geometry
geometric transformation
homothety
circumcircle
reflection
trapezoid
parallelogram

write proof of simson line with homothety :!:  :!:

$ O$ is the circumcenter, $ H$ is the orthocenter of a $ \triangle ABC$. $ (O_a), (O_b), (O_c)$ are reflections of the circumcircle $ (O)$ in $ BC, CA, AB$, respectively, intersecting at the orthocenter $ H.$ $ P \in (O)$ is any point on the circumcircle.  $ D, E, F$ are the feet of perpendiculars from $ P$ to $ BC, CA, AB.$ The perpendiculars $ PD, PE, PF$ cut the circumcircle $ (O)$ again at $ A', B', C'.$ $ \angle B'PA' \equal{} \angle BCA \Longrightarrow A'B' \equal{} AB,$ $ ABB'A'$ is isosceles trapezoid, $ AA' \parallel BB'.$ Similarly, $ BB' \parallel CC'.$

The perpendicular $ PA'$ form $ P$ to $ BC$ cuts $ (O_a)$ at $ P_a$ on the opposite side of $ BC$ than $ A,$ $ P_a$ is a reflection of $ P$ in $ BC,$ $ D$ is the midpoint of $ PP_a,$ $ AA'P_aH$ is a parallelogram, $ P_aH \parallel AA'.$
The perpendicular $ PB'$ form $ P$ to $ CA$ cuts $ (O_b)$ at $ P_b$ on the opposite side of $ CA$ than $ B,$ $ P_b$ is a reflection of $ P$ in $ CA,$ $ E$ is the midpoint of $ PP_b,$ $ BB'P_bH$ is a parallelogram, $ P_bH \parallel BB'.$
Combining, $ P_aH \parallel AA' \parallel BB' \parallel P_bH \Longrightarrow P_a, P_b, H$ are collinear. Similarly, $ P_b, P_c, H$ are collinear. Combining, $ P_aP_bP_c$ is a line (through the orthocenter $ H$).

Homothety with center $ P$ and coefficient $ 1/2$ takes the line $ P_aP_bP_c$ into the Simson line $ DEF$ with the pole $ P.$


calculus
integration
trigonometry
real analysis
real analysis unsolved

Find: $ \int xarctg\sqrt{x} \,dx$

P.S. Do not use integration software :-)

$ \int x\arctan\sqrt {x}\ \mathrm {dx}\equal{}F\left(\arctan\sqrt x\right)\ \mathrm {dx}\plus{}\mathcal C$, where

$ F\equal{}\int 2t\cdot\tan^3t\cdot\left(1\plus{}\tan^2t\right)\ \mathrm {dt}\equal{}\frac 12\cdot\int t\cdot\left(\tan^4t\right)'\ \mathrm {dt}$ a.s.o.

$ I_{n}\equiv\int \tan^{n}t\ \mathrm {dt}$ $ \implies$ $ I_{n\plus{}2}\equal{}\int\tan^nt\left[\left(\tan t \right)'\minus{}1\right]\ \mathrm {dt}\equal{}\frac {1}{n\plus{}1}\cdot\tan^{n\plus{}1}t\minus{}I_n\ .$

You may substiture $ u \equal{} \sqrt{x}$ and to integrate by parts.


linear algebra
matrix
linear algebra unsolved

We say that $B \in \mathcal M_n \left( \mathbb C \right)$ is the pseudo-inverse of $A \in \mathcal M_n \left( \mathbb C \right)$ if $A=ABA$ and $B=BAB$.

(a) Prove that all matrices have pseudo-inverses.

(b) Find all matrices which have unique pseudo-inverses.

If $A$ is inversable then $B=A^{-1}$.
If not, let $A=R_1...R_k A' C_1...C_l$, where $A'$ is the hermitian form of $A$ and $R_1,..,R_k$, $C_1,...,C_l$ are elementary matrices corresponding to row, respectively column operations.
Then $B=C_l^{-1}...C_1^{-1} A' R_k^{-1}...R_1^{-1}$ is a pseudo-inverse of $A$.
The matrices with unique pseudo-inverses are the invertible matrices: $\{A\in M_n(\mathbb{C})|\det A\neq 0\}$
I think that's it. ;)

How can you prove that every matix in C has a related Hermitian matrix!

I am not sure that Hermitian is the right name for $A'$. I meant by $A'$ the diagonal matrix containing only the unitary pivots of the reduced $A$ and I believe one can get that by using only elementary row and column operations even if $A$ is from $M_n(\mathbb{C})$.

I think that is called the reduced echelon form of a matrix.

That's right. My bad, sorry. :oops:
Anyway i've only sketched the solution.

$A \in \mathcal M_2 \left( \mathbb C \right)$
and $A = \left( \begin{array}{ccc} i & i\\ i & i\\ \end{array} \right)$

How would u reduce it to the $A'$ u are talking about?

$A=\left(\begin{array}{cc}1&0\\1&1\end{array}\right)\left(\begin{array}{cc}i&0\\0&1\end{array}\right)\left(\begin{array}{cc}1&0\\0&0\end{array}\right)\left(\begin{array}{cc}1&1\\0&1\end{array}\right)$, where $A'=\left(\begin{array}{cc}1&0\\0&0\end{array}\right)$, and in general when we get to a pivot we first normalize it to $1$ before working the next pivot.

[quote="gnosis"]$\left(\begin{array}{cc}i&0\\0&1\end{array}\right)$.[/quote]

Are you sure that is an elementary matrix ?

Yes. A matrix $\left(\begin{array}{cccccccc}1&0&.&.&.&.&.&0\\0&1&0&.&.&.&.&0\\.&0&.&.&.&.&.&.\\.&.&.&\lambda&.&.&.&.\\.&.&.&.&.&.&0&.\\0&.&.&.&.&0&1&0\\0&.&.&.&.&.&0&1\\\end{array}\right)$(on the diagonal you have only ones and a $\lambda\in\mathbb{C}^*$ and 0 in rest) is an elementary matrix which

interchanges the $i^{th}$ row $r$ of a matrix with $\lambda.r$ if it corresponds to a row operation and $\lambda$ is on the $i^{th}$ column


LaTeX
inequalities proposed
inequalities

Find the greatest real number $K$ such that for all positive real number $u,v,w$ with $u^{2}>4vw$ we have $(u^{2}-4vw)^{2}>K(2v^{2}-uw)(2w^{2}-uv)$

This problem posted again at http://www.mathlinks.ro/Forum/viewtopic.php?t=2488

I rewrite the solution in $\text{\LaTeX}$

[quote="Lagrangia"]this is siuhohung's solution 4 the 1st problem:

Divide both sides by $u^{4}$
then the inequality becomes
$(1-4\frac{v}{u}\frac{w}{u})^{2}> K(2(\frac{v}{u})^{2}-\frac{w}{u})(2(\frac{w}{u})^{2}-\frac{w}{u}))$

Sub. $a=\frac{v}{u}, b=\frac{w}{u}$

$(1-4ab)^{2}>K(2a^{2}-b)(2b^{2}-a)$
The constaint given is transformed to $ab<\frac{1}{4}$
Then
when $a =\frac{1}{2}\Rightarrow  b<\frac{1}{2}$ and when $b=\frac{1}{2}\Rightarrow  a<\frac{1}{2}$
Therefore
$(a-b)^{2}+(a-\frac{1}{2})^{2}+(b-\frac{1}{2})^{2}>0$
Expand and simplify, we have
$(2a^{2}-b)+(2b^{2}-a)>2ab-\frac{1}{2}$
Let
$x=2a^{2}-b$
$y=2b^{2}-a$
$z=1-4ab$
So we are having,
$x+y>-\frac{z}{2}$ and $z^{2}>Kxy$
$2(x+y)>-z$
$4(x+y)^{2}>z^{2}>Kxy$
$4x^{2}+(8-K)xy+4y^{2}>0$

Obviously, $K=16$[/quote]

[quote="sqing"]WLOG $uw-2v^2>0,uv-2w^2>0$[/quote]
What happens if $uw-2v^2<0$ and $uv-2w^2<0$? :wink:


given the orthocentre H,centroid ,G and a vertex,A how can we construct triangle ABC?

1.O(circuncenter) G and H are aligned and OG=GH/2.So O is determined
2.Circle center O radius OA cut AH on P
3.Half of HP is the foot of height and triangle is determined


Let $ a,b,c\ge 0$. Consider: $ \sum_{cyc}a^{3}\minus{}abc\ge e*\sum_{cyc}ab^{2}\plus{}2a^{2}b\minus{}3abc$.

i) Prove the inequality for $ e\equal{}\frac{1}{3}$.
ii) Does there exist a better constant? [I believe so, but I have not been able to find it...]

[quote="Altheman"]Let $ a,b,c\ge 0$. Consider: $ \sum_{cyc}a^{3}\minus{}abc\ge e*\sum_{cyc}ab^{2}\plus{}2a^{2}b\minus{}3abc$.

i) Prove the inequality for $ e \equal{}\frac{1}{3}$.
ii) Does there exist a better constant? [/quote]
I think, $ e_{max}\equal{}\min_{t>0}\frac{t^{3}\plus{}1}{2t^{2}\plus{}t}\equal{}0.662107916...$


You realize there is a real ftw forum, right?

Who did you get your generation 22 from?

Go to the forum index and scroll down till you get to FTW forum.


Let $a, b$ be two positive co-prime integers such that $a+b$ is odd. Let $S$ be a set of positive integers satisfying the following two conditions.
1) $a, b$ belong to $S$
2) If $x, y, z$ belong to $S$ then $x+y+z$ also belongs to $S$
Show that every integer $n \geq 2ab$ belongs to $S$.

I hope its not a repeat. :maybe:

From (i) and (ii) we have $ka+nb\in S$ if $k+n$ odd and $k,n\in N$.
a or b is odd. Let $c=2a,d=b$ if b is odd, else $c=a,d=2a$. Then $(c,d)=1$ and $kc+nd\in S$ for all $k,n\in N$. Therefore for all $m\ge cd$ exist $k,n\in N$, suth that $m=kc+nd\in S$.

@euvtl : Hm.. Do you mean that $x,y,z$ are not necesarrily distinct ?

If so, let $T = pa+qb$ and $P_{T}= p+q$
Since intially, $P_{a}= 1$ and $P_{b}= 1$ is odd, then if $X \in S$, we have $P_{X}$ as the sum of three odd number which must be odd. And we came to the same conclusion with Rust, where $k+n$ is odd..

@Rust : You wrote a typo, it must be $d = 2b$. But I'm still confused even if $kc+nd \in S$, are all elements of $S$ not less than $cd$ can be stated in the form of $kc+nd$ ? Can you give me the proof ? Thx..

but if $x,y,z\in S$ then we have $x+y+z\in S$ and since $x,y,z$ are positive, we cannot say that if $a,b\in S$ then $a+b\in S$. this depends on the problem statement.

@RDeepMath91: Yes, indeed it is not necessary that $x, y, z$ should be distinct. Otherwise we cannot even begin the problem because we know only two elements $a, b$ initially :)

@Rust: Could you please explain your solution a little more? Why is it true that for $(c,d)=1$, $kc+nd\in S$ for all $k,n\in N$? I think you probably meant to add here that $k+n$ is still odd.
Also, I am unable to see how this implies that for all $m\ge cd$ there exist $k,n\in N$, such that $m=kc+nd\in S$


linear algebra
calculus
calculus computations

$ g: \mathbb{C}\rightarrow \mathbb{C}$

$ \forall z,z' \in \mathbb{C}$  $ g(z\plus{}z') \equal{} g(z) \plus{} g(z')$
$ \forall \alpha \in \mathbb{R}$ $ \forall z \in \mathbb{C}$ $ g(\alpha z)\equal{}\alpha g(z)$

Prove that there exists $ a,b \in \mathbb{C}$ / $ g(z)\equal{}az\plus{}b \bar{z}$

You can express $ g(z)$ in terms of $ g(1)$ and $ g(i)$.

This is essentially a linear algebra problem. The stated conditions are that considered as a mapping from $ \mathbb{R}^2$ to $ \mathbb{R}^2,$ $ g$ is linear.

mlok wants you to look at $ g(1)$ and $ g(i).$ This is the representation in terms of the standard basis for $ \mathbb{R}^2.$ From here on, all we're really trying to do is to express the same mapping in an alternate notation.

I still can't see a proof or even a hint :| .

Let $ g(1)\equal{}\alpha$ and $ g(i)\equal{}\beta.$ Both $ \alpha$ and $ \beta$ are complex.

Now suppose $ x,y\in\mathbb{R}.$

$ g(x\plus{}yi)\equal{}g(x)\plus{}g(yi)\equal{}xg(1)\plus{}yg(i)\equal{}x\alpha\plus{}y\beta.$

We now need to write $ x\alpha\plus{}y\beta\equal{}a(x\plus{}iy)\plus{}b(x\minus{}iy).$

Equating coefficients of $ x$ and $ y$ we get

$ a\plus{}b\equal{}\alpha$
$ ia\minus{}ib\equal{}\beta$

This is two linear equations in two unknowns. The solution is $ a\equal{}\frac{\alpha\minus{}i\beta}{2},$ $ b\equal{}\frac{\alpha\plus{}i\beta}{2}.$

So $ g(z)\equal{}\left(\frac{g(1)\minus{}ig(i)}{2}\right)z\plus{} \left(\frac{g(1)\plus{}ig(i)}{2}\right)\overline{z}.$

As I said, it's essentially a linear algebra problem. For that matter, the use of $ 1$ and $ i$ was merely a convenience - we could have used any two points as long as they were linearly independent over the reals.


Let $ [A, B, C]$ denote the largest of $ A$, $ B$, and $ C$. If $ A$, $ B$, and $ C$ are positive integers and $ A \times B \times C \equal{} 7!$, what is the smallest value $ [A, B, C]$ can have?

Let $ A<B<C$.  To minimize $ C$, we need to make $ C$ as close as possible to the cube root of $ 7!$, which is approximately $ 17.\text{something}$.  Now we just check values for $ C$ and go up, starting with 18
$ C\equal{}18$, then $ AB\equal{}280$, but if $ A<B<18$, the maximum possible product of $ A$ and $ B$ is $ 16\cdot 17\equal{}272$
$ C\equal{}19$, then $ 7!$ is divisible by 19, contradiction
$ C\equal{}20$, then $ AB\equal{}252$, giving us $ A\equal{}14$ and $ B\equal{}18$ as a possible solution, so $ \boxed{20}$

Your solution is correct, but as it doesn't say they are distinct, you can't assume that when C=18 the maximum B can be is 17.

well, this can be corrected by noting that if $ C\equal{}B\equal{}18$, then the product $ ABC$ has at least 4 factors of 3, while $ 7!$ only has 2.  probably should have specified that earlier...

[quote="5849206328x"]Let $ A < B < C$.  [/quote]

That's what I refer to.


Does this hold?
$f$ is defined on $[a,b]$ so that $|f(x_1)-f(x_2)|\leq|x_1-x_2|$ and $f(a)=b,f(b)=a$. Show that $f(x)=-x+a+b$ for $x\in [a,b]$.

sorry for this one. nothing to do with calculus.
by the way it holds.


combinatorics proposed
combinatorics
unsolved

Let $ G$ be a simple graph on n vertices such that $ G$ contains no $ k$-complete subgraph but do contain some $ r$-complete subgraph, given $ r < k < 2r$. Prove that: there exists a vertex $ A$ from $ G$ fulfilling the property:
    $ d(A) \leq [n\vdots\frac {2r - k}{r}]$

This question has been not confirmed, yet!  :o . Any ideas will be highly appreciated !

This problem reminds me of Zarankiewicz's theorem...


Post a rule. The next person posts another rule that follows the first rule. The third person posts another rule that follows both rules, and so on.

Scoring: If you post a rule 1, you get 1 pt. If you post a rule 3, you get 3 pts, and so on.

Global rules:

1. You must not make a rule that alters the rule number.
2. You must not make the rules difficult to read.
3. You must not alter the number of points that you get in a rule.
4. Breaking a rule on purpose gives you a loss of 5 pts.

Breaking any of the global rules 1-3 gives a loss of 10 pts.

Let's begin!

Rule 1: All rules must contain the number 1337.

Scores:
lotsofmath: 1

Does rule 1 have to follow rule 2, 3, 4, etc??
(as in when you make a new rule, does it have to also be applicable to all the previous rules)?

I'm not going to keep track of score in my posts. If you feel they're needed, you can keep track when it's your post.

Rule 1: All rules must contain the number 1337.
Rule 2: If any of the scores reach 1337, the rule must mention each person in the scores.

Rule 1 does not have to follow rule 2.

Rule 1: All rules must contain the number 1337.
Rule 2: If any of the scores reach 1337, the rule must mention each person in the scores.
Rule 3: All rules up to rule 1337 must contain the letter "x".

Rule 1: All rules must contain the number 1337.
Rule 2: If any of the scores reach 1337, the rule must mention each person in the scores.
Rule 3: All rules up to rule 1337 must contain the letter "x".
Rule 4: I've said this 1337 times, all rules must end with an exclamation mark!

Scores:
cf249: 4
lotsofmath: 4
ProtestanT: 2

Rule 1: All rules must contain the number 1337.
Rule 2: If any of the scores reach 1337, the rule must mention each person in the scores.
Rule 3: All rules up to rule 1337 must contain the letter "x".
Rule 4: I've said this 1337 times, all rules must end with an exclamation mark!
Rule 5: I really hate 1337. So, you must hate something (or someone) in your rules!

pacman2812: 5
cf249: 4
lotsofmath: 4
ProtestanT: 2

Rule 5 breaks Rule 3.

Rule 1: All rules must have a spelling error.

Rule 1: All rules must have a spelling error.
Rule 2: You must speek like a Canadian (aka add eh after every sentence), eh?
Rule 3: Eh, now we must euse a number (like 9, eh?) after each rule, eh?

pacman2812: 5
cf249: 4
lotsofmath: 4
ProtestanT: 3
james4l: 3
kimmystar94: 2

We should retract the points won for those who accidentally break rules.

Proposal to Global rules:
5. Your rule must have a restriction. (Kinda obvious, but by rule ~20, this is necessary)

Rule 1 broke itself. Also, if you break a rule, you don't keep your points. Nobody gets points due to Rule 1 breaking itself.
Global Rule 5 accepted.

RULE 1: RULES MUST BE WRITTEN IN ALL CAPS.

Scores:
lotsofmath: 5
cf249: 4
ProtestanT: 2

RULE 1: RULES MUST BE WRITTEN IN ALL CAPS.
RULE 2: ZOMGGG!!!!!!! FROM NOW ON THERE MUST BE AT LEAST TEN EXCLAMATION POINTS!!!!!!!!!!

Scores:
lotsofmath: 5
cf249: 4
ProtestanT: 2
james4l: 2
[/quote]

RULE 1: RULES MUST BE WRITTEN IN ALL CAPS.
RULE 2: ZOMGGG!!!!!!! FROM NOW ON THERE MUST BE AT LEAST TEN EXCLAMATION POINTS!!!!!!!!!!
RULE 3: WHAT?!?!?!?!?!?!?!?!?!?! HEY! DON'T IGNORE QUESTION MARKS AND PUT AT LEAST 10 OF THEM IN EACH RULE! OK?????
Scores:
lotsofmath: 5
cf249: 4
pacman2812: 3
ProtestanT: 2
james4l: 2

RULE 1: RULES MUST BE WRITTEN IN ALL CAPS.
RULE 2: ZOMGGG!!!!!!! FROM NOW ON THERE MUST BE AT LEAST TEN EXCLAMATION POINTS!!!!!!!!!!
RULE 3: WHAT?!?!?!?!?!?!?!?!?!?! HEY! DON'T IGNORE QUESTION MARKS AND PUT AT LEAST 10 OF THEM IN EACH RULE! OK?????
RULE 4: GAHHHH OMGOMGOMGOMGOMGOMG!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!?!?!?!?!?!?!?!!!!! ARE YOU SERIOUS? YOU HAVE TO PUT IN OMG IN EVERY RULE AFTER THIS ONE?????????? OMG!!!!!!!!!!!!

Scores:
lotsofmath: 5
cf249: 4
pacman2812: 3
kimmystar94: 3
ProtestanT: 2
james4l: 2

RULE 1: RULES MUST BE WRITTEN IN ALL CAPS.
RULE 2: ZOMGGG!!!!!!! FROM NOW ON THERE MUST BE AT LEAST TEN EXCLAMATION POINTS!!!!!!!!!!
RULE 3: WHAT?!?!?!?!?!?!?!?!?!?! HEY! DON'T IGNORE QUESTION MARKS AND PUT AT LEAST 10 OF THEM IN EACH RULE! OK?????
RULE 4: GAHHHH OMGOMGOMGOMGOMGOMG!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!?!?!?!?!?!?!?!!!!! ARE YOU SERIOUS? YOU HAVE TO PUT IN OMG IN EVERY RULE AFTER THIS ONE?????????? OMG!!!!!!!!!!!!
RULE 5: I CALL THIS SEGREGATION OF PARENTHESES! FROM NOW ON YOU MUST USE 10 LEFT PARENTHESES ("(") AND TEN RIGHT PARENTHESES!!!! (")") OMG, I'M TELLING YOU!!!!!!!!!!!!!!!!!!!!!!!! WHAT????????????????????????????????? YOU DON'T THINK THAT'S CONSTITUTIONAL ((((WELL OF COURSE)))) THEN LOOK AT AMENDMENT ONE (CONSTITUTION) AND CHECK (FREEDOM OF SPEECH)! YOU KNOW WHAT! (!!) THIS IS A LONG RULE! OMG! I'M FAINTING!!!
Scores:
james4l: 7
lotsofmath: 5
cf249: 4
kimmystar94: 4
pacman2812: 3
ProtestanT: 2

RULE 1: RULES MUST BE WRITTEN IN ALL CAPS.
RULE 2: ZOMGGG!!!!!!! FROM NOW ON THERE MUST BE AT LEAST TEN EXCLAMATION POINTS!!!!!!!!!!
RULE 3: WHAT?!?!?!?!?!?!?!?!?!?! HEY! DON'T IGNORE QUESTION MARKS AND PUT AT LEAST 10 OF THEM IN EACH RULE! OK?????
RULE 4: GAHHHH OMGOMGOMGOMGOMGOMG!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!?!?!?!?!?!?!?!!!!! ARE YOU SERIOUS? YOU HAVE TO PUT IN OMG IN EVERY RULE AFTER THIS ONE?????????? OMG!!!!!!!!!!!!
RULE 5: I CALL THIS SEGREGATION OF PARENTHESES! FROM NOW ON YOU MUST USE 10 LEFT PARENTHESES ("(") AND TEN RIGHT PARENTHESES!!!! (")") OMG, I'M TELLING YOU!!!!!!!!!!!!!!!!!!!!!!!! WHAT????????????????????????????????? YOU DON'T THINK THAT'S CONSTITUTIONAL ((((WELL OF COURSE)))) THEN LOOK AT AMENDMENT ONE (CONSTITUTION) AND CHECK (FREEDOM OF SPEECH)! YOU KNOW WHAT! (!!) THIS IS A LONG RULE! OMG! I'M FAINTING!!!
RULE 6: ((((((((((OMG! YOU MUST USE AN EMOTICON IN EACH RULE?!?!?!?!?!?!?!?!?!?!?!?!?!?!?!?! OMG!!!!))))))))))
Scores:
james4l: 7
lotsofmath: 5
cf249: 4
kimmystar94: 4
pacman2812: 9
ProtestanT: 2

(((((((((((((((OMG!!!!!!!!!!!!!!!!!! RULE 6 BROKE ITSELF!!!!!!!!!!!!????????????????))))))))))))))))))

Sorry.  :ninja:

Rule 1:  No rule after this one may contain the letter "h".

Scores:
james4l: 7
lotsofmath: 5
cf249: 5
kimmystar94: 4
pacman2812: 3
ProtestanT: 2

Rule 1: No rule after this one may contain the letter "h".
rule 2: no caps, period.

Scores:
james4l: 9
lotsofmath: 5
cf249: 5
kimmystar94: 4
pacman2812: 3
ProtestanT: 2

[quote="lotsofmath"]Rule 1 broke itself. Also, if you break a rule, you don't keep your points.[/quote]
Isn't it supposed to be only the rules after the given rule have to follow it?

Rule 1: No rule after this one may contain the letter "h".
rule 2: no caps, period.
rule 3: no rule may contain a misspelling.

Rule 1: No rule after this one may contain the letter "h".
rule 2: no caps, period.
rule 3: no rule may contain a misspelling.
rule 4: all rules must contain a number, like 12.

[quote="ProtestanT"]Isn't it supposed to be only the rules after the given rule have to follow it? [/quote]

No, the rule must follow itself, unless specified in the rule (like the current Rule #1).

Scores:
james4l: 9
cf249: 9
ProtestanT: 5
lotsofmath: 5
kimmystar94: 4
pacman2812: 3

Rule 1: No rule after this one may contain the letter "h".
rule 2: no caps, period.
rule 3: no rule may contain a misspelling.
rule 4: all rules must contain a number, like 12.
rule 5: 1 new rule, a comma must be in the rule

Scores:
james4l: 14
cf249: 9
ProtestanT: 5
lotsofmath: 5
kimmystar94: 4
pacman2812: 3

Rule 5 breaks Rule 1.  (I knew somebody would mess up by using "the"!)  :P

Rule 1:  All rules must contain exactly seven words.

Scores:
cf249: 10
james4l: 9
ProtestanT: 5
lotsofmath: 5
kimmystar94: 4
pacman2812: 3

Rule 1:  All rules must contain exactly seven words.
Rule 2: Oh no, rules must now rhyme - whoa...

Scores:
james4l: 11
cf249: 10
ProtestanT: 5
lotsofmath: 5
kimmystar94: 4
pacman2812: 3[/quote]

Rule 1: All rules must contain exactly seven words.
Rule 2: Oh no, rules must now rhyme - whoa...
Rule 3: Peenut shelling, rooles must contain a misspelling

Scores:
james4l: 11
cf249: 10
kimmystar94: 8
ProtestanT: 5
lotsofmath: 5
pacman2812: 3

Rule 1: All rules must contain exactly seven words.
Rule 2: Oh no, rules must now rhyme - whoa...
Rule 3: Peenut shelling, rooles must contain a misspelling
Rule 4: Quak quack quack quack quack quack "quack".

Scores:
james4l: 11
cf249: 10
kimmystar94: 8
ProtestanT: 5
lotsofmath: 5
1=2: 4
pacman2812: 3

Rule 4 breaks global rule 5, 1=2 loses 10 pts.

Rule 1: All Q's in a rqle mqst be replaced with the letter Q.

james4l: 11
cf249: 10
kimmystar94: 7
lotsofmath: 6
ProtestanT: 5
pacman2812: 3
1=2: -10

P.S. Kimmystar94, you gave yourself an extra point.

Rule 1: All Q's in a rqle mqst be replaced with the letter Q.
Rule 2: All rqles mqst contain a nqmber that is a perfect sqqare, like 4.

cf249: 12
james4l: 11
kimmystar94: 7
lotsofmath: 6
ProtestanT: 5
pacman2812: 3
1=2: -10

Rule 1: All Q's in a rqle mqst be replaced with the letter Q.
Rule 2: All rqles mqst contain a nqmber that is a perfect sqqare, like 4.
Rule 3: NOW LISTEN QP, FOLKS - FROM RQLE 4 AND HIGHER, ALL CAPS MQST BE QSED!

james4l: 14
cf249: 12
kimmystar94: 7
lotsofmath: 6
ProtestanT: 5
pacman2812: 3
1=2: -10

Rule 1: All Q's in a rqle mqst be replaced with the letter Q.
Rule 2: All rqles mqst contain a nqmber that is a perfect sqqare, like 4.
Rule 3: NOW LISTEN QP, FOLKS - FROM RQLE 4 AND HIGHER, ALL CAPS MQST BE QSED!
Rule 4: RQLES MQST NOT CONTAIN MORE THAN 9 WORDS.

james4l: 14
cf249: 12
lotsofmath: 10
kimmystar94: 7
ProtestanT: 5
pacman2812: 3
1=2: -10

Rule 1: All Q's in a rqle mqst be replaced with the letter Q.
Rule 2: All rqles mqst contain a nqmber that is a perfect sqqare, like 4.
Rule 3: NOW LISTEN QP, FOLKS - FROM RQLE 4 AND HIGHER, ALL CAPS MQST BE QSED!
Rule 4: RQLES MQST NOT CONTAIN MORE THAN 9 WORDS.
Rule 5: ALL RQLES MQST CONTAIN ONLY 1 PQNCTQATION MARK YAY.

james4l: 14
cf249: 12
lotsofmath: 10
kimmystar94: 7
ProtestanT: 5
pacman2812: 3
erabel: 1
1=2: -10

Rule 1: All Q's in a rqle mqst be replaced with the letter Q.
Rule 2: All rqles mqst contain a nqmber that is a perfect sqqare, like 4.
Rule 3: NOW LISTEN QP, FOLKS - FROM RQLE 4 AND HIGHER, ALL CAPS MQST BE QSED!
Rule 4: RQLES MQST NOT CONTAIN MORE THAN 9 WORDS.
Rule 5: ALL RQLES MQST CONTAIN ONLY 1 PQNCTQATION MARK YAY.
Rule 6: THE NEXT 16 RQLES MQST CONTAIN THE WORD FOLLOW.

cf249: 18
james4l: 14
lotsofmath: 10
kimmystar94: 7
ProtestanT: 5
erabel: 5
pacman2812: 3
1=2: -10

Rule 1: All Q's in a rqle mqst be replaced with the letter Q.
Rule 2: All rqles mqst contain a nqmber that is a perfect sqqare, like 4.
Rule 3: NOW LISTEN QP, FOLKS - FROM RQLE 4 AND HIGHER, ALL CAPS MQST BE QSED!
Rule 4: RQLES MQST NOT CONTAIN MORE THAN 9 WORDS.
Rule 5: ALL RQLES MQST CONTAIN ONLY 1 PQNCTQATION MARK YAY.
Rule 6: THE NEXT 16 RQLES MQST CONTAIN THE WORD FOLLOW.
Rule 7: THE NQMBERING FROM RQLE 1 MQST FOLLOW THE ORDER.

Broke the new global rule "post restrictions must apply"

start over

rule 1: i hate caps, so don't use them.

cf249: 18
james4l: 14
lotsofmath: 10
kimmystar94: 7
ProtestanT: 5
erabel: 5
pacman2812: 4
1=2: -10

rule 1: i hate caps, so don't use them.
rule 2: rules after this rule must have a misspelling.
rule 3: sdrow tsum eb nettir sdrawkcab, tub nehw detalsnart kcab ot lamron, elur 2 tsum ylppa.

[quote="lotsofmath"]Rule 4 breaks global rule 5[/quote]

THAT IS FALSE! There is no 5.

Rule 3 breaks global rule 2 by making the rules difficult to read.

Rule 1: Rules must not use a word that contains more than nine letters.

cf249: 18
james4l: 14
lotsofmath: 11
kimmystar94: 7
erabel: 5
pacman2812: 4
ProtestanT: -3
1=2: -10
dragon96: -10

Rule 1: Rules must not use a word that contains more than nine letters.
Rule 2: All rules must begin with the word "all".

Global rules:
1. You must not make a rule that alters the rule number.
2. You must not make the rules difficult to read.
3. You must not alter the number of points that you get in a rule.
4. Breaking a rule on purpose gives you a loss of 5 pts.
5. Your rule must have a restriction.

Breaking any of the global rules gives a loss of 10 pts.

Scores:
cf249: 20
james4l: 14
lotsofmath: 11
kimmystar94: 7
erabel: 5
pacman2812: 4
ProtestanT: -3
1=2: -10
dragon96: -10

Rule 1: Rules must not use a word that contains more than nine letters.
Rule 2: All rules must begin with the word "all".
Rule 3: All rules after this one must contain 1 (or more) number(s).

Scores:
cf249: 20
james4l: 14
lotsofmath: 11
kimmystar94: 7
erabel: 6
pacman2812: 4
ProtestanT: -3
1=2: -10
dragon96: -10

[quote="cf249"]Yeah, there is, see post #8.[/quote]

Translating, it reads "Every word must be replaced with 'quack'". So it breaks rule 2, not rule 5...

Rule 1: Rules must not use a word that contains more than nine letters.
Rule 2: All rules must begin with the word "all".
Rule 3: All rules after this one must contain 1 (or more) number(s).
Rule 4: All rules cannot have more than 2 instances of the word "the".

Scores:
cf249: 20
james4l: 18
lotsofmath: 11
kimmystar94: 7
erabel: 6
pacman2812: 4
ProtestanT: -3
1=2: -10
dragon96: -10

Rule 1: Rules must not use a word that contains more than nine letters.
Rule 2: All rules must begin with the word "all".
Rule 3: All rules after this one must contain 1 (or more) number(s).
Rule 4: All rules cannot have more than 2 instances of the word "the".
Rule 5: All rules must contain an equation such as 1+1=2

Scores:
cf249: 20
james4l: 18
lotsofmath: 11
kimmystar94: 7
erabel: 6
BOGTRO: 5
pacman2812: 4
ProtestanT: -3
1=2: -10
dragon96: -10

Rule 1: Rules must not use a word that contains more than nine letters.
Rule 2: All rules must begin with the word "all".
Rule 3: All rules after this one must contain 1 (or more) number(s).
Rule 4: All rules cannot have more than 2 instances of the word "the".
Rule 5: All rules must contain an equation such as 1+1=2
Rule 6: All rules must include at least 2+2=[b]4[/b] "z"s: choice words are dazzle or buzz.

Scores:
james4l: 24
cf249: 20
lotsofmath: 11
kimmystar94: 7
erabel: 6
BOGTRO: 5
pacman2812: 4
ProtestanT: -3
1=2: -10
dragon96: -10

Rule 1: Rules must not use a word that contains more than nine letters.
Rule 2: All rules must begin with the word "all".
Rule 3: All rules after this one must contain 1 (or more) number(s).
Rule 4: All rules cannot have more than 2 instances of the word "the".
Rule 5: All rules must contain an equation such as 1+1=2
Rule 6: All rules must include at least 2+2=4 "z"s: choice words are dazzle or buzz.
Rule 7: All rules must mention exactly 8-7=[b]1[/b] place where food is served, like Little Caesar's.  (Pizza, pizza!)

Scores:
cf249: 27
james4l: 24
lotsofmath: 11
kimmystar94: 7
erabel: 6
BOGTRO: 5
pacman2812: 4
ProtestanT: -3
1=2: -10
dragon96: -10

Rule 1: Rules must not use a word that contains more than nine letters.
Rule 2: All rules must begin with the word "all".
Rule 3: All rules after this one must contain 1 (or more) number(s).
Rule 4: All rules cannot have more than 2 instances of the word "the".
Rule 5: All rules must contain an equation such as 1+1=2
Rule 6: All rules must include at least 2+2=4 "z"s: choice words are dazzle or buzz.
Rule 7: All rules must mention exactly 8-7=[b]1[/b] place where food is served, like Little Caesar's.  (Pizza, pizza!)
Rule 8: All rules must exhibit some form of ADD... (Wait, in Burger King, I saw this really cool problem: if f(z)=z^2, then f'(z)=2z. Wow!)... wait, what was I saying? Oh yes, all rules must show that the poster is imitating what a person inflicted with ADD would post.

Scores:
james4l: 32
cf249: 27
lotsofmath: 11
kimmystar94: 7
erabel: 6
BOGTRO: 5
pacman2812: 4
ProtestanT: -3
1=2: -10
dragon96: -10[/quote]


AMC USA(J)MO USAMO Princeton college Harvard Yale

[quote="bubala"]I got my 3rd quarter report card today, and I had my first non-"A"s ever---in fact I got 2 Bs (not B+s), one in English and one in Spanish.  The one in English was quite a shock as I've gotten As on all the essays/tests/whatnot and my teacher just UBERLY hates me randomly.  I expected a not-so-great grade in Spanish, but I thought I'd end up with higher than a B.

So this brings me to my question: what kind of grades do you need to be admitted to top colleges.  These "B"s will probably make it very hard for me to get an "A" in either subject, I'll probably end up with a final grade of "A-" in both subjects, possibly a "B+" in Spanish just because I'm actually no good at Spanish and it's this random honors course thing that means we actually learn stuff (heresy i know).  Will a pair of "A-"s on my transcript kill me college chances?  What about a "B+"???

If so, is there anything I should do to mitigate these harms, and if not, what kind of grades should I make sure I earn for the rest of my high school career to make sure colleges will not take these bad grades seriously???

Thanks guys.[/quote]

er, do colleges see 3rd quarter report cards? I thought they just took the semester information

No, you can get into great colleges with the occasional B. Especially you can if you have a challenging set of courses in your schedule.

It depends on your school and the courses you take as well.  If you go a top of the line high school and take all of the most demanding courses and get an occasional B you'd be better off than someone who goes to Podunk High, takes the easy classes, and gets 98's in each class.

[quote="pieterminate"]Huh?  Colleges can see A's as A-'s?  :o  :oops:[/quote]Colleges can see whatever your high school tells them.  If your high school reports A-'s on your transcript, then that's what colleges will see.

High school reporting standards vary wildly from state to state and often from district to district within a state.

If you use grammar like "my teacher uberly hates me randomly" in the classroom, then maybe it's that, and not your teacher, that's causing the problems. With that said, if you get a couple of Bs in the middle of the quarter and are worried about not getting into a good college, you really need to find something to do with your time.

Actually, (and those who know me can attest to this), I am a grammar freak when I choose to be, so I almost never, if at all, make grammatical errors.  In fact, I got As on virtually every essay, except for this stupid narrative where we had to imitate the voice of Holden Caulfield.  The primary reason for that B that I can see is that I was admitted to my school's rigorous honors' English program, and my English teacher hates me and is thus trying to revoke said admission, but she won't be able to convince the chair of the department to remove me unless I am earning bad grades.

Hmm... well I hope you guys are right :-/  I'm still awfully concerned though.

hey man chill out....
u can get like 6 b's and get into a great college...

actually i think that to study too much for high school is bad for getting ina good college

Colleges don't care if you don't have a $4.0$ unless you're applying to a college that only admits based on an SAT-GPA formula or something stupid like that.

A decently competitive college is going to care a lot more about extracurriculars ([i]dedicated[/i] extracurriculars, one or two activities that you focus on), challenging coursework, the quality of your application essay and interview... there are so many other things to focus on other than GPA!

For example, a student who gets a few Bs every once in awhile but consistently takes challenging coursework (AP/IB classes, that kind of thing) is going to be viewed a lot more favourably than a student who has a $4.0$ while taking classes like basket weaving and ceramics.

As will a student who has, for example, been doing gymnastics since she was $5$, attends competitions, etc.  Or a student who started a debate club at his school and goes to debate competitions.  That sort of thing actually [i]matters[/i], and it's those kind of things that are going to distinguish the boring candidates from the great candidates.

Basically, a decent $\approx 3.0 - 3.5$ GPA is going to be enough so that colleges will start considering other things a lot more heavily.

The rationale behind this?

[b]There are so many $4.0$ students in the country.[/b]  There are so many people with perfect test scores that colleges don't care anymore.  It's not important.  They can't distinguish between their candidates based on grades and tests.  They distinguish based on the factors that will [b]actually matter in college[/b] - leadership/teamwork ability, emotional/intellectual maturity, passion, dedication.  They want a student who'll contribute to their college, not just another cookie-cutter applicant.

Don't forget those X USAMO qualifications that will be on your resume by the time you apply, where $X \ge 2$.  And hopefully (*knock on wood*), $X >> 2$.

Also, imitating the voice of Holden sounds like it could be kind of fun, at least if you approach it in a satirical manner...

[quote="t0rajir0u"]Basically, a decent $\approx 3.0 - 3.5$ GPA is going to be enough so that ...[/quote]
t0rajir0u has a point, but that line goes too far. My institution (CSU Long Beach) admits students mechanically based on an SAT-GPA formula, with the GPA quite important. We're not that selective. The statewide version of our formula is supposed to make 33% of CA high school students eligible, although for non-local students, we set a noticeably higher standard. And in my experience, a 3.0 GPA is [i]low[/i] - dangerously low. I've had experience with students who had GPA's down around there and pretty good SAT scores - and for the most part, they don't do well in our college. The low grades serve as a warning that the work ethic might not be all one would hope for, and so the student may not have the resources to deal with a class that is more difficult than anything in high school. Just being pretty smart may not be good enough.

There is a large practical difference between 3.0 and 3.5.

Of course, that's irrelvant to bubala's original post - he's probably talking about the difference between a 4.0 GPA and a 3.89 - or maybe between a weighted 4.53 and a weighted 4.38. And no, I really cannot tell any practical difference there.

I just found out recently that a kid from my school who was accepted to Princeton has about a 3.0.  He has a lot of other things going for him, of course (ROTC, MOCK trial (his hook...he got offered a scholarship from princeton for this), track, cross country, some other clubs).  He's also a minority (black) so that helped him a little bit as well.

Getting [b]one[/b] B won't hurt you any.  If that happens to be the only chink in your application when you apply, heck, it might actually help you.  Additionally, the 10th, 11th, and the beginning of 12th are more heavily looked at since they show your more recent past.

Yo... chill out.  You're worried about 1 B, that's not even a semester grade?  I've gotten 7, all from freshman-sophomore year...
a 3.8+ is already pretty darn sufficient for a "top college"

[quote="joml88"]I just found out recently that a kid from my school who was accepted to Princeton has about a 3.0.  He has a lot of other things going for him, of course (ROTC, MOCK trial (his hook...he got offered a scholarship from princeton for this), track, cross country, some other clubs).  He's also a minority (black) so that helped him a little bit as well.[/quote]

I don't want to retread the race/college admissions topic again, but you are crazy if you don't think that minority status is a huge factor in admissions here.  I don't care what other intangibles he has going here; that list of extracurriculars is not that impressive and a 3.0 GPA is pathetically low for a school of that caliber.

My school technically calculates class rank, but never releases it (you can ask if you want, but it will never appear on a transcriptor be given to a college).  The main reason they do it is for scholarship purposes, and since I'm in the top 10-15% anyways, I'm no worse off than #1 (actually maybe better, since I've been instead focusing on things that are more relevant).

I think getting a 92 in English (we display numerical grades, btw, so that actually looks better than someone with straight A's at a "90-100 = A" school) but making the US Physics Team and taking math courses up through abstract algebra looks better than someone who simply gets straight A's taking regular class and even a few AP's, at least in my case, since I plan to major in physics and/or math.

[quote="Silverfalcon"]How much does your score on Final matters?

I usually get 95-100 in my finals. But in my history final, I think I got B (it's not definite yet..). I think Final counts as about 10% of semester grade so is it going to hurt me? I don't shoot for really difficult schools but I'm going for schools with medical programs so that's pretty tough too..[/quote]

If you get an A in the class, nobody cares about what you actually got on the final (by nobody I mean no colleges). And even if you get a B in a history class, it's not going to kill (or heavily affect) your chances of getting into a good school. Unless your school has widespread grade inflation (like mine), chances are a B won't do you much harm. Of course you can always make yourself look that much better by getting all A's, but there's no point in worrying about it much.

Huh, at our school finals counts for 25% of the semester grade.

OMG!!! B??? Are you serious??? That means you fail at life and have to go to community college!!!!!! You have shamed your family and everybody you have ever met!!!! You better go bribe the registrar to change your grades or else you have no shot at any ivy since we know that anybody who doesn't go to an ivy will be shot for stupidity.

On a more serious note, if you don't go to a competitive school, and for the most part, high schools aren't competitive, not having a 4.0 when other people at your school applying to the same colleges do have 4.0s will hurt. It's kinda like SAT II scores if only a letter and no pluses or minuses is put on your transcript. The B indicates that maybe your As were all borderline and your teachers were nice and bumped you up, while having a 4.0 could mean that you had over 100 percent in all of your classes. With SAT II scores, if you have all 800s, the admissions officer doesn't know if you just barely made the cutoff or were perfect. For a lot of people, it's all in the illusion of intelligence.

But I wouldn't worry about it. It seems as if you have a lot of things going for you besides your grades.

Maybe your school is a bit harder than most, but at mine getting A's is just a matter of trying. If I pay attention during class, study small amounts every day, and put effort into my homework, I'm sure to get an A and do very well on the tests. Your friend is probably doing something wrong if he has to stay up til 2. That's unhealthy.

I had several Bs and still got into a top college...but I took a difficult courseload, and the Bs were still above-average in those difficult courses.

[quote="stupidkid"]I have a friend who stays up until 2AM everyday to work on homework to maintain his 4.0 GPA (only in 10th grade). On average he gets less than 5 hours of sleep every night. You can see from his red eyes that he stays up late everyday. He also brags about how his lowest grade is a 98%.
[/quote]

We have 30+ or so kids in my grade alone who do that.  It won't help them in the long run though, as most of them focus mainly on grades and not extracurriculars, also taking relatively easy courses in which they know they can get A's.

Besides, since he's only in 10th grade, it won't hurt him to get a B or 2.  And by that, I mean colleges like to see improvement, so he's now got nowhere to go but down.

Also, now in college interviews, they want to get the impression that you'll be able to relax with an outside activity that you enjoy doing, so that you don't get overly stressed.

As long as you have mostly A's, and nothing below a B, then the minor differences beyond that are almost inconsequential.

[quote="stupidkid"]I have a friend who stays up until 2AM everyday to work on homework to maintain his 4.0 GPA (only in 10th grade). On average he gets less than 5 hours of sleep every night. You can see from his red eyes that he stays up late everyday. He also brags about how his lowest grade is a 98%.

Ok enough said about him. On the other hand a senior at my school recently got accepted into MIT. He didn't have perfect SATs or GPA or anything, but rather focused on research and did pretty well in the Siemens competition.

I'd rather get sleep and not kill myself over a letter on a sheet of paper.[/quote]

Staying up till 2 should not give you signs of red eyes or anything o_O I've been going to bed around 3 for this past month and I have to wake up at 7:30 and while I'm groggy and tired, it wears off by midday and my eyes aren't red at all.

I stay up to 2 and wake up at 6 am since I take zero period...
Well, I don't stay up to 2 because of lots of homework...
(I stay up to around 2 during summer too!)..
I do "extracurricular activities" such as tennis, clubs... after school   .
and do my homeworks at night.....

[quote="mtms5467"]I think that people shouldn't stress so much about grades. Grades are just a measurement, while the material and the process of learning are much more important. If you put effort and always try your best, you shouldn't be constantly calculating averages.[/quote]

easy to say.  everyone who works so hard can easily brush off their hard work with a modest grin.  grades do matter, its knowing what grade will be challenging, or going to the next step, in your mind.  grades matter in college, and else where.

if someone asks you if you are doing well in school, you can't proudly say "oh, i get c's, b's, but i understand everything very well, and i really like math, and i'm really good at it, and i do extracurricular stuff and blah blah blah"

you can say "i get straight a's

well, basically, even though you can convince your self that grades are not important, try convincing your parents or teachers, .....hey, try convincing your college! :roll:  you'll get this:  :huuh:  :ewpu:

Grades matter less in college than you think.  Maybe for something like med school or law school they would be important (probably the MCAT or LSAT would still have more weight) but for grad school there are more important things.  This is not to say that you should neglect your grades in college :)

My point was that grades don't matter much unless they're bad.  For example, if you had 1 B and the rest A's and got rejected from a certain college, I doubt you would have gotten in if you had gotten straight A's.


Suppose that m,n are two positive integers such that
\[ \sqrt{7}-\frac{m}{n}\geq 0 \]
Prove that
\[ \sqrt{7}-\frac{m}{n}>\frac{\beta}{mn}\ \ for\ \ \beta=1 \]
Find the largest number $\beta$ such that the above inequanlity still holds

Just for the record: for $\beta =1$ the problem was proposed in 1978 at the Romanian Math Olympiad, grade 9.
The proposer, Radu Gologan, is now the leader of the Romanian IMO team.

Denote
(1)        $\sqrt{7}-\frac{m}{n}>0$
(2)        $\sqrt{7}-\frac{m}{n}>\frac{1}{mn}$
From (1) we get $7n^2-m^2>o$. We show first that $m^2\in \{7k, 7k+1, 7k+2, 7k+4/k\in N\},\ \forall m\in N$. Then prove that  $7n^2-m^2$ cannot be $1$ or $2$, so $7n^2-m^2\geq 3$. From here it's easy to obtain relation (2).

$\beta =1$ is the only number to match the conditions in the statement, hence also the largest.
Indeed, for $\beta =2$ there exist $m,n\in N^*$ s.t.
         $\sqrt{7}-\frac{m}{n}>0$
and
         $\sqrt{7}-\frac{m}{n}<\frac{\beta}{mn}$.
Take, for instance, $m=5$ and $n=2$ (I choosed them s.t. $7n^2-m^2= 3$) .
On the other hand, if a number $\beta>2$  had fitted, then $\beta=2$ would have also fitted, which is false.

A somewhat more interesting question (possibly that originally intended, nmtruong1986?) is when $\beta$ is real rather than natural.  Raluca, your example shows that $\beta$ is less than about 1.457 but doesn't prove it has to be 1.

Raluca, you are misunderstanding. The question is that Find out the largest number, not natural number

Oh yes, you're both right !  :)  I assumed $\beta \in N$ (probably because I had in mind only rational approximations of $\sqrt{7}$.

If there are no wrong, $sup\{\beta\}=\frac{6}{\sqrt7+2}$ and sup can be attained with the equality when $m=2, n=1$


find all function $f: \mathbb{R}{*}\to\mathbb{R}{*}$ satisfying $f(xf(y))f(y)=f(x+y),\ \forall x,y\geq 0$
$\mathbb{R}^{*}$ is a set all numbers real non-negative.

This works for only (integer) powers of $f(0)$.
Let $f(0)=C$
Setting $y=0$ gives $Cf(Cx)=f(x)$
$f(x)=Cf(Cx)=C^{2}f(C^{2}x)=...C^{n}f(C^{n}x)$
Let $x=1$ to obtain $f(C^{n})=f(1)/C^{n}$

So, $f(x)=K/x$ for arbitary non-negative $K$ and positive $x$.
$f(0)$ can be obtained by letting $x$ and $y$ be zero; we get $f(0)=0$ or $f(0)=1.$

Wait...does this work?...


calculus
integration
logarithms
trigonometry
calculus computations

Caculate:
$ \int\limits_{0}^{1} \frac {\ln (1 \plus{} x)}{x^2 \plus{} 1} dx$

With the substitution $ x = \tan t$, we find that the given integral is equal to $ I = \int_{0}^{\frac{\pi}{4}} \log(1 + \tan t) \, dt$. Again apply the substitution $ t = \frac{\pi}{4} - u$ to obtain

$ I = \int_{0}^{\frac{\pi}{4}} \log(1 + \tan(\tfrac{\pi}{4} - u)) \, du$.

Suppose $ \alpha + \beta = \frac{\pi}{4}$. Then $ (1 + \tan \alpha)(1 + \tan \beta) = 2$ by the addition formula for tangent. Thus

\begin{eqnarray*}
2I & = &  \int_{0}^{\frac{\pi}{4}} \log(1 + \tan t) \, dt +  \int_{0}^{\frac{\pi}{4}} \log(1 + \tan (\tfrac{\pi}{4} - t)) \, dt\\
& = &  \int_{0}^{\frac{\pi}{4}} \log(1 + \tan t)(1 + \tan (\tfrac{\pi}{4} - t)) \, dt\\
& = &  \int_{0}^{\frac{\pi}{4}} \log2 \, dt\\
& = &  \frac{\pi}{4} \log2.
\end{eqnarray*}

Therefore we have $ I = \frac{\pi}{8} \log 2$.


why does a moving charge create a magnetic field around it?
can you please explain the clear mechanism?

thanks in advance

Are you familiar with the [url=http://en.wikipedia.org/wiki/Ampere-Maxwell_equation]Ampere-Maxwell equation[/url]?

Basically, a time-varying electric field gives rise to a magnetic field (so does current, but we don't quite have a current here). If you have a lone charge that's moving, the electric field at any point of observation is changing in either direction, magnitude, or both (why?). Thus, as long as the charge is moving, there will be magnetic field at the point of observation.

In terms of vector calculus, the relevant equations are:

$ \nabla \cdot \mathbf{B} \equal{} 0$
$ \nabla \times \mathbf{B} \equal{} \mu_0\mathbf{J} \plus{} \frac {1}{c^2} \frac {\partial \mathbf{E}} {\partial t}$

Where $ \mu _0$ is the magnetic constant, $ c$ is the speed of light, $ \mathbf{E}$ is the electric field, $ \mathbf{B}$ is the magnetic field, and $ \mathbf{J}$ is the current density through that point.


Suppose $n \in \mathbb{N}$.Prove there exist infinite numbers like $m$ that we can't find  $a_1,a_2,...,a_n \geq 0$ that : \[ \displaystyle m=a_1^n+...+a_n^n\]

[i]Edited by Myth[/i]

Hmm... I invented the following approach.
Let $N=M^n-1$, where $M\in\mathbb{N}$. Then there are exactly $M$ perfect $n$-th powers, which are less than $N$. We can form from them at most $M^n$ sets of $n$ summands, but in this way we count some sums many times, for example
\[0+0+0+...+0+a=0+0+...+0+a+0=...,\]
where $1\leq a\leq M-1$. So we counted at least $(n-1)(M-1))$ superfluous sums.
It follows that we can form at most $M^n-(n-1)(M-1)$ distinct sums from perfect powers $0$, $1^n$, ..., $(M-1)^n$. So at least $M^n-(M^n-(n-1)(M-1))=(n-1)(M-1)$ numbers form set $\{0,1,2,...,M^n-1\}$ cannot be expressed as sum of $n$ $n$-th powers.

Note, that this proof will not work for $n+1$ summands.

Well done. My solution was also this ;)  ;)

:)  :)  :)
Let's move it to Solved Section!

Chinese great scientist Loo-Keng Hua's(just like Vinogradov in Russia) work:

[url=http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Hua.html]www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Hua.html[/url]


An astronaut drops a rock from the top of a crater on the moon.  When the rock is halfwa down to the bottom of the crater, its speed is what fraction of its final impact speed?

When the rock is halway down the crater, it's potential energy (with respect to the bottom of the crater) is $mgh/2$ where $m$ is the mass of the rock, $g$ is the gravitational acceleration on the Moon and $h$ is the depth of the crater (this expression holds if $g$ is constant, ie $h$ is much smaller than the radius of the Moon). The potential energy of the rock on top of the crater was $mgh$, so the kinetic energy of the rock halfway down the crater is $\frac{mv_1^2}2 = mgh-\frac{mgh}2=\frac{mgh}2$. At the bottom of the crater the potential energy is zero, so the kinetic energy is $\frac{mv_2^2}2 = mgh$. From these two equations we get $\frac{v_1^2}{v_2^2} = \frac 1 2$, so $v_1 = v_2/\sqrt{2}$.


linear algebra
matrix
geometry
geometric transformation
rotation
linear algebra unsolved

Let X and Y are unknowns and $ X , Y \in M_n(R)$ Slove system equation :
         $ \begin{cases} X^{T}Y^3X \equal{} I_n \\
Y^TX^3Y \equal{} I_n \end{cases}$
$ I_n$ is identity matrix .
$ X^T$ is transpose  of $ X$
:wallbash_red:  :wallbash:

Little ideas.
1) Necessarily $ det(X)=det(Y)=1$ (easy).
2) $ Y^3=X{}^T{}^{-1}X^{-1}$ is a positive definite symmetric matrix and $ Y^3=Pdiag(\lambda_1,\cdots,\lambda_1,\cdots,\lambda_k,\cdots\lambda_k)P^{-1}$ where $ P$ is an orthogonal matrix and $ \lambda_1,\cdots,\lambda_k$ are positive. $ Y^3$ and $ Y$ commute then $ Y=Pdiag(U_1,\cdots,U_k)P^{-1}$ where $ U_j^3=\lambda_jI$.
By the same way $ X=Qdiag(V_1,\cdots,V_l)Q^{-1}$ where $ Q$ is orthogonal, $ V_j^3=\mu_jI$, $ \mu_j>0$.
After I don't know.....
3) A particular case:  $ X^3=Y^3=I$
 then $ X,Y\in{O^+(n)}$. Therefore $ X=Qdiag(I_p,U,\cdots,U,V,\cdots,V)Q^{-1}$ where $ Q$ is an orthogonal matrix, $ U,V\in{O^+(2)}$ are rotations of angles $ 2\pi/3$ and $ -2\pi/3$.


function
induction
inequalities solved
inequalities

Let f be a funtion so that we have $f:R -> (0, \infty)$ a strictly increasing function and $a_{1} \leq a_{2} \leq \ldots \leq an$. Prove that we have:

$ \frac{f(a_{1})}{f(a_{2})}+ \frac{f(a_{2})}{f(a_{3}}+ \ldots + \frac{f(a_{n})}{f(a_{1})} \geq \frac{f(a_{2})}{f(a_{1})}+ \frac{f(a_{3})}{f(a_{2})}+ \ldots + \frac{f(a_{1})}{f(a_{n})}$.

made by Lascu M.

Lemma :
Let a,b,c be three positive real such that a \leq b \leq c. Then :
a 2 c + b 2 a + c 2 b  \geq  c 2 a + b 2 c + a 2 b.

Proof.
It is equivalent to (b-a)(c-a)(c-b)  \geq 0.

Now, we prove the result by induction on n \geq 2.
For n = 2 we clearly have an equality.
Let's suppose that the result holds for n.
Let b_i = f(a_i) > 0.
Since f is increasing we then have b_1  \leq  b_2  \leq ... \leq  b_(n+1).
And, from the induction hypothesis, it suffices to prove that :
b_n/b_(n+1) + b_(n+1)/b_1 - b_n/b_1  \geq  b_(n+1)/b_n + b_1/b_(n+1) - b_1/b_n.
Clearing the denominators, this result is given by the lemma with a = b_1, b = b_n and c = b_(n+1).

Pierre.


[quote="iniquitus"]The last number is $ 23^4$[/quote]

Since  (the second no.) +1 =the last no.

$ 23^4\minus{}1\equiv 0 \pmod {10}$

If $ 23^4\minus{}1\equal{}p^3$ , how can we get this prime no. $ p$?

[quote="iniquitus"]the first and second are multiples of cubes of primes ($ 2^3$, $ 7^2$)

if you're asking how i got the number,
just solve for $ a^4 \equiv 2 mod 7$ and check the first few values[/quote]

[code]Find 3 consecutive integers, the first being a multiple of the square of a prime number, the second being a multiple of the cube of a prime number, and the third being a fourth power of a prime number.
[/code]
In my mind,3 consecutive integers mean n,n+1,n+2 ,$ n\in\mathbb {N}$do you agree?

Your choice ($ 2^3$, $ 7^2$) not the author's need:
the first being a multiple of the square of a number,
the second being a multiple of the cube of a prime number

uh yeah sorry about the confusing post

i meant

$ 23^4 \minus{} 2$ (multiple of $ 7^2$)
$ 23^4 \minus{}1$ (multiple of $ 2^3$)
$ 23^4$

[quote="iniquitus"]uh yeah sorry about the confusing post

i meant

$ 23^4 \minus{} 2$ (multiple of $ 7^2$)
$ 23^4 \minus{} 1$ (multiple of $ 2^3$)
$ 23^4$[/quote]

haha, thanks.I got some confusion.You'are right!! :D  :blush:


Let $a,b,c$ be nonnegative real numbers, no two of which are zero. If $p\ge 1$, then

$\frac 1{(a+b)(a+pb)}+\frac 1{(b+c)(b+pc)}+\frac 1{(c+a)(c+pa)}\ge \frac 9{2(1+p)(ab+bc+ca)}$.

I have a solution only for $p \ge 4$.

Sorry, we have only proved the original inequality holts for p>=1  if a,b,c are the sides of an triangle.

Just out of curiosity, what does cg stand for?

It seems like you have a nice system for getting these expression larger than zero!


limit
real analysis
real analysis unsolved

Show \[if \mathop{\lim na_{n}}_{n \to \infty }= a \ne 0 \Rightarrow \sum_{n = 0}^\infty{a_{n}}, divergence\]

If $\lim_{n \to \infty}na_{n}= a \ne 0$, show $\sum_{n=0}^{\infty}a_{n}$ diverges.

For some $N$, $\forall n \ge N, na_{n}> \frac{a}{2}$, in which case $\sum_{n=0}^{\infty}a_{n}> \frac{a}{2}\sum_{n=N}^{\infty}\frac{1}{n}= \infty$. This is for $a$ positive; for $a$ negative, a similar argument works by reversing the inequality.


without using congruence,Prove that $ 4n \plus{} 3 \nmid 4k^2 \plus{} 1$ for all $ n,k \in \mathbb{N}$

Yes it is quite easy.
$ (\frac {\minus{} 1}{4n \plus{} 3}) \equal{} \minus{} 1$ so $ 4n \plus{} 3$ not divide $ 4k^2 \plus{} 1$

[quote="TTsphn"]Yes it is quite easy.
$ (\frac { \minus{} 1}{4n \plus{} 3}) \equal{} \minus{} 1$ so $ 4n \plus{} 3$ not divide $ 4k^2 \plus{} 1$[/quote]
$ (4n\plus{}3)|(4n\plus{}3)^2\equal{}1\mod 8$.
Therefore we must consider prime divisor $ p|4n\plus{}3$ form $ p\equal{}3\mod 4$.

[quote="Rust"][quote="TTsphn"]Yes it is quite easy.
$ (\frac { \minus{} 1}{4n \plus{} 3}) \equal{} \minus{} 1$ so $ 4n \plus{} 3$ not divide $ 4k^2 \plus{} 1$[/quote]
$ (4n \plus{} 3)|(4n \plus{} 3)^2 \equal{} 1\mod 8$.
Therefore we must consider prime divisor $ p|4n \plus{} 3$ form $ p \equal{} 3\mod 4$.[/quote]
No it is Jacobi rule not Legende rule.


圣诞节了啊……
祝大家圣诞快乐，在新的一年中取得更高的成就！、
Merry Christmas & Happy New Year!
不知大家如何过这个圣诞呢?


Să se arate că numărul $20^{n}+1$ nu poate fi scris ca sumă de două pătrate de numere intregi, dacă $n$ este impar.

Fie $n=2k+1$.

Avem $20^{2k+1}+1 \equiv (-1)^{2k+1} + 1 \equiv 0 \left( \bmod ~ 7 \right)$ şi $20^{2k+1} + 1 \equiv 400^k \cdot 20 + 1 \equiv 8^k \cdot 20 + 1 \not\equiv 8^{k+2} \cdot 5 + 1 \left( \bmod ~ 49 \right)$. [Într-adevăr, $2 \neq 8$.]

Fie $\hat f : \mathbb Z \to \mathbb Z \slash 49 \mathbb Z$, $\hat f (k) = 8^{k+2} \cdot 5 + 1$, $\forall k \in \mathbb Z$. $\hat f$ este periodică de perioadă $7$ deoarece $8^7 \equiv 1 \left( \bmod ~ 49 \right)$.

Calculând, obţinem următoarele: $\hat f(0) = 27, \, \hat f(1) = 13, \, \hat f(2) = 48, \, \hat f(3) = 34, \, \hat f(4) = 20, \, \hat f(5) = 6, \, \hat f(6) = 41$, adică $f(k) \neq 0, \, \forall k \in \mathbb Z$.

[i]De fapt obţinem următoarele:[/i]

$\hat f(0) = 21, \, \hat f(1) = 14, \, \hat f(2) = 7, \, \hat f(3) = 0, \, \hat f(4) = 42, \, \hat f(5) = 35, \, \hat f(6) = 28$, adică $f(k) \neq 0, \, \forall k \in \mathbb Z, \, k \not\equiv 3 \left( \bmod ~ 7 \right)$, care nu prea spune multe.

[i]Wrong: Aşadar, $7^1 \| 20^{2k+1}$, deci $20^{2k+1}$ nu poate fi scris ca suma a două pătrate de numere întregi.[/i]

Am înţeles că ai aplicat teorema de caracterizare a numerelor naturale care se pot scrie ca sumă de două pătrate: [color=blue]Un număr natural $n$ poate fi scris ca sumă de două pătrate daca şi numai dacă fiecare factor prim de forma $4q+3$ din descompunerea lui $n$, apare la putere pară.[/color]
Am incercat şi eu să aplic teorema respectivă, dar "nu merge" cu 7, doarece pentru $n$ impar şi $n\equiv 0(mod49)$, avem că $20^n+1\equiv 0 (mod49)$.

[quote="Marius Damian"]Am înţeles că ai aplicat teorema de caracterizare a numerelor naturale care se pot scrie ca sumă de două pătrate: [color=blue][color=red]Un număr natural $n$ poate fi scris ca sumă de două pătrate daca şi numai dacă fiecare factor prim de forma $4q+3$ din descompunerea lui $n$, apare la putere pară.[/color][/color]
Am incercat şi eu să aplic teorema respectivă, dar "nu merge" cu 7, doarece pentru $n$ impar şi $n\equiv 0(mod49)$, avem că $20^n+1\equiv 0 (mod49)$.[/quote]
     Nu e cumva $\ Marea\ Teorema\ a \lui\ Fermat$?

[quote="stancioiu sorin"][quote="Marius Damian"]Am înţeles că ai aplicat teorema de caracterizare a numerelor naturale care se pot scrie ca sumă de două pătrate: [color=blue][color=red]Un număr natural $n$ poate fi scris ca sumă de două pătrate daca şi numai dacă fiecare factor prim de forma $4q+3$ din descompunerea lui $n$, apare la putere pară.[/color][/color]
Am incercat şi eu să aplic teorema respectivă, dar "nu merge" cu 7, doarece pentru $n$ impar şi $n\equiv 0(mod49)$, avem că $20^n+1\equiv 0 (mod49)$.[/quote]
     Nu e cumva $\ Marea\ Teorema\ a \ lui\ Fermat$?[/quote]
Din cate stiu, Marea Teorema a lui Fermat spune ca [color=blue]pentru $n\geq 3$, ecuatia $x^n+y^n=z^n$ nu are solutii in numere intregi nenule.[/color]

[quote="stancioiu sorin"]Da, stiam si eu de existenta acestei teoreme $\ x^n+y^n=z^n$ n-are solutii intregi pentru $\ n\geq 3$, dar in cartea lui V.Vornicu, la pagina $\ 46$ zice ca se numeste $\ Teorema\ lui\ Fermat$.[/quote]
Asa e, teorema de caracterizare de mai sus este, intr-adevar, a lui Fermat. Dar nu este Marea Teorema a lui Fermat.

OK.Ma scuzati ca am gresit, dar eu chiar credeam ca asa se numeste. :blush:

Nu iti face griji; a gresi este omeneste  :)

si...revenind la problema, are cineva o solutie ?
eu nu prea am gasit :D

Nici eu nu am o solutie, dar cu siguranta stiu ca nu merge cu Teorema lui Fermat.Am incercat pentru $\ 8$ numere prime de forma $\ 4k+3$ si ...fara folos.In general, avem urmatoarea
      [i]Lema[/i]
  Daca numarul $\ k^2+p^2$ este divizibil cu un numar prim de forma $\ 4x+3$, atunci acel numar prim divide si
numerele $\ k,p$.
  Cu asta, vom obtine numai numere prime de forma $\ 4x+3$ numai la exponenti pari. :(

Am gasit eu ceva ce ar putea ajuta la rezolvarea problemei.Avem
  [i]Lema[/i]
  Ecuatia $\ x^2+36=y^5$ nu are solutii in numere intregi.(A.Makowski)
  Demonstratie:
  Presupunem prin absurd ca realtia data e adevarata.Daca $\ x$ este par,
atunci $\ y$ trebuie sa fie si el par.Fiindca $\ x=4m$ sau $\ x=4m+2$, rezulta ca $\ x^2$ congruent cu $\ 0$ sau $\ 4$ modulo $\ 16$ si avem
$\ x^2+36$ congruent cu $\ 4$ sau $\ 8$ modulo $\ 16$.Dar $\ y=2k$
implica $\ y^5$ congruent cu $\ 0$ modulo $\ 16$ si deci ecuatia nu are solutii intregi daca $\ x$ e par.In cazul in care $\ x$ e impar, $\ x=2m+1$
rezulta ca $\ 4$ divide $\ x^2-1$ si $\ 4$ divide $\ x^2+35$.Deoarece trebuie ca $\ y^5$ sa fie congruent cu $\ 1$ modulo $\ 4$, rezulta usor ca
$\ y$ da restul $\ 1$ la impartirea cu $\ 4$ si deci $\ y=4k+1$.Cum ecuatia se mai scrie $\ x^2+4=y^5-2^5$, rezulta ca $\ y-2|x^2+4$ si cum $\ y-2$ are forma $\ 4m+3$, iar $\ (x,2)=1$, aceasta este imposibil.
   Generalizarea lemei de mai sus este urmatoarea:
$\ x^2+4(2^{2n-1}+1)=y^{2n+1}$, cu $\ n\geq 2$, ce se solutioneaza analog ca mai sus.In problema noastra, putem lua $\ y=20$, si cum numarul $\ 2^{2n-1}+1$ poate fi patrat perfect(si $\ 4$ este patrat), am obtibe exact ca $\ 20^{2n+1}$ nu poate fi scris ca suma de patrate, caci $\ x^2$ este patrat perfect.Mai ramane sa demonstram ca pentru acel
$\ n$ de mai sus, $\ n\leq 2$, numarul dat nu poate fi scris ca suma de doua patrate perfecte.
     Mi-a luat cam mult sa gasesc soluita de mai sus, dar nu stiu cat de corecta este.L-as ruga pe dl Marius Damian sa observe greselile din aceasta solutie.

[quote="stancioiu sorin"] Cum ecuatia se mai scrie $\ x^2+4=y^5-2^5$, rezulta ca $\ y-2|x^2+4$ si cum $\ y-2$ are forma $\ 4m+3$, iar $\ (x,2)=1$, aceasta este imposibil.[/quote]
Fii mai explicit...nu am inteles prea bine cum ai obtinut contradictia.
Nu cred ca "lema" aceasta ajuta in rezolvarea problemei initiale.
In schimb, tin sa te felicit pentru perseverenta.

cred ca daca alegem $p$ prim de forma $\mathcal{M}_{4}+3$ care sa il divida pe $y-2$ (acesta exista), atunci avem $p|x^{2}+4$, si cum $4$ este patrat, avem ca $p|x$ si $p|2$. cum $(x,2)=1$, rezulta $p=1$, contradictie. sper ca este bine ce am zis...

[quote="maky"]citez :
$8^k \cdot 20 + 1 \equiv 8^{k+2} \cdot 5 + 1 \left( \bmod ~ 49 \right)$
$8$ si $20$ sunt inversabile in inelul $\mathbb{Z}_{49}$, deci relatia este echivalenta cu $4\equiv8^{2}(\bmod ~ 49)$, si nu prea e chiar asa...
iar $49|(20^{7}+1)$.

edit :
(e posibil sa fie gresit ce zic io, ca sunt cam obosit)
[b]nu[/b] iese modulo 3,7,43, insa obtin ca n este divizibil cu 3,7,43.[/quote]Nu pot sa cred ca ai stat sa calculezi pentru $43^2$ :)

Marius, daca cumva solutia este cu modulo 127 (sau ceva analog) poti sa ne spui de pe acuma :D

Intr-adevar se vede imediat ca $3 \mid n$, si $127 \mid 20^{3n}+1$. Cum $127 \equiv 3 \pmod 4$ ne gandim ca poate poate merge ... da' cine sta sa verifice ca pentru $n\equiv 0 \pmod {3\cdot 7\cdot 43}$ nu putem avea si $20^n + 1 \equiv 0 \pmod {127^2}$? (si probabil ca se poate)

[quote="Valentin Vornicu"]
Marius, daca cumva solutia este cu modulo 127 (sau ceva analog) poti sa ne spui de pe acuma :D [/quote]
Valentin, solutia mea nu foloseste teorema lui Fermat. :)

referitor la mai sus
pai $20^{43\cdot 21}\equiv -1 (\bmod ~ 43^{2})$
cum $n$ este divizibil cu $21$ si $20^{n}\equiv -1 (\bmod ~ 43^{2})$ (pentru ca $20^{21}+1$ este divizibil cu $43$), rezulta ca si $20^{d}\equiv -1 (\bmod ~ 43^{2})$, cu $d=(43\cdot 21,n)$. dar cum $21|n$, rezulta ca $d$ este fie $21$, fie $43\cdot 21$. dar $d$ nu este $21$ pentru ca $20^{21}\equiv 1719 (\bmod ~ 43^{2}=1849)$.

[quote="maky"]lema e buna, dar nu are legatura cu problema.
adica, conform lemei ai ca ecuatia :
$20^{2n+1}+1=x^{2}+4(2^{2n-1}+1)+1$ nu are solutii in numere naturale. acest fapt nu ajuta cu absolut nimic la solutie... :(

eu am niste idei de rezolvare, dar cum momentan nu am ajuns la nimic nu voi posta. poate cine stie, am noroc mai incolo :)[/quote]
   Poate ai dreptate...Problema ne cere sa demonstram ca numarul
$\ 20^{2n+1}$ nu se poate scrie ca suma de doua patrate perfecte.Dar am demonstrat ca ecuatia
$\ 20^{2n+1}=x^2+4(2^{2n-1}+1)+1$ nu are solutii intregi.Dar numerele $\ x^2$ si ${\ 4(2^{2n-1}+1}+1$ sunt patrate perfecte.Pentru al doilea numar e usor sa demonstrezi ca exista o infinitate de valori pentru $\ n$ astfel incat acesta sa fie patrat perfect.(i.e.numerele de forma $\ 4k+1$ admit patrate perfecte)Ca sa ma-ntelegi mai bine, am scris numarul dat sub forma $\ x^2+k^2$, unde avem
$\ |k|=\sqrt{4(2^{2n-1}+1)+1}$Ce este gresit?

numarul $4(2^{2n-1}+1)+1$ nu este patrat perfect. radicalu ala de acolo pe care il notezi $k$ este cam irational...
daca ma gandesc mai bine, $N=4(2^{2n-1}+1)+1$ este chiar congruent cu $5$ modulo $8$, deci nu poate fi patrat perfect.

Da, am inteles acum.Totusi,..ai vreo solutie? :(

[quote]Valentin, solutia mea nu foloseste teorema lui Fermat.[/quote]

domnule profesor, va rog sa ne spuneti si noua solutia. eu unul am renuntat, dupa ce m`am gandit destul la problema si dupa ce am discutat`o cu cativa colegi din loturile de OIM/Yacutia.

Nici aia n-au stiut-o?Mai da'chiar e grea.

pai cu totii am avut aceeasi idee... apoi am incercat mici deviatii de la aceasta idee - $p$ prim de forma $4k+3$ care sa il divida pe $20^{n}+1$, cu valori pt $p$, cu $p$ pe cazu general, cu maximalitate/minimalitate, etc...
eu sunt curios sa vad solutia

Facem demonstraţia prin reducere la absurd.
Din faptul că $n$ este impar, deducem că el are forma $n=2k+1, k\in N$.
Presupunem aşadar că există numerele naturale $a$ şi $b$ astfel încât $20^{2k+1}+1=a^2+b^2$.
Relaţia precedentă se scrie echivalent: \[ (*) (10\cdot 20^{2k})^2+a^2+b^2=(10\cdot 20^{2k}+1)^2  \] şi avem nevoie de următoarea:
[color=blue]Lemă.[/color] Toate soluţiile în numere naturale ale ecuaţiei $x^2+y^2+z^2=t^2$ sunt date de $x,y,z,t\in N$ cu $y,z$ pare şi $x=\frac{l^2+m^2-n^2}{n}$, $y=2l$, $z=2m$, $t=\frac{l^2+m^2+n^2}{n}$, unde $l$, $m$ sunt numere naturale arbitrare, iar $n$ este un divizor al lui $l^2+m^2$ mai mic decât $\sqrt{l^2+m^2}$.
Demonstraţia lemei este simplă şi se găseşte, de exemplu, în lucrarea "O introducere în studiul ecuaţiilor diofantiene"(Editura Gil, Zalău, 2002), autori Titu Andreescu şi Dorin Andrica.
[color=blue]Observatie[/color]:
Soluţiile întregi ale ecuaţiei $x^2+y^2+z^2=t^2$ admit şi o scriere de forma:\[ x=l^2+m^2-n^2, y=2nl, z=2mn, t=l^2+m^2+n^2  \] unde $l$, $m$, $n$ sunt numere întregi arbitrare.
Trecem la soluţia problemei noastre.
Ţinând cont de observaţia anterioară, din (*) rezultă că există $l$, $m$, $n$ numere întregi astfel încât: \[ 10\cdot 20^{2k}=2nl, a=l^2+m^2-n^2, b=2mn, 10\cdot 20^{2k}+1=l^2+m^2+n^2.  \]
Scăzând relaţiile $10\cdot 20^{2k}+1=l^2+m^2+n^2$ si $10\cdot 20^{2k}=2nl$, deducem că $(l-n)^2+n^2=1$ şi apar situaţiile:
[color=red]a)[/color]$|l-n|=1, m=0$ care nu convine, deoarece, în acest caz, $b=0$, adică $20^{2k+1}+1$ este pătrat perfect; imposibil (se arată uşor, prin reducere la absurd).
[color=red]b)[/color] $l=n, m=1$ sau $m=-1,$ de unde $10\cdot 20^{2k}=2l^2,$ imposibil.
În concluzie, numărul dat nu se poate scrie ca o sumă de două pătrate de numere întregi.

[color=red]Comentariu[/color]: Ce numere naturale pot fi puse în locul lui 20, astfel încât concluzia să rămână adevărată?

Acuma am inteles de ce nu iesea problema asta de nici o culoare...

Nu este problema mea; am primit-o de la un prieten si asta este solutia pe care i-am dat-o.

wow !
superba solutie ! felicitari domnule profesor
si trateaza si un caz mult mai general...ca $20$ poate fi inlocuit cu o groaza de numere...
sa zicem $r^{2k+1}+1$ nu se poate scrie ca suma de doua patrate unde $r$ nu e patrat perfect, e par, si cam atat (diferit de $2$)..cel putin asa mi se pare mie la prima vedere. e foarte probabil sa fie raspunsul incomplet...


There is a special forum for learning it... :mad:
Please look a bit to the other sections.

Pierre.


Let be a triangle $ ABC$ and a point $ M\in [BC]$ such that $ \frac {MB}{MC}\equal{}k$ . Prove that :

$ (1 \minus{} k)^2AM^2 \equal{} (1 \minus{} k)AB^2 \minus{} k(1 \minus{} k)AC^2 \plus{} k\cdot BC^2$


A puppy and two cats together weigh 24 pounds. The puppy and the larger cat together weigh exactly twice as much as the smaller cat, and the puppy and the smaller cat together weigh exactly the same as the larger cat. How many pounds does the puppy weigh?

$ P\plus{}LC\plus{}SC\equal{}24$
$ P\plus{}LC\equal{}2SC$
$ P\plus{}SC\equal{}LC$

Subtracting the first two, $ SC\equal{}24\minus{}2SC\Rightarrow SC\equal{}8$.  Then $ P\plus{}LC\equal{}16\Rightarrow LC\equal{}16\minus{}P$ and therefore, $ P\plus{}8\equal{}16\minus{}P\Rightarrow 2P\equal{}8\Rightarrow P\equal{}4$.

x=pup
y=small cat
z=large cat
x+y+z=24
x+y=z
z=12
Then, x+z=2y
2y+y=3y=24
y=8
x+y+z=x+12+8=24
x=4.


function
integration
calculus
algebra solved
algebra

find the limit value of
[tex]S_n=(2^(1/n))/(n+1)+(2^(2/n))/(n+1/2)+...+(2^(i/n))/(n+1/i)+...+2^(n/n)/(n+1/n)[/tex]
it can be prove easily by a definition, but What about another Way......?

[quote="liyi"]Do you mean
\[S_n=\frac{2^\frac{1}{n}}{n+1}+\frac{2^\frac{2}{n}}{n+\frac{1}{2}}+\cdots+\frac{2^\frac{i}{n}}{n+\frac{1}{i}}+\cdots+\frac{2^\frac{n}{n}}{n+\frac{1}{n}}\][/quote]
yes , thanks alot, my skill of typing is very bad

The limit is $1/ln(2)$ .

More general if $f$  is a positive integrable function and $u(i,n)$ is a positive sequence  such that for all $i$ we have  ${\frac{u(i,n)}{n}  \longrightarrow  0}$,

then the sequence  $S_n = \displaystyle\sum_{i=1}^n\frac{f(\frac{i}{n})}{n+u(i,n)}$ is convergent and its limit is $\displaystyle\int_0^1 f(x)\,dx$

To prove it, I use the Rieman sums ..

 :cool:  Is this one the old way ?  :cool:

sorry Diogene
I have said that "it can be prove easily by a definition"
and it is the way showed :D
I am finding another way

Wow, Master, I think that u must type""it can be prove easily by definition of integration", if not, we 'll understand that "definition" is the definition of limit.
   And, with the sum like that, integration is the best way, it's very hard to find another way, OK ?


Determine all three-digit numbers $N$ having the property that $N$ is divisible by $11,$ and $\frac{N}{11}$ is equal to the sum of the squares of the digits of $N.$

Let $ N \equal{} 100a\plus{}10b\plus{}c$, where $ a, b, c\in\mathcal{D}_{10}\equal{}\{0, 1,\ldots, 9\}$. Since $ 11\mid N$, then $ b \equal{} a\plus{}c$ and $ N \equal{} 110a\plus{}11c$, yielding $ N/11 \equal{} 10a\plus{}c$. In fact, we need $ 10a\plus{}c \equal{} a^{2}\plus{}(a\plus{}c)^{2}\plus{}c^{2}\equal{} 2(a^{2}\plus{}ac\plus{}c^{2})$. Hence $ c \equal{} 2d$, for a suitable $ d\in\{0, 1,\ldots, 4\}$, and consequently $ 5a \equal{} a^{2}\plus{}2ad\plus{}(4d\minus{}1)\cdot d$. This is impossible if $ d\ge 2$, since then $ \mbox{RHS}\ge a^{2}\plus{}2a\plus{}3 > 5a$. So necessarily $ d \equal{} 0$ and $ a^{2}\equal{} 5a$, i.e. $ a \equal{} 0$ or $ a \equal{} 5$, i.e. $ N \equal{} 0$ or $ N \equal{} 550$. As $ 0 \equal{} 0^{2}$ and $ 50 \equal{} 5^{2}\plus{}5^{2}$, we're done.

[quote="s.tringali"]Let $ N \equal{} 100a \plus{} 10b \plus{} c$, where $ a, b, c\in\mathcal{D}_{10} \equal{} \{0, 1,\ldots, 9\}$. Since $ 11\mid N$, then $ b \equal{} a \plus{} c$ and $ N \equal{} 110a \plus{} 11c$, yielding $ N/11 \equal{} 10a \plus{} c$. In fact, we need $ 10a \plus{} c \equal{} a^{2} \plus{} (a \plus{} c)^{2} \plus{} c^{2} \equal{} 2(a^{2} \plus{} ac \plus{} c^{2})$. Hence $ c \equal{} 2d$, for a suitable $ d\in\{0, 1,\ldots, 4\}$, and consequently $ 5a \equal{} a^{2} \plus{} 2ad \plus{} (4d \minus{} 1)\cdot d$. This is impossible if $ d\ge 2$, since then $ \mbox{RHS}\ge a^{2} \plus{} 2a \plus{} 3 > 5a$. So necessarily $ d \equal{} 0$ and $ a^{2} \equal{} 5a$, i.e. $ a \equal{} 0$ or $ a \equal{} 5$, i.e. $ N \equal{} 0$ or $ N \equal{} 550$. As $ 0 \equal{} 0^{2}$ and $ 50 \equal{} 5^{2} \plus{} 5^{2}$, we're done.[/quote]

How about 803?

[quote="s.tringali"]Since $ 11\mid N$, then $ b \equal{} a \plus{} c$[/quote]

The error is here.  As Peter has demonstrated, it is also possible that $ b \plus{} 11 \equal{} a \plus{} c$.


limit
superior algebra
superior algebra unsolved

$f(n)$ the maximal order of an element of $S_n$, the symetric group.
Prove that $\lim_{n \to \infty} \frac{Log(f(n))} { \sqrt{n Log(n)} } = 1$.

I think we can reduce the algebra by the formula
\[ f(n)=\max\limits_{a_1 + ... + a_k = n, a_i > 1} lcm(a_1 , ... , a_k) \]
;)

We can reduce it further: we can assume the $a_i$ to be coprime: otherwiese we could reduce a common prime factor from all $a_i$ except that one where it occurs the most and 'fill' the sum with ones.
Even more since $ab \geq a+b$ for all $a,b \geq 2$ we can assume each of the $a_i$ to be the power of a prime: otherwise when some $a_i$ is $p^sq$ with $p$ prime and $q \geq 2$ coprime to $p$, we could splitt this into $p^s$ and $q$.

Well, now this is just a problem of additive/combinatorial number theory... ;)

yep, it is not at all a group theory problem (this is why I posted it in the analysis secion  :)  )


geometry
3D geometry
tetrahedron
trigonometry

This might seem easy, but I couldn't figure it out.

There are four rays in space, all have the same endpoint. All angle measures between any two rays are the same. What is that angle?

[hide="Hints"]1) Regular tetrahedron
2) Cosine Law[/hide]

I still don't know. Anyone got a solution?

[hide]The endpoint will be the centroid of a regular tetrahedron that has a vertex on each ray. So suppose a regular tetrahedron has vertices (0,0,0) (0,2,2) (2,0,2) (2,2,0). The centroid is (1,1,1).

The angle between two vertices through the centroid will be the angle between ${\bf u}$ and ${\bf v}$, where ${\bf u}=-{\bf i}-{\bf j}-{\bf k}$ and ${\bf v}={\bf i}+{\bf j}-{\bf k}$. The angle $\theta$ is given by

\[ \cos \theta=\frac{{\bf u}\cdot {\bf v}}{||{\bf u}||||{\bf v}||}=\frac{-1-1+1}{\left( \sqrt{3} \right)^2}=-\frac 13 \]

So the angle is $\boxed{\arccos -\frac 13 \approx 109.5^\circ}$.[/hide]

This is one of the rare times Chemistry comes in handy in math :P. Four atoms bonding around a central one will make the angles between the bonds as close as possible. It's well-known that this angle is approximately $109.5^{\circ}$ and the shape formed is a regular tetrahedron. Of course deriving this mathematically is not difficult either, as scorpius119 has done.


Given 5 points on the plane, no three collinear.
Prove that the number of convex quadrilaterals formed by these points is an odd number.

It is really easy by considering the three cases about the convex hull of the five points... :wink:

Pierre.

oh!
3, 4, or 5 points in the convex hull  :blush:
Thanks :)


It's not hard - I've seen it once so I won't spoil fun - it's obvious to use Cauchy-Schwarz, after all, isn't it ?  :)

I think, it will be very useful for you, Megus, to write complete solution here.

Nice solution , Kunny !
  My solution is similar to Kunny's  :lol:


Claim: Suppose A is a 2 x k matrix (k>2), A does not have rank 0, B is a k x 2 matrix, B has rank 2, and AB is the 2x2 matrix with all entries=0. Then A has rank 1.

I feel like this should be obvious, but I have not been successful putting the proof together. Can someone please help me out, and be a little bit pedantic? Much appreciated.

(Edit: as discovered below, this is mis-stated! The claim should be:

Suppose A is a 2 x k matrix (k>2), A does not have rank 0, B is a k x (k-1) matrix, B has rank (k-1), and AB is the 2 x (k-1) matrix with all entries=0. Then A has rank 1.)

The reason you can't prove this is that the rank of $ A$ can be $ 2$ and the stated result is false.
\[ \begin{bmatrix}1&1&0&0\\0&0&1&1\end{bmatrix} \begin{bmatrix}1&1\\\minus{}1&\minus{}1\\1&\minus{}1\\\minus{}1&1\end{bmatrix}\equal{}\begin{bmatrix}0&0\\0&0\end{bmatrix}\]

You're right, I mis-stated the problem. (sorry.) This should be right:

Claim: Suppose A is a 2 x k matrix (k>2), A does not have rank 0, B is a k x (k-1) matrix, B has rank (k-1), and AB is the 2 x (k-1) matrix with all entries=0. Then A has rank 1.

Build the $ k\times k$ matrix $ C \equal{} \begin{bmatrix}B & x\end{bmatrix}$ where the column vector $ x$ is chosen outside of the span of the columns of $ B.$ That makes $ C$ an invertible matrix.

We have $ AC \equal{} \begin{bmatrix}0 & *\end{bmatrix} \equal{} D.$ That is, $ D$ is a $ 2\times k$ matrix whose first $ k \minus{} 1$ columns are all zero and whose last column is (as far as we currently know) arbitrary.

Then $ A \equal{} DC^{ \minus{} 1}$ and $ \text{rank}\,A\le\text{rank}\,D\le 1.$

Here's the generalization, which can be proved with basically the same argument:

Suppose $ A$ is $ m\times n,$ $ B$ is $ n\times k$ with $ k\le n,$ $ \text{rank}\,B\equal{}n$ ($ B$ has full rank) and $ AB\equal{}0.$ Then $ \text{rank}\,A\le k\minus{}n.$

In order to have $ A\ne 0,$ we'd have to have $ n<k.$


i have learned much from [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=499573#p499573]shyong's post[/url] - thank you!
[hide]
We have to prove
$2(\sqrt{n+1}-1) < \sum_{i=1}^n \frac{1}{\sqrt{i}} < 2\sqrt{n}$
which is equal to
$\sum_{i=1}^n {2(\sqrt{i+1}-\sqrt{i})} < \sum_{i=1}^n \frac{1}{\sqrt{i}} < \sum_{i=1}^n{2(\sqrt{i}-\sqrt{i-1})}$
The left side:
$2(\sqrt{i+1}-\sqrt{i})<\frac{1}{\sqrt{i}}$
$2(i+1-i)<\frac{\sqrt{i+1}+\sqrt{i}}{\sqrt{i}}$
$2<\frac{\sqrt{i+1}}{\sqrt{i}}+1$
$1<\frac{\sqrt{i+1}}{\sqrt{i}}$
The right side:
$\frac{1}{\sqrt{i}}<2(\sqrt{i}-\sqrt{i-1})$
$\frac{\sqrt{i}+\sqrt{i-1}}{\sqrt{i}}<2(i-(i-1))$
$\frac{\sqrt{i-1}}{\sqrt{i}}+1<2$
$\frac{\sqrt{i-1}}{\sqrt{i}}<1$
[/hide]

like in the other post, you can telescoping or use means (in this case HM-QM for the right side and AM-GM); the two main points of solving are: find a closet formula or try to reconduct to a famous one (by means) or, seeing sigmas everywhere, take a sum of elements of a serie in each side, and the simplest one is telescoping (compliments to shyong) :D


How do you write an U with a . inside it? That's the way to write disjoint union in our course... how do you do that? \bigudot or something doesn't work.

Also, what is the official symbol for disjoint union? Is it also this one or is it different?

In the big list of symbols posted by DPatrick a while ago, I didn't find that symbol.  Here's my solution, maybe not the prettiest, but it works.

In text mode:
A \makebox[0pt]{$\$$\quad\!\!\cdot$\$$}$\$$\cup$\$$ B

In math mode
A \makebox[0pt]{\quad \! $\$$\cdot$\$$} \cup B

And you should define new commands to use those, it will be much faster

In text mode:
$A \makebox[0pt]{\quad\!\!\cdot}\cup B$

In math mode
$A \makebox[0pt]{\quad \! \cdot} \cup B$

Doesn't work for me...

It doesn't work on the forum. However just write down disjoint union after the expression. Most people had no idea (including me) that there is a special sign for that, and not mentioning it will lead to confusion.

[quote="Valentin Vornicu"]It doesn't work on the forum. [/quote]

That's right, it doesn't work on the forum, try it in LaTeX.  That's why I didn't display an example.  You may also have to adjust the spacing to have the dot centered.

The symbol for a disjoint union (resp. intersection) as I know it is the square union (resp. intersection) symbol which can be written in LaTex by the math commands \sqcup (resp. \sqcap).


geometry
trapezoid
perpendicular bisector
geometry proposed

Two circles with centers $K,L$ intersect at $A,B$. A line through $A$ meets the circles at $C,D$. If $M$ is the midpoint of $CD$, $N$ is the midpoint of $KL$ and $E$ is the symmetrical point  of $A$ with respect to $N$ , prove that $EM \perp CD$.

 [u] Babis[/u]

[quote="vittasko"]We denote as $K',$ $L',$ $N',$ the orthogonal projections of $K,$ $L,$ $N$ respectively, on the segment line $CD$ $($ the point $C$ is defined on $K$ $)$ and we have that the point $N',$ is the midpoint of the segment $K'L'$ and $NN'\perp CD.$

It is easy to prove that $K'M = AL'.$

[hide]$MK'+K'C = MA+AL'+L'D = AK'-MK'+2(AL')$ $\Longrightarrow$ $2(MK') = 2(AL')$[/hide]
So, we have that the point $N',$ is also the midpoint of the segment $AM.$

Hence, because of $AN = NE$ $\Longrightarrow$ $NN'\parallel EM$ $\Longrightarrow$ $EM\perp CD$ and the proof is completed.

Kostas vittas.

PS. In my drawing the point $M,$ is between $K',$ $A.$[/quote]

[color=red]Kostas , nice and simple solution  :lol: !!![/color]

     The solution that  I have is a little more complicate :( .

   After your proof , we can prove easily this

[b]Problem:[/b]

  If $CAD$ is a variable line through $A$ , then the perpendicular bisector of segment $CD$ passes through a constant(fixed) point.

[u]Answer:[/u]  This point is $E$ , as stated in the above problem.(It was posed in a spanish olympiad).

Dear Mathlinkers,
if we know that the midcircle of (K) and (L) goes through A, B and M and has for center N, we are done.
Sincerely
Jean-Louis

Let $AK$ cuts $(O_{1})$ again at $F$, $AL$ cuts $(O_{2})$ again at $G$, then it is obvious that $F, B, E, G$ are collinear and $E$ is the midpoint of $FG$. Easy to see that $CF\parallel GD$, so $CFGD$ is a right-angled trapezium and it follows that $EM\perp CD$.


let $a_{1}, a_{2}... a_{n}$ be a sequence of real numbers satisfying : $a_{i+j}\geq a_{i}+a_{j}$ forall $i$ and $j$ .
 prove that :
   $a_{1}+\frac{a_{2}}{2}+...+\frac{a_{n}}{n}\geq a_{n}$

Please continuous at http://www.mathlinks.ro/Forum/viewtopic.php?p=456607#p456607  :wink:


Yushimi took three tests each with a maximum possible score of 100.  One of the grades was 84.  She wants her average to be 88.  What is the great difference possible between the other two test grades that will still allow her to attain this average?

[hide](84+a+b)/3=88

84+a+b=264

a+b=180

The way to make the greatest possible difference is to make one number big and the other small. The biggest one could be is 100.  In this case, the other is 80, so the difference is 100-80, or 20[/hide]

mmm, h_s_potter, maybe you want to reread the question?

Even though this has been referenced myriads o'times, read [url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_Mistakes.php]this[/url].

Yeah, it's a typo.  Mea culpa, I shall fix it now.

[hide] Call her test scores a, b, c. One of them, call it a, is 84. Yushimi wants her average to an 88. So we set up the equation: (a+b+c)/3 = 88.

Simplifying it we get: (84+b+c)/3 = 88, thus getting: 84+b+c = 264.

Subtracting 84, we get b+c=180.

Now we want the greatest difference... therefore, one of the grades should be the highest possible... in this case a 100. So the other grade must be 80.

100-80 = 20[/hide]

[hide]One of her test scores is 84 the other test scores are unknown.

we know that 84+x+y=88*3

84+x+y=264

x+y=180

since the highest test score anyone can get is 100

we know that x=100 and y=80

and the difference of them is 20

thats the answer[/hide]


Formulate the negation of the statement

$ \sim \exists x,x\in\mathbb{R}\wedge(1 < x^2< x)$

Will "$ \sim \exists$" become  "$ \exists$"?

Yes. If you translate the statement into english, it reads "There exists no real number $ x$ such that $ 1 < x^2 < x$." The negation of that would be saying that there does exist some real number $ x$ such that $ 1 < x^2 < x$.

So...this is wrong?

$ \sim (\sim\exists x,x\in\mathbb{R}\wedge(1 < x^{2} < x))$
$ \equiv\exists x,x\notin\mathbb{R}\vee[\sim (1 < x^2\wedge x^2 < x)]$
$ \equiv\exists x,x\notin\mathbb{R}\vee[x^2\le 1 \vee x^2\ge x]$

Utterly wrong. You applied a mix of de Morgan's laws et.al. Basically it's as simple as that: no matter how complicated a proposition or predicate $ P$ is, the negation of $ \sim P$ is $ \sim (\sim P) \equiv P$ (the rule of the double negation).


geometry
geometric transformation
homothety
geometry solved

Prove that nine point circle and incircle of a triangle tangent only one point. Need synthetic proof.

If you have enough patience, you can look at http://www.mathlinks.ro/Forum/viewtopic.php?t=14927 .

  Darij

This theorem also has a nice proof using Cayssey's theorem...(and quite a shorter one)

The result can be deduced from the following problem:
(Russia 1998)
[i]Let $D,E,F$ be the feet of the angle bisectors of angles $A,B,C$, respectively, of $\triangle ABC$, and let $K_a,K_b,K_c$ be the points of contact of the tangents to the incircle of $\triangle ABC$ through $D,E,F$ (that is, tangent lines not containing sides of triangle). Prove that the line joining $K_a,K_b,K_c$ to the midpoints of $BC,CA,AB$, respectively, pass through a single point on the incircle of $\triangle ABC$.[/i]
First, you may prove that three three lines are concurrent and then use Ptolemy's Theorem to complete the proof, for details, please ask yptsoi, since this proof is not done by me.......
Secondly, $\triangle K_aK_bK_c\sim \triangle ABC$. Thus, the concurrency point, the midpoints of $\triangle ABC$, are concyclic, ie the concurrency point lies on the nine point circle, but as we've proved that this point also lies on the incircle, so it is intersecting point of the nine point circle and incircle. Conversely, if we carry out the constructions mentioned to the first and second intersecting points of the incircle and nine point circle, this will also satisfies all the properties. But as the concurrency point must be unique, so the first and second intersecting points of the incircle and nine point circle actaully concide with each other, ie the incircle and nine point circle is tangent to each other.

At last, just a little comment, this problem (Russia 1998) can be done easily by using the result that incircle and nine point circle of a triangle are tangent to each other.
I know that I've left out many details, but I'm sure it won't be hard......anyway, this is all because I'm very busy on preparing my school exam :(

[quote="darij grinberg"][quote="mecrazywong"]Secondly, $\triangle K_aK_bK_c\sim \triangle ABC$. Thus, the concurrency point, the midpoints of $\triangle ABC$, are concyclic, ie the concurrency point lies on the nine point circle, but as we've proved that this point also lies on the incircle, so it is intersecting point of the nine point circle and incircle. [b]Conversely, if we carry out the constructions mentioned to the first and second intersecting points of the incircle and nine point circle, this will also satisfies all the properties.[/b][/quote]

Hmm, I am not really convinced by this latter argument... (but rather the [i]homothety[/i] of triangle $K_aK_bK_c$ to the medial triangle of triangle ABC yields the tangency of the two circles...).

  Darij[/quote]

Even myself think that this proof is not so convincing. However, as the [i]position geometrically[/i](?) of the intersecting points of the two circles are just the same, I think my arguement works, and I believe that there is a much convincing proof, may someone work it out?


inequalities
number theory theorems
number theory

How to show that for each $ n \in \mathbb{N} : p_n < e^{1\plus{}n}$ where $ p_n$ is the $ n$-th prime number.

[quote="Sage!"]How to show that for each $ n \in \mathbb{N} : p_n < e^{1 \plus{} n}$ where $ p_n$ is the $ n$-th prime number.[/quote]

Using Bertrand-Chebyshev theorem : $ p_{n\plus{}1}<2p_n$ and so $ p_n\le 2^n<e^{n\plus{}1}$

I thought about elementary solutions that doesn't use this theorem. Have you encountered this problem before? If yes, were the solution presented as a classic solution?

The words "elementary" and "$ e$" don't match.


limit
trigonometry
logarithms
calculus
calculus computations

i made this one up also.... LOL  :D

[b]without[/b] using power series or l'hospital , find the limit

$ \lim_{x\to \frac{\pi}{2} ^{\minus{}} } \; (\tan \; x)^{(\cos\; x)}$

:D I'm alsow glad to see you. Work... and I'm lazy last time :surf:

$ (\sin x)^{\cos x}\to 1$ when $ x\to \frac {\pi}{2} \minus{}$ because $ \sin x\to 1$ and $ \cos x\to 0$ and $ 1^0 \equal{} 1$

[quote="misan"]i made this one up also.... LOL  :D

[b]without[/b] using power series or l'hospital , find the limit

$ \lim_{x\to \frac {\pi}{2} ^{ \minus{} } } \; (\tan \; x)^{(\cos\; x)}$[/quote]

HI,

$ \lim_{x\to \frac {\pi}{2} ^{ \minus{} } } \; (\tan \; x)^{(\cos\; x)} \equal{} \lim_{x\to \frac {\pi}{2} ^{ \minus{} } } e^{cos(x)ln(tg(x))}$

but
$ \lim_{x\to \frac {\pi}{2} ^{ \minus{} } } cos(x)ln(tg(x)) \equal{} \lim_{x\to \frac {\pi}{2} ^{ \minus{} } } (cos(x)ln(sin(x)) \minus{} cos(x)ln(cos(x)))$
and
$ \lim_{x\to \frac {\pi}{2} ^{ \minus{} } } cos(x)ln(cos(x)) \equal{} \lim_{t \to 0^{ \plus{} }} tln(t) \equal{} 0$

I hope power series is not logaritms

[quote="FOURRIER"]

 $ \lim_{x\to \frac {\pi}{2} ^{ \minus{} } } \; (\tan \; x)^{(\cos\; x)} \equal{} \lim_{x\to \frac {\pi}{2} ^{ \minus{} } } e^{cos(x)ln(tg(x))}$

but
$ \lim_{x\to \frac {\pi}{2} ^{ \minus{} } } cos(x)ln(tg(x)) \equal{} \lim_{x\to \frac {\pi}{2} ^{ \minus{} } } (cos(x)ln(sin(x)) \minus{} cos(x)ln(cos(x)))$
and
$ \lim_{x\to \frac {\pi}{2} ^{ \minus{} } } cos(x)ln(cos(x)) \equal{} \lim_{t \to 0^{ \plus{} }} tln(t) \equal{} 0$
 [/quote]

nice.   you forgot to mention

hence, $ L \equal{} e ^ 0 \equal{} 1$


i want to draw a square colored black.

how do you do that?

oh thanks  :D

how about drawing a hexagon filled with gray?

[quote="isabella2296"]You mean like this?

    [asy]size(150);
    int n=6;
    pair[] v;
    for(int k=0;k<n;++k)
    v[k]=dir(360*k/n);
    v[n]=v[0];
    for(int k=0;k<n;++k)
draw(v[k]--v[k+1],gray);[/asy]

I messed around with fedja's code from somewhere else, and I can't fill it gray for some reason.[/quote]

In your code, you only made the lines gray. You have to use fill() to actually fill

Oh, oops. Where did my brain go?  :P

Ph. Ivaldi, when you wrote that code, what does the "black" do? Make the outline black?

[quote="isabella2296"]Ph. Ivaldi, when you wrote that code, what does the "black" do? Make the outline black?[/quote]
Replace "black" by "red" and you'll see.
Incidentally [url]http://asymptote.sourceforge.net/doc/fill.html#fill[/url]
 :|

[quote="herefishyfishy1"][quote="dragon96"][asy]path thesquare=((0,0)--(0,1)--(1,1)--(1,0)--cycle);
draw(thesquare);
fill(thesquare, black);[/asy]

hopefully this works...EDIT:yes it does

The first line just defines "thesquare" as a variable so I don't have to keep retyping the cycle, which is the square.
The second line draws it.
The third line fills thesquare black. Pretty self explanatory.[/quote]

You do not need to define "thesquare" as a variable, because there is a predefined path called "unitsquare," which is equal to (0,0)--(0,1)--(1,1)--(1,0). Also, you don't even need to draw the square at all. In one line of code:

[asy]fill(unitsquare,black);[/asy][/quote]

Oh k. I was afraid if I just did what you did, then it might find another unitsquare, so I just defined another variable


geometry
trapezoid
circumcircle
parallelogram
perpendicular bisector
geometry solved

Let ABCDE be a convex pentagon such that each diagonal is parallel to at least one side of the pentagon. Also it is Known that BC=CD=DE. Prove it is regular!

Two things are clear: $BCDE$ is an isosceles trapezoid, and $A$ lies on the perpendicular bisector of $BE$ (or $CD$; it's the same perpendicular bisector). We have $AD\|BC,\ AE\|BD$, so $\angle CBD=\angle DAE$. From the additional fact that $CD=DE$, we find $A$ to be on the circumcircle of $BCDE$ (which is cyclic because it's an isosceles trapezoid). This means that, for example, $ACDE$ is also an isosceles trapezoid, so $AE=CD$, and this pretty much ends the proof.

Grobber, your proof is very nice. Actually, the problem is funny: it looks like something one has to think about, but it turns out to be a simple angle chase... So here is my proof, really no brain needed:

Since "each diagonal is parallel to at least one side of the pentagon", the diagonal AC must be parallel to the side DE (in fact, it is parallel to at least one side of the pentagon, but it can't be parallel to AB, to BC, to CD, or to AE, hence it must be parallel to DE). Similarly, the diagonals BD, CE, DA, EB are parallel to the sides EA, AB, BC, CD, respectively.

The quadrilateral BCDE is an isosceles trapezoid (in fact, it is a trapezoid since BE || CD, and since BC = CD = DE, it has equal legs, so that it is either an isosceles trapezoid or a parallelogram, but it cannot be a parallelogram, since else we would have BC || DE, what, together with AC || DE, would yield BC || AC, impossible).

Now call t = < DBE. Then,

t = < CDB    (since t = < DBE, and < DBE = < CDB since EB || CD);
t = < DCE    (since t = < CDB, and < CDB = < DCE since the quadrilateral BCDE is an isosceles trapezoid);
t = < CEB    (since t = < DBE, and < DBE = < CEB since the quadrilateral BCDE is an isosceles trapezoid);
t = < BEA    (since t = < DBE, and < DBE = < BEA since BD || EA);
t = < ABE    (since t = < CEB, and < CEB = < ABE since CE || AB);
t = < CBD    (since t = < CDB, and < CDB = < CBD since the triangle BCD is isosceles);
t = < ADB    (since t = < CBD, and < CBD = < ADB since DA || BC);
t = < DAE    (since t = < ADB, and < ADB = < DAE since BD || EA).

Now, we have t = < DBE = < DCE = < DAE. Hence, the points A, B, C, D, E lie on one circle. Each of the chords AB, BC, CD, DE, EA has the same chordal angle in this circle, namely the angle t (this is because t = < ADB = < CDB = < CBD = < DCE = < ABE). Hence, the pentagon ABCDE is regular. Qed..

  Darij


Sad, sad thing: This actually [b]is[/b] a problem at my school :)  :(

Anyways, someone in charge sent an e-mail to all 8 math teachers at my school. The probability of each of the teachers accepting the post as MATHCOUNTS Coach is as follows (estimated): 0, 0, 0, 1/3, 1/3, 1/2, 1/2, 4/5
a.) What are the chances of us getting a coach?

ab/2) What are the chances of us getting only [b]one[/b] coach?
now think about this problem:
b.) What are the chances of us getting more than one coach?

I won't answer the question for you, but I agree that it is very sad. Lots of people have trouble making commitments, and I hate that.

Oh, no, I know the answer. This is both a math "problem" and the school's "problem" :)

Hmmm, this is harder than it looks...I'm not even sure if I'm right.

[hide]

1. The probability of not getting a coach is

2/3 * 2/3 * 1/2 * 1/2 * 1/5 = 1/45

Therefore, the probability of getting a coach is 44/45

2. Still working this one out

[/hide]

Ok, for #2, if I'm right, I'll be incredibly surprised.

[hide]

The number of groups that are more than 1 that can be made are

5C2 + 5C3 + 5C4 + 5C5  = 26

There are 26 groups that can be made. Multiply this by the chance that you'll get all the coaches available, doing so, you get 26/45

You'll have to understand that I'm basically relying on intuition here - I'm horrible at probability questions, so I'll be overwhelmed if I'm right.

[/hide]

The first one you're right. I decided to add number ab/2 :)

the second one is rather tough. southpaw, i don't think you're right, but don't feel bad, i said it's tough. hint: count the probability that you will get 0 or 1 coach and subtract that probability from 1. further hint: [hide]counting the case where you have 0 coaches is not that hard. harder is counting the case where you have exactly one coach. you will need to break into cases. (first, forget about the first three teachers since they just clutter up the problem). okay, now what is the probability that the fourth teacher will accept the post and nobody else will? what is the probability that the fifth teacher will accept the post and nobody else will? etc. [/hide]

answer: dunno if this is right, possible arithmetical error [hide]37/45[/hide]

Yes, this problem's a toughy for a change :)
Most of my problems were real easy, so this is kind of a fraud because it looks real easy, but it's not.

If no one solves this by Sunday, I'll put this in the Round 2 of the competition.

Well, if you do it the slow and tedious way of [hide]adding up all the cases[/hide] then it's not too hard. Just consider all the situations.

[hide]

So for ab/2 I got 1/5, which means for b I get 1 - (1/45+1/5) = 7/9

[/hide]

erm it wouldn't hurt to actually show the cases.

Something that looks like a family tree might help. You start with the big limbs, representing each teacher, then those split into tiny limbs of possibillities. Try to write all things in the same level at the same altitude (in other words, don't make it a circly looking thing: Just make it look neat)
It worked with me, hope it works with you.

Sigh.

[hide]

If we figure out what the probability of having only one coach is, we've figured out the problem, right? So the answer is

(1/3 * 2/3 * 1/2 * 1/2 * 1/5) *2 + (1/2 * 2/3 * 2/3 * 1/2 * 1/5) * 2 + (4/5 * 2/3 * 2/3 * 1/2 * 1/2) =  1/15 + 2/45 + 4/45 = 9/45 = 1/5

The reason you do this is because you're given the probability of what having one of them is, then you multiply it by the probability of not having the rest of the coaches. Yeah.

[/hide]

why is the first term 1/15. shouldn't it be 1/45? sigh

No, because there are 2 1/3's, so we're taking care of both of them there. Just multiply it out.

Yup, yup. "If I had a dollar for every" correct answer you had in your second to last post, "I'd have [b]one [/b]dollar" :)

Heh I got the same answer as Mystic terminator.  I think there might be a calculation error.. Your work looks right though. not sure  :?


geometry
parallelogram
trigonometry
geometry unsolved

In a parallelogram $ ABCD$ with $ \angle A < \frac {\pi}{2}$, the circle with diameter $ AC$ meets the lines $ BC,CD$ again at $ E,F$ respectively, and the tangent to this circle at $ A$ meets $ BD$ at $ P$. Show that $ P,F,E$ are collinear.

Trying to use convers of menelaus in $ \Delta BCD$....
$ \frac {CF}{FD}\equal{}\frac {AC\sin \angle CAF}{AD\sin \angle FAD}$

$ \frac {BE}{EC}\equal{}\frac {AB\sin \angle BAE}{AC\sin \angle CAE}$

$ \frac {DP}{PB}\equal{}\frac {AD\sin \angle DAP}{AB\sin \angle BAP}$

$ \sin \angle DAP\equal{}\sin \angle CAE$
hOW TO PROVE......$ \sin \angle CAF.\sin \angle BAE\equal{}\sin \angle FAD.\sin \angle BAP$....???????????

[quote="hell_ever"]Trying to use convers of menelaus in $ \Delta BCD$....
$ \frac {CF}{FD} \equal{} \frac {AC\sin \angle CAF}{AD\sin \angle FAD}$

$ \frac {BE}{EC} \equal{} \frac {AB\sin \angle BAE}{AC\sin \angle CAE}$

$ \frac {DP}{PB} \equal{} \frac {AD\sin \angle DAP}{AB\sin \angle BAP}$

$ \sin \angle DAP \equal{} \sin \angle CAE$
hOW TO PROVE......$ \sin \angle CAF.\sin \angle BAE \equal{} \sin \angle FAD.\sin \angle BAP$....???????????[/quote]
What is it? :mad:
If you don't solve the problem,why did you post a non-complete solution?
My proof use sine laws and Menelaus too,but it is complete.I'll post it later.

Very sorry.... :blush:

I want $ \frac {CF}{FD}\frac {BE}{EC}\frac {DP}{PB}\equal{}\minus{}1$

$ (ABC)$ denotes area of $ \Delta ABC$ ->

But $ \frac {CF}{FD}\equal{}\frac {(ACF)}{(AFD)}\equal{}\frac {AC.AF \sin \angle CAF}{AD.AF \sin \angle FAD}\equal{}\frac {AC \sin \angle CAF}{AD \sin \angle FAD}$

Similarly for the rest of the ratios.....$ AB,AC,AD$ get cancelled out.....
It can be easily proved that $ \angle DAP \equal{}  \angle CAE$ -> (draw the other tangent at C meeting BD at Q then $ \angle BCQ \equal{} \angle DAP$

But what about other angles??

Can u post your solution Erken??

Done  :lol:
Let $ AB\cap circle\equal{}X$
$ AXCF$ is rectangle.
$ \angle FAD \equal{} \angle XCB \equal{} \angle BAE$
$ \sin \angle BAP \equal{}  \sin(\angle CAP \plus{} \angle BAC) \equal{} \sin(90 \plus{} \angle BAC)
\equal{} \cos \angle BAC$
$ \sin \angle CAF \equal{} \sin (\angle XAF \minus{} \angle BAC)\equal{}\sin(90 \minus{} \angle BAC)
\equal{} \cos \angle BAC$

So converse of menelaus theorem in $ \Delta BCD$ gives the result.

[quote="Virgil Nicula"][color=darkred][b][u]An equivalent old problem.[/u][/b] Let $ ABCD$ be a rectangle with the circumcircle $ w = \mathcal C(O)$ . Let $ M\in (BC)$ , $ N\in (AD)$ be two points so that $ BM = DN$ . Prove that the lines $ MN$ , $ BP$ and the tangent in the point $ A$ to the circle $ w$ are concurrently.[/color][/quote]
[color=darkblue][b][u]Proof.[/u][/b] $ \{\begin{array}{c}
BM = ND = x\\\\
AD = BC = a\\\\
AB = CD = b\end{array}\|$ $ \implies$ $ \{\begin{array}{c}
MC = AN = a - x\\\\
NC = \sqrt {b^2 + x^2}\end{array}$ .

Thus, the power of $ N$ w.r.t. $ w$ $ \implies$ $ NP\cdot NC = NA\cdot ND$ $ \implies$

$ NP\cdot NC = x(a - x)$ $ \implies$  $ CN\cdot CP = CN\cdot (CN + NP) =$

$ b^2 + x^2 + x(a - x)$ $ \implies$ $ \frac {PN}{PC} = \frac {NP\cdot NC}{CN\cdot CP}$ $ \implies$ $ \boxed {\frac {PN}{PC} = \frac {x(a - x)}{b^2 + ax}}\ .$

$ \{\begin{array}{c}
R\in AA\cap MN\\\\
S\in AA\cap BC\end{array}\|$ $ \implies$ $ SB = \frac {b^2}{a}$ $ \implies$ $ SM = SB + BM = \frac {b^2}{a} + x$ $ \implies$

$ \frac {RM}{RN} = \frac {SM}{AN}$ $ \implies$ $ \boxed {\frac {RM}{RN} = \frac {b^2 + ax}{a(a - x)}}\ .$ Observe that $ \frac {RM}{RN}\cdot \frac {PN}{PC}\cdot\frac {BC}{BM} = 1$ , i.e.

(the [b]Menelaus[/b]' theorem) $ B\in PR\ .$ In conclusion, $ MN$ , $ BP$ , $ AA$ are concurrently.[/color]


linear algebra
matrix
search
linear algebra unsolved

Prove that any $ n x n$ matrix $ A$ can be presented as a product of two symmetric matrices $ BC$, where $ B$ is non-degenerate.

I think it is already somewhere on this forum.
You show that $ A$ is similar to its transpose via a symmetric matrix, i.e. $ A\equal{}BA^TB^{\minus{}1}$, where $ B$ is symmetric. http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1360490909&t=73309
Then it follows that $ AB\equal{}C$ is symmetric and yo uare done.