verified for $p=\frac{2}{3} , q=\ln{2}$.
[/b] [/color]


Al and Bo play the following game: there are $ 22$ cards labeled $ 1$ through $ 22$. Al chooses one of them and places it on a table. Bo then places on of the remaining cards at the right of the one placed by Al such that the sum of the two remaining cards is a perfect square. Al then places of the remaining cards such that the sum of the numbers on the last two cards played is a perfect square, and so on. The game ends when all the cards are played or no more cards can be placed on the table. The winner is the one who played the last card. Does Al have a winning strategy?

Al has a winning strategy

In the table below we can see all the possible pairs
If $ a_{ij}$ is not empty this means that $ a_{ij} \plus{} i$ is a perfect square
The attached graph shows all possible paths

For example, number $ 1$ could be adjacent only to $ 3,8$ or $ 15$

As we can see, $ 18$ has only one pair, the number $ 7$. So the key is to force the opponent to play $ 7$

Al plays 2. Then Bo has 2 possible choices: 7 or 14

If Bo play 7 then Al plays 18 and the game ends here

If Bo play 14 then Al plays 11
Then Bo plays 5 and Al plays 20
Then Bo plays 16 and Al plays 9
Then Bo plays 7 and Al plays 18


Imagine a perfectly smooth homogenous liquid or "fluid" that stretches out to infinity. An open universe of infinite space-time radius, is equivalent to a closed finite compression universe. There are as many fractions from zero to one as there are natural numbers from zero to infinity. The fluid has a small distortion or vortice moving in it.

How can the velocity of the "vortice" in the fluid be measured?

What I understand is the velocity of the "vortice" motion can only be relative to another vortice.

Could anyone help me with this?

[quote="Jean-Louis"]There are as many fractions from zero to one as there are natural numbers from zero to infinity.[/quote]
I wouldn't phrrase it like that.  I would say that the set of natural numbers is equivalent to the set of rational numbers between 0 and 1 (actually, it's equivalent to all of $ \mathbb{Q}$).


Let f(x) = 0 if x is irrational, and let f(p/q) = 1/q (if p/q is fully reduced).

What is the fucntion g defined by g(x) = lim as y->x of f(y)

This is a fairly famous function.  Does it have a name?

I'm pretty sure g(x) = 0 but I'm thinking how to make it rigorous.

I can tell you that you are [hide]correct[/hide], and I can even give you a suggestion for proving the answer:

[hide]take the cases of irrational and rational numbers seperately.  Irrational should be easy.  Rational, a little trickier.[/hide]

OK here's how we find our delta (see http://mathworld.wolfram.com/Limit.html).
for an arbitrarily small epsilon. First we find a large enough r such that 1/r < epsilon. Let n = 1*2*3*...*r. Find the largest k such that k/n < x. Let delta = (x+k/n)/2. It is impossible to find an integer q<r and integer j such that k/n < j/q < x because 1/n < 1/r and that would contradict the maximality of k. Therefore, all rational values c < delta have g(c)  < epsilon and g(c) = 0 for all irrational c, so the limit from below is 0. The same argument works for the limit from above.
Wow that was thick, hopefully there aren't any major errors.

Here is how I would have done it:

Look at the region [tex](x - \frac{1}{n}, x + \frac{1}{n})[/tex] intersect [tex]\mathbb{R}-{x}[/tex] for n = 1, 2, ....  Now, consider the set of rationals with denominators less than M for any M.  This set is finite, which means for large enough n, there are no members of this set in the region, so the value of the function on that region is everywhere less than 1/M.  Thus, the limit of the function is 0.

One of my professors calls f(x) the Riemann function, but he likes to make up his own names for a lot of things.

What I would do to show that g(x) is [hide]0 would be to consider all the numbers of the form a/n! for integers n. This breaks the line up into intervals of length 1/n!. Then at each step take [tex]\displaystyle\epsilon<\min \{|x-a/n!|:a\in\mathbb{Z},a\not=n!\cdot x\}[/tex] and then let n get large.[/hide]


geometry
geometric transformation
reflection
analytic geometry
function

Hi,
i'm trying to do this problem:

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

A tympani drum has a billiard ball of mass m resting in the
middle. The billiard ball is displaced only vertically, very slightly
from its equilibrium, and will oscillate vertically around the
equilibrium position. The round rim of the drum is d in diameter, and
the drum-head is made tight by 8 turnbuckles that are each tightened
to a tension T, pulling the rim of the drum-head down around the
‘kettle’ body of the drum to tune it. What is the formula for the
restoring force? What is the oscillation frequency for the billiard
ball? Derive the equation of motion for this system.  You may ignore the mass of the plastic sheet of the drum-head.

--------------------------------------------------------------------
my thoughts so far are as follows:  this system is in simple harmonic motion which has an inertial force and a restoring force.  let x be the distance the particle (in this case the billiard ball) is displaced.  the inertial force is of the form

$F_i = ma = m (d^2x/dt^2)$ where a is the acceleration and m is the constant for inertial force

the restoring force is of the form $F_r = -kx$, where k is a constant measuring the "stiffness" of the oscillating material.

the oscillating frequency would be $w^2 = \sqrt {k/m}$

the equation of motion is simply
$m(d^2x/dt^2) = -kx$

so i know the general form of the equations for simple harmonic motion, but i'm not sure how to apply it to this specific problem.  what is "m" and "k" in this case?  i know that when the ball depresses the drum, the radial tension forces cancel but the downward component doesn't.  the vertical motion should depend on the tension of the drum membrane and the angle formed.  i just don't know where to go from here, how to put this all together.  do i have to consider that the drum is held by 8 turnbuckles and calculate the force at each one somehow?  i'm really confused.  any help is much appreciated.

[quote="daftjaxx"]Hi,
i'm trying to do this problem:

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

A tympani drum has a billiard ball of mass m resting in the
middle. The billiard ball is displaced only vertically, very slightly
from its equilibrium, and will oscillate vertically around the
equilibrium position. The round rim of the drum is d in diameter, and
the drum-head is made tight by 8 turnbuckles that are each tightened
to a tension T, pulling the rim of the drum-head down around the
‘kettle’ body of the drum to tune it. What is the formula for the
restoring force? What is the oscillation frequency for the billiard
ball? Derive the equation of motion for this system.  You may ignore the mass of the plastic sheet of the drum-head.

--------------------------------------------------------------------
[/quote]It is somewhat more difficult than it might seem at first.

I would recommend familiarizing yourself with the derivation of the wave equation.

Armed with that, $z$, the amplitude of the deflection of the drumhead, is $z\left(r,\theta, t\right)$ in cylindrical coordinates.

The wave equation then specifies the motion of the drumhead as: $\nabla^2 z\left(r,\theta,t\right) =\dfrac{1}{v^2} \dfrac{\partial^2 z\left(r,\theta,t\right)}{\partial t^2}$.

The trouble with that expression is that it is a 2-D partial differential equation, in polar coordinates no less.  I'll leave the rest to you.

[quote="bus"]This means that we need to make some approximations - for example that the motion will be simple harmonic motion as daftjaxx have assumed. In that case, $m$ is simply the mass of the ball (right?) and $k$ needs to be calculated. If the ball is $y$ from the eq position, vertical tension force should be

$8\times T\times\dfrac{y}{r}$

for small $y$. In that case,

$k=8\times\dfrac{T}{r}=\dfrac{4T}{d}$.

But I'm not thinking too much about it as write all this so please correct any mistakes.... :)[/quote]It's not that simple. Yes, the mass is only the mass of the billiard ball, but the wave equation itself is not explicitly dependent upon mass. (The mass of the ball serves only to drive the vibration of the drumhead.)

As it is, $\nabla^2 z =\dfrac{1}{v^2}\dfrac{\partial^2 z}{\partial t^2}$ admits Bessel function (of the first kind) solutions. But as I recall the original problem was only to derive the equation of motion. The equation of motion, in this case, is the wave equation, i.e. $\dfrac{1}{v^2}\dfrac{\partial^2 z}{\partial t^2} - \nabla^2 z = 0$.

[quote="bus"]Hm, I didn't like the mass part either....

But anyway, the differential form of wave equation is quite useless if you can't solve it. $z=f(t)$ would be a much more useful form. I don't know too much about diff equations (and I have no idea what is Bessel function of first kind) so perhaps solving this one is an elementary procedure - but I think that if it was, you'd post a solution as well...

So, I'll try to understand what you're writing here. Is that a wave equation for the drum-head? Can't we assume that the drum-head vibration is, uhm... a stationary wave with wavelength $2d$ ? Then the ball would be a simple harmonic oscillator. Do you think this is a bad approximation? Can you find a solution for $z(t)$ using some better aproximations?[/quote]You're headed down a wrong road here, my friend. As it is, $z = f\left(r,\theta,t\right)$. The resulting differential equation is rather complicated. It is a partial differential equation. Partial differential equations are typically quite unwieldy. Plus for me to solve it leaves nothing for students to work out. Furthermore, solving such was not the point of the original problem. The original problem was to derive the EOM - equation of motion, which like it or not is given in differential form: $\dfrac{1}{v^2}\dfrac{\partial^2 z\left(r,\theta,t\right)}{\partial t^2} -\nabla^2 z\left(r,\theta,t\right) = 0$.

One can write the wave equation in different coordinate systems according to whatever geometry is convenient, however, in this case the cylindrical geometry of the drumhead dictates using cylindrical coordinates. Doing so yields Bessel's equation.

[quote="bus"][quote="Dr. No"]As it is, $z = f\left(r,\theta,t\right)$.[/quote]Yes. That's the equation for the drum-head. What I was trying to point out is that we're probably more interested in motion of the billiard ball only. And solving the equation [i]is[/i] important - otherwise you won't find the frequency the problem statement asks about. (I believe...) Really, I don't think this problem was supposed to be that hard.[/quote]I don't believe it is that hard.

We know that $\omega^2 =\dfrac{k}{m}$, and that $v = \text{ const. }\times\omega$.

The value of $\text{ const. }$ is determined by the geometry, i.e. Bessel's equation, the solution for which I don't recall at the moment. The billiard ball and the drumhead are in lock-step with one another.


[quote="pianoforte"]source: KCATM 2004

How many multiples of 3 less than 1000 can be formed from the digits 2,5, 9?[/quote]
[hide]9 can't be in a three digit number with 2 and 5, and 2 and 5 can't be in a number with three in them because they both have remainder 2 when divided by 3 so 3 of them have to be put together to create a multiple of three. 9 is alone:
9, 99, 222, 225, 255, 522, 525, 555, 999

$\boxed{9 \text{ numbers}}$[/hide]

You under counted. Try not to list all possible values in these types of problems. Try multiplying.

[hide]For 9, you listed them right: 9, 99, 999. Three

But for 2s and 5s, you under counted. Since there are two possible values for the hundreds digit (2 or 5) we have to choices. Then the same for the tens digit (2 or 5) and the ones digit. So the total number of digits is 2*2*2=8.

So the total is 11
[/hide]


$ a,b,c$are non-negative numbers,no both of which are 0,and $ a\plus{}b\plus{}c\equal{}2$prove that
\[ \frac {a^2b^2}{c^2 \plus{} 3ab} \plus{} \frac {b^2c^2}{a^2 \plus{} 3bc} \plus{} \frac {c^2a^2}{b^2 \plus{} 3ca}\leq \frac {1}{3}\]

For $ x,y,z\geq 0,p\geq 1$,show that
\[ (x^{p\plus{}1}\plus{}y^{p\plus{}1}\plus{}z^{p\plus{}1})^p\plus{}2(x^py\plus{}y^pz\plus{}z^px)^p\leq (x^p\plus{}y^p\plus{}z^p)^{p\plus{}1}\]

In the PDF I give the answer of Question 1


quick question

I believe i read somewhere that when i sign up for SAT subject test online, i can sign up just one test, but on test day i can take up to three (even if i didn't register beforehand the other two).  and i will be charged later.

am i correct?

Thanks very much

Yes.

I also believe you can also sign up for tests, and then not take some of them on test day.

Erm, why is this in the college forum? :P

Anyway, I have something positive to add!  In addition to Zakary's comment, if you do decided to take 2 or 3 subject tests in one sitting, I think that if you want to cancel one, you have to cancel all of them (to prevent cheating...ie, spending two hours on one test, and canceling the other test)


There're no math reality TV shows though.  :maybe:

WHO WANTS TO BE A SUPERHERO? GO BLACK MOMMA.

Is this like, for a survery or something?

Cause my real one is Big Brother obv obv.

Shouldn't this be moved to the Round Table or G&FF?  It is not related to mathcounts...

mine is the amazing race but this should be moved i think.

I don't like reality TV. It seems messed up...

TAKE THAT BACK

big brother, survivor, and the amazing race are my favorites. the biggest loser is ok, but i don't really enjoy watching really really really obese people run with their shirts off

DOWN WITH ALL TV!!!!!!!

except for south park :D

[quote="anirudh]i don't see anything relevant :maybe:
a mod should move this[/quote]
[quote="kyyuanmathcount"]Well, he's saying it's relavent, so I'll keep it here.  However, I do request that the talk finishes in no more than a week and we get some action  :P .[/quote]

Seriously, 5 days have passed, and so far, this isn't relevant at all.

And you guys aren't helping by bumping up the topic.

Shut up and let the topic die if you think the topic isn't relevant.

I'm only posting this because it's the top topic atm and me posting this won't bump it up any further.

[quote="bpms"][quote="kyyuanmathcount"][quote="perfect628"]What is everyone's favorite reality show, if any? I have an idea...not going to say yet, I want some info first. Trust me, this question is math related.[/quote]

I'm going to have to ask you to either pm for the information, which usually works better, or just give up the idea altogether.  This topic has been around for a while, and it's been quite spammy.  I'll lock it, but you can pm me if you want it moved instead.[/quote]
Is spammy a real word?[/quote]
I don't think so, but you're being spammy. :dry:


Find all real $\alpha,\beta$ such that the following limit exists and is finite: \[\lim_{x,y\rightarrow 0^{+}}\frac{x^{2\alpha}y^{2\beta}}{x^{2\alpha}+y^{3\beta}}\]

If I haven't made a mistake,the set of $(\alpha,\beta)$ that meets the requirements is ${\mathbf{R}}^{2}-(-\infty,0)^{2}$.

Yes, my answer is the same:
-If $\beta>0$, then $lim=0$;
-If $\beta=0$, $\alpha>0$, then $lim=0$;
-If $\beta=0$, $\alpha=0$, then $lim=\frac{1}{2}$;
-If $\beta=0$, $\alpha<0$, then $lim=1$;
-If $\beta<0$, $\alpha\geq 0$, then $lim=0$;
-If $\beta<0$, $\alpha<0$, then let us take two sequences ${x_{n}}$, ${y_{n}}$ which tend to zero and satisfy ${x_{n}}^{2\alpha}={y_{n}}^{3\beta}$.
Then $\frac{{x_{n}}^{2\alpha}*{y_{n}}^{3\beta}}{{x_{n}}^{2\alpha}+{y_{n}}^{3\beta}}$ $\rightarrow$ $+\infty$ as $n\rightarrow+\infty$.


geometry
Euler
geometric transformation
reflection
circumcircle
angle bisector
geometry proposed

A triangle $ABC$ is given. Let $L$ be its Lemoine point and $F$ its Fermat (Torricelli) point. Also, let $H$ be its orthocenter and $O$ its circumcenter. Let $l$ be its Euler line and $l'$ be a reflection of $l$ with respect to the line $AB$. Call $D$ the intersection of $l'$ with the circumcircle different from $H'$ (where $H'$ is the reflection of $H$ with respect to the line $AB$), and $E$ the intersection of the line $FL$ with $OD$. Now, let $G$ be a point different from $H$ such that the pedal triangle of $G$ is similar to the cevian triangle of $G$ (with respect to triangle $ABC$). Prove that angles $ACB$ and $GCE$ have either common or perpendicular bisectors.

Here is something similar to a solution; it would be a solution if it wouldn't use lots and lots of facts not all of which have been synthetically proven.

First let's identify the points we are talking about, with their numbers in the ETC ([3]).

- The point L is the Lemoine point, also known as symmedian point of triangle ABC. It is $X_6$ in the ETC.

- The point F is the Fermat point, or, more precisely, the first Fermat point of triangle ABC. It is $X_{13}$ in the ETC.

- The orthocenter H of triangle ABC is $X_4$, and the circumcenter O is $X_3$ in the ETC.

- The point D was defined as the point where the reflection $l^{\prime}$ of the Euler line $l$ of triangle ABC in the line AB meets the circumcircle of triangle ABC (apart from the point H'). Actually, it is a well-known fact ([4]) that the reflections of the Euler line $l$ of triangle ABC in the lines BC, CA and AB concur at one point, which lies on the circumcircle of triangle ABC; this is the so-called [b]Euler reflection point[/b] of triangle ABC. This point can also be described as the point where the reflection $l^{\prime}$ of the Euler line $l$ of triangle ABC in the line AB meets the circumcircle of triangle ABC (apart from the point H'). Hence, the point D coincides with the Euler reflection point of triangle ABC. This point is $X_{110}$ in the ETC.

- The point G was defined as the point, different from the orthocenter H of triangle ABC, such that the pedal triangle of G is similar to the cevian triangle of G. According to [1], this yields that the point G is the isogonal conjugate of the Parry reflection point of triangle ABC. Actually, the Parry reflection point is $X_{399}$ in the ETC, and its isogonal conjugate G is $X_{1138}$.

Note that here we have made the first gap in our synthetic proof: in fact, we have used the result of [1], which was proven using barycentric coordinates.

- According to [2], the Parry reflection point of triangle ABC is the reflection of the circumcenter O of triangle ABC in the Euler reflection point D. Hence, this Parry reflection point lies on the line OD. On the other hand, according to the ETC ([3]), the Parry reflection point lies on the line FL. Hence, the Parry reflection point coincides with the point of intersection of the lines FL and OD. But this point of intersection is the point E; hence, we see that the point E is the Parry reflection point of triangle ABC. As the point G is the isogonal conjugate of the Parry reflection point of triangle ABC, it thus follows that the points E and G are isogonal conjugates of each other with respect to triangle ABC. Hence, the lines CG and CE are symmetric to each other with respect to the angle bisector of the angle ACB; in other words, the angle bisector of the angle ACB is also an angle bisector of the angle GCE (whether it's the internal or the external angle bisector, we don't know). And the problem is solved.

Here we have made another gap in our synthetic reasoning, namely we have quoted from the ETC the fact that the Parry reflection point of triangle ABC lies on the line FL.

Anybody to close the two gaps? I am tired now...

[b]References[/b]

[b][1][/b] [url=http://forumgeom.fau.edu/FG2003volume3/FG200310index.html]Jean-Pierre Ehrmann, [i]Similar pedal and cevian triangles[/i], Forum Geometricorum 3 (2003) p. 101-104[/url].

[b][2][/b] [url=http://groups.yahoo.com/group/Hyacinthos/messages/6200?viscount=1]Darij Grinberg, [i]Hyacinthos message #6200[/i][/url] ([url=http://groups.google.de/group/geometry.college/browse_thread/thread/8260a080b44efaba/6899f6d12e2809c1]mirror at Googlegroups[/url]).

[b][3][/b] [url=http://faculty.evansville.edu/ck6/encyclopedia/ETC.html]Clark Kimberling, [i]ETC - Encyclopedia of Triangle Centers[/i][/url].

[b][4][/b] [url=http://de.geocities.com/darij_grinberg/]Darij Grinberg, [i]Anti-Steiner points with respect to a triangle[/i][/url].

  Darij


[quote="sugarwergoingdown"]If the reciprocal of (x - 2) is (x + 2), what is the greatest possible value of x? Express your answer in simplest radical form.[/quote]

[hide]$\frac{1}{x-2}=x+2$
$x^2-4=1$
$x^2-5=0$
$(x-\sqrt{5})(x+\sqrt{5})=0$
Maximum value: $x=\sqrt{5}$[/hide]


1.Tangents are drawn to the circle x^2+y^2=12 at the points where it is met by the circle x^2+y^2-5x+3y-2=0 ; find the point of intersection of these tangents.

2.Prove that along with (1,-2) there is another point the polars of which with respect to the below mentioned circles are the same and find its coordinates .
S1=x^2+y^2+6y+5=0
and
S2=x^2+y^2+2x+8y+5=0

3.Prove that the distance of two points P and Q each from the polar of the other with respect to a circle are to one another as the distances of the points from the centre of the circle.

4.Prove that the polar of a given point with respect to any one of the circles x^2+y^2-2kx+c^2=0'where k is a variable ,always passes through a fixed point , whatever be the value of k.

First question:

Given the circle $ x^2 \plus{}y^2 \equal{}a^2$ and an external point T (h ,k). Let the tangents from T touch the circle at P and Q.

Then the chord PQ has equation $ xh\plus{}yk \equal{}a^2$. PQ is called the chord of contact of tangents.

Do you know this result and how to derive it? We will use it to solve your question.

The common chord of the two circles is $ (x^2 \plus{}y^2 \minus{}12)\minus{}(x^2 \plus{}y^2 \minus{}5x\plus{}3y\minus{}2) \equal{}0$ that is $ 5x\minus{}3y\minus{}10 \equal{}0$. Let the two circles cut at P ,Q and let the tangents  to $ x^2\plus{}y^2\equal{}12$ at these two points meet at T ( h, k ). Then by the piece of theory stated above PQ will have equation

$ xh \plus{}yk \minus{}12\equal{}0$ and because this is the line $ 5x\minus{}3y\minus{}10 \equal{}0$  we have $ \frac{h}{5}$ =$ \frac{k}{\minus{}3}$ =$ \frac{\minus{}12}{\minus{}15}$ from which you can solve for h and k.

Question four

We use this result from coordinate geometry: The polar of a point (p ,q ) with respect to the circle  $ x^2 \plus{}y^2 \plus{}2gx \plus{}2fy \plus{}a\equal{}0$ is the line

$ xp\plus{}yq \plus{}g(x\plus{}p) \plus{}f(y\plus{}q) \plus{}a\equal{}0$

In this question  g =-k,   f=0   and a=$ c^2$   so for the given circle   the polar of the point (p,q ) will be

$ xp \plus{}yq \minus{}k(x\plus{}p) \plus{}c^2 \equal{}0$, that is $ xp\plus{}yq\plus{}c^2 \minus{}k(x\plus{}p) \equal{}0$......(*) (p ,q and $ c^2$ are fixed and k is variable)

The lines $ xp\plus{}yq\plus{}c^2\equal{}0$ and x+p=0 meet in the (fixed) point ( -p , $ \frac{\minus{}p^2\minus{}c^2}{q}$) and regardless of k this point always lies on the line(*) .Thus we have shown that the polar always passes through a fixed point.


function
trigonometry
geometry
inequalities proposed
inequalities

Show that [tex]|tan x|\geq |x|, \forall x: -\frac{\pi}{2} < x < \frac {\pi}{2}[/tex].

Since both $x$ and $\tan(x)$ are odd functions of $x$, we need only show that $0 \le x \le \tan(x)$ for $x \in [0, \frac{\pi}{2})$. Since $\tan(0) = 0$, $\frac{d}{dx}\tan(x) = \frac{1}{\cos^2(x)}$ and $\frac{d^2}{dx^2}\tan(x) = \frac{2\sin(x)}{\cos^3(x)} \ge 0$, we have that $\frac{d}{dx}\tan(x) \ge 1$ for all $x$ in $[0,\frac{\pi}{2})$, the result follows immediately.

an elementary proof for $x< \tan x$, $x \in (0, \frac{\pi}{2} )$.
Take $M$ the point corresponding to $x$ on the unit circle.
let $O$ be the origin and $A$ the point corresponding to $0$ on the unit circle.
let $T$ be the intersection point of $OM$ with $d$, where $d$ is a line passing through $A$ and paralel to $Oy$.
By definition, $AT=\tan x$.
We have Area of sector $(OAM)$ < area of triangle $(OAT)$ and hence
$\frac{1^2 \cdot x}{2}<\frac{1 \cdot \tan x}{2}$
So $x< \tan x$, $x \in (0, \frac{\pi}{2} )$. q.e.d.

By Taylor's mean value theorem,

$\tan x>x+\frac {1}{3}x^3$ for $\vert x\vert <\pi/2$, which is enough.


1
[hide]If you choose red from Bag A

$\frac 59 \cdot \frac {3}{10} = \frac 16$

If you choose white from Bag A

$\frac 49 \cdot \frac 25 = \frac 8{45}$

$\frac 8{45} + \frac 16 = \frac {31}{90}$[/hide]

[quote="agolsme"]A few probability exercises:

1. Bag A contains 5 red and 4 white counters. Bag B contains 6 red and 3 white counters. A counter is picked at random from bag A and placed in bag B. A counter is now picked from bag B. Find the probability that this counter is white.

2. The two events A and B are such that $P(A)=0.6$ $P(B)=0.2$ $P(A|B)=0.1$. Calculuate the probability that at least one of the events occur. Calculate the probability that exactly one of the events occur.

I got

[hide]
1. $\frac{8}{45}$

2. $0.8$, $0.78$ respectively.
[/hide]

However, according to the answer sheet, those are wrong. Can someone explain why?[/quote]

I believe I can.

[hide]

1. To begin with in bag A you have 5 red, 4 white; and likewise in bag B you have 6 red, and 3 white.

You move a counter (marble, whatever...) from A to B. This changes the distribution in B. There are now 10 marbles in B, and 8 in A, though the number left in A is irrelevant to the rest of the problem.

If you chose a red marble, there are now 7 red and 3 white in B: $P_B(\text{white}|\text{chose red}) =\frac{3}{10}$

Else you chose white and there are now 6 red and 4 white in B: $P_B(\text{white}|\text{chose white}) =\frac{4}{10}$

The marginal probability of choosing a white marble from bag B is given by:

$P_B(\text{white}) =P_B(\text{white}|\text{chose red})\cdot P_B(\text{chose red})+P_B(\text{white}|\text{chose white})\cdot P_B(\text{chose white}) \\
\\
= \frac{3}{10}\cdot \frac{5}{9} + \frac{4}{10}\cdot\frac{4}{9}\\
\\
=\frac{31}{90}$

2. Given P(A) = 0.6, P(B) = 0.2, and P(A|B) = 0.1

a) Calculate the probability that at least one of the events occurs

The probability that at least one of the events occurs is the probability of their union occuring; i.e., "A or B or both A and B."

$P(A\cup B) =P(A) + P(B) - P(A\cap B)\\
\\
= P(A) + P(B)-P(B)P(A|B)\\
\\
= 0.6 +0.2-(0.2)(0.1)\\
\\
= 0.8 -0.02\\
\\
= 0.78\\
$

b) Calculate the probability that exactly one of the events occurs

This is the probability of the union minus the probability of the intersection; "A or B, but not both A and B."

$P(A\cup B) - P(A\cap B) = 0.78 -(0.2)(0.1)\\
\\
= 0.78 - 0.02\\
\\
\&= 0.76\\
\\
\blacksquare$[/hide]


search
algebra
polynomial
quadratics
function
domain
real analysis

I want to find a numerical solution to the following non-concave optimization problem $\max b_{1}\log(1+\frac{a_{1}p_{1}}{a_{2}p_{2}+1})+b_{2}\log(1+\frac{a_{3}p_{2}}{a_{4}p_{1}+1})-p_{1}-p_{2}$ subject to $p_{1},p_{2}\geq 0$. where $b_{1},b_{2},a_{1},a_{2},a_{3},a_{4}$ are given positive constants.

The problelm seems hard due to lack of concavity. Is there any way to find a global maximum numerially?

How large can the constants $a_{1},\dots,a_{4}$ and $b_{1},b_{2}$ get in the applications you have in mind? I'm especially interested in $b$'s because at the point of maxima we must have $p_{1}< b_{1}$ and $p_{2}<b_{2}$, so, if $b_{1}$ and $b_{2}$ are not too big, you can try to make an extensive search of the corresponding region by some brute force method especially if $a$'s are not too big as well. In general, the problem of finding the critical points reduces to that of finding the intersection points of two cubic curves and I'm not sure about how one might handle it numerically. I'll try to think more of the problem but, to be honest, numerical methods have never been my strong side :(.

[b]Edit:[/b] By the way, if you know how to find real roots of a general $9$-th degree polynomial in one variable on a given interval numerically (there seems to be a lot of software packages that are capable of this but, to my shame, I have only a very vague idea of how it is achieved), then the problem is quite simple: the system of two third degree equations that arises when trying to find the critical points in your problem can be reduced in a standard way to one $9$-th degree equation on $p_{1}$ (and then, to find $p_{2}$, you need to solve one more quadratic equation for each suspicious $p_{1}$). So, you'll have to check at most $21$ points: at most $18$ inside the domain and at most $3$ on the boundary. If you are interested in such an approach, just let me know and I'll post the details.


A number is called a palindrome if it is symmetrical. For example , the number 23,454,321 is a palindrome. What is the 11,000th palindrome? Please explain with solution. Thanks ;)

Similar questions have been asked before; use the "Search" function.

[hide="Hint"]Try calculating the number of palindromes less than 10, then 100, then 1000, and so on.  In particular, try 1 million.[/hide]


When I tried to submit my solutions, the uploader just sat there with "Uploading..." displayed. Now when I go to the page, it says that I've already submitted a solution. Can someone check if my solution actually got submitted (UID: 9156)? Thanks. I'll just mail it in if it didn't work.

[quote="abacadaea"]no, not really. if he finds out his solution didnt actually go through, he wouldnt be able to submitt.[/quote]Yeah, this way I still have time to mail.

if i mail mine on the 13 will they count?

Yes, October 14 is the [b]postmark[/b] deadline.  It's OK if we receive them later as long as they are postmarked on or before the 14th.


In your problem, $x\in\mathbb{C}$?  :maybe:

1. How do you determine they equal one?
2. How do you go from first limit to second?


On a disk with six numbered sectors (1 to 6), a player spins the arrow twice. The fraction [i]a/b[/i] is formed where [i]a [/i]is the number of the sector after the first spin and [i]b[/i] is the number of the sector where the arrow stops after the second spin. On every spin, each of the numbered sectors has an equal probability of being the sector on which the arrow stops. What is the probability that the fraction [i]a/b[/i] is greater than 1?

[hide]

ANS: 5/12

To get the probability that we desire, lets do a case by case look at the first number.  Each case occurs with probability 1/6 so we will factor that in to get the total contribution of any number.

N=1: No choices of the second number allow for a fraction greater than one.  Contribution: 1/6 * 0/6 = 0.

N=2: Only 1 works for the second number.  Contribution: 1/6 * 1/6 = 1/36.

N=3: 2 choices work. Contribution: 1/6 * 2/6 = 2/36.

N=4: Contribution: 1/6 * 3/6 = 3/36.

N=5: Contribution: 1/6 * 4/6 = 4/36.

N=6: Contribution: 1/6 * 5/6 = 5/36.

In general, if the first roll ended with N, the probability that N/b (b the second number) is greater than 1 is:

probability of getting N * fraction of choices for b that work=
1/6 * (N-1)/6 = (N-1)/36.

Summing this from N = 1 to 6:

P(N/b > 1) = 1/36 ( 0 + 1 + 2 + 3 + 4 + 5) = 5/12.

[/hide]

Good, except you should've just noticed that 1, 2, 3, 4, and 5 pattern in the middle and know that sum is 15.

Here's an even faster approach, without casework.

[hide="Hint"]
What's the probability that $a = b$?

Thus, what's the probability that $a \ne b$?

Thus, what's the probability that $a > b$?
[/hide]

I did it Barnacle's way....
it took slightly longer than i wanted it to

Haha I wrote it out in a case by case fashion so that others could follow... it probably took me about 15 seconds to get the answer otherwise  ;) ...

Ravi your solution is brilliant ... I was looking for a simplification like that but could one...

For those who didn't catch on:

P(a equals b) = 1/6, so P(a doesn't equal b) = 5/6, and half the time a is going to be greater, so [b]5/12[/b]

:)

nice work.

[hide]5/12, i used barnacle's way[/hide]


Sorry to be a complainer, but it deleted my stats (badges,lessons,current topics)  [b]<== Big Annoyance[/b]

Is there anyway to recover the lessons, topics, and badges I lost?

You're not complaining. You should report that. There should be a report button somewhere around in Alcumus. You get Karma!!! :) I don't know but that is a biggie glitchie.

When your diagnostic state is done, it deletes all stats, to start over with the normal course of matters.

Hello,

I have just checked your account and all appears normal. You have passed three subjects and have three badges. If you are not seeing as many badges as you use to see, you may wish to check the [url=http://www.artofproblemsolving.com/Alcumus/FAQ.php]FAQ[/url] and read the sections entitles [b]How come I can no longer see subjects on my report page that I use to be able to see?[/b] and [b]How come I can no longer see badges I use to be able to see?[/b]


summer program
Mathcamp
AwesomeMath
PROMYS
email

[quote="climskywell"]is Canada/USA MATHCAMP better

or

AwesomeMath Summer Program(AMSP) at Texas better?[/quote]

A) This has utterly nothing to do with the American Mathematics Competitions.

B) That is a question of opinion  ;)  I would personally recommend Canada/USA MathCamp.

So... Motion to move to Other Contests and Programs?

And, why is it better, I've been wondering myself.

[quote="pieterminate"]So... Motion to move to Other Contests and Programs?[/quote]

Agreed... done.

This will be AwesomeMath's first year so noone can really give you a first hand opinion of it.  However, gathering from the description of the program it will be a lot like MOP.  Both camps offer a really solid program so you really can't go wrong.  However, it seems to me that AwesomeMath will be more geared toward competition math whereas mathcamp will have more recreational math.  I really encourage you to go through both of their websites and formulate your own opinion.  From there you should be able to figure out which one better suits your tastes.  And I really think it will come down to a matter of taste.

Has any one been at AwesomeMath? How will you rate it? :)

This is AwesomeMath's first year.  It's impossible for anyone to have been there before.

Duh... You can tell I read the postings in a rush.  Thank you for the info! :D

[quote="joml88"]However, it seems to me that AwesomeMath will be more geared toward competition math whereas mathcamp will have more recreational math.  I really encourage you to go through both of their websites and formulate your own opinion.  From there you should be able to figure out which one better suits your tastes.  And I really think it will come down to a matter of taste.[/quote]

I would reformulate the first quoted sentence just a bit. I would say MathCamp has more upper-division university-level math and more theoretical math. (The term "recreational math" has a specific meaning. There definitely is recreational math at MathCamp, but it is possible to be involved in other aspects of math entirely while there.)

I agree with the rest of the advice quoted above. Visit both Web sites, and THINK about what kind of program is more appealing to you. I note that the MathCamp math jam said that MathCamp is always scheduled for a city that is cool in the summer, and Awesome Math is in Dallas, TX, which is not known for being cool in the summer. If you have a preference for HOT weather in the summer time, you know which program to choose.

Has everyone finished their application to MathCamp now? The regular application deadline is past, although there may be some chance to get in late.

What about Mathcamp vs. PROMYS?

I just received the admission email from Mathcamp, but I also applied to PROMYS (and I feel like I did better on their qualifying problems :p) so I'm wondering what PROMYS is like.

read all about it [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=83857]here[/url] or [url=http://www.promys.org]here[/url]


E clar ca $(x_{k}-3)(x_{k}-6) \leq 0$; daca aduni expresiile astea si folosesti cele 2 relatii din enunt, iti da ca suma lor e $0$, ceea ce inseamna ca $x_{k}\in \{3,6\}$, deci $3|x_{k}$. Folosind asta si prima relatie iti da imediat ca $18|n$.


[quote="biffanddoc"]If you get a 2.5 on the AMC and somehow its IMPOSSIBLY super duper hard but everyone thinks its easy so they answer all of the problems wrong. Except you, you leave one blank. THen you would qualify for AIME and it could be normal and you would get 3 or bettter[/quote]

Well, you do multiply your aime score by 10 for index, but still................

[quote="tiredepartment"]Is it normal to have an AIME score higher than your AMC score?[/quote]

No. Even with the 10x factor on the AIME score it's not normal. Possible, but doesn't happen often.

I know it happens... it's just not normal. And yes ThAzN1 would be an example. He's special like that :P.


calculus
integration
algebra
polynomial
function
real analysis
real analysis unsolved

If $ f(x)$ is continuous on $ [a,b]$ and for any integer $ n\geq 0$ ,we have
$ \int_{a}^{b}x^{n}f(x)dx \equal{} 0$
, show that: $ f(x)\equiv0$
[color=red]Furthermore[/color], If $ f(x)$ is only Riemann Integrable, can we get the conclusion:
$ f(x) \equal{} 0 a.e.$
a.e. stands for almost everywhere.

1. Use the fact that the set of Polynomials are dense in C[a,b]
2. Then if $ \parallel{}f\parallel{}^{2}_{2} \equal{} \int_{[a,b]}|f|^2dx \equal{} 0$ then $ f \equal{} 0$ a.e

Thanks 1234567a !

The first problem we assume $ f(x) \in C[a,b]$, and use the fact that the set of Polynomials are dense in $ C[a,b]$, so $ \exists P(x) \in C[a,b]$ such that $ |P(x) \minus{} f(x)| < \epsilon$. and we have
$ \int_{a}^{b}|f|^2dx \equal{} \int_{a}^{b}P(x)f(x)dx \minus{} \int_{a}^{b}(P(x) \minus{} f(x))f(x)dx$.
because:
$ |\int_{a}^{b}(P(x) \minus{} f(x))f(x)dx|\leq\int_{a}^{b}|P(x) \minus{} f(x)\parallel{}f(x)|dx\leq\epsilon(b \minus{} a)\sup_{x\in[a,b]}f(x)$
and $ \int_{a}^{b}P(x)f(x)dx \equal{} 0$. so we have:$ \int_{a}^{b}|f|^2dx\rightarrow0$
then we get the conclusion:$ f(x) \equal{} 0$ a.e. and $ f(x)$ is a continuous function ,so we have$ f(x)\equiv0$

[color=red]But if $ f(x)$ is only Riemann Integrable. Can we find the Polynomials satifies the equation$ |P(x) \minus{} f(x)| < \epsilon$ ?[/color]
so if $ f(x)$ is only Riemann Integrable, can we have the same conclusion?
In other word: If$ f(x)$is Riemann Integrable in $ [a,b]$ ,and for any integer $ n\geq0$,we have$ \int_{a}^{b}x^{n}f(x)dx \equal{} 0$
can we get the conclusion:$ f(x) \equal{} 0$a.e.?
(a.e. stands for almost everywhere)


Buondì, visto che il forum ufficiale italiano è down, un po' di gente si è sentita su msn, e dal confronto i risultati dovrebbero essere questi

ABAAC
DEBBE
BADDC
ECEAD
EDBCA

well, if they are true I do $ 5 \plus{} 5!$  :lol:

solo io e gabriel frequentiamo ML? c'è qualche altro italiano? come è andata?

Da oggi ci sono anch'io!  :D
Io ho fatto uguale, mi sono confrontato anche con Giove... perciò anch'io dovrei aver fatto $ 5^3$ punti... :lol:

sbagliata solo la 9 stupidamente  :(

120 punti....dovrebbero comunque bastare per le provinciali  :P


[url=http://www.crazymonkeygames.com/Endless-Zombie-Rampage.html]Click to play[/url]

Post your high scores :D  and what you bought with your 9000 EXP

We're talking survival mode.


trigonometry
inequalities unsolved
inequalities

We can set $a=\sqrt{2}\tan A, b=\sqrt{2}\tan B,\ c=\sqrt{2}\tan C,\ 0<A,B,C<\frac{\pi}{2}.$

[quote="hainam-sp"]$a,b,c>0$. Prove that:
$(a^2+2)(b^2+2)(c^2+2)\ge 9(bc+ca+ab)$[/quote]

This APM0 2OO4 problem 5.
See [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=APMO+2004&t=19555]post #5[/url].

oh , very easy
 we have :
    ah, APMO  have a solution and it like my solution

hi

i hv a dif proof

out of a,b,c 2 must be say a,b s.t. \[ (a-1)(b-1)\geq0 \]

using this fact apply chebbychev's ineq and we get the result easily :!:


induction
AMC
USA(J)MO
USAMO
algebra proposed
algebra

Prove that for every $n\in\mathbb{N}, \ n\geq 3$, there are $n$ natural different numbers $a_1,a_2,...a_n$ such that
    $\frac 1{a_1}+\frac 1{a_2}+...\frac 1{a_n}=1$.

[Well, if $n\geq 4$, one can prove there are at least two different ways of writing number 1 in the form above.]

This is a well known result. Take, for instance $u_1=2,u_{n+1}=u_1u_2\ldots u_n+1$.
Then\[ \frac{1}{u_1}+\frac{1}{u_2}+\ldots +\frac{1}{u_n-1}=1, \]and it is easy to see that denominators are distinct.
(Approximation of unity with Egyptian fractions.)

[quote="spider_boy"]Another problem related to this problem:
Prove that every natural number $n\ge32$ can be written as sum of
natural numbers whose sum of  inverses(i mean the inverse of $a$ is $\frac{1}{a}$)
is $1$.[/quote]
Actually, the statement is:
(USAMO 1978) An integer will be called good if it can be written as the sum of positive integers (not necessarily distinct) the sum of whose reciprocals is $1$. Given that the integers $33$ through $73$ are good, prove every integer greater than or equal to $33$ is good.

[quote="enescu"]This is a well known result. Take, for instance $u_1=2,u_{n+1}=u_1u_2\ldots u_n+1$
Then
$\frac{1}{u_1}+\frac{1}{u_2}+\ldots +\frac{1}{u_n-1}=1$, [/quote]
More precisely,
$\frac{1}{u_1}+\frac{1}{u_2}+\ldots +\frac{1}{u_{n-1}}+\frac{1}{u_n-1}=1$.
I solved it in the same way.

Here is the author's solution:
[hide]Suppose we already found $a_1,a_2,...a_n$ s.t. $\frac 1{a_1}+\frac 1{a_2}+...\frac 1{a_n}=1$
Then $\frac 1{2}+\frac 1{2a_1}+\frac 1{2a_2}+...\frac 1{2a_n}=1$, so the statement is also true for $n+1$.[/hide]

I had no idea about the Egyptian fractions. :oops:  A very interesting subject, anyway!

If you want to amuse yourself converting ordinary fractions into Egyptian ones, see http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fractions/egyptian.html
There is a "calculator" on that page (the second one) which converts a given ordinary fraction into Egyptian sums, finding all the Egyptian sums of shortest length for it.


function
inequalities
inequalities unsolved

Find the minimum value of this function:
A/(B+2*C+3*D) + B/(C+2*D+3*A)+ C / (D+2*A+3*B) + D/(A+2*B+3*C)

Or it will be more apparent.
 　　A　　　　　　　B　　　　　　　　C　　　　　　　  D
__________ +    ___________+     ___________ +   _____________
B+2*C+3*D　　C+2*D+3*A 　　　D+2*A+3*B 　　　A+2*B+3*C

A,B,C,D are all postive real.

[quote="anthony19891019"]Find the minimum value of this function:
A/(B+2*C+3*D) + B/(C+2*D+3*A)+ C / (D+2*A+3*B) + D/(A+2*B+3*C)

Or it will be more apparent.
 　　A　　　　　　　B　　　　　　　　C　　　　　　　  D
__________ +    ___________+     ___________ +   _____________
B+2*C+3*D　　C+2*D+3*A 　　　D+2*A+3*B 　　　A+2*B+3*C

A,B,C,D are all postive .[/quote]
[b]Problem[/b]: Let $a, b, c, d$ be positive real numbers. Find the minimum value of
$\dfrac{a}{b+2c+3d}+\dfrac{b}{c+2d+3a}+\dfrac{c}{d+2a+3b}+\dfrac{d}{a+2b+3c}$
[hide="solution"]By letting $a=b=c=d$, we guess the minimum value is $\dfrac{2}{3}$.
Now we are going to prove the inequality $\dfrac{a}{b+2c+3d}+\dfrac{b}{c+2d+3a}+\dfrac{c}{d+2a+3b}+\dfrac{d}{a+2b+3c}\geq\dfrac{2}{3}2$
By Cauchy-Schwarz Inequality,
$[(a)(b+2c+3d)+(b)(c+2d+3a)+(c)(d+2a+3b)+(d)(a+2b+3c)](\dfrac{a}{b+2c+3d}+\dfrac{b}{c+2d+3a}+\dfrac{c}{d+2a+3b}+\dfrac{d}{a+2b+3c})\geq (a+b+c+d)^2$
So it remains to prove
$\dfrac{(a+b+c+d)^2}{(a)(b+2c+3d)+(b)(c+2d+3a)+(c)(d+2a+3b)+(d)(a+2b+3c)}\geq\dfrac{2}{3}$
$\Leftrightarrow 3(a+b+c+d)^2\geq 2(4ab+4bc+4cd+4da)$
Since $3(a+b+c+d)^2-8(ab+bc+cd+ca)=(a-b)^2+(a-c)^2+(a-d)^2+(b-c)^2+(b-d)^2+(c-d)^2\geq 0$,
the original inequality is proved.[/hide]
PS This is a shortlist problem in IMO 1993. Another solution is provided [url=http://www.kalva.demon.co.uk/short/soln/sh9324.html]here[/url].