Given that the ratio of $3x-4$ to $y+15$ is constant, and $y = 3$ when $x = 2$, then, when $y = 12$, $x$ equals:

A. $\frac{1}{8}$
B. $\frac{3}{7}$
C. $\frac{7}{3}$
D. $\frac{7}{2}$
E. 8

[hide]When x=2 and y=3, the result is $\frac{1}{9}$. THe constant ratio is $\frac{1}{9}$.

Thus when y=12, $\frac{3x-4}{27}=\frac{1}{9}$, $x=\frac{7}{3}$. [/hide]

3(2)-4=2
3+15=18
ratio 1:9

12:108=12/9:12=4/3:12

so 4/3 or 1.3333
which is

not an answer, must have done something wrong.

How did you get $12: 108$? Just plug in $y=12$ and use that value for the ratio.


Σε ένα τρίγωνο $ ABC$, $ AC>BC$ οι $ CL, CK$ είναι εσωτερική και εξωτερική διχοτόμος αντίστοιχα και η $ CM$ διάμεσος. Θεωρούμε σημείο $ P$  της $ CM$ έτσι ώστε τα σημεία $ C, A_{1}, B_{1}, P$ να είναι ομοκυκλικά, όπου $ A_{1}\equal{}AP\cap BC, B_{1}\equal{}BP\cap AC$. Να δείξετε ότι τα σημεία $ C, P, L, K$ είναι επίσης ομοκυκλικά.

[i]Θα παρακαλούσα όποιον μπορούσε αργότερα να βάλει ένα σχήμα γιατί εγώ δεν έχω το απαραίτητο πρόγραμμα[/i]

Βάζω τη λύση σε "hide" γιατί νομίζω έχει περισσότερο ενδιαφέρον να προσπαθήσει να λύσει κανείς την άσκηση παρά να τη δει λυμένη :wink:

[hide]Λοιπόν από θεώρημα Ceva για τισ σεβιανές $ AA_{1}, BB_{1}, CM$ έχουμε ότι $ \frac{CB_{1}}{B_{1}A}\equal{}\frac{CA_{1}}{A_{1}B}$.
Από αυτό προκύπτει ότι $ A_{1}B_{1}$ παράλληλη στην $ AB$ .
Αρα $ \angle BAC \equal{} A_{1}B_{1}C \equal{} A_{1}PC \equal{} APM$ (αφού $ CA_{1}PB_{1}$ εγγράψιμο)
Αρα τα τρίγωνα $ APM$ και $ AMC$ είναι όμοια και ισχύει $ AM^{2}\equal{} MP*MC$ (1)

Επειδή όμως τα $ K,L$ είναι αρμονικά συζυγή των $ A,B$ κι αφού $ M$ το μέσο της $ AB$ σύμφωνα με το θεώρημα Newton θα ισχύει  $ AM^{2}\equal{} MK*ML$

Αρα λόγω και της 1 $ MP*MC \equal{} MK*ML$ το οποίο συνεπάγεται το ζητούμενο  [/hide]

[color=red][b] Moderator[/b][/color]

[color=blue] Προσθέτω  το σχήμα που μας έστειλε παρακάτω ο φίλος μας Κώστας Βήττας. Κώστα , ευχαριστούμε.[/color]

[quote="bilot4"][i]Θα παρακαλούσα όποιον μπορούσε αργότερα να βάλει ένα σχήμα γιατί εγώ δεν έχω το απαραίτητο πρόγραμμα[/i]

Βάζω τη λύση σε "hide" γιατί νομίζω έχει περισσότερο ενδιαφέρον να προσπαθήσει να λύσει κανείς την άσκηση παρά να τη δει λυμένη :wink:

[hide]Λοιπόν από θεώρημα Ceva για τισ σεβιανές $ AA_{1}, BB_{1}, CM$ έχουμε ότι $ \frac{CB_{1}}{B_{1}A}\equal{}\frac{CA_{1}}{A_{1}B}$.
Από αυτό προκύπτει ότι $ A_{1}B_{1}$ παράλληλη στην $ AB$ .
Αρα $ \angle BAC \equal{} A_{1}B_{1}C \equal{} A_{1}PC \equal{} APM$ (αφού $ CA_{1}PB_{1}$ εγγράψιμο)
Αρα τα τρίγωνα $ APM$ και $ AMC$ είναι όμοια και ισχύει $ AM^{2}\equal{} MP*MC$ (1)

Επειδή όμως τα $ K,L$ είναι αρμονικά συζυγή των $ A,B$ κι αφού $ M$ το μέσο της $ AB$ σύμφωνα με το θεώρημα Newton θα ισχύει  $ AM^{2}\equal{} MK*ML$

Αρα λόγω και της 1 $ MP*MC \equal{} MK*ML$ το οποίο συνεπάγεται το ζητούμενο  [/hide][/quote]

Ωραία άσκηση και ωραία απόδειξη.

Επιτρέψτε μου μόνο μία μικρή σημείωση. Για τον προσδιορισμό του σημείου $ P$ επί της διαμέσου $ AM,$ όπως ορίζεται στην εκφώνηση, αρκεί να φέρουμε ευθεία δια της κορυφής $ B,$ που να σχηματίζει με την πλευρά $ AB$ γωνία ίση με την $ \angle BCM.$

Κώστας Βήττας.

[quote="silouan"]Ωραία λύση μιμ . Περίπου έτσι την έκανα και εγώ χωρίς όμως το θεωρ Newton
Na η λύση μου
[hide]Αρκεί να δείξουμε ότι το P ανήκει στον απολλώνιο κύκλο τον Β,C δηλαδή αρκεί νδο $ \frac{PB}{PA}\equal{}\frac{CB}{CA}$
Λοιπόν από θεώρημα Ceva για τισ σεβιανές $ AA_{1}, BB_{1}, CM$ έχουμε ότι $ \frac{CB_{1}}{B_{1}A}\equal{}\frac{CA_{1}}{A_{1}B}$.
Από αυτό προκύπτει ότι $ A_{1}B_{1}$ παράλληλη στην $ AB$ .
Αρα $ \angle BAC \equal{} A_{1}B_{1}C \equal{} A_{1}PC \equal{} APM$ (αφού $ CA_{1}PB_{1}$ εγγράψιμο)
Αρα τα τρίγωνα $ APM$ και $ AMC$ είναι όμοια και ;ara $ \frac{BM}{PM}\equal{}\frac{CB}{PB}$

Με όμοιο τρόπο βρίσκουμε ότι $ \frac{MA}{MP}\equal{}\frac{CA}{PA}$

Είναι $ \frac{BM}{PM}\equal{}\frac{MA}{MP}$  άρα $ \frac{CA}{PA}\equal{}\frac{CB}{PB}$ άρα
$ \frac{PB}{PA}\equal{}\frac{CB}{CA}$

το ζητούμενο[/hide][/quote]

Ωραία σκέψη Σιλουανέ, αλλά προτείνω μία απλούστερη προσέγγιση.

Από όμοια τρίγωνα $ \bigtriangleup APM,$ $ \bigtriangleup AMC$ $ \Longrightarrow$ $ \frac{MA}{CM}\equal{}\frac{PA}{CA}$ $ \Longrightarrow$ $ \frac{MA}{PA}\equal{}\frac{CM}{CA}$ $ ,(1)$

Από όμοια τρίγωνα $ \bigtriangleup BPM,$ $ \bigtriangleup BMC$ $ ($ γιατί είπαμε ότι $ \angle PBM \equal{}\angle BCM$ λόγω κατασκευής του $ P$

και $ \angle BMP\equiv\angle BMC$ $ ),$ $ \Longrightarrow$ $ \frac{MB}{CM}\equal{}\frac{PB}{CB}$ $ \Longrightarrow$ $ \frac{MB}{PB}\equal{}\frac{CM}{CB}$ $ ,(2)$

Από $ (1),$ $ (2)$ $ \Longrightarrow$ $ \frac{PB}{PA}\equal{}\frac{CB}{CA}$ το ζητούμενο.

$ \bullet$ Μια ενδιαφέρουσα νομίζω παρατήρηση, είναι ότι το πρόβλημα αυτό είναι ανεξάρτητο από τις διχοτόμους $ CL,$ $ CK,$ της γωνίας $ \angle C.$

Σημασία έχει μόνο το να είναι τα $ K$ και $ L,$ αρμονικά συζυγή των $ A,$ $ B.$

Έτσι αφού το σημείο $ P$ είναι μοναδικό σημείο επί της διαμέσου $ CM$ $ ($ ώστε το τετράπλευρο $ CA_{1}PB_{1}$ να είναι εγγράψιμο $ ),$ συμπεραίνουμε ότι κάθε κύκλος χορδής $ CP$ που τέμνει την ευθεία $ AB$ στα σημεία έστω $ K,$ $ L,$ αυτά είναι αρμονικά συζυγή των $ A,$ $ B.$

Άρα μια εναλλακτική εκφώνηση του προβλήματος, που θα μπορούσε να εμφανιστεί, είναι:

[b][size=100][color=DarkBlue]ΠΡΟΒΛΗΜΑ 1. – Επί της διαμέσου $ CM$ δοσμένου τριγώνου $ \bigtriangleup ABC,$ ζητείται να προσδιοριστεί σημείο $ P,$ έτσι ώστε κάθε κύκλος χορδής $ CP,$ που τέμνει την ευθεία $ AB$ στα σημεία έστω $ K,$ $ L,$ αυτά να είναι αρμονικά συζυγή των $ A,$ $ B.$[/color][/size][/b]

Στην ειδική περίπτωση που ο κύκλος χορδής $ CP$ εφάπτεται στην ευθεία $ AB,$ είναι φανερό ότι το σημείο επαφής είναι η κορυφή $ B,$ του $ ABC$ και έτσι:

[b][size=100][color=DarkBlue]ΠΡΟΒΛΗΜΑ 2. – Κύκλος με χορδή την πλευρά $ BC$ δοσμένου τριγώνου $ ABC,$ εφάπτεται στην πλευρά $ AB$ και τέμνει την διάμεσο $ CM$ στο σημείο έστω $ P.$ Εάν $ A_{1}\equiv BC\cap AP$ και $ B_{1}\equiv AC\cap BP,$ αποδείξτε ότι το τετράπλευρο $ CA_{1}PB_{1}$ είναι εγγράψιμο.[/color][/size][/b]

Κώστας Βήττας.


Solution: [hide]The solution is a+b=2. Indeed, writing a=x+1, the equation a^3-3a^2+5a -17 =0 is equivalent to x^3 +2x = 14;  similarly, writing b = y+1, the equation b^3-3b^2+5b +11 = 0 is equivalent to y^3+2y = -14, so obviously y = -x is a solution. Adding the two equations gives
x^3+y^3+2(x+y)=(x+y)(x^2-xy+y^2_2)=0, so either x+y=0  (and hence a+b=x+y+2=2) or x^2 -xy +y^2 +2= (x-y/2)^2 +3y^2/4+2 =0, which is clearly impossible. [/hide]

Wow !
good to see a revival after a long time !!

[quote=laegolas]The real numbers $\alpha ,\beta$ satisfy the equations $$\alpha^3-3\alpha^2+5\alpha-17=0$$
$$\beta^3-3\beta^2+5\beta+11=0$$
Find $\alpha+\beta$.[/quote]

We can use Cardan's method (also called Cardano's method) to solve the system of equations and find the result.

[quote=bel.jad5][quote=color]The following is known about the reals $ \alpha$ and $ \beta$

$ \alpha^{3}-3\alpha^{2}+5\alpha-17=0$ and $ \beta^{3}-3\beta^{2}+5\beta+11=0$

Determine $ \alpha+\beta$[/quote]

Let $f(x)=x^3-3x^2+5x-3=(x-1)^3+2(x-1)$ and we have $f$ is strictly increasing. On the other hand:
$f(\alpha)=\alpha^3-3\alpha^2+5\alpha-3=(\alpha-1)^3+2(\alpha-1)=14=-f(\beta)=(1-\beta)^3+2(1-\beta)$
then we get: $\alpha-1=1-\beta$ which leads to $\alpha+\beta=2$.[/quote]

Nice solution


Source: Ireland MO 1999
Find all nonzero reals $x>-1$ which satisfy
$\frac{x^{2}}{(x+1-\sqrt{x+1})^{2}}<\frac{x^{2}+3x+18}{(x+1)^{2}}$.


linear algebra
matrix
IMC
college contests

Let $ n$ be a positive integer, and consider the matrix $ A \equal{} (a_{ij})_{1\leq i,j\leq n}$ where $ a_{ij} \equal{} 1$ if $ i\plus{}j$ is prime and $ a_{ij} \equal{} 0$ otherwise.
Prove that $ |\det A| \equal{} k^2$ for some integer $ k$.

Is it true??

$ k=\{\begin{array}{cc} [{\sqrt{n}}], &\mbox{ if }n\not=t^2 \\ 0, &\mbox{ if }n=t^2 , t \geq 2 \end{array}$

where [ ] means floor.

[quote="Kurt Gödel"]Let $ n$ be a positive integer, and consider the matrix $ A \equal{} (a_{ij})_{1\leq i,j\leq n}$ where $ a_{ij} \equal{} 1$ if $ i \plus{} j$ is prime and $ a_{ij} \equal{} 0$ otherwise.
Prove that $ |\det A| \equal{} k^2$ for some integer $ k$.[/quote]

Hint: It is enough that the matrix $ A$ is symmetric, its entries are integers, and $ a_{ij} \equal{} 0$ for all pairs $ \left(i,j\right)$ with $ i \plus{} j$ even and $ i \plus{} j\neq 2$, and $ a_{ij} \equal{} 1$ for $ i \equal{} j \equal{} 1$.

[hide="Real hint"]Permute the rows of the matrix $ A$ such that the even rows come first and the odd rows come after them. Permute the columns in the opposite way (i. e. the odd columns come first and the even columns come after them). If $ n$ is even, the rest is clear. If $ n$ is odd, you will have to permute a bit more carefully (the $ 1$-st row and the $ 1$-st column should be in the middle).[/hide]

  darij

We shall give a convenient expression for the determinant of a matrix whose $ (i,j)$-th entry is zero whenever  $ i+j$ is even and at least 4.

To begin with, denote by $ B$ the set of bijective maps from $ \{1,3,\ldots , 2n-1\}$ to $ \{2,4,\ldots , 2n\}$,  and fix an $ h\in B.$ For $ f,g\in B,$ define $ \pi(f,g)\in S_{2n}$ to be the permutation given by $ \pi(f,g)(i): =f(i)$ if $ i$ is odd and  $ \pi(f,g)(i): =g^{-1}(i)$ if $ i$ is even; also define  $ \delta(f): =\textrm{sgn}\pi(f,h).$ One checks that $ \pi(f,g)\pi(g,f)=\textrm{Id}$ and that $ \pi(f,g)=\pi(f,h)\pi(h,h)\pi(h,g),$   and so $ \textrm{sgn}\pi(f,g)=\delta(f)\delta(g)\delta(h).$  It is also straightforward to check that $ \delta(h)=(-1)^n.$

First, let $ X=(x_{i,j})$ be a $ 2n\times 2n$ matrix with $ x_{i,j}=0$ whenever $ i+j$ is even and at least 4.  For  $ \sigma\in S_{2n}$ the product
$ x_{1,\sigma(1)}\ldots x_{2n,\sigma(2n)}$ vanishes unless $ \{\sigma(1),\sigma(3),\ldots , \sigma(2n-1)\}=\{2,4,\ldots , 2n\}$ i.e. $ \sigma=\pi(f,g)$ for some $ f,g\in B.$ Hence
\[ \textrm{det}X=\&\sum_{\sigma\in S_{2n}}(\textrm{sgn}\sigma )x_{1,\sigma(1)}\ldots x_{2n,\sigma(2n)}=\sum_{f,g\in B}\textrm{sgn}\pi(f,g)x_{1,f(1)}x_{3,f(3)}\ldots x_{2n-1,f(2n-1)}x_{g(1),1}x_{g(3),3}\ldots x_{g(2n-1),2n-1}=(-1)^n(\sum_{f\in B}\delta(f) x_{1,f(1)}x_{3,f(3)}\ldots x_{2n-1,f(2n-1)})(\sum_{f\in B}\delta(f) x_{f(1),1}x_{f(3),3}\ldots x_{f(2n-1),2n-1}).\]

Now let $ X=(x_{i,j})$ be a $ 2n+1 \times 2n+1$ matrix with $ x_{i,j}=0$ whenever $ i+j$ is even and at least 4.  One checks that $ x_{1,\sigma(1)}\ldots x_{2n+1,\sigma(2n+1)}$ vanishes unless $ \sigma(1)=1$ and $ \{\sigma(2),\sigma(4),\ldots , \sigma(2n)\}=\{3,5,\ldots , 2n+1\}.$  The determinant of $ X$ is $ x_{1,1}$\times the $ (1,1)$-th minor. (The minor is of course of the form considered in the preceding paragraph.) For a closed expression, set $ y_{i,j}: =x_{i-1,j-1}$ to obtain
\[ \textrm{det}X=(-1)^nx_{1,1}(\sum_{f\in B}\delta(f) y_{1,f(1)}y_{3,f(3)}\ldots y_{2n-1,f(2n-1)})(\sum_{f\in B}\delta(f) y_{f(1),1}y_{f(3),3}\ldots y_{f(2n-1),2n-1}).\]

If in addition $ X$ is symmetric and $ x_{1,1}=1$ we obtain $ \textrm{det}X= \pm1\times \textrm{a square}.$


function
number theory unsolved
number theory

Find all integers[b][i] k[/i][/b] such that there exist a function[b] [i]F[/i][/b] such satisfying:
[b][i]1.f(2) isn't 0.
2.for all integers x,y:
f(x,y)=f(x)+f(y)+kf((x,y)).[/i][/b]

 :alien:

I think you mean something like that, am I wrong?

http://www.mathlinks.ro/Forum/viewtopic.php?highlight=taiwan+1997&t=18691

Pierre.

[quote="RobertuX"]I suppose you try to say that:

[quote="kvmedya"]Find all integers [b]$k$[/b] such that there exist a function [b]$f$[/b] such satisfying:

\[ f(2)\ne 0, \quad {\rm and }\quad
  f(x\cdot y)=f(x)+f(y)+kf((x,y)) \quad \forall \, x,y\in \mathbb{Z}.
\]

 :alien:[/quote]
right?[/quote]

Are you well?                  :rotfl:  :rotfl:
You most solve the problem not ...          :D

[quote="RobertuX"][quote="kvmedya"]Find all integers [b]$k$[/b] such that there exist a function [b]$f$[/b] such satisfying:

\[ f(2)\ne 0, \quad {\rm and }\quad
  f(x\cdot y)=f(x)+f(y)+kf((x,y)) \quad \forall \, x,y\in \mathbb{Z}.
\][/quote]
[/quote]
Well, if $f(4)=0$ then must be $k+2=0$, so $k=-2$ and, if $f(1)\ne 0$ then $k+1=0$, so $k=-1$, then if this function exists must be
1. If $f(1)\ne 0$ then only is possible consider $k=-1$,
2. if $f(1)=0$ and $f(4)=0$ then $k=-2$,
3. if $f(1)f(4)\ne 0$ then for every $k$ is defined a function $f$ because for every prime $p$ the function can be defined as we want and for every $n$ we descompose and use the properties of $f$ to define in this value, so every $k$ is possible to define an $f$.

Best regards.

RX TCM


a circle intersects sides $BC,AC,AB$ of triangle $ABC$ at $A_1,A_2,B_1,B_2,C_1,C_2$ . tangents at ($A_1$ and $A_2$), ($B_1$ and $B_2$), ($C_1$ and $C_2$) to the circle intersect at $A',B',C'$.
Prove that $AA',BB',CC'$ are concurrent.

The point A' is the point where the tangents to the circle at $A_1$ and $A_2$ intersect, i. e. the pole of the line $A_1A_2=BC$ with respect to the circle. Similarly, the points B' and C' are the poles of the lines CA and AB with respect to the circle. Thus, your problem becomes a consequence of http://www.mathlinks.ro/Forum/viewtopic.php?t=38235 .

  Darij

darij grinberg, I have a question  :huh: I think the geometric problem in China mo 2005 of course, at the first round, is the polar of this problem, am I wrong?  :huh:


probability
calculus
integration
conditional probability
probability and stats

All cards are drawn without replacement from a full deck of 52 cards.

What is the probability for exactly two queens to be drawn consecutively?

Is it $ \frac {4}{52} \frac{3}{51}$ ?

That is the probability that you do it in the first two cards. The probability that you do it at some point in the process is considerably more complicated.
By my calculations, its numerical value is about $ 0.22$
(I'd rather not give away the exact version yet)

damn.
Does it make the problem a bit more simple if im to look for the probability for the queen of hearts and the queen of diamons drawn consecutively and the queen of spades and queen of clubs drawn consecutively?

Thanks.

Btw: Those questions require knowledge of distributions and advanced probability ? We only reached the end of conditional probability and the begining of random variables...

More specifically, I get the following approximate probabilities:

$ 0.782624434$ that no two queens are ever consecutive.
$ 0.204162896$ that there is exactly one pair of consecutive queens, and the  other queens occur singly.
$ 0.008687783$ that there are two separated pairs of consecutive queens.
$ 0.004343891$ that three queens occur together, with the other one separate.
$ 0.000180995$ that all four queens occur together.

jmerry seems to have given the sum of the last four numbers as his answer. That would be simpler to compute (as the complement of the first number) than any of the other numbers.

I don't particularly see this as an advanced probability question - more of a tricky counting question. My tools were from elementary combinatorics. (Don't confuse "elementary" with "easy.")

If you prefer fractions, these numbers are

$ \frac{4324}{5525},\frac{1128}{5525},\frac{48}{5525},\frac{24}{5525},$ and $ \frac{1}{5525}.$

Actually, I did it as an inclusion/exclusion sum, but that was the probability I was thinking of.

Here's how I calculated the first number, the probability that the four queens are all separated from each other by other cards.

Label each card with a number from $ 1$ to $ 52,$ according to its position in the deck. Let $ a,b,c,d$ be the locations of the four queens. We have that $ 1\le a<b<c<d\le52,$ and we can choose those locations in $ \binom{52}{4}$ ways.

But to have the four queens separated by other cards, we must have that $ 1\le a<b\minus{}1<c\minus{}2<d\minus{}3\le49.$ We can choose the numbers $ a,b\minus{}1,c\minus{}2,$ and $ d\minus{}3$ in $ \binom{49}{4}$ ways.

Then, the probability that the queens are separated is

$ \frac{\binom{49}{4}}{\binom{52}{4}}\equal{}\frac{\frac{49\cdot48\cdot47\cdot46}{4!}}
{\frac{52\cdot51\cdot50\cdot49}{4!}}\equal{}\frac{49\cdot48\cdot47\cdot46}
{52\cdot51\cdot50\cdot49}\equal{}\frac{48\cdot47\cdot46}{52\cdot51\cdot50}$

The other four quantities are trickier to count.

My expression for the complement was $ 3\cdot \binom{51}{3}\minus{}3\cdot\binom{50}{2}\plus{}\binom{49}{1}$, with the same $ \binom{52}{4}$ denominator. That comes from "gluing together" the queens in various combinations.

Thanks guys.

Do you have suggestion how to get better at these? I can usually solve the easiar problems, but these are extremely hard.

Should I just do hundreds of those untill im sick of it (like I did before to practice limit calculations and integrals) ?

 :blush:


Goofus has 87 dimes and 213 quarters. Doofus has 213 dimes and 87 quarters. How many dollars do Goofus and Doofus have when they combine their money?

A faster way to solve this would be: since the two people have the same amount of coins $(213+87=300)$, we know that they have 300 dimes and 300 quarters. We can go ahead and solve this like ernie's solution, but there is a faster way. We know that $300d+300q=300(d+q)$, and $d+q=0.25+0.10=0.35$. Therefore, our final equation is $300(0.35)=\boxed{\$105}$.


There are ten points marked on a circumference, numbered from $1$ to $10$ and join all points with segments. I color the segments, with red someones and others with blue. Without changing the colors of the segments, renumber all the points from the $1$ to the $10$. Will be possible to color the segments and to renumber the points so that those numbers that were jointed with red are jointed now with blue and the numbers that were jointed with blue they are jointed now with red?

If I understand the question correctly, the answer is no, because $10\choose2$ is odd.


Prove that in any $ \triangle ABC$, we have $ \sin{A}\plus{}\sin{B}\plus{}\sin{C}\leq \frac{3\sqrt{3}}{2}$.

[hide][/$ \text {Since  f(x)\equal{}sin x is concave function. By Jensen's theorem we have}$

   $ sin \frac {A \plus{}B\plus{}C} {3} \ge \frac {sinA \plus{}sinB\plus{}sinC}{3}$

   $ \implies sin \frac {\pi} {3}\ge \frac {sinA \plus{}sinB\plus{}sinC}{2}$

   $ \implies \frac {3\sqrt 2} {3}\ge \ sinA \plus{}sinB\plus{}sinC$

  $ \text {reference : }$

   [url]http://mathcircle.berkeley.edu/trig/node2.html[/url][/hide]

f(x) = sinx is convex so by Jensens
$ f(a) \plus{} f(b) \plus{} f(c) \geq 3 f(\frac {A \plus{} B \plus{} C}{3}) \implies sinA \plus{} sinB \plus{} sinC < 3 sin60 \equal{} \frac{3\sqrt{3}}{2}$

[quote="mathwizarddude"]How to find the minimum?[/quote]
Observe that $ 0 < \sin x \leq 1$ for $ 0 < x < \pi$, which are the bounds for angles in a triangle, so we cannot have $ \sin \alpha \plus{} \sin \beta \plus{} \sin \gamma \leq 0$

Then observe that $ \lim_{t \to 0} \sin t \plus{} \sin t \plus{} \sin (\pi \minus{} 2t ) \equal{} 0$, so we can get arbitrarily close to $ 0$.

This makes the value $ 0$ the unattainable lower bound.

What exactly are you talking about; what do you expect to happen in that case?

Also, a degenerate triangle is not acute because at least one of the non-zero angles must be at least 90º.

[quote="mathwizarddude"]What if the triangle is acute and can be degenerated (i.e., an angle with measure 0)?[/quote]
$ A,B,C < 90$
So assume WLOG $ A \ge B \ge C$.
We know $ B\plus{}C \equal{} 180\minus{}A$
Given $ A$ The smallest possible value of $ sin(B) \plus{} sin(C)$ is when $ B \equal{} A$ and $ C \equal{} 180 \minus{} 2A$. (It's very easy to see if you draw sinus in $ [0;\pi/2]$.
So we want to find the smallest possible value of $ 2sin(A) \plus{} sin(180 \minus{} 2A) \equal{} 2sin(A) \plus{} sin(2A)$
And since $ A \ge 60$. We get the smallest value when $ A$ is very close to $ 90$.(Again. Draw sinus in $ [0;\pi]$). And the lower bound is $ 2$.

In a triangle ABC, for µ > 0 , we have
sinA + µ (sinB+sinC )
≤1/8 {√[2 µ√( µ²+8)+2µ²+4][µ√( µ²+8) - µ²+4]}

it can be solved using Jenson's inequality.if ABC is acute , you can directly prove it using Jenson's inequality.
if it is rt. angled , then it is trivial using standard results. if the triad is obtuse , then a little modification will complete the proof :D

$\sin A+\sin B+\sin C\le\frac {3\sqrt 3}{2}\iff \sum\frac {a}{2R}\le \frac {3\sqrt 3}{2}\iff s\le\frac {3R\sqrt 3}{2}$ .

[b]See[/b] [url=http://www.artofproblemsolving.com/blog/32103][color=darkred][b]here[/b][/color][/url] a proof  of the well-known inequalities $3r\sqrt 3\le s\le\frac {3R\sqrt 3}{2}$ ([b][u]Mitrinovic[/u][/b]).


[img]http://imgs.xkcd.com/comics/nerd_sniping.png[/img]

Humor notwithstanding, the problem is genuinely interesting, and I'd appreciate a solution.  I'd try to arrive at one myself, but

1) my background in E&M is poor, and
2) I can't see a way in which it even makes sense to talk about equivalent resistance in an infinite grid.  The only thing I can think of is a limit of a sequence of equivalent resistances in finite grids of increasing size, and the calculations involved in the resulting recurrence seem to me fairly difficult.

Take a look at Peter Doyle's website:
[url]http://math.dartmouth.edu/~doyle[/url].

There you can download his book with Snell on [i]Random Walks and Electrical Networks[/i].  That book solves related questions.  I'm not sure if it solves the exact problem you posted.

There is a great deal of discussion on the matter in the xkcd fora. You may want to check out this thread

[url]http://fora.xkcd.com/viewtopic.php?f=7&t=16104&start=0[/url]

I recommend the fora there general. :)

It's about solutions to the discrete Laplace equation $ f(x\plus{}1,y)\plus{}f(x\minus{}1,y)\plus{}f(x,y\plus{}1)\plus{}f(x,y\minus{}1)\equal{}4f(x,y)$; if we put a voltage between the two key points, the voltages satisfy this everywhere else. and are bounded at $ \infty$.

A fundamental solution to the continuous version is $ f(x,y)\equal{}\ln(x^2\plus{}y^2)$; we want to find the discrete analogue and subtract two of them.

actually one  could use superposition theorem from electric theory which actually has it's origins in the
properties of laplacians


Hi, is there a general method for finding out whether two matrices A and B are similar,and if yes, then what is the similarity transform P  such that B=PAP^(-1)?
Thanx in advance..

That's a bit much too explain completely here, but in short :

that is what Jordanforms are all about : two matrices are similar if and only if they have the same jordanform


inequalities
inequalities unsolved
algebra
highschoolmath

\[ \Longleftrightarrow\frac{1}{x^2}\plus{}\frac{1}{y^2}\plus{}\frac{1}{z^2}\geq \frac{49}{\sum{(4x\plus{}2y\plus{}z)(4y\plus{}2z\plus{}x)}}\equal{}\frac{49}{35(xy\plus{}yz\plus{}zx)\plus{}14(x^2\plus{}y^2\plus{}z^2)}\]
Let $ p\equal{}x\plus{}y\plus{}z,q\equal{}xy\plus{}yz\plus{}zx,r\equal{}xyz$,then
\[ \Longleftrightarrow \frac{q^2\minus{}2pr}{r^2}\geq \frac{49}{35q\plus{}14(p^2\minus{}2q)}\]
so we should prove $ f(r)\equal{}49r^2\plus{}2pqr\plus{}4p^3r\minus{}q^3\minus{}2p^2q^2\leq 0$,but $ f''(r)\equal{}98>0$,so we only need to check the following two cases:
$ (i)\quad a\equal{}b$
$ (ii)\quad c\equal{}0$

Your Inequality is weaker than the old result: If $ a,$ $ b,$ $ c$ are positive real numbers, then
$ \frac{1}{(2a\plus{}b)^2}\plus{}\frac{1}{(2b\plus{}c)^2}\plus{}\frac{1}{(2c\plus{}a)^2} \ge \frac{1}{ab\plus{}bc\plus{}ca}.$

[quote="Tourish"]
\[ \frac {1}{(2a \minus{} b)^2} \plus{} \frac {1}{(2b \minus{} c)^2} \plus{} \frac {1}{(2c \minus{} a)^2}\geq\frac {23}{11(a^2 \plus{} b^2 \plus{} c^2)}\]
:D[/quote]
How to prove it? Buy the S.O.S theory?

[quote="can_hang2007"]Your Inequality is weaker than the old result: If $ a,$ $ b,$ $ c$ are positive real numbers, then
$ \frac {1}{(2a \plus{} b)^2} \plus{} \frac {1}{(2b \plus{} c)^2} \plus{} \frac {1}{(2c \plus{} a)^2} \ge \frac {1}{ab \plus{} bc \plus{} ca}.$[/quote]
I don't think so
the best constant k for inequality
$ \frac {1}{(2a \plus{} b)^2} \plus{} \frac {1}{(2b \plus{} c)^2} \plus{} \frac {1}{(2c \plus{} a)^2} \ge \frac {k}{ab \plus{} bc \plus{} ca}.$
is $ k\equal{}1,181220153...$
It' can't solve by Cauchy-Schwarz

[quote="Tourish"]
\[ \frac {1}{(2a \minus{} b)^2} \plus{} \frac {1}{(2b \minus{} c)^2} \plus{} \frac {1}{(2c \minus{} a)^2}\geq\frac {23}{11(a^2 \plus{} b^2 \plus{} c^2)}\]
:D[/quote]

This inequality is only true for positive numbers

Now for any real numbers, then we have the following inequality of Pham Kim Hung
$ \frac {1}{(2a \minus{} b)^2} \plus{} \frac {1}{(2b \minus{} c)^2} \plus{} \frac {1}{(2c \minus{} a)^2}\geq\frac {11}{7(a^2 \plus{} b^2 \plus{} c^2)}$

best constant $ k$ for inequality
$ \frac {1}{(2a \minus{} b)^2} \plus{} \frac {1}{(2b \minus{} c)^2} \plus{} \frac {1}{(2c \minus{} a)^2}\geq\frac {k}{7(a^2 \plus{} b^2 \plus{} c^2)}$
is $ k \equal{} 11,60749544...$


[quote="GameBot"]Doug buys some oranges at $ n$ cents per dozen and sells them at $ \frac n8$ cents each. What is the profit, in cents, on each item? Expess your answer as a common fraction in terms of $ n$.[/quote]

$ \frac{n}{8}\minus{}\frac{n}{12}\equal{}\frac{n}{24}$ cents.


This is another Canadian contest  :D . But this one's for grade 12 students (mostly). This is the question I found the most interesting and easiest to do:

http://cemc.uwaterloo.ca/english/contests/past_contest/2006/euclid.pdf
This is the website in case anyone's interested. :idea:

nobody replied  :( . Come on ppl, it's not that hard.

may be there not exist any pakistani peopele here. ;)

its really really easy problem.

because of $DE$, is paralle to $BC$, and $\angle AFC=90$,

then i conclude that : $\angle ADE=90$

and it measn that $AE$ is diameter, and arc $ACE=180$.

$\angle ACE$=arc$\frac{CE}{2}$

$\angle ABC$=arc${\frac{AC}2}$

then if we sum them , we should have:

$\angle ACE+\angle ABC$=arc$\frac{CE}{2}$+arc$\frac{AC}{2}$

=$\frac{180}{2}=90$.

 ;)

[quote="Ashegh"]may be there not exist any pakistani people here. ;)[/quote]
I like your proof, it exactly what I did :) . As for the Pakistani ppl, there are some (me!), but most just don't feel interested (I think).

Here's another harder question. If you are an experienced problem solver, please give other people a chance before posting your solution. :)


One of my speed dials is FTW, but sometimes, on Opera, I click on the link and the site stays blank. If I go to IE, it works perfectly fine. The next day, FTW would work perfectly fine on Opera. Is it something wrong with my comp or is it something else?

Probably an Opera issue, likely related to how it handles the Flash plugin.


induction
pigeonhole principle
number theory
combinatorics unsolved
combinatorics

There is graph $ G_0$ on vertices $ A_1, A_2, \ldots, A_n$. Graph $ G_{n \plus{} 1}$ on vertices $ A_1, A_2, \ldots, A_n$ is constructed by the rule: $ A_i$ and $ A_j$ are joined only if in graph $ G_n$ there is a vertices $ A_k\neq A_i, A_j$ such that $ A_k$ is joined with both $ A_i$ and $ A_j$. Prove that the sequence $ \{G_n\}_{n\in\mathbb{N}}$ is periodic after some term with period $ T \le 2^n$.

I don't have much time to type this up, but here is a somewhat detailed sketch (hopefully correct):

[hide] We prove via induction; the base cases n=1,2,3 are trivial. Assume that a complete subgraph $ K_i$, $ i\ge 3$, exists at some stage $ G_k$. Then this particular subgraph must also exist at any subsequent stage. Hence, the size of the largest subgraph at any stage is monotonically increasing. Since $ i\le n$, at some point this largest size will become constant; assume this size is $ i_0$. Then no member of this $ K_{i_0}$ can be connected to any member outside of it, since otherwise at the next stage we will have a $ K_{i_0\plus{}1}$, contradicting the fact that $ i_0$ is constant. Therefore, we can split the set of $ n$ members into two groups: $ K_{i_0}$ and everyone not in $ K_{i_0}$ and apply the induction hypothesis. Therefore, the period of these n vertices is at most $ 2^{|K_{i_0}|}\times 2^{n\minus{}|K_{i_0}|}\equal{}2^n$, as desired.

Now suppose that no such such subgraph will ever exist. Then at any stage, there must be at most $ n$ segments because otherwise by the Pigeonhole principle, some vertex will have at least three segments coming out of it, at which point the next stage will have a complete $ K_i$, $ i\ge 3$. Then consider the number of segments that exist at any stage $ G_k$ (basic counting argument). It is not hard to see that this value never increases, so at some point it becomes constant. For this point, any vertex with edges coming out of it has exactly 2 such edges; therefore, we can decompose the set of $ n$ vertices into disjoint cycles. If there are at least two cycles, we can apply the induction hypothesis. If there is exactly one, then we have two cases, whether this cycle length is even or odd. If it is even, at the next stage it will decompose into two cycles, so we are done. If it is odd, assume it has size $ r$. If $ r<n$, then we can again use the induction hypothesis. So $ r\equal{}n$; then basic number theory guarantees that this cycle has period at most $ \phi(n)$, which is clearly less than $ 2^n$, as desired.[/hide]

I want to know if the following proof is correct or not:
It's obvious that a cycle in $G_n$ remains in $G_{n+1}$. So any other edge which is not in a cycle might remain or might not remain. And we know that there is at most n-1 of such edges. So there are $2^{n-1}$ possibilities of their existence.  So the Pigeon Hole Principle says that there are two similar graphs among $G_0$ to $G_{2^n-1}$. So obviously there is a cycle. With length at most $2^n$.
I can't see anything wrong with this proof. But I didn't expect this problem to have such an easy proof. So I'm suspicious to my proof. Is it OK or not?

[quote="havaliza"]I want to know if the following proof is correct or not:
[u]It's obvious that a cycle in $G_n$ remains in $G_{n+1}$.[/u] So any other edge which is not in a cycle might remain or might not remain. And we know that there is at most n-1 of such edges. So there are $2^{n-1}$ possibilities of their existence. So the Pigeon Hole Principle says that there are two similar graphs among $G_0$ to $G_{2^n-1}$. So obviously there is a cycle. With length at most $2^n$.
I can't see anything wrong with this proof. But I didn't expect this problem to have such an easy proof. So I'm suspicious to my proof. Is it OK or not?[/quote]

That is wrong , For example take a cycle of length $4$ .
 If you read the problem again you`ll see it said
[quote="April"]$ A_i$ and $ A_j$ are joined [u]only if[/u] in graph $ G_n$ there is a vertices $ A_k\neq A_i, A_j$ such that $ A_k$ is joined with both $ A_i$ and $ A_j$. .[/quote]

The main idea in my solution was looking at the connected components of $G$ .


ratio
AMC
USA(J)MO
USAMO
number theory proposed
number theory

Let $ S$ be set of 2005 positive numbers $ a_1,a_2,..,a_{2005}$ Let $ T_i$ be non empty subset of $ S$, $ s_i$ be sum of the numbers belonging $ T_i$. Prove that set of numbers $ s_i$ can be partitioned into 2005 non-empty disjoint subsets so that the ratio of two arbitrary numbers in a subset does not exceed 2.


Let $ \triangle ABC$ be acute-angled, and let $ P$ and $ Q$ be points on its side $ BC$. Construct a point $ C_1$ in such a way that the convex quadrilateral $ APBC_1$ is cyclic, $ QC_1 \parallel{} CA$, and the points $ C_1$ and $ Q$ lie on opposite sides of the line $ AB$. Construct a point $ B_1$ in such a way that the convex quadrilateral $ APCB_1$ is cyclic, $ QB_1 \parallel{} BA$, and the points $ B_1$ and $ Q$ lie on opposite sides of the line $ AC$. Prove that the points $ B_1$, $ C_1$, $ P$, and $ Q$ lie on a circle.

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=213011#213011

USAMO in intermediate?


Let  ABC  be a triangle. Let  X  be a point on  AB and  Y  be a point on  AC  such that  AX=XY=YB=BC  . If the angle  ABC  measures  132  degrees, find the number of degress in the measure of the angle  BAC .

[img]http://www.part.lt/img/3ee887ae7b60ee95ffdfe81a02235f76859.JPG[/img]

$ \angle BAC\equal{}\angle AYX\equal{}n$

$ \angle AXY\equal{}180\minus{}2n$

$ \angle YBX\equal{}\angle YXB\equal{}2n$

$ \angle XYB\equal{}180\minus{}4n$

$ \angle ACB\equal{}\angle BYC\equal{}4n$

$ n\plus{}4n\plus{}132\equal{}180$

$ n\equal{}\boxed{9,6}$ (degrees)

You're not right.
$ \angle ACB\equal{}\angle BYC\equal{}3n$
$ \angle YBC\equal{}180\minus{}6n$
$ \angle ABC\equal{}180\minus{}4n\equal{}132$
$ n\equal{}12$


geometry
inradius
incenter
Pythagorean Theorem

[quote="Fierytycoon"]I saw the inradius expressed as r = (a+b-c)/2 (where a and b are the legs of a right triangle and c its hypotenuse). How is this formula derived? I can see that the area of the triangle = ab/2 = r((a+b+c)/2) <=> r = ab/(a+b+c), but I do not know whether this is the right path. Suggestions?[/quote]

Let's show these two statements are equivalent for a right triangle...

For any triangle, A=rs, where s is the semiperimeter.  s=(a+b+c)/2.  This can be shown by drawing a line from the incenter to each vertex and from the incenter perpendicular to each side.  We have drawn 6 right triangles, each of which have a height of r, and total combined base of a+b+c.

For a right triangle, A=ab/2.

Substitute and solve for r:
  A = rs
  ab/2 = r*(a+b+c)/2
  ab = r*(a+b+c)
  r = ab/(a+b+c)
[This is the equation you already showed.]

But your other equation says
  r = (a+b-c)/2

Set r=r
  ab/(a+b+c) = (a+b-c)/2
  2ab = (a+b-c)(a+b+c)
  2ab = ((a+b)-c)((a+b)+c)
  2ab = (a+b)^2 - c^2  [diff. of squares]
  2ab = a^2 + 2ab + b^2 - c^2
  c^2 = a^2 + b^2

If you believe the Pythagorean Theorem, we're done.

in the book Proof Without Words, they have a picture proof.  Its nice.

Another way: Let X and Y be the points of tangency of the incircle with AC and BC.  (The incircle is the circle tangent to all three sides of the triangle.)

Find CX in terms of a, b, and c (this is a formula all geometers should know automatically).

What can we deduce iff CX = IX (where I is the center of the incircle)?


Evaluate

\[\cot \left[\sum_{k=1}^n \cot ^{-1}(1+k+k^2)\right]\]

I think that [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=4959]this[/url] (post #4) should help to this hard problem.
Or is that trick unhelpful?

It's not really a trick: you don't need to know this ahead of time to solve the problem. I was able to solve this (it's on the Post-test of the complex/trig class) with no previous knowledge of that identity but was able to derive the identity along the way. (Of course, it took me a MUCH longer time than if I had known it before ;))

Basically, I'll try to motivate how I discovered that arccot(n^2+n+1) = arccot(n+1) - arccot(n). Whenever you see a rather ugly sum, you should immediately be alert to the possibility of telescoping. But how do we split up an inverse cotagent into parts that would telescope? Well, let's try to use the addition formula. We know that cot(a+b) = 1/tan(a+b) = (tan a tan b - 1) / (tan b + tan a) = (1 - cot a cot b) / (cot a + cot b). Right now, this isn't helpful. We want inverse cotagents. So let x = cot a, y = cot b s.t. a = arccot x, b = arccot y. Now we have cot ( arccot x + arccot y) = (1 - xy) / (x + y). If we take arccot of both sides, we now have arccot x + arccot y = arccot ((1-xy)/(x+y)). This is much more helpful. Now we try to get this arccot(k^2+k+1) as resembling the RHS of that identity. Now, either you're good at algebra and you can see that k^2 + k + 1 = (1 - (-k)(k+1)) / (-k + k+1), or you're not that good and you just guess that x = k+1, y = -k, as you want this thing to telescope and those choices work out well. (I did the latter). In any case, you can then find the identity arccot(k^2+k+1) = arccot(k+1) - arccot(k) and finish the problem.


Mr. Smith has three daughters: Anna, Brenda, and Diana. The sum of Anna’s age and Brenda’s age is 11 more than Diana’s age. The sum of Anna’s age and Diana’s age is 25 more than Brenda’s age. The sum of Brenda’s age and Diana’s age is 19 more than Anna’s age. How old is Anna?

[hide]Let the ages of Anna, Brenda, and Diana be $ a$, $ b$, and $ d$, respectively. We have,

$ a\plus{}b\equal{}d\plus{}11$

$ a\plus{}d\equal{}b\plus{}25$

$ b\plus{}d\equal{}a\plus{}19$

Adding them up, we find that

$ a\plus{}b\plus{}d\equal{}55$. Thus,

$ 2a\plus{}19\equal{}55 \implies a\equal{}\boxed{18}$.[/hide]


YOUR INEQUALITY SHOULD BE REVERSED!!!

We have to prove that $ (a \plus{} b \plus{} c)^2(a^2 \plus{} b^2)(a^2 \plus{} c^2)(b^2 \plus{} c^2)\leq8(a^2b^2 \plus{} a^2c^2 \plus{} b^2c^2)^2$ .

Denote this expression by (*). Then,
$ (*)\Longleftrightarrow (a \plus{} b \plus{} c)^2(\frac {a^2 \plus{} b^2}{2})(\frac {b^2 \plus{} c^2}{2})(\frac {c^2 \plus{} a^2}{2})\leq(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2)^2$
$ \Longleftrightarrow (a \plus{} b \plus{} c)^2(ab)(bc)(ca)\leq(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2)^2$ [From the AM - GM ineq]
$ \Longleftrightarrow [abc(a \plus{} b \plus{} c)]^2 \leq(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2)^2$
$ \Longleftrightarrow abc(a \plus{} b \plus{} c) \leq(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2)$
$ \Longleftrightarrow 2a^2b^2 \plus{} 2b^2c^2 \plus{} 2c^2a^2\geq2a^2bc \plus{} 2ab^2c \plus{} 2abc^2$
$ \Longleftrightarrow (ab \minus{} bc)^2 \plus{} (bc \minus{} ca)^2 \plus{} (ca \minus{} ab)^2\geq 0$

Which, of course, is true!! This proves that your inequality should be reversed and also the rectified inequality itself!!  :lol:  :D  :roll:

EDIT: SORRY, yeah, I get it :oops:

You flipped an AM-GM in the second line.

Also, $ a\equal{}2,b\equal{}1,c\equal{}0$ gives $ 180 \geq 128$, so the inequality isn't flipped.

Well, then where is my flaw? I understand that the inequality should be reversed for strictly positive reals::say, $ a \equal{} 1,b \equal{} 2,c \equal{} 3$ or something like that, it gives $ 23400\ngeq117128$. Then perhaps $ a,b,c\in\mathbb{R}^ \plus{}$ is the question which allows the AM GM. Thus, in my opinion, the question should be: PT, for all positive reals $ a,b,c$, $ (a \plus{} b \plus{} c)^2(a^2 \plus{} b^2)(a^2 \plus{} c^2)(b^2 \plus{} c^2)\leq8(a^2b^2 \plus{} a^2c^2 \plus{} b^2c^2)^2$; isn't it so? :huh: Hence I think that my solution is completely correct.

EDIT: Extremely sorry!! :oops:

I'll reiterate: you flipped an AM-GM in the second line. That is where your flaw is.

$ (a \plus{} b \plus{} c)^2(\frac {a^2 \plus{} b^2}{2})(\frac {b^2 \plus{} c^2}{2})(\frac {c^2 \plus{} a^2}{2})\leq(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2)^2$
$ \Longleftrightarrow (a \plus{} b \plus{} c)^2(ab)(bc)(ca)\leq(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2)^2$

This equivalence is false.

For $ (a,b,c) \equal{} (1,2,3)$ the right side is $ 19208$, not $ 117128$.

Well, of course now I get it! Both my assumptions are false?!! :blush:
 I cannot apply AM GM when one of the three $ a,b,c$ are $ \equal{} 0$; and of course, my calculations were all incorrect. Thanks for showing me my flaw :|
I have tried for many values of $ a,b,c$; and in fact all of them yield that my solution is incorrect.

At last, sorry for the inconvenience and also for my pathetic proof.  :(

My proof
$ (a\plus{}b\plus{}c)^2(a^2\plus{}b^2)(a^2\plus{}c^2)(b^2\plus{}c^2)\geq8(a^2b^2\plus{}a^2c^2\plus{}b^2c^2)^2$
leads to
$ \sum\limits_{cyc} {(\frac{{(b \plus{} c)^2 }}{2}}  \plus{} 2bc)(a^2 b^2  \plus{} a^2 c^2  \plus{} b^2 c^2 )(b \minus{} c)^2  \ge 0$
which is obvious :lol:

[quote="TRAN THAI HUNG"]My proof
$ (a \plus{} b \plus{} c)^2(a^2 \plus{} b^2)(a^2 \plus{} c^2)(b^2 \plus{} c^2)\geq8(a^2b^2 \plus{} a^2c^2 \plus{} b^2c^2)^2$
leads to
$ \sum\limits_{cyc} {(\frac {{(b \plus{} c)^2 }}{2}} \plus{} 2bc)(a^2 b^2 \plus{} a^2 c^2 \plus{} b^2 c^2 )(b \minus{} c)^2 \ge 0$
which is obvious :lol:[/quote]
Id est, $ (a \plus{} b \plus{} c)^2(a^2 \plus{} b^2)(a^2 \plus{} c^2)(b^2 \plus{} c^2)$ divisible on $ a^2 b^2 \plus{} a^2 c^2 \plus{} b^2 c^2.$ Are you sure?

My mistake
It should be
$ \sum\limits_{cyc} {((\frac {{(b \plus{} c)^2 }}{2}} \plus{} 2bc)(a^2 b^2 \plus{} a^2 c^2 \plus{} b^2 c^2 )\plus{}2a^4bc)(b \minus{} c)^2 \ge 0$ :blush: Is it true now :oops: ?

Can you please post your solution, Arqady?

Thank you very much. :)

manlio, after expanding it is obviously true, but I don't see a nice proof.

[quote="arqady"]Let $ a,$ $ b$ and $ c$ are non-negative numbers. Prove that:
\[ (a \plus{} b \plus{} c)^2(a^2 \plus{} b^2)(a^2 \plus{} c^2)(b^2 \plus{} c^2)\geq8(a^2b^2 \plus{} a^2c^2 \plus{} b^2c^2)^2
\]
[/quote]

this inequality is written in the book "Secrets in  Inequalities" by Pham Kim Hung, Vietnam

The inequality is by Titu and Gabriel .

The solution I know is to rewrite $ \sum\sqrt {\frac {abc}{b \plus{} c \minus{} a}}\geq a \plus{} b \plus{} c$

This follows from Schur $ abc(a \plus{} b \plus{} c)\geq\sum a^3(b \plus{} c \minus{} a)$ and holder ineq

$ \sum a^3(b \plus{} c \minus{} a)\geq\frac {(\sum a)^3}{(\sum\frac{1}{\sqrt {b \plus{} c \minus{} a}})^2}$

[quote="arqady"]Let $ a,$ $ b$ and $ c$ are non-negative numbers. Prove that:
\[ (a \plus{} b \plus{} c)^2(a^2 \plus{} b^2)(a^2 \plus{} c^2)(b^2 \plus{} c^2)\geq8(a^2b^2 \plus{} a^2c^2 \plus{} b^2c^2)^2
\]
[/quote]
See it at my book here: http://canhang2007.wordpress.com/2008/12/30/new-ebook-of-me-at-mathsvn-forum/ :)
(Solution of problem 12)

[quote="arqady"]Let $ a,$ $ b$ and $ c$ are non-negative numbers. Prove that:
\[ (a\plus{}b\plus{}c)^2(a^2\plus{}b^2)(a^2\plus{}c^2)(b^2\plus{}c^2)\geq8(a^2b^2\plus{}a^2c^2\plus{}b^2c^2)^2\][/quote]
$uvw$ kills it easily. :wink:

[quote=arqady]Let $ a,$ $ b$ and $ c$ are non-negative numbers. Prove that:
\[ (a\plus{}b\plus{}c)^2(a^2\plus{}b^2)(a^2\plus{}c^2)(b^2\plus{}c^2)\geq8(a^2b^2\plus{}a^2c^2\plus{}b^2c^2)^2\][/quote]

We have
\[LHS-RHS=\sum{ab(a^3b+4a^2b^2+2a^2c^2+ab^3+2b^2c^2+2c^4)(a-b)^2}\ge{0}\]


rotation
geometry
geometric transformation

100 points are in the plane, no 3 collinear. Prove that there exists some line in the same plane in which 50 points lie on one side and the other 50 points lie on the other.

[hide="big hint"]rotate the line[/hide]

Let l a line, $B_{1},B_{2},...,B_{100}$ are on l such that $B_{i}\not = B_{j}$ and $A_{i}B_{i}$ perpendicular to l. Then let C is point betwen $B_{50}$ and $B_{51}$, p is line pass C perpendicular to l, and done  :)

[hide]Create a line.  There will be $x$ points on one side and $100-x$ points on the other side where $x\neq 50$.  If you rotate the line $180^\circ$, the number of points on each side will be flip flopped (from less than 50 to more than 50).

Thus, there must be one instant where the number of points on each side is $50$.[/hide]

could you give a rigorous proof of your last statement?  or is it axiomatic/too hard to prove?

they are talking about what happens [i]during[/i] the rotation  :wink:

[quote="mathgeniuse^ln(x)"][quote="13375P34K43V312"]Let's say we have x points on one side of the line and 100-x points on the other side. Then, when we rotate it by 180 degrees, we get 100-x and x. Since no three are collinear, we have to go from x to 100-x one at a time. Since x, 50, and 100-x are all integers, and 50 is between x and 100-x, 50 must show up at some point.[/quote]

Wait won't rotating the line 180 degrees just be the same? :oops:[/quote]

It would look the same, but if we were to keep track of which side was which, then it would be different.

Oh :maybe:

Then yeah I forgot about this question.  It looks the same but it changes...that is so cool

[quote="13375P34K43V312"]Let's say we have x points on one side of the line and 100-x points on the other side. Then, when we rotate it by 180 degrees, we get 100-x and x. Since no three are collinear, we have to go from x to 100-x one at a time. Since x, 50, and 100-x are all integers, and 50 is between x and 100-x, 50 must show up at some point.[/quote]

this is not exactly rigirous, what if you happen to hit two points at the same time..., to finish the problem, all you need is to choose your center of rotation as a point that does not lie on any line determined by two of the ponts


AMC
USA(J)MO
USAMO
AIME
AMC 12
algorithm
AMC 10

i've heard two beleifs of other people and i am wondering if they are correct.

(1)Usamo is qualified for based on an [i]index[/i], which is 1 times the highest amc score (whether it's the 10 or the 12) added to 10 times the aime score (which can only be taken once). then if the index is suffieciently high, one is invited to usamo.
(2) same as (1) if you take the 12. if you take the 10, the index does not matter; the only requirement is to make AIME, and pass a minimum score6-8(usually  on it determined by the amc12 takers.

are either/both/neither of these correct?

You can find your answers [url=http://www.unl.edu/amc/]here[/url], including your question asked in another thread about time limit.

The first one is definitely true.  I'm not sure about the second.

The selection algorithm is at:
http://www.unl.edu/amc/e-exams/e8-usamo/usamo.shtml

Your question is answered in item 4.

Steve Dunbar
MAA Director, American Mathematics Competitions

so, even if you took the 10, and beat the floor value for the aime, you would still need a decent index right?

Yes; I know a person (several AoPsers, actually) that got 132 AMC 10, 7 AIME last year. The floor was 6, cutoff was 202.5 for index.

wow, that must suck to miss it by half a point. now i'm nervous about the amc10 if i don't get a 150.


Hello, everyone,

Microsoft is introducing a new Windows, Vista, and it's coming to stores in 2006.

What do you think is going to be new in this new version of Windows? What do you think Microsoft improved from Windows XP? New programs? Debug fixtures? A new skin? Say your opinions here.

hwenterprise

[quote="solafidefarms"]I've followed the announcements about Longhorn/Vista. From what I've heard, it isn't going to have much more than XP. No Monad(the new shell), a positively ugly WM/skin which is the only one available, few improvements. A new IE which will have very few features - for example, no file-edit-view-bookmarks-etc. bar. Debug features? Wish on. Oh, yes, and a even less dev-friendly file manager. One that has "virtual" folders, meaning that when you move programs around, they don't really move. AHHH!!!

Can you tell I'm not getting Vista?!

When you are installing your new Vista, try Linux before you buy it. You'll be pleased.[/quote]Trust me linux is far, far away from being a (general) good OS. Besides the fact that it's free (some distributions) and it's stable and relatively safe -- network wise, on the rest it basically sucks (multimedia especially, which is why most people have personal computers).

Vista is going to bring many more improvements that you can see at this level. Working on 64 bit processors, different skins (!), a much better IE and so on. I think you heard wrong :D

As for multimedia, what Windows programs are you talking about? Linux has a very good set of programs.

I'd recommend SuSE to get started with; it has practically every program you could need on the install CD's.

[quote="solafidefarms"]As for multimedia, what Windows programs are you talking about? Linux has a very good set of programs.[/quote]I'm not going to get started in a Windows vs Linux home computer. There are tons of programs that are not available on Linux, not to mention drivers and other issues.

PS Let's be serious just configuring the internet access on a computer takes for ever if you are not well versed in Linux. It's just not user friendly, and it will never be, because then it will stop being Linux.

[quote="Valentin Vornicu"]PS Let's be serious just configuring the internet access on a computer takes for ever if you are not well versed in Linux. It's just not user friendly, and it will never be, because then it will stop being Linux.[/quote]

All I had to do to set up the internet on my morphix distribution was to plug in the ethernet cable and run epiphany.

But Windows has been, in general, improving has newer versions are released. I already have windows XP, so I'm probably not getting Vista.

[quote="Derek"]The re-introduction of the blue screen  :D[/quote]I haven't seen a blue screen on my computer in years. However I've seen them often on my mobile phone, running Symbian OS, which is a competitor of Windows Magneto 2005. So the blue screen is not (c)-ed to Microsoft anymore :D

Ok, sure, I'll admit WiFi is hard to do with Linux. But otherwise, internet's a piece of cake. And as for multimedia, try SuSE 9.3 Pro. opensuse.org. Try it. It'll be better then Vista, and by the time there's a new version of Windows, SuSE 10.0 will be out and it's even better, from the beta of it. :) :lol:

And not userfriendly: I think you've gotten Gentoo confused with Linux. SuSE and Ubuntu are perfectly user-friendly. And when you've got the internet, you can just search when you have problems. Not that you will...

I don't know about SuSe, but last year I gave Slackware 9 a try. Worst experience possible. I am far from being a newbie in computers, and I wasn't able to set up the ethernet connection in 6 hours, whereas in windows that takes 3 clicks. I tried searching google, and every effort I made was in vane.

I cannot image the vast majority of people that have no idea what TCP/IP means installing a linux env., and then configuring it. Let's be serious ...

...you tried SLACKWARE?! On the website, it says that it sacrificed user-friendlyness for being like unix. :(

And as for programs not available on Linux that are on Windows, what programs do you speak of that don't have good equivalents in Linux? Sure, there are some, but you can have a dual-boot system.

Hehe, to have a dual-boot you must already know a lot of stuff (to install it yourself). Let's remember that most people have no idea Windows has be installed on a computer! :) And you'ld have to get Vista anyway then, so what's the point in wasting time with another OS?!

You see, the lack of security and stability in Windows systems has been pretty much corrected (compared with Windows 98 and Windows 2000 times). However Linux's lack of popularity hasn't (and for good reasons).

To have adualboot you don't have to do much. The auto-partitioner generally takes care of it.

And I mean to dual-boot XP and Linux, not Vista and Linux.

Besides, not many people are going to upgrade their XP install to Vista; they'll be getting a new computer. Did you hear that Vista will take AT THE MINIMUM a 2.5GHz processor and 512 MB RAM? How many of us have that on our XP computers?(I've got plenty more on my Linux computer, but the family computer has 256 RAM.)

[quote="solafidefarms"]Besides, not many people are going to upgrade their XP install to Vista; they'll be getting a new computer. Did you hear that Vista will take AT THE MINIMUM a 2.5GHz processor and 512 MB RAM? How many of us have that on our XP computers?(I've got plenty more on my Linux computer, but the family computer has 256 RAM.)[/quote]I think that was exactly what they said when they launched Windows XP :D You are "hearing" too much staff that (is most probably) not correct.

About buying a new computer: with what OS do you think a new computer will come? (except if you buy it by parts from newegg :D) Furthermore, Vista is designed for 64-bit processors. Do you know any 64-bit processors that run at less than 2.5 Ghz?! I don't! :)

[quote="solafidefarms"]Obviously Valentin doesn't want to give Linux another try. :mad: And you know that on my Linux computer, I [i]did[/i] get the parts from newegg ;) ...[/quote]Obviously I don't want to waste more time :D Windows XP does everything I need it to do on my home computer, so why should I bother installing another OS? :)

Yes you did get parts, but you sound like the kind of guy that [b]can[/b] build his computer from scrath (this is what I did also, with some parts from newegg). However the vast majority of the people have no idea what is inside a computer case :D

I will admit that I was wrong about the amount of ram. Recommended: 1GB RAM for 32bit, 2GB for 64bit, minimum 512 for 32bit and 1GB for 64bit. http://www.bit-tech.net/news/2005/09/07/vista_hardware_reqs/

You do realize that that link you posted is absolutely presenting a rumour?!  ;) Furthermore Vista is in beta stage, so you have to wait until the final product is out.

I wandering around and looking at the various skins of the new Windows and I found out a download that's an "upgrade" to Longhorn...  :huh:

I downloaded this and it gave me the longhorn sidebar (even though it doesn't work) and some other things of Longhorn. (like the change of icons, change of welcome screen, and other changes) It was 22 mb so I don't think it was a theme or something.

It looks pretty neat, but it doesn't have ANY differences from XP besides the change in icons. *sarcastic wow*

[quote="Drunken_Math"][quote="Valentin Vornicu"]Yeah, I had something similar installed. Had to totally reinstall windows though :) Stuff was not working as it was supposed to after a few months. I kept only the background of the desktop :D[/quote]

Uh-oh...  :(

What happened after a few months?[/quote]As I told you stuff started to be messed up on the computer :)

When all is said, I'm excited for Windows Vista, even if its not going to be a spectacular perfect machine. Technology today isnt up to that (or our programmers arent, whichever) - and we have to accept that Vista wont be perfect - but it WILL be an improvement from XP, and any improvement is good. So yes, I will get Vista.

Also, on Linux, its very hard to compare (practically) Windows to Linux. Linux is a safer alternative, and on some, like SuSe (which i run on one of my systems), come with everything many people will need, and so the fact that other software and drivers arent available isnt much of a hindrance. Linux is a cool system, and, best of all, its free.

I would never ditch my Windows system and just work on Linux though - Windows is the standard, its the architecture of computers. Using only Linux prevents me from using MANY features that are exclusive to Windows - and especially all the apps. So, IMO, Linux is good, but not as a replacement.

Windows XP dies, or crashes, on me a lot. It's either my computer or Windows itself. I hope Vista will have this error fixed. Do you think there'll be a new "Office Vista"???

Windows Vista is probably going to be starndard on PCs of the future. That's what I'm hoping, so my next computer could be Vista.

:D

What about Mac? I just had to get mac osx tiger for school, and its wonderful. If you're getting a new computer to support vista, i probably wouldnt bother. MAC RULES!!!  :lol:  :love:

I don't know. Lots of people don't by Macintosh. I don't know why...

Is Macintosh better than Windows?

I think Windows is better, though that is just me.

Of course Windows is better and easier for the average home user. But for people who have their own servers, Linux is definitely the choice.

Is Linux actually a program like Windows?

I've never really understood Linux...

Linux is an operating system. It's like how Macs are different from Windows.


Al walks down to the bottom of an escalator that is moving up and he counts 150 steps. His friend, Bob, walks up to the top of the escalator and counts 75 steps. If Al's speed of walking (in steps per unit time) is three times Bob's walking speed, how many steps are visible on the escalator at a given time? (Assume that this value is constant.)

[hide]
Let [b]a = 3b[/b] be Al's walking speed and [b]b[/b] be Bob's walking speed. Let [b]D[/b] be the number of steps on the escalator.

I changed this problem a bit to an equivalent problem. When Al is walking down, imagine that the escalator is growing in length at a rate of [b]e[/b]. When Bob is walking up, imagine the escalator is shrinking in length, also at a rate of [b]e[/b].

The left equations mean how far Al and Bob moved from their starting points. The middle equations mean how far the tip of the escalator is from Al's and Bob's starting points. (Time $t_{a}$ and $t_{b}$ are the times they took.) When these equations are equal is when Al and Bob get off of the escalator.
\begin{eqnarray*}3b t_{a}&=& D+e t_{a}= 150 \\ b t_{b}&=& D-e t_{b}= 75 \end{eqnarray*}
Dividing these two equations:
$\frac{3b t_{a}}{b t_{b}}= \frac{150}{75}\Rightarrow t_{a}= \frac{2}{3}t_{b}$
Now subbing $t_{a}$ with $t_{b}$ gives:
\begin{eqnarray*}D+e \left(\frac{2}{3}t_{b}\right) &=& 150 \\ D-e t_{b}&=& 75 \end{eqnarray*}
Add 3/2's of the 1st equation to the second equation to get:
$\frac{5}{2}D = 300 \Rightarrow \boxed{D = 120}$
[/hide]


[quote="liyi"]find the maximum value of $f(x) = \sin x (2+\cos x)$, can it be solved without calculus?[/quote]

Is it sin (x(2+cos x)) or (sin x)(2+cos x)?

[tex]
\begin{eqnarray*}
(\sin x)(2 + \cos x) = 2\sin x + \sin x \cos x\\
&=&2\sin x + \frac{\sin 2x}{2}
\end
[/tex]

There are two things that might make this as large as possible: a large value for sin x and a large value for sin 2x.  The x that makes sin x the larges is pi/2.  Thus if x = pi/2, then

\[
2\sin \frac{\pi}{2} + \frac{\sin \pi}{2} = 2
\]

And if x = pi/4 (so sin 2x will be largest), then we have

\[
\sqrt{2} + \frac{1}{2}
\]

2 is the maximum.

[quote="TripleM"]That wouldn't be enough as a proof - when you have two functions together, their maximum is not necessarily when you have the maximum of one or the other. They could both be just under their maximum, which would give a higher total..[/quote]

I figured that out when checking my answer in a graphing calculator.  It came up with a max of 2.2something.  I've been trying to come up with something better, but I haven't found anything yet.


I got this by email and not sure if this can be solved....

Ms X was dozing in her maths class and woke up just when
the professor said [u][i][b]".. and all the roots are real and positive".  [/b][/i][/u]

Looking at the blackboard/whiteboard she saw a $20th$ degree equation which she tried to copy down hurriedly before the Professor erased it. She could only get the first two terms $x^{20} - 20x^{19}$ before the equation was erased, but remembered that the constant term was $+1$.

Can we help our dreamy student solve the equation and determine all the roots?

Can it be done?

[quote="hydro"]I think we can solve it since the equation was just a Newton bynomial, $(1-x)^{20}=0$. This equation has 20 positive real roots, all $x=1$. The first two terms written on the blackboard agree with this fact, since $(1-x)^{20}=x^{20}-\binom {20}{1}x^{19}+...+1$[/quote]

But is this the only possibility?

[hide]$x_1,x_2,\ldots,x_{20}>0;\quad x_1x_2\cdots x_{20}=1;\quad x_1+x_2+\cdots+x_{20}=20$

[b]AM-GM[/b] inequality gives \[ \frac{x_1+x_2+\cdots+x_{20}}{20}\ge \sqrt[20]{x_1x_2\ldots x_{20}} \] with equality iff $x_1=x_2=\cdots=x_{20}$. But substituting, we have \[ 1\ge 1 \iff \boxed {x_1=x_2=\cdots=x_{20}=1}. \][/hide]

This has appeared here , courtsey of my lazy lab:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=Mahabharata&t=88424[/url]

[quote="Gyan"]This  reminds me of an old story from Mahabharata.

"All  the roots of this equation are real and positive" Guru Dronacharya thundered, and this thundring woke our hero the great  Bhima (who liked to sleep in the class) up. He hurried to copy this 20 degree equation from the  board, but he got only up to the first two terms before Guru D. wiped the black board clean. He however remembered that the constant term was 1. Thus the equation looked like:
$x^{20}-20x^{19} \dots +1 = 0$

Can you help, our hero (lazy as he may be) solve the equation and find all the roots. :)[/quote]

The key point  here is when all the roots are positive real and
  GM = AM = 1, the only possibility is all the roots are equal = 1


On average, how many times must a coin be flipped before there are two more tails than heads or two more heads than tails?

Well, I've figured out that $\displaystyle{\frac{1}{2}}$ of the time it will be after 2 flips, $\displaystyle{\frac{1}{4}}$ of the time it will be after 4, $\frac{1}{8}$ of the time after 6, $\frac{1}{16}$ of the time after 8, etc.

Does that have something to do with it?

Yeah, but all I got from my series was that the probability of getting it at some time is 1, but the fact that it will always happen eventually doesn't help me figure out the average number of times you have to flip the coin to get it...

answer was 4, we'll put solutions up soon.
-kartik


the numbers 1,2,...,n, (n>=5) are written on the circle in the clockwise order. Per move it is allowed to exchange any couple of consecuitive numbers a,b to the couple (a+b)/2 , (a+b)/2.
It is possible to make all numbers equal using these operations?

The sum of the numbers is $n(n+1)/2$ so for them all to be equal, each must be $(n+1)/2$.
I think this problem is solved with invariants but I can't think of a solution right now. Using the n=5 case, I find that after an arbitrary large number of moves we have 3 numbers as the average and the remaining two arbitrarily close to the average, making me think that it is impossible. Now I will think of why it must be so...


I only got 122 on the AMC12A (I had a very bad day) so I'm wondering what kind of score I'll need on the AIME to make it to the AMO.

So, my question is, during a typical year, what AMC12/AIME index (AMC12 score + 10*(AIME score)) is needed to make the cut for USAMO?  Can people who have made it to the USAMO in the past say what their combined index was?

Last year, it was 210.5 (or something similar), two years ago, it was 226.

Wait...on the "Go for the Gusto?" thread, someone said that they rounded the qualifying index score down to the nearest ten. Is this true?


probability
algebra
polynomial
function
inequalities
complex numbers

You may think I am really stupid, but what does it mean if a number is trancendental? Are all trancendental numbers irrational?

Algebraic numbers are those which are solutions to algebraic equations with integer co-efficients, eg $ \sqrt[3]{2}$ is a solution of $ x^{3}\minus{}2 \equal{} 0$.

Numbers that are not algebraic are called transcendental numbers (their existance was first proved by Joseph Liouville in 1844), famous transcendental numbers are $ \pi$ and $ e$.

Transcendental numbers are always irrational, but all irrational numbers are not transcendental. Eg: $ \sqrt{2}$ is irrational, but also algebraic since its a solution to $ x^{2}\minus{}2\equal{}0$.

Almost all real numbers are transcendental numbers (Cantor proved this).

Given the definition, the fact that all rational numbers are algebraic (and thus that all transcendental numbers are irrational) follows because $ \frac{p}{q}$ is a root of $ qx\minus{}p \equal{} 0$.

There are also functions called transcendental functions, which include exponential, logarithmic, trigonometric, inverse trigonometric, hyperbolic, and inverse hyperbolic functions.

[quote="MellowMelon"]Can it actually be said that all transcendentals are irrational? I thought that both the algebraic and transcendental numbers were subsets of $ \mathbb{C}$, not $ \mathbb{R}$. This convention would be because the set of all roots of integer polynomials (i.e. the algebraic numbers) has plenty of complex elements. So in this way you could say that all [i]real[/i] transcendentals are irrational.

Can someone confirm this, or am I being a source of misinformation? :maybe:[/quote]

I dont think this really matters, how many non-real complex numbers do you know that can be expressed as a/b, where a and b are integers? So eitherway, all transcendentals are not-rational regardless of their realness, whether or not you want to distinguish between not-rational and irrational is up to you.

[quote="t0rajir0u"][quote="tabbot"]Almost all real numbers are transcendental numbers (Cantor proved this).[/quote]

To clarify:  in mathematics, "almost" can have a mathematically precise meaning.  To say that something holds for "almost every real" means that the set of reals for which it does not hold have measure zero.  In other words, the "probability" of a randomly chosen real number being transcendental is $ 1$ (equivalently, the probability of a randomly chosen real number being algebraic is $ 0$).  Note that this does not mean that algebraic numbers do not exist!  :)[/quote]
Isn't that sort of like the probability of a strict inequality being almost equal to the probability if the nonstrict inequality?

In this case I don't know if measure theory is the appropriate formalization of "almost all" in this context-- certainly, cardinality will do fine (there are continuum-many transcendental numbers and only countably many algebraic numbers).

Echoing nateharman:
The rationals are a subset of the reals and the transcendental numbers (as well as the algebraic numbers) are not.  It's still true that every transcendental number is irrational (i.e., is not a rational number).

One can show that $ \pi$ and $ e$ are transcendental numbers using some field theory. The proof is far from simple, eventhough it seems to be because it is short.

First we need a rather difficult theorem:

[b]Lindemann-Weierstrauss Theorem:[/b] Let $ \alpha_{1},...,\alpha_{n}$ be distinct algebraic numbers. Then the exponents $ e^{\alpha_{1}},...,e^{\alpha_{n}}$ are linearly independent over $ \mathbb{Q}$.

[b]Theorem:[/b] The number $ \pi$ is trancendental over $ \mathbb{Q}$ because if not $ 0 ,\pi i$ are algebraic and so $ e^{\pi i}$ and $ e^{0}$ are linearly independent. But this is impossible. Q.E.D.


linear algebra
matrix
linear algebra unsolved

If $ V$ is a finite dimension inner product space with basis $ B \equal{} \left\{v_1, ..., v_n\right\}$. If $ Q$ is an associated matrix with respect to this basis such that $ Q_{ij} \equal{} < v_i, v_j >$.

1) Show that $ det(Q)\neq0$
2) If $ D$ is another basis of $ V$ and $ Q'$ is the associated matrix in the basis $ D$.  Show that $ Q' \equal{} (_DM_B) Q (_BM_D)$ and $ det(Q') \equal{} |det(_BM_D)|^2det(Q)$

Quick question: are you talking about a real inner product space or a complex inner product space? (I suspect you mean real.) The notation works out a little differently in the two cases.

Second issue: what's $ A?$ Is it the same as $ _CM_B?$

[quote="Kent Merryfield"]
 what's $ A?$ Is it the same as $ _CM_B?$[/quote]
I believe that $ A \equal{} _DM_B$ (not $ _CM_B?$) is matrix of $ B$  follow basis $ D$
and $ Q' \equal{} _DM_BQ _DM^t_B$ or $ Q' \equal{} AQA^t$ (not $ Q' \equal{} _DM_BQ _BM_D$)
and as Kent Merryfield (I suspect you mean real)!
 Prove 1)
Put $ C_j \equal{} \left[\begin{array}{c} < v_1,v_j > \\
< v_2,v_j > \\
< v_3,v_j > \\
... \\
< v_n,v_j > \end{array}\right]$ .Suppose  exisist $ (\lambda_1,\lambda_2,..,\lambda_n) \in R^n$ such that :
  $ \lambda_1C_1 \plus{} \lambda_2C_2 \plus{} ... \plus{} \lambda_nC_n \equal{} 0$ Implies  $ \lambda_1v_1 \plus{} \lambda_2v_2 \plus{} ... \plus{} \lambda_nv_n \equal{} 0$ , Hence $ \lambda_1 \equal{} \lambda_2 \equal{} ... \equal{} \lambda_n \equal{} 0$ Implies {$ C_1,C_2,..,C_n$} is linear Independent.
Then $ \det(Q) \neq 0$
2)Put $ D \equal{}${$ e_1,e_2,..,e_n$} is basis of $ V$  perform :
 $ e_i \equal{} \sum_{k \equal{} 1}^{n}{a_{ik}v_k}$
  then $ Q'_{ij} \equal{} < e_i,e_j > \equal{} < \sum_{k \equal{} 1}^{n}{a_{ik}v_k},\sum_{p \equal{} 1}^{n}{a_{jp}v_p} > \equal{} \sum_{p \equal{} 1}^{n}\sum_{k \equal{} 1}^{n}{a_{ik} < v_k,v_p > a^t_{pj}}$,($ a^t_{pj} \equal{} a_{jp}$)
 From here easy See that $ Q' \equal{} AQA^t$ then $ \det(Q') \equal{} |\det(A)|^2\det(Q)$


If $ x_1$ and $ x_2$ are the roots of the equation $ x^2\minus{}3x\plus{}1\equal{}0$ then the product $ x_1 x_2$ is:
$ (A)$ $ 4$
$ (B)$ $ 0$
$ (C)$ $ 2$
$ (D)$ $ 3$
$ (E)$ $ 1$

$ x^2\minus{}3x\plus{}1\equal{}(x\minus{}x_1)(x\minus{}x_2) \equal{} x^2\minus{}(x_1\plus{}x_2)x\plus{}x_1x_2$

$ \therefore x_1x_2 \equal{} 1$

The product of the roots of a quadratic can also be found by $ c/a$ so that would be $ 1$

No it's exactly one, approximately 0.9918

This kind of problems can be solved using [color=orange][b]Vieta[/b][/color]'s relations. So, suppose that the equation is $ ax^2\plus{}bx\plus{}c\equal{}0$ where $ a \not\equal{} 0$. Now, the equation may be rewritten in the form: $ x^2\plus{}\frac{b}{a} \cdot x\plus{}\frac{c}{a}\equal{}0$. If we denote with $ x_1,x_2$ the roots of this equation, then we obtain:

$ \rhd$ $ x_1\plus{}x_2\equal{}\minus{}\frac{b}{a}$

$ \rhd$ $ x_1 x_2\equal{}\frac{c}{a}$


let $a,b,c>0$ prove that this inequality
$\frac{1}{a+b}\ + \frac{1}{b+c}\ + \frac{1}{c+a}\ \leq\ \frac{1}{2}\left( \frac{1}{a}\ + \frac{1}{b}\ + \frac{1}{c}\right)$

very easy :

Let $\sum {\frac{2}{a+b}} \leq \sum {\frac{1}{\sqrt{ab}} \leq \sum \frac{1}{a}}$

I have another solution:

$\frac{2}{a+b}= \frac{2}{\frac{1}{\frac{1}{a}}+\frac{1}{\frac{1}{b}}} \leq \frac{\frac{1}{a}+\frac{1}{b}}{2}$ (HM-AM)

So, $\frac{2}{a+b}+\frac{2}{a+c}+\frac{2}{b+c} \leq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$

And the result follows.

[quote="arqady"]ehsan2004, if a=1, b=c=4

then $\sum {\frac{1}{\sqrt{ab}}> \sum \frac{1}{a}}$[/quote]I think you have miscalculated ;)

Valentin, thx!

 ehsan2004, if a>0, b>0, c>0,  then $\sum {\frac{1}{\sqrt{ab}}\leq \sum \frac{1}{a}}$ . sorry.

Or just use $\frac{1}{x}+\frac{1}{y}\geq \frac{4}{x+y}$ which is equivalent with $(x-y)^{2}\geq 0$ and we are done. :P


[quote="idontthinkso"]x,y,z are positive numbers. Please prove that
6xyz+sigma_sym(x^6y^3)>=4sigma(x^5y^2z^2)[/quote]

The inequality does not hold.

try

$ x\equal{}z\equal{}1,y\equal{}1.5$


If a,b,c are real numbers such that a \geq b \geq c and a+b+c=3 prove that

 ab^2+bc^2+ca^2 \leq 27/8

I have a solution, but I prefer not to post it since it is extremely ugly. I would like to know your clever solution, Manlio. Thanks in advance.

This solution is not mine. This problem was on my notesbook but I don't remember where I found it. It was too much difficult for me to solve so I copied the solution.

Let f(a,b,c)=ab^2+bc^2+ca^2

Then f(a,b,c)+f(a,c,b)=(a+b+c)(ab+bc+ca) -3abc=3(ab+bc+ca-abc)=

        =3[(1-a)(1-b)(1-c) +(a+b+c)-1]=3[(1-a)(1-b)(1-c)+2]

Because c \leq 1 \leq a we have (1-a)(1-b)(1-c) \leq 0 if b \leq 1

If b>1 then

(1-a)(1-b)(1-c) \leq ((a+b)/2-1)^2*(1-c) \leq (3/2-1)^2*(1-0)=1/4

Therefore f(a,b,c)+f(a,c,b) \leq 27/4

Noting that f(a,c,b) -f(a,b,c)=(a-b)(b-c)(a-c) \geq 0

we then have f(a,b,c) \leq 27/8

Indeed, it is nice. My solution was also based on the same identity f(a,b,c)+f(a,c,b)=3(ab+bc+ca-abc), but I have complicated too much everything. That's the funny part: just when you think you know something about inequalities another on appears and shows you're wrong.


Show that sec^2 a=4xy/(x+y)^2
is only possible when x=y
AM Gm inequality holding when both x,y are both +ve or -ve. So proved.
Anybody have any other methods of proving?

hello, we have $ \cos(a)^2\equal{}\frac{(x\plus{}y)^2}{4xy}$ and $ (x\plus{}y)^2\geq4xy$ so we got $ \frac{(x\plus{}y)^2}{4xy}\geq1$ if $ xy>0$,
Sonnhard.


Determine two polynomials  $ A$ et $ B$  $ \in\mathbb{C}[X]$ such that $ \sum_{\omega^{n}\equal{}1}\frac{\omega}{(X\minus{}\omega)^{2}}\equal{}\frac{A}{B}$

[quote="erdos2009"]Determine two polynomials  $ A$ et $ B$  $ \in\mathbb{C}[X]$ such that $ \sum_{\omega^{n} = 1}\frac {\omega}{(X - \omega)^{2}} = \frac {A}{B}$[/quote]

Maybe I misunderstood :

${ A=\prod_{\omega^n=1}(X - \omega)^{2}}    \sum_{\omega^{n} = 1}\frac {\omega}{(X - \omega)^{2}}$

${ B=\prod_{\omega^n=1}(X - \omega)^{2}}$


This test was administered Saturday, February 3rd, in Columbus, GA, and I wrote it.

It's not too hard, but I think if you follow the time limit, then it is really good practice for intermediate level.

Right-click and select "save as."

[quote="drunner2007"][quote="cincodemayo5590"]A couple of the problems were a bit predictable, but others were pretty enjoyable and useful for practice. Oh, and the solution for #1 is shorter if you just find the coefficient on $x^{3}$ and whip out Vieta's formula. But this is good to practice for the AMC (coming up soon!) and just for Intermediate Topics, and probably was difficult for those who hadn't seen some of the tricks used on AoPS. Mark, thanks for sharing this!   :)[/quote]

I'm glad you enjoyed it.

Yeah I guess the way I designed #1, I completely forgot about using Vieta's formulas.

Did you go through the whole thing or just skim it?

Has anyone else been through it (that wasn't actually at the tournament).[/quote]

I went through most of the exam, actually working out the more interesting problems but just skimming over the easier ones which I would be able to do in my head in an actual test situation. And I'm glad that your solutions are clear enough for me to follow easily; I've read some answer manuals to tests where I'm lost after the first, oh, ten words or so.  :P

[quote="sapphyre571"]I thought 33 wasn't a great question - maybe I just didn't know enough F&G to know about that graph's max/min proerties (especially the ones outside the zeroes), but I did really like 34. The wording on 35 was fairly ambiguous - I couldn't really understand exactly how the system worked...
19 was pretty good, 27 was good, 29 was way too obvious based on the given numbers, 32 was nice, 9 was pretty good, 18 was good, 29 was nice, 26 was too obvious, just to comment on a few.[/quote]

Number 33 was based off of an idea from ARML a few years back. I thought it was a good and creative problem, but it's true that the solution I wrote is not 100% rigorous. What do you mean by F&G?

35 is a really good question to try (if you haven't already read the solution). basically you flip a coin to see if a digit is a 1 or 0; then IF you do in fact get a 1, you flip the coin a second time to see IF you should move on to the next digit. if this 2nd flip is heads, you go ahead and move on to the next digit. then you take what i just wrote in this paragraph and apply it again, as if you never even knew about the digits before.

yeah 26 was a little obvious, but not if you haven't seen that type of problem before.

i thought 19 was pretty cool/easy too.

several people said that 9 was interesting. i didn't expect many people to comment on it.

just a general question, what did you guys think about the difficulty of 28, 29, and 30? i have a feeling that they are easier than i thought they would be.

I thought #28,29,30 were too easy for where they were placed. #28 is just plain obvious; for 29 it's obvious that we can't do anything with those numbers we are given, and the solution follows; and 30 is pretty easy as well since it's just looking at the most even distribution of $\sum^{10}k$, which is just taking the floor of the sum over 2. So yes, I agree that these three are too easy (if the test was to be ordered based on difficulty).


[quote="csbonny"][b]Let [i]ABC[/i] be a triangle and [i]P,Q[/i] points on the side [i]AB[/i] such that the inradii of triangles [i]APC[/i] and [i]BQC[/i] are the same. Prove theat the inradii of thriangles [i]AQC[/i] and [i]PBC[/i] are also the same[/b][/quote]
Look here: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=180108


The angles $ \alpha$ and $ \beta$ are such that

$ \tan \alpha \equal{} m \plus{} 2$ and $ \tan \beta \equal{} m$

where $ m$ is a constant.

(i) Given that $ \ sec^{2} \alpha \minus{} \sec^{2} \beta \equal{} 16$, find the value of $ m$.

(ii0 Hence find the exact value of $ \tan(\alpha \plus{} \beta)$.

Quite direct Q

[hide="(i)"]

 using $ sec^2\theta \equal{} 1 \plus{} tan^2\theta$

gives $ tan^2\alpha \minus{} tan^2\beta \equal{} 16 \ \Rightarrow 4m \plus{} 4 \equal{} 16$

[b]m = 3[/b]

[/hide]

[hide="(ii)"]

$ tan(\alpha \plus{} \beta) \equal{} \frac{tan\alpha \plus{} tan\beta}{1 \minus{} tan\alpha.tan\beta}  \equal{} \frac{\minus{}4}{7}$

[/hide]


p+q+r+s+t=8

p^2+q^2+r^2+s^2+t^2=16 [p,q,r,s,t all are real numbers]

Find the maximum possible value of p.

[quote="umesh123"]$ p\plus{}q\plus{}r\plus{}s\plus{}t\equal{}8$

$ p^2\plus{}q^2\plus{}r^2\plus{}s^2\plus{}t^2\equal{}16$ [p,q,r,s,t all are real numbers]

Find the maximum possible value of p.[/quote]

\[ q\plus{}r\plus{}s\plus{}t\equal{}8\minus{}p\]

\[ q^2\plus{}r^2\plus{}s^2\plus{}t^2\equal{}16\minus{}p^2\]

From Cauchy:

\[ (1\plus{}1\plus{}1\plus{}1)(q^2\plus{}r^2\plus{}s^2\plus{}t^2) \ge (q\plus{}r\plus{}s\plus{}t)^2\]

\[ 4(16\minus{}p^2) \ge (8\minus{}p)^2 \implies p \leq \frac{16}{5}\]

Therefore \[ p_{max}\equal{}\frac{16}{5}\]


trigonometry
function
calculus
logarithms
limit
real analysis
inequalities proposed

If $n$ is a natural number find his biggest value such that the inequlity

$(sinx)^n+(cosx)^n>\frac{1}{2}$ holds for each $x\in [0,\frac{\pi}{2}]$     :)

Put $x=\frac{\pi}{4}$ then we have $n<4$

so the [u]expected[/u] min value of  $n$ is $3$

then we prove the given ineq for $n=3$

Probably the  $>$ sign can also be a    $\geq$    one.... :)

[quote="socrates"]

then we prove the given ineq for $n=3$
[/quote]

How do you prove it ?

I'm not sure but i think
cauchy works:

$(\sin x+\cos x)(\sin^3x+\cos^3x)\geq 1$

$1\le \sin x+\cos x,\ x\in \left[0,\frac{\pi}{2}\right]$. Let $n\in N,\ n\ge 3$ and the function
$f(x)=\sin ^n x+\cos^n x,\ x\in \left[0,\frac{\pi}{2}\right]\Longrightarrow f'(x)\ .s.s.\ \left(x-\frac{\pi}{4}\right)\Longrightarrow$
$\left(\forall\right)x\in\left[ 0,\frac{\pi}{2}\right],f(x)\ge f\left(\frac{\pi}{4}\right)=2^{1-\frac n2}\Longrightarrow 2^{1-\frac n2}>\frac 12\Longrightarrow n=3.$

[u]Remark.[/u] $(A\ .s.s.\ B)\equiv$ the real numbers $A,B$ have the [b]s[/b]ame [b]s[/b]ign, i.e. $AB>0\ \vee\ A=B=0.$

[u]The other examples (with calculus - real analysis):[/u]

$1.$ ([u]classic[/u]) $\left(1+\frac{1}{n+1}\right)^{n+1}-\left(1+\frac 1n\right)^n<\frac{1}{n+1},\ \left(\forall\right)n\in N^*.$

$2.$ ([u]classic[/u]) $\ln \left(1+\frac 1n\right)<\frac{1}{ne}\left(1+\frac 1n\right)^n<\frac 1n,\ \left(\forall\right)n\in N^*.$

$3.$ ([u]own[/u]) The equation $\sqrt [n] {x+1}-\sqrt [n] x=\frac 1n,\ n\in N,\ n>1$ has
an alone solution $x_n\in (0,1)$ and $\lim_{n\to \infty} x_n=\frac{1}{e-1}.$


search function calculus Support \/closed

I'm just wondering if the search function is not working properly for anybody else? No matter what I type it for the last couple days, it just brings up this huge list of threads that have nothing to do with what I searched for.

Try the following:

-Use more specific and less common words.
-Use quotes and AND/OR operators.
-Search only for threads with what you're searching for in the title.

For me, the search function is still working properly.

I don't think it's a mistake on my part. I've been using this search function for a while, and it's been working perfectly fine.
It's just a few days ago, I'll search for something like "Three blocks of identical mass" and it'll bring up 1000 results that have nothing to do with this.

OK. But I'd still like to know why searching the Calculus Computations and Tutorials forum for any of "L'Hôpital", "L'Hopital", "L'Hospital", "LHopital", "LHospital", or "*pital" and my user name returns zero results. I assure you that the number on that should not be zero.

Apparently the search engine that our forum software uses does not support that particular search. I assume this is due to the fact that the ' character is a special character in MySQL and is stripped for security reasons.

The search does not support wildcards such as *opital as we do not have the server capacity to handle such searches (we don't have thousands of dedicated servers like major search engines). You might want to try the following on google:
[code]site:artofproblemsolving.com/Forum +"l'hopital"[/code]
It's certainly not the best solution as it does not show topic names, but it may help.

[quote="levans"]We have implemented a fix. Please let us know if you continue to experience issues.[/quote]

The search works awesome now, thanks! What was the fix, exactly?


modular arithmetic
quadratics
algebra
polynomial

[quote="Phelpedo"]Find all $x$ such that $x^{2}+2$ is a prime.[/quote]

There are infinitely many $x$ that satisfy.  Do you mean integers $x$?

[hide="partial solution"]For all even integers x(excluding 0), x^2+2 is not prime. For all odd integers that aren't multiples of 3, x^2+2 is a multiple of 3.For a bunch of odd multiples of 3, x^2+2 is prime.[/hide]

Could someone prove it for all odd multiples of 3???????

[quote="Phelpedo"]Find all $x$ such that $x^{2}+2$ is a prime.[/quote]
[hide]$(x+1)(x-1)+3$ Consider x being even. The entire result would be even. In order for that to be two, x=0 is one solution. Lets look at x being odd. Make the substitution $x=2n-1$ . $(2n-2)(2n)+3$ is our new expression. This expression is always odd. Now I'm stuck...[/hide]

[quote="Phelpedo"]Find all $x$ such that $x^{2}+2$ is a prime.[/quote]

edit: nvm this is harder than i thought, i think

[quote="Phelpedo"][quote="chess64"][quote="Phelpedo"]Find all $x$ such that $x^{2}+2$ is a prime.[/quote]

[hide]
For $x=2$ we get $6$, which is not prime. Therefore $x$ must be odd. Modulo 3, we must have $x^{2}+2\equiv \pm1\pmod{3}\iff x^{2}\equiv 0,2$. 2 is not a quadratic residue mod 3, so $x^{2}\equiv 0\pmod{3}\iff x\equiv 0\pmod{3}$. Thus $x=3$ is the only possibility, and we get $3^{2}+2=11$, so it works.
[/hide][/quote]

Not sure what you are doing here. Try odd multiples of three; you'll find that many work.[/quote]
I think he thought that x being 0 mod 3 meant that x is 3.

Woops, I accidentally thought that $x$ was supposed to be prime. Let me fix it...

[quote="Phelpedo"][quote="chess64"][quote="Phelpedo"]Find all $x$ such that $x^{2}+2$ is a prime.[/quote]

[hide]
For $x=2$ we get $6$, which is not prime. Therefore $x$ must be odd. Modulo 3, we must have $x^{2}+2\equiv \pm1\pmod{3}\iff x^{2}\equiv 0,2$. 2 is not a quadratic residue mod 3, so $x^{2}\equiv 0\pmod{3}\iff x\equiv 0\pmod{3}$. Thus $x=3$ is the only possibility, and we get $3^{2}+2=11$, so it works.
[/hide][/quote]

Not sure what you are doing here. Try odd multiples of three; you'll find that many work.[/quote]

That's what I said in my partial solution....

Would someone prove that all odd multiples of 3 work? or do they...

They don't. $51$ is the earliest example I think.

But 57 works.....I'm beginning to think that there are ∞ solutions....

[quote="Αρχιμήδης 6"]Isn't it an open problem? :D[/quote]

The problem $x^{2}+1$ is open, but I don't think this one is.

ok. for this question, we just need to find out if x^2+2 for n odd multiples of 3 for x, that is a composite. We just need to find n.

edit:

I just realized that all odd multiples of 3 are 6 apart from each other, so we just need to try to factor x^2+12x+38

[url=http://www.research.att.com/~njas/sequences/A067201]Here[/url] is the list of the first few such $x$ from the OEIS, if anyone is interested.

[quote]I just realized that all odd multiples of 3 are 6 apart from each other, so we just need to try to factor x^2+12x+38[/quote]

No.  For a polynomial $P$ with integer coefficients, the fact that $P$ factors shows that there are finitely many possible $n$ such that $P(n)$ is prime.  However, given a polynomial $P$ which doesn't factor, we learn nothing about whether or not (or how often, or where) it takes prime values.


Let there be a curve y=25+2.5sin(36:theta:-15). Connect two adjacent crests. Connect one of the crests you connected with the trough in between the two crests. Find the area surrounded by the two lines you've drawn and the curve (the little thin area surrounded by the second line and the curve is also included)

Blah, I said I would answer this:

[hide]

crest to crest distance = period.

period = 360/36 = 10

height = 2(amplitude) = 5/2 *2 = 5

Since a parallelogram is formed it is A=B*H = 10*5 = 50.

[/hide]

I dont really know what thin area you are talking about, but you want to connect crest-crest on the top, crest-trough and crest-trough, and crest-crest on the bottom to make a parallelogram, right?[/hide]

A diagram would help, but if I'm not mistaken, the area averages out to be a triangle... but then again, I'm not sure what is being asked.

If what you say is right, then yes, it [i]does[/i] average out to become a triangle (that is half the parallelogram).  Also, I think what he means by the thin strip is the part of the sine curve that goes into the parallelogram.

Is this the region you're talking about?

JS1527 wrote:Okay I can't edit, so I will just redo my solution.

[hide]

All you have to do is make a parallelogram with CREST-CREST and TROUGH-TROUGH for one pair of sides and then two CREST-TROUGH for the other sides.  This parallelogram has part of the sine curve cutting it into two equal parts.  He is asking for the area of one of those parts.  So you want to find the area of the parallelogram and divide it by two.  I dont get what "small part" of the curve he is talking about though.

crest to crest distance = period.

period = 360/36 = 10

height = 2(amplitude) = 5/2 *2 = 5

Since a parallelogram is formed it is A=B*H = 10*5 = 50.

50/2 = 25

[/hide]

Sorry it took me a while, yup you got it right (obviously )

I was just talking about the triangle because curbed regions is too hard...

Oh so you are talking to me now?

Yea, too hard unless you want to use calc I guess, which makes the whole trick to the problem no fun.


Solve for $ x\in [0,2\pi)$:

$ \sin^3{x}\plus{}\sin{x}\plus{}\cos^3{x}\equal{}2$

hello, use the substitutions
$ \sin(x)\equal{}\frac{2\tan(\frac{1}{2}x)}{1\plus{}\tan^2(\frac{1}{2}x)}$
$ \cos(x)\equal{}\frac{1\minus{}\tan^2(\frac{1}{2}x)}{1\plus{}\tan^2(\frac{1}{2}x)}$.
Sonnhard.

One can also solve it by going for the substutions  $ sin^2 x \equal{} \frac {1 \minus{} cos2x}{2}$ and $ cos^2x \equal{} \frac {1 \plus{} cos2x}{2}$ which will yield us a quadratic in $ cos 2x$ and then solve for $ x$.

hello, using my substitution and setting $ \tan(\frac{1}{2}x)\equal{}t$ you will find the equation
$ (3t^4\plus{}4t^3\plus{}8t^2\plus{}1)(t\minus{}1)^2\equal{}0$.
Sonnhard.

[quote="Aravind Srinivas L"] $ sin x \equal{} \frac {1 \minus{} cos2x}{2}$ and $ cosx \equal{} \frac {1 \plus{} cos2x}{2}$ .[/quote]

Is this typo?
I believe $ \sin^2{x} \equal{} \frac {1 \minus{} \cos{2x}}{2}$ and $ \cos^2{x} \equal{} \frac {1 \plus{} \cos{2x}}{2}$.

Try to find more simple ways :)

hello, yes, this isn't true because $ \sin^2(x)\equal{}\frac{1\minus{}(\cos^2(x)\minus{}\sin^2(x))}{2}\equal{}\frac{1\minus{}(1\minus{}2\sin^2(x))}{2}\equal{}\sin^2(x)$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, using my substitution and setting $ \tan(\frac {1}{2}x) \equal{} t$ you will find the equation
$ (3t^4 \plus{} 4t^3 \plus{} 8t^2 \plus{} 1)(t \minus{} 1)^2 \equal{} 0$.
Sonnhard.[/quote]

Hi,Sonnhard.

well,eventually what solutions for $ x$ have you gotten?

hello, see my substitution and equation, we have $ t\equal{}1$ and from this $ x\equal{}\frac{\pi}{2}$ in the given interval.
Sonnhard.

[quote="Dr Sonnhard Graubner"] we have $ t \equal{} 1$ and from this $ x \equal{} \frac {\pi}{2}$ in the given interval.[/quote]

That's right :)

[hide="another idea"]
Note that
$ \sin{x}\le 1$ and $ \cos{x}\le 1$
[/hide]

Sorry that was a typo. Yes. But still it fetches you the solution. Let me think of a more elegant way.


The Benton Hotel has six floors and the same number of stairs between floors. It takes Peter 30 seconds to climb the stairs from the first floor to the third floor. At this rate, how many seconds will it take Peter to climb the stairs from the first floor to the sixth floor?

if you climbed 1st floor to 3rd floor, you had climb 2 floors.

and that took 30 secs.

so it takes 15 secs to climb 1 floor.

and the prob says 'how many secs to climb 1st floor to 6th floor?'.

you had climb 5 floors, so the answer is 15*5=75.

answer : 75


geometry
integration
calculus
calculus computations

Prove that:
$ \sum_{k \equal{} n}^{ \plus{} \infty}\frac {1}{k^2} \equal{} \frac {1}{n} \plus{} \frac {1}{2n^2} \plus{} O\left(\frac {1}{n^3}\right)$

In less geometric terms: since $ \frac {1}{k^2} \minus{} \int_{k}^{k \plus{} 1}\frac {dx}{x^2} \equal{} \frac {1}{k^2(k \plus{} 1)}$, it follows that $ \sum_{k \equal{} n}^{\infty}\frac {1}{k^2} \equal{} \frac {1}{n} \plus{} \sum_{k \equal{} n}^{\infty}\frac {1}{k^2(k \plus{} 1)}$. This can be repeated: $ \frac {1}{k^2(k \plus{} 1)} \minus{} \int_k^{k \plus{} 1}\frac {dx}{x^3} \equal{} \frac {1}{2k^2(k \plus{} 1)^2}$, hence $ \sum_{k \equal{} n}^{\infty}\frac {1}{k^2(k \plus{} 1)} \equal{} \frac {1}{2n^2} \plus{} \sum_{k \equal{} n}^{\infty}\frac {1}{2k^2(k \plus{} 1)^2}$. And so on...

I got to $ \frac {1}{n} \plus{} \frac {1}{2n^2} \plus{} \frac {1}{6n^3} \minus{} \frac {1}{30n^5} \plus{} \frac {1}{42n^7} \minus{} \frac {1}{30n^9} \plus{} \frac {5}{66n^{11}} \minus{} \dots$
Looks like we have only odd powers after $ n^2$.  :maybe:

[quote="fedja"]I have always thought that in the formula for the area of a triangle $ 2$ is in the denominator ;)[/quote]
Sorry for the typo :blush:

To mlok: Those are Bernoulli numbers, shifted a bit. The alternating signs and skipped powers are characteristic. The coefficients also go to $ \infty$ faster than any exponential, so the series diverges for all $ n$.