quadratics
algebra
polynomial
absolute value
superior algebra
superior algebra unsolved

Since 2 is not a quadratic residue mod 5, the polynomial $ x^2 \minus{} 2$ is irreducible over $ \mathbb{Q}_5$. If $ a^2 \equal{} 2$, then how do we extend the valuation to $ a$?

Do we just have $ v(a^2) \equal{} v(a) \plus{} v(a) \equal{} v(2) \equal{} 0$, so that $ v(a) \equal{} 0$? And this means that the absolute value of $ a$ is $ 1$, right?

And if we have $ b^2\equal{}5$, then is it true that $ v(b) \equal{} \frac{1}{2}$, and that the absolute value of $ b$ is $ \frac{1}{\sqrt{p}}$, right?

But then how would we find the valuation of say, $ 1 \plus{} a$ or $ 1\plus{}b$? That is, the root of $ x^2\minus{}2x\minus{}1$ or $ x^2 \minus{} 2x \minus{} p\plus{}1 \equal{} 0$. And how would we find the valuation of the root of some irreducible polynomial $ f(x) \equal{} 0$ in general?

Yes, $ v(a)\equal{}0$.
In general, one has $ v(c) \equal{} \frac{v \left( \mathrm{N}_{L|K}(c) \right)}{[L: K]}$, where $ L$ is a finite extension of the local field $ K$ (e.g. the $ p$-adics) and $ \mathrm{N}_{L|K}$ is the norm from $ L$ to $ K$ of that extension. Note two things:
- This term defines a valuation (that's the hard part).
- It extends $ v$ and for given $ c$ algebraic over $ K$, it gives the same value for all finite extensions $ L$ containing $ c$.

Especially in your case, $ v(1\plus{}a) \equal{} \frac{v\left((1\plus{}a)(1\minus{}a)\right)}{2}\equal{}\frac{v\left(1\minus{}a^2\right)}{2}\equal{} \frac{v\left(\minus{}1\right)}{2} \equal{}0$.


geometry
3D geometry
sphere
complex analysis
function
algebra
domain

Hello
Let $ f: \Omega \rightarrow \mathbb C$ a holomorphic function, for some open set $ \Omega \supset \overline{\mathbb D}$ (closed unit disk) such that $ f(S^1) \subseteq S^1$.
Prove that $ f$ can be extended to a meromorphic function on the riemann sphere.
My question:
How can I say that $ f$ has some poles?
And btw if I trust there are some, the extension should be some $ F: \Omega \rightarrow \hat{\mathbb C}$ defined only on $ \Omega$ right? The idea should be to define $ F(p) \equal{} \infty$ for every pole $ p$ of $ f$ of order >0, right??

This is the same question asked [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=308004]here[/url]. Extend it with by reflection.

In general, "extend" means that you want a larger domain (in this case, the Riemann sphere or extended plane); a larger codomain is meaningless.
Oh, and "has poles" is not required for a meromorphic function. The requirement is "doesn't have anything worse than poles".

thanks, but reflection even if I don't know the shape of $ \Omega$ ? I can't suppose it is biolomorphic to the unit disk.

First, throw away all of the values outside the disk. Then, reconstruct them with reflection. $ \Omega$ is irrelevant.

I do understand why I can throw away the values in $ \Omega \setminus \mathbb D$ but I don't understand the reflection made in the topic you linked.
The function $ \hat f (z) \equal{} 1/\overline{f(\overline z)}$ does not coincide with $ f$ on $ S^1$ but coincides with $ f(1/z)$, so the extension function is $ \hat f(z) \equal{} 1/ \overline{f(1/ \overline z)}$ right?


Let $ABCD$ a quardilateral and a point $O$ inside  $ABCD$ .Also let $S$ the area

of $ABCD$. If  $(OA)^2+(OB)^2+(OC)^2+(OD)^2=2S$  prove that $ABCD$ is sqrure

and its center is $O$.

Let $a,b,c,d$ denote the lengths of OA, OB, OC, OD.  Note that

$2S \leq ab+bc+cd+da \leq a^2+b^2+c^2+d^2$

which follows from $area(\triangle ABC) = \frac{1}{2}ab\sin C \leq \frac{1}{2}ab$ and $x^2+y^2 \geq 2xy$.  Equality holds iff $a=b=c=d$ and all the angles are right, i.e. if O is the center of square ABCD.


function
logarithms
inequalities
inequalities theorems

$\mathbf{\text{A new generalization of Holder's Ineuqlity}}$
If $f(x),g(x)$ are two increasing and convex function defined on $[0,+\infty)$ and $f(t)=g(t)=0$.Let $F(t)=\ln f(e^{t})$, $G(t)=\ln g(e^{t})$.
Such that:(1)$F(t),G(t)$ are convex for real number t.
(2)$\frac{1}{F(t)}+\frac{1}{G(t)}=\frac{1}{t}$(Then we may call that $f(t)$ and $g(t)$ are conjugated function if they such that condition (2)).
Then we have: \[\sum_{i=1}^{n}a_{i}b_{i}\le f^{-1}\left(\sum_{i=1}^{n}f(a_{i}) \right)g^{-1}\left(\sum_{i=1}^{n}g(b_{i}) \right)\] Where $a_{i},b_{i}>0$,$f^{-1},g^{-1}$ are the inverse function of $f,g$.

Notice:
1.And the Holder's inequality is the special case when $f(x)=x^{p},g(x)=x^{q}$,when $\frac{1}{p}+\frac{1}{q}=1,p,q>0$
2.The idea is come up by me,but I can't sure whether it has been appear somewhere,If you know,please tell me.
3.I didn't have a solution for it,hope somebody can interested in this messy inequality and give a solution. :)

I don't have my pen with me now, so I'm not sure if I've made wrong calculations, but it seems to me that it is true if f,g are multiplicative. I think we can imitate the old proof to develop a young-like inequality, then do it.

[quote="Soarer"]I don't have my pen with me now, so I'm not sure if I've made wrong calculations, but it seems to me that it is true if f,g are multiplicative.[/quote]
Hi,what do you mean by "multiplicative",I can't understand it very well. :?:

 [quote="Soarer"]I think we can imitate the old proof to develop a young-like inequality, then do it.[/quote]
I have tried it,but failed,Can you post the detailed?

It means $f(xy) = f(x)f(y)$ for any $x,y$

Thanks!
But if f,g are multiplicative then we can get $f(x)=x^{p},g(x)=x^{q}$ from the conditions.


[quote="b555"]Let a and d be positie integers. for any positive integer `n`, the number a+nd conntains a block of consecutive digits which constitute the number `n`. Prove that `d` is a power of 10.[/quote]

Let $ d\equal{}\overline{d_1xx...xxd_20...0}$ with $ d_1\neq 0,d_2\neq 0$, $ q$ trailing zeroes and $ p\plus{}q$ total digits.
Let $ a\equal{}\overline{d_3xx...xx}$ with $ r$ total digits.

Let $ k\equal{}1000\plus{}\max(p\plus{}q,r)$

Let $ N\equal{}a\plus{}10^kd$ $ \equal{}\overline{d_1xx...xxd_200...00d_3xx...xx}$
The left part ($ \overline{d_1xx...xxd_2}$) has $ p<k$ digits and so cant contain the representation of $ 10^k$
The right part ($ \overline{d_3xx...xx}$) has $ r<k$ digits and so cant contain the representation of $ 10^k$
The middle part contains $ k\plus{}q\minus{}r$ zeroes

In order the middle part contains the representation of $ 10^k$, we need $ d_2\equal{}1$ and $ k\plus{}q\minus{}r\geq k$ $ \iff$ $ q\geq r$

Let then $ M\equal{}a\plus{}(10^k\plus{}1)d\equal{}N\plus{}d$ $ \equal{}\overline{d_1xx...xxd_200...00xxx...xx}$
The left part ($ \overline{d_1xx...xxd_2}$) has $ p<k$ digits and so cant contain the representation of $ 10^k\plus{}1$
The right part is $ a\plus{}d$ and so has $ p\plus{}q<k$ or $ p\plus{}q\plus{}1<k$ digits and so cant contain the representation of $ 10^k\plus{}1$
The middle part contains $ k\plus{}q\minus{}(p\plus{}q)\equal{}k\minus{}p$ or $ k\plus{}q\minus{}(p\plus{}q\plus{}1)\equal{}k\minus{}p\minus{}1$ zeroes.

In order the middle part contains the representation of $ 10^k\plus{}1$, we need $ k\minus{}p\geq k\minus{}1$ or $ k\minus{}p\minus{}1\geq k\minus{}1$ and so $ p\leq 1$

So $ p\equal{}1$ and $ d_1\equal{}d_2\equal{}1$ and so $ d\equal{}\overline{10...0}\equal{}10^q$

Q.E.D.


ratio
Pythagorean Theorem
geometry
number theory
Diophantine equation

The lengths of the sides of a right triangle are three consecutive terms of an arithmetic progression.  Prove that the lengths are in the ratio 3:4:5.

Give:

Right triangle ABC, with sides a, b, and c.
a, b, c is an arithmatic progression.

Using the Pythagorean Theorem, you know that $c = \sqrt{a^2+b^2}$. You also know that since a, b, and c are in an arithmatic progression, than $c-b=b-a$, or $c=2b-a$.

Thus, you can set them equal to each other.

$\sqrt{a^2+b^2}=2b-a$
$a^2+b^2=4b^2-4ab+a^2$
$3b^2-4ab=0$
$3b-4a=0$

Treat the equation as a Diophantine equation where ordered pair (a,b) must have positive integer solutions. You can see that possible solutions include (0,0), (3,4), (6,8), ... As the pattern continues, a is always in a ratio with b of 3:4.

Thus, the arithmatic progression is now 3n, 4n, xn, where $n \geq 0$. Since it is an A.P., then

$xn-4n=4n-3n$
$x-4=4-3$
$x=5$

Thus, the arithmatic progression is 3n, 4n, 5n, where $n \geq0$. Thus, the ratios of the triangle are always in ratio of 3:4:5.

Better solution is set the sides equal to a-d, a , a+d. Then use Pythagorean theorem and you get a = 4d . Then substitute, so the sides are now 3d, 4d, 5d.


inequalities
number theory unsolved
number theory

$ \sigma (n)$ is the sum of all positive divisors of $ n$ then prove that:

\[ \frac{\sigma (n)}{n}<\prod_{p|n}\frac{p}{p\minus{}1}\]

It is obviosly because
\[ \frac {\sigma (n)}{n} \equal{} \prod_{p|n} (1 \plus{} \frac 1p \plus{} ... \plus{} \frac {1}{p^{ord_p(n)}}) < \prod_{p|n}\frac {1}{1 \minus{} \frac {1}{p}}.
\]


Today, when my school AMC manager was looking up AMC scores in the email (for 10B), he noticed that my first name went where my last name was, and vice versa. I'm assuming that I bubbled the Scantron answer form incorrectly. My score is right, but what should I do when I bubble in my name on the AIME form? I know (with ~250,000 participants) that it is almost impossible to edit my name, so should I bubble in the wrong name on the AIME?

Most likely you should contact the AMC and have them change it in their database.

On the other hand, they will surely notice when you submit your AIME and your name doesn't match any of the people who qualified, at which point they will certainly figure out what happened.

Bubble in your correct name on the AIME answer form.

Send us a message, or have your school contest manager send us a message noting the correct order and spelling of your name.

Steve Dunbar
Director, American Mathematics Competitions


geometry parallelogram trigonometry Asymptote

This is a pretty lengthy solution, and there is probably a much shorter way:

Answer

Let $ \angle CBP = x^{o}$

$ \angle BCD+\angle ABC = 180^{o}\Rightarrow \angle BCD = 180^{o}-2x^{o}$

Since $ \triangle PBC$ is isosceles with the apex at $ P$ then $ \angle BCP = \angle CBP = x^{o}$

$ \Rightarrow \angle PCD = \angle BCD-\angle BCP = 180^{o}-3x^{o}$

Since $ \angle CDP = 2x^{o}$ then $ \angle CPD = x^{o}$

Using the sine rule in $ \triangle CDP$:

$ \frac{\sin{2x^{o}}}{6}= \frac{\sin{(180-3x)^{o}}}{5}$

$ \Rightarrow 5\sin{2x^{o}}= 6\sin{(180-3x)^{o}}$

$ \Rightarrow 10\sin{x^{o}}\cos{x^{o}}= 6\sin{3x^{o}}$

$ \Rightarrow 5\sin{x^{o}}\cos{x^{o}}= 3\sin{(2x+x)^{o}}$

$ = 3\sin{2x^{o}}\cos{x^{o}}+3\cos{2x^{o}}\sin{x^{o}}$

$ = 6\sin{x^{o}}\cos^{2}{x^{o}}+3\sin{x^{o}}(2\cos^{2}{x^{o}}-1)$

$ \Rightarrow 5\cos{x^{o}}= 6\cos^{2}{x^{o}}+6\cos^{2}{x^{o}}-3$

$ \Rightarrow 12\cos^{2}{x^{o}}-5\cos{x^{o}}-3 = 0$

$ \Rightarrow (4\cos{x^{o}}-3)(3\cos{x^{o}}+1) = 0$

$ \Rightarrow \cos{x^{o}}= \frac{3}{4}$ or $ \cos{x^{o}}=-\frac{1}{3}$

But since $ \angle ABC < 180^{o}$ then $ x < 90^{o}$

$ \Rightarrow cos{x^{o}}= \frac{3}{4}$ is the only solution

Using the cosine rule in $ \triangle CDP$:

$ CD^{2}= 6^{2}+5^{2}-2 \cdot 5\cdot 6 \cos{x^{o}}= 36+25-60 \cdot \frac{3}{4}= 16$

$ \Rightarrow CD = 4$

But $ AB = CD \Rightarrow AB = 4$

[quote="mountain.dew"]In parallelogram $ ABCD$, the bisector of $ \angle ABC$ intersects $ AD$ at $ P$. If $ PD=5$, $ BP=6$, and $ CP=6$, what is the value of $ AB$?[/quote]

@sludethrower: Hide your solution!

Here is a much, much shorter solution.
[hide][asy]import cse5; pair B=MP("B",D((0,0))), A=MP("A",D(4*dir(97.1807557815)),N), C=MP("C",D((9,0))), D=MP("D",D(A+C),N), P=MP("P",D((5*A+4*D)/9),N); D(A--B--C--D--cycle); D(B--P--C);[/asy]
Let $ AP=x$ and $ m\angle ABP=\theta$ With a little angle chasing, we see that:
$ m\angle ABP = m\angle PBC = m\angle PCB = m\angle APB$.

This implies that $ AP=x$ and $ \triangle APB\sim\triangle PCB$, so:
$ \frac{x}{6}= \frac{6}{BC}= \frac{6}{AD}= \frac{6}{x+5}$.

Solving for $ x$, we get
$ 36=x^{2}+5x$
$ (x-4)(x+9)=0$

Since $ x>0$, $ x = \boxed{AB = 4}$.[/hide]
EDIT: Made picture.


geometry
trigonometry
analytic geometry
cyclic quadrilateral
trig identities
Law of Cosines
Pythagorean Theorem

On side $\overline{AB}$ of square $ABCD$, right $\triangle ABF$, with hypotenuse $\overline{AB}$, is drawn externally to the square.  If $AF=6$ and $BF=8$ find $EF$, where $E$ is the point of intersection of the diagonals of the square.

[hide="Hint"]
rePLY TO ME[/hide]

[hide]$\overline{AB} = \sqrt{6^2 + 8^2} = \sqrt{100} = 10$

$\angle AFB = \angle AEB = \frac{\pi}{2}$

$AFBE$ is a cyclic quadrilateral because $\angle AFB + \angle AEB = \pi$.  Therefore:

$\overline{AB} = \overline{EF} = 10$[/hide]

[quote="dakyru"]$\overline{AB} = \overline{EF} = 10$[/quote]
no no, i made the same mistake. rethink your logic on that statement. and try unscrambling joml88's hint (lol, clever hint btw)

here's a solution that didn't use the hint that joml88 was pointing at.
[hide]
Because you know $\overline{AE} = 5\sqrt{2}, \overline{AF} = 6$, and we can find the included angle $FAE$, we can find $\overline{EF}$ using the law of cosines:

By the law of cosines, $\overline{EF}^2 = 6^2 + (5\sqrt{2})^2 - 60\sqrt{2}\cos({\theta + 45^\circ})$, where $\cos\theta = \frac{3}{5}, \sin\theta = \frac{4}{5}$

Using the cosine sum formula, we see this reduces to

$\overline{EF}^2 = 36 + 50 - 60\sqrt{2}\displaystyle \left(\frac{3}{5}*\frac{\sqrt{2}}{2} - \frac{4}{5}*\frac{\sqrt{2}}{2}\right).$

Simplifying more, we get: $\overline{EF}^2 = 86 - 60\sqrt{2}\displaystyle \left(\frac{-\sqrt{2}}{10}\right).$

$\overline{EF}^2 = 86 -\frac{-120}{12}.$

Therefore, $\overline{EF} = 7\sqrt{2}.$

This answer agrees with the one obtained by using joml88's hint as well.  His way is much more simpler though.[/hide]

Well i was lazy, but now i'll just post my solution...
[hide]Draw the altitude $FD$.  We know that it's $\frac{24}{5}$ because altitude*hypotenuse=leg1*leg2 (each is twice the area of the triangle).  Then, we calculate  $AD=\frac{32}{5}$.  Let $G$ be the midpoint of $AB$.  Then $EG=\frac{7}{5}$.  Using the pythagorean theorem, we find taht $EF=\sqrt{\frac{49+49*49}{25}}=\frac{7}{5}\sqrt{50}=7\sqrt{2}$.[/hide]

[hide="Answer"]Obviously, since $AF=6$ and $BF=8$, $AB=10$. Also, we know that the area of $\triangle ABF=24$, so we can draw the altitude $AG$ to side $AB$, with length $4.8$, and draw $GH=5$, where $H$ is directly underneath $E$. Because $AGF\sim ABF$ and $FG=4.8$, $AG=3.6$, so $EH=1.4$, since $3.6+1.4=5$, which is $\frac{1}{2}AB$. Therefore, $EF=\sqrt{(4.8+5)^2+(1.4)^2}=\sqrt{98}=7\sqrt{2}$.[/hide]

[quote="xxreddevilzxx"]wow. a very nice hint. You rock!

[hide="solution"]

I only need two words for this.

[b]Ptolemy's Theorem.[/b]

[/hide][/quote]
exactly what dakyru did

[quote="Iversonfan2005"][quote="dakyru"]oh yea, duh.  another way:
[hide]
$\overline{AB} \cdot \overline{EF} = \overline{AF} \cdot \overline{BE} + \overline{BF} \cdot \overline{AE}$

$10 \overline{EF} = 6 \cdot 5\sqrt{2} + 8 \cdot 5\sqrt{2} = 70\sqrt{2} \rightarrow \overline{EF} = 7\sqrt{2}$[/hide][/quote]

Can someone please explain in what situations exactly can you use this formula?[/quote]
when the four points are concyclic, meaning when the four points are on a circle

Here's [url=http://www.cut-the-knot.org/proofs/ptolemy.shtml]Ptolemy's Theorem[/url] with a proof of it included.

[b]Hi, I have a solution:[/b]

[hide]The side of the square is sqr(64-36)=2*sqr(7).
IF you set the points on coordinate plane, point e would be at (sqr(7),sqr(7)) and F would be set at the point (0,-6), using distance formula, we find the distance to be sqr(50+12*sqr(7)), which is about 9
[/hide]

BF is not the hypotenuse, the side of the square is.  So that would make the side equal to 10.


topology
real analysis
real analysis unsolved

Show that the following two definitions of locally compactness in a [b] regular space [/b] are equivalent:
1).a topological space $X$ is locally compact iff every point has a local base of compact neighborhoods
2).a topological space $X$ is compact iff each point has an open neighbourhood whose closure is compact

If wikipedia is right, they are not equivalent for non-Hausdorff spaces.
But they are equivalent for Hausdorff spaces.

yes, i'm sorry for missing that-it is true iff our space is regular.

$(1)\Rightarrow(2)$, since given any point $x$ and any element $D$ of the local base of $x$ made up of compact sets, the closure of an open set around $x$ contained in $D$ is compact, since it's a closed subset of a compact space $D$.

If the space is Hausdorff, $(2)\Rightarrow(1)$: Let $G$ be an open set around $x$ whose closure $\overline G$ is compact. We can now assume that everything takes place in $X=\overline G$. In other words, what we want to prove is this: given a compact Hausdorff space $X$ and a point $x\in X$, vor any open neighborhood $V$ of $x$ there is a neighborhood $V_1$ of $x$ such that $\overline{V_1}\subset V$.

For every point $y\not\in V$, consider disjoint open neighborhoods $x\in V_y,\ y\in U_y$. The open sets $U_y$ cover the compact set $X\setminus V$, so finitely many of them, say $U_{y_i},\ i=\overline{1,n}$, cover $X\setminus V$. It is clear now that the closure of the open set $V_1=\bigcap_{i=1}^nV_{y_i}$ does not intersect $X\setminus V$, meaning that it's contained in $V$.

Edit:

And here's a simple counterexample to $(2)\Rightarrow (1)$ if the space is not Hausdorff: Take, for example, $\mathbb R$ with the cofinite topology, i.e. a set is open iff it's either the empty set or the complement of a finite set. The space itself is compact, so it satisfies condition $(2)$, but the closure of every open set is the entire space, so given an open set $V\ne\mathbb R$, there are no open sets whose closures are contained in $V$.


No matter which fifty five integers may be selected from
${1,2, . . . ,100}$, prove that you must select some two that differ by 9, some two that differ by 10, some
two that differ by 12, and some two that differ by 13, but that you need not have any two
that differ by 11.

[hide="9"]
We have that $55=6 \cdot 9+1$, so there must be at least $7$ numbers that have the same residue $\bmod 9$. But $\left\lceil \frac{100}{18}\right\rceil=6$, and the conclusion follows. [/hide]
[hide="10"]
Similarly, $55= 5 \cdot 10+5$, so there must be at least $6$ numbers in that have the same residue $\bmod 10$. But $\left\lceil \frac{100}{20}\right\rceil=5$, and the conclusion follows.[/hide]
[hide="11"]
Pick the all numbers in the form $a_{n,k}= 11(2n-1)+k$ for $n=1,2, \ldots 5$ and $k=0,1,2,3, \ldots 10$.[/hide]
[hide="12"]
We have that $55=4 \cdot 12+7$, so there must be at least $5$ numbers in each of $7$ residue classes $\bmod 12$. There exist only 4 residue classes in the numbers $\{1,2, \ldots 100 \}$ that contain $9$ elements, the others containing $8$, so the conclusion follows.[/hide]
[hide="13"]
We have that $55=4 \cdot 13+3$, so there must be at least $5$ numbers in the same residue class $\bmod 13$. But $\left\lceil \frac{100}{26}\right \rceil = 4$, and the conclusion follows.[/hide]


Prove that there exists a multiple of 2007 consisting of only 1's and zeroes :D

Well, you can always find 2 integers $ m$ and $ n$, ($ m>n$) such that $ 10^m$ and $ 10^n$ are congruent modulo $ 243$
So, $ \frac {10^m\minus{}10^n}{9} \equiv 0 (mod 27)$

Now try generalising the problem to any natural number n. :D  :lol:

Well that is really easy da befuddlers.
We have done it in class.
I don't how to post in latex.
Tht's the problem now.

aravind srinivas plz dont post the answer here as all chenaiites who attended the amti camp will know the solution. let the others try. :D

@aravind: u can pm ur solution, i don't think it requires latex though :rotfl:

Befuddlers
Can u reveal ur identity?
Are u in chennai?
Did u write AMTI 2nd level exam on 18th october?

i will reveal my identity later :rotfl:
Anyways did u get selected from the amti first level.
i am not living i chennai any longer tthough :rotfl:
So cud u post all the problems of amti 2nd level here  :D
i wrote inmo last year and in 9th but did not get seleced for the camp  :blush:

@ befuddlers I think only G vignesh, sanjith gopalakrishnan and ashwath rabinranath cleared RMO in 9th and 10th for the past two years in Tamil Nadu.
So who r u?

1. The number of students in a university is a perfect square initially. The next year, 200 more students join and the total is 1 more than a perfect square now.
The next year, further 200 more students join and the total is a perfect square. Find the no. of students in the university initially.

2.It is a geometric figure. I can't post it.

3. Suppose P(A) denotes the product of the digits of A. A is a natural number.
Find all A such that A=1.5 P(A).

4. I don't remeber it properly. I will post it soon. I am searching for my Q paper.

5.    same as above

6. I have posted it in Some maths olympiad problems for discussion Problem 2.

7. For positive reals a,b,c,d
prove that
1/a + 1/b + 4/c + 16/d >= 64/(a+b+c+d)

8. I will post it later. Don't remeber properly.

9. Geometric figure. I can't post it.

9.

many others like guna,krishna prasad and sai mali cleared the rmo  :rotfl:

Well Guna seemed to be like an underdog always though he is superb. It is stupid to have forgotten about him.

Wait a minute, are you Krishna Prasad Akkineni from DAV till 10th?
 Sai mali cleared RMO last year. I am talking about 2 years back. And in 2006, there was no RMO in TN. Amti's second level was the rmo separately for Junior and Inter.
But to say the truth, 2006 paper was damn easy. I got 6/8 problems in 3hrs.
You must be Krishna Prasad...........

Well ok not to deviate from the topic.

I will post it here.

Soln: Well, see , instead of using modulos for 10^n, 10^m etc as how hell_ever did,
Consider the sequence:
1, 11, 111, 1111, 11111,................,11111111.........................111111[n ones]
For which n we have to prove the divisibility property.
---->  By PHP, we have atleast one no. div. by n or two nos. leaving same remainder when divided by n which completed the proof.
This proof can also be extended on the generalization lines for any digit ranging from 1-9 along with 0.

Hence generalized.

just because i gave some specific names does not mean no others got.
sry i cant disclose my identity. :rotfl:
consider k.vignesh(now he is in 11th) who went to the imotc last year.  :lol:
there are many who clear the rmo in chennai(nandita bhaskar now in 11th came 3rd in the state for rmo last year.

Look at this very similar problem.  :)
[url=http://www.mathlinks.ro/viewtopic.php?t=235743]From Brazil Olympiad.[/url]


trigonometry
integration
calculus
function
calculus computations

Pleas provide the method or atleast give me a hint. Oh and the question is in the attached file.

[quote="matg"]Please can you explain the last step. How did you take y out of the integral?
Isn't 'y' a function of theta?[/quote]

You were integrating with respect to x so you consider y as a constant.

I would think that the given condition implicitly defines $ y$ [i]in terms of[/i] $ x$ and $ c$, and not as a sovereign variable which can be treated as a constant.

If y is constant can't we straight away take it out in the question itself. In the condition x and y are variables dependant on each other whereas c is a constant. If x is a function of theta so is y. Maybe we have to write y as a function of x and substitute it, etc.......

x= r cos theta, y=r sin theta...

r^4=2c^2r^2 cos 2theta

Please can you post the entire solution?


Just wondering if you guys noticed any increase in site speed starting today :) We've done some *upgrades* :P and, well, it seems to me that it's faster, but it might be only a placebo. So we're anxious to hear what you think & feel about the speed ...  :?

I don't know if it is just me, but lately it is faster 95% or 97% of the time, but it happens something that didn't before (say, a week before), one of each twenty or twenty-five clicks (changing from one page to another) it gets slow, and spend up to twenty seconds to upload the next page.

Also, often when I receive an email saying "Topic Replay Notification - Blah Blah", for some reason, if I click there, I often (say 10-20% of the times) get a "404 - Not Found" message!

But yes, in general, it is faster, but it was fast any way before!

Best,

That usually happens when someone posts something (then the email is instantly sent) and then the person decides to delete their reply, or a moderator does that ... so you get a broken link.

I don't know about the 20-25 clicks thing :D You know where to look to see the page generation times, right?

Well, if this would help you, Valentin: :)

[ Time: 0.4456s ][ Queries: 34 (0.2723s) ][ GZIP enabled - Debug on ]

Not quite, the PGT's show indeed an increase in speed up to 400% :) I just wanted to know if this is being felt ... :)


algorithm
function
parameterization
linear algebra
matrix
algebra
polynomial

Hi,
I wasn't sure if this should go to algebra forums so i just put it here, also excuse me for my bad English...

I am trying to run a few problem solving algorithms and i was wondering if i was missing something in the process.
For description of ANY function (exempt sum functions and diferentials here please for the purpose of my research they are irrelevant (i think :) ) ) i use the following formula

A - constant parameter matrix (which them self can be complex formulas - like A[3,3,1] = 5*sin(Y[3,3,1]) e.t.c.)
Y - source parameter matrix - variables which have no interrelationship
f(x) := A x Y

now, i know that this is just not enough, but i am at a loss here, my math is just to weak. I need some universal formula which can better describe relationship then f(x) = constant operand variable

I hope i made any sort of sense... This is used for seemingly random pattern prediction via use of gene algorithms...

Any help would be greatly appreciated...

regards

Milika Delic

[quote="milika"]
I hope i made any sort of sense... [/quote]
Not to me, sorry :( You may be up to something but I was completely unable to understand what you were doing or what you are asking about. If you can explain your point with some clear simple examples, that might help a bit. Seems like you are trying to build an elementary function of one (?) variable and want some way to "mix" several given functions into a new one. So, tell me what the result will be if you mix $x$ and $e^{x}$. Another problem is that not every function admits an elementary formula. On the other hand, any continuous function on an interval can be approximated arbitrarily well by either algebraic or trigonometric polynomial, so I do not see why you need anything beyond one of these classes. :?

Sorry, i knew this would happen. Let me try again - with examples :)

All the problems i work with can be explained with a matrix of variable parameters and constant operands. My quest is to find these conctant operands...

Like this
f(x) = A x X (Y was en error before - sorry)
where (simple example)
X = [ 5*, 3.2*]
      [ 4*, 7.1*]
so my constant parameters are in a form of Const+Operand(*)
and f(x) = 5*x11+3.2*x12+4*x21+7.1*x22
classic matrix multiplication....
Of course these constant parameters can be in more complex form with multiple const and operand like
4*cos(5.2*(4-x))

Simple example is calculating point of intersection of multiple light beams (if data is not 100% accurate)...
My algorithm - using this approach - will find the most likely origin point...

Hope i made more sense :)

But now i am to tackle a bit more complex problems - like picture/sound frequency prediction and i am afraid that there may be relationships which i can not represent by this model...

So my question is - could i make more complete model (bear in mind that, for example, x11, x12, x21, x31 have no interrelationships!)... It does not matter how complex it is, just to be more complete...

Uf, hope this was not too much...

Thanks for reading...

I would have a much easier time understanding what you've written if it were in LaTeX.  See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690]here[/url] for a LaTeX FAQ for this forum.

I folowed the link to LaTeX and it seems like i will just make it worse in it :)

but i will try to explain symbols better
f(x) = A x X
where A and X are matrices
and operand [b]x[/b] is multiplication of matrices (only for this example - read on to see that operand can be more complex...)

X in
[b]X =
[ 5*, 3.2*]
[ 4*, 7.1*] [/b]
is a 2 by 2 matrix

A is a matrix of complex calculus algorithms (this is perhaps most confusing sorry)
in the example f(x) = 5*x11+3.2*x12+4*x21+7.1*x22 it was just normal multiplication with a const

Now, of course, matrices can be bigger, but X and A are always the same dimensions...

My problem consist in the fact that this model may not represent all posible, or at least, all relevant cases...

Hope this made it all a bit clearer...

you need to put the dollar sign in between your functions.. dollarsignx^2dollarsign

for example $x^{2}$


limit
function
real analysis
calculus
calculus computations

Let $ f(x)\equal{}\frac{1}{q}$ if $ x\equal{}\frac{p}{q}$ where $ p$ and $ q$ are integers with no common factor with $ q>0$ and $ f(x)\equal{}0$ if $ x$ is irrational.  At which points, if any, is $ f$ continuous?

$ \lim_{x \to c} f(c)$ exists nowhere (we can take either rational or irrational paths), so $ f(x)$ is discontinuous everywhere.

No, I think its continuous at every irrational point. :|

this is the Thomae's function.

Hmm.  Now that I think about it, you're right.   :oops:

Firstly, it is necessary to let $ f(0) \equal{} 0$.
Secondly, the above function is so-called Riemann function.
Lastly, I will show you the function discontinuous at every rational different from zero.

Indead, observe first that if $ x \ne x_n \equal{} \frac {p_n}{q_n}$ converges to $ x$ then $ \mathop {\lim }\limits_{n \to \infty } q_n \equal{} \infty$. So, if $ x \notin \mathbb Q$ then $ \mathop {\lim }\limits_{n \to \infty } f\left( {x_n } \right) \equal{} 0 \equal{} f\left( x \right)$. Now if $ \{z_n\} \not \subset \mathbb Q$ converges to $ x$ then $ \mathop {\lim }\limits_{n \to \infty } f\left( {z_n } \right) \equal{} 0 \equal{} f\left( x \right)$. This means that $ f$ is continuous at every irrational.

Likewise, one can show that $ 0$ is a point of continuity of $ f$.

Suppose now that $ \frac {p}{q} \equal{} x \ne 0$. If $ \mathbb Q \not \ni x_n$ converges to $ x$ then $ \mathop {\lim }\limits_{n \to \infty } f\left( {x_n } \right) \equal{} 0 \ne f\left( x \right)$. Consequently, $ f$ is discontinuous at every rational different from zero.

[quote="bookworm_vn"]
Indead, observe first that if $ x \ne x_n \equal{} \frac {p_n}{q_n}$ converges to $ x$ then $ \mathop {\lim }\limits_{n \to \infty } q_n \equal{} \infty$. [/quote]

this is nontrivial to me.  Can you prove it?

Let $ x$ be irrational. Choose any arbitrary $ M.$ Within 1 unit on either side of $ x,$ how many numbers $ \frac{p}{q}$ are there with $ q\le M?$ It's a finite union of finite sets, hence finite. Since it's a finite set, there is a minimum distance from that set to $ x.$ Let $ \delta$ be a positive number less than that minimum distance. Then if $ \left|\frac{p}{q}\minus{}x\right|<\delta,$ we have that $ q>M.$ That proves the assertion in the previous post.

---

There's a theorem that says that a function whose discontinuities comprise a set of Lebesgue measure zero is Riemann integrable. Hence this function is Riemann integrable. The challenge: over some bounded interval - say, $ [0,1],$ just to be specific - show directly that this function is Riemann integrable without quoting that theorem.

Let $ [a,b]$ be arbitrary fixed. Observe that, given a positive integer $ N$, there are only finitely many rational $ p/q$ in $ [a,b]$ such that $ q \leq N$. Let $ k_N$ denote the number of such rationals. Consider a partition of $ [a,b]$ of mesh $ \delta >0$. Then there are at most $ 2k_N$ subintervals $ [x_{i\minus{}1}, x_i]$ containing at least one of the rationals mentioned above. On the other subintervals $ [x_{j\minus{}1}, x_j]$ we have $ M_j \minus{}m_j < 1/N$. Consequently
\[ U\left( {P,f} \right) \minus{} L\left( {P,f} \right) < 2k_N \delta  \plus{} \frac{{b \minus{} a}}
{N}.
\]
Given $ \varepsilon >0$, take $ N>2(b\minus{}a)/\varepsilon$. Then for any partition of mesh $ \delta < \varepsilon/(4k_N)$ we get
\[ U\left( {P,f} \right) \minus{} L\left( {P,f} \right) < \varepsilon .
\]


1: Easy. Prove every even number bigger than 2 and less than $2^n$ is the sum of $n$ primes.

2: Harder. Prove every even number bigger than 2 and less than $4^n$ is the sum of $n$ primes.

No one interested? Even the first one? There is always a prime between n and 2n?

The first uses Bertrands' postulate

Bomb


Farmer John has $ 5$ cows, $ 4$ pigs, and $ 7$ horses. How many ways can he pair up the animals so that every pair consists of animals of different species? (Assume that all animals are distinguishable from each other.)

[hide="Solution"]
Start with the horses, since they are the most numerous:
You can pair the 7 horses with 5 cows and 2 pigs, with 2 pigs left over, but this fails because the 2 pigs can't be paired.
You can pair the 7 hourses with 3 cows and 4 pigs, but this also fails because the two leftover cows can't be paired.

This means that you must pair the horses with 4 cows and 3 pigs, with one pig and one cow left over.

There are 4 ways to choose a pig to leave out and 5 ways to choose a cow to leave out, for 20 combinations of animals to pair with the horses.

Since all animals are distinguishable, let's line up the horses and assign each horse one of the 7 distinguishable animals. There are 7! ways of doing this.

Thus, the answer is $ 4 \cdot 5 \cdot 7! \equal{} 100800$
[/hide]

[hide]of the 9 cows and pigs, 7 will pair up with a horse, so 1 pig pairs up with one cow. There are (5)(4) ways to choose the pig-cow pair. Then there are 7! ways that the horses can get paired up with the remaining 7 animals. The final answer is 7!(5)(4) =$ \boxed{100800}$[/hide]


trigonometry
calculus
algebra
polynomial
induction
search
real analysis

using [i]elementary [/i]trigonometry and [i]elementary [/i]calculus only,
show   :D

$ \textbf{(i)}\;\;\;\sum_{m=1}^{n-1}\;\sin\left( \frac{m\pi}{n}\right)\;=\;\boxed{\cot\left(\frac{\pi}{2n}\right)}$

$ \textbf{(ii)}\;\;\;\prod_{m=1}^{n-1}\;\sin\left( \frac{m\pi}{n}\right)\;=\;\boxed{\frac{n}{2^{n-1}}}$

[quote="misan"]

$ \textbf{(ii)}\;\;\;\prod_{m=1}^{n-1}\;\sin\left( \frac{m\pi}{n}\right)\;=\;\boxed{\frac{n}{2^{n-1}}}$[/quote]

This one was posted here before, I just cant find it.

[hide="Hint"]
Consider the cyclotomic polynomial $ \Phi_{n}(z) = 1+z+z^{2}+...+z^{n}$ then factor it by roots of unity.
[/hide]

It's easy to show (for example by induction) that $ \sum_{m=1}^{n-1}\sin m\xi=\frac{\cos\frac{\xi}{2}-\cos\left(n-\frac{1}{2}\right)\xi}{2\sin\frac{\xi}{2}}$.
If we put  $ \xi=\frac{\pi}{2n}$ we'll got what we want :P

[quote="perfect_radio"][quote="sylow_theory"]Consider the [b]cyclotomic polynomial[/b] $ \Phi_{n}(z) = 1+z+z^{2}+...+z^{n}$ then factor it by roots of unity.[/quote]
That's not the [url=http://mathworld.wolfram.com/CyclotomicPolynomial.html]cyclotomic polynomial[/url] ;) They are quite different. I don't think yours has a name.

Anyway, the idea is correct; also, for related problems, you can check [url=http://mathlinks.ro/Forum/viewtopic.php?search_id=1442391955&t=37627]this[/url] out.[/quote]
That's strange cuz my algebra book define that polynomial by that name.

And if fact, if you go to the link on MathWorld it says,
$ \Phi (z) = \prod_{k=1}^{n}\left( z-e^{2\pi ik/n}\right)$

It starts from $ k=1$ [b]not[/b] front $ k=0$.  :wink:
(So it cant be $ z^{n}-1$. )

[quote="sylow_theory"]That's strange cuz my algebra book define that polynomial by that name.[/quote]
Are you sure? What book are you talking about (title, author)? It surely does not respect the "mainstream" terminology :D

[quote="sylow_theory"]And if fact, if you go to the link on MathWorld it says,
$ \Phi (z) = \prod_{k=1}^{n}\left( z-e^{2\pi ik/n}\right)$

It starts from $ k=1$ [b]not[/b] front $ k=0$.  :wink:
(So it cant be $ z^{n}-1$. )[/quote]
If you click again on that link and actually read what is written there (besides that formula), you'll see that you're wrong:
\[ \Phi_{n}(x) = \prod_{\gcd (k,n)=1}\left( x-\zeta^{k}\right) , \]
where $ \zeta = \exp \left( \frac{2 \pi i}{n}\right)$. But the notation used on Mathworld is very misleading, indeed.
Maybe you'll find it better explained [url=http://en.wikipedia.org/wiki/Root_of_unity#Cyclotomic_polynomials]here[/url].


Find all integer triplets $ (a, b, c)$ that satisfy $ abc \equal{} 4(a \plus{} b \plus{} c)$ with $ a < b < c$.

EDIT: All [b]positive[/b] integers

[hide="hint"]
What happens when $ a\ge 4$?

What happens when $ a\le \minus{}5$?

[/hide]

i think you meant [b]positive[/b] integers??
otherwise , there are infinite solutions to the equation.
take, for example:
$ a \equal{} \minus{}n , b\equal{}0, c\equal{} n$
:)

[quote="hash_include"]otherwise , there are infinite solutions to the equation.[/quote]
That an equation has infinitely many solutions does not preclude you from finding them.

yes, i agree. my bad.
let me rephrase then: one of the generic solutions to the equation is when $ a \equal{} \minus{}n , b\equal{}0, c\equal{} n$

Sorry. I meant positive, non-zero integers. :)

-bump-

Does anyone have a solution for this?

$ abc\equal{}4a\plus{}4b\plus{}4c<4c\plus{}4c\plus{}4c\equal{}12c$, so $ ab<12$. The rest is brute force?

Yeah, a lot of brute force...so we do cases.

When a=1, $ bc\minus{}4b\minus{}4c\minus{}4\equal{}0$ so with SFFT, we get
$ (b\minus{}4)(c\minus{}4)\equal{} 20$ and since b<c, the solutions (a,b,c) are $ (1,5,24), (1,6,14), (1,8,9)$

When a=2, we get $ (b\minus{}2)(c\minus{}2)\equal{}8$, so solutions (a,b,c) are $ (2,3,10), (2,4,6)$

When a=3, we get $ (3b\minus{}4)(3c\minus{}4)\equal{}20$ and we realize that no solutions would result since all the b's we get are less than 3, which contradicts a<b.

Since $ a\equal{}\frac{4b\plus{}4c}{bc\minus{}4}$, we realize that as $ a$ increases, $ b$ has to decrease. Therefore, if a=3 produces no solutions, then any a>3 produces no solutions. The only solutions are $ (1,5,24), (1,6,14), (1,8,9), (2,3,10), (2,4,6)$


geometry
perpendicular bisector
geometry proposed

This isn't about solving a difficult problem (it's not difficult :)), it's about finding a nice solution.

Given four distinct points $A_1,A_2,B_1,B_2$ in the plane, show that if every circle through $A_1,A_2$ meets every circle through $B_1,B_2$, then $A_1,A_2,B_1,B_2$ are either collinear or concyclic.

Suppose that $A, B, C, D$ are neither collinear or concyclic.

If $A_1A_2$ is parallel to $B_1B_2$, then we can easily find two circles through that do not meet.

Otherwise, let the perpendicular bisector of $A_1A_2$ meet the perp bisector of $B_1B_2$ at $O$. Then $OA_1 \neq OB_2$, so the two concentric circles with centre $O$ through $A_1$ and $B_2$ respectively satisfy the properties.

Nice :). I considered an inversion of pole $A_1$. This should make it simpler because you only have one thing to prove: the images $A_2',B_1',B_2'$ of $A_2,B_1,B_2$ through the inversion must be collinear.

[quote="billzhao"]Suppose that $A, B, C, D$ are neither collinear or concyclic.[/quote]

I guess you meant $A_1$, $A_2$, $B_1$, $B_2$ rather than $A, B, C, D$.

Actually, the same solution was given in

  H. S. M. Coxeter, S. L. Greitzer, [i]Geometry Revisited[/i], Toronto - New York 1967,

chapter 5 1. It's somehow irritating that Coxeter and Greitzer use this problem for their introduction of inversion, but don't solve it using inversion. Coxeter and Greitzer prove a stronger result:

[i]If $A_1$, $A_2$, $B_1$, $B_2$ are four distinct points in the plane, then the following two assertions are equivalent:

[b](a)[/b] The points $A_1$, $A_2$, $B_1$, $B_2$ either lie on one circle such that each of the two arcs $A_1A_2$ contains exactly one of the points $B_1$ and $B_2$, or lie on one line such that the segment $A_1A_2$ contains exactly one of the points $B_1$ and $B_2$.

[b](b)[/b] Each circle passing through the points $A_1$ and $A_2$ intersects each circle passing through the points $B_1$ and $B_2$.[/i]

  Darij


Can someone  please  tell me  how  can  i write in  the center of a  page ? Or at  the  right  part of the  center? (a title  for example.) Thanks!

[quote="zabelman"]To center your text, use [color=darkred]\begin{center} TEXT GOES HERE \end{center}[/color]. Likewise, [color=darkred]\begin{flushright}\end{flushright}[/color] aligns on the right, and [color=darkred]\begin{flushleft}\end{flushleft}[/color] aligns on the left.

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

If you want to put a title on your article, there is a built-in way to do it. Put the following lines in your preamble

[color=darkred]\title{Continued Fractions of Tails of Hypergeometric Series}
\author{Jonathan Michael Borwein, et. al.}[/color]

and then put [color=darkred]\maketitle[/color] in the body of your document where you want the title to appear.[/quote]

Ι  would like to make my self  little  bit  more clear:
I  would like  to have a  word  (example: Solutions) on the  center of a  page (not center of a  line), and  a  dedication  on the right  side  of the  centers  page.
Thank  you very much!


How many of you are writing the full test tomorrow? This question is directed specially at computer students.

Ofc shrids will becos he learnt all this up or rather taught all this to the inventor of C++ in his previous janma when he was 3 yrs old and then rediscovered his passion again now :D

Well to the not so fortunate people $ \in$ Set of Fools like me, doubtfulness is the answer :D

please do not compare me with Stroustrup. he is too low level for me :D
and what i learn is "beyond the scope of the school syllabus". *chews paan while saying so*  :rotfl:  :rotfl:

I am taking it. More than CSC exam, it is just two weeks for iit..

[quote="Chapplikka"][quote="rituraj007"]correction--2 weeks & 1 day :rotfl:[/quote]
He means two weeks from FT 9, not two weeks from today.. TOtal kandravi nv...[/quote]

i think its FT 10 :huh: :P :rotfl:

okkkkkkk

but what is FT?((which time zone))

No, fiitjee messed his test namings.. Its supposed to be ft 9 and the ones we got home are supposed to be 6,7,8.. And ft is full test...

Can we see the result or will it come Fiitjee style

Maybe it will come only after the JEE :rotfl: :D :D

Thats Better because if we by mistake(though we shdnt) get an ok(not good) mark then the sensitive people might take it seriously

Open test was supposed to be an open test and is not supposed to come within the realms of the aits.. There is exactly where Fiitjee messed the whole thing up..

What do you mean by open test  is not open?? You expect ppl to write the open test nude or something like that??!! In that context, yea it isnt open...

[quote="rituraj007"]Fiitjee~~the real messer
neways it will give a good practice even without results
but open test isn't really open or is it???[/quote]'

Well you need to pay rs.300 to write it. And its held in a closed shed and you are perfectly allowed to wear any clothing that you like. Don't believe bullies and EKS's like Chapplikka.

Is it possible to teleport between Tirupathi and Maharishi Vidya Mandir? Or even better, defy death? If I write, I am going to end up getting murdered by my family, thats for sure. "Govinda, govinda!"

(It was meant to be religious,  no jabs people)

I'm sure FIITJEE being the perfect gentlemen they are, will release the results on April 12th night or even better 13th morning just to give us a good boost that day. :)

they r just TOOO gentle---have u ever attended their regular classes?????

Please post your expected marks here guys :)

And also continue your discussions here :)

Bet the paper was d a m n easy :D

PS I'm d a m n bored and wish i could have written too

[quote="shruthibaskaran"] Maharishi Vidya Mandir[/quote]
r u  talking about VMC?????????


superior algebra
superior algebra unsolved

Well by multiplicativity of the degrees it must be a multiple of each number ($12$, $40$ here), and smaller than their product also. Thus only the possibilities above.


We have to prove that:
$ \sum_{cyc}\frac {(x \plus{} 1)(y \plus{} 1)^2}{3\sqrt [3]{z^2x^2} \plus{} 1} \geq x \plus{} y \plus{} z \plus{} 3$

But:(By MA-MG and Cauchy)
$ \sum_{cyc}\frac {(x \plus{} 1)(y \plus{} 1)^2}{3\sqrt [3]{z^2x^2} \plus{} 1}\geq\sum_{cyc}\frac{(y\plus{}1)^2}{x\plus{}1}$

$ \sum_{cyc}\frac{(y\plus{}1)^2}{x\plus{}1}\geq x\plus{}y\plus{}z\plus{}3$
as claimed.


Say: $F = (a+b)(b+c)(c+a) = a b^2 + a^2 b + b c^2 + b^2 c + c a^2 + c^2 a + 2abc$
Since $\frac{(a-b)(b-c)(c-a)}{(a+b)(b+c)(c+a)} = \frac{1}{11}$ which is rational, we know $F \neq 0$.

No we have $R = 11(a-b)(b-c)(c-a) - (a+b)(b+c)(c+a)$

$= 10a b^2 + 12a^2 b + 10b c^2 + 12b^2 c + 10c a^2 + 12c^2 a + 22abc$

Now, our desired quantity:

$D = \frac{a}{a+b} + \frac{b}{b+c} + \frac{c}{c+a}$

$\frac{a b^2 + 2a^2 b + b c^2 + 2b^2 c + c a^2 + 2c^2 a + 3abc}{F}$

It's very evident now, that:

$2DF + 8F = R = 0$

Since $F \neq 0$, we divide by $F$ to get: $2D + 8 = 0$ and we end up with our desired quanity $D = -4$.

Denote $\frac a{a+b}=x, \ \frac b{b+c}=y,\ \frac c{c+a}=z$. We have to calculate $s=x+y+z$.
From the statement we have
$(2x-1)(2y-1)(2z-1)=\frac 1{11}$,
hence
(1)   $4(2xyz-(xy+xz+yz))+ 2(x+y+z)-1=\frac 1{11}$.
OTOH, $\frac 1{x}-1=\frac b{a}, \ \frac 1{y}-1=\frac c{b}, \ \frac 1{z}-1=\frac a{c}$. This yields to
$(\frac 1{x}-1)(\frac 1{y}-1)(\frac 1{z}-1)=1$.
We obtain
$1-(x+y+z)+xy+xz+yz-xyz=xyz$,
hence $2xyz-(xy+xz+yz)=1-(x+y+z)=1-s$.
Replace in (1) and obtain a simple equation in $s$.

[quote="Pork_Chop8"][color=red]No we have $R = 11(a-b)(b-c)(c-a) - (a+b)(b+c)(c+a)$

$= 10a b^2 + 12a^2 b + 10b c^2 + 12b^2 c + 10c a^2 + 12c^2 a + 22abc$[/color]
[/quote]

[b]Pork_Chop8[/b] i think you did something wrong there.

[b]Raluca[/b] you got  :first:


The [i]tower function of twos[/i] is defined recursively as follows: $ T(1) \equal{} 2$ and $ T(n \plus{} 1) \equal{} 2^{T(n)}$ for $ n\ge1$. Let $ A \equal{} (T(2009))^{T(2009)}$ and $ B \equal{} (T(2009))^A$. What is the largest integer $ k$ such that
\[ \underbrace{\log_2\log_2\log_2\ldots\log_2B}_{k\text{ times}}
\]is defined?

$ \textbf{(A)}\ 2009\qquad \textbf{(B)}\ 2010\qquad \textbf{(C)}\ 2011\qquad \textbf{(D)}\ 2012\qquad \textbf{(E)}\ 2013$

We just look at the last three logarithms for the moment, and use the fact that $ \log_2 T(k) \equal{} T(k \minus{} 1)$. We wish to find:
\[ \log_2(\log_2(\log_2 ({T(2009)^{T(2009)}}^{T(2009)})))
\]\[ \log_2(T(2009)\log_2(T(2009)\log_2 T(2009))))
\]\[ \log_2(T(2009)\log_2(T(2009)T(2008)))
\]\[ \log_2(T(2009)(T(2008) \plus{} T(2007)))
\]Now we realize that $ T(n \minus{} 1)$ is much smaller than $ T(n)$. So we approximate this, remembering we have rounded down, as:
\[ \log_2(T(2009)) \equal{} T(2008)
\]We have used $ 3$ logarithms so far. Applying $ 2007$ more to the left of our expression, we get $ T(1) \equal{} 2$. Then we can apply the logarithm $ 2$ more times, until we get to $ 0$. So our answer is approximately $ 3 \plus{} 2007 \plus{} 2 \equal{} 2012$. But we rounded down, so that means that after $ 2012$ logarithms we get a number slightly greater than $ 0$, so we can apply logarithms one more time. We can be sure it is small enough so that the logarithm can only be applied $ 1$ more time since $ 2012 \plus{} 1 \equal{} 2013$ is the largest answer choice. So the answer is $ \mathrm{E}$.

[hide="Solution"]This is a pretty ugly sum, so we just start taking logs. Use $ \log_2^k$ to denote $ \log_2$ iterated $ k$ times.
\begin{align*} \log_2 B & = A\log_2 T(2009) \\
& =  AT(2008) \\
\log_2^2 B & = \log_2 A + T(2007) \\
& = T(2009)T(2008) + T(2007)
\end{align*}We can't easily take any more logs, so we try and bound the expression; we can continue taking logs as long as $ \log_2^k B > 0$. Note that we can take the log of both sides of an inequality as it is an increasing function.
\[ T(2009) < \log_2^2 B < T(2009)^2 < T(2010)
\]\begin{align*} T(2008) < & \log_2^3 B < T(2009) \\
& \ \qquad\vdots \\
T(1) < & \log_2^{2010} B < T(2) \\
1 < & \log_2^{2011} B < 2 \\
0 < & \log_2^{2012} B < 1 \\
& \log_2^{2013} B < 0 \\
\end{align*}Once the quantity is negative, we can no longer compute its logarithm, so we can take exactly $ 2013\rightarrow\boxed{\textbf{(E)}}$ logs.[/hide]

We know that taking $ log_2$ of $ T(n)$ exactly $ n \plus{} 1$ times will give 0.

Then $ B \equal{} T(2009)^{\left(T(2009)^{T(2009)}\right)} > 2^{\left(2^{T(2009)}\right)} \equal{} T(2011)$.
Therefore taking $ log_2$ of $ B$ 2012 times will give us something positive, which means we can take $ log_2$ at least 2013 times, so the answer must be E.

[quote=dogseatcheese][quote="Duelist"]
Now we realize that $ T(n \minus{} 1)$ is much smaller than $ T(n)$. So we approximate this, remembering we have rounded down, as:
\[ \log_2(T(2009)) \equal{} T(2008)
\]
We have used $ 3$ logarithms so far. Applying $ 2007$ more to the left of our expression, we get $ T(1) \equal{} 2$. Then we can apply the logarithm $ 2$ more times, until we get to $ 0$. So our answer is approximately $ 3 \plus{} 2007 \plus{} 2 \equal{} 2012$. But we rounded down, so that means that after $ 2012$ logarithms we get a number slightly greater than $ 0$, so we can apply logarithms one more time. We can be sure it is small enough so that the logarithm can only be applied $ 1$ more time since $ 2012 \plus{} 1 \equal{} 2013$ is the largest answer choice. So the answer is $ \mathrm{E}$.[/quote]

When you say we approximate "this" I am not sure what this is... I think it's the
\[ \log_2(T(2009)(T(2008) \plus{} T(2007)))
\]
However, how do you approximate it, because isn't
\[ \log_2(T(2009)) \equal{} T(2008)
\]
exact and not an approximation?

I'm sort of confused...help? I understand everything before and after this step.[/quote]

Yea, I'm a bit confused too


integration
real analysis
real analysis unsolved

Dear all,

I am reading a book and the following is an argument adapted from the book: Since\[ \mathop {\sup }\limits_n \int_{ \minus{} \infty }^\infty  {\left( {\sum\limits_{k \equal{} 1}^n {\left| {{f_{k \plus{} 1}}\left( t \right) \minus{} {f_k}\left( t \right)} \right|} } \right)dt}  < \infty\]there is, by Levi's Lemma, a measurable set $ E$ of measure zero such that for any $ t \in \mathbb R \minus{} E$,\[ \mathop {\sup }\limits_n \sum\limits_{k \equal{} 1}^n {\left| {{f_{k \plus{} 1}}\left( t \right) \minus{} {f_k}\left( t \right)} \right|}  < \infty , \quad \forall t \in \mathbb{R}\backslash E.\]Can anybody tell me what the statement of the Levi's Lemma is? Thank you  :) .

let $ F_n(t) \equal{} \sum_1^n |f_{k\plus{}1}(t) \minus{} f_k(t)|$ (notice that $ n \to F_n(t)$ is increasing positive, and that's the only property used here) and $ E \equal{} \{ t: \sup_n F_n(t) \equal{} \plus{}\infty\}$. If $ 0 < \lambda(E) < \infty$ (the case $ \lambda(E)\equal{}\infty$ is exactly similar), for every constant $ C>0$ you can find $ N$ large enough such that the set $ E_N \equal{} \{ t: F_N(t) > C\}$ has a measure at least equal to $ \frac{\lambda(E)}{2}$:
\[ \int_{\mathbb{R}} F_N(t) \, dt \geq \frac{\lambda(E)}{2} C\]
which gives you the contradiction you wanted, isn't it ?

[quote="alekk"]...
\[ \int_{\mathbb{R}} F_N(t) \, dt \geq \frac {\lambda(E)}{2} C\]
which gives you the contradiction you wanted, isn't it ?[/quote]

Thanks alekk, I got the point I needed :roll:

In addition, the problem is: Let be given $ \{f_n\} \subset L^1(\mathbb R)$, $ f \in L^1(\mathbb R)$ is such that
\[ \int\limits_{ \minus{} \infty }^{ \plus{} \infty } {\left| {{f_{n \plus{} 1}}\left( t \right) \minus{} {f_n}\left( t \right)} \right|dt}  \leqslant \frac{1}
{{{n^2}}},\quad \forall n \geqslant 1.\]Then $ f_n \to f$ a.e. with respect to Lebesgue measure.

Having all the discussion above, we see that
\[ {f_n}\left( t \right) \equal{} {f_1}\left( t \right) \plus{} \sum\limits_{k \equal{} 2}^n {\left( {{f_k}\left( t \right) \minus{} {f_{k \minus{} 1}}\left( t \right)} \right)}  \to f\left( t \right)\]almost everywhere.  :wink:


[quote="Iversonfan2007"]I was just wondering, I faxed my entry form to USAMTS about 3 days ago, but on my profile, it says that they haven't gotten the form yet. Should I fax it again or should I just wait a few days (since I'm sure a bunch of ppl have faxed them this week).[/quote]

We received over 2000 pieces of paper in our fax machine in the last 4-5 days.  It'll be a little while before we have them all sorted out.  Don't re-send the form unless it says we don't have it a couple weeks from now.

I sent in the answers via e-mail on Sunday, but only as TEX, not as PDF (with the USAMTS files attached).  The fact that I sent it was not logged on my USAMTS account.  Is there a way of making sure that they received it?

[quote="doubleloop277"]I faxed in my entry form on October 9, and the website still says that I have not sent it in.  Should I re-fax it or are you still sorting through the forms you received?[/quote]
Still sorting, is my guess. The recent sticky at the top of the forum states that they probably will not get all the scores logged until Wednesday. My understanding is that they do not update [i]anything[/i] for [i]anybody[/i] until all the grading is done. If it still says that they have not received your entry form by Friday, and other people have confirmed that their usamts.org recognizes that their forms have been received (in other words, it says other people's forms are in but not yours) [i]then[/i] I would send it again.

As of now, the website says that it has not received most people's entry forms. Unless you sent it in way before the due date.


In a right triangle, the lengths of the legs are 333 and 444. What is the length of the hypotenuse?

[hide]555 with proportions
think 3-4-5 triangles[/hide]

555:
333 squared + 444 squared = 555 squared


It's rather easy, but interesting.
lets solve:

$ \frac{1}{x} \plus{}\frac{1}{x\plus{}1} \plus{}\frac{1}{x\plus{}2} \plus{}\frac{1}{x\plus{}3} \plus{}\frac{1}{x\plus{}4} \equal{}0$. :)

$ x\equal{}\minus{}2\pm\sqrt{{3\over 2}\pm\sqrt{29\over 20}}$
How is that interesting?  :huh:

[quote="olorin"]$ x \equal{} \minus{} 2\pm\sqrt {{3\over 2}\pm\sqrt {29\over 20}}$
How is that interesting?  :huh:[/quote]

Correct :D


trigonometry
ARML
inequalities
calculus
function
trig identities
Law of Cosines

How do you solve the following problems? (please show work)

1. THe lengths of the sides of a triangle are consecutive integers, and the largest angle is twice the smaller angle. Compute the cosine of the smallest angle.
A. $\frac {3}{4}$

B. $\frac {7}{10}$

C. $\frac {2}{3}$

D. $\frac {9}{14}$

E. None of these

2. Let $x$, $y$ and $z$ be three positive real numbers whose sum is $1$. If no one of these numbers is more than twice any other, then the minimum possible value of the product $xyz$ is:

A. $\frac{1}{32}$

B. $\frac{1}{36}$

C. $\frac{4}{125}$

D. $\frac{1}{127}$

E. None of these

3. The number of pairs of positive integers $(x,y)$ which satisfy the equation $x^2+y^2 = x^3$ is:

A $0$

B $1$

C $2$

D $\infty$

E None of these

Enjoy these problems.  :)

[hide]
1)  A random fact I remembered from last year's ARML practices: it's a 4-5-6 triangle.
Use the law of cosines to get [b]A[/b]

2) I just plugged in a couple values for (x,y,z) to get (2/5,2/5,1/5), so xyz=4/125
[b]C[/b]

3)  $y^2=x^2(x-1)$ so $(x-1)$ is a square.  There are infinitely many $x$ for which $x-1$ is a square, so [b]D[/b]

Are these right?
[/hide]

Danbert's answer for #1 is correct.  If you didn't know that fact, here is a possible solution:

[hide]Let the sides have lengths x-1, x, and x+1.  Also, let the angle opposite the x-1 side be $\theta$, so that the angle opposite the x+1 side is $2\theta$.  From the law of sines, we have

\[\frac{x+1}{\sin 2\theta} = \frac{x-1}{\sin \theta}\]
\[\frac{x+1}{2\sin\theta\cos\theta} = \frac{x-1}{\sin \theta}\]
\[\frac{x+1}{2\cos\theta} = x-1\]
\[\frac{x+1}{x-1} = 2\cos\theta.\]

But from the law of cosines,
$\cos\theta = \frac{(x+1)^2 + x^2 - (x-1)^2}{2x(x+1)} = \frac{x^2+4x}{2x(x+1)} = \frac{x+4}{2(x+1)}$
so

\[\frac{x+1}{x-1} = \frac{x+4}{x+1}\]
\[(x+1)^2 = (x+4)(x-1)\]
\[x = 5\]

Thus $\cos\theta = \frac{9}{12} = \frac{3}{4}$, A.[/hide]

Danbert's number two is incorrect(sorry for putting it so bluntly).

[hide] I'm not sure if my solution is correct, but it's definently not 4/125. You want them to be as seperate as possible so the largest to be as large as possible. For anything over 1/2, the sum of the other two numbers must be less than 1/2. That would mean that atleast one of them is less than half of 1/2. Therefore, the largest number is 1/2. The other two are 1/4 and 1/4. This gives 1/32. This might not be correct, but it's smaller than 4/125.[/hide]

Well, for number 2, I almost know that the problem requires AM-GM equality for a quick solution. The problem just looks almost perfectly set up for that theorem. However, also this is like a #28 on an ASHME, it obviously must use some abstract theorem. Try labeling three points x, y, z, and use AM-GM on them.

I can pretty much guarantee you there is no AM-GM solution.  The inequality sign for AM-GM points the wrong way; it can potentially give you a maximum value for $xyz$, but not a minimum.

For number 2...

[hide]
... if $x\le y\le z<2x$, then $xyz-x(y+z-2x)(2x)=x(2x-y)(2x-z)>0$, so replacing $x,y,z$ with $x,y+z-2x,2x$ decreases the product. Now we can assume that $z=2x$, so the numbers are $x,1-3x,2x$, where $\tfrac15\le x\le\tfrac14$ to guarantee they are in the correct order. Now we just need to minimize $(x)(1-3x)(2x)$ on this interval. Do this however you want (calculus, graphing, whatever), and you'll see that $x=\tfrac14$, so the answer is $\mathbf{\tfrac1{32}}$.

EDIT: Actually, the cubic $(x)(1-3x)(2x)$ has the roots $0,0,\tfrac13$, so the point of inflection is at the average of these roots, i.e. at $\tfrac19$. After this, the function is concave down. Therefore, the minimum over our interval must be at an endpoint, and checking $x=\tfrac14,x=\tfrac15$ is easy.
[/hide]

Qn 3: $x^3 - x^2 = x^2(x-1)$ As long as x-1 is a square, we can put $y = x \sqrt{x-1}$, so there are infinite solutions.


calculus
integration
logarithms
trigonometry
algebra
partial fractions
calculus computations

$ \int{\frac{2t+1}{t^2-t+2}} dt$

For this one I realize that if you substitute
$ u = t^2-t+2$
then, $ du = 2t-1$
I'm not sure how to get it to be $ 2t+1$ so it covers the numerator.

$ \int{\frac{5}{3e^x-2} dx}$

This one I tried adding and subtracting an $ e^x$ on the numerator and splitting it up but I couldn't get anywhere...

$ \int{\frac{2x-3}{x(x-1)^2}} dx$

For this one I realize you have to use partial fractions but I'm not sure how to set it up.

Thanks.

for the first one in the numerator u have $ 2t\plus{}1$  and the differential coefficient of the numerator is $ 2t\minus{}1$

so wat u do is write the numerator as $ 2t\minus{}1\plus{}2$      and now split it.  now 2/t^2-t+1  can be done  by using a standard formula.   other  part is too obvious.

hello, for the first one, writing your integral in the form $ \int\frac{2t\plus{}1}{(t\minus{}\frac{1}{2})^2\plus{}\frac{7}{4}}\,dt$ and setting $ t\minus{}\frac{1}{2}\equal{}u$ we get $ \int\frac{2u}{u^2\plus{}\frac{7}{4}}\,du\plus{}\int\frac{1}{u^2\plus{}\frac{7}{4}}\,du$.
Sonnhard.

hello, for the second one set $ t\equal{}3e^x\minus{}2$, then you will get $ x\equal{}\ln\left(\frac{1}{3}(t\plus{}2)\right)$ and $ dx\equal{}\frac{dt}{t\plus{}2}$.
Sonnhard.

hello, for the last one you must write $ \frac{2x\minus{}3}{x(x\minus{}1)^2}\equal{}\frac{A}{x}\plus{}\frac{B}{x\minus{}1}\plus{}\frac{C}{(x\minus{}1)^2}$.
Sonnhard.

[quote
$ \int{\frac {5}{3e^x - 2} dx}$

.[/quote]

assume :  \[ y = 3e^x \Rightarrow dy = 3e^x dx \Rightarrow  dx = \frac{1}{y} dy\]

\[ now\]

\[ \int \frac{5}{y - 2}. \frac{1}{y} dy  = \int \frac{5}{y\left ( y - 2 \right )}dy
= \int \left [ \frac{\frac{-5}{2}}{y} + \frac{\frac{5}{2}}{y - 2} \right ] dy\]

\[ =- \frac{5}{2}ln\left | y \right | + \frac{5}{2}ln\left | y-2 \right | + c ==- \frac{5}{2}ln\left | e^x \right | + \frac{5}{2}ln\left | e^x-2 \right | + c\]

1. $ \int \frac {2t \plus{} 1}{t^2 \minus{} t \plus{} 2}\ dt \equal{} \int \left(\frac {2t \minus{} 1}{t^2 \minus{} t \plus{} 2} \plus{} \frac {2}{t^2 \minus{} t \plus{} 2}\right)\ dt$

$ \equal{} \ln (t^2 \minus{} t \plus{} 2) \plus{} 2\int \frac {1}{\left(t \minus{} \frac {1}{2}\right)^2 \plus{} \frac {7}{4}}\ dt$

$ \equal{} \ln (t^2 \minus{} t \plus{} 2) \plus{} 2\cdot \frac {2}{\sqrt {7}}\tan ^ { \minus{} 1} \frac {2}{\sqrt {7}}\left(t \minus{} \frac {1}{2}\right) \plus{} C$

$ \equal{} \boxed{\ln (t^2 \minus{} t \plus{} 2) \plus{} \frac {4}{\sqrt {7}}\tan ^{ \minus{} 1} \frac {2t \minus{} 1}{\sqrt {7}} \plus{} C.}$

2. $ \int \frac {5}{3e^x \minus{} 2}\ dx \equal{} \int \frac {5e^{ \minus{} x}}{3 \minus{} 2e^{ \minus{} x}}\ dx$

$ \equal{} \frac {5}{2}\int \frac {(3 \minus{} 2e^{ \minus{} x})'}{3 \minus{} 2e^{ \minus{} x}}\ dx$

$ \equal{} \boxed{\frac {5}{2}\ln |3 \minus{} 2e^{ \minus{} x}| \plus{} C.}$

3. $ \int \frac {2x \minus{} 3}{x(x \minus{} 1)^2}\ dx \equal{} \int \left( \minus{} \frac {3}{x} \plus{} \frac {3}{x \minus{} 1} \minus{} \frac {1}{(x \minus{} 1)^2}\right)\ dx$

$ \equal{} \boxed{3\ln |x \minus{} 1| \minus{} \ln |x| \plus{} \frac {1}{x \minus{} 1} \plus{} C.}$


group theory
abstract algebra
superior algebra
superior algebra solved

Let (G,.)be a group with p*q elements,p<q ,p,q prime numbers(bigger than 1);if G has only one subgroup with p elements,prove that there is x in G so that x^(p*q)=e and x^k is not equal to e when k=1,pq-1.[/quote]

I think the text says the group has exactly one subgroup of order p and exactly one of order q.

Let |G1|=p , |G2|=q be the subgroups mentioned above. For a in G, define f_a(x) = axa^(-1). f_a is an internal automorphism of G so it takes G1 to another subgroup of order p. Thus f_a(G1)=G1, for all a. Equivalently :

(1) aG_1 = G_1a for all a in G

In a similar manner, it is easy to prove that :

(2) aG_2 = G_2a, for all a in G

Let u in G1 be a generator of G1 and v in G2 be a generator of G2. (1) + (2) imply that uv=vu. Then (uv)^k = e ==> u^k v^k = e ==> we prove that it's a necessity that u^k=v^k=e. Suppose it wasn't, then v^(q-k) = u^k (3) ==> Let k' be the residue of k mod p ==> Because k' isn't 0, k' is invertible in Z_p so we get u = v^t for some t, by raising (3) to the k''-th power, where k'' is the inverse of k' in Z_p. Therefore since u=v^t, u is in G2, a contradiction, unless u=e (which is not a generator of G1!). So we must have u^k=v^k=e. Then p|k, q|k ==> pq|k, and since k=pq satisfies (uv)^k = e, then it is the smallest number and we are done.

We know that G has only one subgroup with p elements,so it must be proved that it has only one subgroup with q elements;

[quote]SAT math proves no theorem.[/quote]

Let H with |H|=p be the only subgroup of order p of G. Then let u be an element of H which generates H. Consider v in G\H. Since ord(v) | ord(G)=pq ==> ord(v) is p,q or pq. If ord(v)=pq then we are done. If ord(v)=p then it's easy to see that <v>  \cap  H = {e}, and so G has two distinct subgroups of order p. The only possibility is that ord(v)=q. Consider the automorphism f(x) = v x v^(-1) . Since f(H)=H it follows that uv=vu. It's easy to see like in the previous solution that uv is a generator of G.

To prove that there's only one subgroup of order q:

Such a subgroup does exist (Cauchy's thm). Suppose we have x, y of order q s.t. y isn't in <x>. Then  the group <x,y> has at least q2 >pq elements (the elements of form xm*yn with n and m from 0 to q-1), but it's a subgroup of G, which has pq elements, so there's a contradiction (a subgroup of G has more elements of G).


Is there a partition of the real line into countably many compact sets?

I believe I posted a proof on the forum of a theorem due to Sierpinski, stating that a compact connected topological space cannot be partitioned into more than one but at most countably many non-empty compact subsets. The negative answer follows immediately from that theorem.


geometry
circumcircle
Euler
analytic geometry
vector
geometric transformation
homothety

Let $O$ be the circumcenter and $H$ the orthocenter of an acute triangle $ABC$. Prove that the area of one of the triangles $AOH$, $BOH$ and $COH$ is equal to the sum of the areas of the other two.

We can prove a huge generalization.

First, since G is collinear with H and O (Euler line), it suffices to show that [AGO]+[BGO]+[CGO]=0, where [RST] is the signed area of triangle RST (two vertices form a reference axis, and if the third vertex is below this axis, the area is negative and if it is above, the area is positive).
We prove much more generally that if X=(x,y,z) in barycentrics, then
x[AXP]+y[BXP]+z[CXP]=0 for all points P in the plane of ABC.

Let X be the origin and XP be the x-axis (reference axis) of this coordinate system.  Then the vector sum xXA+yXB+zXC=(x+y+z)X=0.  In particular, the y-components of these vectors, say y_a, y_b, and y_c are the components of A, B, and C repsectively, must satisfy 0=x*y_a+y*y_b+z*y_cimplying 0=x*|XP|*y_a+y*|XP|*y_b+z*|XP|*y_c=x[AXP]+y[BXP]+z[AXP], as claimed.  Since G=(1,1,1), the result follows immediately.

Of course, without any barycentrics or more general theory, a contestant could prove this in a completely analogous manner knowing simply that G=(A+B+C)/3.

Well, there is actually a very simple synthetic proof which only involves the line passing through G.

Are we allowed to post APMO questions now, technically?

Of course we are, they are even on the Canadian MO website.

vectors
OA+OB+OC=OH

AHxOA+BHxOB+CHxOC
=(OB+OC)xOA+(OA+OC)xOB+(OB+OC)xOA
=[OBxOA+OAxOB]+[OAxOC+OCxOA]+[OBxOC+OCxOB]
=0+0+0=0

so there are |OA|=|OB|+|OC| or |OB|=|OA|+|OC|or ...

We see that AH=2RcosA, <OAH=|C-B|, OA=R so S(AOH)=R 2 cos A sin |C-B|. For every ordering of the angles of the triangle we transform the product into sum and we have an identity.

[quote="vinoth_90_2004"][i]maybe this question can be generalized to isogonal conjugates?
(i.e. if P and Q are isogonal conjugates in triangle ABC, then the area of one of APQ, BPQ and CPQ is the sum of the other two.)[/i][/quote]

Not really; actually, if P and Q are two [i]arbitrary[/i] points in the plane, then the area of one of the triangles APQ, BPQ and CPQ equals the sum of the areas of the other two if and only if the line PQ passes through the centroid of triangle ABC. But, of course, not for any two isogonal conjugates P and Q, the line PQ passes through the centroid.

  Darij

Beautiful problem, but I'm under the impression that it's been discussed before.

What we must prove is that one of the distances $d(A,OH),d(B,OH),d(C,OH)$ is the sum of the other two.

Consider the homothety $\mathcal H$ centered at $G$ (the centroid of $ABC$, as usual) of ratio $-\frac 12$. It turns $OH$ into itself, and it turns $A,B,C$ into the midpoints $D,E,F$ of $BC,CA,AB$ respectively. Moreover, it shrinks all distances by a factor of $\frac 12$. Suppose $OH$ cuts the segments $AB,AC$. The distance $d(D,OH)$ is half the distance $d(A,OH)$, because $OH,D$ are the images of $OH,A$ respectively through $\mathcal H$. At the same time, we clearly have $d(D,OH)=\frac{d(B,OH)+d(C,OH)}2$. From these two facts we deduce $d(A,OH)=d(B,OH)+d(C,OH)$, as desired.

Edit:

Oops! Darij was here first :).

Only in the [i]acute[/i] triangle: let $a\le b\le c.$ Thus, $(AOH)=\frac {AO\cdot AH}2\cdot\sin (\widehat {OAH})=$

$R^2\cos A\sin (C-B)=-R^2\cdot \cos (B+C)\sin (C-B)=$ $\frac {R^2}2\cdot (\sin 2B-\sin 2C)$ a.s.o.

$16S\cdot [AOH]=(b^2+c^2-a^2)\cdot \left|b^2-c^2\right|$ a.s.o. Prove easily that $a\le b\le c\Longrightarrow [BOH]=[AOH]+[COH]\ .$

[b]sorry, my English is poor.[/b]

I think this is easy.

wlog $B$, $C$ are on the same side of $OH$. Then we will show that $d(A,OH)=d(B,OH)+d(C,OH)$. But $G$ is on $OH$, It is obvious!

http://www.cut-the-knot.org/wiki-math/index.php?n=MathematicalOlympiads.APMO2004Problem2

Vo Duc Dien

very quick
[hide=Solution]
Suppose $A \ge B \ge C$ for angles $A, B, C$ of the triangle. Other cases are similar. We will show $[OBH]=[OAH]+[OCH]$.

Chase $\angle OBH=A-C, \angle OAH=B-C, \angle OCH=A-B$ in this case. Since $AH=2R \cos A$ and similarly for $BH, CH$ it is enough to show with $\sin$ area formula that
\[ \cos B \sin (A-C) = \cos A \sin (B-C) + \cos C \sin (A-B) \]
\[ \cos B (\sin A\cos C-\sin C \cos A)=\cos A(\sin B \cos C - \cos B \sin C)+\cos C(\sin A \cos B - \cos A \sin B) \]
which is true, so done.
[/hide]

When [hide=cartesian coordinates]
Let $A=(x_1,y_1)$, $B=(x_2,y_2)$, $C=(x_3,y_3)$, and $O=(0,0)$. From this, we can obtain that $H=(\sum x, \sum y)$, where $\sum x=x_1+x_2+x_3$ and $\sum y=y_1+y_2+y_3$. From this, the shoelace formula immediately yields that the signed area
$$[AHO]=\frac{1}{2}
\left|
\begin{array}{ccc}
x_1 & y_1 & 1 \\
\sum x & \sum y & 1 \\
0 & 0 & 1
\end{array}
\right|=\frac{1}{2}(x_1\sum y-y_1 \sum x).$$
Similarly, we can obtain that $[BHO]=\frac{1}{2}(x_2\sum y-y_2\sum x)$ and $[CHO]=\frac{1}{2}(x_3\sum y-y_3\sum x)$. Finally, summing these areas yields
$$[AHO]+[BHO]+[CHO]=\frac{1}{2}(\sum x\sum y-\sum x\sum y)=0$$
which immediately implies the desired conclusion.
[/hide] are actually usable on an olympiad question...

It's well known that $OH$ passes through $G$.
Lemma $:$ let $l$ be a random line passing through $G$ and let $d_1,d_2,d_3$ be distances of $A,B,C$ from $l$ then one of $d_1,d_2,d_3$ is sum of two others.
Proof : Let $M$ be midpoint of $BC$. Let $H_1,H_2,H_3$ be perpendiculars from $B,M,C$ to $l$ then we have $BH_1H_3C$ is trapezoid so $MH_2 = \frac{BH_1 + CH_3}{2} = \frac{d_2+d_3}{2}$. Also we have $AG = 2GM \implies d_1 = 2MH_2 \implies d_1 = d_2 + d_3$.
Now Note that our problem is just to prove that one of $A,B,C$ has distance from $OH$ equal with sum of two others which is the result of our lemma.
we're Done.

[quote=e_plus_pi][b][u][i]SOLUTION[/i][/u][/b]
[b]WLOG[/b] assume $ \angle A \ge \angle B \ge \angle C $

[u][b]CLAIM[/b][/u] $(AOH) + (COH) = (BOH) $
[i][b]PROOF[/b][/i]
Note that .....
$ (1) $ $ (AOH) = ( AH * AO * sin\angle OAH)  = (2R cos A)(R)(sin \angle OAH) = (2R cos A)(R)(sin (B-C))$
$ (2) $ $ (COH) = ( CH * CO * sin\angle OCH) = (2R cos C)(R)(sin \angle OCH)= (2R cos C)(R)(sin(A-B)) $
$ (3) $ $ (BOH) = ( BH * BO * sin\angle OBH ) =(2R cos B)(R)( sin \angle OBH) = (2R Cos B)(R)(sin (A-C)) $

[b]We Show That[/b] $ (1) + (2) =(3).$
$ \implies  2R cos A sin(B-C) + 2R cos C sin (A-B) = 2R cos B sin (A-C) $
Which is true by simple expansion.

My Solution is quite similar to tree3' s solution :)[/quote]

[quote=math_pi_rate]My solution: Let $M$ be the midpoint of $BC$. Then it is well known that $AH=2OM$. Let $A,B,C$ denote the angles of $\triangle ABC$. Notice that $\angle HAO=\angle CAH-\angle CAO=|(90^{\circ}-C)-(90^{\circ}-B)|=|B-C|$. Also, Let $a,b,c$ denote the sides of $\triangle ABC$ and $R$ be its circumradius. Then, $$AH=2OM=2 \sqrt{BO^2-BM^2}=2 \sqrt{R^2-\left(\frac{a}{2} \right)^2}=2 \sqrt{R^2-(R \sin A)^2}=|2R \cos A|$$
$$\Rightarrow [AOH]=\frac{1}{2}(AH)(AO) \sin \angle HAO=|R^2 \cos A \sin (B-C)|=\frac{1}{2}R^2|\sin (A+B-C)-\sin (A-B+C)|=\frac{1}{2}R^2|\sin 2C-\sin 2B|$$

Similarly, we get that $[BOH]=\frac{1}{2}R^2|\sin2A-\sin2C|$ and $[COH]=\frac{1}{2}R^2|\sin2B-\sin2A|$.

Now, WLOG, assume that $\sin2A \geq \sin2B \geq \sin2C$. Then one gets that $[AOH]+[COH]=[BOH]$. Hence, proved. $\blacksquare$[/quote]

[quote=e_plus_pi][b][u][i]SOLUTION[/i][/u][/b]
[b]WLOG[/b] assume $ \angle A \ge \angle B \ge \angle C $

[u][b]CLAIM[/b][/u] $(AOH) + (COH) = (BOH) $
[i][b]PROOF[/b][/i]
Note that .....
$ (1) $ $ (AOH) = ( AH * AO * sin\angle OAH)  = (2R cos A)(R)(sin \angle OAH) = (2R cos A)(R)(sin (B-C))$
$ (2) $ $ (COH) = ( CH * CO * sin\angle OCH) = (2R cos C)(R)(sin \angle OCH)= (2R cos C)(R)(sin(A-B)) $
$ (3) $ $ (BOH) = ( BH * BO * sin\angle OBH ) =(2R cos B)(R)( sin \angle OBH) = (2R Cos B)(R)(sin (A-C)) $

[b]We Show That[/b] $ (1) + (2) =(3).$
$ \implies  2R cos A sin(B-C) + 2R cos C sin (A-B) = 2R cos B sin (A-C) $
Which is true by simple expansion.

My Solution is quite similar to tree3' s solution :)[/quote]

How can you understand that AH=2RcosA...
And angle OAH=B-C...

WLOG assume that $A$ lies on the opposite side of the euler line than $B$ and $C$.

Let $M$ be the center of $BC$. Let $'$ denote the projection of a point onto the euler line, and let $G$ be the centroid. It is well-known that $AGM$ is collinear.

It remains to show that $AA' = BB' + CC'$ or that $2 \cdot MM' = BB' + CC'$. This follows since $MM'$ is the midline of the trapezoid $BB'CC'$.

Solved using hints from [b]EGMO.[/b] WLOG assume that B and C are on the same side of the Euler line. If M is the midpoint, we claim that AD=BE+CF (note that this is preserved cyclically), where D, E, F are the projections of A, B, and C onto that line. Indeed, by sim triangles GDA and GMM' (projection of M) we have AA'=2MM'=BE+CF since MM' is the midline of right trapezoid BEFC. $\blacksquare$

Apply Cartesian coordinates with $OH$ as the real axis. The vectors $\overrightarrow{OA}$, $\overrightarrow{OB},$  and $\overrightarrow{OC}$ sum to $\overrightarrow{OH}$, so the y-coordinates of $A,B,C$ sum to 0. Thus, the absolute value of one of them is the sum of the absolute values of the other two, hence done since all three triangles have $OH$ as the base and some height away from the line.

Without loss of generality, we assume $A$ lies on the opposite side of the Euler line, shown in blue, as $B$ and $C$. (Note it is impossible for $A$, $B$, and $C$ to all be on the same side as the centroid is always inside the triangle.) We claim \[[AOH] = [BOH] + [COH].\]

Let the projections of $A$, $B$, and $C$ onto the Euler line be $A_1$, $B_1$, and $C_1$, respectively. The problem reduces to proving \[AA_1 = BB_1 + CC_1.\]

Denote $M$ as the midpoint of $BC$ and $M_1$ as the projection of $M$ onto the Euler line. As $MM_1$ is the midline in trapezoid $BB_1C_1C$, we have \[MM_1 = \frac{BB_1 + CC_1}{2}.\]

Similar triangles $\triangle AA_1G$ and $\triangle MM_1G$, where $G$ is the centroid and on the Euler line, give \[AA_1 = \frac{AG}{GM} \cdot MM_1 = 2MM_1 = BB_1 + CC_1. \text{ } \blacksquare\]

[asy]
size(250);

pair A, B, C, H, A1, B1, C1, M, G, M1;
A = dir(131);
B = dir(200);
C = dir(340);
H = orthocenter(A, B, C);
A1 = foot(A, H, 0);
B1 = foot(B, H, 0);
C1 = foot(C, H, 0);
M = .5B + .5C;
G = centroid(A, B, C);
M1 = foot(M, B1, C1);

markscalefactor=.005;
fill(A--A1--G--cycle, mediumgreen);
fill(G--M1--M--cycle, mediumgreen);
draw(rightanglemark(G, A1, A));
draw(rightanglemark(B, B1, A1));
draw(rightanglemark(G, M1, M));
draw(rightanglemark(M1, C1, C));

draw(A--B--C--cycle);
draw(B--B1);
draw(C--C1);
draw(M1--M--A--A1);
draw(B1--C1, blue);

label("$A$", A, N);
label("$B$", B, SW);
label("$C$", C, SE);
label("$A_1$", A1, SE);
label("$B_1$", B1, NW);
label("$C_1$", C1, NE);
label("$M$", M, S);
label("$G$", G, NE);
label("$M_1$", M1, N);

dot(A);
dot(B);
dot(C);
dot(A1);
dot(B1);
dot(C1);
dot(M);
dot(G);
dot(M1);
[/asy]

Without loss of generality assume $[AOH] = [BOH] + [COH]$. Now proceed with complex numbers. Then the area of $[AHO]$ is given by,
\begin{align*}
\frac{i}{4}\begin{vmatrix}a&\frac{1}{a}&1\\ a+b+c&\frac{1}{a}+\frac{1}{b}+\frac{1}{c}&1\\ 0&0&1\end{vmatrix} = \frac{i}{4abc}\left [(a^2 - bc)(b+c)\right ]
\end{align*}
Similarly we can calculate the areas of $[BOH]$ and $[COH]$ and adding gives the conclusion.

I will use complex numbers to solve
lets o=0 and h is reel number
X,Y,Z is the foot altitude from A,B,C to OT respectively ,where t=1
WLOG AX>=BY>=CZ
h=a+b+c=a'+b'+c' and
x=(a+a')/2 and y=(b+b')/2 and z=(c+c')/2
AX=│a-a'│/2 BY=│b-b'│/2 CY=│c-c'│/2
because a+b+c-a'-b'-c-=0 so
AX=BY+CZ
==> [AHO]=[BOH]+[COH] .

WLOG let $A$ be on the opposite side of $HO$ as $B$, and $C$. We claim that $[AOH]=[BOH]+[COH]$ Let $G$ be the centroid of $\triangle ABC$ Note that $G$ lies on $HO$ since $HO$ is the Euler line of $\triangle ABC$. Let $H_{A}$, $H_{B}$, and $H_{C}$ be the projections of $A$, $B$, and $C$ respectively onto $HO$. It suffices to show $BH_{B}+CH_{C}=AH_{A}$. Let $M$ be the midpoint of $BC$ and $H_{M}$ be its projection onto $OH$. Note that $MH_{M}$ is the midline of trapezoid $BH_{B}H_{C}C$. By centroid $2:1$ ratios we have $\tfrac{BH_{B}+ CH_{C}}{2}= MH_M=\tfrac{AH_A}{2}$ So $BH_{B}+CH_{C}=AH_{A}$, as desired.

[hide=My proof of the equality given in #2]Let $P=(p,q,r)$, then\begin{align*}x\frac{[AXP]}{[ABC]} & =x\begin{vmatrix}1&0&0\\x&y&z\\p&q&r\end{vmatrix}=xyr-xzq, \\
y\frac{[BXP]}{[ABC]} & =y\begin{vmatrix}0&1&0\\x&y&z\\p&q&r\end{vmatrix}=yzp-yxr, \\
z\frac{[CXP]}{[ABC]} &=z\begin{vmatrix}0&0&1\\x&y&z\\p&q&r\end{vmatrix}=zxq-zyp,
\end{align*}so$$x\frac{[AXP]}{[ABC]}+y\frac{[BXP]}{[ABC]}+z\frac{[CXP]}{[ABC]}=0,$$and therefore$$x[AXP]+y[BXP]+z[CXP]=0,$$as wanted.[/hide]

[hide=My solution]Use complex numbers with $(ABC)=\mathbb{S}^1$, then\begin{align*}\sum \limits _{P\in \{A,B,C\}}[POH] & =\sum \limits _{p\in \{a,b,c\}}\frac{i}{4}\cdot \begin{vmatrix}p&\frac{1}{p}&1\\0&0&1\\a+b+c&\frac{1}{a}+\frac{1}{b}+\frac{1}{c}&1\end{vmatrix} \\
& =\sum \limits _{p\in \{a,b,c\}}-\frac{i}{4}\cdot \begin{vmatrix}p&\frac{1}{p}\\a+b+c&\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\end{vmatrix} \\
& =-\frac{i}{4}\cdot \sum \limits _{p\in \{a,b,c\}}\begin{vmatrix}p&\frac{1}{p}\\a+b+c&\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\end{vmatrix} \\
& =-\frac{i}{4}\cdot \begin{vmatrix}a+b+c&\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\\a+b+c&\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\end{vmatrix} \\
& =-\frac{i}{4}\cdot 0 \\
& =0.
\end{align*}Done.[/hide]

Let $D$, $E$, and $F$ be the foots of the perpendiculars from $A$, $B$, and $C$ to $OH$, respectively. WLOG, let $A$ and $B$ lie on the same side of $OH$ and $C$ lie on the other. Then, we want to prove that $AD+BE=CF$. Since the centroid lies on $OH$, this must be true by coordinates.

WLOG $[AOH] = [BOH] + [COH]$.
By Euler Line, the centroid $G$ lies on $OH$.
Let the projections of $A$, $B$, and $C$ onto the Euler Line be $D$, $E$, and $F$ respectively. Then it is equivalent to show that $AD = BE + CF$.
Define the midpoint of $BC$ to be $M$, and the projection of $M$ onto $OH$ being $N$.
Then notice that $AD = BE + CF = 2MN$. This follows easily from noticing that $\triangle ADG \sim MNG$(by homothety) with a $2:1$ ratio, which finishes.


blogs
binomial coefficients
Pascal\u0027s Triangle

I want to see how many proofs of this we can find...

$\binom{n}{0}+\binom{n}{1}+\cdots+\binom{n}{n} = 2^n$

See my [url=http://wangsblog.com/jeffrey/]blog[/url] for one, and the basics behind another.

well, by the binomial expansion theorem:

$(a + b)^n = \binom{n}{0}a^n + \binom{n}{1}a^{n-1}b + \dots + \binom{n}{n}b^n$

plug in $a = 1$ and $b=1$, we get:

$(1 + 1)^n = 2^n = \binom{n}{0} + \binom{n}{1} + \dots + \binom{n}{n}$

inductive proof:

[hide][b]Base case:[/b] $n=0$, which is just 1=1.

[b]Inductive step:[/b] $\sum_{k=0}^{n+1} \binom{n+1}{k}=\binom{n+1}{0}+\sum_{k=1}^n \binom{n+1}{k}+\binom{n+1}{n+1}$.

$\binom{n+1}{0}=\binom{n}{0}=1=\binom{n+1}{n+1}=\binom{n}{n}$, so the sum is $\binom{n}{0}+\sum_{k=1}^n \binom{n+1}{k} +\binom{n}{n}$.

Using Pascal's Identity on the center terms, the sum in the center equals $\sum_{k=1}^n \left( \binom{n}{k}+\binom{n}{k-1}\right)=\sum_{k=0}^{n-1} \left( \binom{n}{k}+\binom{n}{k+1}\right)$.

Grouping identical binomial coefficients and adding in the $\binom{n}{0}$ and $\binom{n}{n}$ from the outside,
\[ \sum_{k=0}^{n+1} \binom{n+1}{k}=2\sum_{k=0}^n \binom{n}{k}=2*2^n=2^{n+1} \][/hide]

I already posted the proof based on subsets of a set of size n.  Here's another:

Given that ${n \choose r} + {n \choose r+1} = {n+1 \choose r+1}$ (can be proven, or can be taken to mean that Pascal's triangle generates binomial coefficients), we notice that the sum of the first row of Pascal's triangle is $2^0 = 1$.  We then notice that each term in the next row is given by the sum of two terms of the row above it.  Each term in the row above is used once as the left side of a sum, and once as the right side of a sum, so each term is used exactly twice, which implies that the sum of each successive row of Pascal's triangle is twice the sum of the previous, giving us the desired result.

(This is more or less identical to scorpius' proof, but reasoned with words.)

Another "subset" method :
${n \choose 0}$ is the number of subsets with 0 elements
${n \choose 1}$ _________________________1 element
${n \choose 2}$ _________________________2 elements
                      .
                      .
                      .
${n \choose n}$ _________________________n________

So, summing, we're counting the sets of a set with cardinal equal to n, which directly gives $2^n$.

:)

If you set it to $(1+1)^n$, then it expands to just the binomial coefficients. $(1+1)^n = 2^n$

That's [b]exactly[/b] pkerichang's solution  :P

Its probably the most common proof for that. I remember seeing it in one of my school textbooks.

I get to it through combinatorics.
We are to prove
\[ \binom{n}{0}+\binom{n}{1}+\cdot\cdot\cdot+\binom{n}{n}=2^n \]
where both the sides are the number of ways in which any number of things can be selected from $n$ different things.
One way to count it is,
for every object we have 2 possibilities, i.e. selected or rejected. Thus total number of ways inwhich any number of object is selected is $2^n$
Other way is,
number of ways to chose $0$ objects + number of ways to chose $1$ object + ...

Hence Proved. :ninja:


function
real analysis
geometry
rectangle

I'm taking a class in measure theory right now. Do you have specific questions about it?

I was wondering what it means when somebody say a function is Lebesgue measurable.

A function f that defined on a subset E of R^n (n-dimensional real numbers) is Lebesgue measurable if, given a constant c, {x element of E | f(x) > c} is a measurable set in R^n.

I don't know if you know what a Lebesgue measurable set is, so I'll define that too. A subset F of R^n is a Lebesgue measurable set if, given a subset A of F and B of R^n\F, the outer measure of (A U B) is equal to the outer measure of A + the outer measure of B.

The outer measure of a set (it doesn't matter whether or not this set is actually Lebesgue measurable) is the infimum of the sums of the volumes (in n dimensions) of all coverings of the sets by finitely many rectangles (in n dimensions) of that set.


induction
number theory open
number theory

Let a,b are real numbers such that ab=1
Exists k is a positive interger such that a^k+b^k,a^(k+1)+b^(k+1) are rational numbers .Prove that a+b is a rational number

Note that $b = \frac 1 a$. Let $S_n = a^n + \frac 1 {a^n}.$

Then $S_nS_p = S_{n+p} + S_{n-p}$ and $S_{2n} = S_n^2 + 2$ for all integers $n,p$.

Assume that $S_k$ and $S_{k+1}$ are rational numbers. We may assume that $ k > 1$ (otherwise we are done).
From above and with an easy induction, we deduce that $S_{nk}$ and $S_{n(k+1)}$ are rational for all positive integers $n$.
Thus :
$S_1 + S_{2k+1} = S_kS_{k+1} = s$ is rational, and $S_1S_{2k+1} = S_{2k} + S_{2(k+1)} = p$ is rational too.
Moreover, $S_1$ is a root of $X^2 - sX + p = 0$. (1)

Similarly :
$S_1 + S_{2k^2 - 1} = S_{2k^2}S_{2(k^2-1)} = s'$ is rational, and $S_1S_{2k^2-1} = S_{2^2 + S_{2(k^2-1)} = p'}$ is rational too.
And $S_1$ is a root of $X^2 - s'X + p' = 0$. (2)

From (1) and (2), we deduce that $S_1(s'-s) = p'-p$.
Assume that $a > 0$ (the other case is similar) and $a \neq 1$ (otherwise the conclusion holds trivially). Then $S_{n+1}-S_n = S_n(S_1-2) + S_n - S_{n-1}$, thus by induction the sequence $S_n$ is increasing. It follows that $s' > s$.
And then $S_1 = \frac {p'-p} {s'-s}$ is rational, and we are done.

Pierre.


For how many positive integers $ n$ less than or equal to $ 24$ is $ n!$ evenly divisible by $ 1 \plus{} 2 \plus{} \dots \plus{} n$?

$ \textbf{(A)}\ 8\qquad
\textbf{(B)}\ 12\qquad
\textbf{(C)}\ 16\qquad
\textbf{(D)}\ 17\qquad
\textbf{(E)}\ 21$

[hide="Solution"]
We must have $ \frac {n!}{\frac {n(n \plus{} 1)}2} \equal{} \frac {2(n \minus{} 1)!}{n \plus{} 1}$ an integer. If $ n \plus{} 1$ is an odd prime, then the numerator will not be divisible by the denominator. Otherwise, it is divisible. Since there are $ 8$ possible odd prime values for $ n \plus{} 1$, our answer is $ 24 \minus{} 8 \equal{} \boxed{\textbf{(C)}\ 16}$.
[/hide]

How do you know that it will not be divisible when it is an odd prime? As in, is there a proof?
Thanks :)

[quote="PhireKaLk6781"]If $n+1$ was not an odd prime, all of its factors would have been contained in $2(n-1)!$.[/quote]

Yes, but that excludes the case $n+1=p^2$ for some prime $p$, in which case "all its factors" would be just $p$.

More complete:
If $n+1=2^2$, then we see that $2^2\mid 2\cdot 2!$.
If $n+1=p^2$ for some prime $p>2$, then $2p<p^2-2=n-1$ and $p,2p$ will be in $(n-1)!$.
Otherwise, for any $d\mid (n+1)$ with $1<d<n+1$, both $d$ and $(n+1)/d$ will be distinct terms in $(n-1)!$.

[quote=BakedPotato66]Here is my solution, similar to math154's solution.

From the problem statement, we know that
\begin{align*}
\tfrac{n(n+1)}2&|n!\\
\tfrac{n+1}2&|(n-1)!\\
n+1&|2(n-1)!\\
\end{align*}
If $n+1$ is an odd prime, the divisibility does not hold. $n+1$ ranges from 2 to 25 inclusive, and there are 8 odd primes in that range (3, 5, 7, 11, 13, 17, 19, 23), thus the answer is $24-8=\boxed{16}.$[/quote]

You switched the divisibility signs. $a|b\implies a=kb\forall a, b, k\in\mathbb{Z}^+$.
Also, that is quite literally the exact same solution as math154's solution


induction
number theory proposed
number theory

Let $ a$ be the positive integer and $ A\equal{}\{a^2\plus{}a\minus{}1,a^3\plus{}a^2\minus{}1,a^4\plus{}a^3\minus{}1,....\}$  Prove that there are infinetely many members of $ A$ such that any 2 of them are coprime.

$ x_n\equal{}a^{n}\plus{}a^{n\minus{}1}\minus{}1$
We will indution on k.
Let $ k\equal{}2$ then we chose $ \gcd(a_1,a_2)\equal{}1$
Suppose for exist k term satisfy we prove for $ k\plus{}1$. Call k term is $ a_{f(i)},\forall i\equal{}1,..,k$
Let $ t\equal{}\prod_{i\equal{}1}^k a_{f(i)}$
Chose $ n\equal{}\varphi (t)\plus{}1$
$ x_n\equiv a (\mod t)$
But $ \gcd(a,x_n)\equal{}1$ so $ \gcd(a,t)\equal{}1$ .
So $ \gcd(x_n,t)\equal{}1$
This problem claim.


I only got a 129 :(
Gaahhh for stupid mistakes!!! :mad:
And gahhh for only 6.5 hours of sleep the night before the test! :mad:


A traveler went 10 miles south, 7 miles east, 2 miles south, and 2 miles west. How many miles is the traveler from the starting point?

The traveler is displaced $ 10\plus{}2\equal{}12$ miles south and $ 7\minus{}2\equal{}5$ miles east. Completing the Pythagorean triple, the traveler is $ \fbox{13}$ miles from his/her starting point.


Let $A$ be the set of positive integers that do not contain the digit 9 in their decimal expansions. Prove that \[\sum_{a\in A}\frac1a<\infty\]

 :?: (Does it mean that $197\notin A$ but $18\in A$?)

[quote="ifai"]:?: (Does it mean that $197\notin A$ but $18\in A$?)[/quote]

Yes, it does.

[quote="Kalle"]How many elements of A are there among the first $10^{k}$ natural numbers?[/quote]
$9^{k}$. (Thank you!  :lol: )

I think it can be proved like this:
For $k\geq0$, suppose that
\[S_{k}=\sum_{\substack{10^{k}\leq a<10^{k+1}\\ a\in A}}\frac1a < (9^{k+1}-1-9^{k})\cdot\frac1{10^{k}}<8\left(\frac9{10}\right)^{k}\]
Then
\[\sum_{a\in A}\frac1a=\sum_{k=0}^{\infty}S_{k}< 8\sum_{k=0}^{\infty}\left(\frac9{10}\right)^{k}\]
Hence the series converges.


Find all functions $ f: \mathbb{Z} \rightarrow \mathbb{Z}$ such that
\[ f(x^2 \plus{} f(y) \minus{} 1) \equal{} (x \plus{} 1)f(x) \plus{} y\]
holds for all $ x,y \in \mathbb{Z}$.

$ f(x^{2}\plus{}f(y)\minus{}1)\equal{}y$  $ \implies$  $ f$ is surjective

Let $ f(y_{1})\equal{}f(y_{2})$  $ \implies$  $ y_{1}\equal{}y_{2}\equal{}1$ ,$ f$ is injective

$ f(x^{2}\plus{}f(u)\minus{}1)\equal{}(x\plus{}1)f(x)\equal{}u$  $ \implies$

Let $ f(u)\equal{}1$  $ \implies$  $ f(x^{2})\equal{}(x\plus{}1)f(x)\equal{}f(x^{2}\plus{}f(y)\minus{}1)$  $ \implies$  $ f(x)\equal{}1$

But if $ f(x)\equal{}1$  $ \implies$ $ y\equal{}1$  Contradiction!

I'm sorry. The last =-sign should have been a +-sign..  :blush:

Let $ f(y_{1})\equal{}f(y_{2})$  $ \implies$  $ y_{1}\equal{}y_{2}$  $ \implies$  $ f$ is injective

if $ x\equal{}\minus{}1$  $ \implies$  $ f(something)\equal{}y$  $ \implies$  $ f$ is surjective

$ f(f(y))\equal{}2f(1)\plus{}y$,$ f(f(y))\equal{}y$  $ \implies$  $ f(1)\equal{}0$

$ f(f(y))\equal{}y$  $ \implies$  $ f(0)\equal{}1$

$ \implies$  $ f(f(y)\minus{}1)\equal{}1\plus{}y$  $ \implies$  $ f(y\plus{}1)\equal{}f(y)\minus{}1$

$ \implies$  $ f(x\plus{}k)\equal{}f(x)\minus{}k$  $ \implies$  $ f(k)\equal{}1\minus{}k$  QED


he's wrong, $a=b=c=1$ :!: :D it's probably a $6$ instead of a $9$ on $RHS$.hope it's fixed then...
and why would you need the condition $a+b+c=3$??? it's homogenous..

[quote="Yosh..."]If $a,b,c \in R^{+}$ and $a+b+c=3$

Prove $a^{3}+b^{3}+c^{3}+3(a^{2}b+b^{2}c+c^{2}a)+6abc\geq 9(ab^{2}+bc^{2}+ca^{2})$[/quote]

Oops.. Sorry..

I'm not sure this inequality is right..

So the problem is
prove or disprove $a^{3}+b^{3}+c^{3}+6(a^{2}b+b^{2}c+c^{2}a)+6abc\geq 9(ab^{2}+bc^{2}+ca^{2})$

[quote="Yosh..."]
So the problem is
prove or disprove $a^{3}+b^{3}+c^{3}+6(a^{2}b+b^{2}c+c^{2}a)+6abc\geq 9(ab^{2}+bc^{2}+ca^{2})$[/quote]
It's still not true. Try $a=b=1$ and $c=0.$

[quote="Harry Potter"]Oh argady $a,b,c \in R^{+}$  :wink:
I think it is true . And i am finding a solution ( I think we must use Schur 's inequality )[/quote]
Try again $a=b=1$ and $c\to 0$ :P

[quote="Harry Potter"]If $a=b=1$ then $c=0$ . $c$ don't come to 0  :wink:
And We have : $a+b+c=3$  :([/quote]

Take $a=b$ and $c\to 0$, your inequality is wrong...

[quote="Yosh..."]Why you think your inequality can be proven?
Is there any trick to see that?

thx..[/quote]

I tried to find some constan $k\in\mathbb{R}$ for which inequality

$a^{3}+b^{3}+c^{3}+3(a^{2}b+b^{2}c+c^{2}a)+6abc\geq k(ab^{2}+bc^{2}+ca^{2})$ holds.

So, I made up a program and he found that $3.757\geq k$. And I took $k=\frac{15}{4}$


Hi,

My question is could you take the SAME test(like two AMC 10's) on test day A and B?  If so how will scores be counted?  I can't seem to find a direct answer on the FAQ's.

Thank You!

Do you mean like take both the AMC10 A and B on the same day? If so, then I don't think so...(I might be wrong)

You can take the AMC 10A and 10B. You cannot take the AMC 10A and 12A.

This question is answered completely in Frequently Aslked Question AMC 10/12 #13
and in Frequently Asked Question AIME #2 at
[url]http://www.unl.edu/amc/f-miscellaneous/faq.shtml[/url]

Steve Dunbar
MAA Director, American Mathematics Competitions

The A's are on the same day, the B's are too. For example, take AMC 10A and AMC 12B

For scores, I believe that you can choose any one of the two. Obviously you would want to choose the higher score.  :P

OK Thanks guys. (Sry, Mr. Dunbar, I sorta just skimmed through the FAQ's cuz it was so long. :blush:  :fool: )


If $ a \div b \equal{} 2$ and $ b \div c \equal{} \frac{3}{4}$, what is the value of $ c \div a$? Express your answer as a common fraction.

Um? Just multiply the two and take the reciprocal to get $ \frac1{2\cdot\frac34}\equal{}\boxed{\frac23}$?

a = 2b  &   b = a/2

c = 4b/3  plug in (b = a/2) and you get c = 4a/6

c/a = 4a/6 divided by a

this gives you 4/6 = 2/3 which is your answer

Math154is correct. Think of it this way: $a/b*b/c=a/c$. Now we want $c/a$. To get this we take the reciprocal of $a/c$. This would result in an answer of $2/3$ like Math154 said.


calculus
derivative
calculus computations

ok here is the problem
find dy/dx at x=1 if  y=(t^2+2)/(t^2-2)  and t=x^3

i did it two ways and dont know which one is right cuz i got different answers

1st:
(((t^2-2)(2t)-(t^2+2)(2t))) / (t^2-2)^2 =
= (2*t^3-4t - 2*t^3-4t) / (t^2-2)^2
= -8t / (t^2-2)^2
= (-8*x^3) / (x^6- 2)^2    "substitute x for the t"
then when u put 1 on the place of x that equals to "-8"

2nd
(((x^6-2)(6*x^5)-(x^6+2)(6*x^5))) / (x^6-2)^2=
= (6*x^11-12*x^5-6*x^11-12*x^5) / (6*x^5)^2
= -24x^5 / (x^6-2)^2
then after substituting 1 i get -24

so which way is right???or am i doing somethin wrong???

The second one is right. The reason the first one is wrong is that if you take the derivative with variable t, you're getting dy/dt, not dy/dx. You're asked about dy/dx. So you'd have to multiply by dt/dx, which is 3x^2, and that'll give you the same result.


Hi.
I don't know, where to post this "question"?
I think here.
I'm Georian and the flag wich is showing here, is my countries old flag.
If anybody can to change my flag, [b][color=red]PLEASE CHANGE[/color][/b]

Can you please provide an appropriately sized gif (or point us to a gif) of the new flag?

There is one [url=http://commons.wikimedia.org/upload/8/83/Flag_of_Georgia.png]here[/url], but it is too big.


IS ANY1 GOING?

OR IS EVERYBODY GOING TO THE SONATINA COMPETITION?

uh neither....

Soccer in Jackson :D  :D

u should

we're making fun of weird888 in ciphering

W00T
im going and i think farhan is getting first...
am am going to get first...
wow, im bipolar...

btw, yaofan???
are u taking the test too?

I'm going

I'm only posting this so I can see what time aops says it is so I can know when to discuss the AIME II. Which I took.

I informally took the test, but. I'd probably get a 80-ish

chestnut
u should have went. we had a ciphering problem that had u, dynamo, and me in a game show called countdown showdown and in the first round so-and-so go this many points. the question was the ratio of how many points u, nick, and me had and i ended up winning :D  :D

i had it right, and than thay called time XD...

i got like..., 8 points XDDDD :blush:  :oops:

and u guys still got 1st place team...

i got annoyed of saying pizitz middle school

how does tony do his little pencil spinning thingy?

yeah.... i go to the same school as him.... i live in the same house as him.... i sleep in the same bed as him..... you get the picture

chestnut.....
r u married to tony??
lol, i didnt know tony have a wife...
ur prbbly like bo or wendy trying to do something...

so how do u do the pencil spinning thingy?

do u guys have to go to school on the 17th?
everybody should just transfer to berry or at least spain park for next year

eww weird......

he's magic yaofan... magic....

grow up... :D

i wonder if he'd like to tell me... :roll:  :roll:


Perhaps related to [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1316802652&t=21738[/url]?

[quote="Virgil Nicula"] [color=darkred]Let $ ABCXDY$ be a cyclical convex hexagon for which $ CX\parallel AY\parallel BD$ . Prove that
$ ABCD$ is harmonically $ \Longleftrightarrow\ AX\cap CY\cap BD\ne\emptyset\ \Longleftrightarrow\ \frac {AY}{CX} \equal{} \left(\frac {BA}{BC}\right)^2$ .[/color] [/quote]

[color=darkblue][b][u]Proof.[/u][/b] Denote $ \left\|\ \begin{array}{c} AB \equal{} a\ ;\ BC \equal{} b\ ;\ CD \equal{} c \\
 \\
DA \equal{} d\ ;\ AC \equal{} e\ ;\ BD \equal{} f \\
 \\
E\in AX\cap BD\ ;\ F\in CY\cap BD\end{array}\ \right\|$ . Observe that $ \left\|\ \begin{array}{c} DY \equal{} AB \equal{} a \\
 \\
DX \equal{} BC \equal{} b\end{array}\ \right\|$ and

$ \left\|\ \begin{array}{ccc} CX\parallel BD & \Longrightarrow & \widehat {CAB}\equiv\widehat {DAE}\equiv\widehat {CDF} \\
 \\
AY\parallel BD & \Longrightarrow & \widehat {ACB}\equiv\widehat {ADE}\equiv\widehat {DCF}\end{array}\ \right\|$ $ \Longrightarrow$ $ \left\|\ \begin{array}{ccccc} ABC\sim DFC & \Longleftrightarrow & \frac {AB}{DF} \equal{} \frac {CA}{CD} & \Longrightarrow & DF \equal{} \frac {ac}{e} \\
 \\
ABC\sim AED & \Longleftrightarrow & \frac {BC}{ED} \equal{} \frac {CA}{DA} & \Longrightarrow & DE \equal{} \frac {bd}{e}\end{array}\ \right\|$ .

Hence $ AX\cap CY\cap BD\ne\emptyset\ \Longleftrightarrow\ E\equiv F\ \Longleftrightarrow\ DE \equal{} DF\ \Longleftrightarrow\ ac \equal{} bd$ , i.e.  $ ABCD$ is harmonically.

I"ll apply to the hexagon  $ ABCXDY$ the following well-known property :[/color]

[quote] [color=darkred]Let $ ABCDEF$ be a cyclic convex hexagon. Then exists the equivalence

$ AB\cdot CD\cdot EF \equal{} BC\cdot DE\cdot FA\ \Longleftrightarrow\ AD\cap BE\cap CF\ne\emptyset$ .[/color] [/quote]
[color=darkblue]Therefore, $ AX\cap CY\cap BD\ne\emptyset\ \Longleftrightarrow\ AB\cdot CX\cdot DY \equal{} BC\cdot XD\cdot YA\ \Longleftrightarrow\ \frac {AY}{CX} \equal{} \left(\frac ab\right)^2$ .

[b][u]Remark.[/u][/b] In this case ( $ E\equiv F$ ), prove easily that $ ED^2 \equal{} EA\cdot EC$ .[/color]