HI, My granddaughter just started 8th grade. She has a problem I cannot help her with :( .     I'm hoping someone can...the problem is as follows to the word....

 Jake's locker number has four digits and is not divisible by three.The second digit is the sqaure of the first digit.The final two digits name  the square of the second digit. What is Jake's locker number?...
 They also have to show their work......HELP.....thank you

There are only three possibilities since when the first digit exceeds 3, its square will have 2 digits.
1101
2416
3981
And only 2416 is not divisible by 3. So the locker number is 2416.
[hide]Please put this problem in classroom math or getting started or somewhere else, but not the algebra section.[/hide]


induction
algorithm
function
combinatorics unsolved
combinatorics

Let $ x_1,x_2,...,x_n$ be distinct natural numbers such that every subset of $ \{x_1,x_2,...,x_n\}$ has different sum of elements (so $ 2^n$ different sums). Prove that
\[ x_1^2 \plus{} x_2^2 \plus{} ... \plus{} x_n^2 \geq \frac {4^n \minus{} 1 }3 .
\]

Well, I feel it is not too hard.

Without loss of generality let us assume that $ x_1<x_2<\ldots<x_n$. It is known that every natural number can be obtained as the sum of powers of 2 in a unique way.So, the first instinct would be including all the powers of 2 starting from $ x_1\equal{}2^0$.

Verifying the claim, set $ x_1\equal{}1$, in order to make the sequence grow slowly the choice of $ x_2$ would be 2. This generates another number 3. Hence the choice of $ x_3$ would be 4.We would get $ x_n$ by
$ x_n\equal{}x_{n\minus{}1}\plus{}\ldots\plus{}x_1\plus{}1$
By this process we would get all the natural numbers and the instinct is proved right. Then the obvious recurrence relation holds,
$ x_n\equal{}x_{n\minus{}1}\minus{}1\plus{}x_{n\minus{}1}\plus{}1$
$ \Rightarrow x_n\equal{}2x_{n\minus{}1}$

$ \Rightarrow {x_1}^2\plus{}\ldots\plus{}{x_n}^2\equal{}4^0\plus{}4^1\plus{}\ldots\plus{}4^{n\minus{}1}$

and we are done.

[quote="srikanth"]Well, I feel it is not too hard.

Without loss of generality let us assume that $ x_1 < x_2 < \ldots < x_n$. It is known that every natural number can be obtained as the sum of powers of 2 in a unique way.So, the first instinct would be including all the powers of 2 starting from $ x_1 \equal{} 2^0$.

Verifying the claim, set $ x_1 \equal{} 1$, in order to make the sequence grow slowly the choice of $ x_2$ would be 2. This generates another number 3. Hence the choice of $ x_3$ would be 4.We would get $ x_n$ by
$ x_n \equal{} x_{n \minus{} 1} \plus{} \ldots \plus{} x_1 \plus{} 1$
By this process we would get all the natural numbers and the instinct is proved right. Then the obvious recurrence relation holds,
$ x_n \equal{} x_{n \minus{} 1} \minus{} 1 \plus{} x_{n \minus{} 1} \plus{} 1$
$ \Rightarrow x_n \equal{} 2x_{n \minus{} 1}$

$ \Rightarrow {x_1}^2 \plus{} \ldots \plus{} {x_n}^2 \equal{} 4^0 \plus{} 4^1 \plus{} \ldots \plus{} 4^{n \minus{} 1}$

and we are done.[/quote]

You say in order to make it grow slowly, so you choose 1,2,4,8.... but that doesn't proved that it is really the "slowest", right?

[quote="greentreeroad"]
You say in order to make it grow slowly, so you choose 1,2,4,8.... but that doesn't proved that it is really the "slowest", right?[/quote]

It is evident that the following sequence is the slowest, I thought this was obvious from my description.

$ x_1 \equal{} 1$ is the minimum possible.Then $ x_2 \equal{} 2$ is the next element of the sequence so that growth is slowest. Now $ x_1 \plus{} x_2 \equal{} 3$ is generated. Hence the next element of the sequence is $ 4$,again slowest.All numbers from $ 4 \plus{} 1$ to $ 4 \plus{} 2 \plus{} 1$ are generated.So we can only choose $ 4 \plus{} 2 \plus{} 1 \plus{} 1 \equal{} 8$ as the next element of this sequence. At each step $ 1$ is added and that is the least number that can be added.

This can also be verified as induction argument from now.

I don't think this solution is quite right... First of all, if $ x_1 \plus{} x_2 \plus{} ... \plus{} x_k>y_1 \plus{} y_2 \plus{} ... \plus{} y_k$ this doesnt't mean that sum of squares of $ x_i$ is bigger then $ y_i$. You have to prove that $ x_i\equal{}2^{i\minus{}1}$ is really best choice, in other words, if you have some sequence $ y_i$ with the same property then $ x_1^2 \plus{} ... \plus{} x_k^2 < y_1^2 \plus{} ... \plus{} y_k^2$. I am going to think some more about this solution, but you do too. Try to make it formal, there is no telling about what is the "worst case" in real solutions.
Bye

[quote="JBL"]Yes, but the least overall sum might not arise from a greedy process.[/quote]

What is this mean?

Also, I think this problem is a bit similar to Chinese TST 1987 problem 3
see here http://www.mathlinks.ro/viewtopic.php?p=234089#234089
and we encounter almost the same problem

[quote="JBL"]A [i]greedy algorithm[/i] for making several choices (e.g. for choosing a sequence $ x_1 < x_2 < \ldots < x_n$ of positive integers under certain conditions, seeking to minimize some function of the sequence) is an algorithm that makes the best possible choice at each step.  Greedy algorithms do [i]not[/i] necessarily result is best possible global solutions.[/quote]

You are right

I wish good luck to anybody attacking this problem (alas, the contestants did not succeed).

And I just want to specify that the problem is from 239-Olympiad, not from Saint-Petersburg City Olympiad.

[quote="SpongeBob"]I don't think this solution is quite right... First of all, if $ x_1 \plus{} x_2 \plus{} ... \plus{} x_k > y_1 \plus{} y_2 \plus{} ... \plus{} y_k$ this doesnt't mean that sum of squares of $ x_i$ is bigger then $ y_i$. You have to prove that $ x_i \equal{} 2^{i \minus{} 1}$ is really best choice, in other words, if you have some sequence $ y_i$ with the same property then $ x_1^2 \plus{} ... \plus{} x_k^2 < y_1^2 \plus{} ... \plus{} y_k^2$. I am going to think some more about this solution, but you do too. Try to make it formal, there is no telling about what is the "worst case" in real solutions.
Bye[/quote]

Thank you for your suggestion SB. I know neither my proof is formal nor it is complete. I dont have the complete proof now. But I feel I have a point from where we can build upon:

$ 2^i\plus{}2^i\equal{}2^{i\plus{}1}$
$ \Rightarrow 4^i\plus{}4^i>4^{i\plus{}1}$

This gives a kind of clue, if $ y_i$ is written in form of sum of powers of 2 and say it has a $ 2$ raised to some $ k$. When $ y_{i\plus{}1}$ is constructed, we better have $ 2^{k\plus{}1}$ in it, than having $ 2^k$ + something.

Now being a little over enthusiastic,
let me construct {$ y_i$} ,assume $ y_i<y_{i\plus{}1}\quad \forall 1\leq i<n$.
If $ y_i$ contains the highest power of $ 2$ say $ 2^k$, then the BEST gamble for $ y_{i\plus{}1}$ would be having $ 2^{k\plus{}1}$. Else if we use $ 2^k$ again we will be maximizing the sum of the squares. Else if we use $ 2^{k\plus{}2}$ or more, that can be decomposed into $ 2^{k\plus{}1}\plus{}2^{k\plus{}1}$ and again we are maximizing the sum of the squares.

Considering this, $ x_i\equal{}2^{i\minus{}1}$ does the best job. Any efforts to formally write out the proof would be appreciated.

It seems extremely unlikely to me that you will be able to build upon this to any meaningful extent.  What you have, essentially, is the observation that a greedy approach does as well as can be expected (assuming the problem is correct ;) ) -- you haven't introduced any idea to suggest that there is no better global approach, and so you haven't done anything that resembles a complete solution.

Let $ s_1,s_2,s_3,....,s_{2^n}$ be the $ 2^n$ different sums of all subsets of {$ x_1,x_2,x_3,....,x_n$}.
Note that in the case of {$ 1,2,4,8,...$} $ s_1,s_2,...,s_{2^n}$ takes all the values from 0 to $ 2^n\minus{}1$. Now we have to show that this case takes the least value of $ x_1^2\plus{}x_2^2\plus{}...\plus{}x_n^2$. In that case we need to express $ x_1^2\plus{}x_2^2\plus{}...\plus{}x_n^2$ in terms of $ s_1,s_2,s_3,....,s_{2^n}$.
That is actually easy.
$ 2^{2n\minus{}2}\sum_{i\equal{}1}^n x_i^2\equal{}2^n\sum_{i\equal{}1}^{2^n} s_i^2 \minus{} (\sum_{i\equal{}1}^{2^n} s_i)^2$
now without loss of generality we assume that $ s_1<s_2<...<s_{2^n}$. we can say
$ s_1\equal{}0,$
$ s_2\equal{}1\plus{}q_2,$
$ s_3\equal{}2\plus{}q_2\plus{}q_3,$
$ s_4\equal{}3\plus{}q_2\plus{}q_3\plus{}q_4,$
$ ...$
$ s_{2^n}\equal{}2^n\minus{}1\plus{}q_2\plus{}q_3\plus{}...\plus{}q_{2^n}$
 where $ q_2,q_3,...,q_{2^n}$ are non negetive integers since all sums are different.
we see that
$ 2^n\sum_{i\equal{}1}^{2^n} s_i^2 \minus{} (\sum_{i\equal{}1}^{2^n} s_i)^2<$

$ 2^n\sum_{i\equal{}1}^{2^n} s_i^2 \minus{} (\sum_{i\equal{}1}^{2^n} s_i)^2\plus{}2^n\sum_{i\equal{}m}^{2^n} (2s_iq_m\plus{}q_m^2)\minus{}(\sum_{i\equal{}m}^{2^n} q_m)^2\minus{}2(\sum_{i\equal{}m}^{2^n} q_m)(\sum_{i\equal{}1}^{2^n} s_i)$
(where $ m$ is a positive integer $ <\equal{}2^n$)
because
$ 2^n\sum_{i\equal{}m}^{2^n} (2s_iq_m\plus{}q_m^2)>(\sum_{i\equal{}m}^{2^n} q_m)^2\plus{}2(\sum_{i\equal{}m}^{2^n} q_m)(\sum_{i\equal{}1}^{2^n} s_i)$
that is actually
$ 2^{n\plus{}1}\sum_{i\equal{}m}^{2^n} s_i >2\sum_{i\equal{}1}^{2^n} s_i$
which is clear($ s$ is increasing).note that we have assumed $ q_m>0$ above.So the minimum value is attained when all $ q\equal{}0$.That is when the set is {$ 1,2,4,8,....$}(it is not difficult to show right?)

I understand your idea, it's very natural :) I am going to read your solution later for details :)

Bye

[quote="sunset"]another proof:
use induction to show that a[i]>=2^(i-1)
have a look at this: [[http://www.mathlinks.ro/viewtopic.php?t=56414]][/quote]
What is a[i] :?:
Assuming that u meant $ x_i$:
{$ 1,3,5,7$} is a valid sequence. But $ 7<2^{4\minus{}1}$

i'm sorry because of not being clear.
a[i] = x[i]
but if i don't make a mistake,{1,3,5,7} is not a valid set because 3+5=1+7=8 ? :maybe:

[quote="sunset"]another proof:
use induction to show that a[i]>=2^(i-1)
have a look at this: [[http://www.mathlinks.ro/viewtopic.php?t=56414]][/quote]

There are nothing in this link :?:

[quote="Akashnil"]
$ 2^{2n \minus{} 2}\sum_{i \equal{} 1}^n x_i^2 \equal{} 2^n\sum_{i \equal{} 1}^{2^n} s_i^2 \minus{} (\sum_{i \equal{} 1}^{2^n} s_i)^2$
[/quote]

How to check this?

Just compare coefficients: $ x_i^2$ appears $ 2^{n \minus{} 1}$ times in $ \sum_{i \equal{} 1}^{2^n} s_i^2$ and appears $ 2^{2(n \minus{} 1)}$ times in $ \left(\sum_{i \equal{} 1}^{2^n} s_i\right)^2$.  For $ i \neq j$, $ x_i\cdot x_j$ appears $ 2\cdot 2^{n \minus{} 2}$ times in $ \sum_{i \equal{} 1}^{2^n} s_i^2$ and appears $ 2 \cdot 2^{n \minus{} 1} \cdot 2^{n \minus{} 1}$ times in $ \left(\sum_{i \equal{} 1}^{2^n} s_i\right)^2$.

[quote="Akashnil"][quote="sunset"]another proof:
use induction to show that a[i]>=2^(i-1)
have a look at this: [[http://www.mathlinks.ro/viewtopic.php?t=56414]][/quote]
What is a[i] :?:
Assuming that u meant $ x_i$:
{$ 1,3,5,7$} is a valid sequence. But $ 7 < 2^{4 \minus{} 1}$[/quote]

There is a counterexample.
{3,5,6,7}

[quote="Akashnil"]$ 2^n\sum_{i \equal{} m}^{2^n} (2s_iq_m \plus{} q_m^2) > (\sum_{i \equal{} m}^{2^n} q_m)^2 \plus{} 2(\sum_{i \equal{} m}^{2^n} q_m)(\sum_{i \equal{} 1}^{2^n} s_i)$
that is actually
$ 2^{n \plus{} 1}\sum_{i \equal{} m}^{2^n} s_i > 2\sum_{i \equal{} 1}^{2^n} s_i$
[/quote]

Isn't it this?

$ 2^{n\plus{}1}\sum_{i\equal{}m}^{2^n} s_i > 2(2^n \minus{}m\plus{}1)\sum_{i \equal{} 1}^{2^n} s_i$

Then it's not trivial imo.

[quote="maybe"][quote="Akashnil"]$ 2^n\sum_{i \equal{} m}^{2^n} (2s_iq_m \plus{} q_m^2) > (\sum_{i \equal{} m}^{2^n} q_m)^2 \plus{} 2(\sum_{i \equal{} m}^{2^n} q_m)(\sum_{i \equal{} 1}^{2^n} s_i)$
that is actually
$ 2^{n \plus{} 1}\sum_{i \equal{} m}^{2^n} s_i > 2\sum_{i \equal{} 1}^{2^n} s_i$
[/quote]

Isn't it this?

$ 2^{n \plus{} 1}\sum_{i \equal{} m}^{2^n} s_i > 2(2^n \minus{} m \plus{} 1)\sum_{i \equal{} 1}^{2^n} s_i$

Then it's not trivial imo.[/quote]

Thank you "maybe".
I was more interested in the basic idea. So I overlooked that.I think we can assume $ s_{m \minus{} 1} \equal{} m \minus{} 2$ and $ q_m \equal{} 1$(This will not harm the rest of the proof.because we can assign values to $ q_x$ one by one when $ x$ is decreasing from $ 2^n$ to $ 2$ to get a desired set.)Then I think it will not be hard.

[quote="Little Gauss"]How about this?
$ \sum (\pm x_{1} \pm x_{1} \pm \cdots \pm x_{n})^{2}$[/quote]

You mean $ \sum (\pm x_{1} \pm x_{2} \pm \cdots \pm x_{n})^{2}$, right?

$ \sum (\pm x_{1} \pm x_{2} \pm \cdots \pm x_{n})^{2} \equal{} 2^n \sum{x_i ^2 }$
and $ \pm x_{1} \pm x_{2} \pm \cdots \pm x_{n}$ are all the same modulo 2, all different each other, we know when the mimimum is achieved.

Thank you very much, it's a very nice hint indeed. :)

[quote="maybe"]
You mean $ \sum (\pm x_{1} \pm x_{2} \pm \cdots \pm x_{n})^{2}$, right?

$ \sum (\pm x_{1} \pm x_{2} \pm \cdots \pm x_{n})^{2} \equal{} 2^n \sum{x_i ^2 }$
and $ \pm x_{1} \pm x_{2} \pm \cdots \pm x_{n}$ are all the same modulo 2, all different each other, we know when the mimimum is achieved.

Thank you very much, it's a very nice hint indeed. :)[/quote]
 yes:lol:

[quote="maybe"]

$ \sum (\pm x_{1} \pm x_{2} \pm \cdots \pm x_{n})^{2} \equal{} 2^n \sum{x_i ^2 }$
and $ \pm x_{1} \pm x_{2} \pm \cdots \pm x_{n}$ are all the same modulo 2, all different each other, we know when the mimimum is achieved.[/quote]

Are you sure they're all distinct?

You might get something like $ x_1\plus{}...\equal{}\minus{}x_1\plus{}...$, and thus $ 2x_1\plus{}...\equal{}...$, and you don't get contradiction.

Yes, but on both sides you have $ 2x_i$, and then you just divide both sides with 2.

I liked this problem a lot.
Of course the above solutions ,if any, are wrong.
I won't say this problem is great but it is very insightful (at least for me). I have a nice solution.
https://www.artofproblemsolving.com/community/c6h1386508_different_subset_sums_of_a_subset_of_naturals_nice_identity


geometry
circumcircle
rectangle
angle bisector
geometry proposed

Let the incircle (center at $ I$) of triangle $ ABC$ touch $ BC$ at $ D$, let $ M$be the second intersection of $ AI$ with its cirumcircle. denote by $ E,F,$  the the tangency points of the $ A$-mixtilinear incircle of the triangle $ ABC$ with $ CA,AB$ , respectively. Line $ EF$ intersects $ BC$ at $ N$, $ AN$ cuts its circumcircle again at $ P$. show that $ M,D,$ and $ P$ are collinear.

You wrote : 'A-mixtilinear circle of triangle ABC'

What on the earth does it mean??? Pls. clarify.

Not true. Line $ PD$ as defined does not cut $ (O)$ at the A-mixtilinear incircle tangency point $ M,$ but at the midpoint of the circumcircle arc $ BC$ opposite to $ A.$ Please fix your problem.

I'm sorry for my typo, I have corrected it!

$ X \equiv AI \cap BC$ is foot of the internal bisector of the $ \angle A.$ The incircle touches $ CA, AB$ at $ U, V.$ Inversion in $ (I)$ with power $ r^2$ takes $ A, B, C$ into the midpoints $ A', B', C'$ of $ UV, VD, DU,$ $ \triangle A'B'C'$ is the medial triangle of the intouch $ \triangle DUV.$ The triangle sidelines $ AB, BC, CA$ go into 3 congruent circles $ (O_a), (O_b), (O_c)$ with diameters $ ID \equal{} IU \equal{} IV \equal{} r$ (the inradius). $ X$ goes to the intersection $ X'$ of $ IM$ with $ (O_a)$ other than $ I,$ so that the angle $ \angle IX'D$ is right. The circumcircle $ (O)$ of the $ \triangle ABC$ goes into the circumcircle $ (O')$ of the $ \triangle A'B'C',$ i.e., the 9-point circle of the $ \triangle DUV,$ with radius $ \frac {r}{2},$ congruent to the circles $ (O_a), (O_b), (O_c).$ $ M$ goes to the intersection $ M'$ of $ IM$ with $ (O')$ other than $ I.$

The mixtilinear A-incircle $ (K_a)$ in the $ \angle A,$ tangent to $ CA, AB,$ and internally to $ (O)$ at $ K$ goes to a circle tangent to 3 congruent circles $ (O'), (O_b), (O_c)$ intersecting at $ A'.$ So does the mixtilinear A-excircle $ (L_a)$ at the angle $ \angle A,$ tangent to $ CA, AB,$ and externally to $ (O)$ at $ L.$ No other circles are tangent to $ CA, AB, (O)$ are centered on $ AIM.$ One circle tangent to  $ (O'), (O_b), (O_c)$ is the circle $ (A')$ centered at $ A'$ with radius $ r.$ The mixtilinear excircle has both intersections with the angle bisector $ AIM$ on the ray $ (IM,$ its image also has both intersections with $ AIM$ on the ray $ (IM,$ it is centered on the ray $ (IM,$ it cannot be identical with $ (A')$ centered on the opposite ray $ (IA$ $ \Longrightarrow$ $ (A')$ is the image of the mixtilinear A-incircle.

Tangency points $ E', F'$ of $ (A')$ with $ (O_b), (O_c)$ form an isosceles $ \triangle A'E'F'$ with midline $ O_bO_c$ $ \Longrightarrow$ the line $ E'F'$ goes through the inversion center $ I,$ stays in put in the inversion, $ EF$ also goes through $ I.$ Moreover, $ (A')$ touches $ (O')$ at a point $ K'$ diametrically opposite to $ A'$ $ \Longrightarrow$ $ \angle IM'K' \equiv \angle A'M'K'$ is right. Since $ IM' \cdot IM \equal{} IK' \cdot IK \equal{} r^2,$ the angle $ \angle IKM$ is also right. From congruence of the right $ \triangle A'K'M' \cong \triangle IDX',$ we conclude that $ DX'M'K'$ is a rectangle, cyclic, which means that $ DXMK$ is also cyclic.

[hide="well known"]$ I \in EF$ and the right angle $ \angle IKM$ are well known and can be proved by other ways. Given this, the cyclic $ DXMK$ can also be deduced by an angle chase, using isogonality of the Nagel point and the external similarity center of the incircle and circumcircle, which lies on $ AK.$[/hide]
$ EF$ is tangent at $ I$ to a circle $ (M)$ with center $ M$ and radius $ MB \equal{} MC.$ Inversion in $ (M)$ takes the circumcircle $ (O)$ into the line $ BC$ and the other way around. Because of the right angle $ \angle IKM,$ the line $ MK$ cuts $ BC$ at the pole $ N$ of $ IK,$ hence $ IN$ is a tangent to $ (M)$ at $ I,$ i.e., $ E, F, N$ are collinear. Let $ MD$ cut $ (O)$ again at $ P.$ Then $ P, A, N$ are images of $ D, X, K$ under the inversion in $ (M).$ Since the circumcircle of $ DXMK$ goes through the inversion center $ M,$ the points $ P, A, N$ are collinear.

[quote="Zhang Fangyu"]Let the incircle (center at I) of triangle ABC touch BC at D, let K be the second intersection of AI with its cirumcircle (O). denote by E,F,M the the tangency points of the A-mixtilinear incircle of the triangle ABC with CA,AB,(O) , respectively. KD meets (O) at P. BC meets KM at N. Prove that EF,AP pass through N. [/quote]
PI meets (O) at Q.
$ KD.KP\equal{}KC^2\equal{}KI^2 \Rightarrow \triangle KID\sim\triangle KPI$
$ \Rightarrow\angle OAK\equal{}\angle OKA\equal{}\angle KID\equal{}\angle KPI\equal{}\angle KPQ\equal{}\angle KAQ$, so A,O,Q are collinear.
MI passes through the midpoint of the arc BC which doesn't contain M of (O), so IM is perp to KM and MN is the external bisector of angle BMC.
Thus, $ \frac{NB}{NC}\equal{}\frac{MB}{MC}$ and apply Menelaus's theorem, we have that E,F,N are collinear.
Easy to show that AP,KM,EF,BC all pass through N.


you need more parenthesis or latex, make it clearer - does 18sin A/2sin A simplify to 9??

[quote="krnboy32"]simplify the expression

3sin(A)[cos(A)]^2-15sin(A)cos(A)+18sin(A)/2sin(A)-sin(A)cos(A)[/quote]
[hide="solution"]
Assuming what you wrote was
$\frac{3\sin A\cos^{2}A-15\sin A\cos A+18\sin A}{2\sin A-\sin A\cos A}$

We factor out the $\sin A$ to get:

$\frac{\sin A (3\cos^{2}A-15\cos A+18)}{\sin A(2-\cos A)}$.

So, we cancel the $\sin A$ and we are left with:

$\frac{3\cos^{2}A-15\cos A+18}{2-\cos A}$.

This factors to: $\frac{3(\cos A-2)(\cos A-3)}{2-\cos A}$

This is simplified to: $-3(\cos A-3) = 9-3\cos A$
[/hide]


topology
real analysis
real analysis unsolved

does there exists hausdorf space with countably infinite
 topology?

Here's how I interpreted this, but I'm not sure whether this is what you meant (it's a bit unclear):

Does there exist a set $A$ such that the cardinality of the collection of Hausdorff topologies on $A$ is [i]exactly[/i] $\aleph_{0}$?

If this is indeed the question, the answer is 'no'. If $A$ is finite, the number of topologies on $A$ is, of course, finite, and if it's infinite, then, for each subset $B$ of $A$ whose complement is countably infinite, let each singleton $\{b\}\subset B$ of $B$ be open, and endow $A\setminus B$ with the unique topology which makes a certain, fixed bijection $\phi_{B}: A\setminus B\to\mathbb Q$ a homeomorphism. For each $B$ as above we now have a topology $\tau_{B}$ such that the singletons $\{b\}\subset B$ and the open subsets of $A\setminus B$ described bove form a basis.

It's easy to see that if $B_{1}\ne B_{2}$ such that $\mid A\setminus B_{i}\mid =\aleph_{0},\ i=1,2$, then $\tau_{B_{1}}\ne\tau_{B_{2}}$ (because the collections of open singletons differ in the two topologies). Since the collection of such subsets $B$ of $A$ is uncountable, we're done.


[quote="math_music"]Using $ (1.41)^{2}<2<(1.42)^{2}$ show that $ 6.1<\frac{50}{1+\sqrt{50}}<6.3$

Thanks for any help[/quote]
Using the given information, we have $ 1.41<\sqrt2<1.42$, i.e. $ \frac{1}{1.42}<\frac{1}{\sqrt2}<\frac{1}{1.41}$.  Note $ \frac{50}{1+\sqrt{50}}=\frac{50}{1+5\sqrt{2}}$.  Use these last two facts and some arithmetic to obtain your result.  :)

[hide="solution"]$ 6.1<\frac{50}{1+5(1.42)}< \frac{50}{1+5\sqrt{2}}< \frac{50}{1+5(1.41)}<6.3$[/hide]


Prove that among any 39 consecutive positive integers it is always possible to find one whose sum of digits is divisible by 11.

oops i forgot its differrent from smaller numbers

okay.... I think I am making some mistake but the key idea is right.

So the each time the units digit changes, the digit sum, taken mod 11, increases by 1. When we change the tens digits (or the hundreds or whatever else) we change the digit sum again. We have to end up changing it more than 11 times, so we end up going through all the residues mod 11, one of them being 0, making the sum be divisible by 11.

That sounds to me a little bit like, "Hey, look, it works!"

Where'd you get this (nice) problem?

[hide]

From the 39 integers there is a string of 20 consecutive integers such that the units digit and tens digits are the only ones that change and such that the smallest number in this string of 20 integers is an integer with the units digit being 0.

(obvious if there is no numbers ending in -00, and if there is, we can pigeonhole our 39 numbers into two groups, all numbers less than the one ending in -00, and all numbers  :ge:  to it, and thus one group will the desired 20 consecutive integers)

So the sum of the digits of these 20 integers are:

n, n+1......n+9, n+1, n+2....n+10

and this covers all mod 11 residues.

[/hide]


Prove that $\sum sin^{2}A\geq (\sum cosA)^{2}$

_______________________
$VietNam is always in my heart$

I think it only true for acute triangle!
If with acute triangle we have
$2\cos A\cos B=\sqrt{\sin 2A\cot A\sin 2B\cot B}\le\frac{1}{2}(\sin 2A\cot B+\sin 2B\cot A)$
similar and add it and we are done :D

See also http://www.mathlinks.ro/Forum/viewtopic.php?t=14567 beginning with post #15, and see http://www.mathlinks.ro/Forum/viewtopic.php?t=27130 for an equivalent form.

  darij


say i was given 5 keys and was storing each one on a  peg. for some reason, the order of the keys gets misplaced. what is the  probability that  when i put the keys back, none of the keys are in their original spot?

my solution was a bit long and required some casework.  eventually, i got 11/30 as my answer.

does anyone know a faster method to solving this problem?

thank you ^_^

Yes, that answer is correct. The number of "successful outcomes" is called a [url=http://en.wikipedia.org/wiki/Derangement]derangement[/url] and is denoted $ !5$.

Since $ !5$ is equal to $ 44$, our answer is $ \frac {44}{120} \equal{} \boxed{\frac{11}{30}}$.

The formula for $ !n$ which may be found on the link, can be derived from PIE, or the principle of inclusion-exclusion.

EDIT: Fixed LaTeX error.

odd formula, do you think memorizing the 1st 5 or so dearrangemants would be useful for contests and stuff?

They don't show up that often, so not really. Just know the formula $ !n \equal{} n!\sum_{k \equal{} 0}^{n}\frac {( \minus{} 1)^k}{k!}$.

Edit: Yes, for some reason I know this formula as $ !d$ ($ d$ for derangements I guess), but I changed it ;)

(where $ d \equal{} n$.)

By the way, I don't know who invented this notation for the derangement number, but s/he should be taken out back and scolded sternly -- it's just awful, and a simple functional notation like $ D(n)$ or sequence notation like $ D_n$ would be much nicer.

Also, the derangement numbers satisfy two nice recursions.  The easier-to-prove one is that $ D_{n \plus{} 1} \equal{} n(D_n \plus{} D_{n \minus{} 1})$ -- rather than memorizing the numbers themselves, just remember how to prove this recursion and you can compute as many values as you like, probably more quickly than you can compute the formula that comes from inclusion-exclusion (i.e., the one in the preceding post).


inequalities
geometry
3D geometry
inequalities proposed

Define the following quantities: $ X \equal{} a^3 \plus{} b^3 \plus{} c^3$, $ Y \equal{} a^3 b^3 \plus{} b^3 c^3 \plus{} c^3 a^3$, and $ Z \equal{} abc$.  Then, the inequality to be proved can be written as:
$ \sqrt{Y^2 \plus{} 3Z^2 \plus{} XZ} \geq Z \plus{} \sqrt[3]{Z(2Z^2 \plus{} Y^2 \plus{} XZ)}$
Then, moving $ Z$ to the left-hand side and taking cubes,
$ (Y^2 \plus{} 3Z^2 \plus{} XZ)^{3/2} \minus{} 3Z(Y^2 \plus{} 3Z^2 \plus{} XZ) \plus{} 3Z^2 (Y^2 \plus{} 3Z^2 \plus{} XZ)^{1/2} \minus{} Z^3 \geq 2Z^3 \plus{} Y^2 Z \plus{} XZ^3$
By separating terms with and without radicals, this works out to:
$ \sqrt{Y^2 \plus{} XZ \plus{} 3Z^2} (Y^2 \plus{} XZ \plus{} 6Z^2) \geq 4Z(Y^2 \plus{} XZ \plus{} 3Z^3)$
The $ \sqrt{Y^2 \plus{} 3Z^3 \plus{} XZ}$ term cancels, so we can square both sides to get:
$ (Y^2 \plus{} XZ \plus{} 6Z^2)^2 \geq 16Z^2(Y^2 \plus{} XZ \plus{} 3Z^2)$
Finally, this comes out to be:
$ Y^4 \plus{} X^2 Z^2 \plus{} 2 Y^2 XZ \geq 4Y^2 Z^2 \plus{} 4XZ^3 \plus{} 12Z^4$
By applications of AM-GM, the following are true:
$ Y^4 \geq 3 Y^2 Z^2$
$ X^2 Z^2 \geq 3 X Z^3$
$ \frac{1}{3} Y^2 X Z \geq Y^2 Z^2$
 $ \frac{1}{3} Y^2 X Z \geq X Z^3$
$ \frac{4}{3} Y^2 X Z \geq 12 Z^4$
Adding, we get the desired inequality.

[quote="shoki"]if $ a,b,c > 0$ prove that :

$ \sqrt {(a^2b \plus{} b^2c \plus{} c^2a)(ab^2 \plus{} bc^2 \plus{} ca^2)} \ge abc \plus{} \sqrt [3]{(a^3 \plus{} abc)(b^3 \plus{} abc)(c^3 \plus{} abc)}$

(sorry if it has been posted before  :blush: )[/quote]
Dividing by $ abc$ in both sides,the inequality becomes :

$ \sqrt {(\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a})( \frac {a}{c} \plus{} \frac {b}{a} \plus{} \frac {c}{b} ) } \ge 1 \plus{} \sqrt [3]{( \frac {a^2}{bc} \plus{} 1 )(\frac {b^2}{ac} \plus{} 1 )( \frac {c^2}{ab} \plus{} 1 )}$ ,

Let $ \frac {a}{b} \equal{} x,\frac {b}{c} \equal{} y$ and $ \frac {c}{a} \equal{} z$,the inequality is equivalent to :

$ \sqrt {(x \plus{} y \plus{} z)(xy \plus{} yz \plus{} zx)} \geq 1 \plus{} \sqrt [3]{(x \plus{} y)(y \plus{} z)(z \plus{} x)}$ With $ xyz\equal{}1$ ,

Or,equivalently :

$ \sqrt {(x \plus{} y)(y \plus{} z)(z \plus{} x) \plus{} 1} \geq 1 \plus{} \sqrt [3]{(x \plus{} y)(y \plus{} z)(z \plus{} x)}$ ,

Let $ (x \plus{} y)(y \plus{} z)(z \plus{} x) \equal{} \alpha^3$,the inequality is reduced after squaring to:

$ (\alpha \minus{} 2)(\alpha \plus{} 1) \geq 0$ ,

Which is clearly true,since $ \alpha \equal{} \sqrt [3]{(x \plus{} y)(y \plus{} z)(z \plus{} x)} \geq \sqrt [3]{8\sqrt {xyz}} \equal{} 2$

[quote="shoki"]if $ a,b,c > 0$ prove that :

$ \sqrt {(a^2b \plus{} b^2c \plus{} c^2a)(ab^2 \plus{} bc^2 \plus{} ca^2)} \ge abc \plus{} \sqrt [3]{(a^3 \plus{} abc)(b^3 \plus{} abc)(c^3 \plus{} abc)}$

(sorry if it has been posted before  :blush: )[/quote]
See here: http://canhang2007.wordpress.com/2009/11/25/inequality-79-unknown-author/

using Minkowski's inequalily we obtain:
$ A\equal{} \sqrt[3]{(a^2c\plus{}bc^2)(b^2a\plus{}ca^2)(c^2b\plus{}ab^2)}\plus{} \sqrt[3]{ab^2 \cdot bc^2 \cdot ca^2}   \le (a^2c\plus{}b^2a\plus{}c^2b)$
and similarily
$ A\le (ac^2\plus{}ba^2\plus{}cb^2)$
hence the inequality holds

[quote="Dumel"]using Minkowski's inequalily we obtain:
$ A \equal{} \sqrt [3]{(a^2c \plus{} bc^2)(b^2a \plus{} ca^2)(c^2b \plus{} ab^2)} \plus{} \sqrt [3]{ab^2 \cdot bc^2 \cdot ca^2} \le (a^2c \plus{} b^2a \plus{} c^2b)$
and similarily
$ A\le (ac^2 \plus{} ba^2 \plus{} cb^2)$
hence the inequality holds[/quote]

Unexpected very short solution :)

Also can_hang2007 solution is very nice and very short. :)

This inequality has a lot of solutions. Two of them are in Hojoo Lee e-book "Infinity" and another very interesting is on this forum. You can try to find it :wink:

Thank you very much :)


Does anyone know what is the best test prep company to study with for the AP? I know about Kaplan, Barrons, and Princeton Review. Can anyone give me advice on what company should I study with.

Also, how can you do really well on the AP exams? Do you just do practice problems from the test prep books?

When looking for a good book to study from, I often go to Amazon.com and read many reviews.

Actually I bought some books for first-year college!

the best book is going to vary depending on the subject. In addition to the reviews on Amazon, good discussion of specific books for specific subjects can be found on the testing discussion forum at http://www.collegeconfidential.com.  You may want to buy two books (look for them used) so that you can decide which one you like better for review, and use the other one for practice exams.

I personally like Barrons. I got a 5 in ap world with it. It's a lot longer than Princeton Review's, though. (At least for ap world)

As for how to do well... well there are many ways, such as studying with a friend, do some practice exams, etc. etc.


Ok how about, next time we make predictions, we state which year it's for?

Kevin's from Diamond. Anyway, this poll will be wrong anyway, so why have it?

Idk.
I'm bored.
Let's discuss something...
I still think Alan's going to make Nationals.

dude, Amy Zhang is almost definitely not being the top individual.  maybe on the nats team, but it'll take a lot of work from her.

peijin, why is everyone on the poll from lexington?

Guys, come on.  Vote for who you want or ignore it.  Anyone could win, Kevin went from like 13th at Chapter to 4th at states last year, so anything could happen.  Let's try to stay positive and wish everyone the best of luck.

[quote="CatalystOfNostalgia"]Why am I not on the poll?

You should go practice doing Mathcounts instead of talking about other people doing Mathcounts.[/quote]

Good call, I actually have been doing MATHCOUNTS while I keep gettnig interupted by notifications that this board was updated.

I hope Alan and Peijin have been too!!!

Hmmm, looks like everyone favors Alan, well so do I. ZhangPeijin, it looks like you don't that many people in Massachusetts except for people in Lexington. How about the people in other parts of Massachusetts? I heard Acton is good. Good luck Jonathan, get best sixth grader!
I am still wondering why alternates are not allowed to go to states.....Oh well there is always next year. :)

I bet I know who that "someone else" is.

Hao, you don't even know who's on the Acton Team, you just said that because you overheard us talking about Acton during practice.

lol
This is funny.

But really, who is Amy Zhang?

Silly Carl, you forgot to say that the numbers are in degrees.

[quote="CatalystOfNostalgia"]hmmmm if you insist on wasting time then find arctan(tan65-2tan40) without a calculator

then do mathcounts i guess[/quote]

assuming it's degrees, is it $ 25\plus{}180k$?  eh, I did kinda waste my time, didn't I?

Silly Alan, you forgot to say that $ k$ is an integer.

Silly me, I forgot to say that this is from HMMT.

[quote="frost13"]YES!!!!!!!!!!![/quote]

hmmmm is that to the correct answer or to the question for wasting time

actually it's pretty incorrect as 0<arctan x<180

because if you don't want to waste time, prove that n! is never a perfect power for integer n>1

Silly Carl, stop with the trivial things that follow directly from Bertrand's.

[quote="CatalystOfNostalgia"][quote="frost13"]YES!!!!!!!!!!![/quote]

hmmmm is that to the correct answer or to the question for wasting time

[/quote]

To the wasting time part.  I am pretty sure that arctan won't be on the MC states test!!!

If it doesn't, that defeats the point of the problem. Therefore it does. QED

[quote="CatalystOfNostalgia"]hmmmm i am quite sure however it will be on the AIME

anyway show that x^4+3x^3-11x^2+3x+1 has 4 real roots[/quote]

BRUTE FORCE :P

nah, actually, I suppose you could prove there's no way to have 4 complex roots, then 2 complex roots, and obviously you can't have 1 or 3, so there's 0.  I don't really feel like it right now.  Then again, there's probably a nicer way.

[quote="CatalystOfNostalgia"]hmmmm i am quite sure however it will be on the AIME

anyway show that x^4+3x^3-11x^2+3x+1 has 4 real roots[/quote]

You never said that you can't use a graphing calculator. QED.

[quote="Moldytape"][quote="CatalystOfNostalgia"]hmmmm i am quite sure however it will be on the AIME

anyway show that x^4+3x^3-11x^2+3x+1 has 4 real roots[/quote]

You never said that you can use a graphing calculator. QED.[/quote]

hmm, I'm pretty sure you did not mean to say that

eh, I'd rather go with the proof that doesn't require you (or anything else) to find the roots.

[hide="common idea"]Divide by $ x^2$, then let $ k\equal{}x\plus{}\frac1x$. The result follows.[/hide]

You could guess until you bound the roots XD


[quote="confuted"]I have the following:

[code]\documentclass[11pt]{book}
\usepackage{amsmath}
 \usepackage{fancyhdr}

\pdfpagewidth 8.5in
\pdfpageheight 11in

\pagestyle{fancy}
\headheight 35pt

\rhead{\thepage}

\begin{document}
test
\newpage
test
\end{document}
[/code]
I'm sure LaTeX is doing exactly what I'm asking it to do, but that doesn't change the fact that it's not what I want ;) Is there a way for me to make the page numbers alternate sides, like in a book?[/quote]

well try this

\fancyhead[LE,RO]{\thepage}

it will put the page number on the left for even pages and right for odd pages... like on a book.


Let it be is a wooden unit cube. We cut along every plane which is perpendicular to the segment joining two distinct vertices and bisects it. How many pieces do we get?

[quote=rem]Is the anwser $2^{9}=512$?(I am not sure if my solution is right)
Can someone please post the solution?[/quote]

Actually, the official answer is $96$.


At the fish market, a whole fish costs $ \$4.20$.  If the body costs $ \$1$ more than the tail, and the body costs $ \$2$ more than the head, how much does the head cost in dollars?

Let the cost of the body be $ x$. Then, we have

$ x\plus{}x\minus{}1\plus{}x\minus{}2\equal{}4.2 \implies 3x\minus{}3\equal{}4.2 \implies x\equal{}2.4$

Therefore, the head costs $ 2.4\minus{}2\equal{}\boxed{\$0.40}$.


1. Fie f:[0,1]->$\mathbb{R}$ o functie continua. Calculati limita sirului ($ I _n $)  definit prin $ I_n $=$\int _{0}^{1}{x^n*f(x)}dx$.

2. Fie functia f:$\mathbb{R}$ -> $\mathbb{R}$, f(x)=$\frac {x^ {5n-1} + x^ {n-1}} {1+x^{6n}}$, n natural. Sa se gaseasca o primitiva a lui f.

[quote="catalin_dumitru"]1. Fie f:[0,1]->$\mathbb{R}$ o functie continua. Calculati limita sirului ($ I _n $)  definit prin $ I_n $=$\int _{0}^{1}{x^n*f(x)}dx$.

[/quote]

$g_n(x)=x^nf(x)$ is continous, converge simply to

$0$ on $[0,1[$
$f(1)$ for $x=1$

$|g_{n}(x)|\leq ||f||_{\infty}$

Lebesgue's dominated convergence gives $\lim \int_{[0,1]}g_n=\int_{[0,1]}\lim g_n=0$

[quote="catalin_dumitru"]1. Fie f:[0,1]->$\mathbb{R}$ o functie continua. Calculati limita sirului ($ I _n $)  definit prin $ I_n $=$\int _{0}^{1}{x^n*f(x)}dx$.

  Aceasta problema este banala.Ea poate fi demonstrata si daca $f$ este integrabila Riemann.Cum orice astfel de functie este marginita,rezulta ca $lim I_n\leq \frac{M^n}{n+1}=0 $ unde $f(x)\leq M$. ;).teorema clestelui rezolva problema.

[quote="flip2004"][quote="catalin_dumitru"]1. Fie f:[0,1]->$\mathbb{R}$ o functie continua. Calculati limita sirului ($ I _n $)  definit prin $ I_n $=$\int _{0}^{1}{x^nf(x)}dx$. [/quote]
[b]Propunere:[/b] Cred ca putin mai dificil, dar mai intersant,  este de a calcula, in aceleasi conditii, urmatoarea  limita:
\[
\lim\limits_{n\to \infty} n\int\limits_{0}^1 x^n f(x)\; dx  \; \; \; ,\; \; \;  f\in C[0,1] ,
\]
iar in cazul  $ f\in C^2[0,1] $ , sa se calculeze
\[ \lim\limits_{n \to \infty} n\left(f(1)- n\int\limits_{0}^{1} x^nf(x)\; dx  \right) .
\][/quote]
[color=red](1.)
\[
\mathop {\lim }\limits_{n \to \infty } \int\limits_0^1 {x^n f\left( x \right)dx = 0}
\]
(2.)
\[
\mathop {\lim }\limits_{n \to \infty } n\int\limits_0^1 {x^n f\left( x \right)dx = } f\left( 1 \right)
\]
(3.)
\[
\mathop {\lim }\limits_{n \to \infty } n\left( {f\left( 1 \right) - n\int\limits_0^1 {x^n f\left( x \right)dx} } \right) = f\left( 1 \right) + f'\left( 1 \right)
\]
[/color]


Prove the inequality

$ (\sin x + 3x)(x\cos x + 4\sin x)(\sin x + \sin 2x) \geq 30x \sin \2x \sinh$

$ x \in \left[0, \frac{\pi}2\right)$

There's a mistake in that; just try $ x\equal{} \frac{\pi}{6}$


Hi,

I was wondering is this year's non-calculator test pretty much the same difficulty or is it easier than last year?

Thanks!

See the posts in the AMC forum about this... some think it was harder, but I personally think it was about the same.

When we lock the AMC forum, this does not mean that you should start discussing the AMC in other forums.  It means that you [i]should not discuss the AMC[/i].

Besides, this topic has already been discussed rather extensively, both on the AMC forum (which you can still read and search while it is locked) and by the AMC themselves on their website.


Souce: 360 PMC
I need help on this pbm.
Find all pairs $(x,y)$ of nonnegative integers such that $x^2+3y$ and $y^2+3x$ are simultaneously perfect squares.

Since the expressions $x^{2}+3y$ and $y^{2}+3x$ are symmetric in $x$ and $y$, we may without loss of generality assume $x \geq y$. Consider the Diophantine system of equations

$(1) \;\; x^{2}+3y = a^{2}$

$(2) \;\; y^{2}+3x = b^{2}$.

where $a$ and $b$ are natural numbers. For $y=0$ we obtain the solutions $(x,y)=(3t^{2},0).$ Suppose $y > 0$. Then

$x^{2}< x^{2}+3y < x^{2}+3x < (x+2)^{2}$,

hence $a = x+1$ by (1). This implies that

$x^{2}+3y = (x+1)^{2}$,

i.e.

$3y = 2x+1.$

Consequently $x = 3s+1$ and $y = 2s+1$ for a non-negative integer $s.$ So according to (2)

$b^{2}= y^{2}+3x = (2s+1)^{2}+3(3s+1) = 4s^{2}+13s+4,$

which is equivalent to

$(8s+13-4b)(8s+13+4b) = 105.$

Therefore

$(3) \;\; 8s+13-4b = d,$

$(4) \;\; 8s+13+4b ={\textstyle \frac{105}{d}},$

where $d \leq \sqrt{105}$ is a positive divisor of $105 = 3 \cdot 5 \cdot 7$, i.e. $d \in \{1, 3, 5, 7\}.$ Adding (3) and (4), the result is

$2(8s+13) = d+{\textstyle \frac{105}{d}},$

thus

$s = \frac{d+{\textstyle \frac{105}{d}}-26}{16}.$

This formula gives $(d,s) = (1,5), \: (3,{\textstyle \frac{3}{4}), \: (5,0), (7,-{\textstyle \frac{1}{4})}}$ as possible solutions. Seeing that $s$ is an integer, we are left with two solutions,

$s=0 \;\; \Leftrightarrow \;\; (x,y) = (1,1),$

$s=5 \;\; \Leftrightarrow \;\; (x,y) = (16,11),$


calculus
integration
trigonometry
calculus computations

Help me to solve these integrals
$\int {\cos x\cos 2x\cos 3xdx}$

$\int {(x^2 + 2x + 2)e^{ - x} } dx$

1st integrl: Rewrite the product of integrand into sum

2nd integral: Use integral by parts 2 times


Find the max. and min. value of the function f(x):
$f(x)=x+\sqrt{(x^{2}+6x+9)(x^{2}+2x+1)}$ $(x \in [-4;\frac{-5}{4}])$

Another problem, but easy too, but a bit harder than Link_Park problem  :rotfl:  :roll:
Find the max. and min. value of the function f(x):
$f(x)=x\sqrt{(x^{2}+6x+9)(x^{2}+2x+1)}$
$(x \in [-4;\frac{-5}{4}])$


induction
geometry
search
factorial
number theory unsolved
number theory

If a and c are positive numbers and b and d are positive integers such that ab=cd, then there exists a positive number r and positive integers s, t, and u such that the following equations hold : a=rs, b=tu, c=rt, d=su

Prove this using induction on |b| instead of geometry.

You may want to tell us the difference between positive numbers and positive integers  :wink:
And there is no geometry involved (I've no clue how you came to that).

Nonetheless, Darij will search the link to PEN's version  :D

Geometry???

http://www.mathlinks.ro/viewtopic.php?t=150434

For extra fun, apply this theorem to some other factorial rings (UFD's) than $ \mathbb{Z}$ (for instance, $ \mathbb{Z}\left[i\right]$) and rewrite the conclusions in terms of integers.

  darij


calculus
integration
logarithms
calculus computations

Evaluate $ \int_1^e \frac{(x^2\ln x\minus{}1)e^x}{x}\ dx.$

$ \int \frac{x^{2}\ln x\minus{}1}{x}e^{x}dx$

$ \equal{}\int e^{x}x\ln x\minus{}\int \frac{e^{x}}{x}dx$

$ \equal{}\int e^{x}x\ln xdx\minus{}\int \frac{e^{x}}{x}dx$

$ \equal{}\allowbreak e^{x}x\ln x\minus{}\int e^{x}(\ln x\plus{}1)dx\minus{}\int \frac{e^{x}}{x}dx$

$ \equal{}\allowbreak e^{x}x\ln x\minus{}\int e^{x}\ln xdx\minus{}\int e^{x}dx\minus{}\int \frac{e^{x}}{x}dx$

$ \equal{}\allowbreak e^{x}x\ln x\minus{}\left( e^{x}\ln x\minus{}\int \frac{e^{x}}{x}dx\right) \minus{}\int e^{x}dx\minus{}\int \frac{e^{x}}{x}dx$

$ \equal{}e^{x}\allowbreak \left( (x\minus{}1)\ln x\minus{}1\right)$

Hence

$ \int_{1}^{e}\frac{x^{2}\ln x\minus{}1}{x}e^{x}dx\equal{}\allowbreak \allowbreak (e\minus{}2)e^{e}\plus{}e$


analytic geometry
graphing lines
slope
combinatorics unsolved
combinatorics

$ 2n\plus{}1$ lines are drawn in the plane, in such a way that every 3 lines define a triangle with no right angles. What is the maximal possible number of acute triangles that can be made in this way?

I know a weak bound that if in a plane there are finite points then atmost $2/3$ of the triangles formed by them will be acute angled.

according to an old IMO problem 6 [the year I dont remember] , at most 70% of the triangles are acute. I dont know f the bound is attainable. :blush:

If 3 lines define a triangle, then whether it is acute or not depends only of their slopes. So we can replace a line by a parallel one (avoiding three concurrent lines) without changing the acuteness of the triangles. This means we can assume wlog that all lines are tangents of a common circle. It [i]should be [/i]obvious now (note that I haven't said it [i]is [/i]obvious :blush:) that the best solution is obtained if the lines are the sides of a regular (2n+1)-gon, (equivalently. to give the whole picture, the longest diagonals of a regular (2n+1)-gon). In such a constellation, there are $\frac13(2n+1)\binom{n+1}{2}=\frac14 \binom{2n+2}{3} $ acute triangles, which makes their proportion $\frac{n+1}{2(2n-1)}$. This is roughly $\frac14$ as $n$ grows, far from the promised 70%. :huh:

Equally the problem can be stated as follows in terms of the slopes:

[b]Given a set $S$ of (2n+1) distinct nonzero reals such that $ab\ne-1$ for all $a,b\in S$, what is the biggest possible number of sets $\{a,b,c\}\subset S$ such that $ab>0>ac>-1>bc$?.[/b]


What is the smallest positive integer n such that if S is any set containing n or more integers, then there must be three integers in S whose sum is divisible by 3?

Why hasn't anyone done this problem?

[hide="Hint 1"]Experiment.[/hide]
[hide="Hint 2"]Mods.[/hide]

This problem isn't really intended for me since I just did it, but since no one else is going to post a solution, I might as well.
[hide]
We can look at the numbers in $S$ as remainders $R$ such that $0\leq R\leq 2$, and any three numbers in which its $R$s add up to $0,3,6$ means that their sum is divisible by $3$. So let's start out with the possible sets which I can disprove as having $n$ integers being divisible by $3$

$S(1)=1$
$S(2)=1,1$
$S(3)=1,1,0$
$S(4)=1,1,0,0$
$S(5)=1,1,1,0$ or $0,0,0,2$ or $2,2,1,2$ etc.

So I can tell that no matter what I do, there will always be three integers whos $R$s will add up to either $0,3,$ or $6$. So the answer is $\boxed{5}$[/hide]


Let $ f: N\mapsto R$ be a function such that $ f(0) \equal{}0$; $ f(1)\equal{}1$
and $ f(n) \equal{}f(n\minus{}\frac{m(m\minus{}1)}{2})\minus{}f(\frac{m(m\plus{}1)}{2}\minus{}n)$ for all $ m,n\in Z_{\plus{}}$
$ m\geq 2$ and $ \frac{m(m\minus{}1)}{2}< n\leq\frac{m(m\plus{}1)}{2}$.Find the minimum natural number$ n$ such that $ f(n)\equal{}5$


function
complex analysis
complex analysis theorems

Hey, simple question...

How do we find roots of purely complex numbers such as -6i

Thanks  :)

convert to polar form :
$z=-6i = 6 \exp{-\pi/2 i} \Longleftarrow \sqrt{z} = \sqrt{6} \exp{3\pi/4 i} = \sqrt{6} (-1/2\,\sqrt{2}+1/2\,i\sqrt{2})$

Roots are multiple branched functions in the complex plane. If $z=r(\text{cos }\varphi+i\text{ sin }\varphi)$ then
\[ \sqrt[n]{z}=\sqrt[n]{r}\left (\text{cos }\frac{\varphi+k2\pi}{n}+i\text{ sin }\frac{\varphi+k2\pi}{n}\right ), \]
where $k=0,1,\ldots,n-1$.


Show that for all $ n$, $ {2^{3}}^{n}$ + 1 is divisible by $ 3^{n+1}$.

[quote="rodneymckay"]Could you explain the 1st line of your proof? I don't understand how did you get 2^(3n+1) + 1 = (2^3n)^3 +1. Well, if n=1 than 17 = 513...[/quote]
There is $ 2^{3^{n+1}}$ (2^{3^{n+1}}) not $ 2^{3n+1}$, so:
$ \left(2^{3^{n}}\right)^{3}=2^{3^{n}\cdot 3}=2^{3^{n+1}}$

Blah Blah induction

$ f(n)=3^{n+1}$
$ g(n)=2^{3^{n}}+1$

Manipulate:

$ g(n+1)=(g(n))^{3}-3(g(n)-1)(g(n)))$
$ f(n+1)=3f(n)$

f(n+1) divides g(n+1) as a consequence of odd powers of 2 $ mod{3}$


Let $ a,b,c$ are positive real numbers such that $ a\plus{}b\plus{}c\equal{}1$.Prove that
$ 8(ab\plus{}bc\plus{}ca)^2\plus{}12abc\plus{}1\geq 7(ab\plus{}bc\plus{}ca)$

Try $ a \equal{} 0.2, b \equal{} c \equal{} 0.4$  :(


[hide="Some interesting stuff..."]
[url=http://www.mathpages.com/home/kmath404.htm]A Link.[/url] [/hide]


Here are two questions...

$Question 1$
Start with the number by contactinating all digits from $1$ to $100$ in the order... it would like like: $123456\cdots979899100$
Delete exactly $100$ digits from this number ( closing the gaps) so that the resultant number has the greatest possible value.

Similarly, delete exactly $100$ digits from this number ( closing the gaps) so that the resultant number has the least possible value.
What are the resultant numbers?

$Question 2$
There are $27$ balls weighing
$2007^{2}= 4028049$;
$2008^{2}= 4032064$;
$2009^{2}= 4036081$;
$2010^{2}= 4040100$;
$\cdots$
$\cdots$
$2032^{2}= 4129024$;
$2033^{2}= 4133089$;

Show how these weights can be divided into three sets of nine balls each such that the sum of the weights of the balls in each set is the same.

[quote="sapphyre571"][hide="1"]
we know the largest possible number starts with as many 9's as possible
the first 9 comes after eliminating the first 8 digits, so we have 92 more to elim.
we elim 1011...181 (19 digits) to get another 9 (73 left). elim 19 more, and 19 more, and 19 more, and now you have 16 left, and your current number is 999995051...57585960...99100. Eliminate 50512535455565 (15 digits, 1 left), and then the 5 after the 7 to maximize value, meaning the number is
999997896061...99100

Same principle applies for the other one
100000123406162...99100

I think that's right
[/hide]

[hide="2"]
let us group the balls into groups A, B, and C, with each having a weight of a=b=c. Because a=b=c, a, b, and c must all have the same last digit. let this last digit be the single digit k. If we consider all these weights in mod 10, we have 6 9's, 4 6's, 2 5's, 6 4's, 6 1's, and 3 0's.
6x9+4x6+2x5+6x4+6x1+3x0=118, which is not a multiple of three. Since in adding numbers only the last digits of the addends can affect the last digit of the sum, a, b, and c cannot all have the same last digit. therefore it should be impossible to divide the balls into 3 groups so that the sums of the weights in the groups is all the same.

I know that's probably wrong, so a correct solution would be nice

[/hide][/quote]

My comment for $Question 1$
[hide]When you say that "meaning the number is
$999997896061...99100$" ---- You have left out a $5$ here. in fact to reach the next $9$ you have to cross $100$ which is not permitted.

When you say that "Same principle applies for the other one
$100000123406162...99100$" --- I had asked for a number with least possible value, so you can start with leading zeros which later on can be ignored.[/hide]


superior algebra
superior algebra solved

let p,q be distinct odd primes.find all possible class eq. of group of order pq.

This must have been discussed before: if $p<q$ and $p$ does not divide $q-1$, then all such groups are cyclic, otherwise there exist exactly two types of groups, one cyclic and the other one a non-direct semi-direct product of two cyclic groups of order $p,q$.


MATHCOUNTS
quadratics
ratio
logarithms
LaTeX
algebra
quadratic formula

If $x = \sqrt{1 + \sqrt{1 + \sqrt{1 + \sqrt{1 + \cdots}}}}$, then:
\[ \text{(A)}\ x = 1 \qquad \text{(B)}\ 0 < x < 1 \qquad \text{(C)}\ 1 < x < 2 \qquad \text{(D)}\ x\ \text{is infinite} \]
\[ \text{(E)}\ x > $\ \text{but finite} \]

If x exist, it verify x 2 =1+x. So x=(1+ :sqrt: 5)/2. Gold Number

[hide="hint"]square both sides and solve for $x^2-x-1=0$[/hide]
edit-i didn't see your solution afaouzi. :blush:

[quote="eryaman"][hide]
$x=\sqrt{1+\sqrt{1+\sqrt{1+...}}}$

$x=\sqrt{1+x}$
[/hide][/quote]

Very clever! I like it!  :P

The answer is[hide]x =2 -  :varepsilon:
Therefore (C) is correct. [/hide]

I once saw a similar mathcounts Problem....
[hide]
x=sqrt(1+x)
x^2=1+x
x^1-x-1=0
if we apply Mr. Quadratic Formula, we get:
x=1(plus or minus)sqrt(5)/2$

[/hide]

This is actually extremely easy... you don't even have to analyze it, since it's multiple choice. Just see that it's obviously $>1,$ but it can't be $>2,$ because that means it would be $>3,$ since $\sqrt{1+\sqrt{1+\sqrt{1...}}}$ would have to be $>3$ for the same thing to be $>2,$ which just doesn't make sense...

[quote="IntrepidMath"]I once saw a similar mathcounts Problem....
[hide]
x=sqrt(1+x)
x^2=1+x
x^1-x-1=0
if we apply Mr. Quadratic Formula, we get:
x=1(plus or minus)sqrt(5)/2

[/hide][/quote]

your answer should only be $\frac{1+\sqrt{5}}{2}$ because the negative gives a result  of a negative number, wihch cannot occur in a square root.

[quote="JesusFreak197"]This is actually extremely easy... you don't even have to analyze it...[quote]

Yes, that's true, and during a time situation, you shouldn't analyze it. But this isn't timed, and the result is beautiful  :) --I never knew that the golden ratio was found by adding nested square roots of 1. (Is that what it's called?)

That's true; when you have the time, it's always nice to work out a cool solution like this and get the exact answer. I remember when I saw a MathCounts problem that had:

$\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6+...}}}} = ?$

Once I figured out how to solve it, I loved how the answer was so nice...

$x = \sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6+...}}}}$
$x^2 = 6 + \sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6+...}}}} = 6+x$
$x^2-x-6 = (x-3)(x+2)$
$x=3$

:lol:

Quote from MC forum:
[quote="Xantos C. Guin"]...I rot my brain with that game.
about
$\sqrt{20+\sqrt{20+\sqrt{20+\sqrt{20+....}}}}$ hours a day
Yay!
My $LaTeX$ worked[/quote]

$x=\sqrt{20+\sqrt{20+\sqrt{20+\sqrt{20+...}}}}$
$x^2=20+x$
$x^2-x-20=0$
$(x-5)(x+4)=0$
$x=5$

So, about five hours a day? That's quite a bit, I wish I had that much time on school days... I only have that much on weekends.

Once you know how to do these it becomes very easy. I've also seen a logarithm problem just like this...

$x=\sqrt{2^n+1}^\sqrt{2^n+1}^\sqrt{2^n+1}\mbox{...}$

First time using $\mbox{Latex}$, I used Mbox for the "..." but how do you usually do it? And how bout the "Latex" in cool font? (how everyone usually does it)

Actually not sure if the problem is right... please feel free to edit the problem  :blush:  you know what I mean though right?

mean this:?
$\sqrt{2^n+1\sqrt{2^n+1\sqrt{2^n+1\sqrt{...}}}}$?
or
$\sqrt{2^n+1+\sqrt{2^n+1+\sqrt{2^n+1+\sqrt{...}}}}$
type it like this(with dollor signs around it...):
\sqrt{2^n+1\sqrt{2^n+1\sqrt{2^n+1\sqrt{...}}}}


quadratics
number theory
algebra unsolved
algebra

S is the set of all ($a$, $b$, $c$, $d$, $e$, $f$) where $a$, $b$, $c$, $d$, $e$, $f$ are integers such that $a^2 + b^2 + c^2 + d^2 + e^2 = f^2$. Find the largest $k$ which divides abcdef for all members of $S$.

For $1,1,1,2,3$ we get $4^2$ so $k \leq 24$

Let's try to prove that $k = 24$

A square is $0$ or $1$ modulo 3. If none of the $a,b,c,d,e$ is divisible by $3$ we get $5 \times 1 \mod 3 = 2$ which is not a square. So at least one of the $a,b,c,d,e$ is divisible by $3$

A bit more tricky we $\mod 8$
A square is $0, 1$ or $4 \mod 8$.
- If all $a,b,c,d,e$ are odd, we get $5 \mod 8$ not a quadratic residue $\mod 8$.
- If,wlog, $a,b,c,d$ are odd we get $0 \mod 8$ : ok
- If,wlog, $a,b,c$ are odd we get $3 \mod 8$ not a quadratic residue.
- ....

In any case, we get a multiple of $8$

This problem would perhaps find its place in "number theory" sub-sub-sub-forum (but never mind)

By checking for minimum values, we get that the least tuple is $(1,1,1,2,3,4)$, as $1^2+1^2+1^2+2^2+3^2=4^2$.
Thus, if $k\vert abcdef$ for all $a,b,c,d,e,f$ satisfying the equation, $k\vert 24\Rightarrow k\leq 24$.
Claim: $k=24$ is the required solution.

Proof:
$(i)$ If none of the numbers $a,b,c,d,e,f$ are divisible by $3$, we get that $LHS \equiv 5$ $(mod$ $3)$, but $RHS \equiv 1$ $(mod$ $3)$. Thus, we get a contradiction and at least one of the numbers is divisible by $3$ $\Rightarrow \boxed{3\vert abcde}$

$(ii)$ We check modulo 8. Any odd perfect square $\equiv 1$ $(mod$ $8)$. An even perfect square can be either $\equiv 0$ or $4$ $(mod$ $8)$.
$*)$ $Case-1:$ If $a,b,c,d,e$ are odd, $LHS \equiv 5$ $(mod$ $8) \Rightarrow f^2 \equiv 5$ $(mod$ $8)$, $\boxed{Contradiction!}$

$*)$ $Case-2:$ If $a,b,c,d$ are odd, $LHS \equiv 4+e^2$ $(mod$ $8)$. As $e$ is even, $e^2 \equiv 0$ or $4$ $(mod$ $8)$. If $8\vert e^2 \Rightarrow 16 \vert e^2$. Thus $4\vert e$ and $f$ is even $\Rightarrow 8 \vert abcdef$. If $e \equiv 4$ $(mod$ $8) \Rightarrow f^2 \equiv 0$ $(mod$ $8)$. Thus $8\vert f^2 \Rightarrow 16\vert f^2 \Rightarrow 4\vert f$ and $e$ is even $\Rightarrow \boxed{8 \vert abcdef}$

$*)$ $Case-3:$ If $a,b,c$ are odd, $LHS \equiv 3+d^2+e^2$ $(mod$ $8)$. Also, $d$ and $e$ are even $\Rightarrow f$ is odd. Thus $RHS \equiv 1$ $(mod$ $8)$. $\Rightarrow 3+d^2+e^2 \equiv 1$ $(mod$ $8) \Rightarrow d^2+e^2 \equiv 6$ $(mod$ $8)$. But any even square $\equiv 0$ or $4$ $(mod$ $8) \Rightarrow d^2+e^2 \equiv 0$ or $4$ $(mod$ $8) \not\equiv 6$. $\boxed{Contradiction!}$

$*)$ $Case-4:$ If $\leq 2$ numbers on $LHS$ are odd $\Rightarrow \geq 3$ numbers on the $LHS$ are even $\Rightarrow \boxed{8\vert abcdef}$

From $(i)$ and $(ii)$, we get that $\boxed{24\vert abcdef}$, and our claim is proved.


geometry
circumcircle
geometric transformation
rotation
homothety
ratio
geometry proposed

Let ABC be a triangle and D the foot of the altitude from A. Let E and F be on a line passing through D such that AE is perpendicular to BE,AF is perpendicular to CF , and E and F are different from D.Let M and N be the midpoints of the line segments BC and EF respectively. Prove that AN is perpendicular to NM.

call that line from D d. call d' a line from A parallel to BC. Call B',C',M' the projection of B,C,M on d'. The locus of point E when d moves is the circumcircle of ABD and of point F when d moves is the circumcircle of ACD, then the perpendicolar from E to d is pass in B', the perpendicolar from F to d pass in C', then the perpendicolar from N to d pass in M', then the locus of N is a circle with diameter M'D that is the circumcircle of MDA then $ \angle MNA \equal{} \frac{\pi}{2}$.

Obviously, E and F move on the circles (ABD) and (ACD) respectively. <ADE = <EDF = x. As A, D, E and B are concyclic, same for A, D, C and F, we easily get that the triangles ABE and ACF are similar, so we conclude that a rotation of angle (90-x) around A, followed by a homothety of ratio sin x will map the triangle ABC to AEF, hence AN/AM = sin x and, since the angle <MAN is (90-x) we conclude that the triangle AMN is similar to ABE, therefore AN and MN are perpendicular.

Best regards,
sunken rock

Dear Mathlinkers,
the result involves in the Gabriel's message is knowed under the name of "the mid circle theorem".
Any reference?
Jean-Louis


induction
inequalities
limit
real analysis
real analysis unsolved

For $ k>0$, let $ s_{1}\equal{}\sqrt{k}$, and $ s_{n\plus{}1}\equal{}\sqrt{k\plus{}s_{n}}$ for $ n\ge 1$
Show that $ s_{n}$ is bounded above.

$ s_n<k\plus{}1$ by induction (all you need for the induction step is the inequality $ 2k\plus{}1<(k\plus{}1)^2$).

For $ \alpha$ such that $ \alpha \equal{}\sqrt{k\plus{}\alpha}\Longleftrightarrow \alpha ^2\equal{}k\plus{}\alpha\ \cdots [*]$

$ \therefore s_{n\plus{}1}\minus{}\alpha \equal{}\sqrt{k\plus{}s_n}\minus{}\alpha \equal{}\frac{k\plus{}s_n\minus{}\alpha ^2}{\sqrt{k\plus{}s_n}\plus{}\alpha}\equal{}\frac{s_n\minus{}\alpha}{\sqrt{k\plus{}s_n}\plus{}\alpha}\ \because\ [*]$

$ \therefore |s_{n\plus{}1}\minus{}\alpha|\equal{}\frac{|s_n\minus{}\alpha|}{\sqrt{k\plus{}s_n}\plus{}\alpha}\leq \frac{1}{\alpha}|s_n\minus{}\alpha|$.

Ohh.. that easy. Thanks fedja!
So $ s_{n}$ is an increasing sequence and bounded above so converges;  let $ y\equal{}\lim_{n\to \infty}s_{n}$.
Then $ y$ satisfies $ y^2\minus{}y\equal{}k$ so $ y\equal{}\frac{1\plus{}\sqrt{1\plus{}4k}}{2}$ is the required solution.
Am I right?

Yes. Just do not forget to explain why it is not $ y\equal{}\frac{1\minus{}\sqrt{1\plus{}4k}}{2}$ if you intend to present this argument to anybody. ;)

In Japanese University entrance exam, the solution values 6 points in 20 full marks, but in university mathematics, the solution is usual solution.

As for my solution in #3 may be 8 points in 20 full marks.


Find all functions $f: \mathbb R \to \mathbb R$ such that
\[ f(x \plus{} f(y)) \equal{} x^2 \plus{} f(y)^2\]
for all $ x,y \in \mathbb{R}.$

[quote="mathmen"]find all function such that ;
$ (\forall x,y \in \mathbb{R}); f(x \plus{} f(y)) \equal{} x^2 \plus{} f(y)^2$[/quote]
Let $ P(x,y)$ be the assertion $ f(x \plus{} f(y)) \equal{} x^2 \plus{} f(y)^2$

$ P(0,x)$ $ \implies$ $ f(f(x))\equal{}f(x)^2$ and so $ f(x)\equal{}x^2$ $ \forall x\in f(\mathbb R)$

Let then $ y\in\mathbb R$ and $ a\ne 0$ such that $ a>f(y)^2\minus{}f(y)$. Let $ z\equal{}\sqrt{a\plus{}f(y)\minus{}f(y)^2}$ : $ P(z,y)$ $ \implies$ $ f(z \plus{} f(y)) \equal{} z^2 \plus{} f(y)^2\equal{}a\plus{}f(y)$

So $ a\plus{}f(y)\in f(\mathbb R)$ and so $ f(a\plus{}f(y))\equal{}(a\plus{}f(y))^2$ and so $ a^2 \plus{} f(y)^2\equal{}(a\plus{}f(y))^2$ and so $ af(y)\equal{}0$ and so $ f(y)\equal{}0$

Hence the unique solution : $ f(x)\equal{}0$ $ \forall x$

[quote="pco"][quote="mathmen"]find all function such that ;
$ (\forall x,y \in \mathbb{R}); f(x \plus{} f(y)) \equal{} x^2 \plus{} f(y)^2$[/quote]
Let $ P(x,y)$ be the assertion $ f(x \plus{} f(y)) \equal{} x^2 \plus{} f(y)^2$

$ P(0,x)$ $ \implies$ $ f(f(x)) \equal{} f(x)^2$ and so $ f(x) \equal{} x^2$ $ \forall x\in f(\mathbb R)$

Let then $ y\in\mathbb R$ and $ a\ne 0$ such that $ a > f(y)^2 \minus{} f(y)$. Let $ z \equal{} \sqrt {a \plus{} f(y) \minus{} f(y)^2}$ : $ P(z,y)$ $ \implies$ $ f(z \plus{} f(y)) \equal{} z^2 \plus{} f(y)^2 \equal{} a \plus{} f(y)$

So $ a \plus{} f(y)\in f(\mathbb R)$ and so $ f(a \plus{} f(y)) \equal{} (a \plus{} f(y))^2$ and so $ a^2 \plus{} f(y)^2 \equal{} (a \plus{} f(y))^2$ and so $ af(y) \equal{} 0$ and so $ f(y) \equal{} 0$

Hence the unique solution : $ f(x) \equal{} 0$ $ \forall x$[/quote]

$ f(x)\equal{}0$  verify the initial condition?

[quote="pco"][quote="mathmen"]find all function such that ;
$ (\forall x,y \in \mathbb{R}); f(x \plus{} f(y)) \equal{} x^2 \plus{} f(y)^2$[/quote]
Let $ P(x,y)$ be the assertion $ f(x \plus{} f(y)) \equal{} x^2 \plus{} f(y)^2$

$ P(0,x)$ $ \implies$ $ f(f(x)) \equal{} f(x)^2$ and so $ f(x) \equal{} x^2$ $ \forall x\in f(\mathbb R)$

Let then $ y\in\mathbb R$ and $ a\ne 0$ such that $ a > f(y)^2 \minus{} f(y)$. Let $ z \equal{} \sqrt {a \plus{} f(y) \minus{} f(y)^2}$ : $ P(z,y)$ $ \implies$ $ f(z \plus{} f(y)) \equal{} z^2 \plus{} f(y)^2 \equal{} a \plus{} f(y)$

So $ a \plus{} f(y)\in f(\mathbb R)$ and so $ f(a \plus{} f(y)) \equal{} (a \plus{} f(y))^2$ and so $ a^2 \plus{} f(y)^2 \equal{} (a \plus{} f(y))^2$ and so $ af(y) \equal{} 0$ and so $ f(y) \equal{} 0$

Hence the unique solution : $ f(x) \equal{} 0$ $ \forall x$[/quote]

Let $ P(x,y)$ be the assertion $ f(x \plus{} f(y)) \equal{} x^2 \plus{} f(y)^2$

Let $ f(0) \equal{} a$

$ P(x \minus{} a,0)$ $ \implies$ $ f(x) \equal{} (x \minus{} a)^2 \plus{} a^2$

Hence $ f(0) \equal{} a$ $ \implies 2a^2 \equal{} a$ (edited since previous error)

If $ a \equal{} 0$ $ \implies f(x) \equal{} x^2$ but it does not verifiy the inital condition

If $ a \equal{} \plus{} \frac12$ $ \implies f(x) \equal{} x^2 \minus{} x \plus{} \frac12$ but it does not verifiy the inital condition

So there are no solution...

[quote="reason"][quote="pco"][quote="mathmen"]find all function such that ;
$ (\forall x,y \in \mathbb{R}); f(x \plus{} f(y)) \equal{} x^2 \plus{} f(y)^2$[/quote]
Let $ P(x,y)$ be the assertion $ f(x \plus{} f(y)) \equal{} x^2 \plus{} f(y)^2$

$ P(0,x)$ $ \implies$ $ f(f(x)) \equal{} f(x)^2$ and so $ f(x) \equal{} x^2$ $ \forall x\in f(\mathbb R)$

Let then $ y\in\mathbb R$ and $ a\ne 0$ such that $ a > f(y)^2 \minus{} f(y)$. Let $ z \equal{} \sqrt {a \plus{} f(y) \minus{} f(y)^2}$ : $ P(z,y)$ $ \implies$ $ f(z \plus{} f(y)) \equal{} z^2 \plus{} f(y)^2 \equal{} a \plus{} f(y)$

So $ a \plus{} f(y)\in f(\mathbb R)$ and so $ f(a \plus{} f(y)) \equal{} (a \plus{} f(y))^2$ and so $ a^2 \plus{} f(y)^2 \equal{} (a \plus{} f(y))^2$ and so $ af(y) \equal{} 0$ and so $ f(y) \equal{} 0$

Hence the unique solution : $ f(x) \equal{} 0$ $ \forall x$[/quote]

$ f(x) \equal{} 0$  verify the initial condition?[/quote]

 :blush:  :rotfl:  How stupid I am !!!

So ... no solution.  :blush:

Thanks for the correction


FTW
probability
videos
geometry
number theory
AMC 10

I was browsing around the FF, and I say this thread called 'Self-Identity' It's about how you think people see you, and how you see yourself. I kind of wanted to make a revival like it, so I'll explain the rules.
So you list 5 things about you. I'll show you me, let's call it an example.

1. I'm a Filipino-American
2. I'm studying Number Theory at the moment
3. I've been on AoPS since May or June of this year
4. I keep in touch with my Algebra I buddies
5. I am a total spaz.

Like that.
Not like,
I'm nice.
I'm cool.
I have three toes on my left foot. (That's great and all.... but too much information.)
Okay? Let's start :)
*refers back to example for mine

1. I am Japanese-American
2. I am studying AMC 10 preparation at the moment
3. I have been on AoPS since about a year I think
4. I keep touch in all my class mates
5. I make people at my school go to the AoPS website

6. I have 3 toes on one foot (jk)

1. Cuban-American. I win for non-azn nerdiness.
2. Been on AoPS for about 2 years, but I really only go on for the FF.
3. AP Calc AB is what I study now, but not on AoPS.
4. Chuck Norris > Star Wars > Pokémon > Link > Smash Bros. > Rock Music > Everything Else > Country Music > Disney Music.
5. FALCON...............................PAWNCH!!!!!!!!!!!

That's 5, right?

1. No idea
2. Swims
3. Does Geo. Hon.
4. Likes astronomy (facisnating)
5. Plays Mario Kart. :)

1. Japanese-Chinese-American
2. Studying Algebra I as review right now.  :(
3. Plays piano (rag music FTW)
4. I'm a Christian
5. I keep in touch with some of my AoPS Algebra I classmates

1. I'm Korean
2. I'm studying Introduction to Counting and Probability and MATHCOUNTS/AMC 8 Basics at the moment.
3. I've been on since May 30, 2009.
4. I also keep in touch with my Algebra I buddies just like Meewhee.
5. I'm a video game expert.

1. I choose my username because my friends think I act like a monkey.
2. I am addicted to AoPS.
3. When I am not on AoPS or doing math I play tennis, games and read.
4. I enjoy dark chocolate.
5. I LOVE TO SHOP.

6. I am a real monkey (jk)

1.) SORT OF OBVIOUS([size=75]indian) [/size]
2.) Aops is addicting
3.) I play tennis, swim, uh dance, and other stuff
4.) I like chocolate.
5.) I AM k00l (not exactly)

1. I'm a Chinese-American....well....sort of. I'm 1/4 Taiwanese and 1/4 Singaporian and 1/2 Shanghainese. If people ask, I say I'm Chinese.
2. I'm studying Intro to Geometry
3. I've been on AoPS for about a week I think
4. I don't have a lot of buddies yet
5. I secretly learned English thanks to Dora the Explorer, (who speaks Chinese and English on a Chinese broadcast.)

So.... Now you know about me.  :)

Wah.... the topic died!
It hasn't been too long to bump it up, has it?
I think this is a really cool thread!

1.  I am Chinese American  100%(as far as I know)
2  I'm taking Amc8/mathcount Advanced
3.  An AOPSer for about a year
4.  I guess I know a couple people, only one from my town
5.  I am a tennis playing, swimming, semi-pro GAMER

1. Asian
2. Concert pianist
3. Biggest reader
4. no friends
5 has a ferret

1.Chinese-American
2.Trys to be a nerd but fails epically
3.can eat practically anything. (even poo) (jk)
4.Am pessimist.
5.2 guinea pigs

@butler you have friends.
You making me sad :(

1. chinese american
2. currently takin AP Calc BC at my school
3. plays three instruments, albeit at a fail skill level
4. officially hates shakespeare's macbeth (what we're studying in english)
5. has the attention span of a squirrel and finds tha--ooh shiny

1. Taiwanese American
2. Spaz too much
3. Math teacher fails at Geo- thought that (4,4), (-2,4), and (-2, 0) were collinear
4. Fail at piano
5. pessimist

1.  I'm a Christian
2.  I have blue eyes
3.  I like math, and am almost half way through algebra 2, having recently completed Hon. geo.
4.  I like football A LOT
5.  I am a competetive swimmer in the summer
6.  I am very competetive overall.
7.  Oh yeah, and I'm oart italian, irish, german, swedish and all american.

More about me.
6. I am a Christian. (if you are to visit my bloggie)
7. 17 is my favorite number ever.
8. I have a  :rotfl: (laughing) problem.
9. Every time I prove a theorem I find that some other person bet me to it 100s of years ago.
10. I play the Clarinet.

[color=violet][/color]
1.im crazy
2. im a christian! ! ! ! ! ! ! ! ! ! ! ! !
3.im a read-a-holic
4. I luv mah buddies
5.Im a champ in mario kart
6.I have a dog with a super curly tail :o

I don't want this to
DIE DIE DIE.
So here's moar!
1) I purposely use bad grammar and spelling when I feel tipsy. :P
2) I used the word 'tipsy' purposely.
3) I have an addiction to Chap Stick
4) I'm am a dweeb muffin
5) I like techno/electronic music.

okay me

1. I fight with all my friends because they call me a nerd
2. I have been called a human calculator, dictionary, thesaurus and computer.
3. i play soccer
4. i play a lot of super smash bros. brawl
5. uhhh i am Irish

1. i am a girl

2. I was born in USA, but i have german and polish herritage

3. i am in algebra 2 and intro to number theroy

4. i ski and play soccer and tennis

5. i am 10

1. I am Korean
2. I am a boy
3. Butler is my brother...(I hacked his account)
4. I am 19 years old
5. I love animals...

EDIT: Come on!!!! Sorry, he does this sometimes....

1. I am cocasian (if thats how you spell it if not dunno)
2. I am a boy
3. I am lucky not to have a sibling who cares about my AoPS account
4. I am 14
5. I run environmental service projects with my friends


binomial coefficients
number theory open
number theory

I'm familiar with this bound for $ \binom{2n}{n} > \frac {4^n}{2n \plus{} 1}$. Do we have a simmilar lower bound for $ \binom{3n}{2n}$ for instance. Or for $ \binom{an}{bn}$ where $ a > b$ and $ a,b$ relatively prime.

Tnx in advance

Take a look at this paper:
Worsch, Thomas. [url=http://digbib.ubka.uni-karlsruhe.de/volltexte/181894]Lower and upper bounds for (sums of) binomial coefficients[/url].


inequalities
trigonometry
geometry
incenter
angle bisector
area of a triangle
Heron\u0027s formula

let $ ABC$ be a triangle , prove that :

$ \tan\bigg(\frac {A}{2}\bigg)\sqrt {\tan\bigg(\frac {B}{2}\bigg)\tan\bigg(\frac {C}{2}\bigg)}$ $ \plus{}$ $ \tan\bigg(\frac {B}{2}\bigg)\sqrt {\tan\bigg(\frac {C}{2}\bigg)\tan\bigg(\frac {A}{2}\bigg)}$ $ \plus{}$ $ \tan\bigg(\frac {C}{2}\bigg)\sqrt {\tan\bigg(\frac {A}{2}\bigg)\tan\bigg(\frac {B}{2}\bigg)}$ $ \: \le \: 1$

prove that in any acute triangle :

$ \tan(A)\tan(B) \plus{} \tan(B)\tan(C) \plus{} \tan(C)\tan(A) \geq 9$

[hide="1"]Notice that $ \tan\frac {A}{2} \equal{} \tan\left(\frac {\pi}{2} \minus{} \left(\frac {B}{2} \plus{} \frac {C}{2}\right)\right) \equal{} \frac {1 \minus{} \tan\frac {B}{2}\tan\frac {C}{2}}{\tan\frac {A}{2} \plus{} \tan\frac {B}{2}}$.  Thus the left hand simplifies to

$ \displaystyle\sum\left(1 \minus{} \tan\frac {B}{2}\tan\frac {C}{2}\right)\frac {\sqrt {\tan\frac {B}{2}\tan\frac {C}{2}}}{\tan\frac {A}{2} \plus{} \tan\frac {B}{2}}$

Since $ \sqrt {\tan\frac {B}{2}\tan\frac {C}{2}}\le\frac {1}{2}\left(\tan\frac {A}{2} \plus{} \tan\frac {B}{2}\right)\Leftrightarrow \frac {\sqrt {\tan\frac {B}{2}\tan\frac {C}{2}}}{\tan\frac {A}{2} \plus{} \tan\frac {B}{2}}\le\frac {1}{2}$  We have

$ \displaystyle\sum\left(1 \minus{} \tan\frac {B}{2}\tan\frac {C}{2}\right)\frac {\sqrt {\tan\frac {B}{2}\tan\frac {C}{2}}}{\tan\frac {A}{2} \plus{} \tan\frac {B}{2}}\le\displaystyle\sum\frac {1}{2}\left(1 \minus{} \tan\frac {A}{2} \plus{} \tan\frac {B}{2}\right) \equal{} 1$.

We show the validity of the last equality:

$ \cos\frac {A}{2}\sin\frac {A}{2} \equal{} \sin\frac {A}{2}\cos\frac {A}{2}$

$ \Rightarrow\cos\frac {A}{2}\cos\left(\frac {B}{2} \plus{} \frac {C}{2}\right) \equal{} \sin\frac {A}{2}\sin\left(\frac {B}{2} \plus{} \frac {C}{2}\right)$

$ \Rightarrow\cos\frac {A}{2}\cos\frac {B}{2}\cos\frac {C}{2} \minus{} \cos\frac {A}{2}\sin\frac {B}{2}\sin\frac {C}{2} \equal{} \sin\frac {A}{2}\sin\frac {B}{2}\cos\frac {C}{2} \plus{} \sin\frac {A}{2}\cos\frac {B}{2}\sin\frac {C}{2}$

$ \Rightarrow\cos\frac {A}{2}\cos\frac {B}{2}\cos\frac {C}{2} \equal{} \sin\frac {A}{2}\sin\frac {B}{2}\cos\frac {C}{2} \plus{} \sin\frac {A}{2}\cos\frac {B}{2}\sin\frac {C}{2} \plus{} \cos\frac {A}{2}\sin\frac {B}{2}\sin\frac {C}{2}$

$ \Rightarrow1 \equal{} \tan\frac {A}{2}\tan\frac {B}{2} \plus{} \tan\frac {A}{2}\tan\frac {C}{2} \plus{} \tan\frac {B}{2}\tan\frac {C}{2}$

$ \Rightarrow\displaystyle\sum\frac {1}{2}\left(1 \minus{} \tan\frac {A}{2}\tan\frac {B}{2}\right) \equal{} 1$

Nice problem!  :lol:

[/hide]

[hide="2"]We have by Jensen's

$ \frac{\cos A\plus{}\cos B\plus{}\cos C}{3}\le\cos\left(\frac{A\plus{}B\plus{}C}{3}\right)\equal{}\frac{1}{2}$

Thus we have by AM-GM

$ \cos A\cos B\cos C\le\frac{1}{8}$

$ \Rightarrow8\cos A\cos B\cos C\le\sin^2A\plus{}\cos^2A$

$ \Rightarrow8\cos A\cos B\cos C\le\sin A\sin\left(B\plus{}C\right)\minus{}\cos A\cos\left(A\plus{}B\right)$

$ \Rightarrow8\cos A\cos B\cos C\le\sin A\sin B\cos C\plus{}\sin A\sin C\cos B\minus{}\cos A\cos A\cos B\plus{}\cos A\sin B\sin C$

$ \Rightarrow9\cos A\cos B\cos C\le\sin A\sin B\cos C\plus{}\sin A\sin C\cos B\plus{}\cos A\sin B\sin C$

$ \Rightarrow9\le\tan A\tan B\plus{}\tan A\tan C\plus{}\tan B\tan C$

Again, nice problem  :lol:

[/hide]

[hide="1"]
Notice that $ \tan\left(\frac{A}{2}\right)\equal{}\frac{r}{s\minus{}a}$ (because the incenter lies on the angle bisector of $ \angle A$). Thus, \[ \prod\tan\left(\frac{A}{2}\right)\equal{}\frac{r}{s}\]
The inequality is therefore equivalent to\[ \sum\frac{r}{\sqrt{s(s\minus{}a)}}\leq1\iff\sum\frac{\sqrt{(s\minus{}b)(s\minus{}c)}}{s}\leq1\](Heron's formula being used extensively).
Now using Cauchy Schwarz,\[ 1\equal{}\sum\frac{s\minus{}b}{s}\equal{}\left(\sum\frac{s\minus{}b}{s}\right)\left(\sum\frac{s\minus{}c}{s}\right)\geq\left(\sum\frac{\sqrt{(s\minus{}b)(s\minus{}c)}}{s}\right)^2\]proves it.:)
[/hide]

hi
nice ssolutions , thank's

for the firt problem
$ ab \plus{} bc \plus{} ca \geq a\sqrt {bc} \plus{} b\sqrt {ca} \plus{} c\sqrt {ab}$

and we now that in any triangle

$ \tan\bigg(\frac {A}{2}\bigg)\tan\bigg(\frac {B}{2}\bigg) \plus{} \tan\bigg(\frac {B}{2}\bigg)\tan\bigg(\frac {C}{2}\bigg) \plus{} \tan\bigg(\frac {C}{2}\bigg)\tan\bigg(\frac {A}{2}\bigg) \equal{} 1$

 $ \tan\bigg(\frac {A}{2}\bigg)\sqrt {\tan\bigg(\frac {B}{2}\bigg)\tan\bigg(\frac {C}{2}\bigg)}$ $ \plus{}$ $ \tan\bigg(\frac {B}{2}\bigg)\sqrt {\tan\bigg(\frac {C}{2}\bigg)\tan\bigg(\frac {A}{2}\bigg)}$ $ \plus{}$ $ \tan\bigg(\frac {C}{2}\bigg)\sqrt {\tan\bigg(\frac {A}{2}\bigg)\tan\bigg(\frac {B}{2}\bigg)}$ $ \le 1$

for the second

by Cauchy schwartz

$ (a\plus{}b\plus{}c)(\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}) \geq 9$

$ (a\plus{}b\plus{}c)\bigg(\frac{ab\plus{}bc\plus{}ca}{abc}\bigg) \geq 9$

we now that in any acute triangle

$ \tan(A)\plus{}\tan(B)\plus{}\tan(C) \equal{}\tan(A)\tan(B)\tan(C)$

$ a\equal{}\tan(A)$ , $ b\equal{}\tan(B)$ , $ c\equal{}\tan(C)$

$ \tan(A)\tan(B) \plus{} \tan(B)\tan(C) \plus{} \tan(C)\tan(A) \geq 9$

in the second
Why ABC is acute?

Can it be any triangle? :?:

$ \tan(\frac {\pi}{2}) \equal{} ?$

it must be

$ A$ , $ B$ , $ C$ $ \neq \frac {\pi}{2}$


probability
number theory proposed
number theory

Try and prove the Prime Number Theorem using methods of probability by doing the following questions below:

1. If we randomly pick a number from 1 to N for large N, then determine (using methods from basic probability theory) the approximate probability that it is squarefree.

2. Apply the same methods/ideas to try and determine that a randomely chosen number from 1 to N is prime (note: This formula will have a significant error in it at this time).

3. Determine why the method used above doesn't work 100%. Specifically, determine the cummulative error term that one gets and add this error term to your formula.

4. Use this overall formula (which determines the probability that a number chosen from 1 to N is prime) to prove the Prime Number Theorem.

[quote="LeoYard"]1. If we randomly pick a number from 1 to N for large N, then determine (using methods from basic probability theory) the approximate probability that it is squarefree.[/quote]

For a given prime p, the probability that $ p^2\nmid x$ is $ \displaystyle 1\minus{}\frac{1}{p^2}$ so the probability we are looking for is $ \displaystyle\prod_{p\in\mathbb{P}} \left( 1\minus{}\frac{1}{p^2}\right) \equal{}\frac{6}{\pi ^2}$, but I really can't see how this idea can help us for question number 2...

I think what he's driving at in 2) is that the probability should be about $ \prod_{p<\sqrt N}\left(1\minus{}\frac 1p\right)$ but I have no idea how this very approximate estimate can possibly lead to the prime number theorem :?


inequalities
n-variable inequality
IMO Shortlist

Let $ n$ be an integer,$ n \geq 3.$ Let $ x_1, x_2, \ldots, x_n$ be real numbers such that $ x_i < x_{i\plus{}1}$ for $ 1 \leq i \leq n \minus{} 1$. Prove that

\[ \frac{n(n\minus{}1)}{2} \sum_{i < j} x_ix_j > \left(\sum^{n\minus{}1}_{i\equal{}1} (n\minus{}i)\cdot x_i \right) \cdot \left(\sum^{n}_{j\equal{}2} (j\minus{}1) \cdot x_j \right)\]

Effectively, we have to prove
\[ \dfrac{n(n\minus{}1)}{2}\sum_{i<j} x_i\,x_j  \minus{}\left(\sum_{i\equal{}1}^{n\minus{}1} (n\minus{}i) x_i\right)\cdot\left(\sum_{j\equal{}2}^n (j\minus{}1)\, x_j\right)>0
\]
For each $ i$ satisfying $ 1\leq i\leq (n\minus{}1)$, let $ a_i \equal{} x_{i\plus{}1}\plus{}x_{i\plus{}2} \plus{} \ldots \plus{} x_n$ and let $ b\equal{}\sum_{j\equal{}2}^n (j\minus{}1)\, x_j$, and $ c_i \equal{} \dfrac{n(n\minus{}1)}{2} a_i \minus{} (n\minus{}i)b$.
The LHS
$ \equal{}\dfrac{n(n\minus{}1)}{2}\sum_{i<j} x_i\,x_j  \minus{}\left(\sum_{i\equal{}1}^{n\minus{}1} (n\minus{}i) x_i\right)b$
$ \equal{}\dfrac{n(n\minus{}1)}{2}\sum_{i\equal{}1}^{n\minus{}1}x_ia_i \minus{}\left(\sum_{i\equal{}1}^{n\minus{}1} (n\minus{}i) x_i\right)b$
$ \equal{}\sum_{i\equal{}1}^{n\minus{}1}\left(\dfrac{n(n\minus{}1)}{2}a_i \minus{} (n\minus{}i)b\right)x_i$
$ \equal{}\sum_{i\equal{}1}^{n\minus{}1}c_ix_i$
So, it remains now to prove that $ \sum_{i\equal{}1}^{n\minus{}1}x_ic_i >0$. Since $ \sum_{i\equal{}1}^{n\minus{}1}a_i \equal{} b$ and obviously, $ \sum_{i\equal{}1}^{n\minus{}1} (n\minus{}i)\equal{}\dfrac{n(n\minus{}1)}{2}$, we obtain, $ \sum_{i\equal{}1}^{n\minus{}1}c_i \equal{} 0$. Also, since $ x_i<x_{i\plus{}1}$, we get
\[ b \equal{} \sum_{j\equal{}2}^n (j\minus{}1)\, x_j < \sum_{j\equal{}2}^n (j\minus{}1)\, x_n \equal{}\dfrac{n(n\minus{}1)}{2}x_n
\]
So, $ c_{n\minus{}1}\equal{}\dfrac{n(n\minus{}1)}{2}x_n \minus{} b >0$. Further, we have
\[ \dfrac{c_{i\plus{}1}}{n\minus{}(i\plus{}1)}\minus{}\dfrac{c_i}{n\minus{}i} \equal{} \dfrac{n(n\minus{}1)}{2}\left(\dfrac{a_{i\plus{}1}}{n\minus{}(i\plus{}1)} \minus{}\dfrac{a_i}{n\minus{}i}\right)>0
\]
which implies that
\[ \dfrac{c_1}{n\minus{}1}<\dfrac{c_2}{n\minus{}2}<\cdots < \dfrac{c_{n\minus{}1}}{1}
\]Therefore, there exists a $ k$,  such that $ c_1$, $ c_2$, $ \ldots$, $ c_k\leq 0$ and $ c_{k\plus{}1}$, $ c_{k\plus{}2}$, $ \ldots$, $ c_{n\minus{}1} > 0$. But then $ c_i(x_i \minus{} x_k)\geq 0$ for all $ i$, i.e. $ c_ix_i \geq c_ix_k$ for all $ i$. So,
\[ \sum_{i\equal{}1}^{n\minus{}1} c_i x_i > \sum_{i\equal{}1}^{n\minus{}1} c_i x_k \equal{} x_k\sum_{i\equal{}1}^{n\minus{}1} c_i\equal{}0
\]
as required.

Here is a solution by induction and shifting.

Let P(n) be the problem statement.

[b]Base Step:[/b] $3(x_1x_2+x_1x_3+x_2x_3) > (2x_1+x_2)(x_2+2x_3) \to (x_3-x_2)(x_2-x_1) > 0$ which is true.

[b]Inductive Step:[/b] Assume that P(n) is true. We will prove that P(n+1) is true.

Let $f(x_1,x_2,\dots,x_{n+1}) = \frac{n(n-1)}{2} \sum_{i < j} x_ix_j - \left(\sum^{n-1}_{i=1} (n-i)\cdot x_i \right) \cdot \left(\sum^{n}_{j=2} (j-1) \cdot x_j \right)$

It is easy to show that $f(x_1+k,x_2+k,\dots,x_{n+1}+k) = f(x_1,x_2,\dots,x_{n+1})$ So we can assume that $x_1 = 0$ and $x_2,x_3,\dots,x_{n+1} > 0$

It remains to show that $\frac{n(n+1)}{2} \sum_{2\leq i < j \leq n+1}x_ix_j > ((n-1)x_2+(n-2)x_3+\cdots+x_n)(x_2+2x_3+\cdots+nx_{n+1})$

From P(n) is true, we have $\frac{n(n-1)}{2} \sum_{2 \leq i < j \leq n+1}x_ix_j > ((n-1)x_2+(n-2)x_3+\cdots+x_n)(x_3+2x_4+\cdots+(n-1)x_{n+1}$

So, it is sufficient to show that $(n+1)(x_3+2x_4+\cdots+(n-1)x_{n+1}) > (n-1)(x_2+2x_3+\cdots+nx_{n+1})$

which is true because of $x_2 < x_3 < \cdots < x_{n+1}$


find minimal of expression $P = \frac{ab+bc+ca}{a^{2}+b^{2}+c^{2}}+\frac{a^{3}+b^{3}+c^{3}}{abc}$ where $a,b,c > 0$


Let a, b, c be positive real numbers such that ab + bc + ca =1. Prove that 1/(a+b) + 1/(b+c) + 1/(a+c) >=  sqrt(3) +  ab/(a+b) + ac/(a+c) + bc(b+c).

the inequality is equivalent with:
\sum (1-ab)/(a+b)\geq \sqrt(3) <=>
\sum (bc+ca)/(a+b)\geq \sqrt(3) <=>
a+b+c\geq sqrt(3)|^2<=>
\sum a^2+ 2\sum ab=\sum a^2+2>=3=>
\sum a^2 >=1=ab+bc+ca which is obvious by AM-GM

Since a, b, and c are positive reals satisfying ab+bc+ca=1, there exists a triangle ABC such that a=tanA/2, b=tanB/2, and c=tanC/2.

The inequality sumcyc((1-ab)/(a+b))>=3^.5 becomes sumcyc(cot((A+B)/2))>=3^.5.  But this follows easily by Jensen since cot is convex on (0, pi/2) implying sumcyc(cot((A+B)/2))>=3cot((A+B+C)/3)=3^.5 (since A+B+C=pi).

I thought this interesting variant was worth sharing.


"Suppose a circle is centered at the origin with radius larger than one, and the arc length of the arc of the circle above the line y = 1 is $\frac43\pi$. Find the radius of the circle."

This problem was on a math team test I had, and I'm unsure of how to solve it. I don't know how to relate the arc length to the whole circle. I took a fraction as the arc length relates to that of a semicircle, and somehow found the arc to be 120 degrees (making the radius be 2), but it wasn't really a solution. Could you anyone help me with this problem? Thanks! :)

Could you tell us the source of the problem, please? During competition season we are trying to be really sure where problems come from. (Hint to other participants: let's be sure where this problem comes from before posting answers.)

Sure, but it's more of a local contest (not national). It's the Hoover (that's the school that hosted it), 2005 Geometry test, though, if that means anything to anyone. It's just a math team contest in Alabama, really.

Thanks for letting us know the problem source.


Steven  owned a private outdoor car park. His charges were £1 per hour or £5 for the whole day.

One snowy winter evening, Steven  arrived to lock up for the night. He asked his attendant, “So how much did we take today, Jerry?”

Jerry replied, “Sorry boss, we only took £50. We had just 10 cars in and they parked all day. The last one has just left.”

Steven quickly deduced that Jerry was lying to him about the receipts.

How was it possible for Steven to arrive at the above conclusion without asking Jerry to show him all the receipts?

[hide="Possible answer"]The last car just left, so 9 cars left before that.  So they didn't stay for the whole day.[/hide]

[quote="AiW"]It was snowy evening, so Steven could see by the tracks that there has been more than 10 cars.[/quote]
Or maybe...
[hide]Since Jerry said "We had just 10 cars in and [b]they parked all day[/b]," Steven would have noticed 10 quadruples of tire tracks in the snow of the parking lot. However, he didn't, so he immediately knew that Jerry was lying.[/hide]


calculus
integration
limit
logarithms
calculus computations

Let $ a,\ b$ be postive real numbers. Find $ \lim_{n\to\infty} \sum_{k\equal{}1}^n \frac{n}{(k\plus{}an)(k\plus{}bn)}.$

[quote]kunny
Today's calculation of Integral 450
2009 Hirosaki University entrance exam
Let a,\ b be postive real numbers. Find $ \lim_{n\to\infty} \sum_{k\equal{}1}^n \frac{n}{(k\plus{}an)(k\plus{}bn)}$[/quote]

$ \lim_{n\to\infty} \sum_{k\equal{}1}^n \frac{n}{(k\plus{}an)(k\plus{}bn)}$ = $ \lim_{n\to\infty} \sum_{k\equal{}1}^n \frac{n}{n^{2}(\frac{k}{n}\plus{}a)({\frac{k}{n}\plus{}b)}}$

=$ \lim_{n\to\infty}$  $ \int_{1}^{n}\frac{dk}{n(\frac{k}{n}\plus{}a)({\frac{k}{n}\plus{}b)}}$

put $ \frac{k}{n} \equal{} t$

= $ \int_{0}^{1}\frac{dt}{(a\plus{}t)(b\plus{}t)}$

= $ \frac{1}{b\minus{}a} ( \log(\frac{1\plus{}a}{a} ) \minus{}\log(\frac{1\plus{}b}{b} )  )$

For $ a\neq b,$

$ \lim_{n\rightarrow \infty }\frac{1}{n}\sum_{k\equal{}1}^{n}\frac{1}{\left( a\plus{}k/n\right) \left( b\plus{}k/n\right) }$

$ \equal{}\frac{1}{a\minus{}b}\lim_{n\rightarrow \infty }\frac{1}{n}\sum_{k\equal{}1}^{n}\left( \frac{1}{b\plus{}k/n}\minus{}\frac{1}{a\plus{}k/n}\right)$

$ \equal{}\frac{1}{a\minus{}b}\int_{0}^{1}\left( \frac{1}{b\plus{}t}\minus{}\frac{1}{a\plus{}t}\right) dt$

$ \equal{}\frac{1}{a\minus{}b}\ln \frac{a\left( b\plus{}1\right) }{b\left( a\plus{}1\right) }$

For $ a\equal{}b$

$ \lim_{n\rightarrow \infty }\frac{1}{n}\sum_{k\equal{}1}^{n}\frac{1}{\left( a\plus{}k/n\right) ^{2}}$

$ \equal{}\int_{0}^{1}\frac{1}{\left( a\plus{}t\right) ^{2}}dt\equal{}\allowbreak \frac{1}{a\left( a\plus{}1\right) }$


trigonometry
ratio
geometric series
absolute value

find the sum of the infinite series
\[ \frac{1}{2}\plus{}i\minus{}1\plus{}\frac{1}{4}\minus{}2i\plus{}\frac{1}{8}\plus{}2\plus{}2i...\]

i apologize if it's obvious to others, but i cannot figure out the pattern. perhaps parenthesis would be of help?

I couldn't figure out the pattern either. :wink:

are you sure it's not a 3i at the end???

It actually works, it seems.
If it does converge:
\[ S \equal{} \sum_{i \equal{} 1}^\infty \frac {1}{2^i} \plus{} (i \minus{} 1)^i \equal{} \sum_{i \equal{} 1}^\infty \frac {1}{2^i} \plus{} \sum_{i \equal{} 1}^\infty (i \minus{} 1)^i \equal{} \frac {1/2}{1 \minus{} 1/2} \plus{} \frac {i \minus{} 1}{1 \minus{} (i \minus{} 1)}
\]

\[ \equal{} 1 \minus{} 3/5 \plus{} 1/5i \equal{} 2/5 \plus{} 1/5i
\]
Does this actually converge? I have major qualms because the norm of 1-i is $ \sqrt {2} > 1$.

EDIT: No, it doesn't converge. Disregard the Latex.

By De Moivre's Theorem, $ (\cos{135^\circ} \plus{} i\sin{135^\circ})^n \equal{} \left( \minus{} \frac {\sqrt2}2 \plus{} \frac {i\sqrt2}2\right)^n \equal{} \cos{135n^\circ} \plus{} i\sin{135n^\circ}$

$ \implies(i \minus{} 1)^n \equal{} \frac {\cos{135n^\circ} \plus{} i\sin{135n^\circ}}{2^{ \minus{} \frac n2}} \equal{} (\sqrt2)^n\cos{135n^\circ} \plus{} i(\sqrt2)^n\sin{135n^\circ}$, so

$ \sum_{k \equal{} 1}^\infty (i \minus{} 1)^k \equal{} \sum_{k \equal{} 1}^\infty(\sqrt2)^n\cos{135n^\circ} \plus{} i(\sqrt2)^n\sin{135n^\circ}$

can be broken up into geometric series (since $ \cos{135n^\circ}$ and $ \sin{135n^\circ}$ are periodic (cycles of $ 8$ for integer $ n$)).

However, the ratio would still be $ (\sqrt2)^8 > 1$, so the sum doesn't converge.

It's generally a bad notational idea to use both $ i \equal{} \sqrt { \minus{} 1}$ and $ i$ as the name of an integer variable all within the same symbol.

But yes, $ \sum_{k \equal{} 1}^{\infty}(i \minus{}1)^k$ is a geometric series. A geometric series converges if and only if the absolute value of its ratio is less than $ 1.$ The ratio in this series is $ i \minus{}1,$ and $ |i\minus{}1| \equal{} \sqrt {2} > 1,$ so the series diverges.

We should also mention the following theorem: if an infinite series converges, then its terms tend to zero in the limit. This is most useful in its contrapositive form: if the terms of a series do not tend to zero, then the series must diverge.

In the series as described in the original post, there is no evidence that the terms are tending to zero. For that reason, the series should diverge.

Note the complex norm (or magnitude) is multiplicative, so the norm of each term is $ |(i\minus{}1)^n|\equal{}(\sqrt{2})^n\rightarrow \infty$ as $ n\rightarrow\infty$. Just wanted to point that out.

yeah, im sure there is no 3i in the problem (and no parentheses); however, if it helps the answer supposedly is
\[ \frac {2}{5} \plus{} \frac {1}{5}i
\]
..

(btw, the highest score on the test this problem came from was a 96, and i think no one actually got this one right, as in no one knew the pattern (excluding the people that blindly guessed)

That's what I had above in my first, post, but are we sure that the series actually converges? Where was this from?

[quote="someperson01"]..., but are we sure that the series actually converges? [/quote]
Excuse me, please.

At least 3 of the last 5 posters were giving information
toward its divergence (as I looked back on them).


Let $ a,$ $ b$ and $ c$ are non-negative numbers. Prove that:
\[ \sum_{cyc}\sqrt{2(a^2\plus{}b^2)}\geq\sqrt[3]{9\sum_{cyc}(a\plus{}b)^3}\]

[quote="arqady"]Let $ a,$ $ b$ and $ c$ are non-negative numbers. Prove that:
\[ \sum_{cyc}\sqrt {2(a^2 \plus{} b^2)}\geq\sqrt [3]{9\sum_{cyc}(a \plus{} b)^3}
\]
[/quote]

Thank you Can for your interest! :)
Here is my proof:
$ \sum_{cyc}\sqrt {2(a^2 \plus{} b^2)}\minus{}\sqrt [3]{9\sum_{cyc}(a \plus{} b)^3}\equal{}$
$ \equal{}\sum_{cyc}\left(\sqrt {2(a^2 \plus{} b^2)}\minus{}a\minus{}b\right)\minus{}\left(\sqrt [3]{9\sum_{cyc}(a \plus{} b)^3}\minus{}2(a\plus{}b\plus{}c)\right)\equal{}$
$ \equal{}\sum_{cyc}\frac{(a\minus{}b)^2}{a\plus{}b\plus{}\sqrt{2(a^2\plus{}b^2)}}\minus{}\frac{\sum(18a^3\plus{}27a^2b\plus{}27a^2c)\minus{}8(a\plus{}b\plus{}c)^3}{\left(\sqrt [3]{9\sum(a \plus{} b)^3}\right)^2\plus{}2(a\plus{}b\plus{}c)\sqrt [3]{9\sum(a \plus{} b)^3}\plus{}4(a\plus{}b\plus{}c)^2}\equal{}$
$ \equal{}\sum_{cyc}(a\minus{}b)^2\left(\frac{1}{a\plus{}b\plus{}\sqrt{2(a^2\plus{}b^2)}}\minus{}\frac{5a\plus{}5b\plus{}8c}{\left(\sqrt [3]{9\sum(a \plus{} b)^3}\right)^2\plus{}2(a\plus{}b\plus{}c)\sqrt [3]{9\sum(a \plus{} b)^3}\plus{}4(a\plus{}b\plus{}c)^2}\right)\geq$
$ \geq\sum_{cyc}(a\minus{}b)^2\left(\frac{1}{a\plus{}b\plus{}1.5(a\plus{}b)}\minus{}\frac{5a\plus{}5b\plus{}8c}{\left(\sqrt {3\sum(a \plus{} b)^2}\right)^2\plus{}4(a\plus{}b\plus{}c)^2\plus{}4(a\plus{}b\plus{}c)^2}\right)\equal{}$
$ \equal{}\sum_{cyc}(a\minus{}b)^2\left(\frac{2}{5(a\plus{}b)}\minus{}\frac{5a\plus{}5b\plus{}8c}{\sum(14a^2\plus{}22ab)}\right)\equal{}$
$ \equal{}\sum_{cyc}\frac{(a\minus{}b)^2(4c(7c\plus{}a\plus{}b)\plus{}3(a\minus{}b)^2)}{10(a\plus{}b)\sum(7a^2\plus{}11ab)}\geq0.$


Find all triples of positive integers $(a,b,c)$ such that

\[\left(1 + \frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right) = 2\]

[hide="My work"]

First of all, I'm not sure if there exists such triple. I'm saying this because I saw couple Olympiad problems that looks like it has solutions and it doesn't. I'm not sure how I can confirm my "belief" though.

Anyway, expanding what we have gives:

$\frac{a+b+c+ab+bc+ac+abc+1}{abc} = 2$

The upper part can be expressed as $(a+1)(b+1)(c+1)$.

So, the problem now asks if there is any triple (integers) of $a,b,c$ making:

$\frac{(a+1)(b+1)(c+1)}{abc} = 2$. [/hide]

I trid few values but it didn't work. Maybe there is a solution but I don't know how to get it. Any help?

Assuming $a\ge b\ge c$ yields $c\le 3$. Analyzing the three cases, we find the triples $(7,6,2),(9,5,2),(15,4,2),(8,3,3),(5,4,3)$.

Since $a\ge b\ge c$, it follows that $2\le \left( 1+\frac1c \right) ^3$, hence $c\le 3$. The case $c=1$ yields
\[
\left(1+\frac1a \right) \left(1+\frac1b \right)=1
\]
which is impossible. If $c=2$ we obtain
\[
\left(1+\frac1a \right) \left(1+\frac1b \right)=\frac43
\]
therefore (recall that $a\ge b$)
\[
\left( 1+\frac1b \right)^2 \ge \frac43
\]
implying $b<7$. Also, since $1+\frac1a>1$ it follows $b>3$. By inspection, we find the triples $(7,6,2),(9,5,2),(15,4,2)$.
A similar approach in the case $c=3$ gives the other triples.

[quote="peeta"]watch out the inequality sign turns when u take the reciproc ($x^{-1}$) therefore $2\le \left( 1+\frac1c \right) ^3$
u would lose points on a contest, and may get a problem wrong. u were lucky since u did a wrong twice  :roll:

(watch out: this only is true if both sides are positve or negative)[/quote]
Yes, thanks, I posted it in a hurry. Also thanks for your concern, but, at my age, I can hardly attend any math competition as a contestant :(

Sorry I'm new on here.  :)

I don't understand how you got $a \leq b \leq c$ and $c \leq 3$.

I'll be really thankful if someone can explain this to me.  :lol:

ill be glad to explain it to u :D
watch this:

$2\le \left( 1+\frac1c \right) ^3$
$\Leftrightarrow \sqrt[3] 2 \le 1+ \frac 1c$
$\Leftrightarrow c\le \frac 1{\sqrt[3] 2 - 1}\approx 3.847$
since c is an integer $c\le 3$

Thanks to peeta.  :lol:

Just for sidenote, calculators are forbidden.  :P So I wonder how would you find $c \leq 3$.

Also, I have one last not-getting part.

$1+\frac{1}{a} > 1$ is clear since $a > 0$ by the problem. But how does that imply $b > 3$? :?  :huh:

[quote="enescu"][quote="Silverfalcon"] I wonder how would you find $c \leq 3$.[/quote]
This gets boring. Yet: suppose $c\ge4$. Then $\frac1c \le\frac14$ so $\left(1+\frac1c \right)\le\frac54$, that is,$\left(1+\frac1c \right)^3\le\left( \frac54 \right)^3=\frac{125}{64}<2$(no calculator needed for this!)

[quote="Siverfalcon"]Also, I have one last not-getting part.

$1+\frac{1}{a} > 1$ is clear since $a > 0$ by the problem. But how does that imply $b > 3$? :?  :huh:[/quote]
Since $\left(1+\frac1a \right)\left(1+\frac1b \right)=\frac43$ and $1+\frac1a>1$ it follows that $1+\frac1b<\frac43$, or $\frac1b<\frac13$, that is, $b>3$.
Excuse me for saying this, but some paper and pencil might be useful sometime..:)[/quote]

Thank you for great explanation!  :)  :lol:


logarithms
analytic geometry
graphing lines
slope
calculus
derivative

Find the number of solutions to the equation $ x\equal{}3^{x\minus{}1}$, where $ x\in R$.

[hide="Quick and dirty"]
$ 3^{x \minus{} 1} \minus{} x$ is obviously concave up, and thus has 0, 1, or 2 roots.

Now, $ \frac {d}{dx} 3^{x \minus{} 1} \equal{} \ln 3 \cdot 3^{x \minus{} 1}$, which equals 1 when $ x \equal{} 1 \minus{} \frac {\ln \ln 3}{\ln 3}$

At this point, $ 3^{x \minus{} 1} \equal{} \frac {1}{\ln 3} < \frac {\ln 3 \minus{} \ln \ln 3}{\ln 3} \equal{} x$

Hence, the tangent line of slope 1 is below the line $ y \equal{} x$, and thus there are two solutions.
[/hide]

Why do you find where the slope is $ 1$? How can you deduce that there are two solutions?

[hide="Explanation 1"]
Where the slope is 1, the tangent line will be of the form $ x\plus{}a$.

Since $ 3^{x\minus{}1}$ is concave up, $ 3^{x\minus{}1} \geq x\plus{}a$,with equality at the point of tangency.

If $ a>0$, then the whole curve is above the graph of $ x$, and there are no solutions.
If $ a\equal{}0$, then the line $ y\equal{}x$ is tangent to the curve, and there is exactly one solution.
If $ a<0$, then the graph starts out over the line $ y\equal{}x$, dips below it, and rises back above, creating 2 solutions.

We test what $ a$ is by finding the $ x$ coordinate of the point of tangency. $ y\equal{}x\plus{}a \iff a \equal{} y\minus{}x \equal{} 3^{x\minus{}1}\minus{}x$.
[/hide]

[hide="Explanation 2"]$ 3^{x \minus{} 1} \minus{} x$ is concave up. This means there are three possible shapes of the graph:
1. It is monotone decreasing
2. It is monotone increasing
3. It decreases, then increases.

In cases (1) and (2), there is exactly one solution. In case (3), there are:
- 0 solutions if the graph is always above the x axis
- 1 solution if the graph is tangent to the x axis at its root
- 2 solutions if the graph goes below the x axis.

Clearly, we are dealing with case 3.

The point when the derivative of $ 3^{x \minus{} 1}$ is 1 is the point where the derivative of $ 3^{x \minus{} 1} \minus{} x$ is 0: in other words, the relative minimum. If here, $ 3^{x \minus{} 1} \minus{} x > 0$, there are no solutions. If $ 3^{x \minus{} 1} \minus{} x \equal{} 0$, there is one solution. If $ 3^{x \minus{} 1} \minus{} x < 0$, there are two solutions.[/hide]