Alright everyone, I had to do a big chemistry project and now I have to do research on it and write up a report, but I am having a terribly hard time finding information on it, mostly because the experiment does not have a name (or not one that I know at least) so I am asking you for help!

This is what my experiment was...

Put 15 mL of clean mercury in a small beaker. Pour 50 mL of half-saturated NaCl solution on top of the mercury. Using a graphite anode and the mercury as the cathode, electrolyze the salt solution for 30 minutes at a potential of approximately 10 volts. Turn off the current. Transfer the mercury to a beaker containing 50 mL of 0.1 M NaOH solution. Insert a fresh peice of graphite into the alkaline solution, lowering it until it touches the mercury. Why was a gas evolved from the liquids in the second beaker?

Anyone know where I can find information on why a gas was evolved, any help is appreciated.

Thank you in advance.

first of all, it sounds as if the experiment is rather dangerous... considering mercury=toxic.

secondly, keep in mind that when you electrolyze, two things happen.  in your second beaker, H2 should be formed at the mercury cathode, since water has a higher reduction potential than sodium.

at the graphite anode, something must be oxidized... this turns out to be hydroxide, which has a significantly lower reduction potential (and thus higher oxidation potential) than chlorine, oxygen, or mercury.  this will yield oxygen gas.

so it looks like you have a mix of hydrogen and oxygen gases.  try a burning splint test and see what happens.  hope that was helpful

(i'd check that if i were you, i just quickly scanned through the problem and could be missing something.)

As far as the gas from the second beaker is concerned, it must have something to do with the graphite ($\text{C}$).  Therefore, the simplest guess to me would be $\text{CO}_2$.  In the first beaker we had (at the graphite anode) $2\text{H}_2\text{O}\rightarrow \text{O}_2+4\text{H}^++4e^-$.  Perhaps $\text{O}_2$ was somehow sorbed into $\text{Hg}$ and combined with the $\text{C}$ in the second beaker?  Was there an evolution of heat?

Hmm, good idea... I think graphite's supposed to be pretty inert though.  breaking a its network covalent bonds requires a lot of energy.

True.  Hence my somewhat-strong dissatisfaction with my solution, but such is life.

Well... I have already done the actual experiment, what I have to find out now (mostly) is why a gas was formed.

Thank you very much for your replies they are very helpful, keep them coming, if you have ANY websites at all that will help me understand what this experiment did I would appreciate it, since I need 4 sources, hehe.

Also, when I was electrolyzing the NaCl solution and the mercury, it smelled a lot like chlorine. So was the chlorine released as a gas in the first beaker, leaving sodium?

-edit-

Also... whenever I look up electrochemistry experiments it shows salt bridges, was I using the NaCl solution as that, and also electrolyzing it?

Salt bridges should be more important for voltaic cells.

A $10\text{V}$ potential could definitely have oxidized some $\text{Cl}^-$ in addition to $\text{H}_2\text{O}$.  In fact, I believe that $\text{Cl}^-$ is often oxidized along with, or even instead of, $\text{H}_2\text{O}$, a fact that was never well-explained to me.

Can graphite act as a catalyst (to some as-of-yet undetermined reaction), or is it a reactant?

Taken from [url=http://www.corrosion-doctors.org/CP/semi-consum.htm]http://www.corrosion-doctors.org/CP/semi-consum.htm[/url]:

[quote]if oxygen formation predominates, as in low chlorine media, graphite is oxidized to carbon dioxide.[/quote]
So I think it's possible.  How likely, of course, is harder to say.

So you're saying the gas given off at the end was CO2?

During electrolysis chlorine evolved at anode and sodium was created at cathode - it was immediately dissolved in mercury (google mercury sodium amalgam). Such an amalgam is sometimes used as a reducing agent - but it is not very stable.

Now, I don't know exact chemistry of sodium amalgam, so I am not sure about the second part of the experiment. My guess is that graphite electrode was used to speed up the decomposition - sodium was oxidized on the mercury surface, but the charge was transferred to graphite surface were hydrogen was reduced. IIRC reduction of hydrogen on the mercury is slow due to high overpotential needed.

Best,
Borek

Smarkotan oz gluthozmaz

I don't know what an amalgam is  :?

And are you saying it was... hydrogen given off at the end?

Amalgam is an alloy of mercury.

And yes, I think it was hydrogen.

Best,
Borek

so which is it? hydrogen or carbon dioxide?  :?

So let me get this straight, graphite is being used as a catalyst, so are two different gases being given off, one from the graphite and one from the sodium?

And also, why is the gas being given off anyway?

Also wanted to ask if any chemical theories describe what is going on and what "question of chemical principle" is being asked of me.

[quote="Bodom28"]So let me get this straight, graphite is being used as a catalyst, so are two different gases being given off, one from the graphite and one from the sodium?[/quote]

First, I wuld rather not refer to the graphite as catalyst - in a classical chemical sense it is not a catalyst, as it is not taking part in the reaction. It is just an electrode.

Sodium reacts with water on the mercury surface, but this reaction doesn't give any gas. What happens is soidum atoms give off an electron and go to the solution as Na+ cation. Excess electrons are used on carbon electrode to reduce hydrogen from water. Hydrogen evolves on the carbon (graphite) electrode and thats the only gas in the second phase on the experiment.

[quote]And also, why is the gas being given off anyway?[/quote]

While it happens in not-to-straight way (with mercury and graphite help) it is nothing else as reaction between metallic sodium and water. Remember that sodium amalgam is nothing else as solution of elemental sodium in mercury.

[quote]Also wanted to ask if any chemical theories describe what is going on and what "question of chemical principle" is being asked of me.[/quote]

No idea how to help... Many priniciples, many theories here... Usually you should know what to expect as it correlates with lectures.

Best,
Borek

[quote="Borek"][quote="Bodom28"]First, I wuld rather not refer to the graphite as catalyst - in a classical chemical sense it is not a catalyst, as it is not taking part in the reaction. It is just an electrode.
[/quote][/quote]

But it's in the second beaker too... and not being electrolyzed.

And thanks very much for the responses borek  :)

[quote="Borek"]Hydrogen evolves on the carbon (graphite) electrode and thats the only gas in the second phase on the experiment. [/quote]

Don't forget that something must be oxidized as well.  The species present with the highest oxidation potential is hydroxide, which yields oxygen gas upon oxidation.  Thus oxygen gas will also be produced.

[quote="gregx007"]Don't forget that something must be oxidized as well.  The species present with the highest oxidation potential is hydroxide, which yields oxygen gas upon oxidation.  Thus oxygen gas will also be produced.[/quote]

No. Read carefully my explanation. Sodium is oxidized. There is no need for the second oxidation reaction.

Chemically it is the same as if you will put a piece of sodium directly into the water. Soidum is oxidized, hydrogen evolves. The fact that sodium is dissolved in mercury and that the hydrogen evolves on the carbon surface makes it more interesting, but the reaction is still the same.

2Na + 2H2O -> 2Na+ + 2OH- + H2

Best,
Borek

[quote="gregx007"]Borek:  where does the sodium metal come from?  Remember, he put in NaOH, not Na(s).[/quote]

Sodium was reduced to metallic form during electrolysis (first stage of experiment). It is dissolved in mercury, so it can react with the surrounding water (or air) only on the amalgam surface. That gives is some relative stability.

Check this out: http://en.wikipedia.org/wiki/Sodium_amalgam

Best,
Borek


4.	Solve for x:5^4x - 3 * 5^2x + 1 = 250

members please submit more sums of same type with solutions
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ONLINE COACHING FOR SSC CBSE ICSE BOARD MATHS SITTING AT YOUR HOUSE
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[quote="sanbittu"]4.	Solve for x: $ 5^{4x} \minus{} 3 * 5^{2x} \plus{} 1 \equal{} 250$

members please submit more sums of same type with solutions
STUDENTS FROM INDIA
ONLINE COACHING FOR SSC CBSE ICSE BOARD MATHS SITTING AT YOUR HOUSE
FOR DETAILS MAIL sanbittu@hotmail.com[/quote]
I Latexed it for you; I hope this is the equation you wanted.  If so, then here is the solution:
[hide="Solution"]
Substitute $ 5^{2x} \equal{} y$.  Therefore, we see that $ y^2 \minus{} 3y \plus{} 1 \equal{} 250\implies y^2 \minus{} 3y \minus{} 249 \equal{} 0\implies y \equal{} \frac {3\pm \sqrt {1005}}{2}$.  We omit that the minus sign since, otherwise, $ x$ would be forced to be imaginary.  Hence,
\[ \boxed{x \equal{} \frac {\log_5{\frac {3 \plus{} \sqrt {1005}}{2}}}{2}}
\]
[/hide]


Solve the following:

y' = 4(4x + y - 2)^2 + 21


In downtown, there was a major auto accident involving 2 trucks and 6 cars. Both truck drivers were hurt and were taken to the hospital for treatment. However, none of the drivers from any of the cars were found at the accident. Why?

[hide]The trucks were towing the cars.[/hide]

Yep, that's it. (must...post...harder...riddles...)


can U tell me some sites about some known theorems in graphs which i can use in olympiad problems?????

hi,

here a french site about graphs :

[url]http://gilco.inpg.fr/~rapine/Graphe/[/url]


calculus
integration
trigonometry
calculus computations

Please show me how you would solve this problem. PLEASE!!!!

$ \int sin^2xcos^4xdx$

$ (\sin x \cos ^ 5 x)'\equal{}\cos ^6 x\minus{}5\sin ^2 x \cos ^ 4x.$

um...how did you get that?

one more time?  :blush:

$ (\sin x \cos ^ 5 x)' \equal{} \cos ^6 x \minus{} 5\sin ^2 x \cos ^ 4x.$

$ \equal{}\cos ^ 4 x\minus{}6\sin ^ 2 x\cos ^ 4 x.$

$ \therefore \int \sin ^2 x\cos ^ 4 x\ dx\equal{}\frac{1}{6}\int \cos ^ 4 x\ dx\minus{}\frac{1}{6}\sin x\cos ^ 5 x \plus{}Const.$

A more straightforward approach:  write $ \sin^2 x \equal{} \frac{1 \minus{} \cos 2x}{2}, \cos^2 x \equal{} \frac{1 \plus{} \cos 2x}{2}$ and expand.  Another substitution later...

my question is!!!! how did you even think of getting $ sinxcos^5x$?

t0rajir0u, you know what's funny?

i just remembered that and I was like, wait!! i think i found a lot simpler way

then you just stole my post :P


probability
floor function
probability and stats

There are $ k$ male and $ j$ female rabbits in a total population of $ 2n \equal{} k \plus{} j$. They are then divided into two groups each of size $ n$. Each generation, the rabbits mate with the opposite group, and $ 2$ children are born. The probability that the born child is female is $ 0.6$. At the end of a cycle both parents die, and the children are separated arbitrarily in different groups. In the next cycle the children will participate under exactly the same conditions.

What is the estimated number of cycles before all male rabbits are gone?

This is not a textbook problem, just something I've been thinking about. I'm not sure I have stated the problem correctly. But I hope you see what directions I want to go.

we have $ 2n=k+j$ so for some integer d
$ k=n+ d, j=n - d$ if these are seperated into groups A,B of size n,
$ P(r$ males in $ A)= \frac{ {}_kC{}_r {}_jC{}_{n-r}}{_{2n}C_n}=\frac{ _{n +d}C_r _{n-d}C_{n-r}}{_{2n}C_n}$
$ =P(n-r$ males in $ B)=\frac{ _{n +d}C_{n-r} _{n-d}C_r}{_{2n}C_n}=\gamma (n,r,d)$
$ =P(n-r$ females in $ A)=P(r$ females in $ B)$
lets assume d is non negative first
if r males are in group A, then the number of babies born is
$ \beta(n,r,d) =\min (r,n-r)+\min (n+d-r,r-d)$

$ E($babies born$ )=\sum_{r=0}^{n} \beta (n,r,d) \cdot \gamma(n,r,d)$

$ =2\sum_{r=1}^{d-1} \beta \cdot \gamma+2\sum_{d}^{\lfloor \frac{n}{2}\rfloor} \beta \cdot \gamma=2\sum_{d}^{\lfloor \frac{n}{2}\rfloor} [2r-d]\cdot \frac {(n+d)!(n-d)!n!n!}{r!(n+d-r)!(n-r)!(r-d)!(2n)!}$
$ =2\sum_{d}^{\lfloor \frac{n}{2}\rfloor} [2r-d]\cdot \frac {_n C_r {}_nC{}_{r+d}}{_{2n}C_{n+d}}$

closed form?

let  $ (M_1(k),M_2(k))$ be the number of Male in the group $ 1$ and $ 2$ respectively at time $ t\equal{}k$, and $ T_k \equal{} M_1(k) \plus{} M_2(k)$. Then
\[ M_{k\plus{}1}(1) \equal{} \sum_{i\equal{}1}^{A_k} B_{i,k\plus{}1,1}\]
\[ M_{k\plus{}1}(2) \equal{} \sum_{i\equal{}1}^{A_k} B_{i,k\plus{}1,2}\]
where $ B_{i,k,l}$ are independent Bernoulli random variable $ B$ with $ P[B_{i,k,l}\equal{}1]\equal{}p\equal{}0.4$ and
\[ A_k \equal{} \min(M_1(k), n\minus{}M_2(k)) \plus{} \min(M_2(k), n\minus{}M_1(k))\]
\[ A_k \equal{} T_k 1_{T_k \leq n} \plus{} (2n\minus{}T_k) 1_{T_k>n}\]
Hence $ T_k$ is a Markov process that can be described by:
\[ T_{k\plus{}1} \equal{} \sum_{i\equal{}1}^{2 A_k} B_{i,k} \]
where $ A_k\equal{}T_k$ if $ T_k \leq n$ and $ A_k \equal{} 2n\minus{}T_k$ if $ T_k > n$ and $ B_{i,k}$ are i.i.d. Bernoulli random variables.
One are trying to compute the first time that $ T_k\equal{}0$. If $ \tau\equal{}\inf\{t: T_t\equal{}0\}$ and $ E_k\equal{}E[\tau | T_0\equal{}k]$ then:
\[ E_k \equal{} 1 \plus{} \sum_{i\equal{}0}^{2n} P[T_1\equal{}i|T_0\equal{}k] E_i\]
which gives you a linear system to solve since the quantities $ P[T_1\equal{}i|T_0\equal{}k]$ are easily computable.

This is the naive brute force approach, and there must be a clever way to find the answer  :blush:


geometry
circumcircle
geometric transformation
reflection
trigonometry
ratio
homothety

Let $K (O, R)$ and $K_{A}(O_{A}, R_{A})$ be circumcircle and excircle opposite to $A$ in triangle $ABC$. Points $P$ and $Q$ are points where two external tangents touches circle $K$. Now consider families of triangles with these circumcircle and excircle and points of intersections of internal bisectors with opposite sides by $B$ and $C$. Prove that set of all such points is segment $PQ$.

P, Q are the tangency points of the common external tangents of the fixed circumcircle (O, R) and incircle $(I_{a}, r_{a})$ against A. Denote s the triangle semiperimeter and $d_{a}= OI_{a}$ the circumcenter-excenter distance. Let F, E, be the feet of the internal bisectors of $\angle C, \angle B,$ let M, N be the midpoints of AB, AC, and let S, T be the tangency points of the excircle $(I_{a})$ with AB, AC. Let $C' \in AB,\ B' \in AC$ be the reflections of C, B the external bisectors $BI_{a},\ CI_{a}$ of the angles $\angle C,\ \angle B.$ From the triangle $\triangle AC'B'$ and from the cyclic quadrilateral $AMI_{a}N,$

$\angle AC'B'+\angle AB'C' = 180^\circ-\angle A = \angle SI_{a}T = \angle OI_{a}S+\angle OI_{a}T$

Since BC = BC', CB' = CB, we have $AC' = AB+BC = c+a$ and $AB' = AC+CB = b+a$

$MS = s-\frac{c}{2}= \frac{a+b}{2},\ \ NT = s-\frac{b}{2}= \frac{a+c}{2}$

$\sin \widehat{OI_{a}S}= \frac{MS}{OI_{a}}= \frac{a+b}{2d_{a}},\ \ \sin \widehat{OI_{a}T}= \frac{NT}{OI_{a}}= \frac{a+c}{2d_{a}}$

and using the sine theorem for $\triangle AC'B',$

$\frac{\sin \widehat{OI_{a}S}}{\sin \widehat{OI_{a}T}}= \frac{a+b}{a+c}= \frac{\sin \widehat{AC'B'}}{\sin \widehat{AB'C'}}$

As both the sums and sine ratios of the angles $\angle AC'B',\ \angle AB'C'$ and $\angle OI_{a}S,\ \angle OI_{a}T$ are the same, these angles are pairwise equal, $\angle AC'B' = \angle OI_{a}S,\ \angle AB'C' = \angle OI_{a}T.$ Since the angles formed by the lines C'B', AC' and $OI_{a},\ I_{a}S$ are the same and equally oriented and since $I_{a}S \perp AC',$ it follows that $C'B' \perp OI_{a}.$ $\frac{AC'}{AB'}= \frac{a+c}{a+b}= \frac{AF}{AE},$ the triangles $\triangle AFE \sim \triangle AC'B'$ are centrally similar with similarity center A, hence their corresponding sides $EF \parallel B'C'$ are parallel. Since $B'C' \perp OI_{a},$ we also have $EF \perp OI_{a}.$ In addition, since $PQ \perp OI_{a},$ we also have $EF \parallel PQ.$

The common external tangents of $(O),\ (I_{a})$ meet at the external homothety center X of these 2 circles. Let U, V be the tangency points of XP, XQ with $(I_{a}).$ Let PQ meet $XO \equiv OI_{a}$ at Y. From the similar right angle triangles $\triangle POY \sim \triangle XOP \sim \triangle XI_{a}U,$ we obtain $\frac{OY}{OP}= \frac{OP}{OX},\ \frac{I_{a}X}{OX}= \frac{I_{a}U}{OP}.$  Assuming $r_{a}> R,$ $OY = \frac{R^{2}}{OX},\ \frac{OX+d_{a}}{OX}= \frac{r_{a}}{R},$ and eliminating OX yields $OY = \frac{R(r_{a}-R)}{d_{a}}.$ (If $r_{a}< R,$ change the sign of $d_{a}$ in the last 2 expressions.) As the vertex A of $\triangle ABC$ moves on the arc PQ of the fixed circumcircle (O) opposite to the excircle $(I_{a}),$ while AB, AC remain tangent to the fixed excircle $(I_{a}),$ EF remains parallel to PQ. When A $\longrightarrow P$ or $A \longrightarrow Q,$ $EF \longrightarrow PQ.$ The distance of these 2 lines is maximum when the triangle $\triangle ABC$ becomes isosceles with AB = AC and $BC \perp OI_{a}.$ Let the line EF of this isosceles triangle meet $OI_{a}$ at Y'. Then $AY' = AE \sin B = \frac{b^{2}\sin B}{a+b}$ and $OY' = OA-AY' = R-\frac{b^{2}\sin B}{a+b}.$ Using $r_{a}-R = R(2\cos B-\cos A)$ and $d_{a}= r_{a}+R \cos A = R (2 \cos B+1),$ we can rewrite the formula for OY as

$OY = \frac{R(r_{a}-R)}{d_{a}}= R\ \frac{2 \cos B-\cos A}{2 \cos B+1}= R \left(1-\frac{1+\cos A}{2 \cos B+1}\right) =$

$= R\ \left(1-\frac{2\cos^{2}\frac{A}{2}\sin B}{\sin 2B+\sin B}\right) = R \left(1-\frac{2\sin^{3}B}{\sin A+\sin B}\right) = R-\frac{b^{2}\sin B}{a+b}= OY'$

Thus the points $Y' \equiv Y$ are identical which means that the line $EF \parallel PQ$ does not move, so that the lines $EF \equiv PQ$ are also identical at all times.

A synthetic solution can be found here:
http://olympiads.mccme.ru/lktg/2006/6/6-1en.pdf
http://olympiads.mccme.ru/lktg/2006/6/6-3en.pdf

Thanks for the links. These also provide alternative synthetic proofs for the following problems:

1. [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=107271]Collinear with IO[/url]
2. [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=105048]The line through the incenter and the circumcenter[/url]
3.[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=14655]OI = DE[/url]

I"ll present soon a "long" metrical proof of the following problem :

[color=darkred][b]A related problem.[/b] Let $w_{0}=C(I,r)$ , $w=C(O , R)$ and $w_{a}=C(I_{a}, r_{a})$ be the incircle, circumcircle and the $A$- excircle for the triangle $ABC$.

Denote $B_{1}\in BI\cap AC$ and $C_{1}\in CI\cap AB\ .$ Points $P$ and $Q$ are points where the common external tangents of the circles $w$ , $w_{a}$ touch the circle $w$.

Prove that : $\{\begin{array}{cc}1\blacktriangleright & I_{a}Q^{2}-I_{a}I^{2}=2\cdot(OQ^{2}-OI^{2})\ .\\\\ 2\blacktriangleright & B_{1}C_{1}\equiv PQ\ .\end{array}$[/color]

[quote="Virgil Nicula"][color=darkred][b]A related problem.[/b] Let $w_{0}=C(I,r)$ , $w=C(O , R)$ and $w_{a}=C(I_{a}, r_{a})$ be the incircle, circumcircle and the $A$- excircle for the triangle $ABC$.

Denote  $B_{1}\in BI\cap AC$ and $C_{1}\in CI\cap AB\ .$ Points $P$ and $Q$ are points where the common external tangents of the circles $w$ , $w_{a}$ touch the circle $w$.

Prove that : $\{\begin{array}{cc}1\blacktriangleright & I_{a}Q^{2}-I_{a}I^{2}=2\cdot(OQ^{2}-OI^{2})\ .\\\\ 2\blacktriangleright & B_{1}C_{1}\equiv PQ\ .\end{array}$ [/color][/quote]

[color=darkblue]$1\blacktriangleright$ Choose the tangent $QT\ ,$ where $T\in w_{a}\ .$ Thus, $I_{a}T=r_{a}$ , $OQ=R$ , $IA=\frac{bc(p-a)}{p}$ , $OI^{2}=R^{2}-2Rr$ , $OI^{2}_{a}=R^{2}+2Rr_{a}\ .$

$\{\begin{array}{c}\frac{AI}{p-a}=\frac{AI_{a}}{p}=\frac{II_{a}}{a}\Longrightarrow II_{a}=\frac{a}{p-a}\cdot AI\Longleftrightarrow II^{2}_{a}=\frac{a^{2}}{(p-a)^{2}}\cdot \frac{bc(p-a)}{p}=\frac{a\cdot abc}{p(p-a)}=\frac{a\cdot 4Rpr}{p(p-a)}\Longrightarrow\boxed{\ II^{2}_{a}=\frac{4aRr}{p-a}\ }\ .\\\\ I_{a}Q^{2}=I_{a}T^{2}+TQ^{2}=r^{2}_{a}+OI^{2}_{a}-(r_{a}-R)^{2}=r^{2}_{a}+R^{2}+2Rr_{a}-(r^{2}_{a}-2Rr_{a}+R^{2})\Longrightarrow\boxed{\ I_{a}Q^{2}=4Rr_{a}\ }\ .\\\\ I_{a}Q^{2}-I_{a}I^{2}=4Rr_{a}-\frac{4aRr}{p-a}=\frac{4RS-4aRr}{p-a}=\frac{4Rpr-aRr}{p-a}\Longrightarrow\boxed{\ I_{a}Q^{2}-I_{a}I^{2}=4Rr\ }\ .\\\\ 2\cdot(OQ^{2}-OI^{2})=2[R^{2}-(R^{2}-2Rr)]\Longrightarrow\boxed{\ 2\cdot (OQ^{2}-OI^{2})=4Rr\ }\ .\end{array}$

$2\blacktriangleright$ Suppose w.l.o.g. $b>c\ .$ Denote : the middlepoint $M$ of the side $[BC]$ ; the point $N\in BC\cap w_{a}$ ; the intersection $S\in B_{1}C_{1}\cap BC\ .$

$\{\begin{array}{c}MB=\frac{a}{2}\ ,\ NC=p-b\\\\ \frac{SB}{c}=\frac{SC}{b}=\frac{a}{b-c}\end{array}\Longrightarrow$ $\{\begin{array}{c}SM=SB+BM\Longrightarrow\boxed{\ SM=\frac{a(b+c)}{2(b-c)}\ }\\\\ SN=SC-NC\Longrightarrow\boxed{\ SN=\frac{(b-c)^{2}+a(b+c)}{2(b-c)}\ }\end{array}\ .$

From the above relations obtain $QO^{2}-QI_{a}^{2}=R^{2}-4Rr_{a}\ \ (1)\ .$ Therefore, $\boxed{\ B_{1}C_{1}\equiv PQ\Longleftrightarrow SQ\perp OI_{a}\Longleftrightarrow QO^{2}-QI_{a}^{2}=SO^{2}-SI_{a}^{2}\ }\ .$

$SO^{2}=SM^{2}+OM^{2}=SM^{2}+R^{2}-MC^{2}=$ $R^{2}+(SM-MC)(SM+MC)=R^{2}+SB\cdot SC$ $\Longrightarrow\boxed{\ SO^{2}=R^{2}+\frac{a^{2}bc}{(b-c)^{2}}\ }\ .$

$SI_{a}^{2}=SN^{2}+NI_{a}^{2}=$ $\frac{[(b-c)^{2}+a(b+c)]^{2}}{4(b-c)^{2}}+r_{a}^{2}$ $\Longrightarrow SO^{2}-SI_{a}^{2}=$ $R^{2}-r_{a}^{2}+\frac{a^{2}bc}{(b-c)^{2}}-\frac{(b-c)^{4}+2a(b+c)(b-c)^{2}+a^{2}(b+c)^{2}}{4(b-c)^{2}}=$

$R^{2}-r_{a}^{2}+\frac{4a^{2}bc-(b-c)^{4}-2a(b+c)(b-c)^{2}-a^{2}(b+c)^{2}}{4(b-c)^{2}}$ $\implies$ $\boxed{\ SO^{2}-SI^{2}_{a}=R^{2}-r_{a}^{2}-\frac{a^{2}+2a(b+c)+(b-c)^{2}}{4}\ }\ \ (2)\ .$

$SO^{2}-SI_{a}^{2}=QO^{2}-QI_{a}^{2}$ $\Longleftrightarrow$ $4Rr_{a}-r^{2}_{a}=\frac{a^{2}+2a(b+c)+(b-c)^{2}}{4}\ \bullet\ |\ (p-a)$ $\Longleftrightarrow$ $16RS-4Sr_{a}=(p-a)[a^{2}+2a(b+c)+(b-c)^{2}]$ $\Longleftrightarrow$

$4abc-4p(p-b)(p-c)=(p-a)[a^{2}+2a(b+c)+(b-c)^{2}]$ $\Longleftrightarrow$ $8abc-(a+b+c)(a+b-c)(a-b+c)=$ $(b+c-a)[a^{2}+2a(b+c)+(b-c)^{2}]$

$\Longleftrightarrow$ $8abc-a^{3}+2a(b-c)^{2}-a^{2}(b+c)+(b+c)(b-c)^{2}=$ $a^{2}(b+c)-a^{3}+2a(b+c)^{2}-2a^{2}(b+c)+(b+c)(b-c)^{2}-a(b-c)^{2}$ $\Longleftrightarrow$

$8abc+2a(b-c)^{2}=2a(b+c)^{2}$ $\Longleftrightarrow$ $4bc+(b-c)^{2}=(b+c)^{2}\ ,$ what is truly. In conclusion, $SQ\perp OI_{a}\ ,$ i.e. $B_{1}C_{1}\equiv PQ\ .$

[b]Remark.[/b] Otherwise, can prove directly that $B_{1}Q\perp OI_{a}\ ,$ i.e. $B_{1}I_{a}^{2}-B_{1}O^{2}=QO^{2}-QI_{a}^{2}\ .$ Indeed, $B_{1}T=B_{1}C+CT=\frac{ba}{a+c}+(p-b)$ $\implies$

$\boxed{\ B_{1}T=\frac{(a+c)^{2}+b(a-c)}{2(a+c)}\ }$ and $B_{1}O^{2}=R^{2}-B_{1}A\cdot B_{1}C\ ,$ i.e. $\boxed{\ B_{1}O^{2}=R^{2}-\frac{b^{2}ac}{(a+c)^{2}}\ }\ .$ Therefore, $B_{1}I_{a}^{2}=I_{a}T^{2}+B_{1}T^{2}\ ,$ i.e.

$\boxed{\ B_{1}I_{a}^{2}=r_{a}^{2}+\frac{[(a+c)^{2}+b(a-c)]^{2}}{4(a+c)^{2}}\ }\ .$ Obtain $\boxed{\ B_{1}I_{a}^{2}-B_{1}O^{2}=r_{a}^{2}-R^{2}+\frac{a^{2}+2a(b+c)+(b-c)^{2}}{4}\ }\ .$ In conclusion,

$QI_{a}^{2}-QO^{2}=B_{1}I_{a}^{2}-B_{1}O^{2}\ ,$ i.e. $B_{1}Q\perp OI_{a}$ is truly because the following identity is truly : $\boxed{\ 4r_{a}^{2}+a^{2}+2a(b+c)+(b-c)^{2}=16Rr_{a}\ }\ .$

[b]Remark.[/b] I am tired out and find that I have "callosities ", even "wound" at the second finger ! [/color]

nice problem!!

[b]Problem.[/b]
[color=blue]let $\omega$ be circumcircle of triangle $ABC$ and $P,Q$ be points whre common tangent of $\omega$ and A-excircle intersect $\omega$,let $BI\cap AC\equiv B_{1}\ ,\ CI\cap AB\equiv C_{1}$ .
prove that $P,Q,B_{1},C_{1}$ are collinear. [/color]

[i]proof[/i].
let $S$ be intersection point of A-excircle with circumcircle of $ABC$(the one which is nearer to $B$).let $I_{a}S$ intersect $\omega$ at $R$ and $RO$(O is center of $\omega$) intersect $\omega$ at $T$ let $H$ be foot of perpendicular to a line that is tangent to $\omega$ at $T$.(let $I_{a}H=x$)

we have:

$I_{a}S.I_{a}R=r_{a}.I_{a}R=OI_{a}^{2}-R^{2}=2Rr_{a}\rightarrow I_{a}R=2R\ ,\ RS=2R-r_{a}$ now we calculate $TH$ in two ways:

$(1)$ :  $TH^{2}=TI_{a}^{2}-x^{2}=TS^{2}+r_{a}^{2}-x^{2}=4R^{2}-(2R-r_{a})^{2}+r_{a}^{2}-x^{2}=4Rr_{a}-x^{2}$

$(2)$ : let $H'$ be foot of perpendicular line from $O$ to $I_{a}H$ : $TH^{2}=OH'^{2}=R^{2}+2Rr_{a}-(x-R)^{2}=2Rr_{a}+2xR-x^{2}$
from $(1)$ and $(2)$ w have : $2Rr_{a}+2xR=4Rr_{a}\Longrightarrow x=r_{a}$ hence $T\equiv P$ and PH(TH) is tangent to the A-excirlce.

notice that  $PR=RI_{a}=2R$ let $I_{a}P$ intersect $\omega$ at $X$ cause $RX\perp I_{a}P$ hence $PX=I_{a}X$ like this we have midpoint of $I_{a}Q$ lie on the circumcircle of triangle $ABC$ ,also we know that midpoints of $I_{b}I_{a}$ and $I_{a}I$ lie on the $\omega$, hence a homothecy wrt $I_{a}$ with ratio $2$ makes $\omega$ to the circumcircle of $I_{c},Q,I,P,I_{b}$.

now consider $B_{1}C_{1}$ intersect $\omega$ at $P'$ and $Q'$ , $AICI_{b}$ is a cyclic quadrilateral hence $IB_{1}\cdot B_{1}I_{b}=AB_{1}\cdot B_{1}C=B_{1}Q'\cdot B_{1}P'$ hence $I,P',Q',I_{b}$ are on a circle like this we have quadrilateral $P'IQ'I_{c}$ is cyclic , hence 5 points $I_{c}Q'IP'I_{b}$ lie on a circle.

hence $P'\equiv P\ ,\ Q'\equiv Q$ so we are done.

[img]http://image03.webshots.com/3/4/74/29/2173474290098227821yDFHzD_fs.jpg[/img]


How many positive three-digit numbers can be formed such that the sum of the three digits is a prime number less than 10?

First, consider the cases where the sum of the digits is either $ 2, 3, 5, \text{or } 7$
$ \text{Case 1: Sum of digits is 2}:$
$ 101, 110, 200$

$ \text{Case 2: Sum of digits is 3}:$
$ 102, 120, 201, 210, 300$

$ \text{Case 3: Sum of digits is 5}:$
$ 104, 203, 302, 401, 500, 140, 230, 320, 410, 500$

$ \text{Case 4: Sum of digits is 7}:$
$ 106, 205, 304, 403, 502, 601, 160, 250, 340, 430, 520, 610, 115, 151, 124, 133, 142, 223, 232, 214, 241, 313, \newline331,
322, 412, 421, 511$

From this, we can see there are $ \boxed{45}$ 3-digit numbers.

(can anyone find a solution that doesn't involve listing out all the numbers? :oops: because I can't think of one)

I can't think of one other than using some quick counting and casework, like you did.  This is quite hard, because you are still missing some, and I'm getting kind of dizzy thinking of these  :rotfl: Here are the ones I've found that you are missing:
In case $ 2$, we can have the number $ 111$.
In case $ 3$, we can have the numbers $ 113, 131, 311, 122, 212, 221$
These are all of the ones I've found that you don't have...

[hide=Faster]
We have $a+b+c = n$, where $a \ge 1$, $b\ge 0$, $c\ge 0$, and $n=2,3,5,7$. Let $b'=b+1$ and $c'=c+1$. Then $a+b'+c'=n+2$ in positive integers. By stars and bars, there are $\binom{n+1}{2}$ solutions. Summing over $2,3,5,7$ we have $\binom{3}{2} + \binom{4}{2} + \binom{6}{2} + \binom{8}{2} = 52$.
[/hide]


[quote="Chinaboy"]Many of you are missing questions because you give what the question is NOT asking for.

For example, sometimes you give 1:9 when they ask for the answer in form of common fraction.

Sometimes you give the coordinates when the question asks for the sum of the coordinates.

So, underline what the question is asking! When you read the question, underline the important parts of the question![/quote]
Good tip, Chinaboy!
I'll remember that!

[quote="Chinaboy"]Many of you are missing questions because you give what the question is NOT asking for.

For example, sometimes you give 1:9 when they ask for the answer in form of common fraction.

Sometimes you give the coordinates when the question asks for the sum of the coordinates.

So, underline what the question is asking! When you read the question, underline the important parts of the question![/quote]
This looks like it mainly pertains to my case....

It pertains to everyone. When I'm grading your tests, I get very angry if I see one of those mistakes. So please, follow them? It'll help both you and me. :)


function
number theory unsolved
number theory

Find the least number which is divisible by 2009 and its sum of digits is 2009.

$ 4998\underbrace{9\cdots 9}_{170's}8\underbrace{9\cdots 9}_{49's}$

Could anybody give a solution for this problem? It is quite interesting, I think :oops:

well, we can easily see that this least number requires at least $ 224$ digits.

now we have to check conditions for a number to be divisible by  $ 287$   ($ 2009\equal{}7*287$)

[quote="babyandprettyboy"]Can you solve other problem "Find the least number which is divisible by n and its sum of digits is n.(n>1)."?[/quote] $ 2, n\equal{}2$  :rotfl:

[quote="Mashimaru"][quote="Stephen"][quote="v235711"]now we have to check conditions for a number to be divisible by  $ 287$   ($ 2009 \equal{} 7*287$)[/quote]

Well, I have to admit that feizhenpeng's answer is 224 digits. :oops:[/quote]

Right, but now how will you prove that the evil number is divisible by $ 2009$, i.e. by $ 41$ and $ 49$ :blush: . Beside, I also admit that the sum of the digits of this number is $ 2009$  :lol:[/quote]

I just found a divisibility rule regarding the number $ 41$.A number will be divisible by $ 41$ if and only if the number created by removing it´s last digit ,multiplying this last digit by $ 4$  and subtract it from our remaining number is divisible by $ 41$.for example:

$ a_1a_2...a_m$ will be divisible by $ 41$ if and only if $ 41$ divides $ a_1a_2...a_{m\minus{}1}\minus{}4a_m$.
easy to prove, simply use congruence.

now it rests to use this divisibility criteria to find this least number. :wink:

[quote="v235711"][quote="Stephen"][quote="babyandprettyboy"]Can you solve other problem "Find the least number which is divisible by n and its sum of digits is n.(n>1)."?[/quote]

Hey, my friend, isn't that obvious?[/quote]

 would you mind posting it´s solution?[/quote]

OK. Now, if we take n=2, 2's sum of digits is n=2, which is divisible by n=2. So the answer is 2, 2.

[quote="Stephen"][/quote][quote="v235711"][quote="Stephen"][quote="babyandprettyboy"]Can you solve other problem "Find the least number which is divisible by n and its sum of digits is n.(n>1)."?[/quote]

Hey, my friend, isn't that obvious?[/quote]

 would you mind posting it´s solution?[/quote][quote="Stephen"]

OK. Now, if we take n=2, 2's sum of digits is n=2, which is divisible by n=2. So the answer is 2, 2.[/quote]

he did not mean to find the least number of all the numbers . he meant to find the least number
in function of a given $ n$.

[quote="v235711"][/quote][quote="Stephen"][/quote][quote="v235711"][quote="Stephen"][quote="babyandprettyboy"]Can you solve other problem "Find the least number which is divisible by n and its sum of digits is n.(n>1)."?[/quote]

Hey, my friend, isn't that obvious?[/quote]

 would you mind posting it´s solution?[/quote][quote="Stephen"]

OK. Now, if we take n=2, 2's sum of digits is n=2, which is divisible by n=2. So the answer is 2, 2.[/quote][quote="v235711"]

he did not mean to find the least number of all the numbers . he meant to find the least number
in function of a given $ n$.[/quote]

Ah! Sorry I didn't understand the question properly so much. :blush:

[quote="Rust"]
Consider number $ a = 3*10^{223} - 1 = 498\mod 2009$.
Consider $ a + 10^{223} - 10^n = 1334 - 10^n\mod 2009$.
$ 1334 = 22\mod \41$ and $ 10^n\not = 22\mod 41$ for any n.
Therefore minimal number is $ 5*10^{223} - 10^n - 10^m - 1$ and $ 10^n + 10^m = 161\mod 2009$.
[/quote]

Why? I think you are talking about the minimal number with the form you submitted, there may be other forms :maybe:

Adaptation problem:

[b]Find the least number which is divisible by 209 and its sum of digits is 209.[/b]


The number of 4000 digit palindromes can be expressed as $ p \times n^m$  what is the value of $ p \plus{} n \plus{} m$

[hide]
use patterns

1-digit 1-9 is 10^0 so 1*9=9

2-digit 11-99 is 10^0 so 1*9=9

3 digit 101-191 is 10^1 but all the way from 101-901 so 10*9 or 90.
4 digit is 1001-1991 is 10^1 and then 1001-9009 so 10*9 or 90

5 digit is 10001 to 19991 so 10^2, but 10001 to 90009 so 100*9 or 900
6 digit is 100001 to 199991 so 10^2, but 100001 to 900009 so 100*9 or 900

As you can see, the pattern is, if the amount of digits is even, minus 2, if the digits are odd, minus 1. The number you get is the power of 10 that you have. So for example, 6 digit. 6-2=4 4/2=2 so it is 10^2 or 100
Then the result you get is multiplied by 9. So
10^2*9 is the number of 6-digit palindromes.

So find the number of palinfromes with a 4000 digit number.

So you get 4000-2 since it is even which equals 3998.

3998/2=1999.  So 10^1999 which time 9 is  9 x 10^1999 so the answer is

9+10+1999 or 2018
[/hide]

Edit:
Hexa, your solution isn't mathcounts level, use a simpler one.

first digit can be 1-9,

digits 2001-4000 must match digits 1-2000, so 9*10^1999


Αν εχουμε το συστημα ανισοτητων:

$ a_1\cdot X_1\geq X_{1}\plus{}X_{2}\plus{}...\plus{}X_{n}$
$ a_2\cdot X_2\geq X_{1}\plus{}X_{2}\plus{}...\plus{}X_{n}$
$ .............................................$
$ .............................................$
$ a_n\cdot X_n\geq X_{1}\plus{}X_{2}\plus{}...\plus{}X_{n}$
Με $ a_1,a_2,...,a_n >1$

Πως μπορουμε να λυσουμε το συστημα σε συναρτηση με το $ X_1$?
Αν ξερουμε και οτι $ \frac{1}{a_1}\plus{}\frac{1}{a_2}\plus{}...\plus{}\frac{1}{a_n}\leq 1$
(αυτο αποδεικνυεται οτι ειναι απαραιτητο για να λυθει το συστημα)

[u]Ζηταω λυσεις του τυπου(δεν ειναι τοσο περιπλοκο το τι ζηταω οσο φαινεται:[/u]
$ f_1(X_1)<X_2<g_1(X_1)$
$ f_2(X_1,X_2)<X_3<g_2(X_1,X_2)$
$ ..................................$
$ f_k(X_1,X_2,...,X_{n\minus{}1})<X_n<g_k(X_1,X_2,...,X_{n\minus{}1})$

Οπου $ f_k(X_1),g_k(X_1)$ συναρτησεις του $ X_1,X_2,...,X_n$ [b]αλλα και φυσικα των $ a_1,a_2,...,a_n$ επισης.[/b]

[u]ή αν γινεται ακομα καλυτερα λυσεις του τυπου:[/u]
$ f_1(X_1)<X_2<g_1(X_1)$
$ f_2(X_1)<X_3<g_2(X_1)$
$ ..................................$
$ f_k(X_1)<X_n<g_k(X_1)$
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

Πχ αν εχουμε το συστημα:
$ a_1\cdot X\geq X\plus{}Y\plus{}Z$
$ a_2\cdot Y\geq X\plus{}Y\plus{}Z$
$ a_3\cdot Z\geq X\plus{}Y\plus{}Z$

Με $ a_1,a_2,...,a_n >1$ και $ \frac{1}{a_1}\plus{}\frac{1}{a_2}\plus{}\frac{1}{a_3}\leq 1$

Τοτε αυτο συνεπαγεται(καλο θα ηταν να εχουμε και ισοδυναμια, και ισως εχουμε):

$ \frac{a_3}{a_2a_3\minus{}a_2\minus{}a_3}\cdot X< Y < \frac{a_1a_3\minus{}a_1\minus{}a_3}{a_3}\cdot X$

$ \frac{a_2}{a_2a_3\minus{}a_2\minus{}a_3}\cdot X< Z < \frac{a_1a_2\minus{}a_1\minus{}a_2}{a_2}\cdot X$

Αλλα πως γενικευεται αυτο για ΝxΝ εξισωσεις?

[quote="maniopas"]Αρκεί να βρεις τις ρίζες του συστήματος:
$ R_1 \equal{} X_{1}(1 \minus{} a_1) \plus{} X_{2} \plus{} ... \plus{} X_{n}$
$ R_2 \equal{} X_{1} \plus{} X_{2}(1 \minus{} a_2) \plus{} ... \plus{} X_{n}$
$ .................................................................$
$ .................................................................$
$ R_n \equal{} X_{1} \plus{} X_{2} \plus{} ... \plus{} X_{n}(1 \minus{} a_n)$
Συναρτήσει των $ a_1, a_2, ... , a_n$ και $ R_1, R_2, ... , R_n$
(Θα βγουν οπωσδήποτε όλες γιατί είναι n x n σύστημα).

Θα βγάλεις ανισότητες χρησιμοποιώντας ότι $ R_1, R_2, ... , R_n < \equal{} 0$ και τις άλλες συνθήκες που έδωσες.[/quote]

Εστω οτι τις εβγαλα. Πχ:
$ X_1 \equal{} f_1(R_1,R_2,...,R_n,a_1,a_2,....,a_n)$
$ X_2 \equal{} f_2(R_1,R_2,...,R_n,a_1,a_2,....,a_n)$
$ ................................................$
$ X_n \equal{} f_n(R_1,R_2,...,R_n,a_1,a_2,....,a_n)$

Τωρα τι κανω? Πχ για το 3x3 παραδειγμα που εδωσα αν λυσουμε ως προς X,Y,Z βγαινει:
••••••••••••••••••••••••••••••••••••••••••••
$ X \equal{} \frac {a_3 R_2 \plus{} a_2R_3}{a_2 a_3 \plus{} a_1 ( \minus{} a_3a_2 \plus{} a_2 \plus{} a_3)}$

$ Y \equal{} \frac {a_3R_2 \minus{} a_1 ((a_3 \minus{} 1)R_2 \plus{} R_3)}{ \minus{} a_1 a_2 \minus{} a_3a_2 \plus{} a_1 (a_2 \minus{} 1) a_3}$

$ Z \equal{} \frac {a_1(R_2 \plus{} (a_2 \minus{} 1) R_3) \minus{} a_2R_3}{a_2 a_3 \plus{} a_1 ( \minus{} a_3a_2 \plus{} a_2 \plus{} a_3)}$
••••••••••••••••••••••••••••••••••••••••••••

Πως τωρα μπορουμε να χρησιμοποιησουμε τις:
$ R_1\leq 0$
$ R_2\leq 0$
$ R_3\leq 0$

για να βγαλουμε τις λυσεις που ειχα δωσει πιο πριν, δηλαδη:

$ \frac {a_3}{a_2a_3 \minus{} a_2 \minus{} a_3}\cdot X < Y < \frac {a_1a_3 \minus{} a_1 \minus{} a_3}{a_3}\cdot X$

$ \frac {a_2}{a_2a_3 \minus{} a_2 \minus{} a_3}\cdot X < Z < \frac {a_1a_2 \minus{} a_1 \minus{} a_2}{a_2}\cdot X$

Thanks anyway! :)

Είσαι σίγουρος ότι δεν έκανες κάπου λάθος; Κανονικά τα X, Y, Z πρέπει να παρουσιάζουν κυκλικότητα, αφού και οι αρχικές σχέσεις είναι κυκλικές. Ανεξάρτητα από αυτό πάντως:

Από αυτό το σημείο και μετά μπορείς να χρησιμοποιήσεις οποιαδήποτε ανισότητα θέλεις (από τις γνωστές με δεδομένο ότι $ R_1 < \equal{} 0$).  Το πιο πιθανό είναι να βρεις μια σχέση ακόμα πιο ισχυρή κι από αυτή που βρήκες εσύ.

π.χ. αν αντικαταστήσεις στη σχέση που βρήκε για το X ότι $ R_1 < \equal{} 0, R_2 < \equal{} 0$ βγαίνει $ X < \equal{} 0$ που είναι μια υπερβολικά χαλαρή ανισότητα, αν αντικαταστήσεις (επειδή είναι αρνητικά τα X, Y, Z) $ \minus{} R_1 > \equal{} 3(( \minus{} X(1 \minus{} a_1))( \minus{} Y)( \minus{} Z))^{\frac {1}{3}} < \equal{} > R_1 < \equal{} \minus{} 3(( \minus{} X(1 \minus{} a_1))( \minus{} Y)( \minus{} Z))^{\frac {1}{3}}$ και το ίδιο για το $ R_2$ θα βρεις κάτι πιο ισχυρό.

Όσο πιο ισχυρή ανισότητα χρησιμοποιήσεις, τόσο πιο ισχυρό αποτέλεσμα θα βρεις τελικά.


Find all $ x,y\in N^*$ satisfy $ \frac{x^3\plus{}x}{xy\minus{}1} \in N^*, \quad N^*\equal{}N|\{0\}$

Perhaps I'm being a bit hasty but:

[hide=" click to reveal hidden content"]

$ x \mid x^{3} \plus{} 1$ and $ x \mid xy$ so $ x \mid 1$ so $ x \equal{} 1$

So $ \frac {x^{3} \plus{} x}{xy \minus{} 1} \equal{} \frac {2}{2y \minus{} 1}$

so $ y \equal{} 1$.

So there are no solutions.

[/hide]

[quote="AndrewTom"]Perhaps I'm being a bit hasty but:

[hide=" click to reveal hidden content"]

$ x \mid x^{3} \plus{} 1$ and $ x \mid xy$ so $ x \mid 1$ so $ x \equal{} 1$

So $ \frac {x^{3} \plus{} x}{xy \plus{} 1} \equal{} \frac {2}{2y \plus{} 1}$

so $ y \equal{} 0$.

So there are no solutions.

[/hide][/quote]

With $ x\equal{}3,y\equal{}1$ the equation is satisfy!

[quote="AndrewTom"]Yes, I've done something silly (again): $ (x, y) \equal{} (2, 1), (4, 1), ...$ work as well.[/quote]

Can you given the rule to find $ (x,y)$?

[quote="AndrewTom"]
$ x \mid x^{3} \plus{} 1$ and $ x \mid xy$ so $ x \mid 1$ so $ x \equal{} 1$
[/quote]

Why??  :)

Hi. As I explained above, I made a couple of silly mistakes, besides writing + instead of -.

I also meant to write  that some solutions are

$ (x, y) \equal{} (1, 2), (2, 1,), (3, 1), (3, 2)$ but got confused.

Sorry. I'll  try again later.


calculus
derivative
function
parameterization
calculus computations

Given a differential equation of the form $y''+p(x)y'+q(x)y = r(x)$ with a corresponding homogenous solution of $y_h = C_1y_1+C_2y_2$ and a particular solution of $y_p = u_1(x)y_1+u_2(x)y_2$, why is it "ok" to make the assumption $u_1'y_1+u_2'y_2=0$?  I know that it makes the derivatives easier to deal with,  but why can we do this? Thanks in advance.

Once you have some representation $y_p=u_1y_1+u_2y_2$, you can add $zy_2$ to $u_1$ and subtract $zy_1$ from $u_2$ where $z$ is an arbitrary function. That won't change $y_p$ but the sum $u_1'y_1+u_2'y_2$ will change by $z(y_2'y_1-y_1'y_2)$, i.e., by $zW(y_1,y_2)$. Since the Wronskian $W(y_1,y_2)$ does not vanish, we see that we may change this sum by any function we want. In particular, if we have any representation, we can get a representation with $u_1'y_1+u_2'y_2=0$.

Another way to look at it is an idea that goes back at least as far as the Tartaglia [i]et al.[/i] solution to the cubic equation. We have a single equation for a single unknown. In this case, that unknown is $y_p,$ the particular solution. We arrange to write this single unknown in terms of [i]two[/i] new unkonws - in this case the "paramaters" $u_1$ and $u_2.$ But now we've introduced an extra degree of freedom, and hence we presumably have not a single solution but a whole one-parameter family of solutions. That gives us the opportunity to take away that one excess degree of freedom by imposing one condition - and we get to choose that condition.

Stated that way, this is far from foolproof. One equation for two unknowns usually has a one-parameter family of solutions - but that rule has exceptions. Think of that family as a curve; for us to impose an arbitrary other solution is to find another curve and ask for the intersection. That intersection might fail to exist, or there might be multiple intersections.

In the case of variation of parameters, the true justification is [i]a posteriori[/i]. We try something that might or might not work, and find after the fact that it does work and we do obtain a solution. We then invoke the existence-uniqueness theorem for differential equations to claim that we have found everything.


Hello,

I have an eps figure.
I want to use it in an asymptote picture.
How do I this?
(It is a  picture of a King to be used in playing cards)

Sirinial

Hi,

Take a look to the figure 0157: http://piprim.tuxfamily.org/asymptote/generale/index.html#fig0157


This is very difficult problem ...  :P
The maximum value of above expression is
 \[\frac{11\sqrt{33}-45}{24}\]
And the equal hold on:
\[a=b=\frac{\sqrt{33}-3}{2},c=6-\sqrt{33}\]
I am waiting a nice proof .......


At the Marian County Antique Show, 202 tickets were sold on
Friday, 357 were sold on Saturday, and 408 were sold Sunday.
How many dollars were collected if the Friday tickets cost \$3
each, the Saturday tickets cost \$4 each, and the Sunday tickets
cost \$5 each?

you just evaluate:
$ 202\cdot 3\plus{}357\cdot 4\plus{}408\cdot 5\equal{}606\plus{}1428\plus{}2040\equal{}\boxed{4074}$

Cost of Friday's Tickets: $ 3 \times 202 \equal{} 606$

Cost of Saturday's Tickets: $ 4 \times 357 \equal{} 1428$

Cost of Sunday's Tickets: $ 5 \times 408 \equal{} 2040$

Adding these up gives us $ \boxed{4074}$.

EDIT: Beaten


geometry
geometric transformation
reflection

The mortgage on a house is now $ \$$517.50 per month. That amount reflects a 15$ \%$ increase over the last year's monthly payment. By how many dollars did the payment increase from last year?

We have \[ 517.50\equal{}1.15x\Rightarrow x\equal{}450\] so the payment increased by $ 517.50\minus{}450\equal{}\boxed{67.50}$


I am kinda new to LaTex so please bear with me if I don't make sense.

When ever I try to "build the output" it goes through the regular process and says that i have no errors, boxes, ect. Then I try to view the output and it gives me an error message. It looks like:

[img]http://i487.photobucket.com/albums/rr233/iceflame012/Capture-1.jpg[/img]

I guessing it might have to do with me uninstalling my adobe acrobat.
Suggestions on how to fix this would help.. obviously

thanks

P.S. I deleted it because it came with the computer and kept on asking for an activation key and would freeze up and pdf's I opened.

You need to install [url=http://www.adobe.com/products/acrobat/readstep2.html]Adobe Reader[/url] (free, no activation key required) or another pdf reader* in order to view the output.  When you have done this you must tell TexnicCenter where to find it by going to Build, Define Output Profiles, Viewer and replace C:\PROGRAM FILES\ADOBE\ACROBAT 8.0\ACROBAT\ACROBAT.EXE by the new path.

* Some LaTex users like [url=http://blog.kowalczyk.info/software/sumatrapdf/]Sumatra PDF[/url] but there are other ones like [url=http://www.foxitsoftware.com/pdf/rd_intro.php]Foxit[/url] out there.


calculus
limit
function
real analysis
calculus computations

Is real analysis just a rigorous version of calculus? I plan on learning some calculus over the summer. Should I get a real analysis book or a calculus book?

A first semester Real Analysis course in college typically devotes most of its time to proving (with perfect rigor) all of the theorems from single variable Calculus.

Personally, I think learning Calculus in full rigor the first time through could potentially make some things less clear.

For example, I think it's possible (even easy) to have a very clear intuitive idea of what it means to say $\lim_{x \to a} f(x) = L$ without ever having heard anything about $\epsilon$ or $\delta$.  Many of the basic properties of limits are highly intuitive, and giving strange $\epsilon-\delta$ proofs of these facts may actually make the whole subject less clear.

So the first time someone learns Calculus, I think it's probably best to just assume a lot of facts about limits because they are intuitively very clear and highly plausible, rather than try to prove them all rigorously.

Also, there are certain theorems about continous functions which are very clear intuitively, to the point where they may seem obvious, and yet they are surprisingly difficult to prove with complete rigor.  For example, a continous function on a closed interval achieves a maximum value.  Draw a few pictures and you will probably be absolutely convinced that this is true, on intuitive grounds.  The first time you learn Calculus, I'd say it's better to just assume that theorem is true, and not worry about proving it rigorously.  The Intermediate Value Theorem is another example of a theorem like this that seems pretty obvious and yet is kind of difficult to prove rigorously.

Calculus is filled with a lot of facts that are intuitively obvious or highly plausible and yet are perhaps kind of hard to prove rigorously.  Another example is the fact that a continuous function on a closed interval is Riemann integrable.  Newton didn't know how to prove this with modern standards of rigor, but he knew with certainty that it was true, or let's say that it's such a highly plausible fact that he didn't worry about whether or not it was true.  Should you worry about rigorously proving a theorem like this the first time you learn Calculus?  I don't think so... I think that could make the subject unnecessarily confusing.

However, if you do want to see a rigorous treatment of Calculus your first time through, the book Calculus by Spivak was written for just that purpose.  (Although Spivak was probably assuming his readers would have seen Calculus already in high school.)  It's a very good book.

I disagree with DVO. The theory and proofs in one variable analysis are really very simple.  I'd say a completely rigorous approach to multivariable calculus requires at least ten times the effort. I see no reason why you should learn the theorems first, and how to use them, and then learn how to prove them in a later course. It doesn't sound very time efficient. I think that everything should be proved the first time you study one variable calculus; it just seems like it will be more difficult to learn rigour in a seperate course afterwards. In order to learn how to prove things I think seeing classic proofs is vital.

The point you bring up about Newton just illustrates that mathematics really wasn't precise or careful until those pesky germans made it rigorous. A lot of intuitive things were considered to be true, some of which we know now are false.

Well, it partly depends upon how talented the student is, and how much he cares about rigor.  Breez is probably a very talented student, so if he prefers to learn it rigorously the first time through, perhaps that would be best.

I do think it's possible, for example, for an engineer to have a very solid understanding of Calculus--knowing how to derive the formulas and knowing convincing arguments that things like the Fundamental Theorem of Calculus are true-- without ever having done $\epsilon-\delta$ proofs.  And I do think most math students, perhaps not AoPSers but most math students, would become pretty confused if they tried to learn Calculus in full rigor their first time through... There's a danger of missing the big picture and getting lost in all the rigorous details.

In Breez's case, though, you may be right.

I think it is quite possible for DVO to be right in that a lot of "analysis" books/courses spend way too much time having the students look at rigorous proofs of things that don't need it. For example, no one should EVER have to do an $\epsilon-\delta$ proof that $\lim_{x\to3}x^2-9 = 0$. It's just stupid. However, knowing the $\epsilon-\delta$ definition and its significance has been really helpful for me in a lot of situations -- you don't need it that much for problems where you're given the function, and especially for problems where you're calculating things, but it's essential when you're doing more general stuff in the future.

Long story short, study "calculus" if you want to learn how to use the techniques for the sciences and engineering. Study "analysis" if you want to learn how the math works so that you can go on to higher level math. You really don't want to go into upper-level math classes that assume you know analysis and all you know is calculus. You also don't want to (as Kalle said) waste time when you know you'll want to know analysis eventually.

Thanks for the comments/suggestions.

I'll be in AP Calculus BC the coming school year. I am thinking about buying Spivak's CALCULUS and working through it for a rigorous introduction to calc/real analysis while my AP class can act as a supplement. What do you guys think?

That sounds like a very good mix of the computational and theoretical sides of calculus. Both are important.

Id suggest doing first computational part then the "wrap-up" package. I was very confused by all those "weird" things in the beggining so I just skipped to the chase. When I was thru with all of computational part I went back to do the other part and this time it was very clear.


Now i have a big doubt. Or this problem has something wrong, or it's too hard or i posted this in the wrong room. :help:

[quote="probability1.01"]Well...

1) I don't think the positive integers condition really does anything.
2) I think you need to put some kind of condition on p and k.[/quote]

That's what i thought. Thanks [b]probability1.01[/b], i will check this.

[quote="LoboSolitario"]Prove that ${\frac{\left(\displaystyle \sum_{i=1}^{n}a_i\right)^k}{\left(\displaystyle\sum_{i=1}^{n}b_i\right)^p}\leq{\sum_{i=1}^{n}\frac{a_i^k}{b_i^p}}}$, where $a_i,\,b_i$ are positive integers.[/quote]
I think this ineq is wrong.
It must be like this:
$\sum_{i=1}^n \frac{a_i^p}{b_i^{p-1}} \ge \frac{\sum_{i=1}^n a_i^p}{\sum_{i=1}^n b_i^{p-1}}$
 :D  :)

Errr.. I don't think that's right either (take everything to be 1)... but even if it did meet the equality case, I think it's still not true (suppose that $a_i = b_i, p = 2$)

I think it should be so:
$\sum_{i=1}^{n}\frac{a_i^{p}}{b_i^{p-1}} \ge \frac{\left( \sum_{i=1}^n a_i\right)^p} {\left(\sum_{i=1}^n b_i\right)^{p-1}}$

[quote="LoboSolitario"]Prove that ${\frac{\left(\displaystyle \sum_{i=1}^{n}a_i\right)^k}{\left(\displaystyle\sum_{i=1}^{n}b_i\right)^p}\leq{\sum_{i=1}^{n}\frac{a_i^k}{b_i^p}}}$, where $a_i,\,b_i$ are positive integers and $p, k$ are reals.

EDIT: ($p, k$ are reals).

I don't know if this will help. But that was what my friend told me.[/quote]
I know a same but not like yours. I don't remember it exactly $(\alpha)$. :)
...${\frac{1}{\alpha}\frac{\left(\displaystyle \sum_{i=1}^{n}a_i\right)^k}{\left(\displaystyle\sum_{i=1}^{n}b_i\right)^p}\leq{\sum_{i=1}^{n}\frac{a_i^k}{b_i^p}}}$
Maybe $\alpha = p+1$.

Edit, proof shown

For positive $p,q$ and $p \ge q + 1$
we have:
$\displaystyle\sum{\frac{a^p}{b^q}} \ge n^{q+1-p} \cdot \frac{(\sum{a})^p}{(\sum{b})^q}$

Holder's tells us:
$\left(\sum \frac{a^p}{b^q}\right)^{\frac{1}{p}} \cdot \left(\sum 1 \right)^{\frac{p-q-1}{p}} \cdot \left(\sum b \right)^{\frac{q}{p}} \ge \sum \left[\left(\frac{a^p}{b^q}\right)^{\frac{1}{p}} \cdot 1^{\frac{p-q-1}{p}} \cdot b^{\frac{q}{p}}\right] = \sum a$

Since $\left(\frac{1}{p}\right) + \left(\frac{p-q-1}{p}\right) + \left(\frac{q}{p}\right) = 1$ and from the initial conditions $p \ge q + 1 > 0 + 1 \Rightarrow \frac{1}{p} < 1$, and $p \ge q + 1 \Rightarrow p - q - 1 \ge 0$ with positive $q$ gives $q + 1 > 0 \Rightarrow p > p - q - 1 \ge 0 \Rightarrow \frac{p-q-1}{p} < 1$ and non-negative, and finally $p \ge q + 1 > q > 0 \Rightarrow \frac{q}{p} < 1$ and thus all of the powers are non-negative, less than 1 and sum to 1.

Taking both sides to the $p$ power we have:
$\left(\sum \frac{a^p}{b^q}\right) \cdot \left(\sum 1 \right)^{p-q-1} \cdot \left(\sum b \right)^{q} \ge \left(\sum a\right)^p$
$\Rightarrow \left(\sum \frac{a^p}{b^q}\right) \ge \left(\sum 1 \right)^{q + 1 - p} \cdot \frac{\left(\sum a\right)^p}{\left(\sum b \right)^{q}} = n^{q + 1 - p} \cdot \frac{\left(\sum a\right)^p}{\left(\sum b \right)^{q}}$

For p = 3 and q = 1 with 3 terms (n = 3) we have:
$\displaystyle\sum{\frac{a^3}{b}} \ge 3^{-1} \cdot \frac{(\sum{a})^3}{\sum{b}}$
$\frac{a_1^3}{b_1} + \frac{a_2^3}{b_2} + \frac{a_3^3}{b_3} \ge \frac{\left(a_1 + a_2 + a_3\right)^3}{3(b_1 + b_2 + b_3)}$
which is nifty and will come in handy.

This is the correct form and the other one given by Lovasz would give:
$\frac{p^3}{x} + \frac{q^3}{y} + \frac{r^3}{z} \ge \frac{3\left(p + q + r\right)^3}{x + y + z}$ which if we set $p = q = r = M$ and $x = y = z = N$ for some $M$ and $N$ we get:
$3 \cdot \frac{M^3}{N} \ge \frac{3\left(3M\right)^3}{3N} = 27 \cdot \frac{M^3}{N}$
which is only true when $\frac{M^3}{N} \le 0$ or non-positive, hmmm I forgot what we must have true about variables for Holder, but this case does not work I don't think :| but I must go now, goodbye, I will return.

P.S.
[quote="Lovasz"]
It must be like this:
$\sum_{i=1}^n \frac{a_i^p}{b_i^{p-1}} \ge \frac{\sum_{i=1}^n a_i^p}{\sum_{i=1}^n b_i^{p-1}}$
 :D  :)[/quote]

Lovasz ineq. is just p = p and q = p-1 for our generality. So it is true.

P.S.S. Read the new post

No, Pork_Chop8.
[quote="Pork_Chop8"]For positive $p,q$ and $p \ge q + 1$
we have:
$\displaystyle\sum{\frac{a^p}{b^q}} \ge \frac{1}{n^{q+1-p}} \cdot \frac{(\sum{a})^p}{(\sum{b})^q}$[/quote]

Holder's originally states that for $x_i, y_i \in \mathbb{C}$, $1 < p,q < \infty$ and $\displaystyle\frac{1}{p} + \frac{1}{q} = 1$:
$\displaystyle (\sum_{i=1}^{n} |x_i|^p)^{\frac{1}{p}} \cdot (\sum_{i=1}^{n} |y _i|^q)^{\frac{1}{q}} \ge \sum_{i=1}^{n} |x_i y _i|$

Since $|x|: \mathbb{C}\rightarrow\mathbb{(R^{-})'}$ we can say for all $i$, ($|x_i| = a_i$ and $|y_i| = b_i$), $a_i$ and $b_i$ are non-negative reals and thus:
$\displaystyle (\sum_{i=1}^{n} a_i^p)^{\frac{1}{p}} \cdot (\sum_{i=1}^{n} b_i^q)^{\frac{1}{q}} \ge \sum_{i=1}^{n} a_i b _i$

Now, I will change it around a little bit to show how I have used it. Say $r = \frac{1}{p}$ and $s = \frac{1}{q}$ which tells us $1 = \frac{1}{1} > r,s > \frac{1}{\infty} > 0$ and $r + s = 1$. Now, for $a_i^p = A_i$ and $b_i^q = B_i$ (implies $A_i$ and $B_i$ are positive reals) we have:
$\displaystyle (\sum_{i=1}^{n} A_i)^{\frac{1}{p}} \cdot (\sum_{i=1}^{n} B_i)^{\frac{1}{q}} \ge \sum_{i=1}^{n} (A_i)^{\frac{1}{p}} (B _i)^{\frac{1}{q}}$
which readily becomes:
$\displaystyle (\sum_{i=1}^{n} A_i)^r \cdot (\sum_{i=1}^{n} B_i)^s \ge \sum_{i=1}^{n} [(A_i)^r (B _i)^s]$

In this form it is easy to prove by induction for $t$ sets of variables in the set of positive reals $\{A_{1,j}\}$,$\{A_{2,j}\}$,$\ldots$,$\{{A_{t,j}\}}$ (where $A_{k,j}$ represents the jth member of the kth set) and for powers $p_i$ such that $0 < p_i < 1$ and $\sum_{i=1}^{t} p_i = 1$ we have:
$\displaystyle (\sum_{i=1}^{n} A_{1,i})^{p_1} \cdot (\sum_{i=1}^{n} A_{2,i})^{p_2} \cdots (\sum_{i=1}^{n} A_{t,i})^{p_t} \ge \sum_{i=1}^{n} [(A_{1,i})^{p_1} \cdot (A_{2,i})^{p_2} \cdots (A_{t,i})^{p_t}]$

P.S. Anywhere may offer the inductive proof but also one of the initial Holder for 2 sets would be nice, I have not attempted yet, but it looks real hard.


AMC
AIME
probability
Pascal\u0027s Triangle

6.  A fair die is rolled four times. Find the probability that each number is no smaller than the preceding number.

There's a clever way to solve this using a [hide]one to one correspondence[/hide] but I did it a different way using cases.

Here's what I did:

[hide]Sequence needs to have all terms increasing or staying same.  There are C(6, 4) or 15 ways to have a sequence with all 4 digits different.  There are 6 sequences where all 4 digits are the same like 5555.  When 3 digits are the same and 1 is different like xxxy or 1113, there are 6 choices for the x and 5 for the y, so 30 choices.  When 2 digits are same and other 2 digits are same like xxyy or 2244, there are once again 6 choices for x and 5 for y so 30.  Last case is where we have a sequence like xxyz, which has 6*5*4 choices or 120.  But I get around 40 or 50 too many sequences.  Help anyone?[/hide]

If we have a triple (a,b,c,d), then some permutation of it is sorted properly.

If a,b,c,d are all different, we get 1 way:
There are 6*5*4*3/4! such triples.

If 2 of a,b,c,d are identical, we get 3 ways:
6*5*4/3! triples

If 3 .... 3 ways
6*5/2! triples

If 4 .... 1 way
6/1! triples

Summing all this gives 126

There are 6^4 triples, dividing and simplifying gives 7/72, the desired result.

This kind of plug and chug work is actually pretty fast, but kalva's solution (which numerically agrees with mine, btw) is much better.

EDIT: You know, it's funny.  My highest score ever on the AIME was 7 or 8 (in grade 11 and I scored 3 my final year, but on looking over past AIMEs, I feel that I could definitely score 8-10, now.

in the case xxxy, there really isn't 6 * 5 choices like you said. For example, if x = 6, y cannot be any number. However if x = 1, y can be 5 numbers. If x = 2, y can be 4 numbers... so the total is 5 + 4 + 3 + 2 + 1 = 15. But since you neglected to mention the case of xyyy, which also have 15 ways, both of these cases do add up to 30. no "extra" cases here.

However this doesn't work for the next case, xxyy. With similar logic, there are only 15 ways to sequence x and y. Since the mirror-image (lol, not a math word) of this case is also xxyy, you have added 15 extra cases here.

Then for xxyz, there are (4 + 3 + 2 + 1) + (3 + 2 + 1) + (2 + 1) + 1 = 20 cases. Since there are 3 possibilities: xxyz, xyyz, xyzz, then there are 20 * 3 = 60 cases, not 120.

wait... I found 75 extra cases? what did we forget?

[quote="blahblahblah"]If 3 .... 3 ways
6*5/2! triples[/quote]

correction, there are 2 ways (i believe that's a typo =])

now seeing blah's solution, summing up using my (longer) method, you would also get 126.

[quote="Yrael"][quote="blahblahblah"]If 3 .... 3 ways
6*5/2! triples[/quote]

correction, there are 2 ways (i believe that's a typo =])
[/quote]

No, think Pascal's triangle, 1 3 3 1

[quote="Yrael"]maybe I'm misunderstanding your "ways." I meant cases like: xxxy and xyyy, when 3 numbers are the same and 1 is different. is that what you meant?[/quote]

I don't treat xxxy and xyyy as different.  In fact, I don't even treat xxxy and xxyy as different.  I don't need to.

Give me a triple with only two different digits (2,3,2,3), I write this as (3,3,2,2).  There are 6 ways to write this triple that are different: (3,3,2,2) (3,2,3,2) (2,3,2,3) .... (2,2,3,3).

There are only TWO different digits in this number:

Hence 6*5 for the two digits.

Since my method counts (a,b,b,b) the same as (b,b,b,a), I need to eliminate some of these permutations - divide by 2!

Now, I also can write (a,a,b,b) (a,b,b,b) and (a,a,a,b).  All of these must be counted - multiply by 3.

This gives me the number of tuples with exactly 2 different digits.

[quote="Yrael"]ah i see. thank you.

in my original response, i was trying to fix NightFlarer's reasoning. your method, is it what Night called "one to one correspondence"?[/quote]

No.  See the solution at http://www.kalva.demon.co.uk/ for such a solution.

In general, finding a one-to-one correspondence (a bijection) between two things is the best way to solve counting problems.


Calcular
$ \sqrt {1 \plus{} 2\sqrt {1 \plus{} 3\sqrt {1 \plus{} 4\sqrt {1 \plus{} ...} } } }$

[url=http://mathworld.wolfram.com/NestedRadical.html]Look at equation 28.[/url]

BTW, this question has been posted on AOPS before.  Its solution is accredited to Ramanujan, as far as I know...

:O pero ¿cómo llegó a eso? el resultado es igual a 3

[quote="Babelfish"][quote="xdanielxedgex"]but how it arrived at that? the result is equal to 3[/quote][/quote]

Let $ x=3$.  Then
$ x=2+1=\sqrt{(2+1)^2}=\sqrt{1+2(4)}=\sqrt{1+2\sqrt{(3+1)^2}}=\sqrt{1+2\sqrt{1+3(5)}}$

$ =\sqrt{1+2\sqrt{1+3\sqrt{(4+1)^2}}}=\ldots=\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{\ldots}}}}}$

Well, I could say, let $ x\equal{}5$,

$ x\equal{}5\equal{}\sqrt{25}\equal{}\sqrt{1\plus{}2\sqrt{144}}\equal{}\sqrt{1\plus{}2\sqrt{1\plus{}3*\sqrt{blah}}}$

Therefore, this method does not prove that $ x\equal{}3$, since I could have argued equally well that $ x\equal{}5$.


algebra
polynomial
calculus
integration
algebra proposed

Let $a_1,...,a_n$ be some distinct integers. Prove that $(x-a_1)^4\cdot (x-a_2)^4\cdot...\cdot (x-a_n)^4+1$ is irreducible in $Q[X]$. I would also be interested in a proof of the fact that this remains true if we replace 4 with any even positive integer (I cannot solve the last problem).

[quote="harazi"] this remains true if we replace 4 with any even positive integer [/quote]
Obviously not if that integer has an odd divisor :)

Sorry, of course I wanted to say power of 2.  :blush:

[quote] Source: oral ens[/quote] ...
ENS = Ecole Normale Supérieure, if that's what you wanted to know :)

For admission

http://www.ens.fr/entrer/index_en.php

i have the solution but it's too long and nasty.Here some hints:
first when we consider that the polynomial reducible then the polynomial $P(x)$ could be factorized in the form:
            $P(x)+1=(1-P_{-1}(x)P_{0}(x))(1-P_{1}(x)P_{0}(x))$
Where $P_{0}(x)=(x-a_{1})...(x-a_{n})$ and $P_{-1}(x),P_{1}(x)$ are polynomials with integral coefficients with leading coefficent $-1$.Hence we get
               $P_{-1}(x)+P_{1}(x)=-P_{0}(x).T(x)$
where $T(x)=(P_{0}(x))^2-P_{-1}(x)P_{1}(x)$
We have two cases:
If $deg P_{1}(x)=degP_{-1}(x)$ then easily to prove that $T(x)=2$ for all $x \in R$ so that $(P_{1}(x))^2=2$ for all $x=a_{1},a_{2},...,a_{n}$ (Contradiction)
So we can consider that$deg P_{1}(x) <P_{-1}(x)$ Then We have
$(P_{1}(x))^4+1 \equiv (P_{1}(x))^4++(P_{1}(x))^4(P_{0}(x))^4 \equiv 0 (mod 1-P_{0}(x)P_{1}(x))$
So from it we also have $(P_{1}(x))^4+1=(1-P_{1}(x)P_{0}(x))(1-P_{1}(x)P_{2}(x))$
after that we see that we can construct the sequel P_{m}(x),T{m}(x) as general and satisfy:
                 $(P_{m}(x))^4+1=(1-P_{m}(x)P_{m-1}(x))(1-P_{m}(x)P_{m+1}(x)) P_{m-1}(x)6P_{m+1)(x)=-P_{m}(x)T_{m}(x) (P_{m}(x))^2=T_{m}(x)+P_{m-1}(x)P_{m+1}(x)}$ for all $m=0,1,2,...$
From this we immediately see that
                   $\frac{(P_{m}(x))^2+(P_{m+1}(x))^2}{1-P_{m}(x)P_{m+1}(x)} =T_{m}(x) = \frac{(P_{m-1}(x))^2+(P_{m}(x))^2}{1-P_{m-1}(x)P_{m}(x)}=T_{m-2}(x)=....=T(x)$
And we see that if we call the degree of $P_{m}(x)$ is $n_{m}$ then easily to prove that the sequence {n_{m}} decrease
Hence there must exist a first polynomial $P_{m+1}(x)$ that vanishes identically.Then we set P_{m}(x)=z.So we have that $P_{k}(x)$ become polynomials in y,i.e $P_{k}(x)=H_{k}(y)$ for all $k \in Z^{+}$
And here we will the zeros of polynomial and easy to get that $m=0$or $m=1$
And from this we easy to get the contradition.
AFTER THAT problem we can prove the hard problem:
Find all positive integer A such that the Polynomial
                      $Q(x)=A((x-a_{1})(x-a_{2})...(x-a_{n}))^4+1$ is irreducible
Any idea Harazi? :D  :D  :D
Happy new year to everyone on Mathlinks  :)  :)  :)  :)

Thank you for your solution, it's also the solution I found. They do like difficult irreducibility problems at ENS. About your problem, I only dare to conjecture that all numbers different from $4k^4$ for all $k$. Is it at least true? I'm not going to spend one more day on this problem, it already took me a lot of time to find the solution for this problem. And what about the general case, for powers of 2?


Find $ a$ and $ b$ such $ x^{4}\plus{}ax^{3}\plus{}bx^{2}\minus{}8x\plus{}1$ be a total square.  :)

Ohh.. come on ! No one give any solution

when you say a total square you mean a perfect square right?

Total square I mean perfect square , yes

hmm...i get to a certain point every time i try but get stuck at this one area and just can get past it!
Hint???

[quote="Elio (n)"]Find $ a$ and $ b$ such $ x^{4} \plus{} ax^{3} \plus{} bx^{2} \minus{} 8x \plus{} 1$ be a perfect square.  :)[/quote]

Since the coefficients of $ x^4$ and $ x^0$ are $ 1$, the perfect square you seek is of the form $ (x^2\plus{}px\plus{}1)^2\equal{}x^4\plus{}2px^3\plus{}(p^2\plus{}2)x^2\plus{}2px\plus{}1$.So $ p\equal{}\minus{}4$ ,therefore $ a\equal{}\minus{}8$ and $ b\equal{}18$

The entire form of a polinom $ P(x^{2})$ is $ P(x^{2}) \equal{} mx^{2} \plus{} nx \plus{} c$
And we have $ (mx^{2} \plus{} nx \plus{} c)^{2} \equal{} x^{4} \plus{} ax^{3} \plus{} bx^{2} \minus{} 8x \plus{} 1$

$ c^{2} \equal{} 1$
$ c \equal{} \plus{} 1$

$ 2cn \equal{} \minus{} 8$ so if $ c \equal{} \plus{} \minus{} 1$ we get $ n \equal{} \minus{} \plus{} 4$
$ m^{2} \equal{} \plus{}  1$
$ m\equal{}\plus{}\minus{}1$

If we analyse all the possibles event we get
$ a \equal{} (8, \minus{} 8)$ and $ b \equal{} (14,18)$  :)


We know the line must intersect the y-axis so to find the function we need maximized we have $ f(x) \minus{} y_0(x) \equal{} xf'(x)$ or, $ y_0(x) \equal{} f(x) \minus{} xf'(x)$ where $ y_0(x)$ represents the y-intercept as a function of $ x$. We differentiate to find maximum $ \frac{d}{dx}\left[y_0(x)\right] \equal{} f'(x) \minus{} xf''(x) \minus{} f'(x) \equal{} xf''(x) \equal{} 0 \Rightarrow x(6x \minus{} 12) \equal{} 0 \Rightarrow x \in \{0,2 \}$. We are restricted to positives so $ x\equal{}2$ yields the maximum $ y_0$ value. If you wanted the line it's $ y \minus{} y_0(2) \equal{} 2*f'(2) \Rightarrow y \equal{} \minus{}\frac{3}{4}x \plus{} 9$


I did the 2003 National Round, but I don't have an answer key.   :(

Could someone please post one, PM it to me?

Thanks.  I appreciate your help.

[quote="abacadaea"]do you have the link or the file saved on your computer? or did you order it?[/quote]

I got it from somebody, but the answer key was missing.

I would really appreciate it if someone would provide me with the answer key.  I spent a lot of time on the questions already...


Solve the system of equation
$ x\plus{}y\plus{}z\equal{}xy\plus{}yz\plus{}zx$
$ xyz\equal{}1$
with $ x\plus{}y\plus{}z$ being the minimal possible.

[hide="Solution"]$ \frac{1}{yz}\plus{}y\plus{}z\equal{}\frac1z\plus{}yz\plus{}\frac1y$
$ 1\plus{}y^2z\plus{}yz^2\equal{}y\plus{}z\plus{}y^2z^2$
$ y^2(z^2\minus{}z)\plus{}y(1\minus{}z^2)\plus{}z\minus{}1\equal{}0$
$ (z\minus{}1)(y^2z\minus{}y\minus{}yz\plus{}1)\equal{}0$
$ (z\minus{}1)(y\minus{}1)(yz\minus{}1)\equal{}0$

1) z=1, xy=1
$ x\plus{}y\plus{}z\equal{}1\plus{}x\plus{}\frac1x\ge 3$, with equality for $ x\equal{}y\equal{}z\equal{}1$
2) y=1 we get the same
3) yz=1, x=1 we get the same[/hide]

[quote="Bugi"]Solve the system of equation
$ x \plus{} y \plus{} z \equal{} xy \plus{} yz \plus{} zx$
$ xyz \equal{} 1$
with $ x \plus{} y \plus{} z$ being the minimal possible.[/quote]

By Vieta's formula, $ x,\ y,\ z\in{t^3\minus{}kt^2\plus{}kt\minus{}1\equal{}0}\Longleftrightarrow (t\minus{}1)(t^2\minus{}(k\minus{}1)t\plus{}1)\equal{}0$, where $ x\plus{}y\plus{}z\equal{}xy\plus{}yz\plus{}zx\equal{}k$.

$ \therefore x\plus{}y\plus{}z\equal{}1\plus{}(k\minus{}1)\equal{}k\geq 3\sqrt[3]{xyz} \equal{}3$ with equality $ x\equal{}y\equal{}z\equal{}1$.

[quote="Bugi"]Solve the system of equation
$ x \plus{} y \plus{} z \equal{} xy \plus{} yz \plus{} zx$
$ xyz \equal{} 1$
with $ x \plus{} y \plus{} z$ being the minimal possible.[/quote]
$ x\plus{}y\plus{}z\ge3\sqrt[3]{xyz}\equal{}3$ with equality when $ x\equal{}y\equal{}z\equal{}1$.
Is this a valid solution?

I think the condition $ x\plus{}y\plus{}z\equal{}xy\plus{}yz\plus{}zx$ is useless

The xy+yz+zx is there for checking the $ x\plus{}y\plus{}z\equal{}3$. For example, if there was $ 2xy\plus{}xz\plus{}yz$, then $ (1,1,1)$ wouldn't be a solution...

[quote="Bugi"]The xy+yz+zx is there for checking the $ x \plus{} y \plus{} z \equal{} 3$. For example, if there was $ 2xy \plus{} xz \plus{} yz$, then $ (1,1,1)$ wouldn't be a solution...[/quote]

Look again at mathwizarddude post. At am-gm there is equality, iff x=y=z. Since $ xyz\equal{}1,$ then $ x\equal{}y\equal{}z\equal{}1,$ so $ xy\plus{}yz\plus{}zx\equal{}x\plus{}y\plus{}z$ is useless. I think the proposer of this problem, has something else in his mind, i mean a solution like yours, which was also nice.

It dont think its fair  proposing problems like:
Solve the system $ x,y,z>0,$ such that $ xyz\equal{}1, x\plus{}y\plus{}z\equal{}$minimum and $ x^5\plus{}y^5\plus{}z^5\equal{}3x^2y^2z^2$
We just put a condition that its true anyway, since we find $ x,y,z?$ What would you think if the condition was $ x^y\plus{}y^z\plus{}z^x\equal{}3xyz?$ Just to impress the kids with a condition (which is true after we find $ x,y,z$) its not fair and useless. We could have a conditon like $ ''x,y,z$ are sides of a triangle''. So what?

Again, what would you do if the condition was $ x\plus{}y\plus{}z\equal{}2xy\plus{}xz\plus{}yz$? The condition is to check the equality case, and as $ 3\equal{}3$, it's true. But if the condition was $ 2xy\plus{}xz\plus{}yz$, then you would get $ 3\neq4$. So the condition is needed.

The equality case holds iff $ x\equal{}y\equal{}z.$ So from condition $ xyz\equal{}1,$ we get $ x\equal{}y\equal{}z\equal{}1$ If the condition was little different, for example $ x\plus{}y\plus{}z\equal{}2xy\plus{}yx\plus{}zx,$ then there are no positives $ x,y,z,$. Thats ok with me, but i still believe that the condition $ x\plus{}y\plus{}z\equal{}xy\plus{}yz\plus{}zx$ is useless, because we get the desired numbers from the other conditions. It woud be helpfull if the condition was not satisfied, so in this case no numbers exist.

[quote="nickolas"]The equality case holds iff $ x \equal{} y \equal{} z.$ So from condition $ xyz \equal{} 1,$ we get $ x \equal{} y \equal{} z \equal{} 1$ If the condition was little different, for example $ x \plus{} y \plus{} z \equal{} 2xy \plus{} yx \plus{} zx,$ then there are no positives $ x,y,z,$. Thats ok with me, but i still believe that the condition $ x \plus{} y \plus{} z \equal{} xy \plus{} yz \plus{} zx$ is useless, because we get the desired numbers from the other conditions. It woud be helpfull if the condition was not satisfied, so in this case no numbers exist.[/quote]

With the condition $ 2xy\plus{}yz\plus{}zx$, there are numbers, just AM-GM won't help you so much. We get the desired from the other conditions, but this is another which has to be solved.


It is known that $ \frac{200}{n} > 6$ and $ n$ is a positive integer. What is the
largest possible value for $ n$?

We have $ \frac{200}{6}>n\Rightarrow\frac{100}{3}>n\Rightarrow n<33\frac{1}{3}$.  So we need $ n\leq 33$.

Finishing off the solution, the largest integer value of $n$ is $\boxed{33}$.


trigonometry
inequalities
inequalities unsolved

Let $c = cos(\frac {\pi} {11})$ et $s = sin(\frac {\pi}{11}).$
Clearly $s \neq 0.$
Expanding $(c + is)^{11} = -11$, thus identifying imaginary parts, it leads (using $c^2 = 1- s^2$)
$1024s^10 - 2816s^8 + 2816s^6 - 1232s^4 + 200s^2 - 11 = 0.$ (1)

In another hand, using classical formulas for $tan(3x)$ and $sin(2x)$ we have :
$tan(\frac {3\pi} {11}) + 4sin(\frac {2\pi} {11}) = \frac {3sc^2 - s^3} {c^3 - 3cs^2 } + 8sc = \frac {11s - 44s^3 + 32s^5} {c*(1-4s^2)}.$ (2)

Note that $tan(\frac {3\pi} {11}) + 4sin(\frac {2\pi} {11}) > 0$ (each term is positive). Then squaring and substituing (1) into (2), you get the desired equality.

Pierre.

You got to read this thread:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=12775

There is a general formula:

$\displaystyle(\tan(3s) + 4\sin(2s))^2 = 11 - \frac{(\tan(8s) + \tan(3s))\cos(8s)}{\sin(s)\cos(3s)}$

But why is this in Advanced Section -> Inequalities -> Unsolved Problems?


I know the interval of convergence but after that how do I know if is inclusive or exclusive-I know radius is 3 but (-3,3) or [-3,3] or a combination?

Nothing much can be said in general. You will have to check in each case by simply seeing if the the series converges at the endpoints.

To which I'll add: normally, the ratio test and root test tell you nothing about this. You've already gotten all the information you're going to get from these tests when you found the radius of convergence. Deciding on convergence on the boundary will be a matter of asking whether the terms tend to zero, comparisons to $p$-series, and applications of the alternating series test (or possible generalizations of that test.)


[hide]If they buy the windows together, they will use $1200$ dollars. If they purchase separately, Dave will use $\$600$ dollars and Doug will use $\$700$ dollars. They will save $1300-1200=100$ or $\boxed{A}$ dollars.[/hide]

[hide]Together, they buy 15 windows, which would cost 1200, since 3 are free.
Dave alone buys 7 windows, one is free, costs 600.
Doug alone buys 7 windows, one is free, costs 700.
600+700-1200=100 $\boxed{A}$.[/hide]

[quote="now a ranger"]A store normally sells windows at 100 dollars each. This week the store is offering one free window for each purchase of four. Dave needs seven windows and Doug needs eight windows. How many dollars will they save if they purchase the windows together rather than seperately?

A. 100
B. 200
C. 300
D. 400
E. 500[/quote]

[hide]
If they buy separately:

Dave -> $\$600$
Doug -> $\$700$
Added-> $\$1300$

Together -> $\$1200$

Saved-> $\$100\implies\boxed{A}$
[/hide]

[hide]you could also take the sum of amount they each save separately, 100+100=200, from the saved amount if they by together, which is 300 is you get \[300-200=100\][/hide]

[quote="t0rajir0u"]A more interesting question:  How many people need to buy windows together to save at least $100k$ dollars for $k$ a positive integer?[/quote] 5k if they're each buying one window... I think you left something out.

I didn't specify that they were each buying one window.  I said [b]windows[/b].  :D

But that's the point. If you don't know how many windows each person bought, you can't say how much they saved...

You can't even come up with a general formula without knowing SOMETHING about the relation between # of people and # of windows. I suppose you could come up with cases and try to cover every possibility, but have you even figured out whether it's possible to do so?

Wow this was not a fun question.  It was the only one I missed on this test...grr.

@akkaren: Floor function!

[quote="akkaren"]But that's the point. If you don't know how many windows each person bought, you can't say how much they saved... [/quote]

How about an upper bound, then?   :D

The question I intended is to find the [b]smallest number of people[/b] that need to buy windows together to save $100k$ dollars.


Because $n\ge 5$ it had infinetely many solutions ($x_{5},...,x_{n}$ can be choose any).

Are there only $4$ equations for the $n$ variables?
Or the system goes until $\dots=n^{n-1}+n+2^{n-1}$?

[quote="Rust"]$(1-\sum_{k=5}^{n}t_{k}x_{k})$[/quote]

How do you know whether in the bracket is positive or negative?


[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14649[/img]

 :(

Call $ \overline{AH}\equal{}x$, $ \overline{AI}\equal{}y$, $ \overline{HI}\equal{}z$, then:

$ \frac{x}{y}\equal{}\frac{b \cos A \cos{\frac{B}{2}}}{\sin B \sin{\frac{C}{2}}}\equal{}\frac{\cos A}{2\sin{\frac{B}{2}} \sin{\frac{C}{2}}}\equal{}\frac{\cos A}{\cos { \left ( \frac{C\minus{}B}{2}\right )} \minus{} \sin{\frac{A}{2}}}$

but $ \cos { \left ( \frac{C\minus{}B}{2}\right )} \equal{} \frac{x^2\plus{}y^2\minus{}z^2}{2xy}$ then

$ \frac{x^2\plus{}y^2\minus{}z^2}{2y} \minus{} x \sin{\frac{A}{2}} \equal{} y \left ( 1\minus{} 2 \sin^2{\frac{A}{2}} \right )$

$ 2y \sin^2{\frac{A}{2}} \minus{} x \sin{\frac{A}{2}} \plus{} \frac{x^2\minus{}y^2\minus{}z^2}{2y}\equal{}0$

$ \sin{\frac{A}{2}} \equal{} \frac{x\plus{}\sqrt{4y^2\plus{}4z^2\minus{}3x^2}}{4y}$

so we can costruct the sideline AB and AC. Then let J the projection of I on AB and costruct the incircle that have center I and radius IJ. The line parallel to AH from I meet the incircle on E and F such that E it's nearest A then F. Let the perpendicolar line to AH from F thet meet the side line AB and AC on B and C.

It seems you have posted  the topic before
http://www.mathlinks.ro/viewtopic.php?t=204492

"so we can costruct the sideline AB and AC"

how are these constructions AB and AC ?

 :(

you have $ \frac{A}{2}$ and you have the incenter...

read my first reply please...you have $ \sin x \equal{} \frac{a}{b}$ with 0<x<90 so for costruct the angle x let c circle with diameter AB s.t. AB=b and a circle with center A and radius a that meet this circle on C, then $ \angle ABC \equal{} x$

ok i know A/2 construction

i don't know the [code]\sin{\frac{A}{2}} = \frac{x+\sqrt{4y^2+4z^2-3x^2}}{4y}[/code] construction

 :(

using many times pitagora you can costruct a segment with lengh $ \sqrt{4y^2\plus{}4z^2\minus{}3x^2}$ if you have x,y,z: you have 2y and 2z so you have $ \sqrt{4x^2 \plus{} 4z^2}$, you can have also $ \sqrt{3} x$ and so you have $ \sqrt{4x^2 \plus{} 4z^2\minus{}3x^2}$.


Find the sum of the series:
(3/2)nC0 + (5/3)nC1 + (7/4)nC2 + ... + (n + 1) terms

using Binomial theorem.

NOTE: nC0 is n choose 0.

By the way, "sum of series" is redundant - a series already a summation. Saying "sum of a sequence" or just "series" is correct.


Let $d$ and $e$ denote the solutions to $2x^{2}+3x-5=0$ What is the value of $\left(d-1\right)\left(e-1\right)$

$\left(A\right)-\frac{5}{2}$
$\left(B\right) 0$
$\left(C\right) 3$
$\left(D\right) 5$
$\left(E\right) 6$

[hide="Ready"]$\begin{equation*}\begin{aligned}(d-1)(e-1)&=d \cdot e-(d+e)+1\\ &=-\frac{5}{2}+\frac{3}{2}+1\\ &=-1+1\\ &=0 \end{aligned}\end{equation*}$[/hide]

[hide="A little more explanation"]
When approaching these type of problems, the best thing to do is to simplify the polynomial to an expression whose first coefficient is 1.  These can easily be done in this problem by dividing the whole polynomial by 2 yielding: $x^{2}+\frac{3}{2}x-\frac{5}{2}=0$

Examining this equation (and using the fact that the two roots are $e$ and $d$) using Viete's formula yields:
$e*d=-\frac{5}{2}$ and $-(e+d)=\frac{3}{2}$

This is because the last term in a quadratic is the product of the roots and the term before that is the opposite of the sum of the roots.

Now expanding: $(d-1)(e-1)=e*d-(e+d)+1$

And now we can plug in numbers: $\frac{5}{2}-\frac{3}{2}+1=0$

$\boxed{B}$
[/hide]

[hide="alternate method"]You could simply note that one root is 1 and thus the answer is $0$.[/hide]

I did this round this afternoon.

[hide]Roots are $-5/2$ and $1$. Since one of the roots is 1, the product of (d-1)(e-1)=0, B.[/hide]

[hide]You can factor the equation to get
$2x^{2}+3x-5=(2x+5)(x-1)$
Therefore, one of the roots is 1, and then (1-1)=0 so the product will always be 0[/hide]


[i]Author of the problem is Adrian Satja Kurdija (although, of course, it may be possible that there already exists a very similar or same problem author didn't know about). This was Problem 1 (Algebra) for 9th grade on 1st Balkan Student Mathematical Competition 2008 - a competition 11 schools from Balkan area took part in during November 2008. All information about the competition can be found on http://www.mathlinks.ro/viewtopic.php?t=243614.[/i]

Find all nonnegative real numbers $ a$ and $ b$ for which $ (a^3 \plus{} b^3 \plus{} (a \plus{} b)^3)^2 \equal{} 100a^3b^3$.

We have that $ (a^3 \plus{} b^3 \plus{} (a \plus{} b)^3)^2 \minus{}100a^3b^3\equal{}(a \minus{} b)^2 (4 a^4 \plus{} 20 a^3 b \plus{} 57 a^2 b^2 \plus{} 20 a b^3 \plus{} 4 b^4)\equal{}0$, so $ a\equal{}b\equal{}k$ where $ k\geq0$ are all the solutions, thanks to $ a,b\geq0$ :P

[hide="Also"]AM-GM gives $ 2a^3 \plus{} 3a^2b \plus{} 3ab^2 \plus{} 2b^3 \geq 10 \sqrt{a^3 b^3}$, equality iff $ a \equal{} b$.[/hide]

@Jorge Miranda:
how did you factor that?

First I checked that $ a \equal{} b$ worked. Since it's symmetric, $ (a \minus{} b)^2$ has to divide the difference, and then it's just plain long division :)


In a quadrilateral $ABCD$, it is given that $AB$ is parallel to $CD$ and the diagonals $AC$ and $BD$ are perpendicular to each other. Show that (a) $AD \cdot BC \geq AB \cdot CD$ (b) $AD + BC \geq AB + CD.$

Is there any solution to this sum by geometry?

yes, note that the figure is a rhombus (easy to prove ) and so the inequalities to prove are trivial

@above, no it is not a rhombus.

For a proof, we shall proceed to prove the second part first. Consider the branch of the hyperbola $\mathcal{H}$ which passes through $B$ and has $A, C$ as its foci. Suppose $AC$ meets $BD$ at $X$. Without loss of generality, assume $X$ is closer to $A$, and not $C$ (if the distances were equal, then the quadrilateral would be a rhombus and equality holds). Then by Thales' theorem, $BX < DX$. Thus, since the reflection of $B$ over $X$ (which is also the reflection of $B$ in the transverse axis of the hyperbola) lies on $\mathcal{H}$, and $DX > B'X$, we have $DC - AC < BC - AB$, which rearranges to the second part.

Now to show the first part, note that by the perpendicularity lemma we have $AB^2 + CD^2 = AD^2 + BC^2$, and subtracting this from the square of the previous inequality, we are done.

[quote=AOPSMasters]I have just one another proof by geometry. Should I post it? This proof is by my friend, not by me.[/quote]

Post please


How many ordered pairs $ (x, y)$ satisfy BOTH conditions below? \\
\indent Condition I: $ x \equal{} 1$ or $ y \equal{} 0$ or $ y \equal{} 2$ \\
\indent Condition II: $ x \equal{} 0$ or $ x \equal{} 2$ or $ y \equal{} 1$

if x=1, y=1.

also, if y=0 or y=2, x=0 or x=2.

that's 1*1+2*2.

answer : 5

Wait, I don't understand the OR parts.  Can you elaborate on that please? Thanks.  :lol:


induction
number theory proposed
number theory

Let $ < a_0,a_1,a_2,...,a_n >$ be a finite simple continued fraction. Define
$ p_0 \equal{} a_0$ ,
$ p_1 \equal{} a_1a_0 \plus{} 1$
and, $ p_n \equal{} a_np_{n \minus{} 1} \plus{} p_{n \minus{} 2}$
$ q_o \equal{} 1$ ,
$ q_1 \equal{} a_1$
and $ q_n \equal{} a_nq_{n \minus{} 1} \plus{} q_{n \minus{} 2}$
Prove that ,
$ < a_0,a_1,a_2,...,a_n > \equal{} \frac {p_n}{q_n}$

is it so difficult? how much ever i try i dont seem to near the solution :(  someone help me ..

is this topic very easy?? or known?? if so please provide me a link ..
it has been 10 days and no replies..

do it by induction...
suppose that we have $ \frac{p_{n\minus{}1}}{q_{n\minus{}1}}\equal{}<a_0,...,a_{n\minus{}1}>$ (the length is $ n$ and we want to prove it for length=$ n\plus{}1$)
we define these two sequences which have some similarities with the other two:
$ p'_0\equal{}a_1,p'_1\equal{}a_1a_2\plus{}1,p'_n\equal{}a_np'_{n\minus{}1}\plus{}p'_{n\minus{}2}$
$ q'_0\equal{}1,q'_1\equal{}a_2,q'_n\equal{}a_nq'_{n\minus{}1}\plus{}q'_{n\minus{}2}$
one can easily prove by induction these equalities :
$ p'_{n\minus{}1}\equal{}q_n$
$ q'_{n\minus{}1}\plus{}a_0p'_{n\minus{}1}\equal{}p_n$
now we have
$ <a_0,...,a_{n}>\equal{}a_0\plus{}\frac{1}{<a_1,...,a_n>}$
because of the induction we have $ \frac{p'_{n\minus{}1}}{q'_{n\minus{}1}}\equal{}<a_1,...,a_n>$ since the length is $ n$.
$ \equal{}a_0\plus{}\frac{1}{\frac{p'_{n\minus{}1}}{q'_{n\minus{}1}}}\equal{}\frac{a_0p'_{n\minus{}1}\plus{}q'_{n\minus{}1}}{p'_{n\minus{}1}}$
which is equal to $ \frac{p_n}{q_n}$ .
so we r done!


MIT
college
AMC
AMC 12
AIME
Princeton
Harvard

Let's say I want to go to a nice technology college, such as Cal-Tech or MIT. What would the SAT 'requirement' be? I'm assuming like 2400 but you know.

1) You are in middle school, which is not a time to be thinking about college
2) There is no such "requirement." I mean, if you get a 1200, you probably aren't going to be accepted for other reasons, but a good score and a 2400 are virtually identical to them.

Also, keep in mind that they turn down many apps with perfect SATs.

I second [b]perfect628[/b]'s post completely.  :)

To be perfectly honest, SAT's don't weigh in that much. Focus on trying to develop your talents and trying to get involved in interesting things, or take initiatives that interest you, if you run across the opportunity to do so.

But surely, the SAT's count for something? They are like an indicator. How important that indicator is depends on how it fits into your personal image.

If you are highly involved in math but get an imperfect score here (say, even a 750), that's fine. They won't care. Similarly, if you get an 800, they won't think much of it either. Keep in mind that MIT specifically asks for AMC12 and AIME scores, and any good college will look at them if you include them in your resume. If you do well on these, they will obviously be given much more weight than your SAT math score. Same with other math contests. This puts you in a very comfortable position.

Your writing section is a bit more important, but again, the first indicator they'll see of your writing is your essays (ESSAYS!!!!!!), and if applicable, any paper or essay contests, etc. that you may have done well in.

So that leaves the reading section. I advise you to focus about 3/4 of your SAT prep on this section. An MIT study showed that this section, rather than the combined index, is the best indicator of college performance that the SAT has to offer. I'm sure their own admissions office has taken that to heart. It's also probably true. Even still, don't sweat it too much. Just read a lot, or do some debating, or try to go above and beyond in your English class when you write a paper analyzing a book or something. The best ways to improve your critical reading skills don't involve an SAT prep book or a dictionary.

If I had know that when I was in elementary, I would prep back then...
It's never too early to prep for SAT.
700+ in everything is good enough for both schools. So you don't have to really stress out.

Then you got to prepare for something to create strong application:
1. competitions
2. researches
3. music/art stuff
4. advanced studies
5. find 5 adjectives describe yourself and use it as your advantage.

I cannot stress how important [b]mgccl[/b]'s #5 is.

Many schools try to get a picture of you by seeing how you concisely describe yourself. Dartmouth asks exactly what mgccl wrote: 3-5 words to describe yourself. Princeton will ask you for a list of favorite things (favorite book, favorite movie, favorite word, etc). In general, think of words and things that will describe who you are, and [b]write them down[/b]!

Along these lines, start thinking about who you are and how you can describe yourself concisely. Also, think about life experiences that will portray these qualities. This will help you tremendously in your essays.

Well, just from visiting Cornell and hearing about there engineering departments, their "engineering culture" sounds unbelievable awesome.  They told us about the annual dragon battle.  Really cool.

[quote="Erken"]I wonder if the SAT subject test has really huge influence on decision of MIT or Harvard admission offices,if you have IMO medals.
I mean,the main purpose of the SAT is to point out person's capabilities in Math and Physics, but IMO can be better index of capabilities than the SAT subject test,in case you're going to take math major,right?[/quote]

Erken I cannot tell much about Harvard except that they are less likely to admit students who took the path of a matheletes unless you have something on top like attending a prestigious math research programme, doing part-time research with your local university professor, having several possibly golden IMO medals, being much into humanities etc. I guess.

[url=http://www.mitadmissions.org/Matt.shtml]Matt McGann, MIT admission officer[/url] said in one of the recent [url=http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php]Math Jams:[/url] "For some statistics, you can go to the website, but the bottom line is that test scores won't be the thing that gets you in, and probably won't be the thing that keeps you out, either." [url=http://www.mitadmissions.org/topics/apply/admissions_statistics/]Here[/url] you can find the admission statistics of the class entering fall 2008 at MIT.

Recent MIT Math Jams:

[url=http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=114]MIT Undergraduate Admissions Math Jam 2005[/url]
[url=http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=167]MIT Undergraduate Admissions Math Jam 2006[/url]
[url=http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=217]MIT Undergraduate Admissions Math Jam 2007[/url]

The next [url=http://www.artofproblemsolving.com/Community/AoPS_Y_Math_Jams.php]MIT Undergraduate Admissions Math Jam 2008[/url] will be on 23rd October 2008.

You are right SAT (or ACT) does yield a good discrimination as most possible MIT applicants score in the higher percentiles of those tests. Math olympiads are more helpful in this respect but note that it is not equally "easy" (or "hard") to qualify for IMO for example in all countries, so IMO qualification may not be necessarily sufficient given the huge number of international applicants and the set quota on admissions for international students. For example in the given link above 1,470 international applicants tried their luck and 119 (~ 8%) of them got in (3,052 / 10,344 ~ 30% for US), and also notice that "international" most likely means a much bigger applicant pool than only considering the US.


Solution:  :wink: $ \frac{\frac13\pi4^2\cdot6}{\pi4^2\cdot12}\equal{}\boxed{\frac16}$.

[quote="math154"]Solution:  :wink: $ \frac {\frac13\pi4^2\cdot6}{\pi4^2\cdot12} \equal{} \boxed{\frac16}$.[/quote]

Could you explain this more?

Not really; the volume of a cone is $ \frac13\pi r^2h$ and the volume of a cylinder is $ \pi r^2h$.

The base is the same size, so its (6/3)/12=2/12=1/6. MUch quicker.

That's what I would do in a countdown, but the justification is more important. :wink:

http://en.wikipedia.org/wiki/Volume#Volume_formula_derivation

check under "cone" for a derivation of the volume formula for the cone (requires calculus)


Plz don't go to the website to look up the answers :)

1. Last week I bought a word processor small enough to fit in my pocket. It can write in any language and also add, multiply, subtract, divide. It has a delete facility that will correct any error made and no electricity or batteries are required to operate it. Amazingly, it only cost me 25 cents. Why?

2. On the charts I remain number one,
This position cannot be undone,
If I burn, you may get
Just a little bit wet.
In a zeppelin, I am no fun.

3. A weightless and perfectly flexible rope is hung over a weightless, frictionless pulley attached to the roof of a building. At one end is a weight which exactly counterbalances a monkey at the other end. If the monkey begins to climb, what will happen to the weight - will it remain stationary, will it rise or will it fall?

4. Find the country name in the following sentence: As he was only six inches high, Tom Thumb rode the penguin easily.

5. Kevin's mother has three children. The first was called Alpha, the second was called Beta. What was the name of the third?

[hide="Nr. 4"] Guinea [/hide]
[hide="Nr. 5"] Kevin [/hide]

[hide="1"]It's a notepad?[/hide]

[hide="2"]Hydrogen[/hide]

Yeah, I think it will rise by the same amount the monkey climbs.

[quote="Flashlight"][hide="Number 1"]It's a pencil! It can write in any language as long as you know them...[/hide][/quote]

:rotfl: The eraser is the delete facility. Nice job.

#1 is just a piece of object which does not operate coz we need no bettery or electricity to operate it. and pencil and rubber can't be an answer coz u also need paper to write on. and how will it process the words????
  now its the turn of who put the question to tell what r the answers . we don't go to the web-site as is said to us.

I actually heard that question somewhere else, and the answer really is a pencil... though it doesn't completely make sense.


How do I write a double integral over a general region D, so that the "D" appears below and between the two integral signs?

Thanks for your help

Is this what you want?
[img]http://sixthform.info/gblog/pictures/4aed7bc8c2fefc0a6270ec34f9cafa40.gif[/img]
which is given by
\iint\limits_D
(for some reason this doesn't work on AoPS, but should be fine in your document)


[quote="djshowdown2"]When I make a pdf file through TeXnicCenter, I always end up with about 2 inches of header that i don't want.  How can I make this go away?[/quote]

That is not for the TeXnicCenter, but standards of TeX. I thought that the header by default was 1 inch and not 2. Anyway, try this

[code]\addtolength{\voffset}{-0.5in}
[/code]
in the preamble, and you can put more or less space in the number, that is, if it is still too much, well, you can do
[code]\addtolength{\voffset}{-1.0in}
[/code]
or modify in other ways. I hope it helps.

djimenez's solution will work, but [i]The LaTeX Companion[/i] (for one) advises never changing the values \hoffset and \voffset (for technical reasons that are not worth going into here).

The lengths which control vertical layout (from the top of the page down) are \topmargin, \headheight, \headsep, \textheight, and \footskip.  Probably the one that you want to change is \topmargin.

It's a little confusing, but your text starts at:
1 inch + \voffset (which should always be 0) + \topmargin + \headheight + \headsep

As you say, this is about 2 inches of space in total for a standard document if you don't change the settings (it actually comes out to 1 and 8/9 inches, but who's counting?).

The easiest and best way to move everything on your page up, for example, half an inch would be to put the command:

\addtolength{\topmargin}{-0.5in}

in your document's preamble.  This might leave you with too much space at the bottom, so you'd also want

\addtolength{\textheight}{0.5in}

[quote="djshowdown2"]Error: ! File ended while scanning use of \addtolength

Error: ! Emergency stop[/quote]Did you put the \addlength command(s) before your \begin{document}?  Actually, your error looks more like you forgot a } or something.

Alright, thanks, I fixed the header problem, but now printing.   I remember reading somewhere before that I had to change a 600 number to a 300, but I can't remember where to do this or where I read it  :?

EDIT: ojeez, nevermind, sorry... my printer cable was unplugged  :blush:


$S_{n}= \sum_{1 \leq j, k \leq n}^{n}\frac{1}{\sqrt{j^{2}+k^{2}}}$

$S_{n}\leq Cn$, where C is constant. Find minimal C. I managed to find $C=\frac{3}{2}\sqrt{2}$, but it's not minimal - brute force calculation gives C = circa 1,9.

???$S_{n}$ diverges because it contains at least $\frac{1}{\sqrt{2}}\sum_{k=1}^{n}\frac{1}{k}$.How is it correct??

Of course $S_{n}$ diverges, but $\frac{S_{n}}{n}$ does not, and we need to evaluate the latter, not the former.


AMC
AIME
Euler
function
algebra
binomial theorem

this is aime 1983 #6. determine the remainder when (6^83)+(8^83) is divided by 49. im stuck! 6 :equiv: -1 mod 7, 8 :equiv: 1 mod 7, 6^83 :equiv: -1 mod 7 , 8^83 :equiv: 1 mod 7. i was planning on dividing (6^83)+(8^83) by 7 and then 7 again, but my answer didnt match theirs, which is [hide]35[/hide], btw.

Hmm, has anyone ever noticed how the official AIME solutions (written by AMC) never use mods at all? Sometimes they use overly complicated solutions for problems that can be easily solved using mods. :/ This problem is one such case. The solution they give is just too weird for a #6. I too am interested in a nicer solution.

Write 6^83+8^83 as (7-1)^83 + (7+1)^83 and expand those out by the binomial theorem. However, since you're looking for the residue (mod 49), note that you only need two terms from each expansion, since any term that has at least two factors of 7 may be ignored.

ComplexZeta's solution works out rather nicely, too:

(83*7-1)+(83*7+1)=2*83*7 :equiv: 35(mod49)

[quote="Syntax Error"]i dont understand that darkquantum. can you please explain if you dont mind? :)[/quote]

Sure.

If you remember ComplexZeta's post, you can rewrite 6^83 and 8^83 as (7-1)^83 and (7+1)^83. When you expand those two expressions with BInomial Theorem, you get that most of the terms will have at least 7^2 in them (if you don't mind, try out the first couple of terms of (7-1)^83). Thus, when those terms are divided by 49, the remainder is 0, so you can discard those terms. The only terms left are the last two terms of each of the two expressions, namel 83*7-1 and 83*7+1, respectively. Then, you add those and divided by 49, which leaves a remainder of 35.


I have been planning to write this for a long time because I find it funny.
My hard disk spaces left:

C: 367 MB out of 11.1 GB
D: 297 MB out of 11.1 GB
E: 397 MB out of 14.9 GB
(and then it gets even funnier)
I: 58.4 MB out of 53.7 GB
J: 94.2 MB out of 58.0 GB

This friday I am buying a 160GB HD with a USB case and that should solve the problem for now, but I would like you to share similar experiences.

dunno about you

but i would get rid of some clutter

maybe you've put in too many programs that you dont use anymore

or you're playing heavy games that take 2 giga or more each, thus you quickly ran out of 60 giga spacE?