[quote="Shanku"]Darij, this problem is due to you: For non-negative numbers a, b, c, no two of which are zero, show that,
$ \frac {a^2(b \plus{} c)}{b^2 \plus{} c^2} \plus{} \frac {b^2(c \plus{} a)}{c^2 \plus{} a^2} \plus{} \frac {c^2(a \plus{} b)}{a^2 \plus{} b^2} \equal{} a \plus{} b \plus{} c$

Since the problem is so good-looking, the solution must be even better. Please show me the solution. :?:[/quote]

$ \frac {a^2(b \plus{} c)}{b^2 \plus{} c^2} \plus{} \frac {b^2(c \plus{} a)}{c^2 \plus{} a^2} \plus{} \frac {c^2(a \plus{} b)}{a^2 \plus{} b^2} \geq a \plus{} b \plus{} c$

$ \Longleftrightarrow a^6b\plus{}a^6c\plus{}b^6c\plus{}b^6a\plus{}c^6a\plus{}c^6b\minus{}b^2a^5\minus{}c^2a^5\minus{}b^5c^2\minus{}b^5a^2\minus{}c^5a^2\minus{}c^5b^2\geq0$

true due to Muirhead

SOS is the best option:
\[ LHS \minus{} RHS \equal{} \sum_{cyc}\left(\frac {a^2(b \plus{} c)}{b^2 \plus{} c^2} \minus{} a\right) \equal{} \sum_{cyc}\frac {ab(a \minus{} b) \minus{} ca(c \minus{} a)}{b^2 \plus{} c^2} \equal{} \sum_{cyc}\frac {ab(a \minus{} b)}{b^2 \plus{} c^2} \minus{} \sum_{cyc}\frac {ab(a \minus{} b)}{c^2 \plus{} a^2} \equal{} \sum_{cyc}ab(a \minus{} b)\left(\frac {1}{b^2 \plus{} c^2} \minus{} \frac {1}{c^2 \plus{} a^2}\right) \equal{} \sum_{cyc}ab(a \minus{} b)\cdot \frac {a^2 \minus{} b^2}{(b^2 \plus{} c^2)(c^2 \plus{} a^2)} \equal{} \sum_{cyc}\frac {ab(a \plus{} b)(a \minus{} b)^2}{(b^2 \plus{} c^2)(c^2 \plus{} a^2)}\]

I think Chebyshev+Nesbitt works. We have $ LHS \ge (\sum \frac{a^2}{b^2\plus{}c^2})( \sum b\plus{}c)/3$, but $ \sum \frac{a^2}{b^2\plus{}c^2} \ge \frac{3}{2}$

That's "straight from the BOOK." Thanks so much.

Actually I posted it here long ago:
http://www.mathlinks.ro/viewtopic.php?t=14577
and Harazi was the one who found a nice solution (I only had the Muirhead proof). I was the one who later exploited Harazi's trick to the hilt, however :)

  darij

Wow ! That's great for inequality Pros like you and Hazari, Darij. I prefer to fight from behind these AM-GM/Chebyshev/Cauchy-Schwarz/Holder's/Nesbit shields. Not even Schur or Muirhead are friendly to me. Actually this cyclic Sum symbol scares me and cannot tackle it until I spread it on paper. Please pardon my ignorance.

[quote="dgreenb801"]I think Chebyshev+Nesbitt works. We have $ LHS \ge (\sum \frac {a^2}{b^2 \plus{} c^2})( \sum b \plus{} c)/3$, but $ \sum \frac {a^2}{b^2 \plus{} c^2} \ge \frac {3}{2}$[/quote]

Sorry to tell, but your application of Chebyshev is wrong.

  darij

[quote="Shanku"]Darij, this problem is due to you: For non-negative numbers a, b, c, no two of which are zero, show that,
$ \frac {a^2(b \plus{} c)}{b^2 \plus{} c^2} \plus{} \frac {b^2(c \plus{} a)}{c^2 \plus{} a^2} \plus{} \frac {c^2(a \plus{} b)}{a^2 \plus{} b^2} \geq a \plus{} b \plus{} c$

Since the problem is so good-looking, the solution must be even better. Please show me the solution. :?:[/quote]
See also here: http://canhang2007.wordpress.com/2009/12/03/inequality-90-d-grinberg/

Darij, could you please explain why and how application of Chebyshev is wrong ?
I may sound silly.

He's right, it is wrong.
Chebyshev says if $ a \ge b \ge c$ and $ x \ge y \ge z$, then $ ax \plus{} by \plus{} cz \ge \frac {(a \plus{} b \plus{} c)(x \plus{} y \plus{} z)}{3}$. But in my case, if $ a \ge b \ge c$, then $ a \plus{} b \ge b \plus{} c \ge c \plus{} a$ and $ \frac {c^2}{a^2 \plus{} b^2} \le \frac {b^2}{a^2 \plus{} c^2} \le \frac {a^2}{b^2 \plus{} c^2}$. So the last inequality is not in the right direction.

[quote]He's right, it is wrong.
....But in my case, if $ a\ge b\ge c$, then $ a\plus{}b\ge b\plus{}c\ge c\plus{}a$ and $ \frac{c^{2}}{a^{2}\plus{}b^{2}}\le\frac{b^{2}}{a^{2}\plus{}c^{2}}\le\frac{a^{2}}{b^{2}\plus{}c^{2}}$. So the last inequality is not in the right direction.[/quote]

But is it not $ a\plus{}b\ge c\plus{}a\ge b\plus{}c$ if $ a\ge b\ge c$ ?

Actually, yeah you are right for the first inequality but the second inequality is still backwards.
$ a \ge b \ge c$ is used to determine the other two inequalities but doesn't count itself as an inequality when we use Chebyshev. The two are
$ a \plus{} b\ge c \plus{} a\ge b \plus{} c$ and $ \frac {c^{2}}{a^{2} \plus{} b^{2}}\le\frac {b^{2}}{a^{2} \plus{} c^{2}}\le\frac {a^{2}}{b^{2} \plus{} c^{2}}$
And they aren't in the same direction.


geometry
analytic geometry
linear algebra
matrix
absolute value
area of a triangle

A triangle has vertices at $ (\minus{}3,2),(6,\minus{}2),(3,5)$. How many square units are in the area of the triangle? Express your answer as a decimal to the nearest tenth.

[hide="Not MC level, but the determinant interpretation"]
In general, the area of a triangle with coordinates $ (a_n,b_n)$ for $ 1\le n\le3$ is the absolute value of $ \frac12\begin{vmatrix}a_1&b_1&1\\a_2&b_2&1\\a_3&b_3&1\\\end{vmatrix}$. Through row and column additions and subtractions, the determinant of a matrix will stay the same: we want to find the absolute value of $ \frac12\begin{vmatrix}\minus{}3&2&1\\6&\minus{}2&1\\3&5&1\end{vmatrix}\equal{}\frac12\begin{vmatrix}\minus{}3&2&1\\0&2&3\\0&7&2\\\end{vmatrix}\equal{}\frac12\cdot\minus{}3\begin{vmatrix}2&3\\7&2\\\end{vmatrix}\equal{}\minus{}\frac32(2\cdot2\minus{}3\cdot7)\equal{}\minus{}\frac32\cdot\minus{}17$
$ \equal{}\frac{51}2\implies\left|\frac{51}2\right|\equal{}\boxed{25.5}$.
[/hide]


Given positive real numbers $x_1, x_2, \ldots , x_n$ find the polygon $A_0A_1\ldots A_n$ with $A_iA_{i+1} = x_{i+1}$ and which has greatest area.

Well, of course we need $2x_i<x_1+x_2+\ldots +x_n$ for all $i$ (otherwise no such polygon exists). In this case, there exist a polygon satisfying those condition and such that vertices $A_0,A_1,\ldots, A_n$ are cocyclic. (My construction methods are a bit sloppy, but it should exists.)

Let's prove this polygon has maximal area. Glue on each side of our cyclic polygon an arc of the corresponding circle. Now move all side to get a generic polygon satisfying $A_iA_{i+1}=x_i$. Remembering the previous arcs, we get a "curvilinear polygon", with same perimeter as the original circle. Using the isoperimetric inequality, we know this curvilinear polygon has a smaller area than the original circle, and so our generic polygon has a smaller area than the cyclic one.

Remark : there might be some issues when arcs overlap each other, but I believe/hope this only happens in obviously non-optimal cases, like non-convex polygons.


The sum of two numbers is 13, while their product is 36. What
is the sum of their reciprocals? Express your answer as a
common fraction.


analytic geometry
probability
probability and stats

A little question about statistics.

I have a set of points on the coordinate plain that come from experimental measures, the claim of the experiment is to show that the $y$ coordinate is proportional to $\frac{1}{x^{2}}$. So I used Mathematica to fit the data set (Nonlinear fit of the form $a+\frac{b}{(x+c)^{2}}$) and the fitting curve looks pretty well. But can I perform some test, which says somthing like “there is 95% of probability that ther is that kind of proportionality” or similar, and not just saying that the fitting curve looks “pretty well”. Thanks a lot, and sorry for my ignorace in statistics.

The statistics works in a slightly different way when it comes to checking a single hypothesis: what you usually do is roughly speaking the following: you assume a certain hypothesis about the dependence (say, that $y=1/x^{2}$), you choose some model of the measurement error (say, independent Gaussians with mean $0$ whose variances can be determined by measuring $y$ with the same $x$ many times). Then you construct some quantity (basically you'll need to check that the expectations of your Gaussians are given numbers; I believe that there is some standard test for that but I'll have to look it up in textbooks: my memory of statistics is quite rusty) that is within a certain interval with some large probability (say, 95%) under your hypothesis, after which you compute this quantity for your sample and see if it lies within that interval. If it doesn't, you say that the hypothesis is rejected (because, if it were true, you would observe an event of small probability and you don't believe it; if 5% doesn't sound small enough to be unobservable, repeat the experiment several times. If you fall outside the interval 5 times out of 5 attempts, the probability of the observed event becomes $0.05^{5}$, etc.). If it does, you do not have a really convincing reason for rejection of your hypothesis but that's all that can be said: you cannot claim that it is true with high probability. On the other hand, if there is one or several explicit competing hypotheses (like $y=1/x^{a}$, $a>0$, you may show that yours is the best one by some version of the maximal likelihood method. Again you choose some quantity, compute its distribution under each of the hypotheses in your family and choose the one that gives the maximum of the probability density on your particular sample (I hope I remember it right. If not, I would appreciate somebody correcting me here). This way you may be able to draw some conditional conclusions like [i]If the true law is an inverse power one, then the power is between $1.9$ and $2.1$ with $90\%$ probability[/i] (actually a more correct statement would be "either it is between $1.9$ and $2.1$, or you observed a $10\%$ probability event").

That's the general idea of statistics as far as I remember it. Any corrections will be appreciated :). Usually people do not go that far: they just try to determine the average measurement error in $x$ and $y$, draw the corresponding boxes at each measured point and get happy if the theoretical curve hits each box.


trigonometry
inequalities
inequalities proposed

Suppose $\alpha, \beta, \gamma$ are angles of a triangle and $n\in\mathbb{N},\;n\geq 2.$ Prove the inequality \[\cos^{n}\alpha+\cos^{n}\beta+\cos^{n}\gamma\geq\frac{3}{2^{n}}.\]

as we already know that the sum of cosines of the angles of a triangle is less than or =3/2 hence the sum of each of them being raised to 'n' definitely is less than or=3/2

How does that prove the inequality?  :maybe:

$f(x)=x^{\frac{a}{2}},a\ge 2$ has $f'(x)=\frac{a}{2}x^{\frac{a}{2}-1}\ge 0,f''(x)=\frac{a}{2}(\frac{a}{2}-1)x^{\frac{a}{2}-1}\ge 0,\forall a\ge 2$
then $f(x)$ is incresing and convex
$\sum cos^{a}\alpha=\sum (cos^{2}\alpha)^{\frac{a}{2}}\ge 3(\frac{\sum cos^{2}\alpha}{3})^{\frac{a}{2}}\ge 3(\frac{1}{4})^{\frac{a}{2}}=\frac{3}{2^{a}},\forall a\ge 2$
note:$\sum cos^{2}\alpha\ge \frac{3}{4}$ thus we  don't need $a\in\mathbb{N}$
by the similar menthod we can prove
$\sum sin^{a}\frac{A}{2}\ge \frac{3}{2^{2a}},\forall a\ge 4$  :P


I don't know why I am not getting these problems...

What is the pH of a .844 M solution of $\text{NH}_4 \text{NO}_3?$  The ionization constant of the weak base $\text{NH}_3$ is $1.8 \times 10^{-5}.$

a) 11.0
b) 2.7
c) 4.0
d) 4.7
e) 7.0

Which substance, when dissolved in water, produces a basic aqueous system?

a) H2S
b) SO2
c) Na2O
d) CO2
e) HNO3

Thank you in advance.

[quote="pkothari13"]What is the pH of a .844 M solution of $\text{NH}_4 \text{NO}_3?$  The ionization constant of the weak base $\text{NH}_3$ is $1.8 \times 10^{-5}.$[/quote]

$\text{NH}_4^+$ is a weak acid...

Try to use [url=http://www.chembuddy.com/?left=BATE&right=pH-cheat-sheet]pH cheat sheet[/url].

[quote]Which substance, when dissolved in water, produces a basic aqueous system?[/quote]

Think about all substances - how do they react with water (if they do)? How do they dissociate?

[quote="pkothari13"]I don't know why I am not getting these problems...

What is the pH of a .844 M solution of $\text{NH}_4 \text{NO}_3?$  The ionization constant of the weak base $\text{NH}_3$ is $1.8 \times 10^{-5}.$

a) 11.0
b) 2.7
c) 4.0
d) 4.7
e) 7.0

Thank you in advance.[/quote]

 For this one, you first need to take the given Kb (1.8e-5) and divide it into the autoprotolysis constant, Kw, which is always 1.0e-14. Remember the equation Kw=KaKb so to get Ka you divide Kb into Kw, giving you a Ka value of 5.56e-10. Now I was taught to do dissociation problems using what's called an ICE table, which looks a little like this (hopefully it will line up correctly)

        [color=white] --- [/color]NH4NO3---->NH4+   + [color=white]---[/color]  NO3-
I[color=white]-----[/color]0.844[color=white]----------[/color]0[color=white]-----------[/color]0
C[color=white]------[/color]-y[color=white]------------[/color]+y[color=white]---------[/color]+y
E[color=white]----[/color]0.844-y[color=white]--------[/color]+y[color=white]---------[/color]+y

Now, the values in the E line can be used to solve for Ka=[H+][A-]/[HA], so your equation should look like...

5.56e-10=(y)(y)/(0.844-y)  now you assume that y<<0.844 and get,

5.56e-10=(y)(y)/(0.844) so all you do from there is multiply 5.56e-10 by 0.844 then take the square root to get the value y=2.17e-5. From there, just take the -log of y, and -log(2.17e-5)=4.66 so you have a pH of 4.7

Hope that helps.

Hi Borek,

I know how each reacts in water actually, but I think my main problem is identifying WHAT a basic aqueous sytem is?

[quote="pkothari13"]I know how each reacts in water actually, but I think my main problem is identifying WHAT a basic aqueous sytem is?[/quote]

The one with more OH- than H+?

[quote="lasagnaman"]Basic, not as in simple, but as in "with the properties of a base (as opposed to acidic)".[/quote]

ah, okay. i didn't even think of the OH- concept. thanks borek and lasagnaman for clearing that up


Has my signature been banned or something?

reason?????

EDIT: I get it, but shouldn't you remove the offending link first?

Your website (the "WWW" under your post) is annoying. It has happened to me twice now, with that and aops.tk.

PM me in a month, I'll let you have your signature back then (provided you keep it clean of urls.)


There are 30 equally-weighted questions on Mr. Daven's math
final. If a student must score 68$ \%$ or greater to pass, what is
the minimum number of questions that must be answered
correctly to pass?

$ .68\times 30\equal{}20.4$.  But this means that the student must answer 21 questions correctly.


integration
calculus
function
limit
real analysis
calculus computations

Having $ \int_0^1\frac {dx}{\sqrt x},$ we know this converges, is it convergence actually integrability?

This is a vocabulary question, and I'm not 100% sure how I'd answer, or others would answer.

Is it Riemann integrable? No. The Riemann integral can only be defined for bounded functions on bounded intervals.

We can say that $ \lim_{b\to 0^ \plus{} }\int_b^1\frac1{\sqrt {x}}\,dx \equal{} 2$ exists. I keep trying to figure out ways to say that and I keep not using the word integrable: instead I say that the improper integral exists and is finite. This is an interesting issue of vocabulary.

Is it Lebesgue integrable? Yes. And the Lebesgue integral is $ 2,$ agreeing with the improper integral.

In the end, I probably would call this an integrable function - but I think I'm bowing to the Lebesgue sense when I say that.

What about $ \int_{ \minus{} 1}^1\frac1{x}\,dx?$ In that case, the function is not Lebesgue integrable and the improper integral does not exist. And I'd never use the word "integrable." However,

$ \lim_{b\to0^ \plus{} }\left[\int_{ \minus{} 1}^{ \minus{} b}\frac1x\,dx \plus{} \int_b^1\frac1x\,dx\right] \equal{} 0$ does exist. We say that the principal value of this integral converges.

[quote="Kent Merryfield"]
Is it Riemann integrable? No. The Riemann integral can only be defined for bounded functions on bounded intervals.[/quote]
I was forgetting this important point, thanks a lot!


modular arithmetic
number theory unsolved
number theory

$ (a)$ Prove that $ 91$ divides $ n^{37}\minus{}n$ for all integers $ n$.
$ (b)$ Find the largest $ k$ that divides $ n^{37}\minus{}n$ for all integers $ n$.

If $ p\minus{}1|36$, then for any n $ p|n(n^{36}\minus{}1)$. Else exist n, suth that $ p\not |n^{37}\minus{}n$.
Therefore $ N\equal{}2*3*5*7*13*19*37$.

$ a) n^{6k \plus{} 1}\equiv1\pmod{7}$ for $ k\geq 1$  (LFT)
$ n^{12k \plus{} 1}\equiv1\pmod{13}$   for $ k\geq1$   (LFT)
$ 37\equiv1\pmod{12}, 37\equiv1\pmod{6}$  $ \implies$
$ n^{37} \minus{} n\equiv0\pmod{91}$
$ b)37\equiv1\pmod{a}$ $ \implies$
$ a|36$  $ \implies$  $ a \equal{} 1,a \equal{} 2,a \equal{} 3,a \equal{} 4,a \equal{} 6,a \equal{} 9,a \equal{} 12,a \equal{} 18,a \equal{} 36$
$ (a \plus{} 1)$ is a prime  $ k \equal{} 2.3.5.7.13.19.37$   :wink:


[quote="Big Bolzano"]For some natural n , number $4^n + 2^n + 1$ is prime. Prove that $n=3^k$ , $k \in \mathbb{N}$.[/quote]
It's easy,
\[
1+2^n+4^n=\Phi_3(2^n)=\prod_{t|n}\Phi_{3t}(2).
\]
But $\Phi_3(2)=7$ and for every prime $p\ne 3$ $\Phi_{3p}(2)=(1+2^p^+4^p)/7\in \mathbb{N}$, and greater than $q$, and $\Phi_9(2)=73$ so the unique solution is $n=1$.

Ops, I forget to say that $\Phi_n(x)$ are the cyclotomical polynomials. See my Topic to understand better.

RX TCM

This has been posted before, a very long time ago. I remember solving it then, but it was a very ckumsy solution. I wonder why... :) Anyway, try to show that if $d|n$ and $3\not |\ d$, then $4^{\frac nd}+2^{\frac nd}+1|4^n+2^n+1$.

Edit:

I hadn't seen the post above when I posted. In order to prove that, I wrote $4^n-2^n+1$ as $(2^n-\varepsilon)(2^n-\bar\varepsilon)$ and used the fact that if $d$ is not a multiple of $3$, then we either have $\varepsilon^d=\varepsilon,\bar\varepsilon^d=\bar\varepsilon$ or $\varepsilon^d=\bar\varepsilon,\bar\varepsilon^d=\varepsilon$.

use the fact that iff $a_1,a_2...,a_n$ is a complete system of residues mod $n$ then $1+x+x^2...+x^n$ divides $x^{a_1}...+x^{a_n}$


Let $n>2005$ be an integer and let $x^{n}=2006^{n}+2005^{n}$.Is $x$ an integer?


Find the terms of the sequence  $a_n$   with  $a_n=1!\cdot 2!\cdot ....\cdot n!$

where $n\geq 1$ a natural number such that  $a_n$ is a perfect square

$a_{n} = 2^{n-1} \cdot 3^{n-2} \cdot \cdot \cdot (n - 1)^2 \cdot n$
Look at the power of $2$ and $3$. They must be both even, but they haven't the same parity. So the only terms with such property is $a_{1} = 1$.

There are more factors $2$ and $3$, comming e.g. from $8$!

I think the following works:

Consider $n \geq 6$ Take the largest prime $p$ such that $p \leq n$ then $p$ appears once in $p!$,...,$n!$ (because of Bertrand's postulate). So $p$ appears with exponent $n-p+1 \equiv n \mod 2$ so $n$ must be even. Let $n=2k$ and take the largest prime $q$ such that $q \leq k$. Then $q$ appears with exponent $q+2(2k-2q+1)+\epsilon(2k-3q+1) \equiv 1 \mod 2$ (where $\epsilon$ equals one when $2k > 3q$ and $0$ else) so contradiction.

So we are left with checking $n \leq 5$


There are two two-digit perfect squares whose square roots are the same as their units digits. What is the sum of these squares?


MIT
college
algorithm
abstract algebra
articles

Where could a middle schooler find information aboutCS as applied to IOI. Is there a book for the IOI olympiad similar to the AoPS? I am currently reading the MIT intro to algorithms and know C/C++, can read and write some perl and c#, and have a smige of info about java.

[quote="h_s_potter2002"]There's a website that I thinkis excellent for a middle/hish schooler who wants to get into the IOI stuff, at [url=http://ace.delos.com/usacogate]ace.delos.com/usacogate[/url].  They start off at the beginning, although you'd need to know some programming in Pascal, C, C++, or Java.  And the same people also run the USACO competitions, which you would do to get into the training camp for the IOI.[/quote]
Does that site contain everything I must know?

The best way to learn is to start at the beginning and work your way up. There is a lot of resources on USACO pages, but concerning training for IOI, the amount of problems from past IOI, CEOI, BOI, national competitions ... is so large that you will never exhaust it. Many of them also offer online versions of contests. And there are also some other sites you can train at (TopCoder, ACM, Valladolid University archives, ...), but those are a bit different from the IOI-style.

Of course, there is no such thing as a list of all things you need to know to get a fullhouse at IOI. Most of the knowledge (at least mine) is experience backed up with hours of trainig. I have never encountered anything like a book that teaches you all the techniques you need for the IOs, but there is a lot of useful texts at the USACO, for instance. There are books on agorithms or data structures, but many of them are more for industrial use. I find the "Introduction to algorithms" a very good resource for some things, but of course you cannot find everything there.

USACO pages is really a good place to start at, especially if you already know something about the language you're using. If you have any questions or problems, it's good to have somebody more experiencet at hand, but with all those forums on the Internet, I think it's no problem. We'll try to answer all your questions;)

Good luck!

Ok, Besides unrolling loops, using correct variables, and bit shifts instead of multiplication
or division, how do I make my programs more effeicent  and get them to run faster.

Well, this is a hard question to be answered directly and I don't exactly know what you mean. There are two meanings of "making a program more efficient":

1) Altering the program without changing its idea / algorithm. That's what you said: don't do anything stupid, don't do anything twice, do things in a way that the computer likes because it's how it's built. Those changes can prop up your program, but usually only a few percent. Or maybe make it two times faster. Or so. Usually, you don't need to do those optimalizations on IOI or competitions like this, unless your code is really messy in the first place. Personally, I think that with good compilers, some really small things like unrolling simple loops and replacing *2 by >> 1 doesn't make any difference.

2) Changing the algorithm. I don't know how familiar are you with the efficiency and computational complexity issues, but try reading [url=http://www.topcoder.com/tc?module=Static&d1=tutorials&d2=complexity1]this[/url] (hopefully the link works). This is the most important aspect of making the program fast, when you accomplish this part, you can tidy up your code and make some additional "make-up" changes like those you mentioned.

I hope I've answered your question...

Thank You! So it is all about using the right alg?

Well, you could say so. Of course, making the code itself effective is also a good thing, but I would consider it secondary.

I don't know much about real-world programming, there are surely situations where you prefer speed on large data sets / speed on small data sets / portability / simplicity of code / many other factors and you have to choose according to the situation.

But on an olympiad of an IOI style, you should first make up an effective algorithm, which you are able to code in an hour or so (usually, they won't require you to write fibonacci heaps and RBtrees to get fullscore), then think about how to write it conveniently (so that you can easily write it and see into your code) and at last, you consider the effectiveness of the code in small details. But it always depends on the situation...

Is there some big listing of algs somewhere to learn from?

As I said before, I haven't found an ultimate resource of algotirhms yet. Introduction to Algorithms (MIT press), which you seem to have, is the best single book I've encountered. You can find a lot online, maybe some things on the TopCoder site (see the link in my previous post), something at the USACO forums, ...

And the most valuable resource is YOUR OWN EXPERIENCE. Usually, you use a set of simple algorithms that repeat over and over, but usually you have to make your own algorithm for the specific problem, otherwise the whole computer science would become boring, wouldn't it?;-) Mostly, the algorithms are just common sense and math, and the little set of solutions and techniques that repeat over and over in the problems get their names and you see them as "the algorithms" ...

I don't know how much experience do you have, so I don't know how good it is what I'm telling you now, but it looks like you want to study algorithms without doing any practice ... the best thing to do is to sign up at USACO and start.

Memorizing algorthms for programming is like memorizing formulas for math.  Read [url=http://http://www.artofproblemsolving.com/Resources/AoPS_R_A_Prob_Solv.php]this article[/url], even though it talks about forumulas and not algorthms.

I think he meant this:

[url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_Prob_Solv.php]www.artofproblemsolving.com/Resources/AoPS_R_A_Prob_Solv.php[/url]

it wasn't that hard to figure out ... you shouldn't give up after one unsuccessful click! ;)

the training gate at http://ace.delos.com/usacogate is intended to lead people systematically through the algorithms needed for USACO/IOI. You have to already know the basics of programming before you can use the training material, but then it's designed to teach you everything else. It is much more specific to those competitions than a general algorithms book would be. All of the US people who have made it to the training camp or the IOI (and probably most of the English speakers competing for other countries) concentrated on working through the training gate. It's interactive, so you get immediate feed-back. But if you get hoplessly stuck on a problem, the USACO coaches will give you a hint. It's a great resource. The coaches consider it so important that they consider progress through the training gate in addition to performance on the internet contests when they decide who to invite to camp.

I didn't, but my browser doesn't like me :(.


Hola conejita, veo que estas posteando bastante. Este problema es conocido, me parece que se resuelve con congruencia, pero como no me gusta hacer mucho trazo, lo hice con utilizando trigonometría, a ver q te parece.

[b]Solución:[/b]

[b]Lema[/b]: Sea $ABCD$ un cuadrilátero convexo y P el punto de intersección de sus diagonales, entonces:
$AC\perp BD \Longleftrightarrow AB^{2}+CD^{2}=BD^{2}+AC^{2}$
Sol.- La ida es sencilla por el Teorema de Pitágoras, el regreso, que es la parte que nos interesa se desprende de la ley de cosenos como sigue:
Sea $\theta=m<APB$
$AB^{2}=AP^{2}+BP^{2}-2AP.BPcos\theta$
$CD^{2}=CP^{2}+DP^{2}-2CP.DPcos\theta$
$BD^{2}=BP^{2}+DP^{2}+2BP.DPcos\theta$
$AC^{2}=AP^{2}+CP^{2}+2AP.CPcos\theta$
Reemplazando en la hipótesis, obtenemos: $2BC.ADcos\theta=0\rightarrow\theta=90$. Si el cuadrilátero fuera no convexo(esto para el caso de un triángulo obstuso) la demostración es la misma con una variación en los signos. Esto termina la demostración del lema.
En el problema sean $a,b,c$ las medidas de $BC,AC,AB$ respectivamente. Aplicando la ley de cosenos:
$(O_{1}O_{3})^{2}=(\frac{b}{\sqrt{2}})^{2}+(\frac{c}{\sqrt{2}})^{2}-2(\frac{b}{\sqrt{2}})(\frac{c}{\sqrt{2}})cos(90+A)=\frac{b^{2}}{2}+\frac{c^{2}}{2}+bcsen(A)$
$(O_{1}O_{2})^{2}=(\frac{b}{\sqrt{2}})^{2}+(\frac{a}{\sqrt{2}})^{2}-2(\frac{b}{\sqrt{2}})(\frac{a}{\sqrt{2}})cos(90+C)=\frac{b^{2}}{2}+\frac{a^{2}}{2}+basen(C)$
Además: $(BO_{1})^{2}=\frac{c^{2}}{2}$ y $(BO_{2})^{2}=\frac{a^{2}}{2}$

Por el lema:
$O_{1}O_{2}\perp BO_{3}\Longleftrightarrow (BO_{1})^{2}+(O_{2}O_{3})^{2}=(BO_{2})^{2}+(O_{1}O_{3})^{2}\Longleftrightarrow bcsen(A)=basen(C)$
,esto último es cierto por la ley de senos, con lo cual la prueba concluye.

[quote="conejita"]En un triangulo ABC, se construyen cuadrados exteriores a los lados AB, BC, CA con centro O1, O2, O3 respectivemente. Demuestra que O1O2 es perpendicular a BO3.[/quote]
[color=darkblue][b]The first method.[/b] I"ll use the complex numbers. Denote $A(a)\ ,\ B(b)\ ,\ C(c)$ and $w=\frac{1+i}{2}\ (i\cdot\overline w=w\ ,\ w-\overline w=i)\ .$
Thus, $O_{1}(o_{1}=wa+\overline wb)\ ,\ O_{2}(o_{2}=wb+\overline wc)\ ,\ O_{3}(o_{3}=wc+\overline wa)\ .$
Observe that $o_{2}-o_{1}=i(b-o_{3})$. i.e. $wb+\overline wc-wa-\overline wb=i(b-wc-\overline wa)$ (it is show easily).
Therefore, $O_{1}O_{2}\perp BO_{3}$ and $O_{1}O_{2}=BO_{3}\ .$

[b]The second method ([u]Claudio Espinoza[/u]).[/b] $O_{1}O_{2}^{2}=BO_{1}^{2}+BO_{2}^{2}-2\cdot BO_{1}\cdot BO_{2}\cdot \cos\widehat{O_{1}BO_{2}}=$ $\frac{c^{2}+a^{2}}{2}-2\cdot \frac{ac}{2}\cos (B+90)=$ $\frac{c^{2}+a^{2}}{2}+2\cdot S\ .$
Analogously, $O_{2}O_{3}^{2}=\frac{a^{2}+b^{2}}{2}+2\cdot S$ and $O_{3}O_{1}^{2}=\frac{b^{2}+c^{2}}{2}+2\cdot S\ .$ Therefore,
$BO_{3}\perp O_{1}O_{2}\Longrightarrow$ $O_{3}O_{1}^{2}-O_{3}O_{2}^{2}=BO_{1}^{2}-BO_{2}^{2}\Longrightarrow$ $\frac{b^{2}+c^{2}}{2}+2\cdot S-\left(\frac{a^{2}+b^{2}}{2}+2\cdot S\right)=\frac{c^{2}-a^{2}}{2}$, what is truly.

[b]Remark.[/b] [u]Claudio Espinoza[/u], how show you synthetically that $O_{1}O_{2}=BO_{3}\ ?$[/color]

[quote="conejita"]Ola claudio, tu solucion fue igual que la mia. Mi forma de demostrar el lema, lo ice un poco mas facil.
Usemos vectores,  entonces
AB^2 + CD^2 = BC^2 + DA^2  si y solo si
(A-B)^2 + (C-D)^2 = (B-C)^2 + (D-A)^2 si y solo si
AB + CD = BC + AD si y solo si
(A-C)(B-D)=0 si y solo si
AC perpendicular a BD.
Creo que un poco mas rapida.
Lo demas fue igual que lo que tu hiciste.[/quote]Yes, your solution is the nice and the shortest !

Virgil, could you post in spanish here? I mean, there's no problem in writing in english, but that's not the idea

[img]http://img221.imageshack.us/img221/286/edsonlaurazx2.png[/img] \[\begin{gathered}{\text{1}}{\text{. Teorema : }}\mathop{\text{O}}\nolimits_{3}\mathop F\nolimits_{}= \mathop{F\mathop O\nolimits_{2}}\nolimits_{}{\text{ ; }}\mathop{\text{O}}\nolimits_{3}\mathop F\nolimits_{}\bot \mathop{F\mathop O\nolimits_{2}}\nolimits_{}{\text{ }}\hfill \\{\text{2}}{\text{. Del gr\'a fico: }}\vartriangle \mathop O\nolimits_{1}F\mathop O\nolimits_{2}\cong{\text{ }}\vartriangle BF\mathop O\nolimits_{3}{\text{ }}.......(L.A.L) \hfill \\{\text{ }}\Rightarrow{\text{ }}\mathop{\text{O}}\nolimits_{3}\mathop B\nolimits_{}= \mathop{F\mathop O\nolimits_{2}}\nolimits_{}\hfill \\{\text{3}}{\text{. Propiedad de tri\'a ngulos: }}\mathop{\text{O}}\nolimits_{3}\mathop B\nolimits_{}\bot \mathop{F\mathop O\nolimits_{2}}\nolimits_{}{\text{ }}\hfill \\ \end{gathered}\]

[quote="conejita"]Aki otro problema:

En un triangulo ABC, se construyen cuadrados exteriores a los lados AB, BC, CA con centro O1, O2, O3 respectivemente. Demuestra que O1O2 es perpendicular a BO3.

 :)[/quote]

No agan fiesta sin mi pues

mi solucion es con numeros complejos, bastante parecida a la de Virgin

HOLA QUE TAL
USAREMOS EL TEOREMA DE VON AUBEL PARA CUADRILATEROS.
NOS DICE QUE SI SE TIENE UN CUADRILATERO ABCD CONVEXO ENTONCES AL CONSTRUIR LOS CUADRADOS EN LOS LADOS DEL CUADRILATERO, CUYOS CENTROS DE LOS CUADRADOS DE LADO AB,BC,CD,DA SON O1,O2,O3,O4 RESPECTIVAMENTE => O1O3 PERPENDICULAR A O2O4.
AHORA EN EL PROBLEMA DEGENEREMOS EL CUADRILATERO A UN TRIANGULO ENTONCES UN VERTICE SERA UN CENTRO POR LO TANTO AL UNIRLO CON EL CENTRO DE EL OTRO CUADRADO, SERA PERPENDICULAR A EL OTRO SEGEMENTO FORMADO POR LA UNION DE LOS OTROS DOS CENTROS ADYACENTES AL VERTICE QUE ES CENTRO.

Lo del cuadrilátero ya lo había visto antes, pero no sabía que fuera un teorema. En todo caso su demostración se apoya en el hecho de que al unir los centros de dos cuadrados construidos exteriormente sobre dos lados de un triánglo con el punto medio del tercer lado se forma un triángulo rectángulo isósceles. Luego aplicar una congruencia parecida a la que hizo EDSON_. De la congruencia tambíen resulta O1O3=O2O4. Además no es necesario que el cuadrilátrero sea convexo.

Espero que el cambio de notaciones y el inglés no sea problema. Esta demostración la encontré el 1974.

[quote]Oye se me ace muy interesante esr teorema, podrias postear la demostracion de dixo problema??
Aunk me imagino k va a aser usando el mismo lema pa el problema anterior!!
[/quote]

[b]LA DEMOSTRACION ESTA AQUI[/b]
[url]http://geomtry.mundoforo.com/viewtopic.php?p=139#139[/url]


Consider the following perpetual motion machine:
a u-shaped tube is filled with water on one side and in the bottom.  Two gates (red) are shut on either end of the vertical portion of the water-filled tube.  A small, low-density ball is placed in the tube.  It gains velocity as it rises through the water, enough to that it is catapulted over the u onto the other side where it falls into the water.  The top gate opens to allow it through.  The downward velocity it has gained from falling is enough to push it around the bend of the u.  As it passes up, the lower gate opens, while the top gate stays closed so that the total volume between the two gates does not change (since no water can leave via the bottom without causing a vacuum at the top, and as we know, nature abhors a vacuum :) ).  This process repeats indefinitely.

Why doesn't this work?
Hints:  the answer has nothing to do with viscosity, friction, malfunctions in the gates, or the mysterious change of the ball from travelling left to travelling right.  Even if we discount these things, it still won't work.  Why not?

where does the energy in opening/closing gates come from?

Doesn't matter -- assume that the gates move frictionlessly with respect to the water, so they aren't adding any energy to the system.

I'm not sure about this, but [hide]if the ball is buoyant enough to pop the ball over the lip, then won't this same buoyancy counter the velocity gained by the ball's falling motion, thereby preventing the ball from moving past the lower u?[/hide]

No, because its bouyance propels it upwards over a long stretch of water.  It then enters freefall in the air, and only has to pass downwards through a small stretch of water before the process can repeat again.  The downward momentum from falling should be enough to send it around the bottom part of the U.

I'm not sure, but [hide]since after the ball falls a long way only a short bit of water will bring the speed to 0 so the buoyant force must be greater than gravity.

Since after the ball rises a long way and then only needs a short bit of time under gravity to reduce the speed to 0 so the buoyant force must be less than the force of gravity.

contradiction[/hide]

I could always adjust the dimensions of the tank to deal with that objection, for example by making the U curved part larger at either the top or the bottom.  Good thought, though.  The solution should work regardless of those dimensions of the tank.

Once it moves to the point where it receives no gravity, the bouyant force is same from the surface and the gate on the other side from the picture, therefore it will get stuck there.

I have no idea what you mean, but it's not the solution I'm looking for.  There is not point where it has to stop moving in its cycle due to an interaction of the forces of gravity and bouyancy.

I think that if the bottom gate opens, the water will be on the same level on both side of the U-shape.

Oh!  No, I already commented on that in the problem statement -- the top gate remains closed, so the water can't flow out without creating a vacuum at the top of the tube.  The air can't flow down and around the U, so the water will stay in.

Hm, this is puzzling.

The one thing I'm thinking, is what force is responsible for its sideways acceleration?  Is it propelled by water/air currents?  Or does it collide elastically with the walls?

[quote="JBL"]Why doesn't this work?
Hints:  the answer has nothing to do with viscosity, friction, malfunctions in the gates, or the mysterious change of the ball from travelling left to travelling right.  Even if we discount these things, it still won't work.  Why not?[/quote]

Ooooh, missed that.  Then I'm completely at sea.

When the top gate opens, the ball get out so the volume of the water get lower, which will leave space for air to get in. Then it's no longer vaccum.

You're getting there -- that's part of it.  But the top will close afterwards, and when the bottom opens, you still have the same problem.  But the observation you made is very important.

how will the low-density ball get enough velocity to travel thru the h2o?

Going upwards, because of bouyancy.  Going downwards, because of the acceleration due to gravity.  (Remember, light things fall just as fast as heavy things, neglecting air resistance.)

When the bottom opens, the water between the two gates will move that the water level will be balanced. I have proved that there will not be a vaccum between the two gates.

You have not proved any such thing.  Water would still have to move from a tank closed at one end (the top) into somewhere else.  This would be an absolute decrease in the volume of the tank, resulting in a vacuum.  But you're close to what you actually need.

I am totally confused, what do I actually need?

You have now asked that question 3 times, and it has been answered twice.  Would a moderator please delete the second two times he's asked it, since I answered it the first time?

sry about that JBL.  When i pressed the back button on my browser, it re-sent the post.  the last three times were accidental.

[quote="Alison"]But when the ball re-enters, won't it force air out?

[quote="Scrambled"]when the top gate opens and the ball falls out, there is air in the top gate.  When this repeats, more and more air gets into the chamber.  Then the water is too low to succesfully propel it enough to get it to sink in the water.[/quote][/quote].  No the top gate is still closed.  besides, think, it will travel normally through water anyway.

Scrambled is close -- he might want to say something about the lower gate, too, not just the higher gate.

When the ball falls into the water, the water rises the side with no gates. When the second gate opens, the ball went up and then the second gate closes. Since the ball is gone now, the volume of water is the same as before the ball fell into water

[quote="beta"]This might be a very basic physics question, but why would a vacuum has anything to do with water falling through the second gate? I don't think it's because of lacking air pressure since there is air between the first gate and the second gate after the first gate opens. Why nature abhors vacuum?[/quote]

If you fill a skinny tube (such as a straw) with liquid and pinch the end, the liquid will flow out (down) very, very slowly.  This is because water can only leave if a corresponding volume of air enters the straw.  If you were to take a j-shaped object and fill it with liquid and pinch the end, [i]none would flow out[/i], even though there would be flow if you left the end unpinched.  This is because air cannot make it around the bend of the J, since it would need to pass downwards through water.  Similarly, when the lower gate opens, the water cannot all simply flow to the right side, since the top gate is closed.

Actually, re-reading what scrambled wrote, he is pretty much entirely there.  The one thing he didn't say is that as the ball passes through the bottom gate, it will replace its own volume with that of water -- in other words, the amount of water on the right side will increase by exactly the volume of the ball.  This volume is then replaced by air as the ball exits through the top gate.  So eventually, as Scrambled said, the volumes of water on both sides will be such that the bouyancy does not create enough downwards velocity to propel the ball around the loop any more.

I was close. Thanks for the vacuum thing. It messed me up because I was keep thinking that water will move around.

[quote]Actually, re-reading what scrambled wrote, he is pretty much entirely there.  The one thing he didn't say is that as the ball passes through the bottom gate, it will replace its own volume with that of water -- in other words, the amount of water on the right side will increase by exactly the volume of the ball.  This volume is then replaced by air as the ball exits through the top gate.  So eventually, as Scrambled said, the volumes of water on both sides will be such that the bouyancy does not create enough downwards velocity to propel the ball around the loop any more.[/quote]

well where else would the water go?


In a bag are only red and white marbles. The probability of choosing a red marble is $ \frac{3}{5}$. The probability of choosing two red marbles without replacement is $ \frac{3}{10}$. How many white marbles are in the bag?

$x$ is the probability that we choose a red marble given that we already took a red marble. Because the probability of choosing a red marble and the probability of choosing a white marble are the same ($1-1/2=1/2$), there must be the same number of red marbles as white marbles. Therefore, we have the equation $r=w$. However, we must remember that we took a red marble away, so we have $r=w+1$ to account for the extra red marble.

Thanks for the explanation it helped a lot!


1. Whats the largest prime factor of $ (23!)\plus{}(25!)$

2. Expand: $ (\minus{}2\plus{}\sqrt{3})^5$

3. The road from Town A to Town B is level for half the distance; the other half of the distance
is part uphill and part downhill. A cyclist averages 12 mph when on a level road, 6 mph on
an uphill road, and 18 mph on a downhill road. It takes this cyclist 2 hours and 40 minutes to
travel from A to B, and 2 hours to travel from B to A. How many miles long is the uphill
portion of the road from A to B?

4. The product of the ages of three children in a family is 1872. How old is the oldest of these
children if their ages are evenly spaced?

1. What is the largest prime factor of $ (23!) \plus{} (25!)$?

[hide]
Factor out a $ 23!$.
$ (23!)(1 \plus{} 25\cdot 24)$

Simplify
$ 23!\cdot 601$

601 is prime, and the largest prime in $ 23!$ is 23. Therefore your answer is $ \boxed{601}$.
[/hide]

2. Expand $ ( \minus{} 2 \plus{} \sqrt {3})^5$

[hide]
Use the binomial formula:

$ (x \plus{} y)^n \equal{} \binom{n}{0} x^n y^0 \plus{} \binom{n}{1} x^{n \minus{} 1} y^1 \plus{} \binom{n}{2} x^{n \minus{} 2} y^2 \plus{} ... \plus{} \binom{n}{k} x^{n \minus{} k} y^k \plus{} ... \plus{} \binom{n}{n} x^0 y^n$

$ ( \minus{} 2 \plus{} \sqrt {3})^5 \equal{} \binom{5}{0} ( \minus{} 2)^5 (\sqrt {3} )^0 \plus{} \binom{5}{1} ( \minus{} 2)^4 (\sqrt {3} )^1 \plus{} \binom{5}{2} ( \minus{} 2)^3 (\sqrt {3} )^2 \plus{} \binom{5}{3} ( \minus{} 2)^2 (\sqrt {3} )^3 \plus{} \binom{5}{4} ( \minus{} 2)^1 (\sqrt {3} )^4 \plus{} \binom{5}{5} ( \minus{} 2)^0 (\sqrt {3} )^5$
$ \equal{} \minus{} 32 \plus{} 80\sqrt {3} \plus{} \minus{} 240 \plus{} 120\sqrt {3} \plus{} \minus{} 90 \plus{} 9\sqrt {3}$
$ \equal{} \boxed{\minus{} 362 \plus{} 209\sqrt {3}}$
[/hide]

3. The road from Town A to Town B is level for half the distance; the other half of the distance
is part uphill and part downhill. A cyclist averages 12 mph when on a level road, 6 mph on
an uphill road, and 18 mph on a downhill road. It takes this cyclist 2 hours and 40 minutes to
travel from A to B, and 2 hours to travel from B to A. How many miles long is the uphill
portion of the road from A to B?

[hide]

Let the length of the level road = x
Let the length of the uphill road = y
Since the level road is half the distance of the total road, the length of the downhill road is x-y.

$ speed \equal{} \dfrac{distance}{time}$
Speed on level road: 12 mph
Speed on uphill road: 6 mph
Speed on downhill road: 18 mph

So from town A to town B:

Time on level road: $ \dfrac{x}{12}$
Time on uphill road: $ \dfrac{y}{6}$
Time on downhill road: $ \dfrac{x \minus{} y}{18}$

Adding them together would equal the total time traveled, in hours.
$ \dfrac{x}{12} \plus{} \dfrac{y}{6} \plus{} \dfrac{x \minus{} y}{18} \equal{} \dfrac{8}{3}$

Multiply through by the least common denominator, 36.

$ 3x \plus{} 6y \plus{} 2x \minus{} 2y \equal{} 96$
$ 5x \plus{} 4y \equal{} 96$

From town B to town A:

Time on level road: $ \dfrac{x}{12}$
Time on uphill road: $ \dfrac{x \minus{} y}{6}$
Time on downhill road: $ \dfrac{y}{18}$

Again, add the times to equal the total time traveled.

$ \dfrac{x}{12} \plus{} \dfrac{x \minus{} y}{6} \plus{} \dfrac{y}{18} \equal{} 2$

Multiply through by 36 again.

$ 3x \plus{} 6x \minus{} 6y \plus{} 2y \equal{} 72$
$ 9x \minus{} 4y \equal{} 72$

Now you have a system of two equations which can be solved for the uphill distance, y.

$ 5x \plus{} 4y \equal{} 96$
$ 9x \minus{} 4y \equal{} 72$

Adding the two equations yields
$ 14x \equal{} 168$
$ x \equal{} 12$

Substituting $ x \equal{} 12$ back into either equation gives $ y \equal{} \boxed{9}$
[/hide]

4. The product of the ages of three children in a family is 1872. How old is the oldest of these
children if their ages are evenly spaced?

[hide]
The prime factorization of 1872 is $ 2^4\cdot 3^2\cdot 13$
Try to find 3 numbers that can be formed from that to form a 3-term arithmetic sequence.
These numbers turn out to be $ 2^3, 13, 2\cdot 3^2$ or $ \boxed{8, 13, 18}$.
[/hide]

Wow, thanks for the quick replies, really appriciated


search
function
number theory unsolved
number theory

[quote="comet_vn"]let a and b be positive integers such that a^n-1 is divisible by b^n-1 for all the positive integer n
Prove that a is a power of b[/quote]
For sure there is a typo and actually you meant "prove that $ b$ is a power of $ a$". In any case, you should try to use the search function, before posting a problem on the forum. This, in particular, is really a famous one (try [url=http://www.mathlinks.ro/Forum/viewtopic.php?search_id=139876416&t=4556]here[/url]). :wink:


hm, just stop using them?  AOL might delete them automatically if you stop using them for a really long time (?).

I think they're automatically deleted after ~6 months of inactivity.


A girl bought a dog for 10 dollars, sold it for 15 dollars, bought it back for 20 dollars, and finnaly sold it for 25 dollars.  Did the girl make or lose money, and how much did she make or lose?


In a triangle [b]ABC[/b] satisfying [b]AB+BC=3AC[/b] the incircle has centre [b]I[/b] and touches the sides [b]AB[/b] and [b]BC[/b] at [b]D[/b] and [b]E[/b], respectively. Let [b]K[/b] and [b]L[/b] be the symmetric points of [b]D[/b] and [b]E[/b] with respect to [b]I[/b]. [i] Prove that the quadrilateral [b]ACKL[/b] is cyclic.[/i]

Davron

Davron CHITU,NAG'ZMI? DAVRON IF YOU HAVE SILK ROADMATH OLYMPIAD PROBLEMS 2006 PLEASE SEND ME OR MATHFORUM.(MY E-MAIL: OORIF14@MAIL.RU)

See http://www.mathlinks.ro/Forum/viewtopic.php?t=76458 .

  dg

[b]Remark 1.[/b] $IG\perp BC\Longleftrightarrow b+c=3a$ or $b=c\ .$
[b]Remark 2.[/b] $IG\parallel BC\Longleftrightarrow b+c=2a\ .$

[color=darkred][b][u]The proof of the first remark :[/u][/b] $IG\perp BC\Longleftrightarrow b+c=3a$ and in this case

$G\in EF\ ,$ where the points $E\in (AC)$, $F\in (AB)$ are the points where

the incircle $C(I,r)$ touches these sides. Indeed, $GB^2=\frac{2(a^2+c^2)-b^2}{9}\ ,$

$GC^2=\frac{2(a^2+b^2)-c^2}{9}\Longrightarrow$ $\boxed {\ GB^2-GC^2=\frac 13 (c^2-b^2)\ }\ .$

$IB^2=\frac{ac(p-b)}{p}\ ,\ IC^2=\frac{ab(p-c)}{p}\Longrightarrow \boxed {\ IB^2-IC^2=a(c-b)\ }\ .$

$IG\perp BC\Longleftrightarrow IB^2-IC^2=GB^2-GC^2\Longleftrightarrow c^2-b^2=3a(c-b)\ .$

Therefore, $\boxed {\ IG\perp BC\Longleftrightarrow b=c\ \vee\ b+c=3a\ }\ .$

Denote the middlepoint $M$ of the side $[AC]$ and the intersection $G_1\in BM\cap EF\ .$

The Menelaus' theorem applied to the transversal $\overline {EGF}$ for the triangle $ABM\ ,$ i.e.

$\frac{EM}{EA}\cdot \frac{FA}{FB}\cdot \frac{G_1B}{G_1M}=1\ \ (EA=FA)\Longleftrightarrow\frac{G_1B}{G_1M}=\frac{FB}{EM}\ .$

But $EM=\frac 12 \left|a-c\right|$ and $FB=p-b\ .$ Thus , $G\in EF$ $\Longleftrightarrow$ $G_1\equiv G$ $\Longleftrightarrow$

$G_1B=2\cdot G_1M$ $\Longleftrightarrow$ $FB=2\cdot EM$ $\Longleftrightarrow$ $(p-b)=\left|a-c\right|$ $\Longleftrightarrow$

$3a=b+c$ when $a\le c$ or $3c=a+b$ when $a>c\ .$[/color]

[color=darkblue][b][u]The proof of the second remark : [/u][/b] $IG\parallel AB\Longleftrightarrow b+c=2a\ .$

$IG\parallel BC\Longleftrightarrow \delta_{BC}(I)=\delta_{BC}(G)\Longleftrightarrow r=\frac 13h_a\Longleftrightarrow 3ar=2S\Longleftrightarrow$

$3a=2p\Longleftrightarrow b+c=2a\ .$ Therefore , $\boxed {\ IG\parallel BC\Longleftrightarrow b+c=2a\ }\ .$[/color]

mr. Virgil or someone other do you have a solution to the initial problem ?I tried much but...

[hide="Siluoan, look (forward) to the proof of the proposed problem."][quote][color=darkred][b]A equivalent enunciation.[/b] The incircle $C(I,r)$ of the scalene triangle $ABC$ touches its sides in the points $D\in BC$, $E\in CA$, $F\in AB$. Denote the [u]Nagel's point[/u] $N$ and the intersections: $D'\in BC\cap AN$, $E'\in CA\cap BN$, $F'\in AB\cap CP$, $(DB=D'C=p-b$, $EC=E'A=p-c$, $FA=F'B=p-a) .$ The rays $[NA$, $[NB$, $[NC$ meet the incircle in the points $X$, $Y$, $Z$ respectively. Then the quadrilateral $BYZC$ is cyclically if and only if $N\equiv X$ ([b][u]it is fantastically ![/u][/b]) if and only if $3a=b+c$, i.e. $p=2a$.[/color][/quote][color=darkblue][b]Lemma (well-known).[/b] $I\in XD\cap YE\cap ZF$ ([u]the essence of the my proof[/u]).
[b]Proof.[/b] Denote the projection $P$ of the vertex $A$ on the opposite side $[BC]$. Thus, $ah_a=2pr$ and $I\in DX$ $\Longleftrightarrow$ $\frac{XD}{AP}=\frac{D'D}{D'P}$ $\Longleftrightarrow$ $\frac{2r}{h_a}=\frac{a-2(p-b)}{(p-c)-c\cos B}$ $\Longleftrightarrow$ $\frac ap=\frac{2a(b-c)}{a(a+b-c)-(a^2+c^2-b^2)}$ $\Longleftrightarrow$ $\frac ap=\frac{2a(b-c)}{2p(b-c)}\ ,$ what is truly.

[b]The proof of the proposed problem.[/b]
Suppose $b\ne c$ and denote the length of the A-Nagel's line $AD'=n_a\ .$
=================================================================
$1.\blacktriangleright$ [u]Stewart's relation[/u] : $\boxed {an^2_a+a(p-b)(p-c)=b^2(p-c)+c^2(p-b)}\ .$
$2.\blacktriangleright$ [u]Van Aubel's relation[/u] : $\frac{NA}{ND'}=\frac{F'A}{F'B}+\frac{E'A}{E'C}=$ $\frac{p-b}{p-a}+\frac{p-c}{p-a}$ $\Longrightarrow$ $\frac{NA}{ND'}=\frac{a}{p-a}$ $\Longrightarrow$
$\frac{NA}{a}=\frac{ND'}{p-a}=\frac{n_a}{p}$ $\Longrightarrow$ $\boxed {NA=\frac ap\cdot n_a}\ .$
$3.\blacktriangleright$ [u]The lemma[/u] : $\frac{D'X}{D'A}=\frac{XD}{AP}=\frac ap$ $\Longrightarrow$ $\frac{D'X}{a}=\frac{D'A}{p}=\frac{AX}{p-a}$ $\Longrightarrow$ $\boxed {AX=\frac{p-a}{p}\cdot n_a}\ .$
$4.\blacktriangleright$ $(2)\ \wedge\ (3)$ $\Longrightarrow$ $NX=NA-AX=\left(\frac ap -\frac{p-a}{p}\right)\cdot n_a$ $\Longrightarrow$ $\boxed {NX=\frac{2a-p}{p}\cdot n_a}\ .$
Therefore, $N\equiv X$ $\Longleftrightarrow$ $p=2a$, i.e. $\boxed {\boxed {N\equiv X \Longleftrightarrow b+c=3a}}\ .$
=================================================================
[b]The quadrilateral[/b] $BYZC$ [b]is cyclically[/b] $\Longleftrightarrow$ $NY\cdot NB=NZ\cdot NC$ $\Longleftrightarrow$
$\left(\frac{2b-p}{p}\cdot n_b\right)\left(\frac bp\cdot n_b\right)=\left(\frac{2c-p}{p}\cdot n_c\right)\left(\frac cp\cdot n_c\right)$ $\Longleftrightarrow$
$(2b-p)\cdot bn^2_b=(2c-p)\cdot cn^2_c$ (and the relation $(1)$) $\Longleftrightarrow$
$(2b-p)\left[c^2(p-a)+a^2(p-c)-b(p-a)(p-c)\right]=$
$(2c-p)\left[a^2(p-b)+b^2(p-a)-c(p-a)(p-b)\right]$ $\Longleftrightarrow$
$2bc^2(p-a)+2a^2b(p-c)-2b^2(p-a)(p-c)-$
$c^2p(p-a)-a^2p(p-c)+bp(p-a)(p-c)=$
$2b^2c(p-a)+2a^2c(p-b)-2c^2(p-a)(p-b)-$
$-b^2p(p-a)-$ $a^2p(p-b)+cp(p-a)(p-b)$ $\Longleftrightarrow$
$-2bc(p-a)+2a^2p-(p-a)\left[a(b+c)+(b^2+c^2)\right]+$
$+(b+c)p(p-a)-a^2p+p^2(p-a)=0$ $\Longleftrightarrow$
$-8bc[(b+c)-a]+4a^2[a+(b+c)]-4a(b+c)[(b+c)-a]-4\left(b^2+c^2\right)[(b+c)-a]+$ $2(b+c)\left[(b+c)^2-a^2\right]+[a+(b+c)]\left[(b+c)^2-a^2\right]=0$ $\Longleftrightarrow$
$3a^3+5(b+c)a^2+(b+c)^2a-(b+c)^3=0$
$\Longleftrightarrow$ $[3a-(b+c)](a+b+c)^2=0\ .$
Therefore, $\boxed {\boxed {\ \mathrm {the\ quadrilateral\ BYZC\ is\ cyclically\ }\Longleftrightarrow b+c=3a}}\ .$[/color]

[color=red]Oh, after a long run, appear and [/color]$\ldots\ldots\mathrm {\ The\ END\ .}$[/hide]

[hide=" Only for Silouan."]It is ready. Finally, I put an end to the solution of this problem.
The best is $BYZC\ \mathrm {is\ cyclically\ }\Longleftrightarrow \boxed {N\equiv X}\Longleftrightarrow b+c=3a\ .$[/hide]

Thank you very much mr Virgil. A great metrical solution . :D

[quote][color=darkred][b]Remark (see the my proof).[/b] In the triangle $ABC$
the [u]Nagel's point[/u] $N$ belongs to the its incircle $C(I,r)$ if and only if $(b+c-3a)(c+a-3b)(a+b-3c)=0\ \mathrm {(\ India\ ,\ 1995\ )}\ .$[/color][/quote][b]Proof.[/b]
$1.\blacktriangleright$ (Well-known) $NI=p^2+5r^2-16Rr\ : \ \boxed {\ N\in C(I,r)\Longleftrightarrow p^2+4r^2=16Rr\ }\ .$

$2.\blacktriangleright\prod (a+b-3c)=0\Longleftrightarrow \prod (p-2a)=0\Longleftrightarrow$

$p^3-2(a+b+c)p^2+4(ab+bc+ca)p-8abc=0\Longleftrightarrow$

$p^3-4p^3+4p(p^2+r^2+4Rr)-32Rpr=0\Longleftrightarrow p^2+4r^2=16Rr$, i.e.

$\boxed {\ \prod (a+b-3c)=0\Longleftrightarrow p^2+4r^2=16Rr\ }\ .$ Finally, \[ \boxed {\boxed {\ N\in C(I,r)\Longleftrightarrow p^2+4r^2=16Rr\Longleftrightarrow \prod (a+b-3c)=0\ }}\ .  \]


let {$ a_n$} : $ (1 \plus{} \frac {1}{n} )^{n \plus{} a_n}$ = e
prove that: $ {a_n}$ has limitation.find this limitation!

look at
this expression
\[ \Leftrightarrow e^{(\frac{1}{a_{n}})}\equal{}(1\plus{}\frac{1}{n})^n\]
\[ \Leftrightarrow \frac{1}{a_{n}}\equal{}ln((1\plus{}\frac{1}{n})^n)\]
\[ \Leftrightarrow a_{n}\equal{}\frac{1}{ln((1\plus{}\frac{1}{n})^n)}\]
\[ n\rightarrow \infty  \Rightarrow a_{n}\rightarrow 1\]
do you fell?

Do not double post:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=299777[/url]


calculus
integration
limit
calculus computations

The series $ \sum_{n\equal{}1}^\infty x_n$ converges. Must $ \sum_{n\equal{}1}^\infty x_n/n$ also converge?

According to the Cunha-Bolzano-Cauchy convergence criterion,

$ \sum_{n\equal{}1}^\infty x_{n}$ converges <=> $ \forall \epsilon > 0 \,\, \exists p \in \mathbb{N} : n,k \in \mathbb{N} \wedge n \geq p \equal{}> |x_{n\plus{}1} \plus{} x_{n\plus{}2} \plus{} ... \plus{} x_{n\plus{}k}| < \epsilon$. But

$ \left|\frac{x_{n\plus{}1}}{n\plus{}1} \plus{} \frac{x_{n\plus{}2}}{n\plus{}2} \plus{} ... \plus{} \frac{x_{n\plus{}k}}{n\plus{}k}\right| < |x_{n\plus{}1} \plus{} x_{n\plus{}2} \plus{} ... \plus{} x_{n\plus{}k}|$. $ \Diamond$

Since this isn't very interesting for $ x_n>0,$ I won't make that assumption.

If you've been following me here, you must realize by now that I'll try to use integration by parts or summation by parts any time I can.

Define $ X_n\equal{}\sum_{k\equal{}1}^nx_k.$ By hypothesis, $ \lim_{n\to\infty}X_n\equal{}L$ exists.
\[ \sum_{n\equal{}1}^N\frac{x_n}n\equal{}\frac{X_N}{N}\plus{}\sum_{k\equal{}1}^{N\equal{}1}X_n\left(\frac1n\minus{}\frac1{n\plus{}1}\right)\]
Take the limit as $ N\to\infty.$ We get
\[ \sum_{n\equal{}1}^{\infty}\frac{x_n}n\equal{}\sum_{k\equal{}1}^{\infty}X_n\left(\frac1n\minus{}\frac1{n\plus{}1}\right)\]
But $ X_n$ is bounded by $ M.$ (A convergent sequence is bounded). The series on the right is absolutely convergent by the telescoping convergent series
\[ M\sum_{n\equal{}1}^{\infty}\left(\frac1n\minus{}\frac1{n\plus{}1}\right).\]

In this argument, $ \frac1n$ could be replaced by any decreasing sequence that tends to zero - in fact, it doesn't even have to tend to zero. We could replace $ \frac1n$ by any monotone convergent sequence.

[quote]Thanks Carcul, but what's the conclusion of your argument?[/quote]

The conclusion is that, since the sum S_k of k consecutive terms of x_n can be made as small as we want, and the sum of the corresponding terms of x_n/n is less than S_k, then the sum of k consecutive terms of x_n/n can also be made as small as we want, which means that $ \sum x_n/n$ converges.

Of course, much more easier is to use Abel's test, whose proof Professor Kent Merryfield has provided in his post.


I have a document with marginal notes set up using

\newcommand{\marginal}[1]{%
\leavevmode\marginpar{\tiny\raggedright#1\par
}}

But I've noticed that due to the small width, when there are verses or quotes in the marginal, their left indents are way too much and the line formatting gets off.  I think ideally what would work best is having the verse or quote environment modified such that it removes the indent, but I don't want it to affect the main document verse text layout.  Is it possible to manage to do this?  How?  Any help is greatly appreciated, as I'm stumped (and tired).

It would be best to give a [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=157945]minimal example[/url] to show the problem.
Meanwhile, have a look at the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=verse]verse package[/url] which has indent and margin options.
You can also try the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=marginnote]marginnote package[/url] and if you want to redefine an environment use \newenvironment{[i]name[/i]}[ [i]num[/i] ]{[i]before[/i]}{[i]after[/i]} -see [url=http://tug.ctan.org/tex-archive/info/lshort/english/lshort.pdf]The Not So Short Introduction  LaTeX2e[/url] page 109.

Thanks for the links - I'll be reading them.  In the meantime, here's an example...Note how the verse in the margin looks...

[code]\documentclass{book}
\begin{document}

\setlength{\marginparwidth}{1.4in}
\newcommand{\nodate}{\date{}}\nodate
\newcommand{\marginal}[1]{%
\leavevmode\marginpar{\tiny\raggedright#1\par
}}

\mainmatter

Text here to have a margin note.Text here to have a margin note.Text here to have a margin note.Text here to have a margin note.Text here to have a margin note.\marginal{Normal spacing of text is here.
\begin{verse}
Mary had a little lamb little lamb little lamb\\
Mary had a little lamb its fleece was white as snow
\end{verse} And here's some end text for the marginal note.}Text here to have a margin note.Text here to have a margin note.Text here to have a margin note. Here's what verse looks like in the main section (the formatting doesn't need to change here)
\begin{verse}
Row row row your boat gently down the stream\\
Merrily merrily merrily merrily life is but a dream.
\end{verse}
More text more text More text more text More text more text More text more text More text more text More text more text More text more text More text more text More text more text.
\end{document}[/code]

Look at the verse package but meanwhile you could easily define a margin verse environment with
\newenvironment{marverse}{\vspace{-4ex}\begin{verse} \setlength{\parindent}{0pt}}{\end{verse}\vspace{-4ex}}
and put \noindent on the first line of marverse.

Hi,

Thanks for the pointers.  I did indeed research the {verse} package and followed its recommendations for changing indents - see modified marverse.  However, I came upon a strange problem - whenever the marginal notes are within a verse environment, the marverse settings don't seem to apply.  Example below of a non-working and working marginal note:

[code]\documentclass{book}

\begin{document}

\newlength{\saveleftmargini}
\newenvironment{marverse}{\setlength{\saveleftmargini}{\leftmargini}\setlength{\leftmargini}{0.5em}\begin{verse}}{\end{verse}\setlength{\leftmargini}{\saveleftmargini}}

\setlength{\marginparwidth}{1.4in}
\newcommand{\marginal}[1]{%
\leavevmode\marginpar{\tiny\raggedright#1\par
}}

\mainmatter

Text here to have a margin note.Text here to have a margin note.Text here to have a margin note.Text here to have a margin note.Text here to have a margin note.
And here's some end text for the marginal note.Text here to have a margin note.Text here to have a margin note.Text here to have a margin note. Here's what verse looks like in the main section (the formatting doesn't need to change here)
\begin{verse}
Row row row your boat gently down the stream\\
Merrily merrily merrily merrily life is but a dream.\marginal{Marginal \#1 - in verse.  Normal spacing of text is here.
\begin{marverse}
Mary had a little lamb little lamb little lamb\\
Mary had a little lamb its fleece was white as snow
\end{marverse}}
\end{verse}
More text more text More text more text More text more text More text more text More text more text More text more text More text more text More text more text More text more text.\marginal{Marginal \#2 - in text. Normal spacing of text is here.
\begin{marverse}
Mary had a little lamb little lamb little lamb\\
Mary had a little lamb its fleece was white as snow
\end{marverse}
And here's some end text for the marginal note.}
\end{document}[/code]

Is this something that I will have to be resigned to, or is there something else I'm missing?

Thanks,
Charlene

If you look in the log then you'll see
[i]LaTeX Warning: Marginpar on page 1 moved.[/i]
[quote="charlener"]I did indeed research the {verse} package and followed its recommendations for changing indents - see modified marverse.  [[/quote]So why not use the verse package and its options?

I believe that warning isn't from the verse modifications itself but from the use of marginal notes - which I actually do need...and the last posting is using settings recommended in the margin package for changing the indentation for the entire verse...

But you haven't [i]used[/i] the verse package which may do other things behind the scenes. You should try using it, with the indentation options it provides to see if that helps.

So.  Now I'm confused.  I *am* using the verse package and it gives the same effects.  And I'm using the parameters as mentioned in the verse package to modify the left margin for verses, which I have done.  It's only in the very specific combination I mentioned previously that the left margin change as described in the package documentation does not work.  See again:

\documentclass{book}
\usepackage{verse}
\begin{document}

\newlength{\saveleftmargini}
\newenvironment{marverse}{\setlength{\saveleftmargini}{\leftmargini}\setlength{\leftmargini}{0.5em}\begin{verse}}{\end{verse}\setlength{\leftmargini}{\saveleftmargini}}

\setlength{\marginparwidth}{1.4in}
\newcommand{\marginal}[1]{%
\leavevmode\marginpar{\tiny\raggedright#1\par
}}

\mainmatter

Text here to have a margin note.Text here to have a margin note.Text here to have a margin note.Text here to have a margin note.Text here to have a margin note.
And here's some end text for the marginal note.Text here to have a margin note.Text here to have a margin note.Text here to have a margin note. Here's what verse looks like in the main section (the formatting doesn't need to change here)
\begin{verse}
Row row row your boat gently down the stream\\
Merrily merrily merrily merrily life is but a dream.\marginal{Marginal \#1 - in verse.  Normal spacing of text is here.
\begin{marverse}
Mary had a little lamb little lamb little lamb\\
Mary had a little lamb its fleece was white as snow
\end{marverse}}
\end{verse}
More text more text More text more text More text more text More text more text More text more text More text more text More text more text More text more text More text more text.\marginal{Marginal \#2 - in text. Normal spacing of text is here.
\begin{marverse}
Mary had a little lamb little lamb little lamb\\
Mary had a little lamb its fleece was white as snow
\end{marverse}
And here's some end text for the marginal note.}
\end{document}

Perhaps I should contact the package developer directly, though...

Putting a verse environment inside another verse environment is bound to increase the indentation, since the verse environment is designed to do that so you are doubling it. So what we need to discover is exaxctly what you want to achieve. If you want the marginal comments to be the same both inside and outside the verse environment, then you will need different environments in each case.

But it is much better to [i]use[/i] the verse package options. For example, patverse has customisable indents. The following code makes both marginal notes the same and has no warnings. It may not yet be what you want but there are innumerable verse options such as indentpattern or \vin to use to change it.

[code]\documentclass{book}
\usepackage{verse}
\begin{document}

\setlength{\marginparwidth}{1.4in}
\newcommand{\marginal}[1]{%
\leavevmode\marginpar{\tiny\raggedright#1\par
}}

\mainmatter

Text here to have a margin note.Text here to have a margin note.Text here to have a margin note.Text here to have a margin note.Text here to have a margin note.
And here's some end text for the marginal note.Text here to have a margin note.Text here to have a margin note.Text here to have a margin note. Here's what verse looks like in the main section (the formatting doesn't need to change here)

\begin{verse}
	Row row row your boat gently down the stream\\
	Merrily merrily merrily merrily life is but a dream.\marginal{Marginal \#1 - in verse. Normal spacing of text is here.\\[2ex]
	\begin{patverse}
		Mary had a little lamb little lamb little lamb \\
		Mary had a little lamb its fleece was white as snow \\
	\end{patverse}
	}
\end{verse}

More text more text More text more text More text more text More text more text More text more text More text more text More text more text More text more text More text more text.\marginal{Marginal \#2 - in text. Normal spacing of text is here. \\[2ex]

\begin{patverse}
	Mary had a little lamb little lamb little lamb\\
	Mary had a little lamb its fleece was white as snow \\[2ex]
\end{patverse}

And here's some end text for the marginal note.}

\end{document} [/code]


Tom and Huck paint a fence for four hours, after which Jim helps them and they finish two hours later. If Jim had not helped them, it would have taken them 5 more hours to paint the fence. How long would it take Jim to paint the fence alone?

[hide]Tom and Huck's rate- $\frac{1}{9}$
${\frac{4}{9}+\frac{2}{9}+2x=1}$
$x=\frac{1}{6}$ Jim's rate
So it takes Jim 6 hours to paint the fence alone.[/hide]


[quote="GameBot"]At Joe's Pizza a 16 inch diameter pizza and a 12-inch-diameter pizza cost the same per square inch of top surface area. If the cost of a large pizza is $ \$9.60$, what is the cost, in dollars , of the smaller pizza?[/quote]

The pizza with the 12-inch diameter has an area of $ 36\pi$, while the pizza with the 16-inch diameter has an area of $ 64\pi$. We can then set up a proportion as follows:

\[ \frac{36\pi}{9.60} \equal{} \frac{64\pi}{x} \implies x \approx \boxed{17.07} \]

[hide="Larger pizza, not smaller"]There area of the larger pizza is $64\pi$ and the area of the smaller one is $36\pi$, but we are given that the large pizza is $\$9.60$ dollars, not the smaller one, so we have the ratio as $\frac{64\pi}{9.60}:\frac{36\pi}{x} \implies x=5.40$[/hide]


A right triangle has hypotenuse 7.6 units and one leg measures 3.2 units. Find the measure of the other leg rounded to the nearest tenth.


vector
search
linear algebra
linear algebra unsolved

If $ A\cup B\cup C$ is a subspace then $ B\cup C\subseteq A$ or $ A\cup C\subseteq B$ or $ A\cup B\subseteq C$.

thanks.

Obviously not. Consider  $ R^3$,
$ A: \equal{}t(1,0,0), t\in R$,
$ B: \equal{}t(0,1,0), t\in R$,
$ C: \equal{}t(0,0,1), t\in R$.

Sandor

Sorry, but I think you are wrong, $ A\cup B\cup C$ is not a subspace, for example $ (0,1,0)\plus{}(1,0,0)\equal{}(1,1,0)\not \in A\cup B\cup C$. And I forgot, $ A,B$ and $ C$ are subspaces.

OK, how about this counterexample:
(in $ F_2^2$, where $ F_2$ is the field with two elements)
$ A\equal{}\{(0,0),(1,0)\}$, $ B\equal{}\{(0,0),(0,1)\}$, $ C\equal{}\{(0,0),(1,1)\}$. The union is the whole space.

To rule out jmerry's counterexample, add the hypothesis that the field over which these are vector spaces is infinite. Or maybe for this particular specific problem, the field has at least $ m$ elements for some $ m$ to be determined later. (I suspect $ m\equal{}3$ is enough here.)

I remember a problem posted here once about the union of finitely many proper subspaces not being the whole vector space; so far I haven't found the right thing to search for.

The counterexamples are in field with have characteristic $ 2$ or $ 3$ ..., and in my problem, I forgot to put that the field is $ \mathbb{R}$ that have null characteristic, and the vector space is finite dimensional.

Sorry, I misunderstood something.
I think there is a method which works in this (and similar) situations, but a little bit tedious.
Draw A,B and C as circles. Take the basis of  $ A\cap B\cap C$. This basis can be extend by
basis vectors to $ A\cap B$, $ A\cap C$ and $ B\cap C$, and for A, B, C.
Using these basis and your assumption $ A\cup B\cup C$ is a subspace, you can decide
your question, hopefully.

[quote="Kent Merryfield"]To rule out jmerry's counterexample, add the hypothesis that the field over which these are vector spaces is infinite. Or maybe for this particular specific problem, the field has at least $ m$ elements for some $ m$ to be determined later. (I suspect $ m \equal{} 3$ is enough here.)

I remember a problem posted here once about the union of finitely many proper subspaces not being the whole vector space; so far I haven't found the right thing to search for.[/quote]
But here there is no condition that $ A,B,C$ subspaces.
So I think the problem will still be wrong over any field.
Take $ A \equal{} V \minus{} \{x\}$,$ B \equal{} \{x\}$,$ C$ is the null.where $ x$ is a certain vector.

If we add the condition that $ A,B,C$ subspaces.I think that is what kent said :)

[quote="Kent Merryfield"]To rule out jmerry's counterexample, add the hypothesis that the field over which these are vector spaces is infinite. Or maybe for this particular specific problem, the field has at least $ m$ elements for some $ m$ to be determined later. (I suspect $ m \equal{} 3$ is enough here.)

I remember a problem posted here once about the union of finitely many proper subspaces not being the whole vector space; so far I haven't found the right thing to search for.[/quote]

You might have to search for linear functionals to have any luck:

http://www.mathlinks.ro/Forum/viewtopic.php?p=917042#917042


:rotfl: has a jar with 8 protozoans. 6 are green, 2 are blue. He randomly draws protozoans without replacement. Every time he gets a green protozoan, he will put it in box 1. If he gets a blue protozoan, he will put it in box 2. If he draws a blue protozoan after a green protozoan, he will put the blue protozoan in box 1, put boxes 1 and 2 aside, and repeat the process with boxes 3 and 4. This process continues until  :rotfl: runs out of protozoans. What is the probability that there are 2 blue protozoans in box 1 when he finishes?

[quote="easyas3.14159..."]:rotfl: has a jar with 8 protozoans. 6 are green, 2 are blue. He randomly draws a marbles without replacement. Every time he gets a green marble, he will put it in box 1. If he gets a blue marble, he will put it in box 2. If he draws a blue marble after a green marble, he will put the blue marble in box 1, put boxes 1 and 2 aside, and repeat the process with boxes 3 and 4. This process continues until  :rotfl: runs out of marbles. What is the probability that there are 2 blue marbles in box 1 when he finishes?[/quote]
[hide]6/8*5/7*2/6*4/5*3/4*1/3=1/28[/hide]

This looks familiar   :)

Well, that's not exactly right...you might want to read the fifth sentence again. By the way, sorry for putting in marbles instead of protozoans (now edited).


A Roman boy was given this puzzle:
Take one hundred one, and to it affix the half of a dozen, or if you please six; Put fifty to this, and then you will see what every good boy to others should be.

[hide]
Roman numerals:

CI + VI + L = CIVIL

...and Arabic:

101 + 6 + 50 = 157

Take your pick.  :)
[/hide]


Prove that there are infinite positive integers n such that:

$ \frac {5^{n \minus{} 2} \minus{} 1}{n}$

Is a integer.

(Detalhed your answer)

Take $ n=2p$, for a prime $ p \neq 5$. You'll get $ \frac{5^{2(p-1)}-1}{2p}$.
Obviously, ${ 5^{2(p-1)}-1}$ is even and multiple of $ p$, by Fermat Little Theorem.

[quote="luiseduardo"]Anothers answers ?  :maybe:[/quote]

Why asking ? Raf92's answer is simple, quite Ok, and prove the required statement.


Alguien por ahi, no sabe alguna pagina donde esten colgados los resultados generales de la Olimpiada de Mayo 2009???

Gracias

o si no ( si es que se puede) no podrian colgar los resultados aca mismo????

Bueno no creo que sirva de mucho pero esos resultados deberían de estar en la página de la olimpiada argentina:

[url]http://www.oma.org.ar/internacional/may.htm[/url]

Aunque aparentemente todavía no se han actualizado los resultados de este año...

La página ha sido actualizada. Los resultados están dados [url=http://www.oma.org.ar/internacional/resultados_generales-may15.htm]aquí[/url]


I was going to square the entire expression, but I don't think that's the solution.  :huh:

NO.... [hide]what is inside that counts[/hide] :D

[hide="hint"]
$\sqrt{a+b\pm 2\sqrt{ab}}=\sqrt{(\sqrt{a}\pm \sqrt{b})^2}=\sqrt{a}\pm \sqrt{b}$
[/hide]

With a calculator (you know I like cheap things  :D ), it's 828, but why?

oh! you spoilt the fun, doesnt matter though someone had to get it. :D

well I think frt has explained why 236factorial

I'm looking, I'm looking,
*trying to comprehend*  :D

Well, now I finally got it, I didn't think about splitting up 52 into 43 and 9,  :blush:

sometimes these type of problems appear really hard when they have an easy solution.


linear algebra
matrix
algebra
polynomial
linear algebra unsolved

Notation: Let $ P\equal{}\begin{bmatrix}0&1&0&0&1\\1&0&1&0&0\\0&1&0&1&0\\0&0&1&0&1\\1&0&0&1&0\end{bmatrix}$ and $ Q\equal{}\begin{bmatrix}0&0&1&1&0\\0&0&0&1&1\\1&0&0&0&1\\1&1&0&0&0\\0&1&1&0&0\end{bmatrix}.$

(That is, a pentagon and a pentagram.)

The Petersen graph ([url=http://www.artofproblemsolving.com/Forum/profile.php?mode=viewprofile&u=2975]illustration[/url]) has as its matrix
\[ A\equal{}\begin{pmatrix}P&I\\I&Q\end{pmatrix}.\]
Note that this is a $ 10\times 10$ matrix.

Find, by hand, the minimal polynomial and the characteristic polynomial (and hence the equivalent diagonal form) for $ A.$

A hint:

Compute $ A^2\plus{}A.$

Can you use that as a starting point for finding a polynomial equation that $ A$ satisfies?

Along the way, you probably also need to know the values of $ P^2,Q^2,PQ,$ and $ P\plus{}Q.$

Let $ J$ be a $ 5\times 5$ matrix whose all entries equal $ 1$. This matrix is diagonalizable and its eigenvalues are $ 5$ with multiplicity $ 1$ and $ 0$ with multiplicity $ 4$, giving $ \det\left(J\minus{}\lambda I\right)\equal{}\lambda^{4}\left(5\minus{}\lambda\right)$. It is easy to see that $ PQ\equal{}J\minus{}I\equal{}QP$ (in fact, it is enough to check the first equality, since $ P$, $ Q$ are circulant and thus commute with each other) and $ P\plus{}Q\equal{}J\minus{}I$. Therefore
$ \det\left(A\minus{}\lambda I\right)\equal{}\det\begin{pmatrix}P\minus{}\lambda I & I\\ I & Q\minus{}\lambda I\end{pmatrix}\equal{}\minus{}\det\begin{pmatrix} I & Q\minus{}\lambda I\\ P\minus{}\lambda I & I\end{pmatrix}\equal{}$
$ \equal{}\minus{}\det\begin{pmatrix} I & Q\minus{}\lambda I\\ 0 & I\minus{}\left(P\minus{}\lambda I\right)\left(Q\minus{}\lambda I\right)\end{pmatrix}\equal{} \det\left(\left(P\minus{}\lambda I\right)\left(Q\minus{}\lambda I\right)\minus{}I\right)\equal{}$
$ \equal{}\det\left(PQ\minus{}\lambda\left(P\plus{}Q\right)\plus{}\lambda^{2}I\minus{}I\right)\equal{} \det\left(\left(1\minus{}\lambda\right)\left(J\minus{}I\right)\minus{}\left(1\minus{}\lambda^{2})I\right)\equal{}\left(1\minus{}\lambda\right)^{5}\det\left(J\minus{}I\minus{}\left(1\plus{}\lambda\right)I\right)\equal{}$
$ \equal{}\left(1\minus{}\lambda\right)^{5}\det\left(J\minus{}\left(2\plus{}\lambda\right)I\right)\equal{}\left(1\minus{}\lambda\right)^{5} \left(2\plus{}\lambda\right)^{4} \left(3\minus{}\lambda\right)$.

That is correct.

Another approach: From $ P^2 \equal{} Q \plus{} 2I,Q \equal{} P \plus{} 2I,PQ \equal{} P \plus{} Q \equal{} J \minus{} I,$ we can compute that

$ A^2 \equal{} \begin{pmatrix}Q \plus{} 3I & J \minus{} I \\
J \minus{} I & P \plus{} 2I\end{pmatrix}$ from which $ A^2 \plus{} A \equal{} \begin{pmatrix}J \plus{} 2I & J \\
J & J \plus{} 2I\end{pmatrix} \equal{} J \plus{} 2I.$

Hence $ A^2 \plus{} A \minus{} 2I \equal{} J.$ Multiply by $ A$ and use $ AJ \equal{} JA \equal{} 3J$ to get $ A^3 \plus{} 2A^2 \minus{} 2A \equal{} 3J.$ Eliminate $ J$ between those two equations to get $ A^3 \minus{} 2A^2 \minus{} 5A \plus{} 6I \equal{} 0.$

So a polynomial (which turns out to be the minimal polynomial) for $ A$ is $ x^3 \minus{} 2x^2 \minus{} 5x \plus{} 6,$ which factors as $ (x \minus{} 3)(x \minus{} 1)(x \plus{} 2).$

Hence the only possible eigenvalues are $ 3,1,$ and $ \minus{} 2.$ What remains is to compute the multiplicities. Since the sum, with multiplicities, of the eigenvalues must be the trace and the trace is zero, we only need one multiplicity to determine the others. Hence it would suffice to show that $ 3,$ the dominant eigenvalue, has multiplicity one. (This is true in greater generality than just this problem.) Alternatively, we can row eliminate as Rzeszut did to get the multiplicities of $ 1$ and $ \minus{} 2,$ which are $ 5$ and $ 4,$ respectively.


calculus
integration
trigonometry
function
real analysis
real analysis unsolved

$ \int^{\frac {16n^2}{\pi}}_0 \cos \frac {\pi}{2} \left[\frac {x\pi}{n}\right]dx \equal{} ?$

The integral is from $ 0$ to $ \frac {16n^2}{\pi}$ (for I feel that is not visible)

Rajiv,

kindly tell me are u using $ [.]$ for greatest integer.
thank u

is the integral function included in the cos function? i had some confusion regarding this

[quote="kabi"]
kindly tell me are u using $ [.]$ for greatest integer.
[/quote]

No information provided in the question $ \to 1^{st}$ confusion!

[quote="vivek_n"]
is the integral function included in the cos function?

[/quote]
$ \to 2^{nd}$ confusion!

If it is just a bracket then its a simple substitution question.


geometry
power of a point
radical axis
geometry proposed

Let $ ABC$ be an acute triangle with $ \angle BAC\equal{}60^{\circ}$. Let $ P$ be a point in triangle $ ABC$ with $ \angle APB\equal{}\angle BPC\equal{}\angle CPA\equal{}120^{\circ}$. The foots of perpendicular from $ P$ to $ BC,CA,AB$ are $ X,Y,Z$, respectively. Let $ M$ be the midpoint of $ YZ$.
a) Prove that $ \angle YXZ\equal{}60^{\circ}$
b) Prove that $ X,P,M$ are collinear.

Construct equilateral triangle ABC’ , ACB’
Then we have A,P,B,C’  A,P,C,B’ are concyclic
∠YXZ=∠AC’P+∠AB’P=180-120=60
Denote N=XP∩YZ
ZN/NY=XZsin∠ZXP/XY sin∠YXP  =XZ sin∠PAC/XY sin∠PAB= sin∠PAB/ sin∠120 *
sin∠120/ sin∠PAC * sin∠PAC/ sin∠PAB=1
N=N
Done!

For the second part, can be solved by noticing that $M$ lies on radical axis of 2 circles that intersect in $X$ and $P$.

($MY$ and $MZ$ is tangent to $2$ circles)

hi, can you write more clear on which radical axis(please, write the radius and the center of that circle)


I have heard people say that there will be a Florida site for ARML this year, but I haven't found anything on the ARML site on it. Will there be? Who will go?

A while back, someone wrote in the Wikipedia page about ARML that there would be a fourth site.  I don't know who it was, though; their username is "rjyanco".

[url=http://en.wikipedia.org/wiki/American_Regions_Mathematics_League]ARML Wikipedia page[/url]

[quote="joml88"]There hasn't been any [b]official[/b] word about a Florida ARML site being established yet.   My guess is that the people in charge are still evaluating whether or not opening up the new site will be worth it.[/quote]

I believe this is correct.  I'll announce something here when I hear something.

It is official that the West Coast site will be in Las Vegas.

[quote="Sly Si"]A while back, someone wrote in the Wikipedia page about ARML that there would be a fourth site.  I don't know who it was, though; their username is "rjyanco".
[/quote]

rjyanco is an ARML coach with the Western Mass team.  There was enough talk at Penn State last year about opening a 4th site that it would be easy for someone to have gotten the impression that it was a definite for either 2006 or 2007.

Well, someone probably should lobby for a southeast site for ARML for 2006, which will facilitate more participation. Ideal sites may include Orlando, FL and Atlanta, GA.

Hi sarah-marie.

Actually, the information I had about the fourth site came after Penn State.  While I was [i]at[/i] Penn State I was too busy keeping track of 30 students to hear a single thing about Florida.  But I jumped the gun:  the newest I've heard is that there's no fourth site for 2006, but probably one in 2007.

(The existence of the fourth site affects the number of coaches we'll be bringing to PSU, which is why I've heard things from Bryan.)

Rick

P.S. Did you see the pictures of you and Tom at http://users.wpi.edu/~avolfson/critics/0506ARML.html?  Search the text for HCSSiM and you'll find them.  The webpage isn't incredidbly accurate, but I enjoyed it.

[quote="rjyanco"]Hi sarah-marie.

...

P.S. Did you see the pictures of you and Tom at http://users.wpi.edu/~avolfson/critics/0506ARML.html ?  Search the text for HCSSiM and you'll find them.  The webpage isn't incredidbly accurate, but I enjoyed it.[/quote]

The webpage isn't incredidbly [i]what [/i]  ???

[quote="avolfson"]The webpage isn't incredidbly [i]what [/i]  ???[/quote]
You say the HCSSiM talk was about transformations two years in a row, which it wasn't.  I'm referred to in the plural at least twice.  You spell Hershey "Hersey" and "Heresy."  Cary's name is misspelled throughout.

Look, it's a cute page.  I sent the URL to some of my students and posted it here, because it gives a good feel for the trip.  But I wouldn't publish it in Newsweek.

At this point, it looks like there will not be a Florida site in 2006, though ARML is still looking into Florida for future years.


Circle $ T$ has a circumference of $ 12\pi$ inches, and segment $ XY$ is a diameter. If the measure of angle $ TXZ$ is $ 60^{\circ}$, what is the length, in inches, of segment $ XZ$?

[asy]draw(Circle((0,0),13),linewidth(1));
draw((-12,-5)--(-5,-12)--(12,5)--cycle,linewidth(1));
dot((0,0));

label("T",(0,0),N);
label("X",(-12,-5),W);
label("Z",(-5,-12),S);
label("Y",(12,5),E);[/asy]

Any angle inscribed in a semicircle is a right angle. Thus, $ \triangle{XYZ}$ is a $ 30-60-90$ special right triangle, which means that $ XY=2(XZ)$. Since $ C=\pid$, the diameter of a circle with a circumference of $ 12\pi$ inches is $ 12$, and the diameter in this case just happens to be $ XY$. Therefore, $ XZ$ is half of that, or $ \fbox{6}$.

Alternatively, we could note that TX and TZ are both radii. That means $ \triangle TXZ$ is an isosceles triangles, so the base angles are equal. However, since all three angles are $ 60^\circ$, it's equilateral, so XZ = TX = TZ = $ \boxed{6}$.

the triangle inside is a 30-60-90 triangle so 2xz=diameter so all we need to do is to find the diameter. the circumferance is 12pi and the equation for that is 2piR so the radius is 6 and the diameter is 12 so divide that by 2 and get the answer [size=50][hide="..."]lol you could see this never mind the answer is 6 [/hide][/quote][/size]

[size=200][color=#40FFFF][b][i][u]PLEASE RATE MY POST IF THIS HELPED YOU THANK YOU SO MUCH IT IS GREATLY APPRECIATED!!![/u][/i][/b][/color][/size]

[size=50][color=#0040FF][b][i][u]I HAVE 100% CONFIDENCE THAT IM CORRECT SO PLEASE POST ME IF IM WRONG! THANK YOU SO MUCH IT IS GREATLY APPRECIATED !![/u][/i][/b][/color][/size]


This thread isn't a problem for you to solve, this is a genuine question about something I'm not understanding... sorry  :blush:

I have a question from a course I'm taking, and something doesn't make sense. I'll do a problem here and show you what I don't understand and I'd really really appreciate it if someone could fill me in. Thanks in advance :D

Say we have a company who produces computer chips. They claim that out of every batch, 50 chips will work. with a standard deviation of 3 (this is following a normal, bell-curve distribution). They are willing to be wrong 5% of the time. You take a sample test of 9 batches, and you find that the average of all the batches is actually only 48 working chips. Is the company lying?

Okay, here's how I do it. The null hypothesis, $\displaystyle N_{0}: \mu=50,\alpha=0.05$, and my alternative hypothesis is that $\displaystyle N_{1}: \mu<50$. Now to prove my hypothesis.
First I must find the standard deviation of my test case.
\[\displaystyle \bar{x}=46, n=30\]\[\displaystyle \sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}} = \frac{3}{\sqrt{9}}=1\]
Now that we have all our information, and can find the Z-score of our test case.
\[\displaystyle Z_{t}=\frac{\bar{x}-\mu}{\sigma}\]\[\displaystyle Z_{t}=\frac{48-50}{1}=2\]
This means that the chance of our test case occuring is 2.67 standard deviations away from the median on a bell curve. If we calculate the area under the curve for $\displaystyle P(x<2)$, we get 0.978. \[\displaystyle 1-0.978=0.022\]\[\displaystyle 0.022<0.05\]Therefore, because the probability of my test case occuring is less than $\displaystyle\alpha$, this means that it is enough to raise suspicion, and $\displaystyle H_{0}$ has been rejected. Okay, now to me, this makes no sense. What if the company had said that they're only willing to be wrong [b]1%[/b] of the time, instead of 5%? Then apparently my test case would be within the allowable range and $\displaystyle H_{0}$ would NOT have been rejected! HUH? If the null hypothesis is MORE strict and MORE confident in their findings, and your test case OPPOSES it, it is HARDER to disprove them? Why don't all companies claim that their significance level is 0.000000001% and then no matter WHAT your test case is, it'll NEVER disprove the null hypothesis! This seems to be backwards. Can someone please explain?  :blush:

[quote="Bictor717"]The question is faulty. It doesn't make sense to give the alpha value in the way the question describes. Alpha is the rejection value. You decide which alpha would make your test statistic significant.[/quote]
Can you please explain? I've been suspecting that my definition of $\displaystyle\alpha$ has been wrong all along, that's why I made this thread. His way of looking at it simply doesn't add up. Can you give a more detailed description of what $\displaystyle\alpha$ is?

[quote="Bictor717"]When you say $\alpha=.05$, you're saying that if your test statistic comes out to be less than .05, you would consider that statistically significant evidence against the null hypothesis. If the test statistic is greater than the $\alpha$, the test tells you that your sample mean's difference from the population mean can be attributed to sample variation, and that there is not enough evidence to reject $H_0$.[/quote]
Then how would this $\alpha$ value be determined by whoever made the null hypothesis? Is it arbitrary or based on some sort of empirical evidence? The problem to me seems that "Hypothesis testing" is mainly used to detect dishonesty in statistics, but the measure by which we determine whether the statistic is honest ($\alpha$) is provided by the same people who gave you the statistic in the first place! This seems to me to be like paying a robber to guard your house. Why does the "company" not simply set the alpha value to 0.000001 and then ANY error in any set of test statistics will be attributed to "sample variation"! Therefore, there MUST be some sort of way to argue/determine a correct $\alpha$ value...am I right?

$\alpha$ is an arbitrary number chosen by whoever is doing the experiment.  The lower the value it is, the less likely you are to wrongly reject it.  So if you make it really, really small you are just being extremely cautious that you won't make the mistake of rejecting the null when it is actually true.

Statistical tests have nothing to do with being fair/unfair.  They are just there to help you make the best decision you possibly can based on the facts at hand.

The alpha value is decided by the test-maker. Usually when making the test, you have a specific purpose in mind. You set the alpha value based on what you need. Would 10% be sufficient or do you want to be more cautious and use 1%?
Hypothesis testing is not a test for honesty. You use it to see if a proposed distribution/mean/proportion for the variable can be the actual distribution/mean/proportion.

[quote="Bictor717"]Would 10% be sufficient or do you want to be more cautious and use 1%?[/quote]
But how is 1% more cautious? 1% = 0.01 . Your test cases would have to be MORE distant from your null hypothesis in order to be actually wrong. Let's say you were doing a medical experiment. You predict that this operation procedure will only be fatal to 1 out of every 100 people you perform it on, with a standard variance of 2. I totally made these numbers up for the sake of argument. You want to be very certain you're not going to be wrong here, so you set an alpha of 0.1%, which means 0.001 . Now, just imagine how far away from your mean you would have to be in order to reject this $\displaystyle H_{0}$! The likelihood of the z-score of your test case would have to be 0.00009 or less! That is certainly never going to happen so you will NEVER disprove your $H_{0}$! Now imagine you set an alpha of 10%, now the probability of that SAME TEST CASE cannot be below 0.1 -- it is MUCH MORE difficult for your test case to be viewed as "normal" -- it must be CLOSER to your mean! So I don't see why you're saying "be more cautious and use 1%"... because an alpha value of 0.01 allows you to stray farther away from your mean than a 10% alpha value does!

The point of hypothesis testing is that you [b]want[/b] to disprove the null hypothesis. By being cautious, I mean "Not rejecting the null hypothesis if it is in fact true". If your test value is closer to the mean than the alpha value, you cannot make any conclusion. [i]There is neither evidence against nor for the null hypothesis.[/i]
In the example you just gave, if I pick alpha=.01, then it would take a sample of, say, 20 out of 100 cases to be fatal, in order to reject the null hypothesis. But if the actual proportion of fatal cases is 10 out of 100, I would not reject the null hypothesis even if I get something like 12 out of 100. It would be dangerous to perform the procedure because it is actually more risky than I say it is. However, if I pick alpha=.05, I would reject the null hypothesis that it's fatal for 1 out of 100 and conclude that the actual proportion is higher.
On the other hand, if I think 40 out of 100 cases are fatal (this is the null hypothesis), and I pick alpha=.10, I would reject it if I get, say, 25 out of 100 cases to be fatal. If the actual proportion were 40 out of 100 and I reject that, telling patients that it fatal to only like 30 out of 100, I would be making it sound less risky than it actually is. That is a mistake that no doctor would want to make. So to be more cautious, lower alpha to .01. At that level, I would not reject the null hypothesis, even if my sample gives 25 out of 100 to be fatal.
These numbers were made up and possibly exaggerated to demonstrate a point.


If we have a wooden door with a metallic door knob,which would have a lower temperature,the knob or the door(In normal situation) and why.

Well, here's my guess. I say they would each have the same temperature because they should be in thermal equilibrium. The reason the doorknob feels colder to the touch is that it more readily conducts heat from the body, so you lose heat more quickly than if you were to touch the wooden door.


Does anyone know the relationship between the sector area of a paper circle and the volume of the cone formed from the paper  by doing the wrapping ??

Thanks,
Khem

That depends on the radius of the circle.

sorry, but what do you mean by "paper doing the wrapping"?

If its plainfully simeple to you, please explain. I'm terrible at reading correctly come math problems

Here is the problem :
From a circular piece of paper, a sector ABC is removed from the circle by cutting it off. Where B is the center of the circle and A and C are points on the circumference of the circle. Edge BC is then joined to edge BA (without overlap) to form a cone of radius 12 cm and of volume 432 pi cubic centimeters. What is the number of degrees in the measure of angle ABC of the sector that is not used ?

Thanks.

[quote]From a circular piece of paper, a sector ABC is removed from the circle by cutting it off. Where B is the center of the circle and A and C are points on the circumference of the circle. Edge BC is then joined to edge BA (without overlap) to form a cone of radius 12 cm and of volume 432 pi cubic centimeters. What is the number of degrees in the measure of angle ABC of the sector that is not used ?[/quote]

 The volume of a cone is 1/3 of the cylinder of the same height and width, or $\frac{1}{3} r^2 \pi h$. We know the radius, 12, and the volume, so we can use this equation to find the height of the cone: $\frac{1}{3} 144\cdot \pi h=432\pi \Rightarrow 144h\pi =1296\pi \Rightarrow h=3/2$. We know from the top of the cone to the bottom edge is the radius of the large circle, so the radius of the large circle is $\sqrt{146.25}\Rightarrow \sqrt{\frac{585}{4}}\Rightarrow \frac{\sqrt{585}}{2}$ What we want is the number of degrees in the cut-off portion, or the difference of the circumferences of the 2 circles. The circumference of the large one is $\sqrt{585}\pi $, and the circumference of the large one is $24\pi $, so the difference is $( \sqrt{585}-24) \pi$. This is $\frac{(\sqrt{585} - 24)\pi }{\sqrt{585}\pi} \rightarrow \frac{\sqrt{585}-24 }{\sqrt{585}}$. Multiply this by 360 to get the number of degrees(with a calculator :) ):2.78. That doesn’t look right, but what did I do wrong???

Billy

[quote="solafidefarms"]

 The volume of a cone is 1/3 of the cylinder of the same height and width, or $\frac{1}{3} r^2 \pi h$. We know the radius, 12, and the volume, so we can use this equation to find the height of the cone: $\frac{1}{3} 144\cdot \pi h=432\pi \Rightarrow 144h\pi =1296\pi \Rightarrow h=3/2$.
Billy[/quote]

Your calculation of h is wrong.  h=9

Surface area of the cone = 324pi

Slant height of cone  = radius of circle = 15

Area of the portion of circle used to make the cone = 324pi - 144 pi = 180pi

Fraction of circle used to make the cone = 180pi/225pi = 4/5

Angle of sector not used = pi/5 = 72

Naga

This was a problem on last year's national sprint test, I believe. The part of the circle that makes the cone becomes the lateral area, which has the formula 1/2*circumference*slant height, right?


limit
trigonometry
algebra
binomial theorem
calculus
calculus computations

[quote="HFL"]Prove
$ \lim\limits_{x\to0}{\frac {\sqrt {1 - 2x}.\sqrt [3]{1 - 3x}.\sqrt [n]{1 - nx} - 1}{\sin x}}$[/quote]

I think that is what he wants to find out. There is a typo in his tex commands.

By the extended binomial theorem, $ (1 - nx)^{1/n} \approx (1 - x)$ for $ x$ very close to zero. Hence, your limit is

\[ \lim\limits_{x\to0}{\frac {\sqrt {1 - 2x}\sqrt [3]{1 - 3x}\sqrt [n]{1 - nx} - 1}{\sin x}} = \lim_{x\to 0} \frac {(1 - x)^3 - 1}{\sin(x)} = - 3.\]

I think it's supposed to be: [quote="HFL"]Find
$ \lim_{x\to0}{\frac {(\sqrt {1 - 2x} \cdot \sqrt [3]{1 - 3x} \cdots \sqrt [n]{1 - nx}) - 1}{\sin x}}$[/quote]

In which case, by Jimmy's suggested method, it would be
$ \lim_{x\to0}{\frac {(\sqrt {1 - 2x} \cdot \sqrt [3]{1 - 3x} \cdots \sqrt [n]{1 - nx}) - 1}{\sin x}}$

$ =\lim_{x\to0}{\frac {((1-x+O(x^2))^n - 1}{x-O(x^3)}}$

$ =\lim_{x\to0}{\frac{(1-nx+O(x^3)) - 1}{x-O(x^3)}}$

$ = \lim_{x\to0}{\frac{-n + O(x^2)}{1-O(x^2)}}$

$ = -n$

hi!!
we have :

$ \sqrt{1\minus{}2x}.\sqrt[3]{1\minus{}3x}.\sqrt[n]{1\minus{}nx}\minus{}1\equal{}\frac{(\minus{}6nx^{3}\plus{}(5n\plus{}6)x^{2}\minus{}x(n\plus{}5))}{\sum_{k\equal{}0}^{n}(\sqrt[3n]{(1\minus{}2x)(1\minus{}3x)(1\minus{}nx)})^{3n\minus{}k}}$ so we find:

$ \lim_{x\to0}{\frac{(\sqrt{1\minus{}2x}\cdot\sqrt [3]{1\minus{}3x}\cdots\sqrt [n]{1\minus{}nx})\minus{}1}{\sin x}}\equal{}\frac{\minus{}(n\plus{}5)}{3n}$  :)


calculus
function
inequalities
inequalities unsolved

[quote="Lovasz"]Must it be calculus, arqady?[/quote]
I don't understand. What do you mean? Sorry. :(

He's looking for a solution which does not use calculus. ;)

Just proof that $f(b)=\frac{2a}{b+c}+\frac{2b}{c+a}+\frac{2c}{a+b}$ is a convex function.

Which is [i]exactly[/i] [b]arqady[/b]'s solution...  :roll:

[quote="yookon88"]he's looking for a solution without nigzarot...[/quote]
Wow, yookon88! ? אם אתה יודע עברית  :lol:

I live in Jerusalem. אני לומד בתיכון ליד"ה כיתה י"ב מה אתך?

i have one very ugly solution. let b = a+x, and c = a+x+y, and good luck!  :lol:

[quote="M4RI0"]Could someone explain me why:  $f(b)\leq\max\{f(a),f(c)\}$   :?:[/quote]
Since, $f''(b)>0$ for $b\in[a,c].$
If $f''(x)>0$ for $x\in[a,b]$ then $f$ is convex function on $[a,b].$
Therefore $f$ reaches the maximal value on the borders of $[a,b].$ ;)

[quote="Gibbenergy"][quote="Gibbenergy"][quote="Lovasz"]Let $0<a \le b \le c$, then:
\[ \frac{2a}{b+c}+\frac{2b}{c+a}+\frac{2c}{a+b} \le 3+\frac{(c-a)^2}{a(c+a)}  \][/quote]

   Do u allow to use brutal force?  :D[/quote]
[quote="e.lopes"]i have one very ugly solution. let b = a+x, and c = a+x+y, and good luck! [/quote]

Ok!  Your inequality is equivelent to:
       $2a[a(a+b)(a+c)+b(b+c)(b+a)+c(c+a)(c+b)] \leq 3a(a+b)(a+c)(b+c) +(c-a)^2(a+b)(b+c)$
<-->$(b-a)(2a^3-2a^2c-abc+ac^2-2b^2a+c^3+bc^2) \geq 0$
Now, let consider $F= 2a^3-2a^2c-abc+ac^2-2ab^2+c^3+bc^2$
Put $b=a+x$ , $c=a+x+y$ then [b]expand[/b]
  $F=11axy+4a^2y+4ax^2+5ay^2+5x^2y+4xy^2+2x^3+y^3 \geq 0$ :(
It isn't nice at all. I am sure arqardy's solution is more practical. :ninja:

   That is brutal force. :lol:[/quote]

According to Gibb's solution, brutal force is similar to [i]Ravi substitution[/i] ?!?! :what?:


is this solvable without using induction?

Prove:

$(3+\sqrt5)^{n}+(3-\sqrt5)^{n}$ divisible by $2^{n}$

It is easy to verify that the integer sequence $\{a_{n}\}_{n \in{\bf N}}$ defined by

$a_{n}= 3a_{n-1}-a_{n-2}, \; n \geq 3,$

where $a_{1}= 3$ and $a_{2}= 7,$ is given by the formula

$a_{n}={\textstyle (\frac{3+\sqrt{5}}{2})^{n}\:+\: (\frac{3-\sqrt{5}}{2})^{n}}$.

Hence $(3+\sqrt{5})^{n}\:+\: (3-\sqrt{5})^{n}$ is divisible by $2^{n}.$

what you did uses difference equations right? like you estimate that the integer sequence will behave like a geometric sequence so you say that $a_{n}= r^{n}$ and likewise giving you:

$r^{n}= 3r^{n-1}-r^{n-2}\implies r^{2}-3r+1 = 0$ from which you get $r = \frac{3 \pm \sqrt{5}}{2}$

[quote="gotztahbeazn"]like you estimate that the integer sequence will behave like a geometric sequence [/quote]

You can think about it that way if you'd like, but it's annoyingly non-rigorous.  Here's a much better way to think about it:

Suppose we have a sequence $a_{n}= r^{n}+s^{n}$.  Then

$a_{n+1}= r^{n+1}+s^{n+1}= (r+s)(r^{n}+s^{n})-rs(r^{n-1}+s^{n-1})$

Let $A = r+s, B =-rs$.  Then

$a_{n+1}= A a_{n}+B a_{n-1}$

Moreover,

$x^{2}= Ax+B$

Has roots $r, s$.

what about binomial expansion? Or is that just getting unnecessarily complicated?

Very much so.  You can verify that it works with a binomial expansion, if you'd like, since all the irrational terms cancel, but there's so much structure in the terms that it makes more sense to try to generalize and abstract further than that.