geometry
3D geometry
octahedron
geometric transformation
rotation
reflection
symmetry

let Q8 be the quaternion group of eight elements : 1,-1,i,-i,j,-j,k,-k
what is the automorphism group?
i'd say it has 24 elements
so it probably is S4

but is it correct? and how to prove it?

thx,
fred

You can rewrite $Q8$ as $\{e,a,a^2,a^3,b,ab,a^2b,a^3b\}$ with $ord(a)=ord(b)=4$ and $ba=a^{-1}b$. So to construct an automorphism $f$ you have to consider $f(a)$ and $f(b)$. Because of $ord(f(x))=ord(x)$ there are exactly $3$ possibilities for $f(a)$: $a,a^3,b$ and then the remaining two for $b$. In total there are $3!=6$ automorphisms then. To construct them from $f(a)$ and $f(b)$ just use the linearity of $f$.

Misha

Now you can consider if $Aut(Q8)$ is isomorphic to $S_3$ which is quite easy.

That's certainly wrong, Misha. After all, the quaternion group has six elements of order four, which come in three pairs.

Here's a model: let the elements $i,j,k,-i,-j,-k$ be represented by the vertices of a regular octahedron: $i=(1,0,0)$, $j=(0,1,0)$, $k=(0,0,1)$, $-i=(-1,0,0)$, ...  . Then for any face, the product of the elements representing the vertices is either $1$ or $-1$ depending on orientation. (We know that any automorphism fixes those elements). Therefore, the automorphism group is naturally isomorphic to the group of rotations of the octahedron or its dual cube, which has 24 elements and is isomorphic to $S_4$.

I find geometric automorphism groups easier to understand (and I have a Rubik's Cube sitting next to me as I write this)- I hope the model makes it clear to you.

Sorry that I miscalculated the orders but I think your solution is wrong too. As I said you can identify $Q8$ with $\{e,a,a^2,a^3,b,ab,a^2b,a^3b\}$ with $e=1$, $a^2=-1$, $a=i$, $b=j$, $ab=k$, $a^3=-i$, $a^2b=-j$, $a^3b=-k$. It follows immediately $ord(a)=ord(b)=4$ and $ba=a^{-1}b$. As Jmerry said there are $6$ elements of order $4$: $a$, $a^3$, $b$, $ab$, $a^2b$, $a^3b$. But you can choose every element of these $6$ as $f(a)$ and one of the other $5$ for $f(b)$ and get $6\cdot5=30$  automorphisms.

I hope it is correct now.
Do you agree, jmerry?

Misha

No, you're wrong again. If you send $i$ to $i$, you can't send $j$ to $-i$; inverses go to inverses. It really helps to see the group as a single unified structure; those three inverse pairs are the heart of it.

thx!

your explanation with the cube makes it  very visual
so you see the eight elements as vertices?

however

how to see the group of the cube as S4
is this trivial?

Actually, I see the six elements other than $\pm1$ as vertices of an octahedron/faces of a cube. Since $\pm 1$ are fixed, they're left out of the model.
From there, here's how to show that the group is $S_4$. Consider the action of the rotation group on pairs of opposite faces (of an octahedron). We can easily find transpositions by rotating around an axis in the middle of an edge, 3-cycles by rotating around an axis through the middle of a face, and four-cycles by rotating around an axis through a vertex. The remaining order-2 elements are 180-degree rotations around a vertex-axis. This clearly generates a homomorphism onto $S_4$, and the orders match so the rotation group is isomorphic to $S_4$.

It's not really trivial, but what else would such a group be? We could probably get there by counting too, and working out the grubby details.
If we allow reflections, the full symmetry group of a cube or octahedron is $S_4\times \mathbb{Z}_2$

In general for the platonic solids, why is the composition of two rotations just another rotation? I guess I am having trouble seeing that the group has closure. thx

The rotations of a geometric figure form a group because the rotations of space form a group, and the permutations of the elements of the figure form a group.

For a regular solid, we can easily see that every symmetry of the figure can be extended to an isometry of space; each segment between the vertices is sent to another of the same length.
We can classify the isometries that fix the origin by their matrices. The matrices must be real orthogonal matrices, so their determinants are $\pm1$. If the determinant is 1, we can show that it must have 1 as an eigenvalue, and its action is a rotation around the axis of that eigenvector.
The transformations with determinant -1 can be expressed as a composition of reflection in a plane and rotation around an axis perpendicular to that plane.

Why is the dodecahedron rotational symmetry group isomorphic to A5

Aww sorry guys but I am really weak in algebra.Can you tell me whats A5 :blush:

The symmetric group $S_n$ is the group of all permutations on $n$ symbols. The alternating group $A_n$ is the subgroup of even permutations, which can be expressed as a product of an even number of transpositions. This subgroup has index 2 in $S_n$. For $n>5$, $A_n$ is a nonabelian simple group; $A_5$ is the smallest nonabelian simple group, at order 60.

If you map $S_n$ into a group of invertible $n\times n$ matrices by setting $p_{ij}=1$ if $P(i)=j$ and $p_{ij}=0$ otherwise, $A_n$ is the subgroup of these with determinant 1.

Thanks a lot Jmerry.I have myself used this notation so many times before.How can I forget it? :(


modular arithmetic
induction
number theory unsolved
number theory

Let $ X$ be the set of integers of the form

$ a_{2k}10^{2k}\plus{}a_{2k\minus{}2}10^{2k\minus{}2}\plus{}...\plus{}a_210^2\plus{}a_0$

Where $ k$ is a nonnegative integer and $ a_{2i} \in \{1,2,...,9\}$ for $ i\equal{}0,1,...,k$.
Show that every integer of the form $ 2^p3^q$ for $ p$ and $ q$ nonnegative integers , divides some element of $ X$.

[u]xref4search [/u] : every integer 2^p3^q divides some a_{2k}10^{2k} + a_{2k - 2}10^{2k - 2} + ... + a_{2}10^2 + a_{0}[quote="mustafa-yafes"]Let $ X$ be the set of integers of the form

$ a_{2k}10^{2k} + a_{2k - 2}10^{2k - 2} + ... + a_210^2 + a_0$

Where $ k$ is a nonnegative integer and $ a_{2i} \in \{1,2,...,9\}$ for $ i = 0,1,...,k$.
Show that every integer of the form $ 2^p3^q$ for $ p$ and $ q$ nonnegative integers , divides some element of $ X$.[/quote]

Hope somebody will find something prettier :)

First, get rid of factor $ 3$ :
Let $ x\in X$ such that $ 2^p3^q|x$ and let $ n =$ the number of digits (odd) of $ x$.
Then $ y = x(1 + 10^{n + 1} + 10^{2n + 2})\in X$ and $ 3|1 + 10^{n + 1} + 10^{2n + 2}$ and so $ 2^p3^{q + 1}|y$

So the remaining question is now to prove that $ \forall p$, $ \exists x\in X$ such $ 2^p|x$

[hide="Here is a rather ugly method to build a positive multiple of 4^p which is in X"]

For $ p = 0$ or $ p = 1$, this is immediate since $ 4^0\in X$ and $ 4^1\in X$

Let then $ p > 1$ and let the sequences $ a_k$ and $ b_k$ and $ x_k$ defined as :
$ a_0 = 4^p$ and $ b_0 = 0$ and $ x_0 = 4^p$

For $ k\in[0,p - 1]$ :
Choose $ b_{k + 1}$ as the fewest non negative integer such that $ b_{k + 1}4^{p - k} + a_k\pmod{100}\in{1,2,3,4,5,6,7,8,9\}}$
Let us call $ r = b_{k + 1}4^{p - k} + a_k\pmod{100}$ such that $ r\in{1,2,3,4,5,6,7,8,9\}}$
Define then $ a_{k + 1} = [\frac {b_{k + 1}4^{p - k} + a_k}{100}]$
And $ x_{k + 1} = x_k + (100a_{k + 1} - a_k + r)100^k$

1) $ b_{k + 1}$ exists and $ b_{k + 1} < 25$
$ \gcd(4^{p - k},25) = 1$ and so $ \exists k_1$ such that $ k_14^{p - k} = 1\pmod{25}$ and so $ 4k_14^{p - k} = 4\pmod{100}$
Let $ \overline{uv}$ be the two rightmost digits of $ a_k$. $ \exists m$ such that $ 10u + v + 4m\in[101,104]$
And so $ 4k_1m4^{p - k} + a_k\pmod{100}\in{1,2,3,4\}}$
Hence the existence of $ b_{k + 1}$
If  $ \alpha 4^{p - k} + a_k\pmod{100}\in{1,2,3,4,5,6,7,8,9\}}$ with $ \alpha > 25$, then $ (\alpha - 25)4^{p - k} + a_k\pmod{100}\in{1,2,3,4,5,6,7,8,9\}}$, the fewest number is $ < 25$

2) $ a_k\le 4^{p - k}$ $ \forall k\in[0,p]$
$ a_0 = 4^p\le 4^{p}$

And induction is easy :

$ a_{k + 1} = [\frac {b_{k + 1}4^{p - k} + a_k}{100}]$ $ < \frac {b_{k + 1}4^{p - k} + a_k}{100}$ $ < \frac {25\cdot 4^{p - k} + 4^{p - k}}{100}$

$ a_{k + 1} < 4^{p - k - 2} + \frac 1{100}4^{p - k} < 4^{p - (k + 1)}$
Q.E.D

3) $ x_{k}$ is a positive integer divisible by $ 4^p$
$ x_0 = 4^p$ positive integer divisible by $ 4^p$

And induction is easy :
$ 100a_{k + 1} = b_{k + 1}4^{p - k} + a_k - r$ and so $ 100a_{k + 1} - a_k + r = b_{k + 1}4^{p - k}$ and so $ (100a_{k + 1} - a_k + r)100^k = 25^kb_{k + 1}4^{p}$
So $ x_{k + 1} = x_k + (100a_{k + 1} - a_k + r)100^k$ $ = x_k + 25^kb_{k + 1}4^{p}$ is a positive integer divisible by $ 4^p$
Q.E.D.

4) At the end of the process, $ x_p\in X$ and $ 4^p|x_p$
The fact that $ 4^p|x_p$ has been proved in 3)
The fact that $ x_p\in X$ in a consequence of the fact that $ x_k = \overline{a_k0x0y0z...}$ and that $ a_p = 0$ or $ 1$
Q.E.D.
[/hide]

Example 1 : $ p = 5$
[hide="Short application of the above method"]
$ a_0 = 1024$, $ b_0 = 0$, and $ x_0 = 1024$
$ k = 0$ : $ b_1 = 16$ and $ 16\cdot 1024 + 1024 = 17408 = 8\pmod{100}$ $ \implies$ $ a_1 = 174$ and $ x_1 = 17408$
$ k = 1$ : $ b_2 = 13$ and $ 13\cdot 256 + 174 = 3502 = 2\pmod{100}$ $ \implies$ $ a_2 = 35$ and $ x_2 = 350208$
$ k = 2$ : $ b_2 = 12$ and $ 12\cdot 64 + 35 = 803 = 3\pmod{100}$ $ \implies$ $ a_3 = 8$ and $ x_3 = 8030208$
$ k = 3$ : $ b_3 = 0$ and $ 0\cdot 16 + 8 = 8\pmod{100}$ $ \implies$ $ a_4 = 0$ and $ x_4 = 8030208$
$ k = 4$ : $ b_4 = 0$ and $ 0\cdot 4 + 8 = 8\pmod{100}$ $ \implies$ $ a_5 = 0$ and $ x_5 = 8030208$
[/hide]
And the result : $ 8030208 = 4^5\cdot 7842\in X$

Example 2 : $ p = 6$
[hide="Short application of the above method"]
$ a_0 = 4096$, $ b_0 = 0$, and $ x_0 = 4096$
$ k = 0$ : $ b_1 = 22$ and $ 22\cdot 4096 + 4096 = 94208 = 8\pmod{100}$ $ \implies$ $ a_1 = 942$ and $ x_1 = 94208$
$ k = 1$ : $ b_2 = 11$ and $ 11\cdot 1024 + 942 = 12206 = 6\pmod{100}$ $ \implies$ $ a_2 = 122$ and $ x_2 = 1220608$
$ k = 2$ : $ b_2 = 5$ and $ 5\cdot 256 + 122 = 1402 = 2\pmod{100}$ $ \implies$ $ a_3 = 14$ and $ x_3 = 14020608$
$ k = 3$ : $ b_3 = 3$ and $ 3\cdot 64 + 14 = 206 = 6\pmod{100}$ $ \implies$ $ a_4 = 2$ and $ x_4 = 206020608$
$ k = 4$ : $ b_4 = 0$ and $ 0\cdot 16 + 2 = 2\pmod{100}$ $ \implies$ $ a_5 = 0$ and $ x_5 = 206020608$
$ k = 5$ : $ b_5 = 0$ and $ 0\cdot 4 + 2 = 2\pmod{100}$ $ \implies$ $ a_6 = 0$ and $ x_6 = 206020608$
[/hide]
And the result : $ 206020608 = 4^6\cdot 50298\in X$

Hence the result


[quote="Valentin Vornicu"]Huh?  :?[/quote]
i only have three options in the "jump to" list:

Select a forum
--------------------
None.

Why?? Is that my own problem??

Erase your cookies. Then see if it still happens ...

[quote="shobber"]I just noticed that there is a new "jump to" selection. But it only contains two choices: " select a forum" and "none". But the one we used to have contains every forum and sub-forum of the mathlinks. So why we have this nearly useless function??[/quote]
It's fixed now... still don't know what has happened...

Your cookies mixed up somehow. It's because of your computer, there's nothing wrong with the site :)


Find all polynomials that satisfy the following relation for all real $ x>0$:
\[P^2\bigg(\dfrac1x\bigg)\plus{}P^2(x)\equal{}P(x^2)P\bigg(\dfrac1{x^2}\bigg).\]

[quote="sandu2508"]Find all polynomials that satisfy the following relation:

$ P^2\bigg(\dfrac1x\bigg) \plus{} P^2(x) \equal{} P(x^2)P\bigg(\dfrac1{x^2}\bigg)$ for all real $ x > 0$[/quote]
$ P(x)\equal{}0$ is a trivial solution. Let's now look for non zero solutions

Obviously, we have $ P(1)\equal{}0$. Let then $ m$ the greatest integer such that $ P^2(x)\equal{}P^2(z_mx)$ $ \forall x\in\mathbb C$, $ \forall z_m\in\mathbb C$ such that $ z_m^m\equal{}1$ ($ m^{th}$ roots of unity).
We know that such a $ m$ exists since $ m\equal{}1$ fits and no $ m>2n$ (where $ n\equal{}$degree of $ P$) would fit (else, since $ P^2(\frac 1{z_m})\equal{}P^2(1)\equal{}0$, the polynomial $ P^2(x)$ would be a nonzero polynomial of degree $ 2n$ and would have at least $ m>2n$ complex roots).

$ P^2(z_mx)\equal{}P^2(x)$ $ \forall x$ implies either $ P(z_mx)\equal{}P(x)$ $ \forall x$, either $ P(z_mx)\equal{}\minus{}P(x)$ $ \forall x$ (remember these are polynomials)

Consider then $ k\equal{}2m$ and $ z_k$ such that $ z_k^k\equal{}1$. Using $ z_kx$ in the required equation, we see that $ P^2(\frac 1{z_kx})\plus{}P^2(z_kx)\equal{}P^2(\frac 1{x})\plus{}P^2(x)$ and so clearly $ P^2(z_kx)\equal{}P^2(x)$, hence a contradiction.

So we got a unique solution $ P(x)\equal{}0$


limit inequalities logarithms trigonometry

Find
$ lim_{x\rightarrow 0} \sqrt[x]{2^{x\minus{}1}\plus{}\frac{1}{2}}$

Write as $ \lim_{x \to 0} \sqrt[x]{\frac{2^x\plus{}1^x}{2}}$

This is the limit of the power-mean of $ 2$ and $ 1$ as the power approaches $ 0$, and it is well known that this is the geometric mean, $ \sqrt{2}$.

A demonstration of this limit is shown here: [url=http://en.wikipedia.org/wiki/Power_mean_inequality#Geometric_mean]link to page[/url]

[quote="Dr Sonnhard Graubner"]hello, write $ \lim_{x \to 0}\sqrt [x]{2^{x \minus{} 1} \plus{} \frac {1}{2}}$ as $ e^{\lim_{x \to 0}\frac {\ln\left(2^{x \minus{} 1} \plus{} \frac {1}{2}\right)}{x}}$ and use the L'Hospital rule.
Sonnhard.[/quote]

that is the classical way...  :)

hello, yes of course, but this way was not forbidden.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, write $ \lim_{x \to 0}\sqrt [x]{2^{x \minus{} 1} \plus{} \frac {1}{2}}$ as $ e^{\lim_{x \to 0}\frac {\ln\left(2^{x \minus{} 1} \plus{} \frac {1}{2}\right)}{x}}$ and use the L'Hospital rule.
Sonnhard.[/quote]
Can you post your solution please  :)  Anyway what is L'Hospital rule?  :blush:

hello, using L'Hospitals rule we get
$ e^{\lim_{x \to 0} \frac {\ln\left(2^{x \minus{} 1} \plus{} \frac {1}{2}\right)}{x}} \equal{} e^{\lim_{x \to 0} \frac {2^{x \minus{} 1}\ln(2)}{2^{x \minus{} 1} \plus{} \frac {1}{2}}} \equal{} \sqrt {2}$.
Sonnhard.

[quote="TZF"]He's posted all the relevant information for his solution -- the limit can be computed directly using one application of L'Hopital's rule on his latter form of the expression and then simplifying.

Here's an article explaining [url=http://en.wikipedia.org/wiki/L%27hopital%27s_rule]L'Hopital's Rule[/url].

Basically, if $ \lim_{x \to a} f(x) \equal{} \lim_{x \to a} g(x) \equal{} 0$, or both are divergent, then $ \lim_{x \to a} \frac {f(x)}{g(x)} \equal{} \lim_{x \to a} \frac {f'(x)}{g'(x)}$

For example, $ \lim_{x \to 0} \frac {\sin x}{x} \equal{} \lim_{x \to 0} \frac {\frac {d}{dx} \sin x}{\frac {d}{dx} x} \equal{} \lim_{x \to 0} \frac {\cos x}{1} \equal{} 1$[/quote]
Thx both of you a lots TZF and Dr Sonnhard Graubner
I understand it now  :lol:


I run a Safari Version 3.1.2 (5525.20.1), on a Mac 10.5.5, and when I click "Show Minor Edits" at [url=http://www.artofproblemsolving.com/Wiki/index.php?title=Special:RecentChanges&hideminor=0&limit=5000]this[/url] link, it does absolutely nothing. I use the MonoBook skin.

[quote="1=2"]I run a Safari Version 3.1.2 (5525.20.1), on a Mac 10.5.5, and when I click "Show Minor Edits" at [url=http://www.artofproblemsolving.com/Wiki/index.php?title=Special:RecentChanges&hideminor=0&limit=5000]this[/url] link, it does absolutely nothing. [b]I use the MonoBook skin.[/b][/quote]Sorry not supported. As I said numerous times before do not use that skin.

Used AoPS and Mathlinks skins and didn't work there either.

Try using Firefox for Mac. I don't have a Safari for Mac to test with right now.

Someone has something against me.

I used firefox 3.0.1 and it happened again in all skins, and when I logged out I could see them just fine on the AoPS skin.
And my Dad's account can see them on all skins on both browsers.


Has anyone joined mensa?  I was thinking about it but I don't know if there is any purpose to it or not.

I think a moderator should delete one of the double posts.

As for joining MENSA, I don't really know how to join, and I don't think I'd be suited too well for it, so I'm not part of it.

Quoting the brouchure like thing,(which is free but you've probably read it if your're thinking about joining):

"American Mensa Ltd. is an organization for individuals who have one common trait: a score in the top two persent of the populations on a standardized intelligence test. Over five million Americans are eligible for membership... you may be one of them."

Then it goes on about how you get a free magazine with mental problems and how you can publish stuff in you local newsletter and be in a group with the other intellectual members. They also hav special interest groups with subjects like computer gaming,scuba diving, etc.

They make it sound easy to join, though I did horrible on their practice test.

Actually, i don't have the brochure or any info, I just saw an article about it somewhere online and I was wondering if there is any benefit of being a member.  Has anyone joined?

As for the double post, it was becarse I hit the back key right after submitting the topic.  I hope someone deletes it too.

[quote="corpulus"]As for the double post, it was becarse I hit the back key right after submitting the topic.  I hope someone deletes it too.[/quote]

Heh, no problem. It happens all the time, because of lag, and a lot of other reasons.

Their test is quite easy. I had it all correct, except for 2 mistakes in the language part which is normal cuz I'm not english.

The only thing I don't like about mensa is that you have to pay, to get nothing for it :?

I mean, membership to this site means a lot to me, but if they had made me pay before signing up, I'd probably never have joined :?

I qualify for mensa,  but I haven't actually joined yet.

i dunno wut mensa really does.. it kinda blows.. i signed up and all i get is these very uninteresting and poorly printed "magazines". i think they have activities with the local mensans once a while but i dont get info so i never go

I feel like MENSA is just something you can brag about to your friends, nothing really special to be in.

Has anyone seen the SImpsons where Lisa becomes a part of Mensa?


Does anyone know where i can find more information about Winter Camp?

There's some information on the CMS website (http://www.cms.math.ca)

i don't see any info...
just a list of past camps
is there anything on qualifying, etc?

Well, from what I have heard, they invite IMO hopefuls for that year, and they are selected from the ppl who did really well on CMO the previous year or do really well on COMC.

I believe a previous post mentioned that the CMS is very secretive about the Winter Camp.  There's no website for it or anything.

Getting awfully excited eh guys?  I think my only ticket in is to step up on this next COMC, so I'm going to go full throttle this November.  Hopefully I'll get to meet some of ya'll!


ratio
trigonometry
logarithms
Gauss
calculus
calculus computations

hola amigos  :D

can you show

$ \dfrac{\Gamma\,'\left(\dfrac{x}{y}\right)}{\Gamma\left(\dfrac{x}{y}\right)}\;\equal{}\;\boxed{\sum_{k\equal{}1}^{\dfrac{y}{2}\minus{}1}\left\{\cos\,
 \left(\dfrac{2\pi\,k\,x}{y}\right)\cdot\ln\left[\sin^2\left(\dfrac{\pi\,k}{y}\right)\right]\right\}
\minus{}\gamma\minus{}\ln\,(2y)\minus{}\dfrac{\pi}{2}\cot\left(\dfrac{\pi\,x}{y}\right)}$

this one already was posted. try to find  on planeta math Gauss formula.


[quote="linj"]Phil borrows $ 800\$$ at 9% annual interest, compounded annually.
If he pays off the loan at the end of 6 years, how much interest will he owe?[/quote]
Do It count for 6 year or 7 year as question [b]end[/b] of 6 years?

At the end of $ 6$ years, so its compounded 6 times:

$ 800 \cdot (1.06)^6$

Calculates total amount (plus interest), and then subtract base.

[quote="happyme"]At the end of $ 6$ years, so its compounded 6 times:

$ 800 \cdot (1.0\color{red}6\color{black})^6$

Calculates total amount (plus interest), and then subtract base.[/quote]

Possibly a typo, but I think that $ 6$ should be a $ 9$ since the annual interest is $ 9\%$.


im taking AIME II and i want to attend the math jam for AiME I in order to practice. however, i will be out of town. how do i get the transcript?
also, is the version of the AIME you take taken into account for usamo qualification (i.e. if all of the smart ppl happen to take AIME II and get perfects will it be harder)?

Transcript will be posted [url=http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php]here[/url].

If there is a large disparity between difficulty of the two AIMEs (which is unusual), there will be separate qualifying scores. It happened in 2005.

Also, please at least try to use correct spelling in your posts.

i cant press shift and a letter with one arm


Let $ \alpha,\beta,\gamma$ the angles of a triangle,$ n\in N^*$,prove that:
$ \cot^n(\frac{\alpha}{2}) \cot^n(\frac{\beta}{2})\plus{}\cot^n(\frac{\beta}{2})\cot^n(\frac{\gamma}{2})\plus{}\cot^n(\frac{\gamma}{2})\cot^n(\frac{\alpha}{2}) \geq 3^{n\plus{}1}$

Its equivalent to proving

$ (\cot \frac A2 \plus{} \cot \frac B2 \plus{} \cot \frac C2)^n(\tan ^n \frac A2 \plus{} \tan ^n \frac B2 \plus{} \tan ^n \frac C2) \geq 3^{n \plus{} 1}$
$ \Leftrightarrow 3^n\cot ^n (\frac {A \plus{} B \plus{} C}{6}) \cdot 3 \tan ^n (\frac {A \plus{} B \plus{} C}{6}) \geq 3^{n \plus{} 1}$

the last inequality is due to jensen

donote :
$ A_n=\cot^n(\frac{\alpha}{2}) \cot^n(\frac{\beta}{2})+\cot^n(\frac{\beta}{2})\cot^n(\frac{\gamma}{2})+\cot^n(\frac{\gamma}{2})\cot^n(\frac{\alpha}{2})$

$ B=\tan(\frac{\alpha}{2}) \tan(\frac{\beta}{2})+\tan(\frac{\beta}{2})\tan(\frac{\gamma}{2})+\yan(\frac{\gamma}{2})\tan(\frac{\alpha}{2})$

so by caushy's inequality :  $ A_1 \times B\geq9$

and we have : $ \tan(\frac{\gamma}{2})=\tan(\frac{\pi}{2}-\frac{\alpha+\beta}{2})=cot(\frac{\alpha+\beta}{2})=\frac{1-\tan(\frac{\alpha}{2})\tan(\frac{\beta}{2})}{\tan(\frac{\alpha}{2})+\tan(\frac{\beta}{2})}$
so : $ B=1$ and $ A_1\geq9$
by inequality of means 1 and n :
$ \frac{A_n}{3}\geq(\frac{A_1}{3})^n$  $ \iff$   $ A_n\geq3^{n+1}$


Hello all- I was not sure where to post this, but this is urgent to some extent.

I would like some assistance with the following:

1.) Can anyone prove Ceva's and Menelaus's theorem using something called barycentres (no idea what they are)

Also for those of you who would like to do something to help young aspiring students in the world of math, and have some spare time, would you be able to write a section on positive-definite matrices (generally 4 to 5 pages long with proper citations and any theorems) If so, please submit your document to rohankk5@hotmail.com. Other areas needing a section would be:

1.) Isometries
2.) Quadrics
3.) Barycentres in dimesion 2

remember to send all material to rohankk5@hotmail.com..all help is greatly appreciated

First, this topic is at the wrong forum. It's geometry.
Second, proofs of Ceva's and Menelaus' Theorems can be found all over the internet.
Third, you can consult information about positive definite matrices on the internet.
Fourth, again those geometric transformations can also be found elsewhere. You gotta do your own research. No one is gonna 'milk' you math like a baby.
Fifth, this is a problem-solving forum board, not a homework-solving forum board. I am not particularly hounding you but there is a distinction between problem-solving and homework. You are looking at the wrong place.
 :(

well the main idea is that no one is crazy enough (I think) to write up a document just for you :P

and try the web all those topics are widely covered. for the statements of a certain theorem you can open a topic in the right place.

please take time to carefully read the descriptions of the forums :D

[quote=NoDealsHere][quote=mikimoto12]IS THIS A FIFTEEN YEAR BUMP?????[/quote]

Um yesßßßssss.........[/quote]

:o LEGENDARY BUMP ALERT
NOTIFY THE AOPS BUMP SOCIETY

it’s not very hard to scroll to the bottom of fun Factory tbh
just let the old users sleep in peace guys xd


[quote="chien than"]Prove that:$x+y+z \geq \sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}$
where $a;b;c;x;y;z$ are positive real numbers satisfying $xyz=ax+by+cz$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=1899


geometry
incenter
perpendicular bisector
geometry proposed

[color=darkblue][b]I composed this problem inspired by the April's problem from http://www.mathlinks.ro/Forum/viewtopic.php?t=146794 . If is possibly, you can extend it.

Examine the following geometrical drawing described with words lower down and find how more its remarkable (synthetical or metrical) properties :[/b][/color]

[quote][color=darkred][b]Let $ABC$ be a triangle for which $AB\perp AC$. Denote :[/b][/color]

$\{\begin{array}{c}D\in BC\\\ AD\perp BC\end{array}$ ; $\{\begin{array}{c}E\in (AB)\ ,\ F\in (AC)\\\ AE=AF=AD\end{array}$ ; $\{\begin{array}{c}M\in (AB)\ ,\ BM=BD\\\ N\in (AC)\ ,\ CN=CD\end{array}$ ; $S\in EN\cap FM$.[/quote]

[color=darkred][u][b]Here are some remarkable properties :[/b][/u][/color]  $\{\begin{array}{cc}1\blacktriangleright & m(\widehat{EDM})=m(\widehat{MDN})=m(\widehat{NDF})=45^{\circ}\\\\ 2\blacktriangleright & m(\widehat{SDM})=\frac{C}{2}\ ,\ m(\widehat{SDN})=\frac{B}{2}\\\\ 3\blacktriangleright & m(\widehat{ESF})=135^{\circ}\ ,\ \frac{ES}{EN}+\frac{FS}{FM}=\frac{b+c}{a}\end{array}$ .

[color=darkred][b]Another remarkable properties :[/b]

$4\blacktriangleright$ $\triangle DEM\sim\triangle DNF$  and $\triangle MDF\sim \triangle BAC\sim\triangle EDN$.

$5\blacktriangleright$ The quads $AEDN$, $AFDM$, $DEMS$, $DFNS$ are cyclically and $m(\widehat{ADS})=\frac{1}{2}\cdot |B-C|$.[/color]

[color=darkblue][b]Remark.[/b][/color] $b(a-b)+c(a-c)+(a-b)(a-c)=bc\Longleftrightarrow b^{2}+c^{2}=a^{2}\ !$

[size=150][color=darkred]Thank you, Virgil Nicula for your very nice proposed problem![/color][/size]

[color=indigo][u][b]Proof.[/b][/u] Denote $I$ be the incenter of $\triangle ABC$

$1.\blacktriangleright$ We have: $BC$ is a tangent of the circle $C(A,\,AD)$
$\Longrightarrow\left \{\begin{array}{c}\widehat{CDF}=\frac{1}{2}\widehat{DAF}=\frac{B}{2}\\ \widehat{BDE}=\frac{1}{2}\widehat{DAE}=\frac{C}{2}\end{array}\right\|\Longrightarrow \left \{\begin{array}{c}DF\parallel BI\\ DE\parallel CI\end{array}\right\|\Longrightarrow \left \{\begin{array}{c}DF\perp DM\\ DE\perp DN\end{array}\right\|$
$\Longrightarrow\left\{\begin{array}{c}AMDF\;\textrm{is cyclic}\\ ANDE\;\textrm{is cyclic}\end{array}\right\|\quad (1)$

$2.\blacktriangleright\quad (1)\Longrightarrow\widehat{DMS}=\widehat{DMF}=\widehat{DAF}=\widehat{DAN}=\widehat{DEN}=\widehat{DES}$ $\Longrightarrow MSDE$ is cyclic   $(2)$
Similar $NSDF$ is cyclic.

$3.\blacktriangleright$ We have: $\left\{\begin{array}{c}\widehat{EDN}=\widehat{BDA}\\ \widehat{MDF}=\widehat{ADC}\end{array}\right\|\Longrightarrow\left\{\begin{array}{c}\widehat{ADN}=\widehat{BDE}=\frac{C}{2}\\ \widehat{ADM}=\widehat{CDF}=\frac{B}{2}\end{array}\right\|\quad (3)$

$4.\blacktriangleright\quad (2),\,(3)\Longrightarrow\left\{\begin{array}{c}\widehat{SDM}=\widehat{SEM}=\widehat{ADN}=\frac{C}{2}\\ \widehat{ADM}=\frac{B}{2}\end{array}\right\|\Longrightarrow\widehat{SDA}=\frac{1}{2}\left |B-C\right |$
Similar $\widehat{SDN}=\frac{B}{2},$ thus, $\widehat{MDN}=\widehat{MDA}+\widehat{NDA}=\frac{B}{2}+\frac{C}{2}=45^\circ\Longrightarrow \widehat{EDM}=\widehat{MDN}=\widehat{NDF}=45^\circ$
On the otherhand, from $(2),$ we have: $\widehat{ESM}=\widehat{EDM}=45^\circ\Longrightarrow\widehat{ESF}=135^\circ$

$5.\blacktriangleright\quad (2)\Longrightarrow \widehat{SDE}=\widehat{AMF}=\widehat{ADF}=\widehat{AFD}=\widehat{DME}=\widehat{DSE},$ so $ES=ED$
Similar $FD=FS,$ thus we have: $EF$ is perpendicular bisector of $SD\quad (4)$

$6.\blacktriangleright\left\{\begin{array}{c}\widehat{DEN}=\widehat{DMF}=\widehat{ABC}\\ \widehat{EDN}=\widehat{MDF}=\widehat{BAC}\end{array}\right\|\Longrightarrow\triangle DEN\sim\triangle DMF\sim\triangle ABC\quad (5)$

$7.\blacktriangleright\quad (4),\,(5)\Longrightarrow\left\{\begin{array}{c}\frac{ES}{EN}=\frac{DE}{EN}=\frac{AB}{BC}\\ \frac{FS}{FM}=\frac{DF}{FM}=\frac{AC}{BC}\end{array}\right\|\Longrightarrow\frac{ES}{EN}+\frac{FS}{FM}=\frac{b+c}{a}$[/color]

[quote="April"][size=150][color=darkred]Thank you, Virgil Nicula for your very nice proposed problem![/color][/size]
[color=indigo] [b]Proof.[/b] We have: $CD$ is a tangent of circle $(A,\,AD)$ ... so $ES=ED$ ...  $FS=FD$ ; thus, $EF$ is perpendicular bisector of $DS$[/color][/quote]

[color=darkblue][b]Thanks, April ! You found a new interesting property :[/b][/color]

[color=darkred][b]$6\blacktriangleright$ $ES=ED$ , $FS=FD$ and $EF\perp DS$, i.e. the quadrilateral $DESF$ is a deltoid ![/b][/color]

[color=darkblue][b]I found still four new remarkable properties :[/b][/color]

$7\blacktriangleright$[color=darkred][b]The rays $[DS$, $[DA$ are isogonals in the angle $\widehat{MDN}$.
$8\blacktriangleright$The lines $MN$, $EF$, $BC$ are concurently in the point $T$ for which $\frac{TB}{TC}=\frac{c(a-b)}{b(a-c)}$.
$9\blacktriangleright$The line $p=AS$ is the polar of the point $T$ w.r.t. the angle $\widehat{BAC}$.
$10\blacktriangleright$ The points $\{\begin{array}{c}U\in BN\cap CM\\\ V\in BF\cap CE\end{array}$ belong to the line $AS$, i.e. $p=\overline{ASUV}$.[/b][/color]
[color=darkblue][b]Remark.[/b] The collinearity of the points $S$, $U$, $V$ results and from the [u]Pappus' theorem[/u] applied to the points $\|\begin{array}{c}\{M,E,B\}\subset AB\\\ \{N,F,C\}\subset AC\end{array}$.[/color]

[size=134][color=purple][b]Who and when will present an extension of this problem ?![/b][/color][/size]

[color=indigo]$11.\blacktriangleright$ The incenters of $\triangle ABD,\,\triangle EDN,\,\triangle MDF,\,\triangle ADC$ are collinear.
$12.\blacktriangleright$ Midpoints of $AS,\,MN,\,EF$ are collinear.
Midpoints of $AT,\,EN,\,FM$ are collinear.
Midpoints of $AT,\,BN,\,CM$ are collinear.[/color]


Has the plan 1994 points (distinct), no colinear, such that the distance between two points is always an integer?

There's a proof around here somewhere (on the forum) that for any $n$ we can find $n$ points on a circle with mutual rational distances between them, and this problem follows from that one, since we can scale the figure to make the distances integral.


ok, there's 4 day civic unit
Marcia want to be attorney exactly once during the 4 days.

She had to roll a dice.
1 - she's in
2 3 4 5 6 - she's out.

the probability that Marcia will be a defence attorney on the third day and other roles for the remaining days?
the probabiity that Marica will be a defence attorney exactly once during four days?

Help, thanks.

[hide="Probability Marcia will be an attorney on 3rd day  only"]$\frac{5}{6}*\frac{5}{6}*\frac{1}{6}*\frac{5}{6}=\frac{125}{1296}$[/hide]

[hide="Probability Marcia will be an attorney once only"]Add the probability she will be it 1st day only, 2nd day only, 3rd day only, and 4th day only to get $\frac{500}{1296}$ or $\frac{125}{324}$.[/hide]

[quote]other roles for the remaining days[/quote]

I believe we need more information to complete the first one. Is she automatically assigned different roles if she is not chosen to be an attorney? Or do certain dice numbers correspond to different roles?

Thanks, I also thought so, wanted to check. :)


given right triangle and incircle express the diameter of the incicle interms of sides .ie prove that a+b-c.triangle is right angled at c.

im not sure that i understand what your saying. and yes the inradius of a right triangle is just $ \frac{a \plus{} b \minus{} c}{2}$ and the diameter is just $ a \plus{} b \minus{} c$.

[hide]
inradius*semiperimeter=area
call the inradius r
$ r\cdot\frac{a\plus{}b\plus{}c}{2}\equal{}\frac{ab}{2}$
$ r\equal{}\frac{ab}{a\plus{}b\plus{}c}$

[/hide]

seems to contradict slightly what zolo gave

sorry about that i was thinking of something eles related to circum/in radius centroids something like that and just so you know one could have just googled "inradius of a right triangle" and got this pop up

[url]http://library.thinkquest.org/20991/gather/formula/data/163.html[/url]

Actually, the exradius in respect to $ C$ (dubbed $ I_C$ is $ \frac{2S}{a\plus{}b\minus{}c}$. By symmetry, the other two sides' exradii have similar representations.


[quote="fermat3"]let F a function definide in R SUCH FOR EVERY X Y in R we have  f(x*3 +y*3)=(x+y)((f(x))*2-f(x)f(y)+(f(y))*2). prove that f(2006x)=2006f(x). :?:  :roll:[/quote]

It is hard to read your problem when you write it in this way. Does your problem looks like this:

$f(x^{3}+y^{3})=(x+y)(f(x)^{2}-f(x)f(y)+f(y)^{2})$

or it should be stated in different way?


there are 11 teams performing. each day there are some teams performing. on the days they need not perform, they would sit on the audience stand and watch other teams' performance. it is known that for each team, apart from the days they work on, they see other each team's performance at least once, prove that 6 days is the minimum day for meeting the criteria.

i am curious about how to construct effeciently a case for 6 days...i have been drafting and couldn get a good case for that.. :(

Use Sperner's Theory

5C2 = 10 < 11 < 6C3 = 20
So the answer is 6days,,,


[quote="lotharek"]Try induction.[/quote]
but I'm still not able to prove.... ;)  ;)


難しめの大学では、数列そのものが具体的に与えられているものより、抽象的なものを処理していく問題が多いような気がします。
そんな問題を今まとめて解いているので、いくつか投稿してみることにしました。もしこのtopicをご覧の皆さんもそういった問題が何かあれば、ぜひ教えてください  :lol: そしてバンバン解いていってください :P

$\boxed 1$
(1)  $n$ は自然数とし、実数 $a_1,a_2,a_3 \cdots$ に対して、
　　　　　　$b_{n}=n \cdot\left( \displaystyle\sum_{k=1}^{n}a_{k}^{2} \right) - \left( \displaystyle\sum_{k=1}^{n}a_{k} \right)^2$
によって数列 $\{ b_n \}$ を定める。このとき、$(i)$ $b_{n+1}\ge b_{n}$ $(n=1,2,\cdots)$ , $(ii)$ $b_{n}\ge 0$ であることを示し、$(iii)$ $n\ge 2$ のとき、$b_{n}=0$ となるための $a_1,a_2,\cdots,a_n$ の条件を求めよ。

(2)  $N$ は 3 以上の整数とし、$a_1,a_2,\cdots,a_N$ は
　　　　　　$\displaystyle\sum_{k=1}^{N}a_{k}^{2}=1$
を満たす任意の実数とする。さらに、
　　　　　　$S=\displaystyle\sum_{k=1}^{N}a_{k}$
とし、$c_{n}=S-a_{n}$  $(n=1,2,\cdots,N)$　によって数列$\{ c_n \}$ を定める。このとき、
　　　　　　$C=\displaystyle\sum_{k=1}^{N}c_{k}^{2}$
のとりうる最大値と最小値を求めよ。

$\boxed 2$(2001 大阪大　前)
数列$\{ a_n\}$ において、各項 $a_n$ が $a_n \ge 0$ を満たし、かつ $\sum_{n=1}^{\infty}a_n=\frac{1}{2}$ を満たすとする。
さらに各 $n$ に対し、
　　　　　　　$b_n=\displaystyle\prod_{k=1}^{n}(1-a_k)$ , $c_n=1-\displaystyle\sum_{k=1}^{n}a_k$
とおく。ただし、$n\ge 2$ とする。

(1) すべての $n$ にたいして、$b_n \ge c_n$ が成り立つことを示せ。

(2) ある $n$ について $b_{n+1}=c_{n+1}$ が成り立てば、$b_n=c_n$ となることを示せ。

(3) $b_3=\frac{1}{2}$ であるとき、$c_3=\frac{1}{2}$ であることを示せ。また、$b_3=\frac{1}{2}$となる数列$\{ a_n\}$は全部で何種類あるか求めよ。

$\boxed 3$(京都大 ?)
$n$ を $2$ 以上の整数とする。実数 $a_1,a_2,\cdots,a_n$ に対し、
　　　　　　　$S=a_1+a_2+\cdots+a_n$
とおく。また、$k=1,2,\cdots,n$ について、不等式 $|S-a_k|<1$ が成り立っているとする。
このとき、全ての $k$ $(k=1,2,\cdots,n)$ について、$|a_k|<2$ が成り立つことを証明せよ。

$\boxed 4$
$n$ 個の実数 $a_1,a_2,a_3,\cdots,a_n$ はそれぞれ、$1,\frac{1}{2},\frac{1}{3},\cdots,\frac{1}{n}$ に正または負の記号を付けたものとし、$S_n=a_1+a_2+a_3+\cdots+a_n$ とする。このとき、任意の $n$ について、$|S_n|\le \frac{1}{n}$ となるように $a_1,a_2,a_3,\cdots,a_n$ の符号を選べることを証明せよ。

$\boxed 5$
数列 $x_1,x_2,\cdots,x_n$ は
　　　　　　　$\displaystyle\sum_{i=1}^{n}x_i=0$ , $\displaystyle\sum_{i=1}^{n}|x_i|=1$
を満たす。このとき、ある $n$ 個の実数 $a_1,a_2,\cdots,a_n$ のうち、最大値を $M$、最小値を $m$ とするならば、
　　　　　　　$\displaystyle\sum_{i=1}^{n}a_i x_i \le \frac{M-m}{2}$
が成り立つことを証明せよ。

$\boxed{6}$ (Source: '06 AIME I #15)
正の整数$k$について、$|x_k|=|x_{k-1}+3|$を満たす数列$\{ x_k \}$がある。また、$x_0=0$とする。このとき、$|x_1+x_2+...+x_{2006}|$のとりうる値のうち最も小さいものを求めよ。

一見シンプルなようで、きちんと論理的に正しい値を求めるのはかなり難しいと思います。風変わりな問題ですがさすがはAIME（米国数学オリンピック予選みたいなもの）だけあり、良問ですよ。

$\boxed 7$
$n$ を $3$ 以上の奇数とする。$n$ 個の実数 $a_1,a_2,\cdots,a_n$ があり、この中から任意の $2$ 個を選んで作った積
　　　　　　　$a_i a_j$  $(1\le i<j \le n)$
を全て考えるとき、これら $\frac{n(n-1)}{2}$ 個の数の中にある負の数の個数が奇数ならば、$a_1,a_2,\cdots,a_n$ の中の少なくとも $1$ つは $0$ であることを証明せよ。

>>frt　さん
なかなか思考が進まないですけどまたぼちぼちやっていきます:lol:


calculus
derivative
trigonometry
calculus computations

ok.... the equation is:

y + sin y = x

i need to find the 2nd derivative for this. i can't do it directly because then x becomes 1. then i try to divide y + sin y by x, but that's all messed up right now. any ideas maybe?

$y+\sin y = x$

$\dfrac{dy}{dx} + \dfrac{dy}{dx} \cos y =1 \Rightarrow \dfrac{dy}{dx} = \dfrac{1}{1+\cos y} = \left(1+\cos y\right)^{-1}$

$\dfrac{d^2y}{dx^2} = -1\cdot -\dfrac{dy}{dx}\sin y\left(1+\cos y\right)^{-2} = \dfrac{\sin y}{\left(1+\cos y\right)^3}$


Basically the same as on any other contest ... the problems are designed not to use higher level mathematics, but to test your ability to put what you know together in new ways in a short period of time (like the AMC and somewhat the AIME)

JB

Yes.

As with any contest of this level, you are never required to know anything past precalculus.  The beauty of the contest is knowing how the more mundane, everyday ideas can be thrown together into a problem that looks daunting to finish.

I'd say be somewhat well-rounded, as with any contest--you'll find that all areas of mathematics are involved.

Just be there to have fun.  That's what made the experience great for me when I competed.  Even if I wasn't that strong of a competitor, just looking at those ideas was worth the time.

I remember last year...when I hadn't done competition math for like a month. Not fun...I actually managed to solve them, but not having practice boosts your stupid mistake meter.

[quote="lingomaniac88"]Well, this is going to be my first time at ARML, but hopefully not my last.
What kinds of topics would I need to become familiar with?[/quote]

You can look at the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=30264]Practice Series[/url] from last year to see a bunch of old ARML problems.  Also, check out the current one that you can participate in: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=86168 .

(well, you've apparently already found the practice series...so...this is for anyone who hasn't and are wondering what kind of questions to expect)


Let $a_{n}$ be an arbitrary sequence of real numbers. Can we always find a real number $x$ such that all numbers $x+a_{n}$ are Liouville numbers?

The set of such $x$ forms a dense $G_\delta$ set, actually. This is because the set $\mathcal L$ of Liouville numbers is a dense $G_\delta$ set, and the $x$ which interest us are precisely the elements of $\bigcap_{n}(\mathcal L-a_{n})$. By Baire's Category Theorem, this set is also dense and $G_\delta$.

(just in case it's a bit ambiguous: by $\mathcal L-a_{n}$ I mean $\{\ell-a_{n}\ |\ \ell\in\mathcal L\}$)


trigonometry
geometry
geometric transformation
rotation
linear algebra
matrix
quadratics

Let $ X\equal{}\left(
\begin{array}{cc}
\cos{\theta} & \minus{}\sin{\theta} \\
\sin{\theta} & \cos{\theta} \\
\end{array}
\right)$ such that $ X^4\plus{}\frac83X^3\plus{}X^2\plus{}I\equal{}0$. If $ Y\equal{}X\plus{}X^{\minus{}1}$ and $ Y^2\plus{}pY\plus{}qI\equal{}0$, find the value of $ p,q$ and $ \cos{\theta}$.

[hide="Hidden"]We know $ X$ is a rotation, so $ X \plus{} X^{ \minus{} 1} \equal{} 2 \cos \theta I$

Thus, $ X^4 \plus{} X^2 \equal{} 2 \cos \theta X^3$

$ \left( 2 \cos \theta \plus{} \frac {8}{3} \right) X^3 \equal{} \minus{} I$

Taking the norm of both sides, we have $ 2 \cos \theta \plus{} \frac {8}{3} \equal{} \pm 1$
$ 2 \cos \theta \plus{} \frac {8}{3} \equal{} \minus{} 1$ has no solution.
$ 2 \cos \theta \plus{} \frac {8}{3} \equal{} 1 \implies \cos \theta \equal{} \minus{} \frac {5}{6}$

Now, we plug this back in to get $ X^3 \equal{} \minus{} I$. If you don't already see that this doesn't work, take the cosine of the angle of rotation of each side:
$ \cos 3 \theta \equal{} \cos \pi \equal{} \minus{} 1$.
$ 4c^3 \minus{} 3c \equal{} \minus{} 1$
$ 4c^3 \minus{} 3c \plus{} 1 \equal{} 0$

By the rational root theorem, $ c \neq \minus{} \frac {5}{6}$[/hide]

Therefore, there is no such rotation $ X$.

[quote="jelly119"]Let $ X \equal{} \left( \begin{array}{cc} \cos{\theta} & \minus{} \sin{\theta} \\
\sin{\theta} & \cos{\theta} \\
\end{array} \right)$ such that $ X^4 \plus{} \frac83X^3 \plus{} X^2 \plus{} I \equal{} 0$. If $ Y \equal{} X \plus{} X^{ \minus{} 1}$ and $ Y^2 \plus{} pY \plus{} qI \equal{} 0$, find the value of $ p,q$ and $ \cos{\theta}$.[/quote]

I am sorry that I wrong typed the problem. :blush:  Here is the correct one:
Let $ X \equal{} \left( \begin{array}{cc} \cos{\theta} & \minus{} \sin{\theta} \\
\sin{\theta} & \cos{\theta} \\
\end{array} \right)$ such that $ X^4 \plus{} \frac83X^3 \plus{} X^2 \plus{}\frac83X\plus{} I \equal{} 0$. If $ Y \equal{} X \plus{} X^{ \minus{} 1}$ and $ Y^2 \plus{} pY \plus{} qI \equal{} 0$, find the value of $ p,q$ and $ \cos{\theta}$

As you might imagine, we can do something similar:

[hide="How you wanted us to solve for p and q"]We know that $ X$ is an isomorphism**, so:

$ X^{4} \plus{} \frac {8}3X^{3} \plus{} X^{2} \plus{} \frac {8}3X \plus{} I \equal{} 0$
$ \iff X^{2} \plus{} \frac {8}3X \plus{} I \plus{} \frac {8}3 X^{ \minus{} 1} \plus{} X^{ \minus{} 2} \equal{} 0$

We can rewrite the latter form as:
$ (X^2 \plus{} 2I \plus{} X^{ \minus{} 2}) \plus{} \frac {8}{3} (X \plus{} X^{ \minus{} 1} ) \minus{} I \equal{} 0$
$ Y^2 \plus{} \frac {8}{3} Y \minus{} I \equal{} 0$

It is evident that $ p \equal{} \frac {8}{3}$ and $ q \equal{} \minus{} 1$ works. See the third hidden section of text.

[i]** - Why is this necessary and sufficient? You will need to interpret $ X$ as a linear transformation.[/i][/hide]

[hide="Hint to solving for cos theta"]What is $ X^{ \minus{} 1}$? It will help to think of $ X$ as a rotation matrix, rather than computing it directly.

Therefore, what is $ Y$, as a matrix, in terms of $ \theta$? What is the overall matrix equation in terms of $ \theta$?[/hide]

[hide="Flaw in the problem"]You will have discovered in solving for $ \cos \theta$ that $ Y \equal{} 2 \cos \theta I$, and thus $ Y^2 \equal{} 4 \cos ^2 \theta I$. Thus, the equation for $ Y$ in which we had to solve for the coefficients was equivalent to:
$ (4 \cos ^2 \theta \plus{} 2 p  \cos \theta \plus{} q) I \equal{} 0$
$ \iff 4 \cos ^2 \theta \plus{} 2 p \cos \theta \plus{} q \equal{} 0$

However, you know from solving for $ \cos \theta$ that there is only one value, namely, $ \cos \theta \equal{} \frac{1}{6}$. Thus, we get a single line equation in two variables for $ p$ and $ q$, and you cannot actually solve for $ p$ and $ q$. It happens that the $ p$ and $ q$ we found in the first hidden text do work.

The reason  we ran into this is that there is only one solution $ Y$, so only one of the roots of the quadratic in $ Y$ must be this solution. We can arbitrarily choose any other root, within the constraints of the construction of $ Y$. So, we can't nail down $ p$ and $ q$ to a single number.

[b]Question:[/b] Can you figure out what the bounds on $ p$ and $ q$ are?[/hide]

Please note that linear algebra is not widely included in American contest mathematics, so this might not be the best place to post linear algebra questions.

isn't $ X^{\minus{}1}$ the inverse matrix?

It certainly is.

But I guess what I'm driving at is that it performs the inverse transformation. In other words, while going about the problem, think of $ X$ as a rotation by an angle of $ \theta$ rather than as a matrix.


hi everyone,

here is a rather nice result:
let f a transformation of the plane so that if
d(A,B)=1 then d(f(A), f(B))=1.
Prove that f is an isometry ..

nice isn't it ..

E an affine euclidean space of dimension n >2, f:E-->E such that for any
A,B in E, d(A,B)=1 => d(f(A),f(B))=1
Then f is an isometry.

I have some ideas, but I'm not sure (trying some things out; when I started writing I thought it was solved, but I was wrong). We say that $x$ has property $P$ if $d(A,B)=x\Rightarrow d(f(A),f(B))=x$.

By gluing equilateral triangles of side $1$ to each other as many times as we want we can see that $n,\ n\sqrt 3$ have property $P\ \forall\ n\in \mathbb N^*$ (*). Another important thing is this: if $d(A,B)\in (x-y,x+y)\Rightarrow d(f(A),f(B))\in [x-y,x+y],\ \forall\ x>y>0$ which have property $P$. This is derived from the fact that $d(A,B)$ can be seen as the third side in a triangle with two sides of lengths $x,y$. By using this, (*), and the fact that $\{m+n\sqrt 3|m,n\in \mathbb Z\}$ is dense in $\mathbb R$ I think we can show that $f$ enlarges distances. This is because we can find $\{m_k\},\ \{n_k\}\subset \mathbb N^*$ s.t. $\pm(n_k\sqrt 3-m_k)\to d>0$ (from below) as $k\to \infty$. In that case we have $\pm(n_k\sqrt 3-m_k)<d<n_k\sqrt 3+m_k$ for $k$ large enough, and this shows that $\pm(n_k\sqrt 3-m_k)\le d(f(A),f(B))$, and by going to the limit we get $d(f(A),f(B))\ge d=d(A,B)$.

I don't even know if it's correct or not..

The author of  the problem in dimension 3 is Peter Zvengrowski, you got the
right idea $Z+ \sqrt 3Z$ grobber

I think I can complete it. Ok, we now know that $d(f(A),f(B))\ge d(A,B)$. This means that $x+y$ has property $P$ for each $x,y$ having property $P$, which, in turn, means that $x$ has property $P$ $\forall x\in \mathbb Z[\sqrt 3]$. We apply the same strategy as before, this time taking $x_k,y_k\in Z[\sqrt 3]\cap (0,\infty)$ s.t. $x_k+y_k\to d(A,B)$ from above as $k\to \infty$, having $|x_k-y_k|<d(A,B)<x_k+y_k$ for sufficiently large $k$, thus getting $d(f(A),f(B))<x_k+y_k$, and when we go to the limit we get $d(f(A),f(B))\le d(A,B)$, and combined with $d(f(A),f(B))\ge d(A,B)$ we get the result.

Hope there are no mistakes.


can u wrap the surface of a unit cube with a $3.3$piece of paper?

$3.3$ meaning what? $3.3$ square units? In that case, obviously not because the surface area of a unit cube is 6 square units and $6>3$.

i say no because even if you put it on a diagonal, then there will not be enough to cover a whole edge of the cube

I think no as well, but wouldn't know how to prove it or find out for sure...

how nice disuasion. :D

plz continue ure discussion...

i will see my opinion later...

and thanks to all of u...


I've been thinking; I tried to prove the triangles TPR and triangle RST congruent by using "Side Side Side".

However, I got the problem marked wrong.  Why can't I assume T and R are also right angles?  I mean, S being a right angle should force R to share it's X position, and P being a right angle should force R to share it's Y position.  Same for T!  So why can I not assume they're right angles?!

I've attached a picture..

What is the problem? Drawing a picture does not say it!

To answer your question in short, yes, by CPCTC (corresponding parts of congruent triangles are congruent).

Actually, he can't, [b]ernie[/b]; it's backward logic. By first assuming that the quadrilateral is a rectangle, one is essentially using the notion that the triangles are congruent to prove that the triangles are congruent.

$ \boxed {YES}$, it does imply that the two opposite sides are parallel.
Extend $ PT$ to some point $ 0$.
Because $ TS$ and $ PR$ are parallel,
$ <OTS\equal{}90^{\circ}$.
However, $ <OTS\equal{}<TSR\equal{}90^{\circ}$,
so $ TP$ is parallel to $ SR$.  Therefore,
we know that it is a rectangle, and the
rest follows.
Your solution is valid, although not
thoroughly explained which might've
led to your teacher not giving you credit.

[quote="10000th User"][quote="AIME15"]However, because TS and PR are parallel, the angles TRP and RTS are congruent.  Same goes for angle SRT and angle PTR, so by ASA (RT=RT, obviously) they're congruent.[/quote]You provided the solution that his teacher was most likely looking for... but you did not answer what the OP asked about!

No, you cannot make those assumptions. What theorem says that they have to be right angles? What do you mean by 'forcing' certain points to take some specified positions? Until you have proven these 'intuitive' statements of yours, you cannot use 'unproven' statements and prove other propositions.[/quote]

I think what he means by forcing the points to take certain points is that if there are two right angles in a quadrilateral, it must be a rectangle. However, that's not necessarily true because the angles are opposite each other and you can "rotate" the angles. THe other angles only have to add up to 180.

[quote="Nandubetax"]But metafor, the happens subconciously, so ernie seems to be right[/quote]

The assumption is correct, but it can't be used as part of the proof; that is what I'm trying to say.

What?
If it's correct, why can't it be part of the proof?

[quote="ernie"]Because it was found because of the proof (CPCTC).[/quote]

Can't you prove taht they are congruent and then use it?

But if you just say <STR = <TRP and <PTR=<SRT (all valid points), you can say that <T = <R = 90.

It is possible to have two opposite angles of a quadrilateral be right angle and the two others not be. See for yourself: measures of 75, 90, 105, and 90 degrees don't work out.

NO! Not only are the opposite angles of the quadrilateral $ 90^{\circ}$, there is ALSO a pair of parallel sides.  BECAUSE OF THIS, as I already proved, the quadrilateral must be a rectangle.

THERE IS ALSO A PAIR OF PARALLEL SIDES.  That is the key to the problem.

EDIT: Oh, and ernie, that's not the only way you can prove that the quadrilateral is a rectangle. As I already wrote in my first response, you can prove it is a rectangle without any congruency at all.  To do it the way you guys tried to do it would make the proof circular, but there are other ways to go about it.

[quote="Yongyi781"]Because you're done once you've proved that they're congruent :lol:[/quote]

oh lol i didn't read the question...

EDIT: I read the topic again carefully this time XD.
AIME15 did you mean AAS?
Also, forget what I said in my first post.. I didn't notice there were parallel lines D:

Okay, let's prove that $ \angle STP$ and $ \angle SRP$ are right.  First of all, they sum to $ 180^\circ$, because $ m\angle RST=m\angle RPT=90^\circ$.  Also, $ \angle TRP \cong \angle RTS$ because they are alternate interior angles, and $ \angle TRS \cong \angle RTP$ for the same reason.  By the Angle Addition Postulate, $ \angle STP \cong \angle SRP$, and because $ m\angle STP + m\angle SRP=180^\circ$, we conclude that $ m\angle STP = m\angle SRP=90^\circ$, and by the definition of a rectangle, $ PRST$ is a rectangle. $ \fbox$

[quote="AIME15"]Okay, let's prove that $ \angle STP$ and $ \angle SRP$ are right.  First of all, they sum to $ 180^\circ$, because $ m\angle RST = m\angle RPT = 90^\circ$.  Also, $ \angle TRP \cong \angle RTS$ because they are alternate interior angles, and $ \angle TRS \cong \angle RTP$ for the same reason.  By the Angle Addition Postulate, $ \angle STP \cong \angle SRP$, and because $ m\angle STP + m\angle SRP = 180^\circ$, we conclude that $ m\angle STP = m\angle SRP = 90^\circ$, and by the definition of a rectangle, $ PRST$ is a rectangle. $ \fbox$[/quote]

Exactly. So there's nothing wrong with that assumption.

if the parallel lines are given, then this is definitely true
what's the fuss?


Πανενωσιακες του $ 1972$
Μαντεψτε μονοι σας το αποτελεσμα........
Νομιζω περυσι υπηρχε μια παραποιημενη στον Αρχιμηδη των μικρων

[quote="new_member"]Na breite ton megisto akeraio arithmo $ x$ ,wste o arithmos $ 4{}^2{}^7 + 4{}^1{}^0{}^0{}^0 + 4^x$ na einai teleio tetragono.[/quote]

Βγάζουμε όσα δυάρια κοινό παράγοντα μπορούμε ($ 2^{54}$) = τέλειο τετράγωνο.
Μέσα μένει   $ 1 + 2.2^{1945} + 2^{2x - 54}$ .

Άρα για 2x-54 = 2.1945, δηλαδή x = 1972, σίγουρα έχουμε τέλειο τετράγωνο. Για μεγαλύτερα χ παύει να είναι τέλειο τετράγωνο γιατί περικλείεται μεταξύ των διαδοχικών τετραγώνων $ 2^{2(x - 27)}$ και $ (2^{x - 27} + 1)^2$.

Φιλικά,

Μιχάλης Λάμπρου

(ελπίζω να έγραψα το LATEX σωστά, γιατί
ο υπολογιστής μου δεν τα βλέπει πάντα σωστά,
και τα μπερδεύει!)

[color=red]Σάββατο 10/1/200 : έκανα μία μικρή διόρθωση στο LATEX (όχι στα μαθηματικά).
Πάλι δεν βλέπω σωστά το LATEX, αν και δοκίμασα "τα πάντα". Το βλέπω μόνο την πρώτη φορά που μπαίνω στο site αλλά μετά χαλάει. Έτσι είδα ότι είχα κακή γραφή του 2^54, που τώρα το διόρθωσα. [/color]
(φαινόταν ως 2^5  4)

[quote="new_member"]To biblio sto opoio vrika tin askisi paratirisa oti eixe kanei ena oraio prologo o kyrios Lamprou.[/quote]

Πώς περνάνε τα χρόνια! Πάνε πάνω από 10, από τότε που βγήκε το βιβλίο.
Πραγματικά έγραψα τον πρόλογό του, και τον είχα χαρεί.
Ήταν λίγο μετά από την αξέχαστη για μένα εποχή που είχα πάει ως αρχηγός της
Ελληνικής ομάδας στις ΙΜΟ και ΒΜΟ 1996 και 1997. Η ομάδα είχε
Αλεξάκη, Μαλικιώση, Μιχαλάκη (δύο αδέλφια), Μπρέγιαννη, ...

Έχω τον πρόλογο σε pdf. Υπάρχει τρόπος να φορτωθεί στο site? Αν όχι, θα βρω τρόπο να τον αναρτήσω κάπου για να τον διαβάσει όποιος επιθυμεί.

Φιλικά,

Μιχάλης Λάμπρου.

[quote="Orfikoss"]Κύριε Λάμπρου μάλλον πατήσατε το κουτάκι που λέει: 	Disable BBCode in this post
Αν το ξετικάρετε λογικά θα δουλέψει

Ορφέας[/quote]

Ορφέα, ευχαριστώ αλλά περιέργως δεν είναι εκεί το πρόβλημα. Ακόμα πιο περίεργο είναι ότι και από άλλον υπολογιστή, στο γραφείο μου, έχω το ίδιο πρόβλημα. Τι να πω.

Φιλικά

Μιχάλης Λάμπρου

ΥΓ Είχαμε συναντηθεί τον πέρσι Απρίλιο στην Θεσσαλονίκη. Ελπίζω να σε δω και φέτος. Θα είμαι εκεί.

Κύριε Λάμπρου επειδή στο παραπάνω σας υπήρχε πρόβλημα στο LateΧ μπήκα και σας το διόρθωσα και όντως ήταν πατημένο το Disable BBCode in this post και σας το απενεργοποίησα . Και βλέποντας και στα υπόλοιπά σας ποστ υπήρχε το ίδιο κουμπί πατημένο . Πρέπει να κάνετε μια ρύθμιση στον υπολογιστή σας ώστε να μην είναι disable γιατί από ότι υποθέτω ο υπολογιστής το κάνει disable μόνος του . Ελέγξτε λίγο τις ρυθμίσεις ίντερνετ .
Ευχαριστώ


vector
linear algebra
linear algebra unsolved

Let $ V$ and $ W$ be vector spaces over the same field. For a linear transformation $ T: V \rightarrow W$, define its transpose $ T^{t}: W^{*} \rightarrow V^{*}$ on the dual space by $ T^{t}(f) \equal{} fT$ for all $ f \in W^{*}$.

$ T$ is onto if and only if $ T^{t}$ is one-to-one (Shown)

Show that $ T$ is one-to-one if and only if $ T^{t}$ is onto.

Introduce the double-dual spaces $ V^{**}$ and $ W^{**}$. There are canonical injections $ \phi_V$ from $ V$ into $ V^{**}$ and $ \phi_W$ from $ W$ into $ W^{**}$. The obvious diagram commutes, with $ \phi_W\circ T\equal{}(T^T)^T\circ \phi_V$.

From what you already have, $ (T^T)^T$ is injective if and only if $ T^T$ is surjective. If $ V$ and $ W$ are finite-dimensional, this is already everything, as $ \phi_V$ and $ \phi_W$ are isomorphisms.

A composition $ f\circ g$ is always injective if both $ f$ and $ g$ are injective, and can't be injective if $ g$ isn't injective. Applying this to $ \phi_W\circ T\equal{}(T^T)^T\circ \phi_V$, we have that $ T$ is injective whenever $ T^T$ is, but not necessarily vice versa. That's half of what we need: $ T$ is injective if $ T^T$ is surjective.

That last bit is trickier; I don't see a way to do it with mere diagram chasing. If $ T$ is injective, given some $ g\in V^*$, can we construct $ f\in W^*$ so that $ T^T(f)\equal{}g$? Applying the definitions, we want $ f(T(x))\equal{}g(x)$ for all $ x\in V$. The image $ T(V)$ is a subspace of $ W$, and we can clearly define $ f$ on this subspace to make that work. Any linear functional on a subspace can be extended to the whole space, and any such extension here is the $ f$ we wanted.

For vector spaces with no additional structure, that extension theorem is basically the existence of bases. For normed vector spaces and continuous linear functionals, it's Hahn-Banach. Both versions rely on the axiom of choice.

[quote] That last bit is trickier; I don't see a way to do it with mere diagram chasing.[/quote]

That's right, I tried to do that but failed and I didn't think of defining $ f$ as you suggested.
Thank you.


Let $ ABC$ be a triangle. Its incircle $ C(I)$ touches it in $ D\in BC$ , $ F\in AB$ . Define $ E\in AC$ so

that $ EF\parallel BC$ and $ L\in BC$ so that $ CL\equal{}CE$ . Prove that $ AI$ , $ DF$ , $ EL$ are concurrently.

We have $ \angle FEL\equal{}\angle LEC$ and $ \angle BFD\equal{}\angle DFE$ so $ FD, AI, EL$ concur at A-excircle of triangle $ AEF$


inequalities
geometry
circumcircle
trigonometry
trig identities
IMO Shortlist
Triangle

Let $ABC$ be an acute triangle with circumcenter $O$ and circumradius $R$. $AO$ meets the circumcircle of $BOC$ at $A'$, $BO$ meets the circumcircle of $COA$ at $B'$ and $CO$ meets the circumcircle of $AOB$ at $C'$. Prove that \[OA'\cdot OB'\cdot OC'\geq 8R^{3}.\] Sorry if this has been posted before since this is a very classical problem, but I failed to find it with the search-function.

well,
let $A''=AO\cap BC$ $B''=BO\cap AC$ $C''=CO\cap AB$
then $OA'\cdot OA''=R^{2}$
so we need to prove $R^{3}\geq 8OA''OB''OC''$
which is equivalent to
$(x+y)(y+z)(z+x)\geq 8xyz$ where $x=S_{AOB},y=S_{BOC},z=S_{COA}$ :)
or we can get it from the inequality:
$xyz=(a-x)(b-y)(c-z)\rightarrow 8xyz\le abc$ :wink:

I have generalized the problem 1 year ago:

http://www.mathlinks.ro/Forum/viewtopic.php?t=55206

It is well known ( at least to me ) the generalization which mentioned by zhaoli in his post. So, if P is an arbitrary point inwardly to a given triangle ABC and DEF it’s cevian triangle, then we have that:

(AP / PD) + (BP / PE) + (CP / PF)  >= 6, (1) and (AP / PD)·(BP / PE)·(CP / PF)  >=  8,  (2)

I have found the elementary proofs of (1), (2), in a Hellenic textbook by J. Panakis – Triangle’s Geometry – volume A – page 107 - ''Gutenberg'' publications – Athens 1969.

Some other interest (I think) results, not well known (I don’t know any reference in Hellenic publication), are the follows:

1). - [(AP / PD)·(BP / PE)·(CP / PF)] - [(AP / PD) + (BP / PE) + (CP / PF)] = 2, (3). I named this equality as Van Aubel’s equality.

2). - (AP / PD) + (BP / PE) + (CP / PF) = (AP' / P'D') + (BP' / P'E') + (CP' / P'F'), (4).

3). - (AP / PD)·(BP / PE)·(CP / PF) = (AP' / P'D')·(BP' / P'E')·(CP' / P'F'), (5).

Where P', is the isotomic of P, with respect to the triangle ABC and D'E'F', it’s cevian triangle.

Kostas Vittas.

[quote="Yimin Ge"][color=darkred]Let $ABC$ be an [u]acute[/u] triangle with circumcenter $O$ and circumradius $R$. $AO$ meets the circumcircle of $BOC$ at $A'$,
$BO$ meets the circumcircle of $COA$ at $B'$ and $CO$ meets the circumcircle of $AOB$ at $C'$. Prove that $OA'\cdot OB'\cdot OC'\geq 8R^{3}\ .$[/color][/quote]
[color=darkblue][b]Lemma.[/b] In an [u]acute[/u] triangle $ABC$ exists the inequality $\boxed{\ \cos (A-B)\cdot\cos (B-C)\cdot\cos (C-A)\ge 8\cos A\cdot\cos B\cdot\cos C\ }\ .$

Indeed, using the well-known inequality $\{x,y,z\}\subset (0,\infty )\Longrightarrow (x+y)(y+z)(z+x)\ge 8xyz$ for $x: =\sin 2A>0$ a.s.o.

obtain $\prod (\sin 2A+\sin 2B)\ge 8\prod \sin 2A$ $\Longleftrightarrow$ $\prod \sin (A+B)\cdot\prod \cos (A-B)\ge \prod \sin 2A$ $\Longleftrightarrow$

$\prod \sin A\cdot\prod\cos (A-B)\ge 8\prod\sin A\cdot\prod\cos A$ $\Longleftrightarrow$ $\prod \cos (A-B)\ge 8\prod \cos A\ .$

[b]Proof of the proposed problem.[/b] Prove easily that $m(\widehat{OBA'})=90^{\circ}+C-B$ and $m(\widehat{OA'B})=90^{\circ}-A\ .$

From the triangle $BOA'\ ,$ using the Sinus' theorem obtain $\frac{OA'}{\cos (B-C)}=\frac{OB}{\cos A}\ ,$ i.e. $OA'=R\cdot\frac{\cos (B-C)}{\cos A}\ .$

Therefore, using the above lemma obtain the required inequality : $OA'\cdot OB'\cdot OC'=R^{3}\cdot\frac{\prod \cos (B-C)}{\prod \cos A}\ge 8R^{3\ .}$[/color]

cool inequality zhaoli!
the inequalities mentioned by vittasko are well known also to me but the trick with inversion to change it to the generalisation of this problem is very NICE !

http://www.mathlinks.ro/Forum/viewtopic.php?t=55206:
[quote="zhaoli"]Let $ABC$ be an acute-angled triangle, $P$ be an arbitrary point interior to $ABC$. $AP$ meet the circle $BPC$ again in $A_{0}$, $BP$ meet the circle $CPA$ again in $B_{0}$, and $CP$ meet the circle $APB$ again in $C_{0}$. Prove that

(1) $PA_{0}\cdot PB_{0}\cdot PC_{0}\geq 8PA \cdot PB \cdot PC$.

(2) $\frac{PA_{0}}{PA}+\frac{PB_{0}}{PB}+\frac{PC_{0}}{PC}\geq 6$.[/quote]

Thanks, zhaoli, for this nice generalisation.

I didn't bother to read your solution since I'm not very experienced with inversion (I'll try to find it myself later). This ist my solution for the original problem which also works for your generalisation.

By ptolemy, we have \[PA_{0}=\frac{PB\cdot CA_{0}+PC\cdot BA_{0}}{BC}.\] Let $x=\sin\angle BPC_{0}=\sin\angle CPB_{0},$ $y=\sin\angle BPA_{0}=\sin\angle APB_{0},$ $z=\sin\angle CPA_{0}=\sin\angle APC_{0}$. Then we have, by law of sines: \[\frac{CA_{0}}{BC}=\frac{z}{x}, \frac{BA_{0}}{BC}=\frac{y}{x}.\] so we get \[PA_{0}=\frac{z}{x}PB+\frac{y}{x}PC.\quad (1)\] .

Multiplying $(1)$ cyclically yields \[{PA_{0}\cdot PB_{0}\cdot PC_{0}\geq \prod_{cyc}(\frac{z}{x}PB+\frac{y}{x}PC}) \geq PA\cdot PB\cdot PC\] by AMGM.

Summung $(1)/PA$ cyclically yields \[{\sum_{cyc}\frac{PA_{0}}{PA}\geq \sum_{cyc}(\frac{z}{x}\frac{PB}{PA}+\frac{y}{x}\frac{PC}{PA}}) \geq 6\] by AMGM.
qed

[quote="Yimin Ge"]Let $ABC$ be an acute triangle with circumcenter $O$ and circumradius $R$. $AO$ meets the circumcircle of $BOC$ at $A'$, $BO$ meets the circumcircle of $COA$ at $B'$ and $CO$ meets the circumcircle of $AOB$ at $C'$. Prove that \[OA'\cdot OB'\cdot OC'\geq 8R^{3}.\] [/quote]
We can prove easily (as Mr. Virgil did) that the inequality is equivalent to \[{\prod \cos (B-C)} \geq 8 \prod{\cos A}\]
Let $AA' \cap BC=X$. Sine law in $\triangle ABX$ yields \[\frac{c}{\cos (B-C)}=\frac{BX}{\cos C} \iff \frac{\cos (B-C)}{\cos C}=\frac{c}{BX}\]
Again from sine law we get \[\frac{BX}{XC}=\frac{c\cos C}{b \cos B} \iff \frac{BX}{a}=\frac{c \cos C}{b \cos B+c \cos C} \\ \iff \frac{\cos (B-C)}{\cos C}=\frac{c}{BX}=\frac{b \cos B+c\cos C}{a \cos C } \]
So the required inequality transforms into \[\prod ( b \cos B+c\cos C) \geq 8abc \prod \cos A \]
Now we set $a \cos A=x, b \cos B=y, c \cos C=z$. So we need to prove that \[(x+y)(y+z)(z+x) \geq 8xyz \]
Which is apparently true by AM-GM, and we are done!

My solution (i think it's very simple):
Applyin Ptolemy's theorem for the cyclic quadrilateral $ BOCA' $ we have $ OA' \cdot BC= CO \cdot BA' + BO \cdot CA' =R(BA'+CA') $ or
$ \frac{OA'}{R}= \frac{CA'}{BC} + \frac{BA'}{BC}=\frac{sin2B}{sin2A}+\frac{sin2C}{sin2A}=\frac{sin2B+sin2C}{sin2A} $ and similary we have
$ \frac{OB'}{R}=\frac{sin2C+sin2A}{sin2B} $ and $ \frac{OC'}{R}=\frac{sin2B+sin2A}{sin2C} $. Finally we have $ OA' \cdot OB' \cdot OC'=(\frac{sin2C+sin2A}{sin2B})(\frac{sin2B+sin2A}{sin2C})(\frac{sin2C+sin2B}{sin2A}) R^3 \ge 8R^3 $ , because we have by AM-GM  $ (x+y)(y+z)(z+x) \ge 8xyz $ for positive real numbers $ x,y,z $ . DONE !

My Solution using [color=#00f]Inversion[/color]:

Consider an Inversion centered at $O$ with radius $R$.Easy to see that since $A^{'}=(BOM)\cap AO\Rightarrow X=BM\cap AO$ ($X$ is image of $A^{'}$ under this transformation,similarly $Y$ and $Z$).Now using [color=#00f]Inversion Distance[/color] formula,we have:$AA^{'}=\frac{R^{2}}{OA\times OX}\times AX\Rightarrow OA^{'}=AA^{'}-AO=\frac{R^{2}\times AX}{OA\times OX}-R$.Analogously,we get:

$OB^{'}=\frac{R^{2}\times BY}{OB\times OY}-R$ , $OC^{'}=\frac{R^{2}\times CZ}{OC\times OZ}-R$.

So we have to show that $\prod R(\frac{R\times AX}{OA\times OX}-1)\geq 8R^{3}\Leftrightarrow \prod (\frac{AX}{OX}-1)\geq 8\Leftrightarrow \prod \frac{AO}{OX}\geq 8$

But the last one is obvious and the equality holds when $AX$,$BY$,$CZ$ are all medians in $\triangle ABC$.

$Q.E.D$

Let $H$ be the orthocenter of $\triangle ABC$, let $\triangle DEF$ be the orthic triangle of $\triangle ABC$ and let $K,L,M$ be reflection of $H$ over $D,E,F$, respectively.
Inversion with radius $\sqrt{bc}$ at $A$ swaps $A'$ and $H$, $D$ and antipode of $A$ wrt $(ABC)$, thus we need to show that \begin{align*}
\left(\frac{AD\cdot 2R}{AH}-R\right)\left(\frac{BE\cdot 2R}{BH}-R\right)\left(\frac{CF\cdot 2R}{CH}-R\right)\geq 8R^3,
\end{align*}which is equivalent to \begin{align*}
\frac{AK}{AH}\cdot \frac{BL}{BH}\cdot\frac{CM}{CH}\geq 8.
\end{align*}
Now we consider $\triangle KLM$ as our "base triangle". Indeed $H$ is the incenter of $\triangle KLM$ and $A,B,C$ midpoints of arcs $LM,MK,KL$, respectively. Let $X=\overline{KH}\cap\overline{LM}$, $X=\overline{LH}\cap\overline{MK}$ and $Z=\overline{MH}\cap\overline{KL}$. Thus, we have $$\frac{AK}{AH}=\frac{AH}{AX}=\frac{AM}{AX}=\frac{KL}{LX}.$$
Similarly we get that $\tfrac{BL}{BH}=\tfrac{LM}{MY}$ and $\tfrac{CM}{CH}=\tfrac{MK}{KZ}$. By Ceva, $LX\cdot MY\cdot KZ=XM\cdot YK\cdot ZL$, therefore by AM-GM, \begin{align*}
KL\cdot LM\cdot MK&=(LX+XM)\cdot(MY+YK)\cdot(YZ+ZL)\\&\geq 8\sqrt{LX\cdot XM\cdot MY\cdot YK\cdot YZ\cdot ZL}\\&=8\cdot LX\cdot MY\cdot KZ,
\end{align*}as desired.


Find all functions $ f$ $ :$ $ R$ $ \longrightarrow$ $ R$ that for every $ x$,$ y$ $ \in$ $ R$:

$ f(x \plus{} y)$ $ \plus{}$ $ f(x \minus{} y)$ $ \equal{}$ $ f(x)(y \plus{} 2)$ $ \minus{}$ $ y(x^2 \minus{} 2y)$

Let $ y\equal{}2$ and $ y\equal{}\minus{}2$, respectively, then $ f(x)\equal{}x^2$.

Actually, you reach the same result with $ y\equal{}z$ and then $ y\equal{}\minus{}z$ with any $ z\neq 0$.


inequalities
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inequalities proposed

Let $a$, $b$, $c$ be three positive reals. Prove the inequality

$\frac{a^2\left(b+c\right)}{\left(b^2+c^2\right)\left(2a+b+c\right)}+\frac{b^2\left(c+a\right)}{\left(c^2+a^2\right)\left(2b+c+a\right)}+\frac{c^2\left(a+b\right)}{\left(a^2+b^2\right)\left(2c+a+b\right)}>\frac23$.

  Darij

This one still unsolved. Lets solve this. Darij can you give a hint ...?

Davron

Okay.

[hide="Hint"]Show that $\left(b^2+c^2\right)\left(2a+b+c\right)\leq\left(b+c\right)\left(a^2+2b^2+2c^2-bc\right)$. Remains to prove the inequality

$\frac{a^2}{a^2+2b^2+2c^2-bc}+\frac{b^2}{b^2+2c^2+2a^2-ca}+\frac{c^2}{c^2+2a^2+2b^2-ab}>\frac23$.[/hide]

  Darij

hmmm i am thinking over this and i have a picture of you at IMO 2005 Mexico during the Opening ceremony ...

Thank you very much i can send it if you want...

Davron

here's a solution
[hide]

first we apply the reduction given in darij's hint.

then we just need to prove

\[ \frac{a^2}{a^2+2b^2+2c^2-bc}+\frac{b^2}{b^2+2c^2+2a^2-ca}+\frac{c^2}{c^2+2a^2+2b^2-ab}>\frac23 \]

we first apply cauchy. we get

\[ \frac{a^2}{a^2+2b^2+2c^2-bc}+\frac{b^2}{b^2+2c^2+2a^2-ca}+\frac{c^2}{c^2+2a^2+2b^2-ab} \geq \frac{(a+b+c)^2}{5(a^2+b^2+c^2) - (ab + bc + ac)} \]

Now we apply am-gm

\[ \frac{(a+b+c)^2}{5(a^2+b^2+c^2) - (ab + bc + ac)} \geq \frac{(3(abc)^{\frac13})^2}{5(3(a^2b^2c^2)^{\frac13}) - 3(a^2b^2c^2)^{\frac13}} \]

However, this equals 3/4, which is greater then 2/3, so we are done. [/hide]

@Davron: would be nice, thanks a lot. My email is darijgrinberg=gmail.com, where = means @.

@The Original Pi Guy: It is [i]not[/i] always $\geq\frac34$. In fact,

\[ \frac{(a+b+c)^2}{5(a^2+b^2+c^2) - (ab + bc + ac)} \geq \frac{(3(abc)^{\frac13})^2}{5(3(a^2b^2c^2)^{\frac13}) - 3(a^2b^2c^2)^{\frac13}}  \]

is not a valid application of AM-GM...

  Darij

Pi your solution is incorrect because you cannot aplly minimum value of the denominators try to find another

 Darij i am gonna send them maybe tomoroow, sorry i cant now cause i have them on my MC and my MC is in my friend  :)

Thanks

Davron

Yes, I understand now. Thank you.

I've made the same mistake now twice in the same day....

 :wallbash:

I'll fix it when I find a better way to do it...

it took me many calculations but finally i showed that if $a<b<c$ then the expression is monotonically increasing as a function of $a$ so the minimum is acquired when $a=0$ and so
all we need to prove is that $\frac{b^2}{c(2b+c)}+\frac{c^2}{b(2c+b)}\geq \frac{2}{3}$which is true but we have strict inequality because $a,b,c$ are positive

;)  What do you think about Eagles calculations Darij...?

Davron

[quote="Albanian Eagle"]it took me many calculations but finally i showed that if $a<b<c$ then the expression is monotonically increasing as a function of $a$ so the minimum is acquired when $a=0$ and so
all we need to prove is that $\frac{b^{2}}{c(2b+c)}+\frac{c^{2}}{b(2c+b)}\geq \frac{2}{3}$which is true but we have strict inequality because $a,b,c$ are positive[/quote]
Sorry disregard bump from a year ago but it's a nice problem. I'll post the rest but for now just the easy part:
\[\frac{b^{2}}{c(2b+c)}+\frac{c^{2}}{b(2c+b)}\geq \frac{(b+c)^{2}}{(b+c)^{2}+\frac{(b+c)^{2}-(b-c)^{2}}{2}}\geq \frac{2}{3}\]
with equality at $b=c$ but he uses $a<b<c$

Oops, I see I should have already posted this some time ago. This is my solution of the above inequality.

  Darij


The problem is 3^2x- 12 multiplied by 3^x + 27=0
I let y=3x and I get
y^2 - 12y +27 =0
from there I get
Y= (9,3)
Now in my book it says that the answer is 1 or 2 how is this so? I don't know what to do after this.

There are two ways to parse this: \[ (3^{2x}\minus{}12)(3^x\plus{}27)\equal{}0\ (1)\] \[ 3^{2x}\minus{}12(3^x\plus{}27)\equal{}0\ (2)\] Which one, 1 or 2,  do you mean?

The book says that the answer is 1 or 2 not the first problem or the second problem, but actually 1 or 2.

[quote="someperson01"]
Which one, 1 or 2,  do you mean?[/quote]

He does not mean either of these, but rather:
\[ 3^{2x}\minus{}12\cdot 3^x \plus{} 27\equal{}0\]

dvcalculus, you are right that $ y \in \{3,9\}$, but remember that you want values of $ x$ for your answer. Since you set $ y\equal{}3^x$, how would I get back to $ x$? When you figure this out, you will find your answer to be the same as those in the book.


Q1 What is the probability that the product of two randomly chosen integers is divisible by 12?

Q2 Two integers are multiplied. What is the probability that their product is a multiple of 8?

Q2:

[hide="my solution..."]

A random integer divisible by $ n$ is chosen with probability $ \dfrac{1}{n}$, since it occurs in cycles of $ n$. For the product of 2 integers to be a multiple of 8, we have to have either one divisible by 8 or one divisible by 4 and the other divisible by 2. It is not necessarily true that if it is divisible by 4 it will be divisible by 8, but it is true the other way around. Thus, the probability is 1/8*1=$ \boxed{\dfrac{1}{8}}$. This can be thought as there is a 1/8 probability of 8 appearing, which gives $ \dfrac{1}{8}$.[/hide]

bbgun, that answer is not correct. When you stated "Thus, the probability is..." in your explanation, I don't see how that logical connection was made.

[quote="jackie-chan01"]Q1 What is the probability that the product of two randomly chosen integers is divisible by 12? [/quote]

As a hint, I would like to say something. I'm assuming you might not be acquainted with these 'rules'.

For a number to be divisible by 12, it should be divisible by both 3 and 4 or should be divisible by 6 and 2. For a number to be divisible by 2, the number should end with either 0, 2, 4, 6 or 8 in the units place. Similarly, for a number to be divisible by 3 the sum of the digits of the number should be divisible by 3. I guess you know the divisibility rule for the number 4.

For a number to be divisible by 6, the number should be divisible by both 3 and 2.

Using the above 'rules' it becomes quite evident that for a number (i.e. the product of the two numbers) to be divisible by 12 -- it's units place should contain either 0, 2, 4, 6 or 8, and the number (i.e. sum of the digits) should itself be divisible by 3.

Now, applying some of these 'hints'-- try to find a solution to the problem. It becomes quite obvious that the probability is 'high'.

[quote="jackie-chan01"] Q2 Two integers are multiplied. What is the probability that their product is a multiple of 8? [/quote]

For the number (i.e. product) to be divisible by 8 -- it should also be divisible by 4 and 2.

I hope I'm not blindly spouting the various 'divisibility rules', but I feel that being reasonably acquainted with these 'rules' will almost certainly help in resolving these problems.

There is no way to pick any integer such that every two integers are equally likely to be picked. However, you could try to find the probability that any integers from the range $ [\minus{}n, n]$ would satisfy the conditions as n approaches infinity.

[quote="Gen8"]For a number to be divisible by 12, it should be divisible by ... 6 and 2.[/quote]Not true: 6 is divisible by 6 and 2.

Q1. Probability of getting a multiple of 3: $ 1\minus{}(\frac23)^2\equal{}\frac59$
This is basic complementary counting.
Probability of getting a multiple of 4: $ 1\minus{}(\frac34)^2\equal{}\frac7{16}$
Probability of getting two multiples of 2, but none of 4: $ (\frac14)^2\equal{}\frac1{16}$
Total: $ \frac59\cdot(\frac7{16}\plus{}\frac1{16})\equal{}\frac{5}{18}$
@Gen8: What do you mean? I don't consider this probability to be "high" in these circumstances.

[quote="brainomega"]Not true: 6 is divisible by 6 and 2. [/quote]

mate, For a number to be divisible by 6 it should be divisible by both 3 and 2. I don't know what you meant by: "it is divisible 6 and 2". Going by the divisibility rule we can assume that for a number to be divisible by 6 , the sum of the digits of the number should be divisible by 3.

[quote="Gen8"][quote="brainomega"]Not true: 6 is divisible by 6 and 2. [/quote]mate, For a number to be divisible by 6 it should be divisible by both 3 and 2. I don't know what you meant by: "it is divisible 6 and 2".[/quote]
brainomega was giving a counterexample to this statement:
[quote="Gen8"]For a number to be divisible by 12, it should be divisible by both 3 and 4 or should be divisible by 6 and 2.[/quote]
If a number is divisible by 3 and 4, then it is divisible by 12, but if it is divisible by 6 and 2, it may not be divisible by 12, since the the number 6 is divisible by 6 and 2 but not 12.

I'm not sure I see how your hints lead to a solution. Could you explain how to solve the problem using those divisibility rules?


If $a$ and $b$ are positive numbers such that $a^b = b^a$ and $b = 9a$, find the value of $a$.
\[ \text{(A)}\ \sqrt{3} \qquad \text{(B)}\ 3^{1/4} \qquad \text{(C)}\ 1 \qquad \text{(D)}\ \sqrt[3]{2} \qquad \text{(E) No solution} \]


calculus
geometry
vector
linear algebra
matrix
analytic geometry

Hey all! Over the summer in addition to a ton of olympiad stuff I'm probably going to be going further in so-called "higher mathematics," and I think I want to start doing some stuff in multivariable calculus. For a strong math student, would you say a teacher is not needed? I was thinking of simply using my textbook along with some of the MITOCW stuff, or would I need a course (I'm thinking the EPGY one) on it? Any suggestions for the best way to learn?

Thanks.

Assuming you found single variable calculus fairly easy, I think multivariable calculus is not that much harder. I believe you could learn it without an instructor.

Thanks! Any other thoughts or suggestions from anybody? Which book should I use? I have Larson's [i]Calculus with Analytical Geometry[/i]: is it good for multivariable calculus?

Thanks!

Larson's book is okay for an introduction to Multivariable Calculus. It's not the best book out there, but most of the "big name" calculus books are all about the same - Stewart, Edwards and Penny, Thomas, etc.

My favorite of these books are Stewart's and Thomas' (the book from which I learned calculus) because they have slightly better questions for exercises. But if you already have Larson you might as well continue with it as they're not [i]that[/i] different.

If you want a slightly deeper explanation of topics and ideas, Marsden and Tromba's [u]Vector Calculus[/u] book is good - but at this point I don't think it's necessary for you to go for that depth right now. Just play around with the ideas and calculations. The more theoretical stuff can come later.

To that end, if you just want something that cuts to the chase and shows you some cool computational techniques, try [u]Schaum's Outline of Advanced Calculus[/u] by Robert Wrede and Murray Spiegel. It has lots of cool (single and multi-variable) Calculus techniques that you don't see in normal textbooks.

Thanks for all the replies! I will definitely look into these :)

[quote="#H34N1"]Dude the larson book is what we're using for calc AB[/quote]

As I understand Larson's eight edition comes in many "sizes," so to speak, where there include combinations of the subjects  "Precalculus" "Calculus 1" "Calculus 2" and "Calculus 3" (which is multivariable) to "address the different ways instructors teach." So your Calculus text by Larson may be different than mine by Larson (which is Calc 1-3 w/ Analytical Geometry)

That's the exact book I'm reading from right now.

What do you guys think of Apostol? I hear its really good for theory, which is what I'm most interested in  :).

Thanks!

Apostol is a great book but I wouldn't go for Apostol until you've gotten a good intuitive and computational feel for both single and multivariable Calculus. Theory is only good if it builds from a base of visualization and intuition. I find that even if I can read a book and follow what is going on, it won't stick unless I have a good intuitive feel for it. You could start with Marsden and Tromba (which in my opinion is still too advanced for most people to start with, but more or less starts off where BC Calculus ends), and then move on to Apostol if can do nearly all of the exercises in that book.

Edit: Also realize that multivariable calculus probably won't make as much sense until you take (at least) AP Physics, since most of the motivation and examples in multivariable calculus come from Physics.

Dude Anaxerzia

the calculus book Larson with analytic geometry (the really fat one aka the one I have) fails miserably when explaining concepts...

You might want to take EPGy so you get credit for it

[quote="Anaxerzia"]I'm thinking the EPGY one[/quote]

That is a really good choice. I learned Calc I that way, in less than the time than the average person, I got an A- without reading the recommended book, and I was 12. It even shows proofs of facts, shows detailed solutions to most problems, and multiple ways to do some problems(!!!).

[b]Off-Topic Suggestion:[/b]

I realize my response is very oblique, but I personally think you should be comfortable with linear algebra before going into multivariable calculus or differential equations.

Most people don't do that, and most multivariable classes begin with a little introduction on vectors and stuff. But then they come back and start getting revelations about multivariable calculus and synthesizing concepts in retrospect. Which makes sense, I guess; it might be a good idea to know vectors and matrices like the back of your hand before doing any kind of calculus with them!*

My final argument is that an understanding of linear algebra will probably help you much more in contest mathematics than multivariable calculus will. Examples of problems trivialized with linear algebra include [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1088469]this year's AIMEII #9[/url] and [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1038337]this year's AMC 12A #25[/url]. Specifically, you will be able to find a solution much more quickly, and your solution will be more economical.

If you decide to stick with multivariable, I'm sure you'll have fun with that do. But if you buy my suggestion, I suggest you first pick up any standard introductory Linear Algebra book and introduce yourself to the concepts. Pay particular attention to the meaning, uses, and implications of the matrix determinant, but make sure you are reasonably comfortable with all the concepts. Then, take a look at the following book:

[color=blue]Linear Algebra Done Right
Second Edition
by Sheldon Axler
Springer[/color]

It's a theoretical (vs computational) approach to linear algebra. Not only will it give you a rock-solid understanding (and appreciation) of the subject, it takes a proof-based approach, which might be helpful given that you are interested in olympiads. It is important, though, that you involve yourself with the "usual" approach first, so that you can fully understand and appreciate the rich abstractions this book uses. Also, the book does not place the usual amount of emphasis on the determinant, which is something you'll have to make up for yourself, since the determinant is incredibly powerful.

This whole process took me about 5 months of light involvement (that times worth of math class), and you might be able to do it in less.

[i]
* - There was an embarassing situation in my diff.eq's class last year when about 1/4 of the class had taken linear algebra, and 3/4 had not. Lin.alg was not listed as a suggested background, but those who took lin.alg had an average grade of 14 %-points higher than those that did not.[/i]

My current books are an introduction to both Linear and Multivariable, so ya :)


What is the area, in square units, of a triangle with vertices at $ (0,0)$, $ (0,5)$, and $ (7,12)$? Express your answer as a decimal to the nearest tenth.

Here, we have a triangle with base $ 5$ and height $ 7$. Its area is thus $ \boxed{17.5}$.

The width is $7$,
and you know using the coordinates :)

Here we use the Shoelace Theorem. The answer is 35/2 = 17.5

What is the Shoelace Theorem?
(I solved it without o.O)

o.O is just this emoji that doesn't work here but it works in the classroom :(
Here is what it looks like~
[IMG]http://aops-cdn.artofproblemsolving.com/icons/smilies/boggle.gif[/IMG]
and IMO... either International Math Olympiad (??), or In My Opinion.

oh, right. In my opinion. I know that IMO means international math olympiad (BTW i will not make it in a thousand years)


find all the function $ f: Z\rightarrow Z$ satisfying two condition
$ (i): f(1)\equal{}f(\minus{}1)$
$ (ii): f(x)\plus{}f(y)\equal{}f(x\plus{}2xy)\plus{}f(y\minus{}2xy)$

if $ x \equal{} 2^i (2k \minus{} 1)$ for some integer $ k$, $ f(x) \equal{} t_i$  ($ t_i(i \equal{} 0,1,2,3,....)$ is an integer) and $ f(0)$ is an arbitrary integer.

pf)

[b]step1[/b])if $ x,y$ are odd numbers, $ f(x) \equal{} f(y)$

Substitute like below in $ (ii)$

$ 1 \to x$, then we get $ f(1) \plus{} f(y) \equal{} f(2y \plus{} 1) \plus{} f( \minus{} y)$,
$ y \to x$ and $ \minus{} 1 \to y$, then we get $ f(y) \plus{} f( \minus{} 1) \equal{} f( \minus{} y) \plus{} f(2y \minus{} 1).....(iii)$

we know that $ f(1) \equal{} f( \minus{} 1)$ so we can get $ f(2y \plus{} 1) \equal{} f(2y \minus{} 1)$
this implies proof for step1) is ended.

[b]step2[/b])$ f(x) \equal{} f( \minus{} x)$

At $ (iii)$, because $ f( \minus{} 1) \equal{} f(2y \minus{} 1)$, $ f(y) \equal{} f( \minus{} y)$

[b]step3[/b])$ f((2k \minus{} 1)x) \equal{} f(x)$ for all integer k

At $ (ii)$, substitute $ \minus{} y \to y$.. then we get $ f(x) \plus{} f( \minus{} y) \equal{} f(x \minus{} 2xy) \plus{} f( \minus{} y \plus{} 2xy)$
we know that $ f( \minus{} y) \equal{} f(y)$ and $ f(y \minus{} 2xy) \equal{} f( \minus{} y \plus{} 2xy)$ from step2), so we get $ f((2y \plus{} 1)x) \equal{} f((1 \minus{} 2y)x)$
and because $ f((1 \minus{} 2y)x) \equal{} f((2y \minus{} 1)x)$, $ f((2y \plus{} 1)x) \equal{} f((2y \minus{} 1)x)$ is true.....
this implies proof for step3) is ended....

[b]step4)[/b]if $ x,y$ have same exponents of 2, $ f(x) \equal{} f(y)$

At $ (ii)$, substitute $ 2^i \to x$, then we get $ f(2^i) \plus{} f(y) \equal{} f(2^i \plus{} 2^{i \plus{} 1} y) \plus{} f((1 \minus{} 2^{i \plus{} 1})y)$..
we know that $ f(y) \equal{} f((1 \minus{} 2^{i \plus{} 1})y)$, so we get $ f(2^i) \equal{} f(2^i \plus{} 2^{i \plus{} 1} y)$
this is the end of proof for step4)

From step1~4, proof for this problem is ended...and we can know that this really works by checking....

Criticizing is welcomed....


If three numbers are chosen at random, uniformly and independently, from the interval [0,1], what is the probability that they are the lengths of the sides of a triangle?

This has appeared somewhere between 1 and 10,000 times on the forum -- using the search function will help you determine the exact number, as well as find a solution.


I've an idea ... I've not small collection of different math Olympiad problems ... for example almost all problems from USA, China, Russia, Vietnam are only a part of mine collection but I'm too lazy to post them. So I want to exchange some themes ... for example I don't have Bulgarian Math Olympiad problems III round 1991, 1992, 1993, 1994, 2002 and IV round 1992, 1993, 2002 but I have almost all other problems [1960-1990], [1995-2001] and [2003-2007] so if someone want if he publish or send me some theme that I haven't I may post some of my themes at him/her choice... what do you think about it?

I also collect problems from Polish Math Olympiad and Krasnodar/Russia (there are some in kubsu.ru) as well as II (now known as I-st) round of Bulgarian Math Olympiad. There may be some interesting problems in Languages not well known to me (I know only Bulgarian, Russian and English).

You may use borislav_mirchev@yahoo.com or ICQ:226758339 to contact me, or post the problems directly in mathlinks.ro.

http://www.imomath.com/index.php?options=oth|other&p=9r93083
See here.  :wink:

10 x a lot ... but I know for this site and the themes I'm looking for are missing there ... I've themes missing on this site too.  :(  ... old themes are available in Bulgarian books I've and the Magazines but I don't have the Magazines for these years.


number theory
relatively prime
prime numbers
number theory proposed

3*a^2 + a = 4*b^2 + b.
prove a-b is a perfect square.

I think it was posted before. It's from an Iranian olympiad, if I remember correctly.

A try :

Equation can be rewritten (a-b)(1+3a+3b) = b so it suffices to show that in this case gcd(a-b,1+3a+3b) = 1.

Let p prime, p | a-b, p | 1+3a+3b but then p | b, then p | a, which leads to a contradiction with p | 1+3a+3b.

that looks really nice and simple ... and sorry if im missing something really obvious but i dont understand smthing ... why does p is prime, p|a-b and p|1+3a+3b, then p|b and p|a? as i see it a = b =  -1/6 (mod p)?
My solution was lol *much* longer and used Pell, several computations and solving a nasty recurrence.

>>why does p is prime, p|a-b and p|1+3a+3b, then p|b and p|a?

I may equally miss something obvious (I'm making a lot of stupid mistakes these days), but a-b | b so if p | a-b then p | b. If p | a-b and p|b then p |a.

Is-it correct ?

oh ok ... its correct ... i was missing something obvious as usual ...
really nice and simple solution !   :) i feel so stupid now ...  :blush:

[quote="belenos"]A try :

Equation can be rewritten (a-b)(1+3a+3b) = b so it suffices to show that in this case gcd(a-b,1+3a+3b) = 1.[/quote]

Why does it suffice to show?  The product of two relatively prime numbers can be a square.

to magnara:
if the product of two relatively prime numbers is a square, then those 2 numbers are squares. We wish to prove a-b is a square, and we know (a-b)(3a+3b+1) is also a square, so it suffices to show (a-b,3a+3b+1)=1 from which it follows a-b and 3a+3b+1 are both squares.


How many ways can this be prooved:

Number of ways of filling $n$ boxes with $r$ objects such that no box is empty is $\binom{r-1}{n-1}$ ( it is corect :| )

[quote="varun"]How many ways can this be prooved:

Number of ways of filling $n$ boxes with $r$ objects such that no box is empty is $\binom{n-1}{r-1}$ (Hope it is corect :| )[/quote]

if you have 4 objects (ABCD) and three boxes .  Does

AB | C | D  and C | AB | D  the same ?

moreover if no empty box is allow , $r\geq n$

No they are not same.

[hide="Important Note"]I've not mentioned that the boxes are identical ;) [/hide]

So I saw a mistake in my question. Please read it again after I have editted it $1$ time.

Sorry for this :ninja:


function
superior algebra
superior algebra unsolved

Hi,

I have the the following clauses of boolean variables:

$ (a_1,a_2)$

$ (a_2,a_3)$

$ (\bar{a_2},a_4)$

$ (\bar{a_1},\bar{a_4})$

$ (a_1, \bar{a_2}, a_3)$

$ (\bar{a_1}, a_2, a_4)$

$ (a_1, \bar{a_2}, \bar{a_3})$

where

$ a \equal{} true$

$ \bar{a}\equal{}false$

I need to find out if the set of these clauses is satisfiable. I am not sure about $ (\bar{a_1},\bar{a_4})$. It is false as a clause, but would this mean that it would make the whole set unsatisfiable?

many thanx!

[quote="Antonio_26"]Hi,

I have the the following clauses of boolean variables:

$ (a_1,a_2)$

$ (a_2,a_3)$

$ (\bar{a_2},a_4)$

$ (\bar{a_1},\bar{a_4})$

$ (a_1, \bar{a_2}, a_3)$

$ (\bar{a_1}, a_2, a_4)$

$ (a_1, \bar{a_2}, \bar{a_3})$

where

$ a \equal{} true$

$ \bar{a} \equal{} false$

I need to find out if the set of these clauses is satisfiable. I am not sure about $ (\bar{a_1},\bar{a_4})$. It is false as a clause, but would this mean that it would make the whole set unsatisfiable?

many thanx![/quote]

First, that $ (\bar{a_1},\bar{a_4})$ is "false" has nothing to do with whether the function is satisfiable. It is probably better to think of a clause such as $ (\bar{a_1},\bar{a_4})$ in terms of on or off rather than true or false. For instance, the output of $ (\bar{a_1},\bar{a_4})$ is on when both $ a_1$ and $ a_4$ are off.

Second, I plugged these clauses into a Karnaugh map and it turns out they should all be satisfied by the function: $ f \equal{} a_4 \plus{} a_1a_2 \plus{} a_1a_3\bar{a_4} \plus{} a_2a_3\bar{a_4}$


There are 10 people and some of them know each other (if A knows B, it implies that B knows A). Among any three of them there are at least two of them who know each other. Prove that there are four people such that each one of them knows everybody else in the chosen group of four.

[hide="Hint"]
[url=http://en.wikipedia.org/wiki/Ramsey%27s_theorem]Ramsey's Thoerem[/url]
[/hide]


induction
AwesomeMath
summer program
inequalities
trigonometry
function
geometry

I only know how to do the first two questions because they're easy so here they are.

Could you guys post other solutions too?

Thanks
[hide="1"]
When $n=1$,

$1-\frac{1}{2}=\frac{1}{2\cdot 1}$

When $n=k$,

$\sum_{1}^{2k}(-1)^{a+1}\frac{1}{a}=\sum_{k+1}^{2k}\frac{1}{a}$

When $n=k+1$

$\sum_{1}^{2k+2}(-1)^{a+1}\frac{1}{a}=\sum_{k+2}^{2k+2}\frac{1}{a}$

Subtract them from each other to get

$\frac{1}{2k+1}-\frac{1}{2k+2}=\frac{1}{2k+1}+\frac{1}{2k+2}-\frac{1}{k+1}$.[/hide]

[quote="junggi"]oh maaaan i read the introductory thingies, and dude this is TYPO MADNESS[/quote]

Yeah, it's pretty funny. I particularly liked problem 24 on the introductory problem set

Err...I don't think it is best to just post solutions.

its just to warm us up, not an evaluation test.

i dont think :maybe:

If I'm not mistaken, these are just the Awesomemath Year-Round packets. The deadline for all of them  has passed  except the combinatorics  one, so I don't think there's anything wrong with discussing  solutions, especially since all the AMY enrollers will receive a full set of solutions.

For those of you who did number 2, what is the formula for the sum of 4th powers, I can't seem to find it? Or, do I have to guess what it is?

Yea whatever...I guess I misunderstood the email...

Anybody solve the perfect square one? [#9]

[hide="(useless?) progress"]
well i guess they meant to say $n>1,$ so base case:
For $n=2,$ take $(2,1).$ Then $a_{1}=2$ is not a perfect square.
Inductive step:
Assume that for $n,$ there's a permutation $(a_{1},...,a_{n})$ of $(1,...,n)$ such that $a_{1},a_{1}+a_{2},...,a_{1}+...+a_{n-1}$ are not perfect squares.
Now consider $(a_{1},...,a_{n-1},a_{n},n+1),(a_{1},...,a_{n-1},n+1,a_{n})$
Note that
$1+2+...+n=\frac{n(n+1)}{2}\le a_{1}+...+a_{n-1}+(a_{n}\text{ or }n+1)\le 2+3+...+n+(n+1)=\frac{(n)(n+3)}{2}$
now if we prove that in that interval, there is at most one perfect square, then we are done, because if one of $a_{n}$ or $n+1$ make the sum equal to a perfect square, we can take the other, which would make it not equal to a square, and by the induction hypothesis, none of the previous sums are squares.
I'm a noob at number theory, so I don't know how to prove that yet[/hide]

Okay...yea...that makes more sense. I thought I had some counter example to that method, but I just realized that I was being stupid.

2,1,3,4,5,6
2,1,3,4,6,5
but that last one can just be 2,1,3,4,6,7

Anyway,

Lets suppose that the inequality is true for $q$ and $q+1$...

$\frac{(n)(n+1)}{2}\le q^{2}\le \frac{(n)(n+3)}{2}$

$\frac{(n)(n+1)}{2}\le (q+1)^{2}\le \frac{(n)(n+3)}{2}$

$-n\le 2q+1\le n$

...err lets see what Reduce[] in mathematica says...

[code]Reduce[{(n) (n + 1)/2 - 1 < q^2,
  q^2 < 1 + (n) (n + 3)/2, (n) (n + 1)/2 - 1 < (q + 1)^2, (q + 1)^2 <
   1 + (n) (n + 3)/2}, {n, q}, Integers]

False[/code]

okay...that settles that :P

I probably should figure out how to show that...

#7:

[hide="ok fine by induction"]
One of the numbers $a_{1}+a_{2}, a_{2}+a_{3}, ...a_{n-1}+a_{n}$  had better be positive. Erase it and write its sum.
[/hide]

[hide="Extreme principle is much better!"]

Put a sufficiently large "phantom"  :ninja:  number $x$ somewhere along the circle and watch the sums $x$, $x+a_{1}$, blah blah. Choose the smallest value that this gets to. Obviously, this occurs before some positive number. Since every position is relative to that number, starting there works.

[/hide][/hide]

1,-2,3,-1,-1
1,(-2+3),-1,-1

Of course, we can choose another pair that have a nonegative sum, but your proof did not say that.

Here is my problem

you have say...

$a_{1},a_{2},...a_{n}$

ok so do what you did...take two terms and collapse them into one term

$a_{1},a_{2},....,a_{k}+a_{k+1},....,a_{n}$

by induction hypothesis, we have that

$a_{1}\ge 0$
$a_{1}+a_{2}\ge 0$
...
$a_{1}+a_{2}+...+a_{k-1}\ge 0$
$a_{1}+a_{2}+...+(a_{k}+a_{k+1})\ge 0$
...
$a_{1}+a_{2}+...+a_{n}\ge 0$

You have not shown that $a_{1}+a_{2}+...+a_{k}\ge 0$.

This is precisely what my counterexample shows.

EDIT woops...I edited this post instead of adding a new one...lol

I never said to start with the new number. I only reduced the $n$ numbers, with total nonnegative sum,  to $n-1$  numbers, still with total nonnegative sum. Since we're inducting on $n$, we can fall back on the induction hypothesis.

wth...why would use induction on that problem when a combinatorial argument works so much better

[hide="and now"]
we show
\[ m \binom{n}{m}= (n-m+1) \binom{n}{m-1}\]
using the best proof method for combinatorics...story telling!

So one day, Al decided to choose $ m$ members for his varsity hop-scotch team from $ n$ who tried out. He can this in $ \binom{n}{m}$ ways. He then needed to choose a captain. This can be done in $ m$ ways.

On the other hand, Gore decided to choose $ m-1$ members in $ \binom{n}{m-1}$ ways and then picked the captain from the remaining people in $ n-m+1$ ways. Moral of the story: $ \binom{n}{m-1}(n-m+1)=\binom{n}{m}m$. [/hide]

as you pointed out, it's not that hard to do it by using a combinatorial argument. The challenge lies in induction. Since you guys are working on induction, I thought it would be a good challenge.

[quote="roadnottaken"]I personally feel that story arguments aren't quite as rigorous as algebraic proofs, so I feel that it is important to be able to prove facts that way, as well.[/quote]
They are just as valid, regardless of how you "feel" about them.

Both ways are good, and enlightening to the soul. The combinatorial argument makes you think combinatorially, and makes you better at combinatorics. The induction, which seems to be pretty straightforward in this problem, in this case, just shows that you're good at algebra (possibly messy algebra, I haven't actually done it).

[quote="mdk"]I don't think that the induction method is sraight forward, but it might be for you.[/quote]

Well, true, it's not completely trivial, but it takes less thinking than the combinatorial argument, in general.

Funny how this thread has nothing to do with its title.

[quote="Phelpedo"]You will not think combinatorics is easy if you see the counting in two ways packet. It is HARD![/quote]

Well I found Induction to be much harder but anyway for problem 6 advanced the one with the students and the teachers why couldn't the base case be generalized for more students namely we select the group of people whose friends we want to increment by one and increment by one after incrementing those people we don't want to by 6 like the base case for n=1 m=1?

Also for the second last example (11) why must we have l=<s? also why does f(G)=t+l  ?   what if an edge of G_1 is not in a triangle what if it's in two?  Also does G_1 or G_2 include A?

As well when those functions in the inequalities were defined how did they find the function?

 :( Induction is HARD

Also could someone plz answer my circles post? It's the one about the idea tying the solutions together.

As well does someone have a solution or hint to problem 7 it somehow was not in my solution packet.

Three comments on the induction packet:
1. It is probably the hardest AwesomeMath packet ever released. It takes a long time to understand some of the stuff.
2. It has a lot of typos.
3. Some of the problems, you should not try to tackle using induction. Either they have other proofs which don't require it (H8 for example), or they require induction in a teeny detail of the solution- the major idea is something else. Don't necessarily keep saying to yourself "How do I induct, which variable, what base case...", instead just approach the problem in a more natural fashion. One reason that this packet is so hard is that it doesn't focus on one topic in particular, but a technique, and you have to use stuff from all areas.

@hello123: By the way, most people in this forum don't have the packets from that long ago (and I won't have access to them at college either), so it might be a better idea to just post the problems in the subject area (combinatorics, algebra, etc.).

On H7: My solution is buried in one of my many notebooks, and I don't know which:( But I remember NOT solving it with induction at all, but in a more direct fashion.
On H6: The solution is pretty messed up. I don't know what the correct one is, if any. You're on your own, there!
On hard example 11: $ G_2$ is the graph induced by all vertices except A, $ G_1$ is the graph induced by neighbors of A not including A. $ l\leq s$ because $ G_2$ includes $ G_1$. $ f(G)\equal{}t\plus{}l$ because there are t triangles not including A as vertex and $ l$ triangles including A. An edge of $ G_1$ can only be in one triangle including A.

There is a left-over from the 2007 discussions, as to no two perfect squares existing between $ n(n\plus{}1)/2$ and $ n(n\plus{}3)/2$. Assume there are, and take $ k^2$ the least of them; then $ n(n\plus{}1)2 \leq k^2 < (k\plus{}1)^2 \leq n(n\plus{}3)/2$, so $ 2k\plus{}1 \leq n$. But then $ k < n/2$, so $ n(n\plus{}1)/2 \leq k^2 < n^2/4$, or $ 2n\plus{}2 < n$, absurd.