Δινω και εγω μια συνδιαστικη επιπεδου Ευκλειδη-Αρχιμηδη που μου αρεσει πολυ.
Δινονται $ n$ ευθυγραμμα τμηματα ετσι ωστε το πρωτο να ειναι μηκους $ 1$ το δευτερο μηκους $ 2$ ... και το $ n$-οστο μηκους $ n$. Nα βρεθει ποσα τριγωνα μπορουν να κατασκευαστουν , χρησιμοποιωντας $ 3$ απο τα παραπανω τμηματα.

Ειναι ενα bijection με τις μεγιστες πλευρες των τριγωνων. Πιο συγκεκριμενα αρκει να βρουμε τα τριγωνα που εχουν ως μεγιστη πλευρα τους το k-οστο τμημα μηκους k. Για το τυχαιο $ 1 \leq k \leq n$, επιλεγουμε 2 διαφορετικα μεταξυ τους τμηματα με μηκος μικροτερο απο k, ευκολα απο 3 τριγωνικες ανισοτητες προκειπτει οτι τα τμηματα αυτα μαζι με το τμημα μηκους k οριζουν ενα τριγωνο αν και μονο αν το αθροισμα τους ειναι μεγαλυτερο απο k. Αν το μεγιστο απο τα 2 τμηματα που επιλεγουμε εχει μηκος k-m, θα σχηματιζεται τριγωνο αν και μονον αν το αλλο εχει μηκος απ το συνολο {m+1, ... , k - m - 1} το οποιο εχει k - 2m - 2 στοιχεια. Ο m παιρνει ολες τις τιμες μεγαλυτερες του [k/2], αρα για καθε k εχουμε:
$ \sum_{m \equal{} [\frac {k}{2}] \plus{} 1}^{k \minus{} 1}{k \minus{} 2m \minus{} 2} \equal{} (k \minus{} [\frac {k}{2}])(k \minus{} 2) \minus{} (k \minus{} [\frac {k}{2}] \minus{} 2)(k \minus{} [\frac {k}{2}] \minus{} 1) \equal{} (k \minus{} [\frac {k}{2}])([\frac {k}{2}] \plus{} 1) \minus{} 2$
τριγωνα με μεγιστη πλευρα το τμημα μηκους k

Αρα το συνολικο πληθος ειναι:

$ \sum_{k \equal{} 1}^{n}{(k \minus{} [\frac {k}{2}])([\frac {k}{2}] \plus{} 1) \minus{} 2}$ το οποιο μπορει να γραφτει σε κλειστο τυπο αλλα θελει λιγη δουλεια.

PS: Η λυση ηταν λιγο στο ποδι οποτε μπορει να υπαρχει καποιο λαθος στις πραξεις, αλλα ο συλλογισμος ειναι σωστος.

Για να βρούμε πόσα έχουν μέγιστη πλευρά κ θέλουμε να βρούμε όλα τα ζευγάρια ακεραίων $ (a,b)$ με $ 1 \leqslant a < b \leqslant k\minus{}1$ και $ a\plus{}b > k$. Παρατηρούμε ότι αν $ a\plus{}b < k$ τότε για το ζευγάρι $ (k\minus{}b,k\minus{}a)$ έχουμε $ (k\minus{}a) \plus{} (k\minus{}b) > k$ και αντίστροφα. Επειδή ακριβώς $ [(k\minus{}1)/2]$ έχουν $ a\plus{}b\equal{}k$ τότε ο αριθμός των ζευγαριών που θέλουμε ισούται με $ \frac{1}{2} \left(\binom{k\minus{}1}{2} \minus{} \lfloor \frac{k\minus{}1}{2} \rfloor \right)$.

Άρα ο αριθμός των τριγώνων ισούται με $ \frac{1}{2} \sum\limits_{k\equal{}3}^n \left(\binom{k\minus{}1}{2} \minus{} \lfloor \frac{k\minus{}1}{2} \rfloor \right)$

Μετά θέλει πράξεις. Νομίζω είναι πιο βολικό να χωριστεί πρώτα σε περιπτώσεις αν ο $ n$ είναι άρτιος ή περιττός.

Για να βρεθεί το άθροισμα βοηθάει να παρατηρήσουμε ότι $ \sum\limits_{k\equal{}3}^n \binom{k\minus{}1}{2} \equal{} \binom{n}{3}$. (Γιατί ισχύει αυτό; )

Eτσι πολυ ωραιος Δημητρη.

Η ταυτοτητα που λες βγαινει ευκολα με το τριγωνο του Pascal.

Πιστευω για τις πραξεις που εχεις καταληξει βολευει ακομα περισσοτερο να δουμε τι γινεται αν ο k ειναι αρτιος ή περιττος καθως στην πρωτη περιπτωση εχουμε $ (m \minus{} 1)^2$ τριγωνα και στην δευτερη $ m(m \minus{} 1)$ οπου $ m \equal{} [\frac {k}{2}]$.

Λοιπον ετσι περιπου την ελυσα και εγω οταν την ειδα και μετα παω να δω τι λυση ειχε στο βιβλιο που την βρηκα και περα απο αυτην δινει μια λυση με ακεραια σημεια που ειναι καταπληκτικη και πολυ απλη!

Συνοπτικα η πολυ ομορφη αυτη λυση ειναι η εξης,
Αν θεωρησουμε το τετραγωνο πλευρας k τοτε ουσιαστικα αρκει να μετρησουμε ποσα ακεραια σημεια υπαρχουν μεσα στο πανω τριγωνο που σχηματιζει η κυρια με την δευτευουσα διαγωνιο! qed!

[quote="Athinaios"]nick.. ειναι αργα και συγνωμη αν βλεπω λαθος αλλα αν ο $ m$ παρει τιμες πανω απο το $ [\frac {k}{2}]$ τοτε αντιθετα με την τριγωνικη οι δυο πλευρες εχουν αθροισμα κατω απο $ k$..[/quote]

Σωστος, εγραψα λαθος τους δεικτες στο αθροισμα, θα παει απο 1 μεχρι [κ/2]....

Oντως λογικα φτιαχνει η προσεγγιση με καποιες ομως μικροδιορθωσεις.  :)
Πιο συγκεκριμενα το συνολο που εχεις εχει $ k \minus{} 2m \minus{} 1$ στοιχεια οχι $ k \minus{} 2m \minus{} 2$.
Kαι το αθροισμα θα πηγαινει λογικα απο το $ 1$ μεχρι το $ [\frac {k}{2}] \minus{} 1$ καθως αν παρει το $ m$ το $ [\frac {k}{2}]$ εχεις προβλημα.  :wink:

[quote="Athinaios"]
Η ταυτοτητα που λες βγαινει ευκολα με το τριγωνο του Pascal.
[/quote]

Ναι. Ένας διαφορετικός τρόπος να το δούμε είναι να παρατηρήσουμε ότι υπάρχουν ακριβώς $ \binom{k\minus{}1}{2}$ τρόποι να διαλέξεις τρία στοιχεία του $ \{1,2,\ldots,n\}$ με το μεγαλύτερο να είναι $ k$. (Ακριβώς έτσι προέκυψε αυτό το άθροισμα όταν κοιτάζαμε τα τρίγωνα με μεγαλύτερη πλευρά το κ.)

[quote="Athinaios"]
Πιστευω για τις πραξεις που εχεις καταληξει βολευει ακομα περισσοτερο να δουμε τι γινεται αν ο k ειναι αρτιος ή περιττος καθως στην πρωτη περιπτωση εχουμε $ (m \minus{} 1)^2$ τριγωνα και στην δευτερη $ m(m \minus{} 1)$ οπου $ m \equal{} [\frac {k}{2}]$.
[/quote]

Το άφησα εκεί διότι ήμουν σίγουρος πως αν έκανα τις πράξεις κάπου θα έκανα λάθος.  :D

[quote="Athinaios"]
Λοιπον ετσι περιπου την ελυσα και εγω οταν την ειδα και μετα παω να δω τι λυση ειχε στο βιβλιο που την βρηκα και περα απο αυτην δινει μια λυση με ακεραια σημεια που ειναι καταπληκτικη και πολυ απλη!

Συνοπτικα η πολυ ομορφη αυτη λυση ειναι η εξης,
Αν θεωρησουμε το τετραγωνο πλευρας k τοτε ουσιαστικα αρκει να μετρησουμε ποσα ακεραια σημεια υπαρχουν μεσα στο πανω τριγωνο που σχηματιζει η κυρια με την δευτευουσα διαγωνιο! qed![/quote]

Εντάξει. Η αντιστοίχιση που έδωσα ακριβώς αυτό κάνει. Λέει πως όσα σημεία έχουμε πάνω από την διαγώνιο, άλλα τόσα έχουμε κάτω.

[quote="Demetres"]

[quote="Athinaios"]
Λοιπον ετσι περιπου την ελυσα και εγω οταν την ειδα και μετα παω να δω τι λυση ειχε στο βιβλιο που την βρηκα και περα απο αυτην δινει μια λυση με ακεραια σημεια που ειναι καταπληκτικη και πολυ απλη!

Συνοπτικα η πολυ ομορφη αυτη λυση ειναι η εξης,
Αν θεωρησουμε το τετραγωνο πλευρας k τοτε ουσιαστικα αρκει να μετρησουμε ποσα ακεραια σημεια υπαρχουν μεσα στο πανω τριγωνο που σχηματιζει η κυρια με την δευτευουσα διαγωνιο! qed![/quote]

Εντάξει. Η αντιστοίχιση που έδωσα ακριβώς αυτό κάνει. Λέει πως όσα σημεία έχουμε πάνω από την διαγώνιο, άλλα τόσα έχουμε κάτω.[/quote]

Δεν διαφωνω καθολου! Απλα ειλικρινα πιστευω οτι η αναπαρασταση ενος αποτελεσματος στον χωρο (ή στο επιπεδο) δινει παντα μια καλυτερη και πιο κομψη εποπτεια.


This is a really easy question on 2001 AMC 12. But I got a wrong answer and I kinda think the answer key is wrong. Can anyone please take a look at this?and explain it if u get it? thxxxxxxxxxx

The mean of three numbers is 10 more than the least of the numbers and less than the greatest.The medium of these numbers is 5. What's the sum?
The answer key says 25.
But I got 15. (-5,5,15)
What do u think :?:

You got a bad explanation of the problem. The problem actually states that the mean is 15 less than the greatest number and 10 greater than the smallest number.

[hide="Problem"]
$ \frac{1}{3}((m\minus{}10)\plus{}5\plus{}(m\plus{}15))\equal{}m$

$ \frac{1}{3}(2m\plus{}10)\equal{}m$

$ 2m\plus{}10\equal{}3m$

$ m\equal{}10$
Thus, $ 3m\equal{}30$.
[/hide]

not sure, i obtained the same answer as you.


Find the number of ordered pairs of real numbers (a,b) such that (a + bi)^2002 = a - bi

(i being the   :sqrt: -1 and not another variable of course)

Please include an explanation, if you would be so kind.

There's no need to post this in several different forums -- people will see it wherever you put it if you only post it once.


What is package xy? For some reason, I don't have that package. Is there somewhere that I could get it? (If I actually want it -- it depends what it does :))

You can read a little more about it [url=[url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuidePack.php]Here you go.[/url]]here[/url].

If it's not already installed with your MiKTeX, look at the directions for using MiKTeX to install fancyhdr (on that page linked above) - you may well be able to install xy the same way.


Please help me to solve these questions :

Let $a_1,a_2,...,a_n$ and $b_1,b_2,...b_n$ be real numbers between 1001 and 2002 inclusive. Suppose $\displaystyle \sum^n_{i=1} {a_i}^2 = \sum^n_{i=1} {b_i}^2$
Prove that $\displaystyle \sum^n_{i=1}\frac{{a_i}^3}{b_i} \leq \frac{17}{10} \sum^n_{i=1} {a_i}^2$
Determine when equality holds.

the first version of this problem I saw was proposed in Chinese High School Math Contest in 1998. In that version, $a_i,b_i$ are ranged from 1 to 2.  :lol:

$\frac 12\leq \frac{a_i}{b_i}\leq 2$
$\left(\frac{a_i}{b_i}-\frac{1}{2}\right)\left(\frac{a_i}{b_i}-2\right)\leq 0$
$a_i^2+b_i^2\leq \frac{5}{2}a_i b_i$ (*)
On one hand, we have from (*) that $\sum a_i^2\leq \frac{5}{4}\sum a_ib_i$
On the other hand (from (*) again) , we have $\frac{a_i^3}{b_i}+a_ib_i\leq \frac{5}{2}a_i^2$, thus $\sum\frac{a_i^3}{b_i}+\sum a_ib_i\leq \frac{5}{2}\sum a_i^2$.
Combine them, we obtain that $\sum\frac{a_i^3}{b_i} \leq \frac{17}{10}\sum a_i^2$

The equality holds when $n$ is even, one half of $a_i(b_i)$ are 1001, and the other half are 2002; and $a_i\neq b_i$ for all $i$.


geometry
geometric transformation
dilation
quadratics
function
algebra

$ 1.$ On page 400 in Atkins's Chemical Principles, it explains that for very dilute solutions of weak acids, we must solve for $ x$ in the equation $ x^3 \plus{} K_ax^2 \minus{} (K_w \plus{} K_a[HA]_{initial})x \minus{} K_aK_w \equal{} 0$. How do we know when the solution is dilute enough that we have to use this? Like would we use this for a 0.01 M solution of the weak acid? 0.001? 0.0001? Any general guide lines for when we need to use this cubic?

$ 2.$On page 212 in Atkins, is $ \bigtriangleup U \equal{} \bigtriangleup E$? I see $ \bigtriangleup E$ in past USNCO exams but not $ \bigtriangleup U$. In the book, it says $ \bigtriangleup H \equal{} \bigtriangleup U \plus{} P \bigtriangleup V$

$ 3.$ On page 401 in Atkins, why is $ [H_3O^ \plus{} ]$ really $ [H_3O^ \plus{} ]/c$? That is what the book says. What is c?

1. Only in ideally dilute solutions we can use molar concentrations in place of thermodynamic activities, because the activity coefficients are close to unit. In most problems of general chemistry, you have to solve at most a quadratic equation, because there are several (valid) approximations we can do.

2. U and E are common ways of representing the state function [i]internal energy[/i].

3. That is very likely related to the thermodynamic activity: 1/c is the activity coefficient.

First, thanks for responding to my post :D

1. I googled ideally dilute solution and got: "Dilute solution in which the solute may be regarded as obeying Henry’s law, so that all the solute activity coefficients may be approximated to 1. Physical Chemistry Division, unpublished; see also 1994, 66, 545"

That doesn't tell us much about what low molar concentration or K value is required for us to consider using the cubic equation. How can you tell when the solution is dilute enough?

3. Okay, I'll read about the activity coefficient. So far, the wikipedia explanation seems to be above my head... If anyone has a simple explanation..

[quote="Carcul"]

3. That is very likely related to the thermodynamic activity: 1/c is the activity coefficient.[/quote]

Is it possible that $ c$ stands for standard concentration? I have seen books use that before (and likewise with standard pressure).

Perhaps aufha can post the passage from the book.

I forgot to type in the superscript for c. I still don't know what it is, though:
"Remember that $ [H_3O^ \plus{} ]$ is really $ [H_3O^ \plus{} ]/c^0$.

This was a reminder in the margin near the book's explanation of the cubic equation that I posted above. So I still don't get why, but the $ [H_3O^ \plus{} ]$ in the cubic is really $ [H_3O^ \plus{} ]/c^0$. I actually don't get why $ [H_3O^ \plus{} ]$ is ever $ [H_3O^ \plus{} ]/c^0$ nor do I know what $ c^0$ is.

In that case $ c^{\circ}$ means standard concentration.

Thanks.

Why is H3O+ concentration equal to itself divided by 1 mol dm 3-?
http://goldbook.iupac.org/S05909.html

Do you remember reading that the equilibrium constant is supposed to be unitless and then thinking to yourself how the **** can it be dimensionless when they are using concentration/pressure?

Well the explanation for that is the actual definition of the constant is taking whatever concentration something is and dividing it by the standard concentration and then putting it into the constant expression.


hello, are you sure that this limit exist?
Sonnhard.


function
probability
algebra
domain
real analysis
calculus
integration

Can anyone please take me through the steps to solviing this proof? I would really appreciate the help. Thanks!

E[a.u(x)+b.v(x)]=a.E[u(x)]+b.E[v(x)]

Just to make clear to the other readers - I assume this is related to your previous post about variance. Therefore, $X$ is a random variable, $u$ and $v$ are (nice enough) functions of that random variable, $a$ and $b$ are constants, and $E(\cdot)$ is the expected value.

In other words, you want to prove that the expectation is linear.

The issue of "proof" in this case is wholly dependent on where you're starting from. What do you think the definition of expectation is? (Or, for that matter, the definition of random variable.)

The most general starting place is this:

A probability sample space is a measure space, with the measure of the total space being one. That means that there is a $\sigma$-algebra of measurable sets, and a countably additive measure (conventionally called $P$ for probability) on those sets. A random variable is a (presumably real-valued) measurable function whose domain is the probability sample space.

The expected value of a random variable is the (Lebesgue) integral of that function over the whole probability sample space. Not all measurable functions are integrable, hence not all random variables have expected values.

In that context, your statement is the linearity of the integral, which is one of the first things proved in general integration theory.

Of course, I don't think that where you started from. So what do you have as the definition of expected value?


inequalities
trigonometry
geometry proposed
geometry

Let $ABC$ be a triangle such that $\sum{\sin{2A}}= 4\prod{\sin{A}}$. Prove that:
$4Rr+7r^{2}\geq  5$.

why don't u explain it yourself , mr know-it-all bronze medal guy ? :P

I'm not sure but with only condition of angle we can creat so small triangle (similar the exist triangle for that condition) with $R,r$ is small we will have your inequality is false.

the condition is equivalent with $pr^{2}=1$, so yes the triangle is small, and the inequality holds.

[quote="pohoatza"]Let $ABC$ be a triangle such that $\sum{\sin{2A}}= 4\prod{\sin{A}}$. .[/quote]
Your condition is only for angles is it equivalent to $pr^{1}=1$ ? :?:

Has anyone looked if the problem is correct?The condition of the inequality $\sum \sin{2A}=4\prod \sin{A}$ holds for any triangle.


[quote="hollandman"]Let $ a,b,c > 0$. Prove that

$ \frac {a^2}{a^2 \plus{} (b \plus{} c)^2} \plus{} \frac {b^2}{b^2 \plus{} (c \plus{} a)^2} \plus{} \frac {c^2}{c^2 \plus{} (a \plus{} b)^2}\geq \frac {3}{5}$.[/quote]
$ \sum_{cyc}\frac {a^2}{a^2 \plus{} (b \plus{} c)^2} \equal{}\sum_{cyc}\frac {a^4}{a^4 \plus{} a^2(b \plus{} c)^2}\geq\frac{(a^2\plus{}b^2\plus{}c^2)^2}{\sum(a^4\plus{}2a^2b^2\plus{}2a^2bc)}.$
But $ 5(a^2\plus{}b^2\plus{}c^2)^2\geq3\sum_{cyc}(a^4\plus{}2a^2b^2\plus{}2a^2bc)$ is true even for all reals $ a,$ $ b$ and $ c.$


probability
function
expected value
real analysis
real analysis unsolved

Hi all, here is a question on poisson random variable:

  As we know , if N is a poisson random variable with mean value = p . i.e.  $E[N] = p$  then the probability density function of N is given by:

      $f(n) = e^{-p}*p^n/(n!)$

Now prove that there cannot be a function $g()$ such that
 $E[g(N)] = 1/p$

That doesn't make sense, not without some extra conditions that you haven't told us. We can create any expected value we like. We could just let $g$ be constant: $g(n)=\frac1p.$ Then $E(g(N))=\frac1p.$

Sorry,  forgot to mention that $g(N)$ is not a constant function , but it is a function of $N$

Anyway, again we can easily construct similar example to disprove your statement, can't we? :?

You can easily modify Kent's counterexample so that it isn't a constant either.

Perhaps what you mean is that if g is a function of N (independent of p), then there exists a p such that
\[E_p(g(N)) \neq  \frac{1}{p}.\]

What is really mean is this :

Prove that  there cannot be a function g() , which is not a constant function such that:

$[g(0)*e^{-p} + g(1)*e^{-p}p^2/2! + g(3)*e^{-p}p^3/3! + . . . ]  = 1/p $

[b]symmetricmean wrote:[/b]

[quote]What is really mean is this :

Prove that there cannot be a function g() , which is not a constant function such that:

$[g(0)*e^{-p} + g(1)*e^{-p}p^2/2! + g(3)*e^{-p}p^3/3! + . . . ] = 1/p$
 [/quote]

Here is the proof: it is very simple.  Just take the Limit as p approaches 0 on both sides , you get $g(0)$ =  :inf:  . Hence there cannot be a function g() of N such that $E[g(N)] = 1/p$.

I don't think the proof is complete.  The limit of an infinite sum is not necessarily the sum of the limits.  When you take the limit as p approaches 0 (from above), I agree the right hand side approaches infinity.  Hence the left hand side approaches infinity.  But how do you go from there to $g(0)$ being infinite?  For example, you have to rule out fast-growing functions like $g(n) = n!^2$.

I'm not saying that you can't modify the proof to work, but it does need more argument.


Yay, 2008th G&FF topic!

Post some of your favorite YouTube videos.

[youtube]1t5Mem5ay8s[/youtube]

Cool, I figured out how to post videos! (not that it's the least bit hard)

[youtube]2xHp4otT1rI[/youtube]

lingomaniac88 that video you posted is so cool.  :lol:

[youtube]fBQnqf4ejQM[/youtube].

[hide="Erick Liangs Crazy Asian Mother"]
[youtube]PKnloiM-0Ns[/youtube]
[/hide]

[hide="Master Chief Sucks at Halo 3 (Beta)"]
[youtube]Jz3br1q5wIQ[/youtube]
[/hide]

[hide="Master Chief Sucks at Halo 2"] <-Sequal to one above.
[youtube]0RWQ7R555i4[/youtube]
[/hide]

[hide="Master Chief Sucks at Halo 3"] <-Sequal to one above.
[youtube]LTxO_pgMqys[/youtube]
[/hide]

I don't think that you know it but it is amazing especially at 1:27-1:34

[youtube]2TXPFgRsVUE[/youtube]

I think Elmo Behind the Scenes is pretty funny.

It's the world's FASTEST Rubik's cube solve.

[youtube]aZwE4KKuiws[/youtube]


limit
trigonometry
function
calculus
calculus computations

Evaluate the limit
$ lim_{x\rightarrow0}$=$ \frac{tanx\minus{}x}{x^{2}tanx}$

Using L'Hopital's Rule and Pythagorean identity,

$ \lim_{x\rightarrow 0}\frac{\tan x\minus{}x}{x^{2}\tan x}$

$ \equal{}\lim_{x\rightarrow 0}\frac{\tan ^{2}x}{2x\tan x\plus{}x^{2}\plus{}x^{2}\tan ^{2}x}$

$ \equal{}\lim_{x\rightarrow 0}\frac{1}{2\left( \frac{x}{\tan x}\right) \plus{}\left( \frac{x}{\tan x}\right) ^{2}\plus{}x^{2}}$

$ \equal{}\frac{1}{3}$

Or use Taylor series: $ \tan x \equal{} x \plus{} \frac {x^3}{3} \plus{} O(x^4)$ on top and $ \tan x \equal{} x \plus{} O(x^2)$ on the bottom.

Raghav Grover: use \lim and \tan (and similarly for other common mathematical functions; \max, \sin, etc.).  Note that you have to do \tan x -- \tanx is meaningless in LaTeX.


calculus
integration
logarithms
trigonometry
calculus computations

Evaluate $ \int_{\minus{}1}^{1} \frac{1}{\sqrt{1\plus{}x}\plus{}\sqrt{1\minus{}x}}\ dx$.

That would be 2(Sqrt(2)-Arccoth(Sqrt(2))

indefinite would be

   1/2 (-2 Sqrt(1 - x) + 2 Sqrt(1 + x) + Log(1 - Sqrt(1 + x)) +
   Log(-4 (2 + Sqrt(1 - x) - Sqrt(1 + x))) - Log(-1 + Sqrt(1 + x)) -
   2 Log(1 + Sqrt(1 + x)) + Log(-4 (2 + Sqrt(1 - x) + Sqrt(1 + x))))

lol i forgot to integrate at my last step...

oooo

-Sqrt(2) + pi/2 + arcsinh(1)

anti dervi would be

(Sqrt(1 - x^2) - Sqrt(1 + x^2) + x ArcSin(x) + x ArcSinh(x))/(2 x)

[quote="Stank"][quote="Materazzi"][quote="AndrewTom"]Evaluate $ I \equal{} \int_{ \minus{} 1}^{1} \frac {1}{\sqrt {1 \plus{} x} \plus{} \sqrt {1 \minus{} x}}\ dx$.[/quote]

$ I \equal{} 2\sqrt {2} \plus{} \ln(\sqrt {2} \minus{} 1) \minus{} \ln(1 \plus{} \sqrt {2})$ ;)[/quote]

$ \ln(\sqrt {2} \minus{} 1) \minus{} \ln(1 \plus{} \sqrt {2})$ is equal to 2 (Sqrt(2) - ArcSinh(1))
 :ninja:[/quote]

It's wrong .

hello, why should this be wrong?
Sonnhard.

o wait i meant

2sqrt2 +ln(sqrt2-1)-ln(1+sqrt2)=2 (Sqrt(2) - ArcSinh(1))
 :blush:

hello, this is the same result as mine, $ 2(\sqrt{2}\minus{}arcoth(\sqrt{2}))$
Sonnhard.

As a check, I have the following inelegant method which gives the same answer as Sonnhard et al.

$ I= \int_{-1}^{1} \frac{1}{\sqrt{1+x} + \sqrt{1-x}}\ dx$

Let $ x=\cos 2\theta = \cos^{2}\theta - \sin^{2}\theta$

Then $ I= \sqrt{2}\int_{0}^{\frac{\pi}{2}} \frac{\sin 2\theta}{\sin \theta + \cos \theta}\ d\theta$

Let $ \phi = \theta + \frac{\pi}{4}$

Then $ I=-\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{\cos 2\phi}{\sin \phi} \ d\phi$

$ =\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{2\sin^{2}\phi -1}{\sin\phi}\ d\phi$

$ =\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} 2\sin \phi - \csc \phi \ d\phi$

$ =-[2\cos \phi +\ln \tan\frac{\phi}{2}]_{\frac{\pi}{4}}^{\frac{3\pi}{4}}$

${{ =-[2(\cos \frac{3\pi}{4}} - \cos \frac{\pi}{4}}) + \ln\frac{\tan\frac{3\pi}{8}}{\tan\frac{\pi}{8}}]$

$ =2\sqrt{2} - \ln \tan^{2}\frac{3\pi}{8}$

$ =2\sqrt{2} - \ln \frac{\sin^{2}\frac{3\pi}{8}}{\cos^{2}\frac{3\pi}{8}}$

$ =2\sqrt{2} - \ln\frac{1-\cos\frac{3\pi}{4}}{1+\cos\frac{3\pi}{4}}$

$ =2\sqrt{2} -\ln\frac{1+\frac{1}{\sqrt{2}}}{1-\frac{1}{\sqrt{2}}}$

$ =2\sqrt{2} - \ln(3 + 2\sqrt{2})$.


The three squares $ACC_{1}A''$, $ABB_{1}'A'$, $BCDE$ are constructed externally on the sides of a triangle $ABC$. Let $P$ be the center of the square $BCDE$. Prove that the lines $A'C$, $A''B$, $PA$ are concurrent.

First of all, it's really easy to show that $CA'\perp BA''$, so if $T=CA'\cap BA''$, $BTCP$ is cyclic. Now show that $PBA,CBA'$ are similar, so $\angle APB=\angle TCB$. Smilarly, $\angle APC=\angle TBC$, so $AP$ makes with $PB,PC$ the same angles as the diagonal $PT$ of $PBTC$ does, so, in fact, $T$ must lie on $AP$.

I have a slightly different solution which shows a bit more. But first of all, I will rename the points in the problem, since the notations of the original post make me recall Littlewood's "Miscellany", particularly the remark about Jordan, who is said to have had the habit that, when he had to introduce four variables which were completely analogous and equal in rights - so most of us would call probably call them $a$, $b$, $c$, $d$ -, he denoted them by something like $a$, $M^{\prime}_{3}$, $\epsilon_{2}$, $\Pi_{1,2}^{\prime\prime}$. Anyway, here is how I would write the problem:

[i]The three squares $BCC_{a}B_{a}$, $CAA_{b}C_{b}$, $ABB_{c}A_{c}$ are constructed externally on the sides BC, CA, AB of a triangle ABC. Let P be the center of the square $BCC_{a}B_{a}$. Prove that the lines $CA_{c}$, $BA_{b}$ and AP are concurrent.[/i]

In fact, I will show even more: I will denote by T the point of intersection of the circumcircles of the squares $CAA_{b}C_{b}$ and $ABB_{c}A_{c}$ different from the point A. Then I will show that this point T lies on the lines $CA_{c}$, $BA_{b}$, AP and $B_{c}C_{b}$ and that these four lines form angles of 45° with each other at the point T.

In order to show this, it is enough to prove that

$\measuredangle ATA_{c}= \measuredangle A_{c}TB_{c}= \measuredangle B_{c}TB = \measuredangle BTP = \measuredangle PTC = \measuredangle CTC_{b}= \measuredangle C_{b}TA_{b}$
$= \measuredangle A_{b}TA = 45^{\circ}$.

In fact, $\measuredangle ATA_{c}= 45^{\circ}$ is very easy to prove: Since the point T lies on the circumcircle of the square $ABB_{c}A_{c}$, we have $\measuredangle ATA_{c}= \measuredangle ABA_{c}$, but $\measuredangle ABA_{c}= 45^{\circ}$ is clear because of the square $ABB_{c}A_{c}$. Thus, we have $\measuredangle ATA_{c}= 45^{\circ}$. The proofs of $\measuredangle A_{c}TB_{c}= 45^{\circ}$, $\measuredangle B_{c}TB = 45^{\circ}$, $\measuredangle CTC_{b}= 45^{\circ}$, $\measuredangle C_{b}TA_{b}= 45^{\circ}$ and $\measuredangle A_{b}TA = 45^{\circ}$ are similar.

Remains to prove that < BTP = < PTC = 45°. In fact, we have

$\measuredangle BTC = 360^{\circ}-\left(\measuredangle CTC_{b}+\measuredangle C_{b}TA_{b}+\measuredangle A_{b}TA+\measuredangle ATA_{c}+\measuredangle A_{c}TB_{c}+\measuredangle B_{c}TB\right)$
$= 360^{\circ}-\left(45^{\circ}+45^{\circ}+45^{\circ}+45^{\circ}+45^{\circ}+45^{\circ}\right) = 90^{\circ}$.

Thus, the point T lies on the circle with diameter BC. On the other hand, since the point P is the center of the square $BCC_{a}B_{a}$, we have < BPC = 90°, and consequently, the point P lies on the circle with diameter BC. Now, as we have seen that both points T and P lie on the circle with diameter BC, we get < BTP = < BCP. But since the point P is the center of the square $BCC_{a}B_{a}$, we have < BCP = 45°, and thus, < BTP = 45°. Similarly, < PTC = 45°. This completes the proof of

$\measuredangle ATA_{c}= \measuredangle A_{c}TB_{c}= \measuredangle B_{c}TB = \measuredangle BTP = \measuredangle PTC = \measuredangle CTC_{b}= \measuredangle C_{b}TA_{b}$
$= \measuredangle A_{b}TA = 45^{\circ}$,

and thus the solution of the problem.

  Darij

Let the intersection of $A'C$ and $AB$ be $Q$ and the intersection of $A''B$ and $AC$ be $R$ and the intersection of $AP$ and $BC$ be $S$.

Then $\frac{BS}{SC}= \frac{\triangle{ABP}}{\triangle{ACP}}=\frac{AB \cdot PB \cdot \sin{(B+45^{o})}}{AC \cdot PC \cdot \sin{(C+45^{o})}}=\frac{AB \cdot \sin{(B+45^{o})}}{AC \cdot \sin{(C+45^{o})}}$

$\frac{CR}{RA}=\frac{\triangle{CA''B}}{\triangle{AA''B}}=\frac{CB \cdot CA'' \cdot \sin{(C+45^{o})}}{AB \cdot AA'' \cdot \sin{(A+90^{o})}}=\frac{CB \cdot \sin{(C+45^{o})}}{AB\cdot \sin{(A+90^{o})}}\cdot \frac{\sqrt{2}}{1}$

$\frac{AQ}{QB}= \frac{\triangle{ACA'}}{\triangle{BCA'}}=\frac{AC \cdot AA' \cdot \sin{(A+90^{o})}}{BC \cdot BA' \cdot \sin{(B+45^{o})}}=\frac{AC \cdot \sin{(A+90^{o})}}{BC \cdot \sin{(B+45^{o})}}\cdot \frac{1}{\sqrt{2}}$

Thus $\frac{BS}{SC}\cdot \frac{CR}{RA}\cdot \frac{AQ}{QB}= 1$. So $AP$, $CA'$ and $BA''$ are concurrent.

[quote="darij grinberg"][b]The three squares $BCC_{a}B_{a}$, $CAA_{b}C_{b}$, $ABB_{c}A_{c}$ are constructed externally on the sides BC, CA, AB of a triangle ABC. Let P be the center of the square $BCC_{a}B_{a}$. Prove that the lines $CA_{c}$, $BA_{b}$ and AP are concurrently (in the point $T$).[/b][/quote][color=darkblue][b]I'll add some details at the conclusion of this classical problem.[/b]

$1.\blacktriangleright$ The $A$- median of the triangle $ABC$ is the $A$- altitude in the triangle $AA_{b}A_{c}$.

The $A$- altutude of the triangle $ABC$ is the $A$- median of the triangle $AA_{b}A_{c}$.

If denote the middlepoints $S$, $M$ of the segments $A_{b}A_{c}$, $BC$ respectively and the projection $H$ of the point $A$ to the line $A_{b}A_{c}$, then

$AS\perp BC$, $AM\perp A_{b}A_{c}$ and $A_{b}A_{c}=2m_{a}$, $AS=\frac{1}{2}a$.

$2.\blacktriangleright$ $BA_{b}=CA_{c}$, $BA_{b}\perp CA_{c}$. If  $T\in BA_{b}\cap CA_{c}$, then $T\in B_{c}C_{b}$ and $AT\perp B_{c}C_{b}$.

$3.\blacktriangleright$ The lines $CB_{c}$, $BC_{b}$ and the $A$- altitude of the triangle $ABC$ are concurrently.

$4.\blacktriangleright$ If the points $A_{1}$, $B_{1}$, $C_{1}$ are the centers of the mentioned squares (same order), then the lines $AA_{1}$, $BB_{1}$, $CC_{1}$ are concurrently (the [b]Vecten's point[/b]).[/color]

This problem is a direct application of the below ''isogonic'' theorem, as a generalization of the Jacobi theorem.

Through each vertex of a given triangle, we draw two lines isogonal with respect to the corresponded angle. Prove that the lines connecting each vertex of the given triangle, with the intersection point of the isogonal lines through the other vertices, are concurrent.

Two simple synthetic proofs of the ''isogonic'' theorem, have been posted in Hyacinthos forum.

http://groups.yahoo.com/group/Hyacinthos/message/13680

http://groups.yahoo.com/group/Hyacinthos/message/13685

Kostas Vittas.

Dear Mathlinkers,
an article concerning the Vecten's figure and its developpement can be seeing on my site
http://perso.orange.fr/jl.ayme , la figure de Vecten  vol. 5 p. 70
Sincerely
Jean-Louis

Anyone can help me????

First of all, it's really easy to show that $CA'\perp BA''$

How do we prove it?


[quote="biffanddoc"][hide]
Since $\frac 34$ are less than 6 feet and $\frac 23$ are less than 5 feet, $\frac{1}{12}$ are between 5 and 6 feet.
If we call between 5 and 6 feet $x$, there is $2x$ between 4 and 5 feet and $12x$ total.
Since $\frac 14$ of the people are more than 6 feet, $3x$ people are.
That leaves $6x$ people under 4 feet.
So $6x=6$ and $x=1$
The total is $12(1)=\boxed{12}$[/hide][/quote]

How did you get $12x$ total?

btw, your answer is correct

[quote="ch1n353ch3s54a1l"]Several students are standing in line from shortest to tallest. Three-fourths of them are less than 6 feet tall. Two thirds are less than 5 feet tall. 6 students are not yet 4 feet tall. There are twice as many between 4 and 5 feet as there are between 5 and 6 feet tall. How many students are standing in line?[/quote]

Wow, lots of shorties (for middle school)!  :D  :rotfl:
[hide="my answer"]Well, there are 6 students standing less than 4 ft, 2x students between 4ft and 5ft, x students between 5ft and 6ft, and x+2 [ 1/4 of all students, therefore (3x+6)/(x+2)] students above 6ft. Then, they said that 2 thirds are less than 5ft tall, so 2(2x+2) = 2x+6, and when you simplify the equation, you get that x= $\ 1$. So insert that into the [i]x[/i]s and add all of them ( 6+2+1+3) up to get $\ 12$ as your final answer. [/hide]

I called the 5-6 feet $x$ and since it is $\frac{1}{12}$ of the total, the total is $12x$.

um biffandoc, I think you need to double check; I'm positively sure that the answer is just 12, not 12x.

I called the total $12x$, then solved for $x$
Reread my solution.

o ok, sorry! :oops:  :D  I didn't read the above; I just did the problem lol.


1) Proove that holds $ |\minus{}x\plus{}y\plus{}z|\plus{}|x\minus{}y\plus{}z|\plus{}|x\plus{}y\minus{}z|>\equal{}|x|\plus{}|y|\plus{}|z|$

2) If $ x\plus{}y\plus{}z\equal{}0$ and $ |x|\equal{}|y|\equal{}|z|\equal{}1$ proove that |$ x\minus{}y|\equal{}|y\minus{}z|\equal{}|z\minus{}x|\equal{}\sqrt3$

[hide]
2. Let $ x\equal{} x_1\plus{}i\cdot x_2$, and same for $ y, z$.
Then given says,
$ x_1^2\plus{}x_2^2\equal{}1$(same for $ y,z$)
and
$ x_1\plus{}y_1\plus{}z_1\equal{}0 \rightarrow x_1\plus{}y_1\equal{}\minus{}z_1 \rightarrow z_1^2\equal{}(x_1\plus{}y_1)^2$
$ x_2\plus{}y_2\plus{}z_2\equal{}0 \rightarrow x_2\plus{}y_2\equal{}\minus{}z_2 \rightarrow z_2^2\equal{}(x_2\plus{}y_2)^2$

$ |x\minus{}y|^2$
$ \equal{}|(x_1\minus{}y_1)\plus{}i\cdot (x_2\minus{}y_2)|^2$
$ \equal{}(x_1\minus{}y_1)^2\plus{}(x_2\minus{}y_2)^2$
$ \equal{}2(x_1^2\plus{}y_1^2)\minus{}(x_1\plus{}y_1)^2\plus{}2(x_2^2\plus{}y_2^2)\minus{}(x_2\plus{}y_2)^2$
$ \equal{}2\plus{}2\minus{}z_1^2\minus{}z_2^2$
$ \equal{}4\minus{}1\equal{}3$

So $ |x\minus{}y|\equal{}\sqrt{3}$, as is $ |y\minus{}z|, |z\minus{}x|$
[/hide]


calculus
derivative
calculus computations

Let's say that I'm asked to solve: $ y \equal{} x \cdot \frac{dy}{dx} \plus{} y^2 \cdot \frac{dy}{dx}$. What do I do?

(I've gone through a dozen... hmm... boring methods in my algebra class, but I'm not figuring this one out.)

Write it as:

$ y\equal{}xy'\plus{}y^2y'$

or:

$ dy(x\plus{}y^2)\equal{}ydx$

or:

$ y^2dy\equal{}ydx\minus{}xdy$

Now, what can you multiply both sides by to make each total derivatives?  Then just integrate.


The British Mathematical Olympiad Round 1:
http://www.bmoc.maths.org/home/bmo1-2010.pdf  . Have fun!

Yes I posted them just to let people have a go at solving the problems.


geometry
trapezoid
dissection
cyclic quadrilateral
IMO
IMO 1972

Given $n>4$, prove that every cyclic quadrilateral can be dissected into $n$ cyclic quadrilaterals.

Actually, $n\ge 4$ should work. Our initial quadrilateral will be $ABCD$.

For $n=4$, we do this:

Take $E\in AB,F\in CD,\ EF\|AD$ with $E,F$ sufficiently close to $A,D$ respectively. Take $U\in AD,V\in EF$ such that $AEVU$ is an isosceles trapezoid, with $V$ close enough to $F$ (or $U$ close enough to $D$) that we can find a circle passing through $U,D$ (or $F,V$) which cuts the segments $UV,DF$ in $X,Y$. Our four cyclic quadrilaterals are $BCFE,\ AEVU,\ VFYX,\ YXUD$.

For $n\ge 5$ we do the exact same thing as above, but now, since we have an isosceles trapezoid, we can add as many trapezoids as we want by dissecting the one trapezoid with lines parallel to its bases.


topology
induction
Euler
linear algebra
matrix
invariant
graph theory

[quote="April"]Given a finite tree $ T$ and isomorphism $ f: T\rightarrow T$. Prove that either there exist a vertex $ a$ such that $ f(a) = a$ or there exist two neighbor vertices $ a$, $ b$ such that $ f(a) = b$, $ f(b) = a$.[/quote]

Hmm, does the following make sense:

Consider the tree $ T$ as a topological space $ X$, and extend $ f$ to an automorphism $ g$ of this space $ X$ (by setting $ g\left(v\right) = f\left(v\right)$ for any vertex $ v$ of $ T$, and extending to each edge by linearity). Then, we want to show that this topological automorphism $ g$ has a fixed point (because if such a fixed point is a vertex of the tree, we get a fixed vertex, while if it is an interior point of an edge, we get a fixed edge). By the [url=http://en.wikipedia.org/wiki/Lefschetz_fixed-point_theorem]Lefschetz fixed-point theorem[/url], this will be immediate once we succeed to show that $ \sum_{k\geq 0}\left( - 1\right)^k\mathrm{Tr}\left(g_*\mid H_k\left(X,\mathbb{Q}\right)\right)\neq 0$. This simplifies to $ \mathrm{Tr}\left(g_*\mid H_0\left(X,\mathbb{Q}\right)\right)\neq \mathrm{Tr}\left(g_*\mid H_1\left(X,\mathbb{Q}\right)\right)$ (because the space $ X$ is $ 1$-dimensional). Now, $ \mathrm{Tr}\left(g_*\mid H_1\left(X,\mathbb{Q}\right)\right) = 0$ (since $ H_1\left(X,\mathbb{Q}\right) = 0$, because $ X$ is contractable), but $ \mathrm{Tr}\left(g_*\mid H_0\left(X,\mathbb{Q}\right)\right)\neq 0$ (because $ H_0\left(X,\mathbb{Q}\right)$ is $ \mathbb{Q}$ (since the space $ X$ has one connected component), and $ g_*$ is an automorphism of $ H_0\left(X,\mathbb{Q}\right)$ (because $ g$ is an automorphism of $ X$)), thus $ \mathrm{Tr}\left(g_*\mid H_0\left(X,\mathbb{Q}\right)\right)\neq \mathrm{Tr}\left(g_*\mid H_1\left(X,\mathbb{Q}\right)\right)$, and we are done.

Is this bollocks or a solution? My knowledge in topology has much suffered from topology lectures being too early in the morning, so I am surprised I could even remember how to compute $ g_*$...

[Edit: thanks to Peter Scholze for replacing my idiot ramblings about $ H_1$ with the correct argument ;) ]

[b]EDIT:[/b] Apparently the above was quite an overkill (though a nice overkill...). The problem is an easy induction: Take a vertex $ v_0$ with only one neighbour. The automorphism $ f$ maps it to a vertex $ v_1$ which also has only one neighbour. This vertex is, in turn, mapped to a new vertex $ v_2$ which also has only one neighbour. And so on, until we find an $ n$ such that $ v_n=v_0$. Now, apply the problem to the tree $ T\setminus\left\{v_0,v_1,...,v_{n-1}\right\}$. The only thing one needs to care about is that this tree is nonempty, but if it is empty, then every vertex of $ T$ must have had one neighbour only, meaning that $ T$ is the tree on two vertices.

  darij

can someone explain me what does "isomorphism $ f: T \rightarrow T$" mean? thanks. :blush:
it seems that this problem is almost Problem 27.1.2, An Introduction to Graph Theory, Douglas Brent West.

An isomorphism between graphs is a bijection $ f$ between their vertex sets such that $ \{f(a), f(b)\}$ is an edge in the image if and only if $ \{a, b\}$ is an edge in the preimage.

An isomorphism from a graph to itself is sometimes called an automorphism.  For example, a path graph has exactly two automorphisms, while $ K_n$ has $ n!$ automorphisms (any permutation of the vertices will do).

[hide=Official solution]We use induction on the number of vertices of graph
For $N=1,2$ it is obvious.
Divide the set of vertices into two parts:
In the first part $\Gamma_1$ there are vertices whose degree is $1$ and in second part $\Gamma$ take others.
It is clear that $\Gamma_1$ is nonempty , if otherwise $\implies $ all vertices have degree $\ge 2$ so there is cycle which is a contradiction.For $N\ge 3$ it is easy to see that $\Gamma$ is also nonempty and has less than $N $ vertices.It's easy to see that
If $A\in \Gamma $ then $f (A)\in \Gamma $ and from $A\in \Gamma $ we can go to any $B\in \Gamma $ without traveling to any city in $\Gamma_1$ since all vertices in $\Gamma_1$ have degree $1$ and the whole graph is connected.So $\Gamma $ also satisfies all problem conditions and since it has less than $N $ vertices we are done by induction.

[quote=April]Given a finite tree $ T$ and isomorphism $ f: T\rightarrow T$. Prove that either there exist a vertex $ a$ such that $ f(a)\equal{}a$ or there exist two neighbor vertices $ a$, $ b$ such that $ f(a)\equal{}b$, $ f(b)\equal{}a$.[/quote]

Induct on $|T| \ge 1$; base cases being obvious. Now pick all vertices of $T$ with degree exactly $1$. Call this set $\mathcal{S}$. Note that for any $s \in \mathcal{S}$ we can define $g(s)$ to be the neighbour of $s$. Since $s, g(s)$ are neighbours, so are $f(s), f(g(s))$. Now if $v, f(s)$ are neighbours then $f^{-1}(v), s$ are hence $v=f(g(s))$ so $f(s) \in \mathcal{S}$. Thus $f$ induces a bijection on $\mathcal{S}$ and the embedded $f^{*}$ defined on $T|\mathcal{S}$ is an isomorphism on this sub-tree. Induction hypothesis applies.

Edit: oops same as other induction proofs.

(reposted because the original presentation was somewhat haphazard)

There exists a non-inductive approach.

Choose an arbitrary vertex $v\in \mathcal{V}(T)$ and consider the minimal subtree $\Pi\subseteq T$ who contains both $v$ and $f(v)$, or in other words, the (unique) non-self-intersecting path $\Pi$ connecting $v$ to $f(v)$. Embed $T$ in Euclidean Space so that $\Pi$ lies on a line and each edge of $\Pi$ is of equal length. Observe that there is some minimal positive integer $n$ for which $\underbrace{f\circ f\circ \cdots\circ f}_{n\text{ times}}=\text{id}$.

If $\Pi$ contains an even number of edges, then there will be a vertex $\alpha\in\mathcal{V}(\Pi)$ at the midpoint of $v$ and $f(v)$. If this is not that case—if $\Pi$ has odd length—then there will instead be two vertices $\alpha_1,\alpha_2\in\mathcal{V}(\Pi)$ with the edge between them containing said midpoint. We will show that either $f(\alpha)=\alpha$ or $f(\alpha_1)=\alpha_2\text{ and }f(\alpha_2)=\alpha_1$, as the case may be.

To that end consider the subgraph $\Pi\cup f``\Pi\subseteq T$ obtained by taking the union of with $\Pi$ its image under $f$, looking something like

[center][img]https://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvMi84LzFlNDNkZmQ0MWE0ZjBlNDAzMDI1ZTI1OTA5YjA4ODYzM2QxYzFhLnBuZw==&rn=VHdvIEl0cy5wbmc=[/img][/center]

In particular, the claim is equivalent to the fact that the length of the orange segment (the intersection of $\Pi$ and $f``\Pi$) is at least half the length of $\Pi$. This is necessarily the case, however, as if it weren't, one could repeat the process to produce the graph $\Pi\cup f``\Pi\cup\cdots \cup \underbrace{f\circ f\circ \cdots\circ f}_{n-1\text{ times}}``\Pi$ as in the below example (in which $n=5$).

[center][img]https://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvYi8xLzkxYWE1ZTYzOTgxYmE0OTBiNjNhMjkyOGZmMDY5ZTg2NWQxYTYzLnBuZw==&rn=Q3ljbGUgU2hvdC5wbmc=[/img][/center]

And then by symmetry excise from its center a cycle, violating the assumption of $T$'s treeness $\blacksquare$

[quote=starchan]Here's an interesting solution.
[hide = solution]
Let $d(u, v)$ denote the distance between vertices $u, v$ in $T$. To each vertex $v$, assign a quantity $F(v)$ as: \[F(v) = \sum_{u \in G} d(v, u)\]
Consider the set $\mathcal S$ of vertices for which $F(v)$ is minimised. It's well known that $\mathcal S$ forms a path in $T$. Observe that $f(\mathcal S)$ must be a permutation of $\mathcal S$, since distances between vertices are preserved. Since $\mathcal S$ forms a path, it follows that so does $f(\mathcal S)$. Hence $f(\mathcal S) = \mathcal S$ or $\mathcal S$ reversed. In either case, either the middle vertex or the middle edge of $\mathcal S$ satisfies the problem's conditions.
[/hide][/quote]

Nice! If you modify the quantity evaluated at each node \(v\in \mathcal{V}\left(T\right)\) from your \(\sum_{v' \in \mathcal{V}\left(T\right)} \text{d}\left(v', v\right)\) to \(\max_{v' \in \mathcal{V}\left(T\right)} \text{d}\left(v', v\right)\) instead, then (if I'm not mistaken) the locus of vertices that minimize said quantity is either a single vertex or a pair of adjacent vertices, whence the claimed fixed vertex/edge.


function
calculus
derivative
Asymptote
calculus computations

Hi, I already solved this problem but I wasn't sure if I did it correctly because for the critical numbers I obtained from the first derivative and the second derivative I could not find a local max or min and thought it was unusual. I checked my derivative over too to see if I made any errors but I don't think I did.

The function is f(x) = x/[(x^2)-9]

I obtained f ' (x) = [-(x^2) - 9]/[[(x^2) - 9]^2]
I found the critical numbers of 3 and -3.
So I evaluated f'(x) over the intervals from
(-infinity,-3)  decreasing
(-3,3) decreasing
(3, infinity) decreasing.

Therefore since they all were decreasing, I reasoned that there would be no max or min values but I find that hard to believe.

As well when I tried to check the concavity of the function I ended up with
f ''(x) = [-38x + 4(x^3)] / [(x^2) - 9]^3
I obtained the critical numbers of -3, 3, 0, -sq 19, sq 19.
So again I evaluted f '' (x) over the intervals of
(-infinity, -sq19) concave down
(-sq19, -3) concave down
(-3, 0)concave down
(0,3) concave up
(3, sq19) concave up
(sq19, infinity) concave up

Also by sq19 I mean square root of 19!

I obtained the inflection point of (0,0)

However, again I am worried that I did not find the right values to evaluate the function over because there are so many.

Thank you so much for your  help!

It is indeed decreasing on each of the intervals it's continuous on. Between those intervals, it jumps from $ \minus{}\infty$ to $ \plus{}\infty$.

Your concavity calculations are seriously mixed up. $ \sqrt{19}$ is not an interesting point, and you've got the middle intervals wrong as well.

is that because my second derivative is wrong?

No. I didn't calculate it myself, and my statements were based on the assumption that your second derivative was right.

ok well because there are two vertical asymptotes at -3 and 3, does that mean there is no point of inflection?

No. There is certainly a point of inflection between the two vertical asymptotes, and there might be another pair of inflection points.


Twenty percent less than $ 60$ is one-third more than what number?

$ \textbf{(A)}\ 16\qquad
\textbf{(B)}\ 30\qquad
\textbf{(C)}\ 32\qquad
\textbf{(D)}\ 36\qquad
\textbf{(E)}\ 48$


analytic geometry
trigonometry
email
ratio
geometry
angle bisector
complex numbers

[color=darkblue]Let $ABC$ be a right-angle triangle ($AB\perp AC$). Define the middlepoint $M$ of the side $[BC]$ and the point $D\in (BC)$, $\widehat {BAD}\equiv\widehat {CAD}$. Prove that exists a point $P\in (AD)$ so that $PB\perp PM$ and $PB=PM$ if and only if $AC=2\cdot AB$ and in this case $\frac{PA}{PD}=\frac 35$.[/color]

Denote a = BC, b = AC, c = AB. We also os Put the coordinate origin at A, the x- and y-axes along AB, AC. P is the intersection of the angle bisector AD with a circle (N) centered at the midpoint N of BM, because the angle $\angle BPM = 90^\circ$ is right. The equation of the angle bisector AD is y = x and the equation of the circle (N) is

$\left(x - \frac{3c}{4}\right)^2 + \left(y - \frac b 4\right)^2 = \frac{a^2}{16} = \frac{b^2 + c^2}{16}$

Substituting y = x, we get an equation for the x-coordinate of P, equal to its y-coordinate.

$\left(x - \frac{3c}{4}\right)^2 + \left(x - \frac b 4\right)^2 = \frac{b^2 + c^2}{16}$

$2x^2 - \frac{3c + b}{2}\ x + \frac{c^2}{2} = 0$

The distances of $P = (x, x)$ from $B = (c, 0)$ and $M = (c/2, b/2)$ are equal, hence

$(x - c)^2 + x^2 = \left(x - \frac c 2\right)^2 +\left(x - \frac b 2\right)^2$

$\frac{3c^2 - b^2}{4} = (c - b)x,\ \ x = \frac{3c^2 - b^2}{4(c - b)}$

We assumed $b \neq c$, otherwise the point P with the desired properties clearly does not exist. Substituting this into the equation for x obtained previously, we get an equation for b, c:

$2(3c^2 - b^2)^2 - 2(3c + b)(c - b)(3c^2 - b^2) + 8c^2(c - b)^2 = 0$

Multiplying the first 2 factors of the middle term,

$2(3c^2 - b^2)^2 - 2(3c^2 - b^2 - 2bc)(3c^2 - b^2) + 8c^2(c - b)^2 = 0$

$4bc(3c^2 - b^2) + 8c^2(c - b)^2 = 0$

$b(3c^2 - b^2) + 2c(c - b)^2 = 0$

$2c^3 - 4bc^2 + 2b^2c + 3bc^2 - b^3 = 0$

$2c^3 - bc^2 + 2b^2c - b^3 = 0$

$c^2(2c - b) + b^2(2c - b) = 0$

which clearly has the only non-zero solution b = 2c or $AC = 2AB$. The length of the angle bisector AD of a right angle triangle $\triangle ABC$ is

$AD = \sqrt{bc\left(1 - \frac{a^2}{(b + c)^2}\right)} = \frac{bc \sqrt 2}{b + c}$

while the segment length AP is

$AP = x \sqrt 2 = \frac{(3c^2 - b^2) \sqrt 2}{4(c - b)}$

If b = 2c, then

$AD = \frac{2c \sqrt 2}{3},\ \ AP = \frac{c \sqrt 2}{4},\ \ \frac{AD}{AP} = \frac 8 3$

$\frac{PD}{AP} = \frac{AD - AP}{AP} = \frac{AD}{AP} - 1 = \frac 5 3$

[quote="Davron"]Can anyone solve this not by using coordinate system please.

Davron[/quote]

Sure, but what's wrong with a coordinate system ?

$BD = \frac{ac}{b + c},\ \ PN = BN = \frac{BM}{2} = \frac a 4$

$ND = BD - BN = \frac{ac}{b + c} - \frac a 4 = \frac{a(3c - b)}{4(b + c)}$

From the right angle triangle $\triangle PND$,

$\tan ADB \equiv \tan \angle PDN = \frac{PN}{ND} = \frac{b + c}{3c - b}$

In the triangle $\triangle ABD$, we also know the angle $\angle BAD = 45^\circ,\ \tan \angle BAD = 1$

$\frac b c = \tan \angle ABC \equiv \tan \angle ABD = -\tan(\angle ADB + \angle BAD) = \frac{\tan \angle ADB + \tan \angle BAD}{\tan \angle ADB \cdot \tan \angle BAD - 1} =$

$= \frac{\frac{b + c}{3c - b} + 1}{\frac{b + c}{3c - b} - 1} = \frac{b + c + 3c - b}{b + c - 3c + b} = \frac{4c}{2(b - c)}$

$\frac b c = \frac{2c}{b - c}$

$b(b - c) = 2c^2$

$\left(\frac b c\right)^2 - \frac b c - 2 = 0$

$\frac b c = \frac{1 \pm \sqrt{1 + 8}}{2} = 2\ \text{or}\ -1$

:)  No no everything is ok thank you very much but i dont like the coordinate system because i dont know it well but i want learn it, how can i start and where can i study?Please help me if you can. Thank you very much!

Davron

I am happy see you, Yeti ! Your solution is nicely. Do you like this problem ? I do. I wish your Email (private message) because I want send you some nice things (... and mathematics).

There's also a nice and short solution using complex numbers (I've changed letters so that at $C$ is right angle and $D$ lies closer to $A$ than $M$). We choose origin just as Yetti did - namely $C$. Then CB will be real axis and AC will be imaginary axis. (I'll use lowercase letters for coordinates of points denoted capital letters).

we have: $A=ai$, $B=b$ (where $a,b \in R$). Also (using bisector theorem):

$d= \frac{ab+bai}{a+b}=ab\frac{1+i}{a+b}$ now:

$PM \perp AP$and $PM=AP$ $\iff ai = p +(\frac{ai+b}{2}-p)i \iff 2p = ({\frac{a}{2}-\frac{b-2a}{2}i)(1+i)}$
$\iff \exists _{t \in R} : 2td = ({\frac{a}{2}-\frac{b-2a}{2}i)(1+i) \iff \exists _{t \in R} : 2tab\frac{1+i}{a+b} = ({\frac{a}{2}-\frac{b-2a}{2}i)(1+i)}}$
 $\iff \exists _{t \in R} : 2tab\frac{1}{a+b} = ({\frac{a}{2}-\frac{b-2a}{2}i) \iff b=2a \iff 2AC=AB}$. Also by counting $t$ we can get wanted proportion.

Virgil hello how are you, why dont you post that useful things for us...   please

Davron

Thanks Virgil, I liked the problem, otherwise I would not do it (and quite obviously, you are not asking for help, certainly not from me). My e-mail is in my member profile.

Yetti.

I didn't find the your E-mail in the your member profile !
The beauty of this problem is deed that the ratios $\frac{AC}{AB}$ and $\frac{PA}{PD}$ are two rational numbers.

[quote="Virgil Nicula"]Davron, give me your email and I shall send and you ![/quote]

Thank you very much   :)  I would like to meet more friends like you ...

tha is kostakoz@yahoo.com thank you very very much ... :)  :)

Davron


limit
function
integration
real analysis
real analysis unsolved

[hide="Hint."]
[color=darkred]If the function $ f$ is integrably on the mentioned interval, then :

$ \boxed {\lim_{n\to\infty}\frac 1n\cdot\sum_{k \equal{} 1}^nf\left(\frac kn\right) \equal{} \int_0^1f(x)\ \mathrm {dx}}\ .$ [b]Generally :[/b] $ a < b$ and $ \lambda\in [0,1]$ $ \implies$

$ \boxed {\lim_{n\to\infty}\frac 1n\cdot\sum_{k \equal{} 1}^nf\left(a \plus{} \frac {k \minus{} \lambda }{n}\cdot (b \minus{} a)\right) \equal{} \int_a^bf(x)\ \mathrm {dx}}\ .$

[b]Remark.[/b] Prove easily these limits using the Riemann's sum.[/color] [/hide]

[quote="blue_se7en"]1.$ \lim_{n \to \infty} \sum_{i \equal{} 1}^n \frac {n}{n^{2} \plus{} i^{2}}$[/quote]
We have a Riemann sum here:

$ \lim_{n \to \infty } \sum\limits_{i \equal{} 1}^n {\frac{n}{{n^2  \plus{} i^2 }}}  \equal{} \lim_{n \to \infty } \frac{1}{n}\sum\limits_{i \equal{} 1}^n {\frac{1}{{1 \plus{} \left( {\cfrac{i}{n}} \right)^2 }}}  \equal{} \int_0^1 {\frac{1}{{1 \plus{} x^2 }}\,dx} .$

Now this is routine.

The second one can also be considered a limit of Riemann sums. Do similar things with it.


Rewrite it as $\sqrt{10^{6}-1}+\sqrt{10^{6}+1}$ then square it, we will get 2000 is larger  :P

In general, since $f(x) = \sqrt{x}$ is concave, we have

$f(a_{1})+f(a_{2})+\cdots+f(a_{n}) \le n f\left(\frac{1}{n}\sum a_{i}\right)$,

which follows by Jensen's with equal weights. In this case we have $n = 2$, $a_{1}= 10^{6}-1$, and $a_{2}= 10^{6}+1$.


Could anybody give me any Lester's book or Yaglom book of complex numbers in geometry in pdf or djvu???

THANKS A LOT

I think you can reach June by following the link:
http://www.thejuniverse.org/geometry/LC/LCbiblio.html
She kindly sent me her papers after I emailed her requesting these a few years ago.
Sincerely, Jeff


In rectangle ABCD, H is the midpoint of $\bar{BC}$, E lies on $\bar{AD}$, and F lies on $\bar{AB}$. In rectangle CEFG, H lies on $\bar{FG}$ and HG=3 in. $m\angle DEC=45^o$. What is the difference between the areas of the two rectangles?

I worked it out, though I don't have a solution posted

[hide]

The area of rectangle $ABCD$ to the area of rectangle $CEFG$ is in the ratio of $\sqrt{2} : 1$

[/hide]

It's kinda hard to draw it out and visualize it. I made my drawing pretty big. You just have to find 45-45-90 triangles and compare the sides. I got 27.


MATHCOUNTS FTW AMC USA(J)MO USAMO ARML geometry

Well, time is almost up in our preparation for Nationals.  My group only has three more practices to go.  I want to wish all the people who have qualified for Nationals good luck.  I hope you enjoy yourself and have a great 3-4 days.

All of you are really good and are truly the best of the best!  I hope you realize what a great honor and accomplishment it is to get to Nationals.  It took me as a coach 13 years to finally get a team to Nationals.  I am always amazed at how good and fast all the kids are.

Let the games begins again!!!

thank you very much!

good luck to every1, and especially to me  :D

Good luck to everyone, especially the Massachusetts team!

Please get first, please get first.

if IN wins, then i have an excuse for not making nats.

[quote="mathcrazed"]well erm. I need to seriously go practice now.[/quote]

I probably should start practicing too.

As a MathCounts alum, I'd like to say a few things to first-timers, especially 8th graders.

I've begun to realize, believe it or not, that the competition itself is actually a relatively minor portion of the nationals experience. Although it is important and rewarding to perform well, the bottom line is that the vast majority of competitors do not qualify for the Countdown round. Do your best, but remember--the experience, memories, people you'll meet, friends you might make constitute much greater long-term value than even a national placement. Above all, have fun, go out and talk to people, and don't let everything revolve around the testing and cloud the experience. Meeting new people is the one biggest thing I wish I had done instead of stuck to my team so much.

An opportunity like MathCounts nationals won't come up often, if at all. Moreover, this is last experience in an effort-filled MathCounts career after so much effort. For people who've been practicing really hard, one could say (albeit a bit melodramatically) that nationals closes a chapter in one's life. Make the experience a good one.

Our coach has been coaching 3 yrs.  :D

But then again, the kids at my school are beast, (3 USAMO qualifiers, 3 nats qualifiers, 3 ARML qualifiers.. you name it)

[quote="JBoyd"]...  [b]My group only has three more practices to go.[/b]...[/quote]

how do you fit in so many practices? does everybody on the Texas team go to the same school or something to that effect?

The florida team only met twice. They(the coach and his "assistant") gave us alot of material but still...we only met twice.

how many meets(practices) did other states have?

If MA beats 6th, I'm going to have to conclude that I am detrimental to national teams...obviously that's not a good reason to not want them to beat 6th, so I do.

AL had....... zero. Our coach just sent us stuff that we work on and send back questions to her.

That's right! Remember totally pwning the pants off ur teammates in brawl the day after the test!!!!!

I don't believe in luck, so i hope you alll do good at nats and that an AoPSer wins!!! :P (AL FTW, even tho, uh, we havent won since 1991!)(rhyme)

Ohio had one, where we did 5 team rounds.
We probably would have had more if our state hadn't been delayed about a month.

Good luck, everybody!

[quote="numberdance"]*counts Saturdays* Oklahoma had six meetings.

And we mostly spent them doing team rounds...[/quote]
yeh, florida did saturdays too, but the state competition was on like the 29th(wth?) so we had like no time...somehow we only had 2 practices...

ehh, we also worked on team rounds.i think we did like 16 of them? we still have 2 more :P

we got several 10's! yay!

and i(i dont know about everyone else on the team) did 18 years of sprint and targets, in 6 days! :P
i [b]almost[/b] achieved my goal of doing all of them in a week.

what scores did you all average on nats tests?

I'm sure how they have three left, but the Texas kids all get together every weekend as does the MA team.  Maybe we can squeeze in three more practices while in Denver :lol:

We're heading to Denver tomorrow, so best of luck to everyone competing and see you in a day or two.

If MA beats 6th it will be because Catalyst wrote us some practice problems and his contribution was invaluable!!!

Try to enjoy the whole experience everyone!!

Coach Frost

good luck everyone! i am on the Dodds team, so i have to start flying today (im leaving for the airport in less than half an hour)

[quote="mathking123"]how do you fit in so many practices? does everybody on the Texas team go to the same school or something to that effect?[/quote]

Generally those teams that have participants near each other can practice more together. For most states, that's not the case as kids are too spread out across the state. Massachusetts is probably easy because I would guess most or all the kids come from the Boston area. In Texas, I think most or all the students come from the Houston area - although I could be completely wrong on this.

MO had states on March 1st and from that day on we only had one practice!!! We average 1 practice every 9 weeks! MO will PHAIL. :(

[quote="shtsxc12"][quote="levans"]
In Texas, I think most or all the students come from the Houston area - although I could be completely wrong on this.[/quote]
Actually, for Texas team, two kids from Houston area, one from College Station, antoher from Austin. The parents from Colledge Station had to drive almost 4 hours round trip three times weekly, the kid in Austin actually flew to Houston weekly., Coach Boyd picked him up and drove him from/to airport each time. Texas team does not live together, but they have a lot of dedication and very supportive parents and coach ![/quote]
And how much does the person spend on airplane fees??? :whistling:

[quote="dnkywin"]And how much does the person spend on airplane fees??? :whistling:[/quote]

TSPE (Texas Society of Prof Eng) paid airplane fees.

Since I'm in HS now, I got tips for Nationals...

- Don't pull off an all-nighter day before test

- Don't spend every minute trading pins before test

- Don't sleep through your wakeup call and alarm

- The curfew's there for a good reason

- Eat breakfast

Yeah..

Since this would be that last day for me on aops since Nationals start(I would be in school), I say good luck to everyone and the Massachusetts team and the two people from my school. Good luck!!!!!!!!!!!!!!!!

I know people are probably arriving here now, but good luck all the same! Wish I could be there in person, but schools mysteriously get angered by unexcused absences. Well, I'll be checking out the webcast in any case.

[quote="math154"]LOL Missouri has some potential. Last year, everyone screwed up.[/quote]

I'm sorry, but seeing as how you didn't even represent Missouri last year, you have no right to talk about people messing up, which they didn't.

MA had 2 people from the same school, and the other two are no more than a 30min drive from our school.
I think

I can't believe I have no life and am actually posting this.

Anyway, stuff worth knowing:

TX first team. MD second(with 3 people in top 6!!!). WA third.

Bobby Shen 1st indiv.

Darryn Wu(sorry if mispelled) cd winner. Came back from 3-1 deficit.

Basically, many favorites to make cd didnt make it this year. If you ask me nats was reaaally weird this year(but it might this weird every year. what do I know)

Denver is soooo boring...

p.s. MO got 8th!!!   :D  :D  :D

Maybe I will go to sleep now...

BOBBY SHEN GOT 2ND PLACE!

lol, i can't believe darryl got 1st he's a sixth grader xD and the first to win the title!

I wonder if nationals has ever been weirder than this year...


Prove that for all natural $n$ one can find $n$ consecutive natural numbers, such that neither of them has the form $a^b$, $a>1$, $b>1$

Chinese Remainder Theorem is what you need:
Take an $x$ with $x+i \equiv p_i \mod p^2$ ($p_i$ any primes), then $x+i$ is divisible by $p_i$ but not by $p_i^2$.


Again as with the previous problem I know the answer to the question below however I do not understand it. In fact in general the class I am expects me to understand proofs however no one has ever formally tought me how to write and understand proofs so if someone could try and help me out or point me in the right direction that would be great.

Let $ \sim $ be an equivalence relation on a set $ X$.

Prove that if $ [a]\cap[b]\neq\O$ then $ [a]\subseteq[b]$

Well, what do you know about equivalence relations? Since it's the only thing you are given, you'll only need the definition of equivalence relation ~ and of equivalence class [a].

Perhaps it would be more intuitive for you to prove a "stronger" property: $ [a]\cap[b]\neq\emptyset \Rightarrow [a] \equal{} [b]$.

It can be, if I'm not mistaken, actually shown that these two properties are equivalent as intersection is a commutative operation and $ F\subseteq G \wedge G\subseteq F\Leftrightarrow F \equal{} G$.

If you need any further hints, please, say so...

Since $ [a]\cap [b]\neq\emptyset$, $ \exists z\in [a], [b]$ such that $ a\sim z$ and $ z\sim b$.  But since $ \sim$ is an equivalence relation, it has transitivity.  Hence $ a\sim z\sim b\Rightarrow a\sim b$.  This means that their equivalence classes are the same since, so $ [a]\equal{}[b]$.  Also, the following are equivalent for any sets $ A$, $ B$:
1) $ A\equal{}B$
2) $ A\subseteq B$ and $ B\subseteq A$

Your result follows.


superior algebra
superior algebra unsolved

How many elements of $ S_{8}$ have the same cycle structure as $ (12)(345)(678)$?

Can anyone check my work?

There are $ \binom{8}{2}$ choices for the first 2-cycle. There are $ \binom{6}{3}*2$ choices for the middle 3-cycle, the *2 coming from the distinction between the permutation $ (123)$ and $ (132)$. There are 2 possibilities for the last 3-cycle. However, this makes no distinction between the permutation $ (345)(678)$ and $ (678)(345)$ which are the same, so we divide by 2.

This gives that the number of elements is $ \binom{8}{2}\binom{6}{3}*2*2*\frac{1}{2}\equal{}1120$.


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

Prove the above.

It's #4 from here http://www.iitjee.org/iit-question-papers/iit-jee-2003-main/iit-jee-2003-maths-main.html

The last "-" is actually meant to be a "=".

This is similar to a Vandermonde -- compare with http://www.artofproblemsolving.com/Forum/viewtopic.php?t=318205

@JBL:Could you show a solution with Vandermonde?
[hide=" this is cool"]
$ \[\binom{n}{j} \binom{n\minus{}j}{k\minus{}j} \equal{} \binom{n}{k}\binom{k}{j}$
but $ \binom{n}{k}$ is a constant so we can factor it out
and now we have a very familiar sum i.e. the binomial series whose sum is quite nice :)
[/hide]


My book says the relative bond strength of hybrid orbitals is as follows:

$ s \minus{} 1$
$ p \minus{} 1.73$
$ sp \minus{} 1.93$
$ sp^2 \minus{} 1.99$
$ sp^3 \minus{} 2$

Can someone give me an explanation for this?

What can you comment  on the bond strength of bonds in $ BF_3$ and $ BF_4 ^ \minus{}$ using this or otherwise?

Well, covalent bonds are a combination between sigma and pi bonds. Because pi bonds are weaker than sigma bonds, there is no linear pattern in the strength of the bonds (remember that a single covalent bond is one sigma, a double bond has one sigma + one pi, and a triple has one sigma, 2 pi).

[quote="7h3.D3m0n.117"]covalent bonds are a combination between sigma and pi bonds[/quote]

That's not a very good description of covalent bonds. Also, the explanation for the relative strenght of bonds made up of different hybrid orbitals does not have nothing to do with sigma versus pi bonds.

Now, [b]Conjurer[/b], to answer your question try to answer these:

1) What is the shape of each orbital s, p, sp, sp2, sp3?

2) What is the relation between the shapes of sp, sp2, sp3, and their p character?

3) What is a covalent bond, and what electron localization favours its strenght?

4) Arrange the hybrid orbitals sp, sp2, sp3 in crescent order regarding your answer to 3).

5) Draw the final conclusions.

1) OK I know
2) s character : sp>sp2>sp3
3)Bond formed by sharing electrons , hmm covalent bonds are strong if the bond lengths are small or multiple bonds are present . If hybridisation involves more s character then, I suppose, the electrons are more bound with the nucleus hence lower potential energy.

4)My answer to third suggests sp> sp2>sp3 i,e the reverse order?

Just a few points.

Well firstly Carcul asked for the $ p$ characteristics of the hybrid orbitals so that should imply to you that considering the $ p$ characteristics is important.

Next when we consider these hybrid orbitals we are considering the $ \sigma$-framework of the molecules so when Carcul asked about bond strength, it has nothing to do with multiple bonds (i.e. $ \pi$ bondings). Bearing this in mind, try to answer that question again.

So, yeah hoped that helped, a bit.

I recently learnt hybridisation but I'm still not too sure if I understand it well ...

As for Carcul's questions, I dont know the shapes of sp,sp2 and sp3 orbitals .. I know how they are formed though but not the shapes.. and what is meant by their $ p$ character?

A covalent bond is sharing electrons and the strength depends on the overlap of the orbitals(greeater the overlap, greater the strength). but yeah that does not really help much

The key lies in maximum orbital overlap.

OK since I know s character then p character should vary inversely. But i cant seem to get anywhere.Can you please tell me the answer?

The more p character an hybrid orbital have more it will be directed in space - remember the shape of p and s orbitals. This means that a corresponding bond will have electrons closest to the space between the nuclei and not closest to each nucleus. Electrons in the region between nuclei contribute more to covalent bond than electrons closest to a nucleus.


Let $a,b,c$ be positive real numbers with $a+b+c=1$.Prove that the following inequality occurs:
\[a\sqrt{b}+b\sqrt{c}+c\sqrt{a}\leq\sqrt{a^{2}+b^{2}+c^{2}}\]

$\sum{a^{2}}\cdot \sum{a}\geq  (\sum{a\sqrt{b}})^{2}$

but $\sum{a}$ = 1 and it`s over


vector
inequalities
advanced fields
advanced fields unsolved

How to prove that all Lp norms are equivalent?
That is for every p, q >0 there exists constants a and b such that for every vector x
a * Lq(x)  <= Lp(x)  <= b * Lq(x)

Now, I know the theorem that all the vector norms are equivalent, but here I need to express explicitly a and b..

Clearly, you are talking about norms in a finite dimensional space; in an infintite dimensional space, the norms are not equivalent. So assume we are talking about $\mathbb{R}^n$ (or $\mathbb{C}^n$,) Assume that $0< p < q<\infty.$

In one direction, use Hölder's inequality:

$\sum_{i=1}^n|x_i|^p\le\left(\sum_{i=1}^n1\right)^{1-\frac{p}q} \left(\sum_{i=1}^n|x_i|^q\right)^{\frac{p}q}$

Taking roots, $||x||_p\le n^{\frac1p-\frac1q}||x||_q.$

For the other direction, start by assuming that $||x||_p=1.$ Then for each $i,$ $|x_i|\le1$ and thus $|x_i|^q\le |x_i|^p.$

Thus, $\sum_{i=1}^n|x_i|^q\le\sum_{i=1}^n|x_i|^p=1.$

The general case scales from this by homogeneity, yielding

$||x||_q\le ||x||_p.$

Both of these inequalities are sharp, as can be seen by using the points $(1,0,0,\dots,0)$ and $(1,1,1,\dots,1)$ for $x.$

Both inequalities work with the obvious modifications for $q=\infty.$


geometry
angle bisector
geometry unsolved

In a ABC triangle, with <A = 90, let K be the feet of the perpendicular from A to BC. Let L be  the point where the angle bisector of <AKC intersects to AC, let M and N be the midpoints of KC and AL respectively. Let P be the point where Mn intersects to KA. show that BK = 2AP

Applying [b]Menelaus theorem[/b] to triangle $ KMP$ with the transversal $ ANC$, we have $ \frac {CK}{CM}\cdot\frac {NM}{NP}\cdot\frac {AP}{AK} \equal{} 1$, i.e. $ \frac {2\cdot AP}{AK} \equal{} \frac {NP}{NM}$. Let $ Q$ be the intersection of the line passes through $ M$ and perpendicular to $ BC$ with $ AC$, then we have $ \frac {2\cdot AP}{AK} \equal{} \frac {NA}{NQ}$ $ (1)$

Let $ BC \equal{} a$, $ CA \equal{} b$, $ AB \equal{} c$, we can easily to get that $ AL \equal{} \frac {bc}{b \plus{} c}$, so $ NA \equal{} \frac {bc}{2\left(b \plus{} c\right)}$ and $ NQ \equal{} AQ \minus{} AN \equal{} \frac {b}{2} \minus{} \frac {bc}{2\left(b \plus{} c\right)} \equal{} \frac {b^2}{2\cdot\left(b \plus{} c\right)}$. It follows that $ \frac {NA}{NQ} \equal{} \frac {c}{b} \equal{} \frac {BA}{CA} \equal{} \frac {BK}{AK}$ $ (2)$

From $ (1)$ and $ (2)$, we conclude that $ 2\cdot AP \equal{} BK$, as desired.
[img]http://i213.photobucket.com/albums/cc110/April_ML/t952174.png[/img]

I can´t get it, can you tell me how you get that AL = bc/b+c   :wink:

Because $ KL$ is the angle bisector of $ \angle AKC$, so we have $ \frac {LA}{LC} \equal{} \frac {KA}{KC}$.

On the other hand, $ \triangle KAC\sim\triangle ABC$, it follows that $ \frac {KA}{KC} \equal{} \frac {AB}{AC} \equal{} \frac {c}{b}$.

Hence, we conclude that $ \frac {LA}{LC} \equal{} \frac {c}{b}$, which implies $ \frac {LA}{LA \plus{} LC} \equal{} \frac {c}{c \plus{} b}$, i.e. $ AL \equal{} \frac {bc}{b \plus{} c}$.

Best regards. April.

Maybe a little bit simpler is to use Menelaus for triangle $ AKC$ and transversal $ PNM$.

$ \frac {CM \cdot KP \cdot AN}{MK \cdot PA \cdot NC}\equal{}1$
$ \frac {KA\plus{}AP}{PA}\equal{}\frac{NC}{AN}\equal{}\frac{CL\plus{}LN}{LN} \Rightarrow \frac{KA}{PA}\equal{}\frac{CL}{LN}\equal{}2\frac {CL}{LA}\equal{} 2\frac{CK}{KA}$ $ \Rightarrow KA^2 \equal{} 2CK.PA$ but $ KA^2\equal{}CK.BK$ ad we are done.

Hi all,

  Given solutions are nice, yet there is a feeling
  that it is an Arbelos setup and with all those
  half lengths there must be some solution with
  Archimedes written all over it.

   M.T.


[hide] Kelly's way
kelly drives $ 21 miles$ at $ 42 mph.$ Solving that, that's 30 minutes

Brenda's way
She drive in a straight line, so we can build a 9,12,15 right triangle .
She travels $ 15 miles$ at $ 45 mph$. solving that, it 20 minutes

$ 30\minus{}20\equal{}10 minutes$
[/hide]


trigonometry
number theory unsolved
number theory

[hide]$ (x \minus{} iy)^3 \equal{} 10 \minus{} 30i \Rightarrow |(x \minus{} iy)^3| \equal{} |10 \minus{} 30i|$

$ \Rightarrow x^2 \plus{} y^2 \equal{} \left(10^2 \plus{} 30^2\right)^{\frac {1}{3}}$

$ \Rightarrow x^2 \plus{} y^2 \equal{} 10$[/hide]

thanks much But how can we do if don't use the complex number. And u has something wrong:
$ (x \minus{} iy)^3 \equal{} x^3 \minus{} 3x^2yi \plus{} 3x(iy)^2 \minus{} (iy)^3 \equal{} x^3 \minus{} 3xy^2 \minus{} (y^3 \minus{} 3x^2y)i \equal{} 10 \minus{} 30i$
$ \Rightarrow |(x \minus{} iy)^3| \equal{} |10 \minus{} 30i| \equal{} \sqrt {10^2 \plus{} 30^2} \equal{} 100$
$ \Rightarrow x^2 \plus{} y^2 \equal{} |x \minus{} iy| \equal{} 100^{\frac {1}{3}}$

that should have been $ (x \plus{} iy)^3 \equal{} 10 \minus{} 30i$

[quote="ngoclam1088"]Let
$ \begin{cases} x^3 \minus{} 3xy^2 \equal{} 10 \\
y^3 \minus{} 3x^2y \equal{} 30 \end{cases}$
Find $ P \equal{} x^2 \plus{} y^2 \equal{} ?$[/quote]
my solution : :thumbup:
[img]http://www.mathmontada.net/vb/uploaded/17_1257844656.gif[/img]
with my best wishes
mrmohamed

[quote="Johan Gunardi"][quote="ngoclam1088"]Let
$ \begin{cases} x^3 \minus{} 3xy^2 \equal{} 10 \\
y^3 \minus{} 3x^2y \equal{} 30 \end{cases}$
Find $ P \equal{} x^2 \plus{} y^2 \equal{} ?$[/quote]square both equations, sum them up, take the third root, done.[/quote]
u mean my solution above your post :wink:
 :huh:

Sorry, I used mobile phone to post my solution, so I can't open the image.

[quote="ngoclam1088"]Let
$ \begin{cases} x^3 \minus{} 3xy^2 \equal{} 10 \\
y^3 \minus{} 3x^2y \equal{} 30 \end{cases}$
Find $ P \equal{} x^2 \plus{} y^2 \equal{} ?$[/quote]
$ x \equal{} k \cdot \cos v, y \equal{} k \cdot \sin v$:
$ x^2\plus{}y^2\equal{}k^2$
$ k^3 \cos 3v \equal{} x^3\minus{}3xy^2 \equal{} 10$
$ k^3 \sin 3v \equal{} y^3 \minus{} 3x^2y \equal{} 30$
Then $ k^6 (\cos^2 3v \plus{} \sin^2 3v) \equal{} 10^2 \plus{} 30^2 \equal{} 10^3 \iff k^2 \equal{} 10$ so $ x^2\plus{}y^2 \equal{} 10$


superior algebra
superior algebra solved

Let $x \star y = xy+mx+ny+p,$ for all $x,y \in R$ be a law of compozition on $R$, an internal law. Also $m,n$ and $p$ are real parameters. We know that the law is asociable $(x \star y) \star z = x \star (y \star z).$ Which are the law's non-simetrizable elements?

[quote="amfulger"]who cares?[/quote]

Ok, I do. I want it out of the unsolved list.
By solving the equation implied by the associativity of $\star$, we get $m=n$ and $p=m^2-m$.
So $x\star y=(x+m)(y+m)-m$.
$e=1-m$ is the neutral element of $\star$ and $\star$ is commutative.
$x\star x'=e$ iff $(x+m)(y+m)=1$. So the only non-simetrizable element is $-m$.

This is not a problem. This is just a piece of junk the techer throws at you to make sure you understand the definitions. And smart students usually dodge.


At his usual rate a man rows $15$ miles downstream in five hours less time than it takes him to return. If he doubles his usual rate, the time downstream is only one hour less than the time upstream. In miles per hour, the rate of the stream's current is:

$\text{(A)} \ 2 \qquad \text{(B)} \ \frac52 \qquad \text{(C)} \ 3 \qquad \text{(D)} \ \frac72 \qquad \text{(E)} \ 4$

[hide]Let $s$ be his speed and $c$ be the speed of the current. The problem tells us that

$\frac{15}{s+c}=\frac{15}{s-c}-5$ and $\frac{15}{2s+c}=\frac{15}{2s-c}-1$. Solving, $c=2$. [/hide]


function
geometry
geometric transformation
parameterization
algebra unsolved
algebra
Functional Equations

can a=0??
if can, then f(x)=x^(sqrt2)
 f(f(x))= x^2-0
this solution sounds stupid, but comple with the requirement :D

@chichi: Well $f(-1)$ does not exist ... Hence your function does not satisfy the requirement.

$x^\sqrt{2}$ is barely counted as polynomial...
But for $a=0$ the solution $f(x) = |x|^{\sqrt{2}}$ , $f(0)=0$ should work; but I think the question is to find all $a$ such that such function exists, not to find one for a special $a$.

If $a\le0$ exist continiosly increase function $g: R_+ \to R_+$, suth that $f(x)=g(|x|), g(g(y))=y^2-a$.

[quote="pavel25"]$f(f(x))=x^2-a$
Hence, $f(x)$ is a polinom.
Suppose the highest degree of $x$ in $f(x)$ is $n$.
So, the highest degree of $f(f(x))$ is $n^2$
Hence, $n^2=2,n=\sqrt{2}$ , so $a=0$ and:
$f(x)=x^\sqrt{2}$[/quote]
But pavel25, the question request us to find function, not find $a$, as ZetaX said above.

[quote="spider_boy"][quote="Lovasz"]Sorry. This is the problem with full condition (in my document). But the one I posted above is also exciting :D .

[quote]Does there exist a function $f: \mathbb{R}\to\mathbb{R}$ such that $f(f(x))=x^2-a, \forall x \in \mathbb{R}$ if $a >1$ is real number.[/quote]

The answer is no, try, $\forall body$ :) .[/quote]

Dear Lovasz, is the solution by using fixed points of the function.
I have seen problem " does there exist a function satisfying $f(f(x))=x+1997$( or $f(f(x))=x^2+1997$,I dont remember exactly)."
It's solution used fixed points.
I will think on this problem!!!!! :D  :D[/quote]
Sorry, Tarlan, I hadn't seen your post until I posted this message ("12:20 pm" and "12:21 pm") :)
I will think about it, but also thanks for your proof if I'm impossible.

[quote="spider_boy"]I have seen problem " does there exist a function satisfying $f(f(x))=x+1997$( or $f(f(x))=x^2+1997$,I dont remember exactly)."
It's solution used fixed points.[/quote]
Well I think the problem you meantion originally was in (positive) integers ;)

hmm...I see no idea in applying fixed points :|

Friends, yesterday i looked up my notes.The problem that I tried to guess was
Prove that there doesn't exist a function $f$ satisfying:$f(f(x))=x^2-1997$.When I looked at the solution i was impressed:
the first sentence of the solution:Let us prove that there doesn't exist a function
$f$ satisfying $f(f(x))=x^2-q$ for $q>1$ :o  :o
The main contours of the solution:
Define $g(x)=f(f(x))=x^2-q$
We find x's such that $g(x)=x$,then we find x's such that $g(g(x))=x$
the reason for $q$ being $>1$ is at the last part.there we must have
the  [b]discriminant[/b] positive.
Then after obtaining the four fixed points of $g(g(x))$ we find that there is additional
one which is contradiction.

The problem I said has appeared at Kvant. :D  :) [url=http://kvant.mccme.ru/pdf/1997/01/24.pdf]kvant.mccme.ru/pdf/1997/01/24.pdf[/url]
and its solution M1578 [url=http://kvant.mccme.ru/pdf/1997/04/21.pdf]kvant.mccme.ru/pdf/1997/04/21.pdf[/url]
But in russian!!! :|

[quote="silouan"]We have many Russians in the forum so   I pleased them to put the translation of the article. Please[/quote]

But I have already described the main parts.Try to continue yourself! :)

Yes, spider boy has described the main parts. The solution basically examines (since f must not be continuous) only discrete points on the graph, and examines fixed points on the graph. Clearly there must be two sets, ie. the two fixed points on $g(x) = x$ and $g(g(x))=x$..however if $g(x)=f(f(x))$ then we can find another pair of fixed points unique from the two pairs we found earlier, thus there cannot exist an f(x) for such parameters

Problem 1, Romanian National Olympiad 1991 (grade 9) reads as follows:
Let $a,b$ be real numbers such that $|a-b|>2$. Prove that there is no function $f: \mathbb{R}\longrightarrow\mathbb{R}$ such that $f(f(x))=x^2-(a+b-1)x+ab.$
The official solution starts with the following lemma:
If $g: A\longrightarrow A$ is a function with exactly two fixed points such that $g\circ g$ has exactly four fixed points, then there is no function $f: A\longrightarrow A$ such that $f\circ f=g$.
Problem proposed by Petru Mironescu :)

[quote="The problem"]Does there exist any function $f: \mathbb{R}\to\mathbb{R}$ such that $f(f(x))=x^2-a$, with $a>1$ is constant.[/quote]

What do you think about this solution, man?

Let $g(x)=x^2-a$, then we have:

$g(g(x))=x$
$\iff (x^2-a)^2-a-x=0$
$\iff \Theta(x)=x-a)(x^2+x-a+1)=0$                 $(*)$
Let $m,n$ be the roots of the eqn $x^2-x-a=0$, $p,q$ be the roots of the eqn $x^2+x-a+1=0$
Suppose that there exists the function $f$ satisfy: $f(f(x))=g(x)$

$a)$ If $f(m)=k$ then $f(k)=f(f(m))=g(m)=m$, hence $g(k)=f(f(k))=f(m)=k$
thus, $k=m$ or $k=n$.

$b)$ If $f(p)=r \Rightarrow f(r)=f(f(p))=g(p)=q$

$c)$ If $f(q)=s \Rightarrow f(s)=f(f(q))=g(q)=p$

But: $g(r)=s\Rightarrow f(f(r))=f(q)=s$ and $g(s)=r$
Assume $r=q$ whence
                 $g(r)= g(q)=f(f(q))=f(s)=r=p$
So:$p=q$:unreasonable$\Rightarrow r\neq s$ and $r \neq p$
           $\Rightarrow$ $(*)$ has $\boxed{5}$ roots, while $\deg\Theta(x)=4$

Obviously, there is no function satisfy the question.
Q.E.D.


algebra
polynomial
Vieta
geometry
3D geometry
calculus
derivative

solve:
$ \begin{cases} (x \plus{} y)^3 \equal{} z \\
(y \plus{} z)^3 \equal{} x \\
(z \plus{} x)^3 \equal{} y \\
\end{cases}$

Let $ x \plus{} y \equal{} a,\ y \plus{} z\equal{} b,\ z \plus{} x \equal{} c$, we have $ 2(a^3 \plus{} b^3 \plus{} c^3) \equal{} a \plus{} b \plus{} c$ and

$ (a \minus{} b)(a^2 \plus{} ab \plus{} b^2 \plus{} 1) \equal{} 0$

$ (b \minus{} c)(b^2 \plus{} bc \plus{} c^2 \plus{} 1) \equal{} 0$

$ (c \minus{} a)(c^2 \plus{} ca \plus{} a^2 \plus{} 1) \equal{} 0$.

Let $ a^3 \equal{} x$ $ b^3 \equal{} y$ $ c^3 \equal{} z$

The conditions turn to

$ (a^3 \plus{} b^3)^3 \equal{} c^3$
$ (b^3 \plus{} c^3)^3 \equal{} a^3$
$ (c^3 \plus{} a^3)^3 \equal{} b^3$

Let $ x \plus{} y \plus{} z \equal{} k$

$ (k \minus{} a^3)^3 \equal{} a^3$ <==> $ k \equal{} a^3 \plus{} a$

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

Therefore, a,b,c are 3 roots of $ f(t) \equal{} t^3 \plus{} t \plus{} k$

If we assume none of a,b,c are equal, then by vieta, they sum to 0, so k=0, so a=b=c=0, contradiction!

So WLOG a=b.

$ (a^3 \plus{} b^3)^3 \equal{} c^3$ so $ c \equal{} a^3 \plus{} b^3 \equal{} 2a^3$

Plugging into the second equation,

$ (3a^3)^3 \equal{} a^3$

$ 3a^3 \equal{} a$

$ (3a^2 \minus{} 1)a \equal{} 0$

So either a=b=0, which implies c=0, so x=y=z=0, or

$ a \equal{} b \equal{} \pm1/\sqrt {3}$ and $ c \equal{} \pm2/3\sqrt {3}$ which gives the 2 other solution sets for x,y,z

[quote="rofler"]$ (k \minus{} a^3)^3 \equal{} a^3$ <==> $ k \equal{} a^3 \plus{} a$[/quote]

I can't understand that.

I took the cube root of both sides and moved a term over...

Are $ x,\ y,\ z$ limited to real numbers?

Mkay, I only solved for the reals. You however, didn't justify a traignle substitution explicitly, so :P

Edit: Damn, nvm

KK, if we let x+y+z=k, then

$ (k \minus{} z)^3 \equal{} z$

$ z^3 \minus{} (3k)z^2 \plus{} (3k^2 \plus{} 1)z \minus{} k^3 \equal{} 0$
so x,y,z are solns of $ f(t) \equal{} t^3 \minus{} (3k)t^2 \plus{} (3t^2 \plus{} 1)t \minus{} k^3 \equal{} 0$

If x,y,z distinct, then x+y+z=3k by vieta, but x+y+z=k, so k=0, and

therefore, x,y,z satisfy $ f(t) \equal{} t^3 \plus{} t$ which works out to be $ x \equal{} i$ $ y \equal{} \minus{} i$ $ z \equal{} 0$ and all permutations thereof.

So WLOG, let x=y. Then $ z \equal{} 8x^3$ and

Plugging into the second equation, $ (9x^3)^3 \equal{} x$

$ 9^3x^9 \equal{} x$

$ x(x^8 \minus{} 1/9^3) \equal{} 0$

Giving x=y=0, which then gives z=0, or

$ x\equal{}y\equal{}CiS(k*pi/4)*(1/9^3)^{1/8}$

$ z\equal{}8(CiS(k*pi/4)*(1/9^3)^{1/8})^3\equal{}8Cis(3k*pi/4)*(1/9^9)^{1/8}$

Where k= 0,1,2,3,4,5,6,7

And all possible permutations of x,y,z given by the solns above.

We have $ y \equal{} \sqrt [3] {z} \minus{} x \equal{} \sqrt [3] {x} \minus{} z$, hence  $ \sqrt [3] {z} \plus{} z \equal{} \sqrt [3] {x} \plus{} x$. But $ \sqrt [3] {x} \plus{} x$ is strictly increasing (it's derivative is $ 1/3x^{ \minus{} 2/3} \plus{} 1$, which is the sum of two squares, thus is always positive, no critical points). So $ x\equal{}y\equal{}z$ by symmetry.
It follows $ x\equal{}y\equal{}z\equal{}\sqrt{2}/4$.

Note: I only found the real roots.


quadratics
modular arithmetic
number theory open
number theory

I am familiar with the following result - for a given integer $ n$ (non-square), there are infinitely many primes $ q$ such that $ \left( \frac{n}{q} \right) \equal{} \minus{}1$. Is the following stronger result holds too?
Given distinct primes $ p_1, \cdots, p_n$, then there are infinitely many primes $ q$ such that $ \left( \frac{p_i}{q} \right) \equal{} \minus{}1$ for all $ 1 \le i \le n$.

It will be not very  hard if we use  Dirichlet 's theorem but I think you want an elementary solution . Is it right ?

I didn't notice that, you are right. We just need to choose
\[ q \equiv 1 \pmod 4 \; , q \equiv s_i \pmod {q_{i}} \; 1 \le i \le n\]
where $ q_i$ are prime divisors of the square-free part of $ n$, and $ \left( \frac{s_i}{q_i} \right) \equal{}1$ (almost the same as the proof I read).

The proof probably didn't want to use Dirichlet's Theorem... I assumed that if my claim was true it would have been mentioned  :blush:

So, yes, it would be cool to see elementary approaches (if we exclude the requirement that $ q$ is a prime)


[quote="mc123"]Hi,
I would like to turn your attention to the following URL

http://mathcounts.saab.org/mc.cgi

Enjoy[/quote]

Saab is often posted, often used, but thanks for posting this for those of the members here who haven't used it much.
Quite useful.

Have any one of you realized that some of her solution and/or answers are wrong? just a few, I get it every so often

Usually people just assume the person's a he and put he, when you put "she" it makes people think the person is a definite she.

[quote="SplashD"]Does it really matter? I'm just saying that not of his or her solutions/answers are right.[/quote]

Any specific problems? I haven't bumped into any that are wrong yet but I rarely use it.

[quote="SplashD"]Does it really matter? I'm just saying that not of his or her solutions/answers are right.[/quote]
Well if some are bad solutions and you missed it, then you can post it on this forum and someone will give a nice solution.

It's pretty easy to tell which one's are wrong, and see what was done wrong  :D that's why I don't post any of them.

The only problem I have with that site is that some of the solutions are incredibly inefficient (for example listing when theres an elegant counting solution) or calculation tedious on sprint round problems

and i think that many of them are not labeled correctly (like sprint when it should be target and vice versa)

the tedious solutions bug me too

other wise, its great to practice, see what you need to improve on.


Let ABC is a triangle   Prove
1+$ \frac{R}{r}>\equal{}\sum\sqrt\frac{a}{\minus{}a\plus{}b\plus{}c}$

I don't know this expresstion ;) .If it's true @@ , we can prove it by Am-Gm ^^
<=>$ \sum ( \frac{a}{b\plus{}c\minus{}a}\plus{}1 \minus{}2\sqrt{ \frac{a}{b\plus{}c\minus{}a} } ) \ge 0$


A fun thing here:

What is the state with the best chance of having a team winning the NCAA basketball tourney, considering all teams have an equal chance.  This requires some background knowledge, plus knowing where schools like Gonzaga are located.  But here is a bracket to help you out.  Sorry if this question doesn't belong here, but it is a fun one :lol:

[img]http://www.madduxsports.com/record5.jpg[/img]

it depends on how well the team performs
but i'm for UNC all the way

Since North Carolina has two #1 seeds(Duke and UNC) and at least one more team(NC State), I'd go with them.

You could make the problem more doable(or more tricky) by assigning probabilities based on seeding.

North Carolina cuz they have 5 teams in NCAA ( the max)

[quote="MysticTerminator"]North Carolina as UNC will win. If they don't for some incredibly contrived reason, like all the players contracting malaria, Wake will.[/quote]

you wish........duke will take it easy

I don't think UNC will win if you saw the way they lost to those pathetic teams.  Some teams can just barely make it to the NCAA tournament then knock a few good teams off, then lose to a bad team. When you play a good team, you have more energy. Did you see the Bucknell-Kansas game last night? (Just to prove my theory) I say that UNC won't win, and perhaps not any of the no. 1 seeds in each bracket. :no:

[quote="MysticTerminator"][quote="mysmartmouth"]Let's try it!  Can anyone come up with a method of showing which teams have the best chance of winning, taking ranks into account?[/quote]

UNC: 1 - epsilon
Wake: .1 + epsilon
Illinois: .1
Duke: -.2
All others: 0

as epsilon tends to 0[/quote]

what?!?!?!

[quote="MysticTerminator"][quote="mysmartmouth"]Let's try it!  Can anyone come up with a method of showing which teams have the best chance of winning, taking ranks into account?[/quote]

UNC: 1 - epsilon
Wake: .1 + epsilon
Illinois: .1
Duke: -.2
All others: 0

as epsilon tends to 0[/quote]

How can a team have negative chance of winning?  Maybe, since some believe negative numbers are greater than infinity, Duke has more than infinify chance of winning, which could mean that even in an infinite number of parallel universes,  Duke would win in every single one.   :D

Hmm, you all seem to have problems with sarcasm.

[quote="Mysmartmouth"]Let's try it! Can anyone come up with a method of showing which teams have the best chance of winning, taking ranks into account?[/quote]

A simple method would be: if teams with ranks m, n play each other, their probabilities of winning are given (respectively) by $\frac{n}{m+n},\frac{m}{m+n}$.  This almost certainly gives weak teams too good a chance (number 16 seeds don't beat number 1 seeds, certainly not 1 out of every 17 times), but it's simple to do an adjustment like $\frac{17n-m}{16m+16n},\frac{17m-n}{16m+16n}$ which weights it more heavily for the better team (giving number 1 seeds a 271/272 chance in their first game -- not bad!)

Using my first model on a tournament of 8 teams ranked 1-8 with usual bracketing, the probability the team ranked 1 wins is given by 225148/426465, or approximately 52.8%.  In a 4-team tournament (teams ranked 1-4), the teams have chances of 14/25, 6/25, 22/175, and 13/175 or  56%, 24%, 12$\frac{4}{7}$% and 7$\frac{3}{7}$%.

did y'll see the Vermont mid court 3pt? knocked off syracuse
can't believe WF lost, i picked them for the final two
unc going all the way
duke final four
nc state rule because them upset both charlott and Uconn(3pt at the last secs)
Gonzaga just can't win
utah is on a row
kasas not lucky this yr, maybe next yr
BC almost made it to Illinois
alabama is elite 8 is long forgotten
UAB upsets LSU but can't do it to Arizona
I pick
final four: UNC, Illinois Duke and Washington/West virginia
then UNC wins it all

With all the crazy upsets so far, it wouldn't surprise me at all to see either Duke or Carolina go down today.

state w/ best chance is definitely west virgina. they looked good against wake and anything is possible.

I'll have to say North Carolina because it looks like the Finals will be UNC vs. Illinois. And Sean May is a lot better than Augustine.....

You can look at this question as number of teams, but you have to take another factor into account: place on the braket. North Carolina has 3 teams in the Sweet 16, more than any other state, but they are all on the same side of the bracket, so only 1 could possibly go to the final. Kentucky and Wisconsin both have 2 teams in the Sweet 16, but they are on opposites sides of the bracket, so theoritically we could have an all wisconsin or kentucky final.

However, the prob. of a North Carolina team in the final is far greater than have 2 Wisconsin or Kentucky teams in the final, and perhaps greater than having one team from either state. However, I still think the Illini will beat said North Carolina team.


inequalities
trigonometry
geometry
circumcircle
geometry proposed

[color=purple]Let $ m_{a}, m_{b}, m_{c}$ are three medians of triangle and$ r_{a},r_{b},r_{c}$ are three exradii . Prove that
     $ \frac{r_{a}\plus{}r_{b}}{m_{c}}\plus{}\frac{r_{b}\plus{}r_{c}}{m_{a}}\plus{}\frac{r_{c}\plus{}r_{a}}{m_{b}}\le\frac{2R}{3r}\plus{}\frac{1}{6}$
  My conjecture :  $ \frac{r_{a}\plus{}r_{b}}{m_{c}}\plus{}\frac{r_{b}\plus{}r_{c}}{m_{a}}\plus{}\frac{r_{c}\plus{}r_{a}}{m_{b}}\ge 6$
[/color] :)  :lol:

[color=purple]Sorry, in the my opinion, in the your proposed inequality is something which isn't right ![/color]

[quote="Virgil Nicula"][color=darkred]Prove that in a triangle $ ABC$ exists the following chain of inequalities :
$ \boxed{\frac{r_{a}+r_{b}}{m_{c}}+\frac{r_{b}+r_{c}}{m_{a}}+\frac{r_{c}+r_{a}}{m_{b}}\le 2\cdot\sqrt{\frac{4R+r}{r}}}\le$ ${ 6\cdot\sqrt{\frac{R}{2r}}}\le\frac{2(R+r)}{r}\le\frac{3R}{r}$.[/color][/quote]
[color=darkblue][b][u]Proof.[/u][/b] I"ll apply the well-known relations $ \{\begin{array}{c}S = pr = (p-a)\cdot r_{a}= (p-b)\cdot r_{b}= (p-c)\cdot r_{c}\\ \\ 2r\le R\ ;\ m^{2}_{a}\ge p(p-a)\ ;\ 9R^{2}\ge a^{2}+b^{2}+c^{2}\\ \\ a(p-a)+b(p-b)+c(p-c) = 2r(4R+r)\end{array}$ and [b]C.B.S.'s inequality[/b].

Thus, $ \sum\frac{r_{b}+r_{c}}{m_{a}}=\sum\frac{aS}{m_{a}(p-b)(p-c)}=\sum\frac{ap(p-a)}{Sm_{a}}=$ $ \frac{1}{S}\cdot\sum[\frac{\sqrt{p(p-a)}}{m_{a}}\cdot\sqrt p\cdot a\sqrt{p-a}]$. Therefore,

$ \sum\frac{r_{b}+r_{c}}{m_{a}}\le\frac{\sqrt p}{S}\cdot\sum[\sqrt a\cdot\sqrt{a(p-a)}]\le$ $ \frac{\sqrt p}{S}\cdot\sqrt{\sum a\cdot\sum a(p-a)}=$ $ \frac{2p}{S}\cdot\sqrt{r(4R+r)}=$ $ 2\cdot\sqrt{\frac{4R+r}{r}}$ $ \implies$ $ \boxed{\ \sum\frac{r_{b}+r_{c}}{m_{a}}\le 2\cdot\sqrt{\frac{4R+r}{r}}\ }$.

[b]In the case of the acute or right triangle[/b] $ ABC$, we have $ m_{a}\le R+R\cos A = 2R\cos^{2}\frac{A}{2}=\frac{2Rp(p-a)}{bc}=\frac{2Rap(p-a)}{4Rpr}=\frac{a(p-a)}{2r}$ $ \implies$

$ \frac{r_{b}+r_{c}}{m_{a}}=\frac{a(p-a)}{rm_{a}}\ge 2$, i.e. $ \boxed{\ r_{b}+r_{c}\ge 2m_{a}\ }$ ([b]interesting ![/b]). In conclusion, $ \max\{A,B,C\}\le 90^{\circ}$ $ \implies$ $ \boxed{\ \sum\frac{r_{b}+r_{c}}{m_{a}}\ge 6\ }$.

[b][u]Remark[/u] ("entertainment").[/b] The bisector of the angle $ \widehat{BAC}$ meet the line $ BC$ in the point $ D$ and again the circumcircle in the point $ A'$. Denote $ \{\begin{array}{c}AD = l_{a}\\\\ AA' = L_{a}\end{array}$.

Observe that $ \triangle ABA'\sim\triangle ADC$ $ \implies$ $ AD\cdot AA' = bc$, i.e. $ \boxed{\ L_{a}l_{a}= bc\ }$. .I"ll use the well-known relations $ \{\begin{array}{c}abc = 4RS = 4Rpr\\ \\ \boxed{\ r(r_{b}+r_{c}) = a(p-a)\ }\\ \\ m_{a}\ge l_{a}\ ;\ L_{a}\le 2R\\ \\ \sum a^{2}(p-a) = 4pr(R+r)\end{array}$.

Thus, $ \sum\frac{r_{b}+r_{c}}{m_{a}}\le\sum\frac{r_{b}+r_{c}}{l_{a}}=$ $ \frac{1}{r}\cdot\sum\frac{a(p-a)}{l_{a}}=$ $ \frac{1}{r}\cdot\sum\frac{a(p-a)L_{a}}{l_{a}L_{a}}=$ $ \frac{1}{r}\cdot\sum\frac{a(p-a)L_{a}}{bc}\le$

$ \frac{2R}{r}\cdot\sum\frac{a(p-a)}{bc}=$ $ \frac{2R}{r\cdot abc}\cdot\sum a^{2}(p-a) =$ $ \frac{4pr(R+r)}{2pr^{2}}$ $ \implies$ $ \sum\frac{r_{b}+r_{c}}{m_{a}}\le\sum\frac{r_{b}+r_{c}}{l_{a}}\le\frac{2(R+r)}{r}$.

Prove easily that in any triangle $ ABC$ we have $ \boxed{\ r_{b}+r_{c}\ge 2l_{a}\ }$. Indeed, $ r_{b}+r_{c}\ge 2l_{a}$ $ \Longleftrightarrow$ $ a(p-a)\ge 2rl_{a}$ $ \Longleftrightarrow$ $ a^{2}(p-a)^{2}\ge 4r^{2}\cdot\frac{4bcp(p-a)}{(b+c)^{2}}$ $ \Longleftrightarrow$

$ a^{2}(p-a)(b+c)^{2}\ge 16bc(p-a)(p-b)(p-c)$ $ \Longleftrightarrow$ $ a^{2}(b+c)^{2}\ge 4bc[a^{2}-(b-c)^{2}]$ $ \Longleftrightarrow$ $ a^{2}(b-c)^{2}+4bc(b-c)^{2}\ge 0$, what is truly.

I used the well-known relation $ (p-a)(p-b)(p-c) = pr^{2}$. In conclusion, [b]in any[/b] $ \triangle ABC$ we have $ \boxed{\ 6\le\sum\frac{r_{b}+r_{c}}{l_{a}}\le\frac{2(R+r)}{r}\ }$.[/color]

[color=purple]Thanks alot Virgil ! Sorry because my mistake  :D my inequality must be
  $ \frac {r_{b} \plus{} r_{c}}{m_{a}}\le 3\left(\sqrt {\frac {4R}{r} \plus{} 1} \minus{} 1\right)$
 And your is stronger than mine  :lol: !
 May be in triangle we have the following inequality :
  $ \sum\frac {r_{b} \plus{} r_{c}}{l_{a}}\ge 2\sqrt {\frac {4R}{r} \plus{} 1}$
     And I can prove more from your idea : In triangle let $ c \equal{} \min\lbrace a,b,c\rbrace$ we have $ r_{a} \plus{} r_{b}\ge\frac {2}{3}\sqrt {1 \plus{} \frac {4R}{r}}.m_{c}$ :D ; with all triangle we have :$ r_{a} \plus{} r_{b}\ge 2l_{c}$ :) [/color]