linear algebra
matrix
ratio
linear algebra unsolved

Dear friends,

How does the rank and condition number of a matrix get affected if its rows are removed one at a time or more at a time. Is there any reference for that?

Thanks,
bhupala

[quote="fedja"]The condition number makes sense only for square matrices so the second part of the question is senseless as written.[/quote]

I do not understand this. Condition number of a matrix is defined as the ratio of the highest singular value to that of lowest singular value. Singular values are defined for any matrix. Correct me if I am wrong!

bhupala.

[quote="t0rajir0u"]That's a computation that depends on the row you remove.  The rank either stays the same or decreases by $ 1$ depending on whether the row you're removing is in the span of the rest of the rows or not.  I don't know what you can say about the condition number.[/quote]

Thanks can you give some reference where I can get the information about this?

bhupala.

[quote="bhupala"]
I do not understand this. Condition number of a matrix is defined as the ratio of the highest singular value to that of lowest singular value. [/quote]
With this definition, you may be right. I am just used to $ \|A\|\cdot\|A^{\minus{}1}\|$ :). But some extra clarification is needed: I am also used to the idea that an $ m\times n$ matrix has $ n$ singular values, so, if the number $ m$ of rows is less than the number $ n$ of columns, the lowest singular value is just $ 0$ and the condition number is $ \plus{}\infty$. Do we agree on that?

If we do, then the only interesting case is $ m\ge n$ and here anything can happen: imagine the $ 2\times 2$ identity matrix to which one very small row is added. The singular values are then all about $ 1$ but not all equal, so the condition number is slightly above $ 1$. If you remove that small row, it drops to $ 1$, but if you remove one of the rows in the identity part, it goes up to a large number.


[quote="Kikicoco"]Prove that :
4 Rr 2 ( \sum cotg(A/2)) =abc.
when i found the sol for a homework i found this problem.[/quote].
And then i have a ineq :R 2 ( \sum sin2A.cosA) \geq S(ABC) :D

It is my solution but it isn't nice
R  2   (   \sum sin2A.cosA)  \geq  S(ABC)
<=> \sum ( cos  2 A)/(sinBsinC) \geq 1
We have
sinBsinC \leq cos 2 A/2
Then we have a  inequality which proved ,it is old problems.
Although it is hard to prove


Place three discs with radius $ r$ in a square with sides of length 1 so that the discs do not intersect: as on the figure. What is the greatest possible value of $ r$?

[img]http://i250.photobucket.com/albums/gg265/geometry101/NielsHenrikAbel1997Number8.jpg[/img]

A. $ \frac {1}{3}$
B. $ \frac {1}{4}$
C. $ \frac {\sqrt {2}}{6}$
D. $ 2 \sqrt {2} \minus{} \sqrt {6}$
E. $ \frac {\sqrt {2}}{1 \plus{} 2 \sqrt {2} \plus{} \sqrt {3}}$

Wow I found a really wierd solution heh.
Draw the 3 circles. It can be seen that all the circles' centers are r away from the edges of the square, and therefore are on a square with side length 1-2r. Joining them would create an equilateral triangle with side length 2r.
If we draw a square with the largest possible equilateral triangle inside it (all vertices on the square), we can set up some equations to find the side length. Let the side length of the square be s. Let the square be ABCD, and the equilateral triangle be AEF, with E on BC and F on CD. We can see that ADF is congruent to AEB, by the hypotenuse leg theorem. Therefore, FAD=EAB, and since EAF is 60 and BAD is 90, we know that FAD=EAB=15. we can then also see that ECF is a 45-45-90 triangle.
$ cos15 \equal{} \frac {\sqrt {2 \plus{} \sqrt {3}}}{2} \equal{} \frac {s}{AE}$
We have:
$ \frac {1}{2r} \minus{} 1 \equal{} \frac {2}{\sqrt {2 \plus{} \sqrt {3}}}$
$ \frac {1}{2r} \equal{} \frac {2}{\sqrt {2 \plus{} \sqrt {3}}} \plus{} 1$
$ \sqrt {2 \plus{} \sqrt {3}} \equal{} (4 \plus{} 2\sqrt {2 \plus{} \sqrt {3}})r$
$ r \equal{} \frac {\sqrt {2 \plus{} \sqrt {3}}}{4 \plus{} 2\sqrt {2 \plus{} \sqrt {3}}}$
uh oh that doesn't look like one of the choices..

I got $ \frac {\sqrt {2}}{1 \plus{} 2 \sqrt {2} \plus{} \sqrt {3}}$ as the answer.

First I had to assume that there are only two possible configurations in which the area of the circles could be largest, they are when 2 of the circles are side by side and the other is in between them kind of like in your picture, or when one of the circles is in the corner and the other 2 are taking up the remaining space.  To find what r equals in each of these you want to find how many radius lengths the side length of the square is equal to.

Since 2 circles are side by side in the first configuration the radius would be 1/4.

In the second configuration to find the radius, first I found the distance between the centers of two circles which equals 2r obviously, but since that line is at an angle of 45-30=15 degrees you have to use 2cos(15) which gives you the horizontal distance between the centers, which equals 1.93... then you add the two other radius and you get 3.93...

In other words the side length of the square equals 1 and (3.93...)r  so r equals .254... which after calculating your answers is the same as $ \frac {\sqrt {2}}{1 \plus{} 2 \sqrt {2} \plus{} \sqrt {3}}$.

Hmmm I think I did something wrong, an earlier step could've messed mine up. I think I saw that same denominator on my paper...

[hide="Official Solution by Website"]

I translated and used my logic best as I could..

Let the square be $ ABCD$ with $ AD \perp AB, AD \perp CD$. Now, then we know that $ AC \equal{} \sqrt{2}$. Let the center of the top left circle be $ E$ and the point of tangency between two other circles be $ F$. Note that $ E,F,C$ will be collinear.  Let the radius of each circle be $ r$.

$ AE \equal{} r \sqrt{2}$ (This can be done by dropping altitude from $ E$ to $ AD$.

From here, now, let the center of the bottom middle circle be $ P$. Then, because $ PF$, the radius, is tangent to $ AC$, we have $ \triangle EPF$ as a right triangle. Note that $ EP \equal{} 2r$ and $ PF \equal{} r$. So, $ EF \equal{} \sqrt{(2r)^2 \minus{} r^2} \equal{} r \sqrt{3}$.

Drop altitude from $ F$ to $ BC$. Draw a perpendicular line from $ P$ to this altitude. By 45-45-90 triangle, we know that the height of altitude above $ P$ is $ \frac{r}{\sqrt{2}}$. Since the total altitude is $ \plus{}r$, we know that this altitude has length of $ \frac{r}{\sqrt{2}} \plus{} r$. To find $ FC$, we multiply $ \sqrt{2}$ (another 45-45-90 right triangle). This yields the total length of $ r \plus{} \sqrt{2} r$.

So:

$ AC \equal{} r ( \sqrt{2} \plus{} \sqrt{3} \plus{} 1 \plus{} \sqrt{2}) \equal{} r (2 \sqrt{2} \plus{} \sqrt{3} \plus{} 1) \equal{} \sqrt{2} \rightarrow r \equal{} \frac{\sqrt{2}}{2 \sqrt{2} \plus{} \sqrt{3} \plus{} 1}$. E. [/hide]


algorithm
number theory
greatest common divisor
Euclidean algorithm

uh...euclidean algorithm?

Should be pretty obvious

[b]Generalization:[/b]  Show $(a^{m}-1, a^{n}-1) = a^{(m, n)}-1$.

There is a [b]very[/b] nice way to do this by Bezout's Theorem ;)

rough sketch of proof:

[hide]

Alright, using euclidean algorithm you get (wlog b>a)  So then define , and repeat this with  and , and you get some new  and so on. What other operation does this sort of represent? Eventually you must reduce to either some  such that , or you end up simply with  and then that's no fun cause your GCD is 1.

edit> now that I searched what bezouts theorem is, yea, it probably will help here to use that.

[/hide]

Yeah, the Euclidean algorithm-solution is quite well-known. t0rajir0u, could you post your proof with Bezout?

[hide="Solution"]
Let $d = (m, n)$.  $d | m, d | n \implies a^{d}-1 | a^{m}-1, a^{n}-1$ (elementary).  So $a^{d}-1$ is a common divisor.  Let $k$ be such that $k | a^{n}-1, k | a^{m}-1$ (in other words, $k$ is an arbitrary common divisor).  Then

$a^{n}\equiv a^{m}\equiv 1 \bmod k$

From Bezout's Theorem, $\exists x, y: mx+ny = (m, n)$.  Let $d = (m, n)$.  Then

$a^{mx+ny}\equiv a^{m(x+y)}\equiv a^{n(x+y)}\equiv 1 \bmod k$

Hence $k | a^{d}-1$.  Since every common divisor divides $a^{(m, n)}-1$, it must be the greatest common divisor.  QED. [/hide]


hmm, yeah; also, are you definitely coaching?

It appears as though we are going to Georgia after all  :(

[quote="MysticTerminator"]what's wrong with georgia?[/quote]

Primarily that it is not Penn State.  But it might be ok...

[quote="MysticTerminator"]what's wrong with not being penn state

I'm not quite sure what you guys are getting at here

like, I think y'all still get to go to an amusement park?[/quote]

Carowinds, I think.

BTW, who got an invitation, I did.

I think the top 20 did...

[quote="b-flat"][quote="PI-Dimension"]BTW, who got an invitation, I did.

I think the top 20 did...[/quote]

It is listed in the header of the email who the email was sent to.  Also, it was only sent to 18 people, so I doubt that it was top 20  :roll: Anyway, it was probably top 15 plus some other people...[/quote]

berman and Jeremy Hahn didn't get it. I think they were invited  :wink:

in the email I got, both berman and hahn were included in the "to:" part of the message, though the email specifically asked for hahn to reply (as though he hadn't been getting previous emails).

I have been invited to go too.  But probably I will have to miss the important practice session on the 17th when teams are assigned because I am going to Carnegie Hall to perform on the same day so probably I will have to be an alternate.  Has there been anyone on the B team who did not attend the practice session?


modular arithmetic
algebra
binomial theorem
number theory
number theory unsolved

A good idea: try to post each problem in a different thread.

2) If $ 2^n-1$ is prime then $ n$ itself is prime ($ 2^{pq}-1$ is multiple of $ 2^p-1$, it follows from the factoring $ x^q-1=(x-1)(x^{q-1}+\ldots+x+1)$).

If $ 2^n+1$ is prime then $ n=2^k$ for some k. The idea is the same as above.

(Attention: the converse is not necessarily true!)

So, the only way these terms are prime is $ 2^k$ be a prime. It occurs only if $ k=1$, and $ 2^2-1=3,2^2+1=5$ are both primes.

In fact, the primes of the format $ 2^n-1$ are called [b]Mersenne Primes[/b] and the primes of the format $ 2^n+1=2{}^2{}^k+1$ are called [b]Fermat Primes[/b].

1) by Fermat's Little Theorem, $ a\equiv a^p\equiv \minus{}b^p \equiv \minus{}b \pmod{p}$, so we may write $ b\equal{}\minus{}a\plus{}kp$. then by the binomial theorem, $ a^p\plus{}b^p\equal{}a^p\plus{}(\minus{}a\plus{}kp)^p\equal{}a^p\minus{}a^p\plus{}\binom{p}{1}a^{p\minus{}1}kp\plus{}\cdots\equal{}a^{p\minus{}1}kp^2\plus{}\cdots$, where the terms after the first are all clearly divisible by $ p^2$.

[quote="scoop"]1) if p prime odd, $ p \mid a^p \plus{} b^p$ show that $ p^2 \mid a^p \plus{} b^p$

I've gotten to the part that $ (a \plus{} b)^p \equiv a^p \plus{} b^p (mod\, p)$ but I am stuck there.

2) Show that $ 2^n \minus{} 1$ and $ 2^n \plus{} 1$ cannot be both primes at the same time.[/quote]

[color=blue]1)$ \bullet$ If $ p \mid a$ (or $ p \mid b$) $ \Longrightarrow$ $ p \mid a$ and $ p \mid b \Longrightarrow (a^p \plus{} b^p ) \vdots p^2$

$ \bullet$ If $ p \nmid a$ and $ p \nmid b \Longrightarrow a^p \equiv a \  (mod p)$ and $ b^p \equiv b  \ (mod p) \Longrightarrow (a \plus{} b) \equiv 0 (mod p)$

$ \Longrightarrow a^p \plus{} b^p \equal{} (a \plus{} b)(a^{(p \minus{} 1)} \minus{} a^{(p \minus{} 2)}.b \plus{} .... \minus{} a.b^{(p \minus{} 2)} \plus{} b^{(p \minus{} 1)}) \vdots p^2$

(because : $ (a \plus{} b) \equiv 0 \ (mod p)$ and $ (a^{(p \minus{} 1)} \minus{} a^{(p \minus{} 2)}.b \plus{} .... \minus{} a.b^{(p \minus{} 2)} \plus{} b^{(p \minus{} 1)})$

$ \equiv (a^{(p \minus{} 1)} \minus{} a^{(p \minus{} 2)}.( \minus{} a) \plus{} .... \minus{} a.( \minus{} a)^{(p \minus{} 2)} \plus{} ( \minus{} a)^{(p \minus{} 1)}) \equiv p.a^{(p \minus{} 1)} \equiv 0 \ (mod p) )$[/color]

[quote="Johann Peter Dirichlet"]A good idea: try to post each problem in a different thread.[/quote]
I guess you are correct  :blush: Also thanks for the extra info on those specific types of primes

Nice solution pluristiq. Maybe it confused me a bit that it has as hint to use
$ (a\plus{}b)^{p}\equiv a^{p}\plus{}b^{p}(mod\, p)$ to prove it.

[quote="x164"]The second part can be solved much more easy by calculating mod 3.  :D[/quote]
Yeap, you are correct, by trying some examples you can see that :)

hophinhan, nice and straightforward solution. I tried some factoring of the binomial myself by I got stuck.
My main error in my approach was that I was raising to square to get immediately the term $ p^2$


$k$ be a positive integer such that for all prime $p>k$ then $k$ is perfect square modulo $p$

was proved by vess that k is perfect square.

go to 'search' to find the link

Obviously, all perfect squares work, man, so you can rest assured that $k=1$ is not the only solution :). However, these are all the $k$'s: the perfect squares. In other words, if $k$ is not a perfect square, there are infinitely many primes $p$ s.t. $k$ is not a square modulo $p$. The proof is simple if we employ Dirichlet's Theorem.


http://www.freerice.com/index.php

New and improved, it now has Art, Chemistry symbols, Geography(country capitols), English Grammar and Vocabulary, foreign languages(French, Italian, German, Spanish),  and Math(multiplication tables).

Go for the multiplication tables if you want to donate the most rice. Go to something like Art or German if you actually want to learn. :wink:

"Sorry, we are unable to process rice donations so fast.

Please click BACK on your browser and try again."

 :(

I like the new subjects, I suppose, though they're a lot easier than the vocab.  60 levels of vocab, 2 of math....

The grammar is somewhat like the SAT writing grammar.

[quote="iin77"]"Sorry, we are unable to process rice donations so fast.

Please click BACK on your browser and try again."

 :( [/quote]

lol iin ur too fast. but that has happened to me before. cool they added stuff when i last donated it only had vocabulary.


geometry
circumcircle
geometric transformation
reflection
geometry proposed

let $ H$ be foot of a perpendicular line from $ O$ center of an arbitrary circle to an arbitrary line($ \ell$) .
and $ A$ and $ B$ be two points on $ \ell$ symmetric wrt $ H$ .let $ P$ and $ Q$ be two points on the cirlce let $ AQ$ and $ BP$ intersect the circle again at $ R$ and $ S$ . and $ PR$ and $ QS$ meet $ \ell$ at $ M$ and $ N$ .prove that $ HM \equal{} HN$

Mmm, and why is this nice problem without answer ?
 I think we must solve it.Probably a similar way works .

 Babis.

[quote][color=darkred][b][u]A similar problem.[/u][/b] Let $ ABCD$ be a cyclic quadrilateral with the circumcircle $ w = \mathcal C(O)$ . For a line $ d$

denote the point $ S\in d$ for which $ OS\perp d$ and $ \{\begin{array}{ccc} M_1\in AB\cap d & , & N_1\in CD\cap d\ . \\
\ M_2\in AD\cap d & , & N_2\in BC\cap d\ . \\
\ M_3\in AC\cap d & , & N_3\in BD\cap d\ .\end{array}$

Prove that $ SM_1 = SN_1$ $ \Longleftrightarrow$ $ SM_2 = SN_2$ $ \Longleftrightarrow$ $ SM_3 = SN_3$ .[/color][/quote]
[color=darkblue][b][u]Remark.[/u][/b] The particular case $ O\in d$ is the ex. $ 5.28^*$ pag. $ 51$ from the book

[b]Geometrie plana. Culegere de probleme , Editura GIL , Zalau , 2002 .[/b] [/color]

[hide="Proof."]
[color=darkblue]Denote the power $ p_w(X)$ of the point $ X$ w.r.t. the circle $ w$

and the intersection $ R\in AC\cap BD$ . Therefore $ \{\begin{array}{c} p_{w}(R) = RA\cdot RC = RB\cdot RD\ . \\
 \\
p_{w}(N_{3}) = N_{3}D\cdot N_{3}B\ . \\
 \\
p_{w}(M_{3}) = M_{3}A\cdot M_{3}C\ .\end{array}$

$ \blacktriangleright$ Apply the Menelaus' theorem to the transversals $ \overline{AM_{2}D}$, $ \overline{BN_{2}C}$ for the triangle $ M_{3}RN_{3}$ :

$ \frac {M_{2}M_{3}}{M_{2}N_{3}}\cdot \frac {DN_{3}}{DR}\cdot\frac {AR}{AM_{3}} = 1$ and $ \frac {N_{2}N_{3}}{N_{2}M_{3}}\cdot\frac {CM_{3}}{CR}\cdot\frac {BR}{BN_{3}} = 1$ $ \Longrightarrow$

$ \frac {N_{2}M_{3}}{N_{2}N_{3}}\cdot \frac {M_{2}M_{3}}{M_{2}N_{3}}\cdot\frac {RA}{RD}\cdot\frac {RC}{RB}\cdot\frac {N_{3}D}{M_{3}C}\cdot\frac {N_{3}B}{M_{3}A} = 1$, i.e. $ \boxed{\ \frac {M_{2}M_{3}}{M_{2}N_{3}}\cdot\frac {N_{2}M_{3}}{N_{2}N_{3}} = \frac {p_{w}(M_{3})}{p_{w}(N_{3})}\ }\ \ (1)\ .$

$ \blacktriangleright$ Apply the Menelaus' theorem to the transversals $ \overline{M_{1}AB}$, $ \overline{N_{1}CD}$ for the triangle $ M_{3}RN_{3}$ :

$ \frac {M_{1}M_{3}}{M_{1}N_{3}}\cdot\frac {BN_{3}}{BR}\cdot\frac {AR}{AM_{3}} = 1$ and $ \frac {N_{1}M_{3}}{N_{1}N_{3}}\cdot\frac {DN_{3}}{DR}\cdot\frac {CR}{CM_{3}} = 1$ $ \Longrightarrow$

$ \frac {M_{1}M_{3}}{M_{1}N_{3}}\cdot\frac {N_{1}M_{3}}{N_{1}N_{3}}\cdot\frac {RA}{RB}\cdot\frac {RC}{RD}\cdot\frac {N_{3}B}{M_{3}A}\cdot\frac {N_{3}D}{M_{3}C} = 1$, i.e. $ \boxed{\ \frac {M_{1}M_{3}}{M_{1}N_{3}}\cdot\frac {N_{1}M_{3}}{N_{1}N_{3}} = \frac {p_{w}(M_{3})}{p_{w}(N_{3})}\ }\ \ (2)\ .$

From the relations $ (1)$ , $ (2)$ obtain the conclusion of the our problem.[/color][/hide]
[quote] [color=darkred][b][u]An interesting particular case.[/u][/b] Let $ ABC$ be a triangle. Denote the diameter $ [AA']$

of the its circumcircle $ w = C(O)$  and  $ \{\begin{array}{c} P\in BC\cap AA' \\
\ M\in AB\cap PO \\
\ N\in AC\cap PO\end{array}$ . Prove that $ OM = ON$ .[/color][/quote]

an angle chasing proof.

Let's call the center with center $ O$ , $ \omega$.
let circumcirlce of$ \triangle BNS$ intersect $ \omega$ at $ R'$ .let $ NR'\cap \omega \equiv P'$ and $ BR'\cap \omega \equiv Q'$.
draw a line perpendicular to $ OH$ from $ S$ let it intersect $ \omega$ at $ S'$ we have:$ \angle P'R'S \equal{} \angle SBN \equal{} \angle PSS'$ hence $ P$ is reflection of $ P'$ wrt $ OH$ similary $ Q'$ is reflection of $ Q$ wrt $ OH$.
now notice that $ A$  and $ B$ are symmetric wrt $ OH$ and also $ Q$ and $ Q'$ hence $ R$ and $ R'$ are also symmetric.hence lines $ PR$ and $ P'R'$ are symeetric wrt $ OH$ hence $ HM \equal{} HN$.

I don't think I proved butterfly theorem as simple as its generalization.

How about usin this thorem:

[b]Theorem[/b] Let $ ABC$ be a triangle with circumcenter $ O$, and $ \ell$ a line. Let:
$ X \in \ell \cap BC$
$ Y \in \ell \cap CA$
$ Z \in \ell \cap AB$
Let $ P$ a point on $ l$ so that $ OP \bot \ell$.
If $ X', Y', Z'$ are the reflexions of $ X, Y, Z$ with respect to $ P$.
Then, the lines $ AX', BY', CZ'$ concur on the
circumcircle of triangle $ ABC$.

The  solution of the problem follows imediatelly from the theorem (Can you guess why?). I belive this is the solution Darij gave in one of his papers.

there is also a solution (two actually) using polarity:
[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=960553#960553[/url]


Find all pair of integers (x,y) satisfy: $ x^{2007}\equal{}y^{2007}\minus{}y^{1338}\minus{}y^{669}\plus{}2$

Try solving the equation: find all pair $ (x,y)\in Z^{2}$ such that $ x^{3}\equal{}y^{3}\minus{}y^{2}\minus{}y\plus{}2$

Just a little hint:

[hide]Prove that $ (y\minus{}1)^{3}< x^{3}\leq (y\plus{}3)^{3}$[/hide]

And at my problem, we can't apply this method to solve it! Two problems are distinc!

Why not?

Let $ t \equal{} x^{669}$ and $ k \equal{} y^{669}$ then we have to solve

$ t^{3}\equal{} k^{3}\minus{}k^{2}\minus{}k\plus{}2$

To Mods: Please lock this topic, this is a problem from M&Y, issue 9, and we can't discuss it now

See the solution here  :mad:

http://trungtuan.wordpress.com/2007/09/22/solution-of-problem-t6363-in-my/

Also, you can see here :

http://www.mathlinks.ro/Forum/viewtopic.php?p=934571#934571


geometry
geometric transformation
rotation
radical axis
geometry proposed

Let $ ABC$ be a triangle, and $ D$ be a point where incircle touches side $ BC$. $ M$ is midpoint of $ BC$, and $ K$ is a point on $ BC$ such that $ AK\perp BC$. Let $ D'$ be a point on $ BC$ such that $ \frac{D'M}{D'K}=\frac{DM}{DK}$. Define $ \omega_{a}$ to be circle with diameter $ DD'$. We define $ \omega_{B},\omega_{C}$ similarly. Prove that every two of these circles are tangent.

This problem is interesting. Just an observation:

Denote by $ \Omega_{9}$ the center of the nine-point  circle and by $ L$ the projection
of $ \Omega_{9}$ onto $ BC$.

Clearly $ (KMDD')$ form a harmonic division. Let $ T_{1}$ be the intersection of the circles
$ d_{KM}, d_{DD'}$ -  with diameters  $ KM$ and $ DD'$.  Let $ L'$ be the projection
of $ T_{1}$ onto $ BC$.  Then pairs $ (KM), (DD'),(L',\infty)$ are in the involution
on $ (BC)$ that is determined by the rotation of the angle $ 90^{\circ}$ around the vertex $ T_{1}$.
Thus we have $ (KMDL') \equal{} (KMD'\infty)$, which gives us $ KL' \equal{} ML'$, therefore $ L\equiv L'$.
Hence the radical axes of $ d_{KM}, d_{DD'}$ passes through $ \Omega_{9}$.
In other words $ L$ is the radical center of $ C(\Omega_{9},\frac{R}{2}),d_{KM}, d_{DD'}$.
Hope I calculated everything correct or again I did not understood involution right.

Hard and technical problem :D .  Did anyone solve this in the exam time, Omid?
[hide="Solution"]
Lemma1. Let $ C$ be a circle with center $ O$ and radius $ R$ and $ A,B$ two point collinear with $ O$. If $ AO*BO \equal{} R^{2}$ then any circle passing through $ A, B$ is perpendicular to $ C$.
Lemma2. If $ C_{1}, C_{2}$ are two tangent circles and $ C_{3}$ is circle perpendicular to both, then $ O_{3}$ center of $ C_{3}$ is on the common tangent of $ C_{1},C_{2}$.

Lemmas proof.[i] Exercise ![/i]

Let $ D''$ be the midpoint of $ DD'$. Since $ (D',K,D,M) \equal{}\minus{}1$ then $ D''K*D''M \equal{} (\frac{DD'}{2})^{2}$. So by lemma1, $ \omega_{a}$ is perpendicular to Nine-point circle. Also note that $ \omega_{a}$ is perpendicular to incircle. Hence by lemma2, its center, D'', is on the common tangent of incircle and nine-point circle.

Let the common point of incircle and nine-point be $ T$.

In a similar way, the centers of $ \omega_{b},\omega_{c}$ are also on the tangent line at $ T$. Now we have three circles with their centers all on a line and all 3 of them passing through a point on this line. Since they are not identical, they are all tangent to each other. [/hide]Any mistakes?

not hard at all.

let F be fuerbach point of $ \triangle ABC$ ,let a tanget at F to the incircle intersect BC at N, we have $ NK.NM\equal{}NF^{2}\equal{}ND^{2}$ hence N is midpoint of DD' ,hence a circle with diameter DD' pass through Fuerbach point.
the same is for $ \omega_{b},\omega_{c}$ and as proved their centers lie on a tangent line to incircle at F

as far as I know yeah almost everyone...  :D


Show that there exists an equiangular polygon with side lengths 1^2, 2^2,...,1990^2 (in some order)

[hide="Hint"]See second post from:
http://www.mathlinks.ro/Forum/viewtopic.php?p=366472#366472[/hide]


how do u get to the next level of videos

its so hard to do that if the problems are not what you learned??????

you just have to get more badges to get more videos!!!

please don't revive topics more than one month old


1)  (x-1)(x-3)(x+2)(x+4)+24
2)  9x^4-13x^2+4
please add method

thank you so much

for the second one
you have 9x^4-13x^2+4
you can let y = x^2
that way your equation will become 9y^2 - 13y +4 which is simple enough to solve

This method is probably easier than the other one posted:

[hide="Problem 2"]
$ 9x^4\minus{}13x^2\plus{}4\equal{}0$

$ 9x^4\minus{}9x^2\minus{}4x^2\plus{}4\equal{}0$

$ 9x^2(x^2\minus{}1)\minus{}4(x^2\minus{}1)\equal{}0$

$ (9x^2\minus{}4)(x^2\minus{}1)\equal{}0$

$ 9x^2\equal{}4$, $ x\equal{}\pm\frac{2}{3}$

$ x^2\equal{}1$, $ x\equal{}\pm 1$[/hide]

(Note: I didn't check for extraneous solutions)

1) Multiply the first and the third factors, and the second and the forth:
$ (x^2\plus{}x\minus{}2)(x^2\plus{}x\minus{}12)\plus{}24\equal{}0$
Now, let $ t\equal{}x^2\plus{}x\Rightarrow (t\minus{}2)(t\minus{}12)\plus{}24\equal{}0\Rightarrow t^2\minus{}14t\plus{}48\equal{}0$
with the solutions $ t_1\equal{}6, \ t_2\equal{}8$
You have to solve two quadratics:
$ x^2\plus{}x\minus{}6\equal{}0$ and $ x^2\plus{}x\minus{}8\equal{}0$


Opera will let you have the adless version of their (all-right) browser for FREE while supplies last. Go here and fill it out, and you've got a ad-free version of Opera:http://my.opera.com/community/party/

Thought ya'll might like to know.

Billy

I thought you meant the form of music.....  :(


AMC AIME trigonometry geometry circumcircle incenter inradius

Let the discussion begin.

I need to resolve #13, 14, and #15, so it'll be a bit before I post statistics. The scores looked pretty good though.

I really messed up this mock AIME, I solved 10 problems, and made 3 careless mistakes. There was only 2 problems that I didn't check, and I got both wrong :( . So tomorrow, I'm not gonna be lazy and I'll check EVERY question I solved.

I believe I got a 5.  I made one mistake on factoring and I was so close on the trig one but unfortunately gave up just one easy step too soon  :( .

Great problems. Let's still discuss this mock? (Post Lockdown)

[quote="darktreb"][quote]arguably better than the real one[/quote]

Definitely better than the real one in my opinion.[/quote]

Hehe and in some sense(not in score) I did better on this one than the real one.

Where can I find the anwers to this?  I did not PM you my answers, I am doing it right now.

K some solutions,

My favorite problem out of the first 10 is #7

[hide]#7. Consider triangle PAD, PBC. Drop altitude from P to AD, BC, since those altitudes are equal because P lies on the angle bisector, [PBC]/[PAD] = BC/AD.

[PBC]= 11*24sinCPB /2, [PAD] = 8*13 sinAPD /2, so

[PBC]/[PAD] = 11*24/(8*13) = 33/13=BC/AD. m+n=46[/hide]

#10

[hide]Subtract second and first equation we have
$(p-q)(p+q+r)=48$
subtrace third and second equation we have
$(q-r)(p+q+r)=48$

So we see that $p-q=q-r=d$, so p, q, r is in an arithmetic progression. p+q+r=3q

So $3qd=48$. Now rewrite the first equation as

$(q-d)^2 + q(q-d) + q^2 = 211$
$d^2 - 3qd + 3q^2= 211$
$d^2 + 3q^2 = 259$
$\frac{256}{q^2} + 3 q^2 = 259$
$256 + 3q^4 = 259q^2$
$(3q^2 - 256)(q^2 - 1) = 0$
$q=\frac{16}{\sqrt{3}}$, $d=\sqrt{3}$
so (13*16+16*19+13*19)/3 = 253[/hide]

Come on, solutions people. Every problem but the last two was solved, so I'll post solutions to them. Here's a solution to #14:

Answer: $392$. Let $M_A$, $M_B$, and $M_C$ denote the midpoints of the sides opposite $A$, $B$, and $C$ respectively, and write $P$ for the center of $\Omega$. Finally, let the tangents of $\omega_1$, $\omega_2$, and $\omega_3$ with $\Omega$ be denoted by $T_1$, $T_2$, and $T_3$.

Note that $M_AM_B = 7$, $M_BM_C = 5$, and $M_CM_A = 6$. Take points $U_1, U_2$, and $U_3$ on the extensions of $T_1M_A$, $T_2M_B$, and $T_3M_C$ past $M_A$, $M_B$, and $M_C$ respectively such that $T_iU_i = 9$ for $i = 1, 2, 3$. Because $T_1$ lies on $\omega_1$, $T_1M_A = 5$ so that $M_AU_1 = 4$. Analogously, $M_BU_2 = 3$ and $M_CU_3 = 2$.

Now consider circles $\omega_1'$, $\omega_2'$ and $\omega_3'$ centered at $M_A$, $M_B$, and $M_C$ and of radii $4$, $3$, and $2$ respectively. $\omega_1'$, $\omega_2'$, and $\omega_3'$ are mutually externally tangent and contain points $U_1, U_2$, and $U_3$ respectively. But the circumcenter of $U_1U_2U_3$ is $P$; ergo, the circumcircle $\Omega'$ of triangle $U_1U_2U_3$ is tangent to $\omega_1'$, $\omega_2'$, and $\omega_3'$. Moreover, $R_{\Omega}' = 9 - R_{\Omega}$. Applying the explicit form of the Descartes Circle Theorem for the outer circle, we find

$R_{\Omega}' = \left|\frac{2 \cdot 3 \cdot 4}{2\cdot 3 + 3\cdot 4 + 4\cdot 2 - 2\sqrt{2\cdot3\cdot 4 \cdot (2+3+4)}}\right|
= \frac{12 \cdot (13 + 6 \sqrt{6})}{216 - 169} = \frac{156+72\sqrt{6}}{47}$

from which $R_{\Omega} = 9 - R_{\Omega}' = \frac{267 - 72\sqrt{6}}{47}$. The answer is therefore $267 + 72 + 6 + 47 = 392$.

Here's a solution for #15:

Answer: $037$. A point that is equidistant from two lines lies on one of the two lines that bisect angles formed at the intersection of the two lines. Considering this fact, it is clear that $n = 4$, where $P_1$ is the incenter of $ABO$ (which we will denote $I$) and $P_2, P_3$, and $P_4$ are the three excenters. Let $r$ denote the inradius of $ABO$ and $r_1$, $r_2$, and $r_3$ the three exradii. We have

$r + r_1 + r_2 + r_3 = 4R + 2r = \frac{103}{3}$
$\frac{1}{r} + \frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = \frac{2}{r} = \frac{2}{3}$

This pair of equations is easily solved for $r = 3$ and $R = \frac{85}{12}$, where $R$ is the circumradius of $ABO$. By the Extended Law of Sines, $\sin{\angle{B}} = \frac{AO}{2R} = \frac{14}{2\cdot\frac{85}{12}} = \frac{84}{85}$. Pythagoras yields $\cos{\angle{B}} = \pm \frac{13}{85}$. Then $\cot{.5\angle{B}} = \frac{6}{7}$ or $\frac{7}{6}$.

Let the incircle of $ABO$ be tangent to $\overline{BO}$, $\overline{OA}$, and $\overline{AB}$ at $P$, $Q$, and $R$ respectively. Let $AQ = X$ and $QO = 14 - X$. Since $r = IQ = 3$, we have $\cot{.5\angle{A}} = \frac{X}{3}$ and $\cot{.5\angle{O}} = \frac{14-X}{3}$. Since $\cot{.5\angle{A}}\cot{.5\angle{B}}\cot{.5\angle{O}} = \cot{.5\angle{A}} + \cot{.5\angle{B}} + \cot{.5\angle{O}}$, we have

$X\cdot(14-X) = \frac{42 + 9\cot{.5\angle{B}}}{\cot{.5\angle{B}}} \le 49$

by substitution and AM-GM. $\cot{.5\angle{B}} = \frac{6}{7}$ leads to $58 \le 49$, impossible, so $\cot{.5\angle{B}} = \frac{7}{6}$, which leads to $X = 5$ or $9$. Thus, there are four possible $B$, each obtained by reflecting $B$ over $\overline{AO}$ and the perpendicular bisector of $\overline{AO}$. Since we are maximizing the sum $x+y$, we choose $X = 5$ and assume $y > 0$. Now, $BR = \frac{7}{2}$ and $RA = 5$. Since $\cos{\angle{A}} = 2\cos{.5\angle{A}}^2 - 1 = \frac{25}{17} - 1 = \frac{8}{17}$, it must be that $x = 14 - \frac{17}{2} \cdot \frac{8}{17} = 10$ and $y = \frac{17}{2}\cdot\frac{15}{17} = \frac{15}{2}$, which gives $x+y = \frac{35}{2}$ and an answer of $35 + 2 = 37$.

[quote="darktreb"]Hey Thomas, how did you do on the AIME? Certainly seems like it was well within your range of ability to get a 15. (But of course, things don't always work out the way we plan :))[/quote]

I managed to get myself a 12 on this one. (I must've gotten stupider since last year; both AMC and AIME performance went down!) At least it'll be sufficient for USAMO qualification purposes.

Interesting lesson I learned on this one: #15 isn't necessarily very hard. That means you don't need to apply exotic formulae to kill it. When I determined that $G$ was on the incircle, I immediately decided the solution must use the identitiy

$IG^2 = r^2 + \frac{1}{36}\left(5(a^2+b^2+c^2) - 6(ab+bc+ca)\right)$

which caused me to not even look for simple, more elegant solutions like power of a point.

K, I'll help write up some solutions:

1.)

Call the dimensions of the box $x$, $y$, and $z$.  So from the problem we have that:

\begin{eqnarray*}
 x^2+y^2+z^2 &=& 47^2\\
 y-z &=& \sqrt{2005}\\
 x &=& 2
\end{eqnarray*}

Pluggin $x=2$ into the first we have $y^2+z^2=2205.$  Squaring the second equation we get \[y^2+z^2-2yz = 2005\Rightarrow yz=100.\]  And thus the volume is $xyz=\fbox{200}.$

5.)   The first thing is to notice how much like binomial coefficients these numbers look like.  Thus we can write the expression $\frac{1}{0!10!} + \frac{1}{1!9!} + \frac{1}{2!8!} + \frac{1}{3!7!} + \frac{1}{4!6!} + \frac{1}{5!5!}$ as:

\[\frac{1}{10!}\left(\binom{10}{0}+\binom{10}{9}+\binom{10}{8}+\binom{10}{7}+\binom{10}{6}+\binom{10}{5}\right).\]

Call the sum inside the parentheses $S$.   Now notice that $2S=\left(\binom{10}{0}+\binom{10}{1}+\cdots+\binom{10}{10}\right)+\binom{10}{5}.$  Using a well known combinatorial identity we get $2S=2^{10}+\binom{10}{5}=1276\Rightarrow S=638.$ Thus we get $\frac{638}{10!}=\frac{2\cdot 11\cdot 29}{10!}.$  And our answer is $11+29+2+3+5+7=057.$

#6 is a bit tough to show completely rigorously. It is a particular instance of the fact $(F_p,F_q) = F_{(p,q)}$. Consider the Fibonacci sequence started from $F_0 = 0$ and $F_1 = 1$. By the Euclidean algorithm, we have $(F_{m+1}, F_m) = (F_m, F_{m-1}) = \cdots = (F_2, F_1) = 1$. But if we take the Fibonacci numbers modulo an arbitrary integer $n$, the sequence must be 0 exactly every $k_n$ terms. This is because the Fibonacci sequence depends on only the previous two terms, which assume less than $n^2$ values - thus it repeats. Since $F_0 \equiv 0 \pmod n$, we must have some first non-zero term $F_{k_n}$ divisble by $n$. But after $F_{k_n}$, the sequence (under the modulo) begins with the residue of $F_{k_n - 1} \pmod n$, which is relatively prime to $F_{k_n}$ and thus relatively prime to $n$. This establishes the claim.

Now, $n | F_p$ if and only if $k_n | p$. This means that $n | F_{1989}$ AND $n | F_{1495}$ if and only if $k_n | 1989$ AND $k_n | 1495$, or equivalently, $k_n | (1989, 1495) = 13$, which in turn happens if and only if $n | F_{13} = 233$. Thus, the answer is the 13th Fibonacci number.

For #8 I divided both sides by 2 and replaced $\sqrt{2}/2$ with $\sin 45$. It made things come out almost immediately.

I'll edit this later with a full solution when I have the time.

Note that 2004, 3507, and 4676 are all mulitples of 167, so we scale the triangle by a factor of 167 to get sides of 12, 21, and 28 respectively. It's easy to see that the length of the tangent line for a pair a pair of tangent spheres is $\sqrt{(r_1+r_2)^2-(r_1-r_2)^2}=2\sqrt{r_1r_2}$ where $r_1$ and $r_2$ are the radii of the two spheres. Using this fact we find that the lengths of the sides of triangle ABC are $2\sqrt{12\cdot21}, 2\sqrt{12\cdot28}, 2\sqrt{21\cdot28}.$ Also notice that $12\cdot21+12\cdot28=21\cdot28=588$ thus ABC is a right triangle and its area is $\frac{2\sqrt{12\cdot21}\cdot2\sqrt{12\cdot28}}{2}=336\sqrt{3}$ Now multiply by $167^2$ to get $9370704\sqrt{3}$ and the answer is $704+3=707$

Darktreb, what was your solution for #8?

Let's see I didn't even take this AIME but I can make something up.

$\frac{2\sin 16 \cos 32}{\sin 29}=\sqrt{2}-2x$
$\frac{\sin 16 \sin 58}{\sin 29} = \sin 45 - x$
$2 \sin 16 \cos 29 = \sin 45 - x$
$x = \sin 45 - 2 \sin 16 \cos 29$
$x = \sin (16 + 29) - 2 \sin 16 \cos 29$
$x = \sin 16 \cos 29 + \cos 16 \sin 29 - 2 \sin 16 \cos 29$
$x = \sin 29 \cos 16 - \sin 16 \cos 29 = \sin 13$

WOOHOO I AM A WOOTLER

Edit--Oh yeah it's also time for Silverfalcon to swoop down and gobble I mean  :o :o :o  hungry this guy for digging up an old thread.

[quote="probability1.01"]
Edit--Oh yeah it's also time for Silverfalcon to swoop down and gobble I mean  :o :o :o  hungry this guy for digging up an old thread.[/quote]

Haha..  :D

Nah, as long as it relates to the topic (asking for solution or providing solution), it's fine. I just don't like people reviving old threads for like "LOL, I took this test and got ###."

Sorry for the revive, but can someone point me to a solution of number 11? I'm feeling Newton sums but idk how to proceed.

Minor error: the problem states $x_1^3+x_2^3+x_3^3=-24$ instead. This would make the polynomial $x^3-8x+8$. Considering, the roots, it does not make a difference, really. Alternatively, plugging in any of the roots of $x^3-8x+8$ into $\gamma^6-15\gamma^4+\gamma^3+56\gamma^2$, yields 56, so I am inclined to think that something else interesting is going on.

Excuse me where can I get the Mock problems?

This thread was made before I was born :o

[quote=P_Groudon]
[hide=Solution 11]
Answer: $056$

By squaring the first equation:
$$x_1^2 + x_2^2 + x_3^2 + 2(x_1x_2 + x_2x_3 + x_1x_3) = 0 \Rightarrow x_1x_2 + x_2x_3 + x_1x_3 = -8$$

Let $P(x)$ have roots $x_1, x_2,$ and $x_3$. By Vieta's formulas, $P(x) = x^3 - 8x - c$, where $c = x_1x_2x_3$. We know that $P(x_1) = P(x_2) = P(x_3) = 0$.

Therefore,
$$x_1^3 - 8x_1 - c = 0$$
$$x_2^3 - 8x_2 - c = 0$$
$$x_3^3 - 8x_3 - c = 0$$
Adding these three equations together:
$$(x_1^3 + x_2^3 + x_3^3) - 8(x_1 + x_2 + x_3) - 3c = 0 \Rightarrow c = -8$$
Therefore, $P(x) = x^3 - 8x + 8$. By the Rational Root Theorem, we can find $-2$ as a root of $P(x)$. So $P(x) = (x+2)(x^2 - 2x - 4)$. The other roots are $1 \pm \sqrt{5}$. The root with the smallest absolute value is $1 - \sqrt{5}$ (which is negative), so $\gamma = \sqrt{5} - 1$. Let $Q(x) = x^6 - 15x^4 + x^3 + 56x^2$. We seek the value of $Q(\gamma)$. We attempt to divide $Q(x)$ by $P(x)$. Using polynomial division, we have $Q(x) = (x^3-7x-7)(x^3-8x+8) + 56$. We know that $\gamma^3 - 8\gamma + 8 = 0$, since $\gamma$ is a root of $P(x)$.

Therefore, $Q(\gamma) = 56$.
[/hide]
[/quote]

I think the polynomial factors as $(x - 2)(x^2 + 2x - 4)$, with roots $2, -1 \pm \sqrt{5}$. However, this still results in $\gamma = \sqrt{5} - 1$.

What's the difficulty of this mock?

I think it's on the easy side compared to recent tests.

Been a while, but I changed the solutions in post 30 to be in a PDF document instead of a bunch of hide tags. Also, upgraded several of the solutions with diagrams and such.


When drawing the initiation and propagation steps for this reaction, does it matter which hydrogen I replace with Cl? B/c cyclooctane is symmetrical and all C-H bonds have secondary hydrogens, it shouldn't matter right?

How can I find delta $ H$ (STP) for each of the 2 steps?

How can I calculate heat of formation of chlorocyclooctane?

Yes, all carbons can be considered equivalent.

[quote="Bythecliff"]How can I calculate heat of formation of chlorocyclooctane?[/quote]

From the heat of formation of cyclooctane.


integration
calculus
function
real analysis
real analysis unsolved

Let $ f > 0$ on $ \mathbb{R^{d}}$ and $ E_{k} = \{x : f(x) > 2^{k}\}$ then I know that $ f$ is integrable iff $ \sum_{k = - \infty}^{\infty}2^{k}m(E_{k}) < \infty$

Given $ f$ is non-negative and satisfies $ m(\{x : f(x) > t\})\leq \frac {1}{1 + t^{2}}$ for any $ t > 0$
The question is : determine those values $ p$, $ 1\leq p < \infty$ for which $ f\in L^{p}(\mathbb{R}^{d})$ and find a minimum value of $ p$ for which $ f$ may fail to be in $ L^{p}$

The following is my though :

I have $ m(\{x : |f(x)|^{p} > t^{p}\})\leq \frac {1}{1 + t^{2}}$ for any $ t > 0$

replace $ t^{p} = 2^{k}$  or $ t = 2^{\frac {k}{p}}$ I have

$ m(\{x : |f(x)|^{p} > 2^{k}\})\leq \frac {1}{1 + 2^{\frac {2k}{p}}}$

Now let $ E_{k} = \{x : |f(x)|^{p} > 2^{k}\}$ then I have

$ \sum_{k = - \infty}^{\infty}2^{k}m(E_{k})\leq \sum_{k = - \infty}^{\infty}2^{k}\frac {1}{1 + 2^{\frac {2k}{p}}}\leq \sum_{k = - \infty}^{\infty}2^{k}\frac {1}{2^{\frac {2k}{p}}} = \sum_{k = - \infty}^{ - 1}2^{k}\frac {1}{2^{\frac {2k}{p}}} + \sum_{k = 0}^{\infty}2^{k}\frac {1}{2^{\frac {2k}{p}}}$

for the first series $ \sum_{k = - \infty}^{ - 1}2^{k}\frac {1}{2^{\frac {2k}{p}}} = \sum_{k = - \infty}^{ - 1}2^{k - \frac {2k}{p}} = \sum_{k = - \infty}^{ - 1}2^{k(1 - \frac {2}{p})}$, therefore the series converges iff $ 1 - \frac {2}{p} > 0$  or $ p > 2$

for the second series $ \sum_{k = 0}^{\infty}2^{k}\frac {1}{2^{\frac {2k}{p}}} = \sum_{k = 0}^{\infty}2^{k(1 - \frac {2}{p})}$ therefore the series converges iff $ 1 - \frac {2}{p} < 0$  or $ p < 2$

Could you please let me know where I got wrong.? The answer is $ p > 2$ but I don't know where I am wrong.

thanks so much

The correct answer is that this is in $ L^p$ for $ p < 2.$

As for your sum: you analyzed the situation for $ k > 0$ correctly.

But let's take another look at the situation for $ k < 0.$ We now have that $ 2^{\frac {2k}p}$ is small, not large, and $ 1 \plus{} 2^{\frac {2k}p}$ might as well be $ 1.$

So $ \sum_{k \equal{} \minus{} \infty}^{ \minus{} 1}\frac {2^k}{1 \plus{} 2^{\frac {2k}p}}\le\sum_{k \equal{} \minus{} \infty}^{ \minus{} 1}2^k,$ which is finite no matter what $ p$ is.

Hence the answer is the answer you get from the $ k > 0$ part of the sum.

Here are three useful equivalent formulations you can use.

Define $ \lambda_f(t) \equal{} m\left(\{|f(x)| > t\}\right).$

We have that $ f\in L^p$ iff any of the following are true:

$ (1)\qquad \sum_{k \equal{} \minus{} \infty}^{\infty}2^{kp}\lambda_f(k) < \infty$

$ (2)\qquad \int_0^{\infty}t^p( \minus{} d\lambda_f(t)) < \infty$

$ (3)\qquad \int_0^{\infty}pt^{p \minus{} 1}\lambda(t)\,dt < \infty$

Both (2) and (3) compute $ \|f\|_p^p$ exactly. The sum in (1) isn't an exact computation but is equivalent to that norm in the sense of being between two fixed constant multiples. The integral in (2) is a Stieltjes integral.

(3) is often the most useful of these. In this particular case, we are left looking at $ p\int_0^{\infty}\frac {t^{p \minus{} 1}}{1 \plus{} t^2}\,dt,$ which converges for $ p < 2.$

Note that the condition you have on $ \lambda_f(t),$ if you look at the limit as $ t\to 0,$ forces $ f$ to be essentially supported on a set of finite measure (measure no greater than 1). Over a set of finite measure, it's harder for a function to belong to $ L^p$ for larger $ p.$ Hence, the set of $ p$ for which a particular function does belong to $ L^p$ must be of the form $ p < a$ or $ p\le a.$


trigonometry
inequalities
inequalities unsolved

Let ABC be a triangle with $\hat{A}\geq 90^{o}$. Prove the following inequalities:
$\sin{\frac{A}{2}}+\sin{\frac{B}{2}}+\sin{\frac{C}{2}}\leq \frac{\sqrt{2}}{2}+2\sin{\frac{\pi}{8}}$
$\cos{\frac{A}{2}}+\cos{\frac{B}{2}}+\cos{\frac{C}{2}}\leq \frac{\sqrt{2}}{2}+2\cos{\frac{\pi}{8}}$

No one care about this problem?  :maybe:

[color=red][Moved to the Inequalities section. It will probably receive more attention there. -mathmanman][/color]


floor function
combinatorics proposed
combinatorics

Supposed you have 4 red balls, 4 green balls and 4 blue balls,

You want to arrange them in a straight line such that no 2 adjacent balls are of the same colour.

(i) How many ways are there such that this is possible?
(ii) Is there a generalisation for n balls of a same colour?

have been looking at this for a while.. please help :-)

Thanks

The answer to (ii) is

$2\sum_{k=0}^{\lfloor n/2\rfloor} {n-1\choose k}\left[ {n-1\choose k}{2n+1-2k\choose n+1} + {n-1\choose k+1}{2n-2k\choose n+1}\right]$

In particular, for n=4 the number of ways is 1092.


[color=purple] Solve the following equation [/color]
  $(5+\sqrt{21})^{x}+(5-\sqrt{21})^{x}=8$

Bad title! Locked! http://www.mathlinks.ro/Forum/viewtopic.php?t=144403


Let z be a complex number (different from 0). Say, for each of the complex numbers A, B and C, if it's real or imaginary.

A = $z^2 + \bar{z}^2$

B = $\frac {z-\bar{z}}{z+\bar{z}}$

C = $\frac {z^2 - \bar{z}^2}{z\bar{z} + 3}$

Hello, mathmanman. To solve each of your problems, substitute x+iy for z and x-iy for the conjugate of z and simplify the expression; the answer is real of the coefficient of i equals zero and complex if the coefficient is nonzero.

Hmm ok... isn't there a simpler method?

edit : nvm, this way is simple :)

Note also that if $w \in C$ then we have the following equivalences

$w$ is real iff $w=\bar{w}$  and

$w$ is imaginary (e.g $2i$) iff $w=-\bar{w}$

you can prove these statements for practice!!

Thanks :)
Those are pretty trivial, aren't they?  :?

First one : z = x + iy
$z = \bar{z}$ <==> x+iy = x-iy <==> y = -y <==> y = 0 :)


Suppose that $A$ and $B$ are invertible matrices. Find the inverse of $I+AB$.

There's no reason that $I+AB$ has to be invertible - and if it is, there's not going to be a nice formula.

Just to make a silly example, let $A=I$ and $B=-I.$

I think I found an expression assuming that it is invertible.

If $AB$ has norm less than one, you could do something random like $I-AB+ABAB-ABABAB+...$, but I'm not sure how to approach it in general.

[quote="drunner2007"]Well I used elementary matrices to rewrite it as a single product[/quote]
That's not a very specific answer. If that's all you have to say, then we'll just assume that whatever you did was wrong, and we'll move on to some other problem.

I noticed that my first method was incorrect.

Well I implicitly assumed that $A$ and $B$ are $n$x$n$ square matrices. Then adding $I$ to the sum $A+B$ is equivalent to multiplying $A+B$ on the left by $E_{n}E_{n-1}\ldots E_{1}$, where $E_{i}$ is the elementary matrix arrived at by multiplying the $i$th row of $I$ by $(a_{ii}+b_{ii}+1)/(a_{ii}+b_{ii})$, where $a_{ij}$ and $b_{ij}$ denote the $i,j$-entries of matrices $A$ and $B$, respectively. However, I notice now that this method is wrong.

So then maybe we can write $A+B$ as $[a_{1}+b_{1}\hspace{.1in}a_{2}+b_{2}\ldots a_{n}+b_{n}]$ where $a_{k}$ and $b_{k}$ denote the $k$th column of $A$ and $B$ respectively. Then we can write $I+AB$ as $[E_{1}(a_{1}+b_{1})\hspace{.1in}E_{2}(a_{2}+b_{2})\ldots E_{n}(a_{n}+b_{n})]$ were $E_{i}$ is the elementary matrix arrived at by multiplying the $i$th row $I$ by $(a_{ii}+b_{ii}+1)/(a_{ii}+b_{ii})$. Now if we consider the matrix $[E_{1}(a_{1}+b_{1})\hspace{.1in}E_{2}(a_{2}+b_{2})\ldots E_{n}(a_{n}+b_{n}) \hspace{.1in}I]$ and row reduce it we should find the inverse on the left side. Is there a nice way to find what the right side will look like once we reduce the left side?

The only idea is that there is no nice formula and your method leads nowhere.

[quote]Then we can write $I+AB$ as $[E_{1}(a_{1}+b_{1})\hspace{.1in}E_{2}(a_{2}+b_{2})\ldots E_{n}(a_{n}+b_{n})]$ were $E_{i}$ is the elementary matrix arrived at by multiplying the ith row $I$ by $(a_{ii}+b_{ii}+1)/(a_{ii}+b_{ii}).$[/quote]
As far as I can tell, this is wrong. And even if it weren't wrong, left-multiplying each column of a matrix by a different matrix isn't the same as left-multiplying by a matrix; as such the result would have nothing to do with finding inverses.

[quote="Kent Merryfield"]left-multiplying each column of a matrix by a different matrix isn't the same as left-multiplying by a matrix [/quote]

and where exactly am i left multiplying $AB$ by a matrix?


floor function
quadratics
number theory
relatively prime
number theory proposed

Let $p$ and $q$ be positive odd numbers, write
\[p = 2a+1\qquad q = 2b+1\]
and suppose that $\gcd (p,q)=1$. Let $k$ be the number of lattice points $(x,y)$ such that
\[0<x\qquad 0<y\qquad 2(xp+yq) < pq\]
Show that $ab-k$ is an even number.

Another way to state the problem: Let $p$ and $q$ be relatively prime positive odd integers. In $R^{2}$ consider the triangle whose vertices are
\[(p/2,0)\qquad(0,q/2)\qquad(p/2,q/2)\]
Show that there are an even number of lattice points inside that triangle.

We want to evaluate
$S=\sum_{k=1}^{b}\left\lfloor\frac{q+2kp}{2q}\right\rfloor = \frac{b}{2}+\frac{b(b+1)}2\frac{p}{q}-\frac1{2q}\sum (q+2kp\mod 2q).$

For $1\leq k\leq b$, let $r_{k}=(kp\mod q)$, so $0<r_{k}<q$.  Then

$(q+2kp\mod 2q) = \left\{\begin{array}{cc}2r_{k}+q,&r_{k}\leq b\\2r_{k}-q,&r_{k}>b.\end{array}\right.$

So if $\alpha$ is the number of $k$ for which $r_{k}>b$, then
$S=\frac1{2q}\left(qb+pb(b+1)-2\sum r_{k}-q(b-2\alpha)\right)$.

If for each $k$ with $r_{k}>b$, we were to replace $r_{k}$ with $q-r_{k}$, we would have the numbers 1,...,$b$, which sum to $b(b+1)/2$.  So $\sum r_{k}\equiv b(b+1)/2+\alpha$ mod 2.  So mod 4, we have

$2S \equiv qb+pb(b+1)-(b(b+1)+2\alpha)-q(b-2\alpha) \equiv 0$,

so $S$ is even.  This reminds me of a proof of quadratic reciprocity (note that $\left(\frac{p}{q}\right)=(-1)^\alpha$.)

Congratulations, Eigenray.
This proposition (which took me two weeks to prove) is basically the same thing as quadratic reciprocity, but with a cuter statement, weaker hypotheses, and a stronger conclusion! I'm gradually writing up the whole story for the web. For now, here's what I mean by a stronger conclusion:
Let p and q be distinct odd primes. Define
u = the number of x in [1,(p-1)/2] such that qx is not in that interval mod p
v = the number of x in [1,(q-1)/2] such that px is not in that interval mod q.
Quadratic reciprocity says
(p-1)(q-1)/4 + u + v = 0 mod 2.
But in fact
(p-1)(q-1)/4 + u + v = 0 mod 4.


Find $ x,y \in \mathbb{R^*}$ so that:

$ \sqrt{\frac{1}{2} \cdot (x\minus{}y)^2\minus{}(x\minus{}y)^4}\equal{}y^2\minus{}2x^2$.

and

$ y \ge 4x^4\plus{}4yx^2\plus{}\frac{1}{2}.$

my english is bad, however i hope u can understand me.
here is my solution:
since $ \sqrt{\frac{1}{2}\cdot (x\minus{}y)^{2}\minus{}(x\minus{}y)^{4}}\equal{}y^{2}\minus{}2x^{2}$  follow cosi'sinequality we have y^2<=2x^2+1/4 so that y^2<=1/4(*). from
$ y\ge 4x^{4}\plus{}4yx^{2}\plus{}\frac{1}{2}.$ we have y(1-4x^2)>=1/2+4x^4(1)
if 1-4x^2>=0, (1) <=> y>=1/2 so y=1/2.
if 1-4x^2<0(x^2>=1/4), (1)<=> y<1/2 but since first equation we have y^2>=2x^2>=1/2 combine with (*) so this is unreal.
so the answer is x=0 and y=1/2


Note that if the product of any two distinct members of {1,16,27} is increased by 9, the result is the perfect square of an integer. Find the unique positive integer $n$ for which $n+9,16n+9,27n+9$ are also perfect squares.


LaTeX
inequalities proposed
inequalities

[b][u]The author of this post is: darij grinberg,[/u][/b]
_____________________________________________________________

[quote="Gil"]Let a,b,c be real numbers. Prove that:
a) |a+b|1, |a|1, |b|1 ==> |a|+|b|+|a+b|2.
b) |a+b+c|1, |a|1, |b|1, |c|1 ==> |a+b|+|b+c|+|c+a|2.[/quote]

Neet indeed... and simple!

Part [b]a)[/b] follows from part [b]b)[/b] with c = 0. Hence, in the following, I will only prove part [b]b)[/b].

Let x = b + c, y = c + a and z = a + b. Then, we have to prove | x | + | y | + | z | $\leq$ 2.

But since x = b + c, y = c + a and z = a + b, we have x + y + z = 2 (a + b + c), and thus | a + b + c | $\leq$ 1 yields | x + y + z | $\leq$ 2. Also, y + z - x = 2 a, and thus | a | $\leq$ 1 yields | y + z - x | $\leq$ 2. Similarly, | z + x - y | $\leq$ 2 and | x + y - z | $\leq$ 2. And hence it remains to solve the following problem:

If x, y, z are three real numbers such that | x + y + z | $\leq$ 2, | y + z - x | $\leq$ 2, | z + x - y | $\leq$ 2 and | x + y - z | $\leq$ 2, then we have | x | + | y | + | z | $\leq$ 2.

This trivially follows from another assertion:

For any three real numbers x, y, z, the number | x | + | y | + | z | equals one of the numbers | x + y + z |, | y + z - x |, | z + x - y | and | x + y - z |.

Remains to prove this assertion. Well, WLOG we can assume that at most one of the numbers x, y, z is negative (else, we can replace x, y, z by -x, -y, -z, and nothing really changes). Hence, either all numbers x, y, z are nonnegative, or exactly one of them is negative. If all of the numbers x, y, z are nonnegative, then simply | x | + | y | + | z | = x + y + z = | x + y + z |, and we are done. Hence, the only case to consider is when exactly one of the numbers x, y, z is negative. Let this number be x. Then, | x | + | y | + | z | = -x + y + z = | y + z - x |. And we are done again.

$\blacksquare$

darij

The problem should be:

Let a, b, c be real numbers. Prove that:
[b]a)[/b]
|a + b| $\leq$ 1, |a| $\leq$ 1, |b| $\leq$ 1 ==> |a| + |b| + |a + b| $\leq$ 2.
[b]b)[/b]
|a + b + c| $\leq$ 1, |a| $\leq$ 1, |b| $\leq$ 1, |c| $\leq$ 1 ==> |a + b| + |b + c| + |c + a| $\leq$ 2.

  darij

Oops, sorry.  I've corrected the initial post.

I remember being so glad that Darij's post made so little use of $\LaTeX$ (because it was necessary to retype all the $\LaTeX$ equations when reposting the threads after the hacker attack that I forgot to check for the symbols in the original post. :blush:


There is a triangle whose sides are all integers.
The perimeter of the triangle is twice the area of the triangle.

What are the lengths of the sides of the triangle?

[quote="sen"]by guessing, a 3 4 5 right angle triangle will do the trick.

anybody with proof this only one?

$a+b+c=2\sqrt{(\frac{b+c-a}{2})(\frac{a+c-b}{2})(\frac{a+b-c}{2})(\frac{a+b+c}{2})}$

$2(a+b+c)^2=2(a^2b^2+a^2c^2+b^2c^2)-4(a^4+b^4+c^4)$

any use?[/quote]

how about using AM-GM? will that be useful?


[quote="chess64"]Is it possible to divide a polynomial like $AB - 9AC + BC$ by $A-B$ using long division or another method?[/quote]

Are AB digits or is it product?

You could, but it would be kind of long considering the C is not in the divisor.[/hide]

The C makes it kind of odd, but I'm pretty sure you still can... and I can guarantee that it's possible to divide something like A2 - 9AB + B2 by A - B.


Find all solutions in integers of the equation

x^3 + ( x+1)^3+ ( x+2)^3 +……………+ (x+7)^3 = y^3       :rotfl:

[size=150][color=red]mathematicien@naseej.com[/color][/size]

good works  [size=150]enescu[/size]    :10:

I can't see at the moment what the property is  ;)

Make the substitution x -> x - 3 and the left member will be written as:
L = Σi=1,3[(x-i)3 + (x+i)3] + x3 + (x+4)3 = 8 * x3 + 12*x2 + 132*x + 64 = (2*x + 1)3 + 63*(2*x + 1) = a3 + 63*a,
where a = 2*x + 1 (*).
So we have the ecuation a * (a2 + 63) = y3.
(a = 0, y = 0) is not a solution for this ecuation [it leeds to x = -1/2 not integer].
Then (a,y) with a < 0 is a solution if and only if (-a, -y) is solution.
We suppose a > 0.

For a <> 1, 2, 3 we have: a3 < a3 + 63*a = y3< (a + 3)3.
[the first is obvious and the second turns into: 0 < 9*a2 - 36*a + 27, 0 < a2 - 4*a + 3, 0 < (a - 1)*(a - 3)].
So y = a + 1 or y = a + 2. Both are not good.

Remains:
a = 1 => by (*) x = 0, so in the end x = -3, y = 4.
a = 2 => by (*) x is not integer.
a = 3 => by (*) x = 1, so in the end x = -2, y = 6.

Also the negative ones:
a = -1 => by (*) x = -1, so in the end x = -4, y = -4.
a = -2 => by (*) x is not integer.
a = -3 => by (*) x = -2, so in the end x = -5, y = -6.


geometry
perimeter
inequalities
triangle inequality

How many non-congruent obtuse triangles are there with integer sides lengths and perimeter 11? aops-1, pg 130

I don't get the first part of the solution in the book, how did they find all of the triangles with perimeter 11 in the first place? I don't want to think that I have to count the hard way!

Well, remember that if a triangle is obtuse then $a^2+b^2<c^2$.   They may have used that.

[hide="How to find all triangles of perimeter 11"]
The Triangle inequality states that, for any three sides of a triangle, $a + b > c$.  This is pretty easy to see geometrically - try to construct, for example, a triangle of side lengths $1, 1, 2$.  You'll quickly see that this is impossible because two of the sides are flat on the third.  Trying $1, 1, 3$ produces even worse results.

WLOG (Without Loss Of Generality), $a \le b \le c$.  Because we know that $a + b > c$, and we know that $a + b + c = 11$, it follows that $11 - c > c$.  Because we designated $c \ge b \ge a$, we also know that $2c \ge 2a \ge 2b$, from which it follows that $2c \ge a + b \implies 2c \ge 11-c$.  This gives our only accepted values of $c$ as $c = 4, 5$.  (Plug in different values of $c$ into the equations and see for yourself.)

For $c = 4$, we have $a + b = 7$.  Because we designated $b \ge a$, the only possible values are $b = 4, a = 3$.  For $c = 5$, we have $a + b = 6$, and there are two possible sets of values - $b = 4, a = 2$ and $b = 3, a = 3$.

So the only triangles with integer side lengths of perimeter 11 are $(3, 4, 4), (2, 4, 5), (3, 3, 5)$, as in the text.

Edit:  There is actually one other triangle.  For $c = 5$, we have $a + b = 6$, and we can choose $b = 5, a = 1$ to get the triangle $(5, 5, 1)$ (which AoPS seems to have missed; however, it is not an obtuse triangle, and so does not affect the answer to the problem.)  [/hide]

[hide="How to solve the problem given these triangles"]
Looking at a right triangle, $a^2 + b^2 = c^2$.  If you increase the right angle, and make it an obtuse triangle, keeping $a$ the same results in an increase of $c$, so it is intuitive to state that $a^2 + b^2 < c^2$ for an obtuse triangle.

Now, looking at our triangles, $3^2 + 4^2 \nleq 4^2$, so the first triangle doesn't work.  However, $2^2 + 4^2 \leq 5^2$ and $3^2 + 3^2 \leq 5^2$, so the other two triangles work.  As in the text, there are $\boxed{2}$ obtuse triangles with integer sides and $a + b + c = 11$.[/hide]

thanks for the help, I understood that we needed too use the fact $a^2 > b^2 + c^2$ but counting has always been horrible for me, and i couldn't figure out how to find all the triangles without trying way too many combinations. thanks!


calculus
derivative
integration
real analysis
real analysis unsolved

I am doing some physics when I come across this one.

$ \frac {\partial}{\partial\psi}(\nabla\psi)^{2}$

Apparently, you can calculate it using some kind of integration by parts. Also, the result has a Laplacian term in it. I have no clues how to obtain that.

Your subject title says gradient, but you wrote $ \Delta$. Often times it is denoted that $ \Delta\equal{}\nabla^2$. Do you want $ \Delta$ or $ \nabla$ there?

Oops sorry about that. Typing too fast never does any good.

Oh, my, that physics notation... :roll: Apparently what was meant is the following: consider the functional $ I(\psi)\equal{}\int (\nabla\psi)^2$ (with zero boundary conditions for $ \psi$). Then $ I(\psi\plus{}h)\equal{}I(\psi)\plus{}2\int\langle\nabla\psi,\nabla h\rangle\plus{}\text{second order term}$, so the first variation is the linear functional that maps $ h$ to $ 2\int \langle\nabla\psi,\nabla h\rangle\equal{}\minus{}2\int (\Delta\psi)\cdot h$ (integration by parts). Thus the answer is $ \minus{}2\Delta\psi$.


function
real analysis
real analysis unsolved

Here is a hard question for me:

Construct a convergent sequence of Riemann integrable real-valued functions on $[0,1]$ whose limit function is NOT Riemann Integrable.

Can anyone show me or give me a hint?

The classic function which is not Riemann integrable is $f(x)=\begin{cases}1&x\in\mathbb{Q}\\ 0&x\not\in\mathbb{Q}\end{cases}$

On the other hand, $g(x)=\begin{cases}\frac1q&x=\frac{p}{q}\text{ in lowest terms}\\ 0&x\not\in\mathbb{Q}\end{cases}$ is Riemann integrable. Can you construct variations on this $g$ which converge to $f$ (pointwise)?

jmerry, how can you show that the function $g(x)$ is Riemann Integrable? Also, how about letting $q$ approach $1$ and therefore $x$ approaches $p$? Will a function that that converge pointwise to $f(x)$?

Please help me. I'm thoroughly confused.

In terms of your original question, the function $g$ is a distraction.

Let $\{r_{n}\}$ be an enumeration of the rationals in $[0,1].$ (This is a countable set, which means that it is possible to find a one-to-one correspondence between it and the natural numbers. The enumeration we want is precisely that one-to-one correspondence.)

Now let $f_{n}(x)=\begin{cases}1&x\in\{r_{1},r_{2},\dots r_{n}\}\\0&\text{otherwise}\end{cases}$

Now what can you show about this sequence?

Can't we show $f_{n}(x)$ converges pointwise to $f(x) = 1$ if $x\in Q$ and $f(x) = 0$ if $x \ne Q$? That would seem to make sense.

For each $n\in\mathbb{N}$ is $f_{n}$ integrable?

Then we're done here.

Note that this sequence converges pointwise but not uniformly. Indeed, the uniform limit of Riemann integrable functions would be Riemann integrable.

Just one more question Kent, what is $x_{1}, x_{2}, x_{3}\ldots x_{n}$?

[color=green][A typo which has now been fixed.][/color]

Ah- I had a different version in mind. My example was $f_{n}=g^{\frac1n}$.


Compute the value of A divided by B given that A = 666666 ^6 and B = 222222 ^6


Convert the measure to radians:

$ 1 \frac{2}{3}$ revolutions counter clockwise from the x axis (on a unit circle).

Just note that 1 revolution is $ 2\pi$ radians..

Well do you want the whole entire angle?  $ \frac{2}{3}$ of a revolution is $ \frac{4\pi}{3}$ radians so $ 1\frac{2}{3}$ revolutions is $ 2\pi\plus{}\frac{4\pi}{3}\equal{}\frac{10\pi}{3}$ radians.


If an integer is simultaneosly a square and a cube, then prove that it is of the form $ 7K$ or $ 7K \plus{} 1$. :D  :D

[hide="Yeah, I am bored"]

A number can be congruent $ 0, \pm 1, \pm 2, \pm 3$ $ \pmod{7}$.
If we square it, we have $ 0,1,4,2$ $ \pmod{7}$
If we cube it, we have $ 0,\pm 1, \pm 1, \pm 1$. Conclusion follows.

 [/hide]


A family $L$ of 2006 lines on the plane is given in such a way that it doesn't contain
parallel lines and it doesn't contain three lines with a common point.We say that
the line $l_1\in L$ is  [i]bounding[/i] the line $l_2\in L$,if all intersection points
of the line $l_2$ with other lines from $L$ lie on the one side of the line $l_1$.
Prove that in the family $L$ there are two lines $l$ and $l'$ such that the following
2 conditions are satisfied simultaneously:
[b]1)[/b] The line $l$ is bounding the line $l'$;
[b]2)[/b] the line $l'$ is not bounding the line $l$.

Suppose that $l$ is a line in $L$.
Suppose that $p_1,p_2,\ldots,p_{2005}$ are the intersection points of other lines with $l$.
Also suppose that $p_i$s are arranged in order in some direction on the line $l$.
Then exactly the lines which have the points $p_1$ and $p_{2005}$ are bounding the line $l$.
For the line $l$ let $s(l)=p_1p_{2005}$ be the segment between $p_1$ and $p_{2005}$.
Suppose the wanted result is not correct, so if $l$ bounds $l'$ then $l'$ bounds $l$.
Construct a graph with $L$ as its vertex set.
Draw an edge between two lines if one bounds the other.
In this graph degree of every vertex is $2$ so the graph is composed of some cycles.

Suppose that $l_1,l_2,\ldots,l_k$ is a cycle.
Suppose that $P_i$ is the comon endpoint of $s(l_i)$ and $s(l_{i-1})$ ($l_{-1}=l_k$).
If $k$ is even, then $P_3P_4$ intersects $l_1$ so $P_4$ and $P_3$ are on two sides of $l_1$.
$P_4P_5$ intersects $l$, so $P_5$ and $P_3$ are on the same side of $l_1$.
Using induction we see that $P_{odd}$ and $P_3$ are on the same side of $l_1$ and $P_{even}$ is on the other side.
So $P_k$ and $P_3$ are on opposite sides of $l_1$ so $P_2P_3$ and $P_kP_1$ can not intersect which is contradiction.
If $k$ is odd then $k<2006$ so there is another line like $l$.
Now using similar arguments we see that $P_{odd}$s and $P_1$ are on the same side of $l$ and $P_{even}$s are on the other side, so $P_k$ and $P_1$ are on the same side and so $P1P_k$ can not intersect $l$ which is also contradiction.
So we always arrive at a contradiction, so the problem is proved!


Let $v_1,v_2,v_3$ the distances from an arbitrary point $P$ to the vertices of a triangle $ABC$ and $s_1,s_2,s_3$ to the sides.

Prove that

$v_2v_3 +v_3v_1+v_1v_2 \geq 4(s_2s_3+s_3s_1+s_1s_2) $

This is more an inequality problem.
Once we prove the following:
(ab+bc+ca)(x+y+z)^2 >= (xy+yz+zx)(a+b+c)^2
[where a+b+c >= x+y+z]
we are done, by that theorom that says v1+v2+v3 >= 2(s1+s2+s3).
I think it's true.


limit integration geometry algebra function domain email

yo dawgs..how life in the hood homies:D:D (valentin this means u never went to IMO...cuz u got no proof gangster....and joe can own u in chess anyday anytime right....? so dont even try and claim u went to IMO...cuz iut never happened  :D  :?  :(  :)  \sum  \cup  2  \sqrt  \downto  \pm  \pi  \leq  \lim  \lambda  \cap  \gamma  \varepsilon  \geq  \emptyset  \int

What an immature guest! Outrageous! Blasphemous! Can we erase this coward's post please? Only cowards post as 'Guest'.

this guest has splintercell's ip address :? , splintercell do you have anything you want to add before a banip/username is issued?

I can't believe splintercell did this. By the way Valentin, who is Traian Lalescu? I saw that name as a category in the College-->Undergrad Contest section.

well his ip is the same as the guy who posted, so basically if he doesn't say anything I will block the guests' ip which is the same as splintercells. maybe it was something like his younger brother trying to make a joke or something :D ...

traian lalescu was a known romanian mathematician which lived at the begining of the 20th century. he has some results in geometry, and this contest was named after him in his honour.

Ah, I see... :cool: ... Thx Valentin. Unfortunately, that category is still empty. :(

wait, what? i definitely didnt do anything Valentin this is completly outrageous :( do u rlly think i would do something like this?

yeah, all i can say is this was not me and i dont even know how it could be, and no id ont have a younger brother or any other siblin (only child valentin :( and yes the above post was me i forgot to log on beacuse of the cookies

ya well i once got someone banned from messing on a forum at his pc :D but he deserved it :P

and oh well even if it was you, just erase the stuff and we'll all be happy :)

Among the tons of spam I receive every day, I got a most literary one today:

[quote="Spammer"][b]Subject:[/b] Hello
[b]From:[/b] joe dowd <mathsss2001@yahoo.co.uk>
[b]To:[/b] [my email address]
[b]Date:[/b] 27.03.04 20:04:56

Received: from [217.12.10.51] (helo=web25103.mail.ukl.yahoo.com)
by mx22.web.de with smtp (WEB.DE 4.101 #82)
id 1B7J6u-0004Pm-00
for [my email address]; Sat, 27 Mar 2004 20:04:48 +0100
Message-ID: <20040327190447.11839.qmail@web25103.mail.ukl.yahoo.com>
Received: from [216.195.14.207] by web25103.mail.ukl.yahoo.com via HTTP; Sat, 27 Mar 2004 19:04:47 GMT
Date: Sat, 27 Mar 2004 19:04:47 +0000 (GMT)
From: =?iso-8859-1?q?joe=20dowd?=
Subject: Hello
To: [my email address]
MIME-Version: 1.0
Content-Type: multipart/alternative; boundary="0-1812070801-1080414287=:11706"
Content-Transfer-Encoding: 8bit
Sender: mathsss2001@yahoo.co.uk

I read your site and wanted to tell you that although some aspects of the site were useful and informative, the other sections were a waste of my time. Your proofs had serious flaws in them, diagrams poorly explained and other common mistakes in calculations. If you sincerely want to put up a site (in plain terms) of crap, then don't do it in such a beautiful topic of geometry. It makes me disgusted to see website that are done by people exactly like you, who have no idea about what they are doing[/quote]

Well, concerning flaws there always is a possibility to overlook something, but "mistakes in calculations" is nonsense since I don't make calculations (my site is about [i]synthetic[/i] geometry). All the rest of the mail is nonsense as well.

What is interesting is the question whether the spammer above had any kind of relation to the one who spammed mathlinks before... :D

  Darij

well the address of the mathlinks spammer is 216.195.1.2
your spammer has 216.195.14.207
and splintercell usually uses 216.195.14.207
so ... there is some sort of connection, I cannot find this a random occurence, especially when only splintercell uses the 216.195.x.x prefix :?

[quote="darij grinberg"]...
[quote="Spammer"][b]Subject:[/b] Hello
[b]From:[/b] joe dowd <mathsss2001@yahoo.co.uk>
[b]To:[/b] [my email address]
[b]Date:[/b] 27.03.04 20:04:56

Received: from [217.12.10.51] (helo=web25103.mail.ukl.yahoo.com)
by mx22.web.de with smtp (WEB.DE 4.101 #82)
id 1B7J6u-0004Pm-00
for [my email address]; Sat, 27 Mar 2004 20:04:48 +0100
Message-ID: <20040327190447.11839.qmail@web25103.mail.ukl.yahoo.com>
Received: from [[b]216.195.14.207[/b]] by web25103.mail.ukl.yahoo.com via HTTP; Sat, 27 Mar 2004 19:04:47 GMT
Date: Sat, 27 Mar 2004 19:04:47 +0000 (GMT)
From: =?iso-8859-1?q?joe=20dowd?=
Subject: Hello
To: [my email address]
MIME-Version: 1.0
Content-Type: multipart/alternative; boundary="0-1812070801-1080414287=:11706"
Content-Transfer-Encoding: 8bit
Sender: mathsss2001@yahoo.co.uk
...
[/quote]

...[/quote]
[quote="Valentin"]well the address of the mathlinks spammer is [b]216.195.1.2[/b]
your spammer has [b]216.195.14.207[/b]
and splintercell usually uses [b]216.195.14.207[/b]
so ... there is some sort of connection, I cannot find this a random occurence, especially when only splintercell uses the 216.195.x.x prefix :( [/quote]
Final Verdict Valentin?

Hmm, I find something quite deceiving... the account is from @yahoo.co.uk, in UK, account in UK. If we analyze closely the email's information (header) we can find A LOT of information.

valentin, why does the forum allow guests to post?
maybe it should requires logging in first....

[quote="darij grinberg"]Request on [url=http://www.hexillion.com/whois/]RegistryFusion unified Whois[/url] for 216.195.14.207...

[quote="RegistryFusion unified Whois"]  <?xml version="1.0" encoding="utf-8" ?>
- <ServiceResult>
  <ErrorCode>Success</ErrorCode>
  <Message>The service successfully processed your request.</Message>
- <QueryResult>
  <QueryString>!NET-216-195-8-0-1</QueryString>
  <ServerName>whois.arin.net</ServerName>
  <ErrorCode>Success</ErrorCode>
  <FoundMatch>Yes</FoundMatch>
- <WhoisRecord>
  <CreatedDate>2002-03-27T00:00:00Z</CreatedDate>
  <UpdatedDate>2002-03-27T00:00:00Z</UpdatedDate>
  <DatabaseUpdatedDate>2004-03-26T19:15:00Z</DatabaseUpdatedDate>
- <Registrant>
[b]  <Name>SHREWSBURY ELECTRIC & COMMUNITY CABLE</Name> [/b]
  <Address>100 Maple Avenue</Address>
[b]  <City>Shrewsbury</City> [/b]
  <StateProvince>MA</StateProvince>
  <PostalCode>01545</PostalCode>
[b]  <Country>US</Country>[/b]
  <CountryCode>US</CountryCode>
  </Registrant>
- <Network>
  <CreatedDate>2002-03-27</CreatedDate>
  <CreatedDate>2002-03-27</CreatedDate>
  <UpdatedDate>2002-03-27</UpdatedDate>
  <UpdatedDate>2002-03-27</UpdatedDate>
[b]  <IPRange>216.195.8.0 - 216.195.15.255</IPRange> [/b]
  <CIDR>216.195.8.0/21</CIDR>
  <Name>SHREWS-NET-CABLE1</Name>
  <Handle>NET-216-195-8-0-1</Handle>
  <ParentHandle>NET-216-195-0-0-1</ParentHandle>
  <Status>Reassigned</Status>
  <Comment />
  <DatabaseUpdatedDate>2004-03-26 19:15</DatabaseUpdatedDate>
  </Network>
- <TechContact>
  <Handle>SD826-ARIN</Handle>
  <Name>Dorsey, Scott</Name>
  <Phone>+1-508-841-8648</Phone>
  <Email>sdorsey@ci.shrewsbury.ma.us</Email>
  </TechContact>
  <HeaderText />
  <StrippedText>CustName: SHREWSBURY ELECTRIC & COMMUNITY CABLE Address: 100 Maple Avenue City: Shrewsbury StateProv: MA PostalCode: 01545 Country: US RegDate: 2002-03-27 Updated: 2002-03-27 NetRange: 216.195.8.0 - 216.195.15.255 CIDR: 216.195.8.0/21 NetName: SHREWS-NET-CABLE1 NetHandle: NET-216-195-8-0-1 Parent: NET-216-195-0-0-1 NetType: Reassigned Comment: RegDate: 2002-03-27 Updated: 2002-03-27 TechHandle: SD826-ARIN TechName: Dorsey, Scott TechPhone: +1-508-841-8648 TechEmail: sdorsey@ci.shrewsbury.ma.us # ARIN WHOIS database, last updated 2004-03-26 19:15</StrippedText>
  <FooterText># Enter ? for additional hints on searching ARIN's WHOIS database.</FooterText>
  <RawText>CustName: SHREWSBURY ELECTRIC & COMMUNITY CABLE Address: 100 Maple Avenue City: Shrewsbury StateProv: MA PostalCode: 01545 Country: US RegDate: 2002-03-27 Updated: 2002-03-27 NetRange: 216.195.8.0 - 216.195.15.255 CIDR: 216.195.8.0/21 NetName: SHREWS-NET-CABLE1 NetHandle: NET-216-195-8-0-1 Parent: NET-216-195-0-0-1 NetType: Reassigned Comment: RegDate: 2002-03-27 Updated: 2002-03-27 TechHandle: SD826-ARIN TechName: Dorsey, Scott TechPhone: +1-508-841-8648 TechEmail: sdorsey@ci.shrewsbury.ma.us # ARIN WHOIS database, last updated 2004-03-26 19:15 # Enter ? for additional hints on searching ARIN's WHOIS database.</RawText>
  </WhoisRecord>
- <QueryResult>
  <QueryString>216.195.14.207</QueryString>
  <ServerName>whois.arin.net</ServerName>
  <ErrorCode>Success</ErrorCode>
  <FoundMatch>Yes</FoundMatch>
- <WhoisRecord>
- <Network>
  <TargetHandle>NET-216-195-8-0-1</TargetHandle>
  </Network>
  <HeaderText />
  <StrippedText>SHREWSBURY ELECTRIC & COMMUNITY CABLE SHREWS-NET (NET-216-195-0-0-1) 216.195.0.0 - 216.195.31.255 SHREWSBURY ELECTRIC & COMMUNITY CABLE SHREWS-NET-CABLE1 (NET-216-195-8-0-1) 216.195.8.0 - 216.195.15.255 # ARIN WHOIS database, last updated 2004-03-26 19:15</StrippedText>
  <FooterText># Enter ? for additional hints on searching ARIN's WHOIS database.</FooterText>
  <RawText>SHREWSBURY ELECTRIC & COMMUNITY CABLE SHREWS-NET (NET-216-195-0-0-1) 216.195.0.0 - 216.195.31.255 SHREWSBURY ELECTRIC & COMMUNITY CABLE SHREWS-NET-CABLE1 (NET-216-195-8-0-1) 216.195.8.0 - 216.195.15.255 # ARIN WHOIS database, last updated 2004-03-26 19:15 # Enter ? for additional hints on searching ARIN's WHOIS database.</RawText>
  </WhoisRecord>
  </QueryResult>
  </QueryResult>
  </ServiceResult>[/quote]

But I am not sure whether it really was splintercell: the IPs may be some internet cafs or something similar. Looking forward to hear splinercell's opinion about this all...

  dg[/quote]

ok, i have quite some things to say:

1.) The ip-range around this location is always or usually 216.195.x.x (cable company)

2.) i think if might not have realized, but as 3x.lich said, the account came from @yahoo.co.uk (i live in the us)

3.) My ip is not 216.195.14.207 (my ip is dynamic, and keeps changing-it will reassign itself after some time) If you want to know your own ip address, just go to command prompt in windows, and hit ipconfig, you will see your own ip.

4.) As Mr. Grinberg said, those ips are more lilely internet cafes, as esp here, we have many stores which provide interent access (Borders, Barnes & Noble..etc)

another thing, as admin of this board valentin why can't you disable guest posting-everyone seems to be in aggreement when they say to disable it everywhere on the board?  :?

http://www.shrewsbury-ma.gov/Schools/Central/DIRECTIONS/Directions.html
this is a link to the place that is coming from. the coincidence is so high that is:
1) either its splintercell; or
2) someone who goes to the same high school as splintercell.
either way, it is splintercell's job to either stop this himself or stop his friends from abusing the forums/people here.

yahoo domain does not mean anything, I can just open an email account in Korea's servers and my email address still carries the korean domain, which is yahoo.co.kr, and I LIVE IN THE US as well. BTW, I do have one email address there.
[quote="splintercelll990"]This might be a random topic, but hows the weather in your country/region? From where I am..its snowing!!! :D (just might cancel school..estimated 5-10 inches :))

oh yeah 3x, i live in Massachusets, so i guess we don't
Last edited by splintercelll990 on Tue Mar 16, 2004 7:08 pm; edited 1 time in total[/quote]
[quote="splintercelll990"]actually nope, im not i live in shrewsbury...but if you live in lexingtion you have one of the best high schools (they win at like every math competition :D-WPI, WOCOMAL..everything including the annual Harvard-MIT math competition)[/quote]
all quotes from Non Math Topics->The Babylon Tower->Weather in your country
that's all, and I want to hear a final verdict from Valentin.

I am still awaiting for splintercell's reply :D I don't want to make any harsh and quick decisions about this.


Hi:

I am not involved in any type of math studies or anything but I am just looking for a response to this.  I need to figure out the answer to [i]i[/i] in this equation:

125.5153 = {1-[1/(1+i)]^180} / i

Thanks for everyone's help.

hello, do you meant $ 125,5153\equal{}\frac{\left(1\minus{}\left(\frac{1}{1\plus{}i}\right)^{180}\right)}{i}$?
Sonnhard.

Hi:

No I mean 125.5153 - the i is going to be the monthly interest rate - thanks.

I think he meant to ask if the LaTeX formula is correct -- not all countries use dots to indicate a fractional part, you know... (and this is an equation of degree 180, which is definitely not an easy thing to solve -- I think we'll need to use some sort of clever trick).

Don't expect to go and solve that exactly without a calculator. Depending on how frequently it is being compounded, you can consider approximating it with [url=http://en.wikipedia.org/wiki/Compound_interest#Continuous_compounding]continuously compounded interest[/url].


Given a circle of raidus $2$, there are many line segments of length $2$ that are tangent to the circle at their midpoints. Find the area of the region consisting of all such line segments.

$\text{(A)}\ \frac \pi 4 \qquad \text{(B)}\ 4-\pi \qquad \text{(C)}\ \frac \pi 2 \qquad \text{(D)}\ \pi \qquad \text{(E)}\ 2\pi$

By "the region consisting of all such line segments" I assume you mean the region that would be shaded if you drew all of the lines on a piece of paper:

[hide="Probably wrong"]
Think of a line segment connecting the endpoints of each tangent line to the center of the circle. By the Pythagorean Theorem, each such segment has length $\sqrt{5}$.

Therefore all the tangent lines fall within the region between a circle of radius $\sqrt{5}$ and a circle of radius $2$. Therefore the area of that region is:

$\pi\cdot5-\pi\cdot4=\pi$

The answer is D.
[/hide]

Yes, that is correct. I just found it has already been posted elsewhere. Thanks Mathnerd314. :)
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=376602#p376602


Suppose that I have 100 dice and I roll all of them and sum the values that were rolled. What is the probability that the sum is divisible by 7?

i was actually thinking about posing this problem next year to my arml team, but i guess i should not have posted it...anyway

[hide="hint"]generating functions[/hide]

after that, the problem is still pretty difficult, i don't have a good hint for that one, but here is a solution

[hide="solution"]for an explanation of generating functions, look elsewhere

the generating function is:

$f(x)=(x+x^2+x^3+x^4+x^5+x^6)^n$

where the degree of the x term is the sum of the rolls, we want the sum of the coefficients with degrees $x^{7k}$ with $k\in\mathbb{Z}^+$

let $a_1,a_2,...a_7$ be the distinct solutions to $x^7-1=0$

it can be verified that the desired sum is

$\sum_{i=1}^7\frac{a_i}{7}$

using the fact that the roots of unity taken to any power except a multiple of 7 will sum to $0$

doing this, and noting that for any nonreal root, we have $x^6+x^5...+1=0$, the sum becomes

$\frac{(1+1+1+1+1+1)^n+(-a_2)^n+(-a_3)^n+...(-a_7)^n}{7}$

now we have that $n=100$, so the negatives drop out, and we reduce the exponents using $(a_i)^7=1$

$=\frac{6^{100}+(a_2)^2+...+(a_7)^2}{7}$

note that the sum of the roots of unity squared is $0$, but since the root of 1 is separate, the sum of the nonreal roots is $-1$

$=\frac{6^{100}-1}{7}$, and divide by the total events, or $6^{100}$, to get:

$\frac{1}{7}-\frac{1}{7*6^{100}}$[/hide]


geometry
trapezoid
AMC
AIME
incenter
number theory
relatively prime

Trapezoid $ABCD$ has sides $AB=92$, $BC=50$, $CD=19$, and $AD=70$, with $AB$ parallel to $CD$. A circle with center $P$ on $AB$ is drawn tangent to $BC$ and $AD$. Given that $AP=\frac mn$, where $m$ and $n$ are relatively prime positive integers, find $m+n$.

[hide]Given points of tangency, we draw lines from P perpendicular to sides $AD$ and $BC$ at points $M$ and $N$, respectively. Next, extend $AD$ and $BC$ to meet at a point $X$. Notice that $\triangle MPX\cong \triangle NPX$, by Hypotenuse-Leg congruence. Thus, $\angle AXP=\angle BXP$.

We are now able to apply the angle bisector theorem. Notice that $\frac{AP}{BP}=\frac{AX}{BX}=\frac{AD}{BC}=\frac75$. Therefore, $AP=92\cdot\frac{7}{12}=\frac{161}{3}$ and $m+n=\boxed{164}$.[/hide]

[hide="questions about your solution"]
First, how can you be so sure $\frac{AX}{BX} = \frac{AD}{BC}$?  I didn't see that, and calculated $CX$ and $DX$ and used that with angle bisector theorem.  Let me tell you, it was VERY nasty.  Although, on the AIME I'd gladly trudge through that nastiness to get the answer...

Also, you can prove $\angle AXP=\angle BXP$ by making a line parallel to $AB$ tangent to the circle, and extending $AD$ and $BC$ to meet it.  The triangle with points $X$ and those two intersection points has incenter $P$, and then $XP$ is an angle bisector by the definition of the incenter.
[/hide]

[hide="answers about solution"]From the angle bisector theorem, $\frac{AP}{AX}=\frac{BP}{BX}$. Multiplying both sides by $AX$ and dividing both sides by $BP$ gives the desired result.

I'm not sure what you're trying to say for the second part. If you drew a tangent to the circle, you wouldn't need to extend $AD$ and $BC$, as the tangent already meets those lines. [/hide]

[quote="pkothari13"]SSA isn't a type of congruence.... :huh:[/quote]Oops, my bad.  :blush:

I was thinking of HL congruence and just typed something else. (They are similar, huh?  :P )

ohhhhh yeah, i see it .... sameeee thing  :D

Question:  How did you get that AP = 92(7/12)??

Oh, but then how does that work? Aren't AB, BP, and AP all sides of that bottom triangle?

we're given the large length. The length we want is 7/12 of the large length.

4everwise, what I was trying to say was you could prove the angles are equal since clearly P is the incenter of a triangle with a vertex X and the sides given, and the incenter of the point of concurrency of the angle bisectors.

I got a really nice solution ( I think....)

Let the radius of the circle be $r$ and the height of the tapezoid

be $h$.

So the area of the trapezoid is $\frac{(19+92)*h}{2}=\frac{111h}{2}$.
Now we devide the trapezoid into three triangles: $ADP$, $DCP$, and $BCP$.
 Area of $ADP$ = $35r$
 Area of $DCP$ = $\frac{19h}{2}$
 Area of $BCP$ = $25r$
Adding all the three triangles, we get $60r+\frac{19h}{2}$, and note that this is equal to the area of the trapezoid.
Hence $\frac{111h}{2}=60r+\frac{19h}{2}$
$30r=23h$

Now we look at triangle $ADP$, since $AP$ is what we are looking for.

Area of $ADP$ = $35r$ = $AP*h$
Solving for $AP$, $AP=\frac{35r}{h}$. Also we know that $30r=23h$

Thus $AP=\frac{161}{3}$.

[quote="Adunakhor"]Sorry to revive an old thread, but I have the same question as deej21. How do you know right away that $\frac{AX}{BX} = \frac{AD}{BC}$?

[quote="4everwise"]From the angle bisector theorem, $\frac{AP}{AX}=\frac{BP}{BX}$. Multiplying both sides by $AX$ and dividing both sides by $BP$ gives the desired result.
[/quote]

Doesn't that just give $\frac{AP}{BP} = \frac{AX}{BX}$?[/quote]

The Angle Bisector Theorem on $\triangle XAB$ gives \[\frac{PA}{PB} = \frac{XA}{XB} = \frac{DA + DX}{CB + CX}.\] By the similar triangles $\triangle XAB \sim \triangle XDC$ we know that $\frac{XA}{XB} = \frac{DX}{CX}.$ Therefore, \[\frac{DX}{CX} = \frac{DA + DX}{CB + CX} \implies \frac{DA}{CB} = \frac{DX}{CX}\]\[\implies \frac{PA}{PB} = \frac{DA}{CB} = \frac{7}{5}.\]


[hide]notice that $(x_{i}-m)(x_{i}-M) \leq 0$
so we obtain $x_{i}+\frac{mM}{x_{i}}\leq m+M$
now let $i =\bar{1,n}$ and adding all the inequalities we have
$n(m+M) \geq \sum^{n}_{i=1}x_{i}+mM\sum^{n}_{i=1}(\frac{1}{x_{i}})\geq 2\sqrt{mM}\sqrt{(\sum^{n}_{i=1}x_{i})(\sum^{n}_{i=1}(\frac{1}{x_{i}})}$ and this is what we want :D[/hide]

this is the inequality of P. Schweitzer. Actually it has a generalization which belongs to L.V. Kantorovici:

0<m<xk<M , n>0, n natural, k=1,n and tk is pozitive real, k=1,n. then:

(t1*x1+ t2*x2 +...+tn*xn)(t1/x1 + t2/x2+....+ tn/xn) <= (t1+t2+...+tn)*(m+M)^2\4*m*M

if we take t1=t2=...=tn=1 we obtain Schweitzer

[quote="kunny"]Similar problem from Japan.

Kumamoto University entrance exam/Education 1985

Let $a_{1},\ a_{2},\ b_{1},\ b_{2},\ m,\ M$ be positive real numbers such that $m\leq \frac{b_{k}}{a_{k}}\leq M\ (k=1,\ 2)$.

Prove the following inequality.
\[\frac{(a_{1}^{2}+a_{2}^{2})(b_{1}^{2}+b_{2}^{2})}{(a_{1}b_{1}+a_{2}b_{2})^{2}}\leq \frac{(m+M)^{2}}{4mM}\]
.[/quote]

this is Polya-Szego for n=2 :wink:  in Kantorovici take xk=ak/bk and tk=ak*bk ,k=1,n


look what i did to your sig and avatar  :P

nerd_of_the_ages

STOP SPREADING RUMORS.

also, this topic is spam.


calculus
derivative
inequalities open
inequalities

[hide="a"]
Let $s=a+b$. Then we have
\[F=\frac{(s-b)^{2}+2b^{2}}{2bs}=\frac{s^{2}+3b^{2}}{2bs}-1\]
By AM-GM, $s^{2}+3b^{2}\geq 2bs\sqrt{3}$ so the minimum value of $F$ is $\boxed{\sqrt{3}-1}$. It's attained when $a=\sqrt{3}-1,b=1$.
[/hide]
Part b has a very scary equality case. I wouldn't touch it without calculus, or even write the answer in terms of radicals (would you want to wait [color=cyan]13209434[/color] hours to download the page?) :wink:

[quote="scorpius119"][hide="a"]
Let $s=a+b$. Then we have
\[F=\frac{(s-b)^{2}+2b^{2}}{2bs}=\frac{s^{2}+3b^{2}}{2bs}-1 \]
By AM-GM, $s^{2}+3b^{2}\geq 2bs\sqrt{3}$ so the minimum value of $F$ is $\boxed{\sqrt{3}-1}$. It's attained when $a=\sqrt{3}-1,b=1$.
[/hide]

Why is $bs>0$?

Well, I was assuming $a,b$ positive. Otherwise, $a=-1,b\rightarrow 1^{-}$ makes $F\rightarrow-\infty$ and the problem is silly.

For $a\neq 0$, divide by $a^{2}$ for numerator and deniminator and let $x=\frac{b}{a}$.

Thanks all ; a;b;m >0 and who can help me solve problem 2 ?

to everyone : I'm sorry but : " No one can solve Pro 2 ? " ( Use Derivative  :wink:  )


The base 6 system of counting uses 6 distinct digits to represent any number. The non-negative integral numbers in this system can be listed in ascending order as 0, 1, 2, 3, 4, 5 ,10, 11, 12, 13, 14, 15, 20, 21,.... Find the sum (as expressed in base 6) of the first 100 numbers listed in the series given.

(a) 34530 (b) 34153 (c) 35214 (d) 34560 (e) 34532

[hide=" (c)"]

The sum of the numbers is $ 5050$ in base ten which is $ 35214$ in base six.

[/hide]

I'm afraid that maybe incorrect; The answer is given as option a.

Sorry, I was rushing as usual. You're right. I did $ 1$ to $ 100$ and didn't see that it was $ 0$ to $ 99$. So it's $ 100$ less, that is $ 4950$, which in base six is $ 34530$, that is (a).


Triangle ABC has two cevians from A to BC. Let one cevian intersect BC at D and the other intersect BC at E. We are given that angle DAB is congruent to angle EAC and that $BD=CE$. Prove that triangle ABC is isosceles. (This isn't the exact wording)

EDIT:[hide="Hint"]
areas
[/hide]

[quote="knexpert"]Triangle ABC has two cevians from A to BC. Let one cevian intersect BC at D and the other intersect BC at E. We are given that angle DAB is congruent to angle EAC and that $BD=CE$. Prove that triangle ABC is isosceles. (This isn't the exact wording)[/quote]

[hide]I haven't looked at it for very long but maybe $\frac{\sin (\angle BAD)}{BD}=\frac{\sin (\angle EAC)}{EC}$...so then triangles $ABD$ and $ACE$ are congruent and $AB=AC$...[/hide]

[hide]
Observe that $\triangle BAD$ and $\triangle CAE$ share an altitude, and the bases opposite that altitude are equal. Therefore, the areas of $\triangle BAD$ and $\triangle CAE$ are equal. Now, using the fact that $\angle BAD=\angle CAE$, the equal areas give us
\[AB\cdot AD=AC\cdot AE\]
(this follows from the $\frac{1}{2}ab\sin C$ area formula).

Similarly, we can look at $\triangle BAE$ and $\triangle CAD$, do the same area stuff as above, and get
\[AB\cdot AE=AC\cdot AD\]
Now just multiply the two equations together and divide by $AD\cdot AE$ to get $AB=AC$.
[/hide]


1.Show that for all nonnegative real values $ a,b,c$ an inequality occurs:
$ 4(\sqrt{a^3b^3}\plus{}\sqrt{b^3c^3}\plus{}\sqrt{c^3a^3})\le 4c^3\plus{}(a\plus{}b)^3$
2.Show that for all integer numbers $ k>m\ge 1$ an inequality occurs:
$ \frac{\sqrt[k]{k!}}{ \sqrt[m]{m!}}<\frac{k}{m}$
.

The second ineq is problem 11 from third series of 1. round Polish MO which is still running (till 3rd December) ! See here
http://www.mathlinks.ro/viewtopic.php?t=223905


Suppose we lift an object to a certain height. Now it has gained some potential energy. Now I keep the object down. Where has the potential energy of the object gone now? Disappeared? :!:

the potential energy reduces as we are doing work on the object

Your system is the object and the earth. But energy is not conserved because an external force i.e the person holding it is doing work on it.


AMC
USA(J)MO
USAMO
AIME
email
AMC 12
AMC 12 B

I was wondering what the specifications are for getting an award on the AMC?

I coach a state winner (10B, Minnesota) who appeared not to get any state winner award, although he did get the other awards you mention that fit his situation. Or maybe there is no A/B distinction for each test if there are state awards, or perhaps state awards are only passed out in states that have state have state coordinators, which I think is a vacant position in Minnesota right now.

[quote="tetrahedr0n"]There are a lot of different awards. There is info on them at the amc website. Just to give you a basic idea, you get a diploma for making usamo, a diploma for making aime, and if you are <= 10th grade, a diploma for 90+ on amc 12, if you are <= 8th grade, a diploma for 90+ on amc 10, a pin for making top 1% on either amc 10 or amc 12, a pin for being school winner for amc 10 or 12. if you win school amc 10 2 times-bronze medal, 3times - silver medal, 4times gold medal, same if you replace amc 10 with 12. also award for state winner i think.[/quote]

odd...i haven't gotten any of that...ever...

thats strange also. i have gotten almost all those, excluding state winner. they are sent to your coach usually. i dont know what to say, except if you are missing something, you could email amc?

[quote="tetrahedr0n"]they are sent to your coach usually[/quote]

I'm homeschooled so my coach should be my parents...

When I made AIME and USAMO last year, I didn't get a 'diploma' (by this I assume you mean 'certificate').  And I've been state winner for two years, and never gotten anything.  I think the only awards are pin-bronze-silver-gold, and the AMC12 awards as listed in the results book.

The 90+ award (a certificate) for under eighth grade on the AMC 10 was mailed to our test administrator (not me, but another homeschooling parent), who hand-delivered it the day my son took the AIME. I think based on what is said here in this thread that AT LEAST not every state has state awards, although perhaps some states do.

[quote="ComplexZeta"]If you're the only one in your school taking the AMC, you don't get school winner. There's a minimum number that you need, but I'm not sure what it is.

In California, at least for the last three years, there were state prizes. I got a math book in 2001, 2002, and 2003. (Fortunately, they were different books.)[/quote]

Are you sure?  We only had one guy take the AMC12B, and he got a pin...

Hm. We had that situation before, and the person didn't get any school award. But that was for AMC10B. Maybe that makes a difference?

I was the AMC10 school winner as an 8th grader, but I didn't get anything.  Have they not sent this stuff out yet?

I think that they should give better awards to USAMO qualifiers.  Because for most people, it is easier to be the school winner for 2 or 3 times than to qualify for USAMO.


That is correct, but please simplify :twisted:

No its not actually.. its got a nice trick to it, its not just arithmetic. I assume the actual question (unless it was just made up) would be express it as a fraction in lowest terms..

It was just made up, by moi :twisted:  :twisted:  :twisted:

But it still reduces both significantly and nicely, so maybe I'm not so eval

Hint: [hide]Do something bold and daring with the fraction[/hide]

Super Hint: [hide]The answer is an inverse of an integer, so...[/hide]

Yup, can you explain how you went from the first fraction to the second in one step?

Again, yup yup.
If it was given that the answer is an inverse of an integer, then the numerator becomes 1 in the answer so you need to divide the denominator by the numerator and you get the answer.

Lesson here: Always look for the easy way out

One thing you can do is to start multiplying the periodic part by small integers to see if you can find another part of it elsewhere. For example, if you multiply by 2, are you just shifting it over? Very often, that is the case. (Think of multiplying by powers of 10 and the removing the integer part.) My favorite one (which is fairly hard but very interesting in my opinion), is to write 0.010102030508132134559046... as a fraction. (That's the Fibonacci numbers each taking up two digits, but eventually they run over, and that's why after 55 it's 90 rather than 89.)


\[ 1)\ \left|\begin{array}{c} a,\ b,\ c\in\mathbb{N} \\
\frac {ab \plus{} 1}{c}\in\mathbb{N},\ \frac {bc \plus{} 1}{a}\in\mathbb{N},\ \frac {ca \plus{} 1}{b}\in\mathbb{N}\end{array}\right|\Longrightarrow a \equal{} ?,\ b \equal{} ?,\ c \equal{} ?
\]

\[ 2)\ \left|\begin{array}{c} a,b,c\in\mathbb{Q},\ a,b > 0 \\
a \plus{} \frac {1}{b}\in\mathbb{Z},\ b \plus{} \frac {1}{c}\in\mathbb{Z},\ c \plus{} \frac {1}{a}\in\mathbb{Z}\end{array}\right|\Longrightarrow a \equal{} ?,\ b \equal{} ?,\ c \equal{} ?
\]

\[ 3)\ \left|\begin{array}{c} a,\ b,\ c\in\mathbb{Z} \\
\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}\in\mathbb{Z} \\
\frac {b}{a} \plus{} \frac {c}{b} \plus{} \frac {a}{c}\in\mathbb{Z}\end{array}\right|\Longrightarrow\left|a\right| \equal{} \left|b\right| \equal{} \left|c\right|
\]

Problem 1:

If we have that $ a|bc \plus{} 1 ; b|ac \plus{} 1 and c|ab \plus{} 1$, then we have that:

$ abc|(ab \plus{} 1)(ac \plus{} 1)(bc \plus{} 1) \equal{} a^2b^2c^2 \plus{} a^2bc \plus{} ab^2c \plus{} abc^2 \plus{} ab \plus{} ac \plus{} bc \plus{} 1$
Hence, $ abc|ab \plus{} ac \plus{} bc \plus{} 1$.
Then, $ ab \plus{} ac \plus{} bc \plus{} 1 \geq abc$.
Suppose, without loss of generality, that $ a \equal{} min{a,b,c}$.

$ ab \plus{} ac \plus{} bc \plus{} 1 \geq abc < \equal{} > \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \plus{} \frac {1}{abc} \geq 1$

How $ a \equal{} min{a,b,c}$, if $ a \geq 4$, then the left side is less than 1, contradiction.
So, $ a \equal{} 3$, $ a \equal{} 2$ or $ a \equal{} 1$.

Let now divide in cases:

Case 1)$ a \equal{} 1$:
$ bc|b \plus{} c \plus{} 1$
From here, we get $ b \plus{} c \plus{} 1 \geq bc$.
If $ b \equal{} min{b,c}$, wlog, we have: $ \frac{1}{b} \plus{} \frac{1}{c} \plus{} \frac{1}{bc} \geq 1$.
If $ b \geq 3$, we have that the left side is less than 1, contradiction.
Hence, $ b \equal{} 1$ or $ b \equal{} 2$.
If $ b \equal{} 1$, $ c \equal{} 1$ or $ c \equal{} 2$.
If $ b \equal{} 2$, $ c \equal{} 3$.
Conclusion: In this case, we have the following solutions: $ (1,1,1)$ ; $ (1,1,2)$ ; $ (1,2,3)$

Case 2) $ a \equal{} 2$
By a similar way, we get the following solutions:
$ (2,3,7)$

Case 3) $ a \equal{} 3$
Another time, by a similar way, we don't get solutions here.

Then, the only solutions are:
$ (1,1,1)$ ; $ (1,1,2)$ ; $ (1,2,3)$ ; $ (2,3,7)$ and by the simmetry, all the permutations of these triples are solutions too.

Best regards!


is there a fast way to do the problem:

a ball is dropped from a height of two meters and on each bounce bounces up 3/4 of its pervious height. what is the total(up and down) distance that the ball will travel, in meters?

please include explanation.

[hide]Since for a geometric sequence $\frac{a}{1-r}$, and the first few terms are $2+\frac{3}{2}+\frac{3}{2}+\frac{9}{8}+\frac{9}{8}...$ I took out the doubled terms so that the sum is $\frac{2}{1-\frac{3}{4}}+\frac{\frac{3}{2}}{1-\frac{3}{4}}$ which equals $\boxed{14}$.[/hide]

[hide]
I think infinite geometric series
For the going down part, the series is
2,3/2... so $a=2$ and $r=\frac 34$
$\frac{2}{1-\frac 34}=8$
For going up, it is the same except get rid of the 2
So it is 6.
Total is 14[/hide]

We can convert this to the form $2\times \displaystyle\sum_{k=0}^\infty 2\times (\frac{3}{4})^k -2=2\times\frac{2}{1-\frac{3}{4}}-2=16-2=14$.

This uses the fact that $\displaystyle\sum_{i=1}^n ar^i=\frac{ar-ar^n}{1-r}$.

Samath: Can you express that formula without sigma notation?


I am getting repititive.. hope this problem is more challenging than a similar one I posted earlier.

$\tau(n)$ is the number of divisors of the integer $n$.

Given any positive integer $k$, show that there exist $k$ consecutive positive integers $x_1, x_2, ..., x_k$ such that $\tau(x_i)$ does not divide $x_i$ for all $1 \leq i \leq k$

i have a solution for '$k$ ODD consecutive integers'  :(

Sorry for leaving this hanging:

Here is a hint:
[hide]

Consider the set S ={ $n$ | $\tau(n)$ divides $n$}.

Now, Suppose the contrary, i.e for some k, for every k consecutive integers $x_{1}, x_{2}, ..., x_{k}$ , there is some $1 \le i \le k$ such that $\tau(x_{i})$ divides $x_{i}$.

This would imply that the density of S is at least 1/k.

Show that the density of S is zero.

[/hide]


linear algebra
matrix
calculus
integration
MIT
college

Prove that for any natural numbers $0 \leq i_1 < i_2 < \cdots < i_k$ and $0 \leq j_1 < j_2 < \cdots < j_k$, the matrix $A = (a_{rs})_{1\leq r,s\leq k}$, $a_{rs} = {i_r + j_s\choose i_r} = {(i_r + j_s)!\over i_r!\, j_s!}$ ($1\leq r,s\leq k$) is nonsingular.

OK, I got a really neat combinatorial solution. We're going to prove that this determinant counts something that exists, that is, that this determinant is greater than zero.

Consider the lattice $\mathbb{Z}^2$. Define a [i]path[/i] between two points $P$ and $Q$ of $\mathbb{Z}^2$ as a sequence of lattice points, each equal to the predecessor plus $(0,-1)$ or $(1,0)$, with first term equal to $P$ and te last one equal to $Q$. Define a [i]non-intersecting path system[/i] connecting two subsets $X$ and $Y$ of $\mathbf{Z}^2$, $|X| = |Y| = n$, as a set of $n$ vertex-disjoint paths, each one connecting a point of $X$ to a point of $Y$.

[b]Claim.[/b] $\det(A)$ is the number of non-intersecting path systems connecting $X = \{(0,i_1),(0,i_2),...,(0,i_n)\}$ to $Y = \{(j_1,0),(j_2,0),...,(j_n,0)\}$.

If this claim is true, the problem is trivial, since it's very easy to find one non-intersecting path system.

[i]Proof of the claim[/i]: By the definifion of determinant, $\det(A) =\sum_\rho{\rm sgn}(\sigma)a_{1\sigma(1)}...a_{n\sigma(n)}$, where $\rho$ is the set of all $n!$ permutations of $(1,2,\ldots,n)$. Considering that $a_{rs} = {i_r + j_s\choose i_r}$, $a_{1\sigma(1)}...a_{n\sigma(n)}$ is the number of sets of $n$ paths connecting $(0,i_k)$ to $(j_{\sigma(k)},0)$, that is, a path system, intersecting or not. In particular, all non-intersecting path systems are counted when $\sigma$ is the identity. Then all non-intersecting path systems appears with a plus sign.

But the intersecting path systems cancel out in the determinant: consider the leftmost intersection; if there is more than ore of those, pick the bottom one. Suppose this intersections is between paths connecting the pairs $(0,i_l), (j_m,0)$ and $(0,i_p), (j_q,0)$. This path system is counted in a summand of the determinant with two factors $a_{lm}$ and $a_{pq}$. But we can obtain other path system with the same edges with the same paths, except that we interchange the two paths connecting $(0,i_l), (j_m,0)$ and $(0,i_p), (j_q,0)$ with the two paths connecting $(0,i_l), (j_q,0)$ and $(0,i_p), (j_m,0)$, with the same edges. This path system is counted in another summand of the determinant, with the same factors, except that we interchanged $a_{lm}$ and $a_{pq}$ with $a_{lq}$ and $a_{pm}$. But the sign of this new permutation is exactly the opposite of the first one, so this path system do cancel out. Note that the choice of this inversion has no fixed point and is bijective, thus [b]all[/b] intersecting path systems cancel out. The non-intersecting ones won't though, and the result follows.

Very beautiful approach, indeed. Yet, I bet that it must be done easier using integrals, just as Moubinool did for the matrix $(a_i+a_j)!$.

Probably not. I've actually seen this method before- I think there's a recent Mathematical Monthly article featuring it.

The claim about the determinant and the nonintersecting paths is true in general, if the entries of our matrix are the number of paths and the setup works.

Well, this problem, in fact, originally is proposed in "reverse" way: express $det (A)$ as an combinatoric number.
I have a paper with various combinatorics problems (maybe in a MIT seminar), and in a few days I will put it in Downloads section. This problem is a particular case of one of these.