Tark scored an integer number of points on each of five tests. He then computed the following statistics for those five scores: the mean was 72, the median was 70, and the mode was 80. What was the least possible score he could have had on any test?

[quote="hoshattack"][hide]Not too bad...

The 3 numbers we have to have are 70, 80, and 80.  The average of these 3 is 77.5, and we need to get it down 5.5.  Thus the other 2 numbers should have an average of 5.5 less than 72, or 67.5.  In order to get a .5 at the end, the difference between the number has to be odd, and we want the greatest difference to get the lowest number.  So we make a number 69, the highest possilbe before the middle number. We then quickly go down 1.5 from 67.5, because the difference between 67.5 and 69 is 1.5, and you get 66.  So our numbers are 66, 69, 70, 80, and 80.  The lowest possible score is a $\displaystyle{\boxed{66}}$. :D [/hide][/quote]

Correct me if I am wrong, but isn't (66+69+70+80+80)/5=73?

Isn't it 61,69,70,80,80

$\displaystyle{\boxed{61}}$.

[hide]Sum is 72*5=360.
The last three numbers must be 70,80 and 80.
69 is the highest possible value for one of the remaining tests.
Subtracting, 360-80-80-70-69=61.[/hide]


$\forall n \in \mathbb{N}$, $P(n)$ denotes the number of the partition of $n$ as the sum of positive integers (disregarding the order of the parts), e.g. since $4 = 1+1+1+1 = 1+1+2 = 1+3 = 2+2 = 4$, so $P(4)=5$. "Dispersion" of a partition denotes the number of different parts in that partitation. And denote $q(n)$ is the sum of all the dispersions, e.g. $q(4)=1+2+2+1+1=7$. $n \geq 1$. Prove that

(1) $q(n) = 1 + \sum^{n-1}_{i=1} P(i).$

(2) $1 + \sum^{n-1}_{i=1} P(i) \leq \sqrt{2} \cdot n \cdot P(n)$.

the first part is rather trivial:
 $q(n)=\sum_\lambda \text{disp}(\lambda)$
as $\lambda$ runs over all the partitions of $n$ and $\text{disp}(\lambda)$ counts the number of distinct parts in $\lambda$. clearly this is the same as counting for each $i$, the number of partitions of $n$ that count a distinct part of size $i$.If $i=n$ then of course there is only one such partition. else, the number of partitions that have a part of size $i\  = p(n-i)$, so $q(n)=1+\sum_{i=1}^{n-1} p(n-i)$
and that is what we need.

the second part is easily improved by the fact that $np(n)=\sum_{i=1}^n \sigma(i)p(n-i)$ where $\sigma(i)$ is the sum of all positive divisors of $i$.

I just think the second question should be changed a little.
The original problem should be $ 1\plus{}\sum_{i \equal{} 1}^{n}P(i)\leq\sqrt{2n}P(n)$


Given a positive integer $ n$, let $ b(n)$ denote the number of positive integers whose binary representations occur as blocks of consecutive integers in the binary expansion of n. For example $ b(13) \equal{} 6$ because $ 13 \equal{} 1101_2$, which contains as consecutive blocks the binary representations of $ 13 \equal{} 1101_2$, $ 6 \equal{} 110_2$, $ 5 \equal{} 101_2$, $ 3 \equal{} 11_2$ , $ 2 \equal{} 10_2$ and $ 1 \equal{} 1_2$.

Show that if $ n \leq 2500$, then $ b(n) \leq 39$, and determine the values of n forwhich equality holds.

Let $ b(n) \equal{} b_k(n) \plus{} \cdots \plus{} b_2(n) \plus{} b_1(n)$ where $ b_i(n) \equal{}$ the number of different consecutive blocks of lenght $ i$ in the binary representation of $ n$

If $ n < 2048$ then $ n < 2^{11}$ then $ n$ has at most $ 11$ digits on its binary representation, $ k \le 11$
$ b(n) \le b_1(n) \plus{} b_2(n) \plus{} \cdots \plus{} b_{11}(n)$
$ b_{11}(n)$ There is $ 1$ block with max $ 2^{10}$ possibilities, for $ n \equal{} (1d_{10}d_{9}\cdots d_{1})_2$ then $ b_{11}(n) \le 1$
$ b_{10}(n)$ There are  $ 2$ blocks with max $ 2^{9}$ possibilities then $ b_{10}(n) \le 2$
$ b_{9}(n)$ There are $ 3$ blocks with max $ 2^{8}$ possibilities then $ b_{9}(n) \le 3$
$ \cdots$
$ b_{4}(n)$ There are $ 7$  blocks with max $ 2^{3}$ possibilities then $ b_{4}(n) \le 7$
$ b_{3}(n)$ There are $ 8$  blocks with max $ 2^{2}$ possibilities then $ b_{3}(n) \le 4$
$ b_{2}(n)$ There are $ 9$  blocks with max $ 2^{1}$ possibilities then $ b_{2}(n) \le 2$
$ b_{1}(n)$ There are $ 10$  blocks with max $ 2^{0}$ possibilities then $ b_{1}(n) \le 1$
$ b(n) \le 1 \plus{} 2 \plus{} \cdots \plus{} 7 \plus{} 4 \plus{} 2 \plus{} 1 \equal{} 35$
Similarly If $ 2048 \le n < 2500$ then $ k \equal{} 12$ but $ n \equal{} (100d_{10}d_{9}\cdots d_{1})_2$

What are the values of n for which equality holds?

Well, there is surely something wrong with the offered proof, as 39 is achieved by two numbers and both are less than 2048.  In particular, the numbers are 2008 and 2009 -- a shame they didn't save this problem for a later year ;)

The largest value of $ b(n)$ occuring among $ k$-digit numbers (in binary) is given by {1, 2, 4, 6, 9, 12, 16, 21, 26, 32, 39, 46} for $ 1 \leq k \leq 12$.  The numbers that achieve these values are {{1}, {2, 3}, {6}, {12, 13, 14}, {28, 29}, {56, 57, 58, 59, 60, 61}, {116, 120, 121, 122, 123}, {244}, {488, 489, 492, 500}, {984, 985, 1000, 1001, 1004}, {2008, 2009}, {4016, 4017, 4018, 4019, 4024, 4025, 4056, 4057}}

Ah, and the error: mszew claims $ \sum_{i \equal{} 1}^{11} b_i(n) \leq 1 \plus{} \ldots \plus{} 7 \plus{} 4 \plus{} 2 \plus{} 1$ termwise, but the sum on the left has more terms than the sum on the right  :wink:

Let $ b(n) \equal{} b_k(n) \plus{} \cdots \plus{} b_2(n) \plus{} b_1(n)$ where $ b_i(n) \equal{}$ the number of different consecutive blocks of lenght $ i$ in the binary representation of $ n$

If $ n < 2048$ then $ n < 2^{11}$ then $ n$ has at most $ 11$ digits on its binary representation, $ k \le 11$
$ b(n) \le b_1(n) \plus{} b_2(n) \plus{} \cdots \plus{} b_{11}(n)$
$ b_{11}(n)$ There is $ 1$ block with max $ 2^{10}$ possibilities, for $ n \equal{} (1d_{10}d_{9}\cdots d_{1})_2$ then $ b_{11}(n) \le 1$
$ b_{10}(n)$ There are  $ 2$ blocks with max $ 2^{9}$ possibilities then $ b_{10}(n) \le 2$
$ b_{9}(n)$ There are $ 3$ blocks with max $ 2^{8}$ possibilities then $ b_{9}(n) \le 3$
$ \cdots$
$ b_{5}(n)$ There are $ 7$  blocks with max $ 2^{4}$ possibilities then $ b_{4}(n) \le 7$
$ b_{4}(n)$ There are $ 8$  blocks with max $ 2^{3}$ possibilities then $ b_{4}(n) \le 8$
$ b_{3}(n)$ There are $ 9$  blocks with max $ 2^{2}$ possibilities then $ b_{3}(n) \le 4$
$ b_{2}(n)$ There are $ 10$  blocks with max $ 2^{1}$ possibilities then $ b_{2}(n) \le 2$
$ b_{1}(n)$ There are $ 11$  blocks with max $ 2^{0}$ possibilities then $ b_{1}(n) \le 1$
$ b(n) \le 1 \plus{} 2 \plus{} \cdots \plus{} 7 \plus{} 8 \plus{} 4 \plus{} 2 \plus{} 1 \equal{} 43$

Lets refine some of those numbers:
If $ n \equal{} (0d_{10}d_{9}\cdots d_{1})_2$ then there is one less on each $ b_{11}(n), \cdots, b_{4}(n)$ for $ b(n) \le 35$
If $ n \equal{} (10d_{9}\cdots d_{1})_2$ then there is one less on each $ b_{10}(n), \cdots, b_{4}(n)$ for $ b(n) \le 36$ then $ n \equal{} (11d_{9}\cdots d_{1})_2$
idem for:
If $ n \equal{} (11110d_{6}\cdots d_{1})_2$ then there is one less on each $ b_{7}(n), \cdots, b_{4}(n)$ for $ b(n) \le 39$ then $ n \equal{} (11111d_{6}\cdots d_{1})_2$
Aditional refining:
If $ n \equal{} (1111111d_{4}\cdots d_{1})_2$ then there is one less on $ b_{6}(n)$, two less on $ b_{5}(n)$ and three less on$ b_{4}(n)$for$ b(n) \le 37$
If $ n \equal{} (1111110d_{4}\cdots d_{1})_2$ then there is one less on$ b_{5}(n)$, two less on$ b_{4}(n)$ plus an aditional one less for the $ 0$ on $ b_{5}(n)$ and $ b_{4}(n)$ for $ b(n) \le 38$
If $ n \equal{} (1111100d_{4}\cdots d_{1})_2$ then there is one less on $ b_{6}(n)$, two less on $ b_{5}(n)$ and $ b_{4}(n)$ for $ b(n) \le 38$
then $ n \equal{} (1111101d_{4}\cdots d_{1})_2$ with one less on $ b_{4}(n)$ and additionaly one less for the $ 0$ on $ b_{6}(n)$, $ b_{5}(n)$ and $ b_{4}(n)$ for $ b(n) \le 39$

Checking those $ 15$ numbers to get $ 2008$ and $ 2009$

Similarly If $ 2048 \le n < 2500$ then $ k \equal{} 12$ but $ n \equal{} (100d_{9}\cdots d_{1})_2$

Is there an easier way of doing this question: it seems much more difficult than the first three questions?

Working through mszew's last post on this, I am confused in the Additional refining section. I can see the additional one less for the 0 in the cases ending 10d4...d1 and 01d4...d1, but what about the one ending 00d4...d1; and which are the one, two and three less in the case ending 11d4...d1? Can this be made clearer?

[quote="AndrewTom"]Is this problem well-known?
How did JBL get the results he gives?[/quote]  Hi AndrewTom.  One suggestion: you can go back and edit your previous posts for up to a day or two, and it makes the forum a little "cleaner" if everyone does that instead of making multiple posts in a row.  As for your question, I just wrote a quick computer program to compute the values in question  :oops:  :P  It looks to me like this problem doesn't afford any nicer solution than the sort of brute-force analysis that mszew applied.


Let a quadrilateral $ABCD$ and a point $O$ in its interior such that $\widehat{AOD}=\widehat{BOC}=90$
and $\widehat{ODA}=\widehat{OCB}$ .Let $M,\ N$ the midpoints of $AD,\ BC$ respectively and a line trough $O$ parallel to $MN$ intersects $BD,\ AC$ at $K,\ L$ .
Prove that $OK=OL$

This problem is a good step for the solution of another one...

http://www.mathlinks.ro/Forum/viewtopic.php?t=115173

kostas Vittas.


[hide]Well, there are 10 numbers that have a ones digit of 5, and 10 numbers with atens digit of 5.  SInce one of the numbers with a 5 is in both categories (55), you need to add 10 and 10, and subtract 1, getting 19.

19/100 * 100 = 19%[/hide]

[hide]

19%

10 numbers that end with 5, 10 numbers that begin with 5, one number that has both

10 + 10 - 1 = 19

19/100 = 19%

[/hide]

[hide]19%

after doing MC long enough, you have these things memorised[/hide]

Hello, MCrawford!

Quote:What percent of the natural numbers from 1 to 100, inclusive, have at least one digit which is a 5?

[hide]

Whenever I see an "at least" problem, I tend to go after the complement.

How many numbers do NOT have a 5?

There are 8 one-digit numbers without a 5.

Among the two-digit numbers:

The first digit must not be 0 or 5: 8 choices.

The second digit cannot be 5: 9 choices.

Hence, there are 72 two-digit numbers without a 5.

So there are: 8 + 72 = 81 numbers without a 5,

leaving 100 - 81 = 19 which have at least one 5.

Answer: 19%

[Usually the complement approach is much shorter than the direct approach

. . . but it's certainly not true this time!]

[/hide]

[quote=sleepypuppy]you just bash it.[/quote]

another old bump

[quote=sleepypuppy]you just bash it.[/quote]

Bruh can you just stop


It's been said that no images are allowed in signatures. What's the policy on Asymptote?

The only concern is how large the signatures will actually be. I think in #H34N1's sig, the size doubled...

I mean, is it allowed at all? It technically is an image...

So is the $ \text{\LaTeX}$ code. Image tags (!) are not allowed. Images generated through Asymptote are allowed, as long as they are not offensive and/or very disturbing.


Susan's calculator has a key which replaces the number displayed by its cube. If a $ 2$ is displayed, how many times must Susan press the ``cubing'' key to display a number which is greater than a billion?

1st time: 8
2nd time: 512
3rd time: 134217728
4th time: greater than 1 billion

So Susan must press the cubing key $ \boxed{4}$ times.


[quote="DuongChinh_K41SPHN"]suppose that M, N and P are three point lying respetively on the edges AB, BC, CA of a triangle ABC such that
AM+BN+CP=MB+NC+PA.
prove that S(MNP)<=1\4 S(ABC)[/quote]
To Mods: Please lock this topic, this is a problem from M&Y, issue 10, and we can't discuss it now


Slove the equation with integer roof:
$x^{2}y^{2}-4(x+y)=z^{2}$


1.If $ 3!\cdot 5!\cdot n!\equal{}10!$, then what is the value of $ n$?
[hide="Answer"]$ 7$[/hide]
2.Becca is standing in a hole that is 50 inches deep. She will dig the hole 54 inches deeper so that the top of her head will be the same distance below ground level as it is above groundlevel. What is Becca's height, in inches?
[hide="Answer"]77 [/hide]
3. What is the sum of all positive two-digit integers with exactly 12 positive factors?
[hide="Answer"]402[/hide]
4. Each of the cards shown below has a number on one side and a letter on the other. How many of the cards must be turned over to prove the correctness of this statement for these cards:"If a card has a vowel on one side, then it has a prime number on the other side"?
[hide="Answer"]2[/hide]

[hide="number one"]
3!=6
5!=5*4*3*2*1
10!=10*9*8*7*6*5*4*3*2*1
cross some of the 10! out
10*9*8*7=
5*2*3*3*2*2*2*7
you get 2,3,4,5,6,7 or 7![/hide]

[hide="number 2"]
I am not getting this. How can her head be above and below ground level?[/hide]

[hide="4"]
I think it is 6 though, since the cards can have

A-prime
B-not prime
C-not prime
3-a vowel
4-not a vowel
5-not a vowel

Since you don't know which to pick you might pick the 5 the last to flip over and therefore you would have to flip over 6 cards.[/hide]

[hide="4"]The converse of a statement is not necessarily true,  but the contrapositive is always true.

Thus, we check the one without a prime, and the one with a vowel.

answer: 2[/hide]

[hide="3"]12 =2*6, 4*3, or 12*1 (two factors)
12 = 2*2*3
$ 2^{11}$ is obviously not two-digit.  Thus we consider two cases.

The only case for $ 2*6$ is $ 2^5*3\equal{}96$.

Then for $ 4*3$: $ 2^3*3^2\equal{}72$

For $ 2*2*3$: $ 2^2*3*5\equal{}60$
$ 2*3^2*5\equal{}90$

$ 2^2*3*7\equal{}84$.

The total sum is $ 402$.[/hide]

I think for number two he means as it "was" above groundlevel.

[hide="2"]
So lets say that going down is negative, and up is positive.

So Becca is standing in a hole that is -50. She is x. That makes her y above the top of the hole.

so -50+x-y=0 or -50+x=y

When she digs the hole 54 inches deeper, she is standing in a hole that is
-104. She is still x. That makes her -y below the top of the hole.

So -104+x+y=0 or -104+x=-y

add two equations to get

-154+2x=0
2x=154
x=77

So her height is 77 inches.[/hide]

Was that from a nats sprint because I think I remember those problems.


limit
ratio
quadratics
continued fraction
real analysis
real analysis theorems

Do you guys know where I can find the [i]standard[/i] derivation of the well-known result below?  :maybe:

[b]Let[/b] $ \mathrm{F}_{n}$ [b]denote the[/b] $ n$[b]-th element of Fibonacci's sequence. Prove that[/b] $ \displaystyle \lim_{n \to \infty} \frac {\mathrm{F}_{n \plus{} 1}}{\mathrm{F}_{n}} \equal{} \phi.$

I post this because I recently managed to construct a proof of this fact and I wish to know whether my approach is novel or not. It's not that the result is that big or something... I'm just curious about the originality of my proof.

I thank you in advance for your insightful replies.  :wink:

Regards,
José.

Well which method did you use?  There are many different ways to do this.  In my opinion, one of the simpler ones is to assume that the ration $ \frac{F_{n\plus{}1}}{F_{n}}\equal{}\phi$ exists, and then use the fact that $ F_{n\plus{}2}\equal{}F_{n\plus{}1}\plus{}F_{n}$, divide by $ F_{n\plus{}1}$ to get $ \frac{F_{n\plus{}2}}{F_{n\plus{}1}}\equal{}1\plus{}\frac{F_{n}}{F_{n\plus{}1}}\Rightarrow\phi\equal{}1\plus{}\frac{1}{\phi}$, by assumption.  Now just multiply through by $ \phi$ and solve the quadratic, remembering to take the positive root.

[quote="JRav"]Well, which method did you use?  There are many different ways to do this.  In my opinion, one of the simpler ones is to [b]assume[/b] that the ratio $ \frac {F_{n \plus{} 1}}{F_{n}} \equal{} \phi$ exists... [/quote]

As a matter of fact, the [i]novelty[/i] of my proof resides in the verification of the limit's [b]existence[/b]. That's crucial here, isn't it?

Thanks for taking the time to reply, [color=red][b]JRav[/b][/color].  :wink:

I haven't seen it, but I have doubts as to whether or not your proof is truly novel.  There are many interesting facts about the [url=http://en.wikipedia.org/wiki/Golden_ratio]Golden Ratio[/url], many of which have nothing to do with Fibonacci.

[quote="coquitao"]I'm just curious about the originality of my proof.[/quote]  Then perhaps you should post your proof so that it is possible to judge whether it's original or not :roll:  :wink:

Define standard.  In terms of most generalizable, one proves $ F_n \equal{} \frac {\phi^n \minus{} \varphi^n}{\phi \minus{} \varphi}$ and then the limit is a routine calculation; however, in terms of [i]deepest[/i] I prefer to talk about the [url=http://en.wikipedia.org/wiki/Golden_ratio#Alternate_forms]continued fraction of $ \phi$[/url] and its convergents.  (A very specialized proof is to apply some kind of fixed point theorem to $ f(x) \equal{} 1 \plus{} \frac{1}{x}$ in an appropriate interval.)

[quote="JBL"] Then perhaps you should post your proof so that it is possible to judge whether it's original or not :roll:  :wink:[/quote]

 Before proceeding to the perusal of my file, you may want to take a look at [url=http://mathworld.wolfram.com/CassinisIdentity.html]this[/url].  :D

That's not a novel technique.  It generalizes to a well-known property of the convergents of continued fractions.  See, for example, the discussion [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=217361]here[/url].

[quote="t0rajir0u"]That's not a novel technique.  It generalizes to a well-known property of the convergents of continued fractions.[/quote]

Which well-known property is that, esteemed [b]t0rajir0u[/b]?  :(

You actually have a point there, [b]t0rajir0u[/b].  :o

Thanks for taking the time to take a look at that proof of mine.


iif my school doesnt offer amc 10 or 12 are there separate testing sights that i can go to ? :?:

This is the third time you've asked this question, and it's already been answered.  Please stop posting it.
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=233280[/url]


[quote="the future"]Suppose that $ |x_{i}| < 1$ for $ i \equal{} 1,2,3,....,n$.
Also
$ |x_{1} \plus{} |x_{2}| \plus{} .... \plus{} |x_{n}| \equal{} 16 \plus{} |x_{1} \plus{} x_{2} \plus{} ....x_{n}|$

What is the smallest possible value of n?[/quote]

We want $ \sum_{i\equal{}1}^n|x_i|\equal{}16\plus{}|\sum_{i\equal{}1}^nx_i|$

Let $ X\equal{}\sum_{x_i> 0}|x_i|$ and $ Y\equal{}\sum_{x_i<0}|x_i|$. We have :

$ X\minus{}Y\equal{}a$ and $ X\plus{}Y\equal{}16\plus{}|a|$ for some real $ a$. And so :

$ X\equal{}8\plus{}\frac{a\plus{}|a|}{2}\geq 8$ and $ Y\equal{}8\plus{}\frac{|a|\minus{}a}{2}\geq 8$
So we have at least 9 positive $ x_i$ and at least 9 negative $ x_i$.

So $ n\geq 18$

Let then $ x_1\equal{}x_2\equal{}\ldots\equal{}x_9\equal{}\frac{8}{9}$ and $ x_{10}\equal{}x_{11}\equal{}\ldots\equal{}x_{18}\equal{}\minus{}\frac{8}{9}$. We have $ \sum_{i\equal{}1}^{18}|x_i|\equal{}16\plus{}|\sum_{i\equal{}1}^{18}x_i|$ $ \equal{}16$

So the minimal value for $ n$ is $ n\equal{}18$

[quote="the future"]Thank you very much.

So the general solution would be that if its

$ |x_{1} \plus{} |x_{2}| \plus{} .... \plus{} |x_{n}| \equal{} EVEN \plus{} |x_{1} \plus{} x_{2} \plus{} ....x_{n}|$

and EVEN>=2, its EVEN+2?

and if its

$ |x_{1} \plus{} |x_{2}| \plus{} .... \plus{} |x_{n}| \equal{} ODD \plus{} |x_{1} \plus{} x_{2} \plus{} ....x_{n}|$
and ODD>=1 its ODD+1?

Is that correct[/quote]

Yes, it is.


search
conics
ellipse
geometry
geometry open

Hi,

Could you give me some olympic problems involving Newton's line ?

Thanks.

Friendly advice: use the Search option. At least 15 useful threads can be found searching for Newton's line.

Oh. You see at this link:
http://agutie.homestead.com/files/NewtonTheorem.htm

It's full slove theorem Newton's line.

Dear Mathlinkers,
you can also have a look on

"Méthodes et techniques en Géométrie, A propos de la droite de Newton" editons Ellipse, Paris, ISBN 2-7298-1585-6

where 30 different proofs are presented with historical notes.
Sincerely
Jean-Louis

using complex number, take the incircle as unit circle with origin as center
then let four tangential points be a, b, c, d, then using inversion z -> 1/(conjugate z)
you can invert midpoints (a+b)/2 etc. into vertices of circumscribed quadrilateral.

...I guess after that it would be series of calculations


linear algebra
matrix
algebra
polynomial
inequalities
linear algebra solved

let $M$ be a symmetric positive definite matrix. Prove that $det(M) \leq \prod A_{i,i}$

[b]Lemma[/b]
If $A=(a_{ij})$ is positive definite, then
$f(y_1,y_2,\dots,y_n)=\left|
\begin{array}{ccccc}
a_{11} & a_{12} & \cdots & a_{1n} & y_1\\
a_{11} & a_{12} & \cdots & a_{1n} & y_2\\
\cdots & \cdots & \cdots & \cdots & \cdots\\
a_{n1} & a_{n2} & \cdots & a_{nn} & y_n\\
y_1 & y_2 & \cdots & y_n & 0
\end{array}
\right|
$
is negative definite.
[b]Proof[/b] Let $Y = (y_1,\dots,y_n)^T$ then
$
\left(
\begin{array}{cc}
E & 0\\
-(A^{-1}Y)^T & 1
\end{array}
\right)
\left(
\begin{array}{cc}
A & Y\\
Y^T & 0
\end{array}
\right)
\left(
\begin{array}{cc}
E & -A^{-1}Y\\
0 & 1
\end{array}
\right)
=
\left(
\begin{array}{cc}
A & 0\\
0 & -Y^{-1}A^{-1}Y
\end{array}
\right)
$
Hence  $
\left(
\begin{array}{cc}
A & Y\\
Y^T & 0
\end{array}
\right)
$ is congruent with $
\left(
\begin{array}{cc}
A & 0\\
0 & -Y^{-1}A^{-1}Y
\end{array}
\right)
$
The one on the right-hand side is positive definite (notice that $A^{-1}$ is positive definite), so is the one on the left-hand side.

Now we say that $|A|\leq a_{nn}P_{n-1}$, where $P_{n-1}$ is the upper-left $(n-1)\times(n-1)$ minor.

$|A| = \left|
\begin{array}{cc}
P_{n-1} & \alpha\\
\alpha^T & a_{nn}
\end{array}
\right| =
\left|
\begin{array}{cc}
P_{n-1} & \alpha\\
0 & a_{nn}
\end{array}
\right| + \left|
\begin{array}{cc}
P_{n-1} & \alpha\\
\alpha^T & 0
\end{array}
\right|$ then apply the lemma.

Hence $|A|\leq a_{11}\cdots a_{nn}$.

liyi, I don't believe that your diagonalization of $f$ is correct. Let's look at the $2\times 2$ case:

$\begin{bmatrix} a & y \\ y & 0 \end{bmatrix}$ has as its characteristic polynomial $\lambda^2-a\lambda-y^2.$ Note that $a$ is not an eigenvalue.

Furthermore, the product

$\begin{bmatrix} 1 & 0 \\ -\frac ya & 1\end{bmatrix}\begin{bmatrix} a & y \\ y & 0\end{bmatrix}\begin{bmatrix} 1 & \frac ya \\ 0 & 1\end{bmatrix}=\begin{bmatrix} a & 2y \\ -y & -\frac{y^2}a\end{bmatrix}$

which doesn't look like your calculation.

Okay...I missed typing a minus sign.
Now it has been corrected.

That doesn't make the lemma true; no matrix of the form $\begin{pmatrix}A&X\\Y&Z\end{pmatrix}$ can possibly be negative definite.

[quote="alekk"]Liyi, what is $E$ ?[/quote]
The identity matrix.

Here is a much easier proof: consider the matrix $b_{i,j}=\frac{a_{i,j}}{\sqrt{a_{i,i}a_{j,j}}}$ . It is positively defined and has elements on the main diagonal equals to 1 and thus we may assume from start that $a_{ii}=1$. In this case , since the eigenvalues of A are positive, their product is at most 1 by the Am-GM inequality, since thier sume is TrA, which is n.


[hide]Letting $ a\equal{}\sqrt[3]{2\cos \frac{2\pi}{7}}$, etc. we have $ a^3,b^3,c^3$ the roots of $ t^3\plus{}t^2\minus{}2t\minus{}1$ (we can get this from
\[ \frac{u^6\plus{}u^5\plus{}u^4\plus{}u^3\plus{}u^2\plus{}u\plus{}1}{u^3}\equal{}\left( u\plus{}\frac{1}{u}\right)^3\plus{}\left( u\plus{}\frac{1}{u}\right)^2\minus{}2\left( u\plus{}\frac{1}{u}\right) \minus{}1\]

So we get the equations
\[ a^3\plus{}b^3\plus{}c^3\equal{}\minus{}1,a^3b^3\plus{}b^3c^3\plus{}c^3a^3\equal{}\minus{}2,a^3b^3c^3\equal{}1\Rightarrow abc\equal{}1\]
Let $ x\equal{}a\plus{}b\plus{}c$ and $ y\equal{}ab\plus{}bc\plus{}ca$. Then
\[ a^3\plus{}b^3\plus{}c^3\minus{}3abc\equal{}(a\plus{}b\plus{}c)(a^2\plus{}b^2\plus{}c^2\minus{}ab\minus{}bc\minus{}ca)\Rightarrow x(x^2\minus{}3y)\equal{}\minus{}4\]
In a similar way (with $ ab,bc,ca$ instead),
\[ \minus{}5\equal{}(ab\plus{}bc\plus{}ca)((ab\plus{}bc\plus{}ca)^2\minus{}3abc(a\plus{}b\plus{}c))\equal{}y(y^2\minus{}3x)\]
To solve the system for $ x$, use the first equation to get $ y\equal{}\frac{x^3\plus{}4}{3x}$ and substitute. We get
\[ \minus{}5\equal{}\frac{(x^3\plus{}4)((x^3\plus{}4)^2\minus{}27x^3)}{27x^3}\]
\[ 0\equal{}135x^3\plus{}(x^3\plus{}4)(x^6\minus{}19x^3\plus{}16)\equal{}x^9\minus{}15x^6\plus{}75x^3\plus{}64\equal{}(x^3\minus{}5)^3\plus{}189\]
\[ \Rightarrow (5\minus{}x^3)^3\equal{}7\cdot 3^3\Rightarrow x\equal{}\sqrt[3]{5\minus{}3\sqrt[3]{7}}\]
as desired.[/hide]


calculus derivative function limit algebra polynomial induction

Let $ f(z) \in \mathcal{M}(\mathbb{C})$ be a meromorphic nonconstant function, with poles away from the integers, with the property that $ f^{(k)}(m) \in \mathbb{Z}$ for all $ k \in \mathbb{N}_0, m \in \mathbb{Z}$.

1. Show that $ \lim_{m \in \mathbb{Z}, |m| \to \infty} f(m) \equal{} \infty$.

2.** Must $ f$ be a polynomial?

An answer to the second question: no, it need not be a polynomial. An example is the function $ f(z)\equal{}\sum_{n}{a_n (\sin(2\pi z))^n}$, where $ a_0\equal{}1$ and $ a_n$ is built by induction Assume we have constructed $ a_0,a_1,...,a_{k\minus{}1}$. Then one can easily see that (if the function is really entire, and we will see it will be) $ f^{(k)}(n)\equal{}a_k\cdot k! (2\pi)^k \plus{}g(a_0,...,a_{k\minus{}1})$ and this for all integers $ n$. Here $ g$ is an ugly function of $ a_0,...,a_{k\minus{}1}$, but which has the taste of being independent of $ n$. Now, choose $ a_k$ such that $ f^{(k)}(n)\equal{}[g(a_0,...,a_{k\minus{}})]\plus{}2$. This also gives an estimation of $ a_k$: its absolute value is between $ C_1^n/n!$ and $ C_2^n/n!$ for two absolute constants $ C_1,C_2>0$ and so the function is indeed entire. It's see to see however that it cannot have finite order (if my computations are ok, it grows something like $ \exp(\exp(C|s|))$) and I still have no idea whether there is such a function of finite order which is not a polynomial.

Harazi's construction also gives a counterexample to the first question since $ f\equal{}a_0\not\to\infty$ on $ \mathbb Z$.

Hi, thanks for the post! Nice construction! My "solution" was wrong  :blush:. Completely wrong: it doesn't work in the case where $ f$ is entire of finite order, neither. This becomes the problem: [b]if $ f(z)$ is entire of finite order and satisfies $ f^{(k)}(m) \in \mathbb{Z}$ for all $ k,m$, must it be a polynomial? [/b] (It's not unreasonable to ask, is it?)

There was a similar-looking early result by A. Gel'fand [it is used in his first proof of the transcendence of $ e^{\pi}$]: there exists a constant $ c > 0$ such that, if $ f(z)$ is any nonpolynomial entire function taking the Gaussian integers into the Gaussian integers, then $ \sup_{|z| \equal{} r} |f(z)| > \exp(c r^2)$ for all $ r \gg 0$. ($ f$ can't grow too slowly unless it is a polynomial).

Actually, historically the first (and the simplest) result of this kind is due to Polya: $ 2^z$ is the slowest-growing nonpolynomial entire function taking integer values at the integers. The key is that every entire function $ g(z)$ has an absolutely convergent expansion $ g(z) \equal{} \sum_{k \geq 0} b_k \binom{z}{k}$; and $ g(z)$ takes integer values at the integers if and only the $ b_k \in \mathbb{Z}$. (If $ g(z)$ is integer-valued and not a polynomial, then infinitely many $ b_k$ are nonzero integers, hence of absolute value $ \geq 1$; and this suffices to conclude the existence of arbitrarily large $ z$ with $ |g(z)| \geq 2^{|z|} / |z|$, say). The proof of Gel'fond's result I mentioned is along similar lines. :)


can someone give me a complete breakdown of synthetic division? something like 'the idiot's guide to synthetic division' or 'synthetic division for dummies' or soemthing. I skipped Algebra 2 and completely missed it and now i couldn't do synthetic division to save my life. thanks

ummm...i am bad at explaining perhaps
http://www.purplemath.com/modules/synthdiv.htm
would help you :lol:
tell me if it doesn't :P

If that doesn't help, basically, you have a polynomial divided by $x-a$ Take the first coefficient, multiply it by $a$ and add the second coefficient. Do this again. Multiply it by the $a$ and add the next coefficient.


superior algebra
superior algebra unsolved

Let $G$ be a group such that $[G:Z(G)]<\infty$ prove that $|G'|<\infty$.


Two triangles are contained in a circle, such that each has a vertex at the center, and two on the circle. Their bases are unequal, and their areas are. A right triangle is constructed with the bases as legs. What is the ratio of the hypotenuse to the radius of the circle? Express your answer as a common fraction in lowest terms.

[hide]the ratio is 2/1, because inscribed right triangles always have a diameter as their hypotenuse[/hide]


An acute triangle is inscribed in a circle. The resultingn three minor arcs of the circle are reflected about the corresponding sides of the triangle. Are the reflected arcs concurrent?

they are concurrent at the orthocenter of the triangle, this is a well known fact

Ok, so prove this well-known fact  :wink:

[hide="Hint"]If $M\in\stackrel{\frown}{BC}$, then prove $\angle BHC=\angle BMC=\pi-\angle A$[/hide]

For any triangle [b]the point symmetrical [/b]to the orthocenter  with regard one of triangle sides is on the circumcircle. So the circle symmetrical to the circumcircle with regard one of triangle sides [b]passes through the orthocenter[/b]. So the three circles are always concurent (all three pass thru the orthocenter).


prove that if one choose more than numbers from the set {1,2,3...,2n},then too of them are relatively prime

May be you mean if one choose more thean n numbers from this set then two are relatively prime.But it is famous and obvious becouse by pigeon-hole principle you will have two consecutive numbers which of course are relatively prime.

It is not true for any n numbers because we may take all even numbers no two of them are relatively prime.


trigonometry geometry symmetry calculus derivative AMC USA(J)MO

I was doing this proof(upload below) when my geo teacher walks up to me, looks at my proof and says use law of sines
I don't know why I should so I insisted on the SAS post.
So if anyone better than me at Geo (basically almost everyone here) tell me why I need to use laws of sine?
(My reasoning is displyed in the Geometry Proof.PNG) (Sorry its Paint!! :lol: )
Danke~~!Gracias~~!Merci~~!  :D

err.... by symmetry, you're done. And anyways. all those obtuse triangles with angle 150 degrees are congruent by SAS lol. so you're done.

There is absolutely no reason to use the Law of Sines for a problem like this. You should add a short explanation of why the triangles are, indeed, congruent, of course, but that's it. I surmise that you are just studing the Law of Sines and your teacher wants you to apply it everywhere. If so, he/she should be more inventive with the problems and you should be more patient with him/her :)

look back on your congruence theorems for triangles

um try like sas or somethig

too lazy to work it out =D

umm. for the sake of argument, say one of the segments is longer than the others. then think "what makes that segment special from the other two?". since all the squares are congruent, the answer is nothing. basically spelling out the symmetry. i feel like a cheater.

though if you were doing this on, say, a math test for school, i would definitely use SAS. i don't think they would accept a one-word proof :wink: .

I thought you wanted a solution with congruency. Don't stoop down; use symmetry.

I don't understand your posts...

Didn't he already say that he used SAS?  And why do we need symmetry in this case, seeing as we know that the squares are congruent and that the triangle is equilateral?  No symmetry is needed...

I'll offer a proof:
"It is given that quadrilaterals $ ABCD$, $ DEFG$, and $ BGHI$ are congruent squares.  Therefore, all of the side lengths of the squares are congruent.  So, $ \overline{CD} \cong \overline{DG} \cong \overline{BG} \cong \overline{FG} \cong \overline{IB} \cong \overline{BD}$.  Also, because $ ABCD$, $ DEFG$, and $ BGHI$ are squares, angles $ CDB$, $ IBG$, and $ FGD$ are $ 90^\circ$.  Because it is given that triangle $ BDG$ is equilateral, we know that angles $ BDG$, $ DBG$, and $ BGD$ are $ 60^\circ$.  By the Angle Addition Postulate (that's how I learned it; there are various other names for this mathematical fact), angles $ CDG$, $ IBD$, and $ EDB$ are $ 150^\circ$ (each is composed of two angles: a right angle and a $ 60^\circ$ angle, as already noted).  Therefore, by SAS Congruence, triangles $ CDG$, $ IBD$, and $ BGF$ are congruent.  Because of CPCTC (Corresponding Parts of Congruent Triangles are Congruent, or Chinchilla Poop Covers The Cage), $ \overline{CG}$, $ \overline{BF}$, and $ \overline{ID}$ are congruent."

If I wanted to be really lazy, that proof would just be two sentences (because it's quite obvious to see that the triangles are congruent).  But because you specifically called me out, logic812 (the subtitle is "Please help, MathAnd"), I actually made it somewhat rigorous.

As the others said, the law of sines is not needed.  Using the law of sines on this problem is tantamount to using calculus on an AMC problem (which is, again, not necessary but still doable).

BTW, you, cognos, and I all know a has-been senior at our school who used calculus on an AMC problem (Daluga).
And since I used the word "calculus" twice in this post, I guess I'll have to tell you a play on words I heard about it:
There are thousands of people that are part of a Facebook group called "I Wish I Was Your Derivative So I Could Lie Tangent To Your Curves."  Or something similar to that.  No joke. :rotfl:  :rotfl:  :rotfl:

haha no it's not inappropriate at all. Pretty clever. :D

\cong gives $ \cong$

Chinchilla poop? Does Mr. Warner teach it like that? XD

[quote="MathAndKnowledge"]And why do we need symmetry in this case, seeing as we know that the squares are congruent and that the triangle is equilateral?  No symmetry is needed...
[/quote]
the equilateral triangle and the congruent squares [i]create[/i] the symmetry. your post implies that you are actually using that reasoning, except you didn't know the word for it.  :lol:

@ undefined: Yes, Mr. Warner teaches like that.

@ mwpl: Hmm...  I understand that the triangles/squares create the symmetry, but one does not actually [i]use[/i] symmetry in the proof.  I could say, [i]after[/i] proving $ CG \equal{} ID \equal{} BF$, [i]after[/i] the symmetry has been established by the congruencies, that, by symmetry, $ AG \equal{} HD \equal{} BE$.  Or maybe I just don't understand.  If I still sound wrong, can someone explain this to me...?  I'm pretty sure that I know what symmetry is...
What I mean is, you can't just say as your entire proof the following:
"Because the squares and the equilateral triangle are given, by symmetry, CG=DI=BF."
Or can you...?

[quote="MathAndKnowledge"]
What I mean is, you can't just say as your entire proof the following:
"Because the squares and the equilateral triangle are given, by symmetry, CG=DI=BF."
Or can you...?[/quote]
well, you'd have to say that the squares are congruent, but in my mind i don't see why not. i'm sure a math teacher would frown upon it (unless they actually covered it, which i'm not sure yours did? idk), so, again, i wouldn't use the symmetry approach at school.

@ anyone with a little more usamo proof-writing experience than me: would it work to basically say something like, "because the given squares are congruent and the triangle is known to be equilateral, by symmetry, $ CG \equal{} DI \equal{} BF$" (sorry for the plagiarism, mathand :) )?

[quote="mwpl11"][quote="MathAndKnowledge"]
What I mean is, you can't just say as your entire proof the following:
"Because the squares and the equilateral triangle are given, by symmetry, CG=DI=BF."
Or can you...?[/quote]
well, you'd have to say that the squares are congruent, but in my mind i don't see why not. i'm sure a math teacher would frown upon it (unless they actually covered it, which i'm not sure yours did? idk), so, again, i wouldn't use the symmetry approach at school.
[/quote]
Sorry about not fully explaining the problem (the post took me less than 3 min. w/ paint  :lol: )
but it specifically stated all of the squares are congruent.
Thanks for the replies guys! (and gals!)

whoah.... my last post
the quote was supposed to be from mwpl11   :P  Sorry, it won't let me edit it for some reason. [color=green]Mod: Edited.[/color]
Anyway, I guess one could progress even without the fact that the squares are congruent because all four sides of a squre are congruent. and prove the squares congrent with what? (I can't think of it)  :maybe: -_-...
Uh-oh.

[quote="logic812"]Anyway, I guess one could progress even without the fact that the squares are congruent because all four sides of a squre are congruent. and prove the squares congrent with what? (I can't think of it)  -_-...
Uh-oh.[/quote]

Huh?  The squares each have a side on the equilateral triangle.  The equilateral's side lengths are equal.  Therefore all three squares have equal side lengths.  Therefore, the squares are congruent...

[quote="logic812"]I get your point, but I'm looking for a theorem that explicitly states that squares that have equal side legnth are congruent.
(Is there one?)[/quote]Definition of a square. :) idk

Um... What do you mean, logic?

I'll give you an example.  Let's say we have square $ ABCD$ and square $ EFGH$.  Now, let's say we know that $ AB \equal{} EF$.  By the definition of a square, $ AB \equal{} BC \equal{} CD \equal{} DA$, and $ EF \equal{} FG \equal{} GH \equal{} HE$.  By the Transitive Property (applied many times) we have $ AB \equal{} EF \equal{} BC \equal{} FG \equal{} CD \equal{} GH \equal{} DA \equal{} HE$.  We also know that both squares have four right angles, which are therefore equal to each other.  Thus, corresponding sides and angles are congruent, so the two squares are congruent.  There is no theorem needed...

It's like saying, "What is the theorem that says two equilateral triangles are congruent if they have equal side length?"

P.S.: What type of geometry are you learning...  Because I don't know of any good geometry class that says you need the Law of Sines on a problem as easy as this or makes you think that you need a theorem to prove that two squares are congruent if you know that their sides are congruent.  I must say that I'm disappointed with your teacher...

[quote="MathAndKnowledge"]"I Wish I Was Your Derivative So I Could Lie Tangent To Your Curves."  Or something similar to that.[/quote]

http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=195104

Argh, my last post was posted when I was extremely sleepy, forgive my folly in asking whether how to prove the squares are congruent.
I used SAS, but my teacher is like confused(?) with the SAS inequality, and told me to use laws of Sine.
I better start reviewing for the test... :maybe:  :maybe:  :| meh

wth? How can a geometry teacher be confused with SAS? He's pretty limited if he really keeps on insisting that you use the Law of Sines..

I talked to my Geo teacher (not a formal teacher) and decided that I should go into more in depth trig!!
(Those of you who know how good at trig I am, you know why I added "!!".)
I said that I would prefer working on quadrilaterals than trig-crap (hahaha  :rotfl: )(actually not that funny) so I am still left with a less-than-mediocre trig knowledge.
In pre-calc, does the class go into a lot of trig?

dood, precalc basically = trig, and matrices, and conics and stuff like that. speaking of which, i am taking the test to skip out of pre-calc, in exactly 1 week. wish me luck.

I will wish you luck, but check your PM box please.

The very next problem is to prove that triangle CIF is equilateral
i suppose it is safe to state that since the triangle is equilateral, it is also equiangular
(oh wth, that's obvious)
Any other suggestions on solving that triangle CIF is equilateral?

If anyone could tell me how to get rid of the rating thingy below my avatar...

cognos, here is a problem to test your skills

A,B, and C are three vertices of a cube, as shown in the diagram,. Find the measure of <ACB, to the nearest tenth of a degree.

Fair enough?

The cube is just crudely drawn and <ABC does not form a 45 degree angle.

[quote="logic812"]If anyone could tell me how to get rid of the rating thingy below my avatar...
[/quote]
You mean how to change "not rated yet" to something else?
You should post something that people will be tempted to rate. Usually it has to be either a very high quality post, or an outrageous spam. The second option has a lot of undesirable side effects (like deleting and locking your posts, getting warnings, and, finally, getting banned), so I recommend trying the first :lol:

^ fedja: good call :D

logic: er that's just $ \sin^{\minus{}1}\cfrac{1}{\sqrt{3}}\approx\boxed{35.3^\circ}$..

um.....yeah, that is like a mathcounts level problem. undefined his correct and please edit out my name......

Here is an interesting problem: not really contest-like, but nice:

without a calculator, approximate $ \cos{20^\circ}$

[hide="hard long tedious way"]develop an equation for 1/3 angle formulae (if possible)  and then use the 1/3 angle formula on 60 [/hide]

[hide="hint"]use the taylor series for $ \cos{x}$, which is $ \cos{x} \equal{} \displaystyle\sum_{k\equal{}0}^{\infty} (\minus{}1)^{k} \frac{x^k}{k!}$ but this only works for $ x$ in radians. which is $ \frac{\pi}{9}$   and notice that $ \pi \approx \sqrt{10}$. [/hide]

[quote="cognos599"]
without a calculator, approximate $ \cos{20^\circ}$
[/quote]
Or just say something like $ 1\minus{}\cos x$ is approximately quadratic, so $ 1\minus{}\cos 20^\circ\approx 1/9(1\minus{}\cos 60^\circ)\equal{}\frac 1{18}$. Thus $ \cos 20^\circ\approx 17/18$. The error is about $ 0.005$. If you want to do even better, say something like $ \cos x\approx 1\minus{}ax^2\plus{}bx^4$ and use $ 2$ points: $ 60^\circ$ and $ 90^\circ$ to interpolate (of course, you think of them as of $ 2/3$ and $ 1$ and then $ 20^\circ\equal{}2/9$. The system is
\[ \frac 49 a\minus{}\frac{16}{81}b\equal{}\frac 12,\qquad a\minus{}b\equal{}1\]
Solving, you get $ a\equal{}\frac{49}{40}$, $ b\equal{}\frac{9}{40}$. Now, plugging $ 2/9$ in, we get $ \cos 20^\circ\approx 1\minus{}\frac{49}{810}\plus{}\frac 2{5\cdot 9^3}\equal{}\frac{7290\minus{}441\plus{}4}{7290}\equal{}\frac{6853}{7290}$. The error is now about $ 0.0004$ and, what is extremely nice, it doesn't exceed $ 0.001$ on the entire interval $ [0,1]$ (an attempt to interpolate with just a quadratic polynomial would be much worse; the interpolation using only the $ 60^\circ$ angle would result is $ 1\minus{}\frac 98x^2$ with as large error as $ 0.125$ at $ 90^\circ$). So, unless you want more than 3 decimal places for your $ \cos$, you may put the calculator aside :)

A good approx of a the taylor series is the first 5 terms becuase after that, the factorials get huge and it is unneeded to keep on going.

[quote="logic812"][quote="#H34N1"][quote="logic812"]If anyone could tell me how to get rid of the rating thingy below my avatar...

Anirudh, here is a problem to test your skills

A,B, and C are three vertices of a cube, as shown in the diagram,. Find the measure of <ACB, to the nearest tenth of a degree.

Fair enough?

The cube is just crudely drawn and <ABC does not form a 45 degree angle.[/quote]

some things:

-firstly, ABC doesn't form a 45 degree angle....its quite obvious it doesn't
-that problem is like pwnable by just pytahgoras and like law of cosines
-why did you say ABC doesn't form a 45 degree angle...lol?
-how is that a test of skillz...its like moems lvl[/quote]
I know it isn't 45 degrees, I just put it up so people don't mistake <ABC for an angle on the same square.
I know it was easy, but I'll post a harder one.
Idk, I couldn't think of anything more inventive at the time.[/quote]

also, when'd I challenge your skills


Show that $ \alpha \equal{} \{ r \in \mathbb{Q}: r^{2} < 2 \}$ is not a Dedekind cut.

So let $ r \equal{} 0$. Then $ r \in \alpha$. Now choose $ q \equal{} \minus{}3$. Then $ q < r$ but $ q \not \in \alpha$. Thus $ \alpha$ is not a Dedekind cut.

Is this correct?


Given a triangle $ABC$, take $A'$ on the ray $BA$ (on the opposite side of $A$ to $B$) so that $AA' = BC$, and take $A\"$ on the ray $CA$ (on the opposite side of $A$ to $C$)
so that $AA\" = BC$. Similarly take $B', B\"$ on the rays $CB, AB$ respectively with $BB' = BB\" = CA$, and $C', C\"$ on the rays $AB, CB$.
Show that the area of the hexagon $A\"A'B\"B'C\"C'$ is at least 13 times the area of the triangle $ABC$.

I am quite certain that you can get this one, Mathx, it's really not difficult. It suffices to prove [A'BC"]+[B'CA"]+[C'AB"]+[AA'A"]+[BB'B"]+[CC'C"] >= 15*[ABC]. Write all areas out as the product of two sides and the sine of the angle in between, and use AM-GM.  :)

you´re right¡¡¡¡ anyway this problem is from the iberoamerican mathematical olympiad 92 i think nice problem¡¡

[quote="marko avila"]anyway this problem is from the iberoamerican mathematical olympiad 92[/quote]
Actually, it is much older. It was proposed for the Romanian Selection Test in 1970, but I think it is an even older russian or hungarian problem.


Hi, all!

I intend to score a 120+ on the AMC 12; what are the best resources available to help me do so? I'm bored as hell in school with the childish anecdotes they call "AP Calculus", where everything's just routine, so I'd like a chance to prepare for an exam that really stretches the limits of problem solving.

Thanks, all!

-Arachnotron

[quote="Arachnotron"]Hi, all!

 I'd like a chance to prepare for an exam that really stretches the limits of problem solving.

[/quote]

[horrible joke]lol,if you are taking AP calc. then why are you searching for an exam that stretches the [b]limits[/b] of problem solving. [/horrible joke]


can anybody tell me the definition of the real numbers !

Two examples of ways to define the real numbers

1. Dedekind Cuts. Search on it.
2. The cartesian product of the natural numbers and the power set of the natural numbers ($ \mathbb{N} \times P(\mathbb{N})$). The $ \mathbb{N}$ is to denote what the floor of the real number is. The $ P(\mathbb{N})$ gives the expansion of the number past the decimal point in binary. If, for instance, we consider the real number with the set of all odd natural numbers as the element of $ P(\mathbb{N})$, then our number would have $ .1010101010\ldots$ after the decimal point in binary.

It should be noted that defining the real numbers is no simple task.


probability
symmetry
algebra proposed
algebra

Prove that $ \ \sum_{r\equal{}0}^n C_r^{n\plus{}r}\left(\frac 12\right)^{n\plus{}r}\equal{}1$

Consider an $ (n\plus{}2) \times (n\plus{}2)$ grid of squares, and suppose we are moving a chip starting from the top left. We move it right or down each step with probability $ \frac{1}{2}$, and we stop when it reaches the right or bottom edge. By symmetry the probability we hit either edge is $ \frac{1}{2}$. The ways in which we can reach the right edge are to move right $ n\plus{}1$ times and down $ r$ times, $ r \leq n$, and in each case we have $ {n\plus{}r \choose r}$ ways to pick the downward moves (note the last move has to be right). When we add all of these possibilities up, we get $ \sum_{r\equal{}0}^n {n\plus{}r \choose r} \left( \frac{1}{2} \right)^{n\plus{}r\plus{}1} \equal{} \frac{1}{2}$, which is a multiplication by 2 away from the desired identity.

[quote="MellowMelon"]Consider an $ (n \plus{} 2) \times (n \plus{} 2)$ grid of squares, and suppose we are moving a chip starting from the top left. We move it right or down each step with probability $ \frac {1}{2}$, and we stop when it reaches the right or bottom edge. By symmetry the probability we hit either edge is $ \frac {1}{2}$. The ways in which we can reach the right edge are to move right $ n \plus{} 1$ times and down $ r$ times, $ r \leq n$, and in each case we have $ {n \plus{} r \choose r}$ ways to pick the downward moves (note the last move has to be right). When we add all of these possibilities up, we get $ \sum_{r \equal{} 0}^n {n \plus{} r \choose r} \left( \frac {1}{2} \right)^{n \plus{} r \plus{} 1} \equal{} \frac {1}{2}$, which is a multiplication by 2 away from the desired identity.[/quote]

very nice prove! I also came across this identity in a probability problem a different one (if you will ask I will give my probability model).There is a solution using binomials only. Can any one will give a try? I also want to know alternate ways.

Notice the equivalence to http://www.mathlinks.ro/viewtopic.php?t=63957 .

  darij


ratio
geometry
incenter
geometry unsolved

In a scalene triangle $ ABC$, $ I$ is the incenter and $ AI,BI$ meet $ BC,CA$ at $ A_1,B_1$ respectively. The lines through $ A_1,B_1$ parallel to $ AC,BC$ respectively meet the line $ CI$ at $ A_2,B_2$. Lines $ AA_2,BB_2$ meet at $ N$, and $ M$ is the midpoint of $ AB$. If $ CN\parallel{}IM$, find the ratio $ CN: IM$.

I think maybe this posted before.

anyway i don't recall a solution

My conclusion is that the triangle is a right triangle on angle C and that the relation is 2

Let $K,L$ be midpoints of $CB,CA$. $KI,LI$ intersects $AC,BC$ at $D,E$. Then it is easy to show that $AA_2E,BB_2D$ are straight lines. Let $A_3,B_3, C_3$ be the points of tangency of the excircles with the line segments opposite $A,B,C$. Then it is quite well known that $MI\parallel CC_3$ and similarly for the other 2 sides, and note that $N$ is the intersection of $AE,BD$, thus $AE,BD,CC_3$ are concurrent. Note that $\frac{BE}{EC}=\frac{B_3L}{LC}=\frac{c-a}{b}$. Thus by Cevas, we have
\[\frac{a+c-b}{b+c-a}\frac{c-a}{b}\frac{a}{c-b}=1\] and this miraculously simplifies to $a^2+b^2=c^2$ or $\angle ACB=90$. Now note that
\[\frac{CN}{NC_3}=\frac{CE}{EB}+\frac{CD}{DA}=\frac{b}{c-a}+\frac{a}{c-b}=\frac{(a+b-c)c}{(c-a)(c-b)}\]\[ \implies \frac{CN}{CC_3}=\frac{(a+b-c)c}{ab}
\]Now let $CI$ intersects $AB$ at $F$. Then
\[\frac{IM}{CC_3}=\frac{FM}{FC_3}=\frac{\frac{bc}{a+b}-\frac{c}{2}}{\frac{bc}{a+b}-\frac{a+c-b}{2}}=\frac{c}{a+b+c}
\]Finally just take the ratio and do a little algebraic manipulation and you should get the result 2.


geometry
rhombus
circumcircle
trapezoid
perpendicular bisector
geometry unsolved

[quote="shobber"]Find the least positive integer $n$ ($n\geq 3$), such that among any $n$ points (no three are collinear) in the plane, there exist three points which are the vertices of a non-isoscele triangle.[/quote]

It seems that ti suffices to consider the diameter of the set and then work some cases, am I right?

Daniel

[quote="shobber"]Find the least positive integer $n$ ($n\geq 3$), such that among any $n$ points (no three are collinear) in the plane, there exist three points which are the vertices of a non-isosceles triangle.[/quote]  [i]Proof.[/i] I'll prove that $n = 7$ fits. To show this, consider the following

[b]Lemma.[/b] Assume that the set $M$ consists of $5$ points in the plane such that every triangle composed of three points of $M$ is isosceles. Then there exists a set $N \subset M$ consisting of $4$ points $A$, $B$, $C$, $D$ such that $\overline{AB}=\overline{AC}=\overline{CD}$ and $\overline{AD}=\overline{BC}=\overline{BD}$. Moreover, the fifth point $E \in M \setminus N$ lies on the perpendicular bisectors of the sides $\overline{AC}$ and $\overline{BD}$ (since $AC \parallel BD$).

[i]Proof of the Lemma.[/i] Let $M={A}$, $B$, $C$, $D$, ${E}$ be a given set with the property that every triangle composed of three points of $M$ is isosceles and let $G$ be its graph. Since no three points are collinear, every edge of $G$ appears at least twice. Let without loss of generality $\overline{AB}=\overline{AC}$ and $\overline{BC}=\overline{BD}$.

Suppose that $\overline{AB}=\overline{BD}$. Consider triangle $\triangle ACD$. If $\overline{AC}=\overline{AD}$, then $D$ is the intersection of circles $(A)$ and $(B)$ with diameters $\overline{AB}$. If $\overline{AC}=\overline{CD}$, then $D$ is the intersection of circles $(C)$ and $(B)$ with diameters $\overline{AB}$. In both cases, $ADBC$ is a rhombus composed of two equilateral triangles $\triangle ADB$ and $\triangle ABC$. Now, $E$ does certainly not lie inside $ADBC$ (if so, $E$ must be the circumcenter of $\triangle ADB$ or $\triangle ABC$, a contradiction). Suppose now wlog that $\overline{BE}=\overline{BC}$ or $\overline{CE}=\overline{BC}$ (since $E$ must not lie on every perpendicular bisector). Then $\overline{AE} \ne \overline{BC}$ and $\overline{DE} \ne \overline{BC}$. Hence $E$ lies on the perpendicular bisector of $\overline{AD}$. Therefore, $BC \perp DE$. But then triangle $\triangle AEC$ is not isosceles, a contradiction. If $\overline{AD}=\overline{CD}$, then $D$ is the intersection of the perpendicular bisector of $\overline{AC}$ and circle $(B)$ with diameter $\overline{AB}$. Suppose that $AEBC$ is no rhombus composed of two equilateral triangles, since we already dealt with this case (just switch the roles of $D$ and $E$) and consider triangle $\triangle AEC$. Suppose wlog that $\overline{AE}=\overline{AC}$, since $E$ must not lie on the perpendicular bisector of $\overline{AC}$. Therefore, $\overline{EC}=\overline{EB}$ (triangles $\triangle ACB$ is isosceles), from which we conclude that $E$ is the intersection of the perpendicular bisector of $\overline{BC}$ and the circle $(A)$ with diameter $\overline{AB}$. Now there are just $4$ possibilities for $(D$, $E)$. It's easy to show that each of them leads to a contradiction. Therefore, we may suppose that $\overline{AB}=\overline{AC}$ and $\overline{BC}=\overline{BD}$, but $\overline{AB} \ne \overline{BD}$. Consider two cases:

1. Suppose that $\overline{AD}=\overline{BD}$. Since $\triangle ADC$ is isosceles, we either have $\overline{AC}=\overline{CD}$ or $\overline{AD}=\overline{DC}$. In the latter case, switching $(A$, $B$, $C$, $D) \to (D$, $C$, $B$, $A)$ leads directly to the contradiction obtained one paragraph above. Hence $\overline{AC}=\overline{CD}$ and thus $\overline{AB}=\overline{AC}=\overline{CD}$ and $\overline{AD}=\overline{BC}=\overline{BD}$. A quick angle chase shows that $AC\perp BD$ and moreover $\angle CBD=36^{\circ}$ (this actually determines every angle in the configuration $ADBC$). Suppose that $E$ does not lie on the perpendicular bisector of $BD$ and let wlog $\overline{DE}=\overline{DB}$. If $\overline{EC}=\overline{ED}$, then triangles $\triangle BDC$ and $\triangle EDC$ are concurrent, which means that $E$ is the image of $B$ when beeing reflected on $DC$. But then $A$, $C$, $E$ are collinear (this follows from the angles), which is a contradiction. Hence $\overline{EC}=\overline{DC}$, which means that triangles $\triangle EDC$ and $\triangle ADC$ are concurrent. Consequenty, $B$, $C$, $E$ are concurrent (an angle chase again), which is a contradiction again. Hence $E$ lies on the perpendicular of $\overline{BD}$ which proves the lemma.

2. Suppose that $\overline{AD}=\overline{AB}$. Hence $D$ is the intersection of circles $(A)$ with diameter $\overline{AB}$ and $(B)$ with diameter $\overline{BC}$. Wlog suppose that $E$ does not lie on the perpendicular bisector of $\overline{AD}$ (since $AB$ is an axis of symmetry). If $\overline{AE}=\overline{AD}$, i.e. $E$ belongs to the circumcircle of $\triangle BCD$, we get $\overline{BE}=\overline{DE}$. Let $\angle DAB= \angle BAC= 2\alpha$. An angle chase now shows that triangle $\triangle BEC$ has angles $\angle BEC=\alpha$, $\angle ECB=1/2\alpha$ and $\angle CBE= 180-3/2\alpha$.  Since $\triangle BEC$ is isosceles and $\alpha \ne 90^{\circ}$, we get $\alpha =72^{\circ}$. Thus, if we switch $(A$, $B$, $C$, $D$, $E) \to (E$, $D$, $B$, $A$, $C)$, the conditions of the lemma are fulfilled. Suppose now $\overline{DE}=\overline{AD}$. If $\overline{BE}=\overline{ED}$, $ABED$ is a rhombus. Therefore, $\overline{EC}=\overline{BC}$, since $E \not \equiv A$. This yields $\overline{DC}=\overline{AD}$, since $C$, $A$, $E$ are not collinear, which is a contradiction. If $\overline{BE}=\overline{BD}$, we get $\overline{AE}=\overline{BE}$ as well, since $E \not \equiv C$, $D$.

This finally proves the lemma, since these are the only two cases we have to consider ($\triangle ABD$ is isosceles and $AB \ne BD$). $\square$

Suppose now we've got a set $M$ conisting of $7$ points, such that every triangle composed of three points of $M$ is isosceles. By the lemma, there exist four points $A$, $B$, $C$, $D$ of $M$ such that they form an equilateral trapezoid and $\overline{AB}=\overline{AC}=\overline{CD}$ and $\overline{AD}=\overline{BC}=\overline{BD}$. Moreover, the remaining three points must all lie on the perpendicular bisector of $\overline{AC}$, clearly a contradiciton.

To see that $n=6$ does not fit, take a regular $5$-gon and its circumcenter. Therefore, $n=7$, as required. $\square$

i don't understand. The problem is really easy!
Let $S(n)$ is number of isosceles triangles in $n$ points.
Then $S(n)\le n(n-1)$ and we can find $n<8$ when there isn't a nonisoscele triangle.

The answer stays the same when the condition of points being colinear is removed. The answer is n=7. To see why n=6 is not sufficient, take a regular pentagon along with its center, and this set of 6 points clearly satisfies the condition of a good set. Given 2 points A,B in a set S that satisfies problem's condition, all other points in S are either on the circle (A,B) (meaning, with center A, passing through B), on (B,A) or on [A,B] (meaning the perpendicular bisector of A,B).
Assume A,B,C are coliniar in S. Then, wlog B is the midpoint of segment AC. (A,B) is tangent to (C,B) and [A,B] is parallel to [CB] and therefore every other point in S has to lie on (B,A). But (A,C),(C,A) are tangent to (B,A) at C,A so every other point of S has to lie on [A,C]. Buy [A,C] intersects (B,A) at 2 points that complete the configuration to a square and its center, which are only 5 points. So we may assume that there are no triples of colliniar points in S.


modular arithmetic
analytic geometry
number theory
greatest common divisor
relatively prime
number theory theorems

Can somebody state it exactly and post a few problems to illustrate its use?

http://www.math.sfu.ca/histmath/China/3rdCenturyBC/CRP1.html
http://www.cut-the-knot.org/blue/chinese.shtml

:)

Lol, I looked at both websites...and now I'm completely confused...can someone explain the method? :P Thanks.

[color=blue][b]Chinese remainder theorem[/b]
[i]
Suppose that $n_{1},...,n_{m}$ are pairwise relatively prime positive integers (i. e. $n_{1},...,n_{m}\in \mathbb{N}^{+}$ and for $i\ne j$ we have $\gcd\left( n_{i},n_{j}\right)=1$) and $r_{1},...,r_{m}$ are remainders modulo respectively $n_{1},...,n_{m}$. Then there exists exactly one natural number $n$ that $0\leq n <n_{1}\cdot...\cdot n_{m}$ and $n\equiv r_{i}\pmod{n_{i}}$ for any $i\in\{ 1,...,m \}$.
[/i]
[/color]

I hope that now it is clear for you.

A problem
 Prove that for all possitive Integer n It exists the Set of n distinc integer Terms satisfy that
 a) Product Of n-1 Random terms Mod Other Term=1.

can one show some more examples of applications?

[quote="Rzeszut"][color=blue][b]Chinese remainder theorem[/b]
[i]
Suppose that $ n_{1},...,n_{m}$ are pairwise relatively prime positive integers (i. e. $ n_{1},...,n_{m}\in\mathbb{N}^{\plus{}}$ and for $ i\ne j$ we have $ \gcd\left( n_{i},n_{j}\right) \equal{} 1$) and $ r_{1},...,r_{m}$ are remainders modulo respectively $ n_{1},...,n_{m}$. Then there exists exactly one natural number $ n$ that $ 0\leq n < n_{1}\cdot...\cdot n_{m}$ and $ n\equiv r_{i}\pmod{n_{i}}$ for any $ i\in\{ 1,...,m\}$.
[/i]
[/color]

I hope that now it is clear for you.[/quote]

Another Form of [b][color=orange]Chinese Remainder Theorem[/color][/b]:

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

Suppose we have a set of  $ n$  congruences of the form :
$ \begin{array}{c}x\;\equiv\;a_{1}\;\left(\;\text{mod }m_{1}\;\right)\\ x\;\equiv\;a_{2}\;\left(\;\text{mod }m_{2}\;\right)\\ \vdots\\ x\;\equiv\;a_{n}\;\left(\;\text{mod }m_{n}\;\right)\end{array}$

where $ m_{1},m_{2},\cdots,m_{n}$ are relatively prime. Let:

$ M \equal{}\prod^{n}_{k \equal{} 1}m_{k}$

and for all $ i\in\mathbb{N}$($ \;1\leq i\leq n\;$),let $ y_{i}$ be an integer that satisfies:

$ y_{i}\cdot\frac{M}{m_{i}}\;\equiv\;1\left(\;\text{mod }m_{i}\right)$

then one solution of these congruences is:
$ x_{0}\equal{}\sum^{n}_{i \equal{} 1}a_{i}y_{i}\cdot\frac{M}{m_{i}}$

Now any $ x\in\mathbb{Z}$ satisfies the set of congruences if and only if it satisfies:

$ x\;\equiv\; x_{0}\left(\;\text{mod }M\;\right)$

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

[b][color=green]this can also be generalized to the following:[/color][/b]

Let $ m_{1},m_{2},\cdots,m_{k}$ be positive integers relatively prime. Let $ a_{1},a_{2},\cdots,a_{k},b_{1},b_{2},\cdots,b_{k}$ be integers such that for $ i \equal{} 1,2,\cdots,k$ we have $ gcd\left(b_{i},m_{i}\right) \equal{} 1$. Then the set of congruences:

$ \begin{array}{ccc}b_{1}x\;\equiv\; a_{1}\left(\;\text{mod }m_{1}\;\right)\\ b_{2}x\;\equiv\; a_{2}\left(\;\text{mod }m_{2}\;\right)\\ \vdots\\ b_{k}x\;\equiv\; a_{k}\left(\;\text{mod }m_{k}\;\right)\end{array}$

is always solvable and its easy to give a same solution as in Chinese Remainder Theorem for this one! :)

[quote="SpongeBob"]This is very interesting one:

Point with integer coordinates in plane we will call GOOD if on
segment which connects that point and origin there is at least one
more point with integer coordinates (for instance (4,2) is a good
point but (3,1) isn't). Show that for every $n$ one can find a
square with sidelength $n$ in which every point with integer
coordinates is GOOD point.

Bye[/quote]

Can someone solve this? Thanks.


floor function
limit
real analysis
real analysis solved

The problem was: prove that there exists an ascending sequence of rational numbers that converges to the real $\alpha$.

So I defined $a_n = \lfloor \alpha \rfloor + \sum^n_{i=1} 10^{-i} \cdot c_i$ ($c_i$ being the i-th digit after the decimal point in its decimal representation)

Which is clearly rational, clearly ascending, and in my opinion also clearly converging to $\alpha$. But the last got denied by a teacher here, he claimed that one can never prove that this sequence actually converges to alpha... (and honestly if he says it's wrong I don't expect it to be right...)

But then the question comes to my mind: why?? Why can't it be proved that this converges to alpha? Is it something with the "decimal representation"? This looks fairly trivial to me, since it clearly converges and there can clearly be no other number the limit then alpha... Am I right on this?

Can anyone provide me some insight on why I'm right/wrong? I appear to be seriously in struggle with this :)

Thanks a lot!!!

One could quibble about the meaning of the word "ascending". If that means "non-decreasing", which is to say that $a_n\le a_{n+1}$ for all $n$, then you are correct. If the word "ascending" means "increasing", that is, $a_n<a_{n+1}$, then you have a problem that's annoying (but not impossible) to deal with.

But as I said, if you mean "non-increasing", then you're right. Your definition of $a_n$ can be recast as this:

$a_n=10^{-n}\lfloor 10^n\alpha \rfloor.$

From this we can show that:

$10^na_n\le 10^n\alpha<10^na_n+1$

$a_n\le\alpha<a_n+10^{-n}$, which can be rearranged to

$\alpha-10^{-n}<a_n\le\alpha.$

Since $|a_n-\alpha|<10^{-n}$ and $10^{-n}$ can be made arbitrarily small, this fulfils the definition of the convergence of a sequence.

That is, $\lim_{n\to\infty}a_n=\alpha\iff\forall\epsilon>0\exists N\ni n>N\implies |a_n-\alpha|<\epsilon.$ Quote that.

Yes... I ended up with that epsilon too. I don't know what he's gonna say of it cuz he claimed litteraly it could not be converted to epsilon-form :)

Anyway thanks!


algebra
polynomial
parameterization
abstract algebra
Ring Theory
superior algebra
superior algebra solved

[quote="Moubinool"]$F_2$ field with 2 elements, $R$ the commutative ring $R=F_{2}[t]/{t^3F_{2}[t]}$

1/ How many elements does $R$ contain ?
2/ What is the characteristic of $R$ ?
3/ Find all ring homomorphism $R\rightarrow R$[/quote]

If I understand well, you mean that $R$ is the ring of the polynomials in $\mathbb F_2$ modulo $t^3$. In such a case

[hide="1"]
Well, $R$, as a set, may be seen as the set of polynomial of degree at most 2, therefore there are 8 of them.
[/hide]

[hide="2"]
Rings in general does not have characteristic, fields does, but I believe you mean 2.
[/hide]

[hide="3"]
Well, a non-trivial homomorphism $R\rightarrow R$ needs to fix $1$. Besides, you know that if $\phi(t)=at^2+bt+c$ (where $\phi$ is the homomorphism in question), and $\phi(0)=\phi(t^3)=(ac^2+abc+b^2c)t^2+bc^2t+c^3$, therefore, $c=0$ and the other two parameters are free. Therefore, we have four choises for $\phi(t)$, and those fix the rest of the elements, there are four homomorphisms $R\rightarrow R$
[/hide]

For caracteristic of a ring $R$ is defined by the $q$ positive integer such that
$qZ$ is the kernel of the morphism $f;Z\rightarrow R$ with $f(n)=n.1_{R}$


An elastical, horizontal spring of negligible mass, with $k = 100 \, \text N \slash \text m$ and initial length $\ell_{0}$, is compressed by an object of mass $m_{1}= 2 \, \text{kg}$. The system is set free.
[img]6947[/img]
a. Neglecting friction, determine the maximum height the object reaches on the spherical surface of radius $R = 0.2 \, \text m$.
b. Compute the reaction force in the maximum height point.
c. Suppose that there is friction, $\mu_\text{sliding}= 0.01$ and that in point $A$ the object of mass $m_{1}$ collides elastically with an object of mass $m_{2}= 2 m_{1}$, which is not moving. Determine the speed of the second object right after the collision.


[quote="sonny"][hide]800 miles. The best way is just $1000*4/5=800$[/hide][/quote]
Correct. :)

[quote="i_like_pie"]Steve and Tracy are planning a 1000 mile trip in an automobile with five tires, four of which are in use at a time. They plan to interchange them so that each tire is used for the same amount of miles. For how many miles will each tire be used?[/quote]

[hide="Solution"]

$\frac{1000\cdot4}{5}$, because the car has 4 tires, and a spare one, and each one is then used for $\boxed{800}$ miles.  :) [/hide]

[quote="bpms"][quote="anirudh"][quote="i_like_pie"]Steve and Tracy are planning a 1000 mile trip in an automobile with five tires, four of which are in use at a time. They plan to interchange them so that each tire is used for the same amount of miles. For how many miles will each tire be used?[/quote]

[hide="Solution"]

$\frac{1000\cdot5}{4}$, because the car has 4 tires, and a spare one, and each one is then used for $\boxed{800}$ miles.  :) [/hide][/quote]
[hide=";)"]$\frac{1000*5}{4}$ isn't right. 800 is.[/hide][/quote]

It was a typo.  :blush:  :oops:  I fixed it. :blush:


[quote="saeid"]let$A=\{1,2,\dots,v\},\ B_1,B_2,\dots,B_b \subset A$ such that $|B_i|=k$ and every element of $A$ is in $r$number of $B_i$.
and every two elements of $A$ exist in one of the $B_i$ Maximum.
if $x\in A,\ x \not \in B_j$ then there exists exactly $t$ from $B_i$ such contain $x$ and $B_i \cap B_j=\phi$ prove that:
$v=k(1+\frac{(k-1)(r-1)}{t})\ ,\ b=r(1+\frac{(k-1)(r-1)}{t})$[/quote]
the problem yeild us to the $tarhe bolooki$ ;)

[quote="saeid"]let$A=\{1,2,\dots,v\},\ B_1,B_2,\dots,B_b \subset A$ such that $|B_i|=k$ and every element of $A$ is in $r$number of $B_i$.
and every two elements of $A$ exist in one of the $B_i$ Maximum.
if $x\in A,\ x \not \in B_j$ then there exists exactly $t$ from $B_i$ such contain $x$ and $B_i \cap B_j=\phi$ prove that:
$v=k(1+\frac{(k-1)(r-1)}{t})\ ,\ b=r(1+\frac{(k-1)(r-1)}{t})$[/quote]
assume that $S$ is the set of all pairs such that $x\in B_1,y\notin B_1$ and $x,y$ are together in one of the $B_i$ thus:
$(k-1)(r-1)=|S|=t(v-k)$ and we know that $bk=rv$ from these equalities we can reach the other one ;)
sorry if I didnt translate it good :oops:


Please prove, and show for which case the equality holds.

For a triangle with sides a,b,c
a^2 + b^2 + c^2 >= 4(3)^(1/2) * K

Where K is the area of triangle abc.

This is an old IMO question.

http://www.kalva.demon.co.uk/imo/isoln/isoln612.html

This result is worth remembering. It can be useful.

For example, IMO1991/5 can be solved using this theorem.

It's called Weitzenbck's inequality.

http://mathworld.wolfram.com/WeitzenboecksInequality.html

Oh really? This is an old IMOer? Wow! A friend gave it to me and I had fun tackling it so yeah, I'm glad you guys recognized that!

Thanks!!

As I said it's worth remembering ! This can be very useful.

hello, a more general result is the inequality by Hadwiger-Finsler, see here:
http://en.wikipedia.org/wiki/Hadwiger-Finsler_inequality
Sonnhard.

this is an old thread but ..
I don't know how a general inequality can be apear on IMO.Its like to prove Cauchy or Jensen.This is not OK !

Hello! it's IMO 1961 problem 2. you can see my solution in my blog here: http://mathsmaster.wordpress.com/2009/01/04/imo-1961-probleme-2/


Ok so I would like Texnic Center to remember my open files from last time, but I haven't found an option to do that (similar to WinEdt's feature). Is it even possible?

I've not come across such a feature though that doesn't prove it doesn't exist :) The TeXniCenter authors have recently started a forum which has a [url=http://www.latex-community.org/viewforum.php?f=36]Feature Suggestions[/url] section you may wish to use.


yek seri masaleye jalebe rahat va bedard bokhor (albateh dar tarsim) ro inja bayan mikonam va omidvaram harki har rahi midoone post kone. :D

if we have $a,b,c$ then draw $x$ .($ax^2+bx+c+=0$)

[quote="jensen"]draw a triangle with $(b,c,t_a)$.(without calculating.)[/quote]
what does it mean "$t_a$"?

[quote="mohamadhosein"][quote="jensen"]draw a triangle with $(b,c,t_a)$.(without calculating.)[/quote]
khob choon $b,c$ ro darim pas nesbate $b/c$ ra ham darim.khate $ac$ ra dar nazar begir va do dayere az $a$ be shoahaye $t_a,c$ bezan va hala ye tajanose nesbat bee markaze $c$ va nesbate$(b+c)/b$ bezan har ja mojanes dayereye$t_a$ dayereye digar ra ghat kard rase$b$ mibashad.
vala az masaele dige chizi nafahmidam aya manzoore az x ye nemoodare?[/quote]
na manzooram pare khaty be toole $x$ mibashad.
az rahe hal mamnoon. :D

[quote="mohamadhosein"][quote="mohamadhosein"][quote="jensen"]draw a triangle with $(b,c,t_a)$.(without calculating.)[/quote]
khob choon $b,c$ ro darim pas nesbate $b/c$ ra ham darim.khate $ac$ ra dar nazar begir va do dayere az $a$ be shoahaye $t_a,c$ bezan va hala ye tajanose nesbat bee markaze $c$ va nesbate$(b+c)/b$ bezan har ja mojanes dayereye$t_a$ dayereye digar ra ghat kard rase$b$ mibashad.
vala az masaele dige chizi nafahmidam aya manzoore az x ye nemoodare?[/quote]
masaln mikham bedoonam in masale ro shoma ta didin hal kardin ya modati tool keshid.[/quote]
mitoonam beporsam daghighan ba ki hasti?

in masale ba yek kami mohasebe besiar rahat hal mishe. :D

[quote="BaBaK Ghalebi"]chera hatta faghat pargar ham too tarsimhaye hendesi kafie...
faghat kafie sabet koni ke monakese ye noghte mesle $A$ ro nesbat be ye dayere mafrooze $\sum$ ke noghtei mesle $A'$ hast ro mitooni ba pargar peida koni...
inam asoone dige...
gheir az ine?? :huh:  :?[/quote]
midoonam mishe va midoonam ke to mitooni vali shayad ye seria ba khatkesh bishtar hal konan.

[quote="BaBaK Ghalebi"]chera hatta faghat pargar ham too tarsimhaye hendesi kafie...
faghat kafie sabet koni ke monakese ye noghte mesle $A$ ro nesbat be ye dayere mafrooze $\sum$ ke noghtei mesle $A'$ hast ro mitooni ba pargar peida koni...
inam asoone dige...
gheir az ine?? :huh:  :?[/quote]

man keh gheir az i nagoftam.

enekaso nemishe be onvane ye abzare fiziki dar nazar gereft.

ageh deghat kardeh bashi, manzooram in  bood keh alaveh

bar khat kesh va par gar, hoosh ham lazemeh.

yani nabaiad mesle yek jabeh abzare dar basteh bood. ;)

[quote="BaBaK Ghalebi"][quote="payman_pm"][quote="BaBaK Ghalebi"]$(t_{a},t_{b},t_{c})$[/quote]
in masale tarsimpazir nist yani inke aslan ba khatkesh o pargar rasm nemishe!!
dalilesh ham ine keh khob age tool nimsaz ro bar hasb azla hesab konim
forje 3 mishe.hamoon tor ke ham midoonin hamchin tarsimy nadarim![/quote]
mishe keshid
momkene begi khob pas forje 3 ro mishe keshid dge age beshe in mosalaso keshid...onvaghte ke ye eshtebahe dge anjam daD...eshtebahe avaleto adamaye dgei ham anjam dadan...  :wink:[/quote]
man ke nafahmidam shoma che gofti?
am be har hal ta zamani keh dar safhe mosatah bahs mikonim ghabel rast nist.
in yeh ghazieast!!barayeh mesal ba khatkesho pargar nemishe rishe sevom 2 ro rasm kar.
deghat konid keh kolan  nemishe yani in ghablan esbat shode.esbatesh ham kollan rabti beh hendese nadareh!!!

[quote="payman_pm"][/quote][quote="BaBaK Ghalebi"][quote="payman_pm"][quote="BaBaK Ghalebi"]$(t_{a},t_{b},t_{c})$[/quote]
in masale tarsimpazir nist yani inke aslan ba khatkesh o pargar rasm nemishe!!
dalilesh ham ine keh khob age tool nimsaz ro bar hasb azla hesab konim
forje 3 mishe.hamoon tor ke ham midoonin hamchin tarsimy nadarim![/quote]
mishe keshid
momkene begi khob pas forje 3 ro mishe keshid dge age beshe in mosalaso keshid...onvaghte ke ye eshtebahe dge anjam daD...eshtebahe avaleto adamaye dgei ham anjam dadan...  :wink:[/quote][quote="payman_pm"]
man ke nafahmidam shoma che gofti?
am be har hal ta zamani keh dar safhe mosatah bahs mikonim ghabel rast nist.
in yeh ghazieast!!barayeh mesal ba khatkesho pargar nemishe rishe sevom 2 ro rasm kar.
deghat konid keh kolan  nemishe yani in ghablan esbat shode.esbatesh ham kollan rabti beh hendese nadareh!!![/quote]
man darbare keshidane forje 3 hich harfi nazadam...manzooram in bood ke forje 3 hich rabT be masale ma nadare va masale ma ghabele tarsime...

dar halate kolli keshidane mosalas ba dashtane toole 3 nimsaze on gheire momken ast.

[quote="payman_pm"][quote="BaBaK Ghalebi"]$(t_{a},t_{b},t_{c})$[/quote]
in masale tarsimpazir nist yani inke aslan ba khatkesh o pargar rasm nemishe!!
dalilesh ham ine keh khob age tool nimsaz ro bar hasb azla hesab konim
forje 3 mishe.hamoon tor ke ham midoonin hamchin tarsimy nadarim![/quote]

Chi forje 3 mishe??? :?:  :?:

[/quote]
man darbare keshidane forje 3 hich harfi nazadam...manzooram in bood ke forje 3 hich rabT be masale ma nadare va masale ma ghabele tarsime...[/quote]
eybaba bebin agar tool yek zelo ro bar hasb nimsaz mosalas hesab konim az forje 3 mishe rah standardesham estefadeh az ghazieh stowart
ast.
hala bia farz kon ye hamchin mosalasi ro rasm kardi be ebarat digeh
age betooni rasm koni toonesti ba dashtan yeh tool moshakhas rishe sevomesho rasm koni.in ham hamintoor ke ghablan toozih dadam nemisheh.
mored badi ham ine ke age rah halli bara masale dari khoshhal misham ono bebinam chon be har hal har ki in masalaro TARIM kone kheili chiza ro beh ham mirize :lol:
beh har hal age bazam harfam ro motevaje nashodi lotfan raheto post kon  :wink:

[quote="Amir.S"]man hanooz nafahmidam chi forje 3 mishe?[/quote]
$t_{a}^{2}=\frac{bc((b+c)^{2}-a^{2})}{(b+c)^{2}}$
$t_{a}^{2}=\frac{b^{3}c+bc^{3}+2b^{2}c^{2}-a^{2}bc}{(b+c)^{2}}$
$t_{a}^{2}b^{2}+t_{a}^{2}c^{2}+2t_{a}^{2}bc=bc^{3}+cb^{3}+2b^{2}c^{2}-a^{2}bc$
$t_{a}^{2}b^{2}+t_{a}^{2}c^{2}+2t_{a}^{2}bc-bc^{3}-cb^{3}-2b^{2}c^{2}+a^{2}bc=0$$(I)$
$t_{a},t_{b},t_{c}$ adad sabet hastand chon tool anha ro darim.
agar baraye $t_{b},t_{c}$ ham rabete $(I)$ ro benevisim yek dastgah 3 moadele 3 majhool az daraje 3 darim.
yani $a,b,c$ rishe haye yek moadel az daraje 3 hastand.(va ehtemalan
be soorat rishe sevom neveshte mishan)
chenin moadelati dar safhe ghabel tarsim nistand albate dar halat KOLLI.
momkene agar $t_{a},t_{b},t_{c}$ adad khasi bashand masale tarsim beshe
albate motmaen nistam.

[quote="saeid"]khob ma midoonim ye moadele daraje 3 dar halat kolli reeshehash ghabele rasm nist , dar morede 3 moadele 3 majhool sohbat nakardim.[/quote]
pesar khoob!

avvalan age dastgah ro hal koni(keh pishnehad mikonam in karo nakoni!! :lol: )$a,b,c$ be soorat rishe sevom mishan keh ghabel rasm nist!

sanian hamin keh dastgah darje $3$ ast mishe har moadele ro bar hasb
do moade dige nevesht ke bazam natije mishe keh har kodom az azla
rishe ye moadele darje $3$ ast ke bazam hamoon natije ghably bedast miayad!

Be tore kolli sabet mishe ye adad tarsimpazire (ba dashtane parekhat be tool 1) iff chand jomlei minamale oon az daraje $2^{k}$ bashe.


AMC
AIME
USA(J)MO
USAMO
geometry
ARML
number theory

hi. i'll try to make this as brief.

so i started the preparation for the amc 12/ aime 2-3 months ago iwth little knowledge on AMC/AIME but good mathocunt type knowledge.

with 1 prac test each. the score difference then and now is.
AMC 12.  90-105  ---->  105-120    <--- they were like 16 right 9 left out
AIME.  1   -----> 4 (just tried the first 7 since the rest was too mind boggling)

My questions:

does amc 12 and aime cover the same topics.?
if not what should i study for AMC 12 (in AoPS)?   <--what r the imporantt chapters in aops for amc prep like important chapters for aime mr rruszyk previously posted.

what would have more score rise AMC or AIME?

what's the usual USAMO qualifying score?? would 125 AMC 7 aIME be fine?

any advices suggestions would be appreciated.

thanks

Higher than 100 on AMC 12 gets you in AIME, and I believe that USAMO would require a grade of at least 10 on AIME.

thanks
no, i'm a junior. so this is my last chance.

ignore the above questions. :D

but could you tell me which is better considering i get about 115 on AMC and 3 on AIME.
i only did 2 AIME and 3 AMC so far.

i went over AoPS 2 imporatnt chapters. (but not thorougly).

so would practicing the old AMC/ AIme tests increase my score or should i just reread the AoPS and do some tests?

This is my personal view on it but I'd say do anything you can get your hands on.  You don't need to restrict yourself to only AMC problems but try working on ANY (e.g. ARML, MAO) problem you don't know how to solve.

[quote="Elemennop"]What grade are you in? I'm assuming you're an underclassman since you mentioned MathCounts. If you're in grade 10 and below, you'll probably only need a 7 or 8 on the AIME to make the USAMO (AMC score is not counted for underclassmen). Otherwise, you'll probably need a 210-220 index (AMC + 10xAime).

The AIME and AMC do tend to cover similar topics (that is, they both encompass the major topics such as probability/combinatorics, algebra, geometry, number theory, and so on). However, only the last 10-15 questions of the AMC get near the same group as the AIME.

For preparation, I'll just quote what Rrusczyk posted in another thread:

[quote]Here's how I'd roughly prioritize the chapters for AIME preparation:

High: 1, 2, 3, 4, 6, 7, 9, 13, 15, 16, 17, 18, 19, 23

Medium: 5, 10, 11, 12, 14, 24

Low: Everything else. [/quote][/quote]

If I'm in grade 10 but taking AMC 12, would they use the index method, or 7 or 8 on AIME?

They use floor score regardless of what test you take if you're an underclassman.


Can someone help me understand how to solve this question ---
One or more of five siblings have eaten some cakes without their mother's permission.When questioned , they gave the following answers --
1st child: Only one of us ate the cake
2nd child:Exactly two of us ate the cake
3rd child:Exactly two of us did not eat the cake.
4th child : Only one of us did not eat the cake.
5th child :All of us ate the cake.
Their mother knew from past behaviour that the guilty children lied while te others told the truth.The number of children who ate the cakes were ?
1 , 2 , 3 , 4 or 5 ???

I'm clueless as to how to solve this one.... :?

[hide]Well if only 4 ate it, it works. Well actually, you see that only one statement can be true, and thus it has to be 4.[/hide]

[quote="karthik_k"]Could you explain that ?????
Please......... :)[/quote]

Sorry. Well suppose two people were telling the truth. Then we know that 2 is telling the truth, but nobody else can or they would contradict what 2 had said. So only one person can be telling the truth.

So suppose 1 person is telling the truth. Well, then the person who says that 1 person is telling the truth is telling the truth. So 4 is telling the truth because he says "Only one of us did not eat the cake" => "Only one of us is telling the truth" which coincides with what we wanted.


function
factorial
Pascal\u0027s Triangle

I have just started to C when I came upon this problem.

With very limited knowledge that I possess( very minimal i.e. i dont know anything about arrays, strings, structures, functions, pointers or case control structure)

Write a program that produces the Pascal's triangle.

It may be to your best interests to learn about arrays; it's the easiest way to construct Pascal's triangle (with a row-by-row iteration).

Of course, you can define a factorial function and iterate through $ {n choose k}$ that way, or write a recursive method to construct single elements leading up to that point, etc.

(You should start with JAVA if you are just beginning to learn programming; then move on to an assembly language or C after you become familiar with the programming process (Essentially the problem-solving portion in thinking of the code).

But anyways, google "C Pascal Triangle code" and you should find some results; I expect that most of them will  use arrays of some sort. I'll give some specific help if that doesn't work out.

[hide=abellows2021 (#3)]
[url="https://artofproblemsolving.com/community/p22321812"]Original Post[/url]
Yes here it is in py

[pywindow]
#Pasclacs triangle
def pascal(row):
  masterRow = []
  srow = [1,1]
  for i in range(row):
    thisRow = [1]
    for dig in range(i+1):
      dig += 1
      thisRow.append(srow[dig]+srow[dig-1])
    thisRow.append(1)
    print(thisRow)
    srow = thisRow
    masterRow.append(thisRow)
    thisRow = []

rows = input("Howb many rows?: ")
pascal(int(rows))
[/pywindow]
[/hide]

This is about C, not python. There was already a thread for this, what's the point of bringing back one from 2009?


Prove or disprove:

Given three non-collinear points, one and only one circle can be drawn through those three points.
[hide="hint"] Don't use coordinates....  [hide="hint"]equidistance[/hide][/hide]

Maybe I'm missing something, this seems too easy:

[hide] The points form a triangle. Every triangle has one circumcircle.[/hide]

obviously, you can always show that the perpendicular bisectors of the sides are always concurrent for 3 noncollinear points, furthermore, this is the center of the circle, that will pass through all 3 points


I am no expert in linuistics or languages. I just happened to make this curious observation, curious atleast to me. Now in any language there are words which have different meanings in different contexts. One such English word is "draw"

Two different meanings are:

1. I draw water from the well.

2. I draw pictures.

Now in Hindi, the word for "draw"(as in 1) and the word for "draw"(as in 2) are the same (the hindi word would be keenchna).  Though the meanings are different, the word is the same in Hindi as well as English.

This struck me as being curious. Is it just a coincidence?

It's interesting that Latin duco has the same two meanings (though not too much of the second), even though neither duco nor keenchna is etymologically related to draw - traho and dhrajati are. Yet apparently the English meaning of draw as in "draw a picture" is only 800 years old. Since there doesn't appear to be any obvious connection between the two meanings, I would guess that the similarity is either REALLY old (this seems doubtful) or English/Hindi developed the meanings from contact with Latin (maybe English in Hindi's case). Just a guess, though.

People pull on pens or brushes to make pictures, so the semantic connection could exist readily in a lot of different languages.

In Chinese, the verbs are completely distinct between those two meanings.

Thanks people. It was news to me that Latin too shared this feature. Interesting and the explanations are an eye-opener.

[quote]I would guess that the similarity is either REALLY old (this seems doubtful) [/quote]

It's not necessarily too far-fetched.  Indo-European languages often share the same poetic formulas, with even the same  cognates-- if you want examples, you can read Calvert Watkins' book  [i]How to Kill a Dragon[/i] (You should read it anyway, if you like historical linguistics, because the book is written beautifully.)  I wonder if this is indeed another example.

[quote]or English/Hindi developed the meanings from contact with Latin (maybe English in Hindi's case).[/quote]

It would be easier to postulate that Hindi adopted the equation of "drag" with "write" from Sanskrit.  The Sanskrit verb root [i]likh-[/i] can mean either draw (as in, plough a furrow) or write, draw, engrave.  ([url]http://webapps.uni-koeln.de/tamil/[/url])

[/i]

In swedish this would be "dra vatten från brunnen". Curiously, you can only "dra" lines, but no other figure or picture as far as I know.

Actually dra can be used exactly like that in swedish too. Except dra as "lead" has the connotation of the person being lead not totally enthusiastic to the idea, and for cheating someone we say "bedra", but the be- thing is just an old thing (the exact meaning of which is unclear to me) that we've dropped from most verbs, but not all.

[quote="fedja"]In Russian you can провести (draw) a line (but no other figure), провести (lay) a furrow or a pipeline, провести (lead) somebody to some destination, and провести (successfully cheat) somebody. I guess the links between every two adjacent meanings in this chain are clear but if I gave you just the endpoints, it would seem really strange :).[/quote]

Wow!! This is really interesting. True, one has to trace back the usage exactly to get a complete understanding.

[quote]Wow!! This is really interesting. True, one has to trace back the usage exactly to get a complete understanding.[/quote]

Here's another.  The Sanskrit word "satī" (or "suttee") is the practice of a widow immolating herself on her husband's funeral pyre.  It's really the present participle (nom. s. f.) of the verb root "as-" "to be."   How did this happen?  The participle "being" took on the meaning of "being good," or, in the feminine, "being a good woman/wife."  One of the cultural practices that would make you a good woman was throwing yourself on the funeral pyre with your dead husband, and so "satī" took on that meaning.

Coincidentally, the verb "as-" plus the prefix "su," "good," gives us "swastika."

If you like this sort of thing, you should check out Calvert Watkin's [i]American Heritage Dictionary of Indo-European Roots[/i]-- look up the entry for "dragon."


This is semi-stolen from an IChO Forum. This may very well be spam (and it is). Hey, we all need to plan ahead :roll: . Please don't send me hate mails.

A year ago it would have been a struggle between Chemistry or Maths. However, now it has to be Maths!

math! math! math! math! math! math! *chorus joins in* math! math! math! math! math! math! math! math! math! math! math! math!

Biochem or math. Probably math. But biochem is always fun  :) .

Wow, no one picked physics! I thought heaps of people would vote that.

Never tell your physics teacher that, 'I only took up Physics so I can understand Chemistry better.'

Math!!! I would be taking Pre-Calc this year (8th grade) but my school system is stupid and I would run out of math courses by senior year. So I'm stuck taking a freaking awesome AoPS course!!! W00t!

[quote="math_explorer"]Math, of course. There's 7,000,000 USD waiting out there![/quote]

You mean for those problems?  One of them was already solved.

And I'd take math, because I feel like it

Well, American culture seems to think that spelling is more important than math.
Just compare the prizes for the national spelling bee to MC...

Yes, I agree. But what ****ing use is spelling? I mean, who the **** cares that you can memorize the spellings of a million or so words?

Back on topic....

Definitely chemistry! Spend your last day plotting how to develop a destructive bomb that would blow up the entire galaxy! Nah, jk. I chose math.

Solved? Really? Which?

P=NP
Riemann Hyp.
Hodge
Yang-Mills
Birch+Swinnerton-Dyer
or what? I can't remember...

I had no idea...
(I'm a newbie and can't figure out how I'm supposed to use the quote feature.)

Well that makes it 6,000,000 USD.

Still math.

I'm running out of courses already. My school was nice and let me take math 7 grades ahead, but next year their grades are gonna be exhausted, so I end up learning how to make math fair things.

The one you can't remember is the solved one: Poincare Conjecture.

BanishedTraitor, that can't be right. I forgot two. There're seven.

So, someone solved Poincare. There's one more.

Ahh... Navier-Stokes. Just looked round and saw it. Thanks Klebian for letting me notice...  :rotfl:

Ok, who?

I wouldn't do any of those. I would read a book.

I honestly don't know what I would do. Probably start a petition to ban the use of oxygen-dihydride (lol changing the name should do the job  :P ).


According to an ancient belief, when a friend visits a sick person, $ \frac {1}{70}$ of his sickness is taken away. What is the least amount of friends that need to visit to take away at least $ 86$(percent) of a person's illness?

I was wondering if there was a way to do this without a calculator/without logarithms?

EDIT:This was a MC problem. I wasn't sure what round it was though...and it got moved :( I guess it's a Target/Team Round question.

[hide]137 friends[/hide]

hmm...good question...ill think about it but i think that you need logs and that it isnt mathcounts level!

No, you definitely need logs, unless you want to guess and check.

123456789, is right.

You'd have to enter log(14/100) / log(69/70) on your calculator and round up.

[hide]
\[ \left(\frac {69}{70}\right)^n \le 0.14,\ n \in \mathbb{Z}\implies n \ge 137
\]
[/hide]

Yes i totally aree but why couldnt you do...

[hide]$ ({\frac{1}{70}})^n \ge .87$

yeilding $ n \ge 137$ --i think!!![/hide]

zolojetto, that's not going to work. $ \frac {1}{70} < 0.87$, so repeated multiplications of $ \frac {1}{70}$ on the LHS will never bring the LHS to be greater than the RHS.

Your equation means to repeat multiplication on $ \frac{1}{70}$. However, the problem says that $ \frac{1}{70}$ of the illness is TAKEN AWAY. Therefore, you must do $ 1\minus{}\frac{1}{70}\equal{}\frac{69}{70}$.


[quote="Iversonfan2005"]4 :pi: -8[/quote]

PLEASE EXPLAIN AND HIDE ANSWERS!!!!

Hm, I got a different and weird answer.  I really like the question though.

[hide]
The important part is to find how much of the band is touching the semicircles.  If we extend the band to create a larger square outside the semicircles, the semicircles are obviously tangent to this square, and the lengths between a vertex of the square and the points of tangency of the closest semicircle are equal.  If these two are equal, we know the band is 90 degrees of the circumference.  To prove this, we can create a kite with points: vertex of large square, center of semicircle, and both points of tangency of the semicircle.  The vertex has an angle of 90 degrees, and the two points of tangency, by definition, have angles measuring 90 degrees, so the final angle (the one of the center of the circle) is 90 degrees.  Now we can find the length the band takes on each semicircle, which is $\frac \pi 4$, so the total length of all four semicircles is $\pi$.
Now we need to find the length the straight parts occupy.  This length is the same as the distance between the centers of two adjacent semicircles.  Since these centers are the midpoints of the smaller square, we can create a 45-45-90 right triangle with side lengths of 1, and a hypotenuse of $\sqrt 2$, which is our length.  There are four of these, so the total length of the straight parts is $4\sqrt 2$
So the total length of the band is $\pi + 4 \sqrt 2$
[/hide]

Not sure if this is right.
[hide]I put the figure on a coordinate plane, with the center of the circle at $(0,0)$.  To find the length of the band, we can split it up into two "types".  One which is over the circle, and one which is straight.  The one over the circle will be over parts of the four circles, and the straight parts will be in between them.

For the circle parts, each semicircle will have half of it over the circle part, and since it is radius one, each of those parts is $\frac{\pi}{2}$, making the length of all circle parts $2\pi$.

The straight parts will come between those.  So we must find the coordinates of two of those starting locations.  Consider the top semicircle.  If a 45-45-90 triangle is drawn from the center of the semicircle to the semicircle, then the two legs of it will be $\frac{\sqrt{2}}{2}$.  So the coordinates of one starting point is $(\frac{\sqrt{2}}{2},1+\frac{\sqrt{2}}{2})$.  Now consider the semicircle on the right.  Using similar logic as above, the coordinates of the endpoint of the line connecting the point on the top circle to this point are $(1+\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})$.  So using the distance formula the length of one of these four straight segments is $\sqrt{1+1}=\sqrt{2}$, so the total lenght of the straight segments is $4\sqrt{2}$.

Adding up the circular and straight parts gives us $2\pi+4\sqrt{2}$.[/hide]

Ah, crap.  I did the circumference wrong.  You're right, the part of the band on each semicircle is $\frac \pi 2$ not $\frac \pi 4$.....  :(

wait, deej21, i think your first solution might be right, because you notice the band does not stretch all around the the semicircle, and for an equilateral triangle, i know that if you stretch a band around it, the circle parts add up to the circumference of one circle, or each corner being 1/3 of a whole circle, and with two circles tangent in one place, there should be one circle (i think) because there will be two semiclrcles plus 2 diameters...so with "2" sides, we have each "corner" being a semicircle, or 1/2 of the circumference, and with 3 sides, we have each corner being 1/3 of the circumference, so if the patter continues each of these should be 1/4 (i'm not completely sure on this), so i think your first answer is right...although if i'm wrong i won't be surprised :D

[quote="deej21"]My SOLUTION is right.  At least I still think it is.  I just made a calculation error when calculating how much area 90 degrees of the circumference takes up.  I made the circumference $\pi$, thus making 90 degrees of it $\frac \pi 4$, when really the circumference is $2\pi$ and 90 degrees of it is $\frac \pi 2$.

And if you look at my solution, I prove that each corner is $\frac 1 4$ (thus 90 degrees) of the circumference of a whole circle.[/quote]

o, oops  :blush:  i see what happened,  :D, dumb mistake on both our parts  :blush: , yea, i think that the second answer is right now


If $ C$ and $ P$ are Hermitian matrices of the same order with $ P>0$, then $ CPC\plus{}P^{\minus{}1}\geq 2C$.
  Here $ P>(\geq)0$ means $ P$ is positive definite (semidifinite).

[b]Very well[/b] !
[hide] $ P$ is Hermite matrix and $ P > 0$
Show that the exist $ (U,D) \in U_nxD_n(R)$, (Here $ U$ : unita matrix)
   $ P \equal{} UDU^{*},D \equal{} diag(\lambda_1,\lambda_2,....,\lambda_n), \lambda_k > 0 , \forall k \in${$ 1,2,...,n$}
Put $ y \equal{} U^{*}Cx \equal{} \left[\begin{array}{c} y_1 \\
y_2 \\
... \\
y_n \end {array} \right]$ ,$ z \equal{} U^{*}x \equal{} \left[\begin{array}{c} z_1 \\
z_2 \\
... \\
z_n \end {array} \right]$
 Then $ \forall x \in M_{n1}(C)$ We have :
    $ x^{*}CPCx \plus{} x^{*}P^{ \minus{} 1}x \equal{} (x^{*}CU)D(U^{*}Cx) \plus{} (x^{*}U)D^{ \minus{} 1}(U^{*}x) \equal{} y^{*}diag(\lambda_1,\lambda_2,...,\lambda_n)y \plus{} z^{*}diag(\frac {1}{\lambda_1},\frac {1}{\lambda_2},...,\frac {1}{\lambda_n})z \equal{} \lambda_1|y_1|^2 \plus{} \lambda_2|y_2|^2 \plus{} ... \plus{} \lambda_n|y_n|^2 \plus{} \frac {1}{\lambda_1}|\overline{z_1}|^2 \plus{} \frac {1}{\lambda_2}| \overline{z_2}|^2 \plus{} ... \plus{} \frac {1}{\lambda_n}| \overline{z_n}|^2 \equal{} \sum_{k \equal{} 1}^{n}{(\lambda_k|y|^2 \plus{} \frac {1}{\lambda_k}|\overline{z_k}|^2)} \geq 2 \sum_{k \equal{} 1}^{n}{|y_k\parallel{}\overline{z_k}|} \geq 2|\sum_{k \equal{} 1}^{n}{y_k \overline{z_k}}| \equal{} 2|x^{*}Cx| \geq 2x^{*}Cx$ [b](1)[/b]
[b] By[/b] $ U^{*} \equal{} U^{\minus{}1},y \equal{} U^{*}Cx ,z \equal{} U^{*}x$ ,then $ z^{*} \equal{} x^{*}U$, and $ z^{*}y \equal{} x^{*}Cx \equal{} \sum_{k \equal{} 1}^{n}{y_k\overline{z_k}} \in R$ Then implies $ |x^{*}Cx| \geq x^{*}Cx$
[b](1)[/b] Show that $ CPC \plus{} P^{ \minus{} 1}\geq 2C$[/hide] :welcomeani:


a set of prime numbers have the following properties:
median is 7, mode is 11 and the mean is 9.
find the largest possible number in this set

uhmm is this in the right forum? :maybe:


ratio
trigonometry
arithmetic sequence
geometry
trig identities
Law of Sines

If the sides of a triangle are in Arithmetic Progression and the greatest angle exceeds the smallest by α, show that the sides are in the ratio 1 - x : 1 : 1 + x , where x = √(1 - cos α)(7 - cos α)?

How to prove this? I am totally stumped by this question..

[hide="The relevant lemma"] The [url=http://mathworld.wolfram.com/LawofSines.html]extended Law of Sines[/url] implies that the sides of any triangle are in the ratio $ \sin A : \sin B : \sin C$.    [/hide]

Thanks for replying. But can you please explain a little more in detail about how does the extended sine rule imply that the sides are in the ratio sin A : sin B : sin C?

$ a\equal{}2R\sin \alpha$
$ b\equal{}2R\sin \beta$
$ c\equal{}2R\sin \gamma$

so they are in the ratio $ \sin \alpha : \sin \beta : \sin \gamma$


Figure $ ABCD$ is a rhombus with diagonals of length 20 and 24. Find the area of rhombus $ ABCD$.
[asy]pair A,B,C,D;
unitsize(1inch);
A = (.1,.9);
B = (1,1);
C = (.9,.1);
D = (0,0);
draw(A--B--C--D--A);
label("$A$",A,NW);
label("$B$",B,NE);
label("$C$",C,SE);
label("$D$",D,SW);[/asy]

T area of a rhombus is the product of the diagonals divided by two. Thus, the area is 20 * 24/2 = 20 *12 = 240


Jane bought six items from a basket of 25-cent and 50-cent items.  Kyle bought six items from a basket of 50-cent and 75-cent items.  What is the difference between the lowest possible cost to jance and the highest possible cost to Kyle?

[hide]the lowest price for Jane would be $.25 \times 6$, which is $1.50$.  the highest price for Kyle would be $.75 \times 6$, which is $4.50$.  Thus, the difference between them is $4.50 - 1.50 = 3$.[/hide]

[size=75][color=darkred][Hide your answers!][/color][/size]

[hide="Click to see my answer"]
[b]Jane's lowest amount possible is [/b]$\$1.50 (\$0.25\times6=\$1.50)$
[b]Kyles highest amount possible is[/b] $\$4.50 (\$0.75\times6=\$4.50)$
$\$4.50-\$1.50=\$3.00$

$\$3.00$ [b]is the answer[/b][/hide]

[quote="apusey"]Can there be 6 .25 items or are the items .25 .26 .27 .28 .29 .30?[/quote]
the items are 6, 25 cent stamps so only 25 cents.