For positive real numbers $ a,b,c,d$ prove that $ [\frac{c}{(a \plus{} 2b)} \plus{}\frac{d}{(b \plus{} 2c)} \plus{}\frac{a}{(c \plus{} 2d)} \plus{}\frac{b}{(d \plus{} 2a)}][{2(ab\plus{}bc\plus{}cd\plus{}da\plus{}ac\plus{}bd)}] \ge (a\plus{}b\plus{}c\plus{}d)^2$

For example, $ \frac {c}{a \plus{} 2b} \equal{} \frac {c^2}{ac \plus{} 2bc}$, then Cauchy in engel form finishes it. If you don't know engel cauchy, note the right parentheses is the sum of all the denominators after multiplying like I said and split it up that way, then use cauchy, that's how engel cauchy is derived.

Yeah; it's a direct application of cauchy-Schwarz; notice that:

$ \frac{a^2}{x}\plus{}\frac{b^2}{y}\plus{}\frac{c^2}{z}\plus{}\frac{d^2}{t} \geq \frac{(a\plus{}b\plus{}c\plus{}d)^2}{x\plus{}y\plus{}z\plus{}t}$

Which Zhautykov problem are you refering? I think this is too easy. :lol:


I'm not really sure what you're asking... Could you explain what you mean?

[quote="brady00n"]A number x is divisble by another number y iff x is congruent to 0 (mod y).  I know that it means x is divisble by y when there is no remainder left, but what does the mod specifically mean here?  I don't think it means that the divisor becomes the base...  If 12 is divisble by 6, that would mean 12 is congruent to 0 (mod 6)...

Help me!!  AMC10 is tomorrow![/quote]

$12$ IS congruent to $0 \bmod 6$.  All that means is that $12$ expressed in base $6$ ends with a $0$, which it does:  $12_{10} = 20_{6}$.

That also means that if you divide $6$ into $12$, your remainder is $0$, but the actual meaning of $\bmod$ is more general.  Loosely, $x \equiv y \bmod z$ means that if you add or subtract $z$ a lot of times to one of the numbers, you will eventually get the other one.  Looking at $12 \equiv 0 \bmod 6$ this is obviously true, but we can also write $12 \equiv 12 \bmod 6$ or $12 \equiv 144 \bmod 6$ even though these are clearly not remainders.

Or, if you want to think about it another way, $x \equiv y \bmod z$ means $x-y$ is divisible by $z$ (or, in mathematical notation, $z | (x-y)$ ).

iff means if and only if.

So, a number x is divisble by another number y iff x is congruent to 0 (mod y) means if x is divisible by y, then x congruent to 0 mod y, and if x is congruent to 0 mod y, then x is divisible by y.


inequalities
quadratics
floor function
algebra
polynomial

Let $ a > 2$ be a constant.If there are just $ 18$ positive integers $ x$ satisfying the inequality

$ (x \minus{} a)(x \minus{} 2a)(x \minus{} a^2) < 0$ ,then find the range of $ a$.

The range of positive integers which satisfy the inequality holds is $ 0<x<a,2a<x<a^2$, i.e., $ 1\le x\le a\minus{}1,2a\plus{}1\le x\le a^2\minus{}1$. Hence
\[ a\minus{}1\plus{}a^2\minus{}1\minus{}2a\equal{}18
\]
whence we get a quadratic equation $ a^2\minus{}a\minus{}20\equal{}0$, i.e., $ (a\plus{}4)(a\minus{}5)\equal{}0$. Since $ a>2$, we have $ a\equal{}5$.

joh,$ a$ is not always integer,the given conditions for $ a$ is $ a\in \mathbb{R}$ and $ a>2$.

By the definition
$ a>x>0, a^2>x>2a$
then we get following equation
$ (\lfloor{a^2}\rfloor\minus{}\lfloor{2a}\rfloor\plus{}\lfloor{a}\rfloor)\equal{}\lfloor{a^2}\rfloor\minus{}\lfloor{a\plus{}\frac{1}{2}}\rfloor\equal{}18$ then $ 18>a^2\minus{}1\minus{}a\minus{}1$ then$ 5>a>2$
then a>4 $ \lfloor{a\plus{}\frac{1}{2}}\rfloor>4$ then $ \lfloor{a^2}\rfloor>22$
If a=4+r 1>r>0 $ \lfloor{16\plus{}8r\plus{}r^2}\rfloor>22$ then $ 8r>5$ then
$ \lfloor{a\plus{}\frac{1}{2}}\rfloor\equal{}5$
then
$ \lfloor{8r\plus{}r^2}\rfloor\equal{}7$
then solving the inequality $ 8>8r\plus{}r^2\ge7$ we get the answer.

hello, i have found that the lower bound for a would be $ a>\left(1\plus{}\sqrt{29\minus{}4\sqrt{7}}\right)$
 :lol:

EDIT: Wow I thought there was a square root sign around the 29 :facepalm:

That cant be a complex number.
@math486 plz show how you got that bound

[quote="nikeballa96"]@mew, how is that a complex number? $ \sqrt {29 \minus{} 4\sqrt7}$ would be a complex number if $ 4\sqrt7$ was greater than 29, which would mean that $ \sqrt 7$ would have to be greater than 7. am i misunderstanding you somewhere?[/quote]
Reals are complex numbers too...

hello.  my previous answer for the lower bound for $ a$ is wrong, i missed some details that was given in the problem.  Here is the solution:

First of all, since $ a>2$, then obviously $ 0<a<2a<a^2$.
Hence, the solution set for the polynomial inequality is the interval $ (\minus{}\infty,a)\cup(2a,a^2)$

Now we find that if $ a\equal{}5$, the positive integer solutions are exactly 18.  They are $ \{1,2,3,4,11,12,13,...24\}$  Obviously, if $ a$ were to be a little bit greater than $ 5$, then there would now be more than $ 18$ positive integer solutions.  In particular, say if $ a\equal{}5.0001$, then in addition to the original solutions, $ 5,25$ will now also be solutions.  Hence $ a$ cannot exceed $ 5$.  Also, if $ a$ were to be a little bit less than 5, say $ a\equal{}4.999$, then there will be exactly $ 19$ positive integer solutions since $ 10$ will enter the solution set.

We notice that $ a\equal{}4$ will yield less than $ 18$ positive integer solutions.  So there must be an $ a>4$ such that there will be exactly 18 positive integer solutions.  Thus, $ \{1,2,3,4\}$ will always be in the solution set and the remainder of the solution set must be either $ \{9,10,11,...22\}$ or $ \{10,11,12,...23\}$ or $ \{11,12,13,...24\}$.  Clearly the first and the last ones are impossible since for the first one, we must have $ a<4.5$ and for the last one, we must have $ a>5$.  So the remainder of the solution set is $ \{10,11,12,...23\}$ and we will have this if $ \sqrt{23}<a\leq\sqrt{24}$.

Thus, the only possibilities for $ a$ are $ \sqrt{23}<a\leq\sqrt{24}$ or $ a\equal{}5$

 :lol:


topology
pigeonhole principle
geometry
3D geometry
sphere
combinatorics unsolved
combinatorics

For a set S in R_3 whereby the mutual distance of any pair of points in S is rational, prove or disprove that S is countable.

The idea is to prove the assertion by contradiction, by first finding a two-dimensional manifold which contains an uncountable subset of $ S$, then a one-dimensional manifold which contains an uncountable subset of $ S$, and finally a finite set of points which contains an uncountable subset of $ S$.

Suppose $ S$ is uncountable.  Since there are countably many rationals but $ S\times S$ is uncountable, by the pigeonhole principle at least one element $ s\in S$ is equidistant from uncountably many points (countable union of countable sets is countable); that is, an uncountable subset  $ S_1\subset S$ lies on a sphere of some finite rational radius $ r$ centered at $ s$.  By the same argument, there is at least one element $ s_1\in S_1$ that is equidistant from uncountably many points on $ S_1$, and hence a subset $ S_2\subset S_1$ lying on an intersection of two spheres (a circle).  Repeating the argument once more gives an uncountable subset that lies on a finite set of points, a clear contradiction.  Thus, $ S$ is countable.

I think the argument generalizes fairly easily to $ \mathbb{R} ^n$


geometry
incenter
circumcircle
trigonometry
angle bisector
geometry unsolved

Let $ I$ denote the incenter of an acute triangle $ ABC$. The incircle $ (I)$ touches $ BC,CA,AB$ at $ D,E,F$ respectively.
The angle bisector of $ BIC$ meets BC at M and meets EF at P. Prove that $ PD\geq\frac {1}{2}\sqrt {4DE.DF \minus{} EF^2}$

In the your figure is a mistake : must $ P\in IM\cap EF$ ?!

No, P is the intersection of AM and EF!
Why do you think P is the intersection of IM and EF? In this case, it's a weaker problem! But how's your solution?

[quote="ricardokaka"]
The angle bisector of $ BIC$ meets BC at M and meets EF at P.   [/quote]
Read upper what wrote you !? Thus, $ P\in IM\cap EF$ .

Oh, i'm so sorry!
I edited my problem!

[quote="ricardokaka"]$ I$ is the incenter of an ancute triangle $ ABC$. The incircle $ (I)$ touches $ BC,CA,AB$ at $ D,E,F$ respectively.
The angle bisector of $ BIC$ meets BC at M and [b]AM[/b] meets EF at P. Prove that $ PD\geq\frac {1}{2}\sqrt {4DE.DF \minus{} EF^2}$[/quote]

it 's very easy

[b]solution[/b]

we use stewart in trangle MEF :[color=darkred] $ MF^{2} PE\plus{} MF^{2}PE \equal{} EF MP^{2} \plus{} PE PF EF$[/color](okie)

I can't understand your solution! It's not exact and it's so short!

[quote=" ricardokaka "]@hữu quân & all: Please help me! I don't understand your solu![/quote]

sorry ricardokaka

now I'll help you
I have  mistake in my proof we use stewart in trangle PEF : $ PD^{2} \equal{} \frac{ DE^{2}*PF \plus{} DF^{2}*PE \minus{} PE*PFEF}{EF}$

we have lemma

PD is bisector of trangle EPF .

I think proof lemma very easy for you  . then

we observe $ PF^{2} \equal{} ( PF \plus{} PE )^{2}$ >= $ 4PF*PE$ (1)

in view of lemma we have $ PD^{2} \equal{}DE^{2}*PF \plus{} DF^{2}*PE \equal{} DF*DE*EF$ (2)

we use (1) , (2) so that solve .ok

Huu quan!
I think there are some mistakes in your solution!
I think that you want to say that

[i]"we have lemma

[b]DP is bisector of trangle EDF [/b].

I think proof lemma very easy for you. then

we observe $ PF^{2} \equal{} ( PF \plus{} PE )^{2}$ >= $ 4PF*PE$ (1)

in view of lemma we have $ PD^{2} \equal{} DE^{2}*PF \plus{} DF^{2}*PE \equal{} DF*DE*EF$ (2)

we use (1) , (2) so that solve "[/i]

But i cann't prove the lemma "[b]DP is bisector of trangle EDF[/b]".
Continue to help me, please!

[quote="ricardokaka"][color=darkred]Let $ ABC$ be a triangle and the its incircle $ (I)$ touches  $ BC$ , $ CA$ , $ AB$ at $ D$ , $ E$ , $ F$

respectively. The angle bisector of $ BIC$ meets $ BC$ at the point $ M$ and $ AM$ meets $ EF$

at the point $ P$. Prove that exists the relation $ 4\cdot PD^2\ge 4\cdot DE\cdot DF \minus{} EF^2$ .[/color] [/quote]
[quote][color=darkred][b][u]Lemma I[/u][/b] Let $ ABC$ be a triangle and the bisector of the angle $ \widehat {BAC}$ meets

the side $ [BC]$ at the point $ L$ . Then $ AL^2 \equal{} AB\cdot AC \minus{} LB\cdot LC$ .[/color][/quote]
[hide="Proof of lemmma I."]
[color=darkblue]Denote the diameter $ [AM]$ of the circumcircle of the triangle $ ABC$ . Observe that

$ LA\cdot LM \equal{} LB\cdot LC$ and $ \triangle ABM\sim\triangle ALC$ $ \implies$ $ \frac {AB}{AL} \equal{} \frac {AM}{AC}$ $ \implies$

$ (AL \plus{} LM)\cdot AL \equal{} AB\cdot AC$ $ \implies$ $ \boxed {AL^2 \equal{} AB\cdot AC \minus{} LB\cdot LC}$ .[/color][/hide]
[quote][color=darkred][b][u]Lemma II.[/u][/b] In $ \triangle\ ABC$ consider the points $ M\in (BC)$ , $ E\in (CA)$ , $ F\in (AB)$ .

Denote the intersection $ P\in AM\cap EF$ . Then $ \frac {PF}{PE} \equal{} \frac {MB}{MC}\cdot\frac {AC}{AB}\cdot\frac {AF}{AE}$ . [/color][/quote]
[hide="Proof of lemma II."] [color=darkblue]$ \left\|\begin{array}{c} \{U,V\}\subset AM \\
 \\
BU\parallel CV\parallel EF\end{array}\right\|$ $ \implies$ $ \left\|\begin{array}{c} \frac {PF}{UB} \equal{} \frac {AF}{AB} \\
 \\
\frac {VC}{PE} \equal{} \frac {AC}{AE} \\
 \\
\frac {UB}{VC} \equal{} \frac {MB}{MC}\end{array}\right\|\ \bigotimes$ $ \implies$ $ \boxed {\frac {PF}{PE} \equal{} \frac {MB}{MC}\cdot\frac {AC}{AB}\cdot\frac {AF}{AE}}$ .[/color] [/hide]
[color=darkblue][u][b]Proof of the proposed problem.[/u][/b]

I'll use at first the [b]lemma II[/b] in the our case : $ \left\{\begin{array}{c} AF \equal{} AE \\
 \\
\frac {MB}{MC} \equal{} \frac {IB}{IC}\end{array}\right\|$ $ \implies$ $ \boxed {\frac {PF}{PE} \equal{} \frac {IB}{IC}\cdot \frac {AC}{AB}}\ \ (1)$ .

The segments $ [DF]$ , $ [DE]$ are chords in the circles with the diameters $ [IB]$ , $ [IC]$ respectively.

Therefore, $ \left\{\begin{array}{ccc} DF \equal{} IB\cdot\sin B \\
\ DE \equal{} IC\cdot \sin C\end{array}\right\|$ $ \implies$ $ \frac {DF}{DE} \equal{} \frac {IB}{IC}\cdot\frac {\sin B}{\sin C}$ , i.e. $ \boxed {\frac {DF}{DE} \equal{} \frac {IB}{IC}\cdot\frac {AC}{AB}}\ \ (2)$ .

Thus, $ (1)\ \wedge\ (2)\implies$ $ \frac {PF}{PE} \equal{} \frac {DF}{DE}$ $ \implies$ the ray $ [DP$ is the bisector of the angle $ \widehat {EDF}$ .

Now I'll use the [b]lemma I[/b] for the triangle $ EDF$ , i.e $ DP^2 \equal{} DE\cdot DF \minus{} PE\cdot PF$ .

Thus, $ 4\cdot PD^2 \plus{} EF^2 \equal{}$ $ 4\cdot DE\cdot DF \minus{} 4\cdot PE\cdot PF \plus{} (PE \plus{} PF)^2 \equal{}$

$ 4\cdot DE\cdot DF \plus{} (PE \minus{} PF)^2$ $ \implies$ $ 4\cdot PD^2 \plus{} EF^2 \ge 4\cdot DE\cdot DF$ , i.e.

$ \boxed {\ 4\cdot PD^2\ge 4\cdot DE\cdot DF \minus{} EF^2\ }$ . We have equality iff $ PE \equal{} PF$ $ \Longleftrightarrow$

$ DE \equal{} DF$ $ \Longleftrightarrow$ $ AB \equal{} AC$ , i.e. the triangle $ ABC$ is $ A$ - isosceles.[/color]

[quote="Virgil Nicula"][color=darkred][b][u]Extension.[/u][/b] Let $ ABC$ be a triangle and [b]let $ X$ be a point so that the ray $ [AX$ is the bisector of the angle[/b] $ \widehat {BAC}$ . Define : the projections $ D$ , $ E$ , $ F$ of the point $ X$ on the sidelines $ BC$ , $ CA$ , $ AB$ respectively ; the point $ M\in (BC)$ so that the ray $ [XM$ is the bisector of the angle $ \widehat {BXC}$ ; the intersection $ P\in AM\cap EF$ . Prove that $ 4\cdot PD^2\ge 4\cdot DE\cdot DF - EF^2$ .[/color]      [/quote]
[quote][color=darkred][b][u]Lemma I[/u][/b] Let $ ABC$ be a triangle and the bisector of the angle $ \widehat {BAC}$ meets

the side $ [BC]$ at the point $ L$ . Then $ \{\begin{array}{c} AL^2 = bc - LB\cdot LC \\
 \\
4\cdot AL^2 \ge 4bc - a^2\end{array}$ .[/color][/quote]
[hide="Proof of lemmma I."]
[color=darkblue]Denote the diameter $ [AM]$ of the circumcircle of the triangle $ ABC$ . Observe that

$ LA\cdot LM = LB\cdot LC$ and $ \triangle ABM\sim\triangle ALC$ $ \implies$ $ \frac {c}{AL} = \frac {AM}{b}$ $ \implies$

$ (AL + LM)\cdot AL = bc$ $ \implies$ $ \boxed {AL^2 = bc - LB\cdot LC}$ . Therefore,

$ 4\cdot AL^2 + a^2 =$ $ 4bc - 4\cdot LB\cdot LC + (LB + LC)^2 =$ $ 4bc + (LB - LC)^2\ge 4bc$.

In conclusion, $ \boxed {4\cdot AL^2 \ge 4bc - a^2}$ , with equality iff $ LB=LC$ $ \Longleftrightarrow$ $ b=c$ .[/color][/hide]
[quote][color=darkred][b][u]Lemma II.[/u][/b] In $ \triangle\ ABC$ consider the points $ M\in (BC)$ , $ E\in (AC$ , $ F\in (AB$ .

Denote the intersection $ P\in AM\cap EF$ . Then $ \frac {PF}{PE} = \frac {MB}{MC}\cdot\frac {AC}{AB}\cdot\frac {AF}{AE}$ . [/color][/quote]
[hide="Proof of lemma II."] [color=darkblue]$ \|\begin{array}{c} \{U,V\}\subset AM \\
 \\
BU\parallel CV\parallel EF\end{array}\|$ $ \implies$ $ \|\begin{array}{c} \frac {PF}{UB} = \frac {AF}{AB} \\
 \\
\frac {VC}{PE} = \frac {AC}{AE} \\
 \\
\frac {UB}{VC} = \frac {MB}{MC}\end{array}\|\ \bigotimes$ $ \implies$ $ \boxed {\frac {PF}{PE} = \frac {MB}{MC}\cdot\frac {AC}{AB}\cdot\frac {AF}{AE}}$ .[/color] [/hide]
[color=darkblue][u][b]Proof of the proposed problem.[/u][/b]

I'll use at first the [b]lemma II[/b] in the our case : $ \{\begin{array}{c} AF = AE \\
 \\
\frac {MB}{MC} = \frac {XB}{XC}\end{array}\|$ $ \implies$ $ \boxed {\frac {PF}{PE} = \frac {XB}{XC}\cdot \frac {AC}{AB}}\ \ (1)$ .

The segments $ [DF]$ , $ [DE]$ are chords in the circles with the diameters $ [XB]$ , $ [XC]$ .

Thus, $ \{\begin{array}{ccc} DF = XB\cdot\sin B \\
\ DE = XC\cdot \sin C\end{array}\|$ $ \implies$ $ \frac {DF}{DE} = \frac {XB}{XC}\cdot\frac {\sin B}{\sin C}$ , i.e. $ \boxed {\frac {DF}{DE} = \frac {XB}{XC}\cdot\frac {AC}{AB}}\ \ (2)$ .

Hence $ (1)\ \wedge\ (2)\implies$ $ \frac {PF}{PE} = \frac {DF}{DE}$ $ \implies$ the ray $ [DP$ is the bisector of the angle $ \widehat {EDF}$ .

Now I'll use the [b]lemma I[/b] for the triangle $ EDF$ , i.e $ \boxed {\ 4\cdot PD^2\ge 4\cdot DE\cdot DF - EF^2\ }$ .

We have equality iff $ PE = PF$ $ \Longleftrightarrow$ $ DE = DF$ $ \Longleftrightarrow$ $ AB = AC$ .[/color]

Oh, it's wonderful! Thank you very much, [b]Virgil Nicula[/b]. And now, please help me to solve some new problems in our forum!

[quote="ricardokaka"]
But i cann't prove the lemma "[b]DP is bisector of trangle EDF[/b]"[/quote]
Maybe i'm too late,but anyway i would like to post one of the shortest proof of this fact:
Let $ \zeta(I)$,be the incircle.Denote $ \zeta\cap AP\equal{}N$,then since $ AF\equal{}AE$,then $ \angle FNP\equal{}\angle ENP$,so $ MP$ is an angle bisector of $ FNE$.
Since $ MFNE$ is harmonic quadrilateral,then $ NF\cdot DE\equal{}NE\cdot DF$.So $ DP$ is an angle bisector of $ \angle EDF$.And we are done. :)

Hi [b]Virgil Nicula[/b]:
In your solution for the problem of extension, there is a conclusion: [i]"The segments [DF] , [DE] are chords in the circles with the diameters [XB] , [XC]"[/i].
I think that it's not exact! Please check it!

... the [b]projections[/b] $ D$ , $ E$ , $ F$ of the point $ X$ on the sidelines $ BC$ , $ CA$ , $ AB$ respectively...

Hence he segments $ [DF]$ , $ [DE]$ are chords in the circles with the diameters $ [XB]$ , $ [XC]$ respectively.


[quote="aadil"]a natural no is said to be good if it can be represented as the sum of 2 consecutive natural nos. as well as the sum of 3 comnsecutive natural nos. find the  no of good nos < 2009[/quote]

Clearly the no has to be an odd multiple of 3.
but it has to be greater than 6.

first such number is 9 last is 2007

so the number is (2007-9)/6 +1=333+1=334


Does anyone have any idea how to sort a sequence of numbers using Fibonacci Heaps? Or does anyone have the algorithm of Fibonacci Heaps implemented in C or C++?

thank you very much!

I found something searching the web.. so no need for help! thanks anyway!  :lol:

it is well explained and given full algorithm in CLRS. u can have a look there!


[quote="aadil"]just substitute $ n$ as $ mk$ where $ k\in\mathbb{Z^ \plus{} }$ and simplify[/quote]
Huh? Are you assuming that m divides n to prove that m divides n?

[hide="hint"]Prove that $ gcd(a^m \minus{} 1,a^n \minus{} 1) \equal{} a^{(gcd(m,n))} \minus{} 1$[/hide]

[quote="math4y"]Let $ m,n,a\in N^*$and :$ a^m \minus{} 1 \vdots a^n \minus{} 1$
Prove that:$ m \vdots n$[/quote]
It's actually quite easy....and Aadil...you got a wrong argument there...
Let $ m \equal{} np \plus{} q,0\le q \le n \minus{} 1$
Therefore,
\[ a^q \cdot (a^n)^p \minus{} 1 \equiv 0 (mod \ a^n \minus{} 1)\]

\[ \Longrightarrow a^q \cdot (1)^p \minus{} 1 \equiv 0 (mod \ a^n \minus{} 1)\]

\[ \Longrightarrow a^q \minus{} 1 \equiv 0 (mod \ a^n \minus{} 1)\]
This implies $ a^n \minus{} 1|a^q \minus{} 1$
If $ a^q \minus{} 1$ is positive...Therefore $ a^q \minus{} 1 \ge a^n \minus{} 1$ which is impossible since $ q < n$
$ \Longrightarrow a^q \minus{} 1 \equal{} 0 \Longrightarrow q \equal{} 0$ and the result follows...
We are done... :lol:


abstract algebra
Ring Theory
superior algebra
superior algebra unsolved

$ X$ is a fixed set, and $ P(X)$ is the set of subsets of $ X$. We define : $ A\triangle B \equal{} (A\setminus B)\cup (B\setminus A)$
Then $ (P(X),\triangle,\cap )$ is a commutative ring which has unity, with $ \triangle, \cap$ are the unity of addition and multiplication respectively.

For $ a\in X$ let $ I_a\equal{}\{A\in P(X): a\notin A\}$.
Prove that $ I_a$ is the maximal ideal of $ P(X)$. If $ X$ is finite, show all ideals of $ P(X)$ and prove that all maximal ideals of $ P(X)$ have the form $ I_a$ for $ a\in X$.

the ring is isomorphic to $ \mathbb{F}_2^X$, I think it is easier to work with this. each $ a \in X$ yields a projection $ \mathbb{F}_2^X \to \mathbb{F}_2$ with kernel $ I_a$. hence $ I_x$ is a maximal ideal. if $ X$ is finite, an ideal corresponds to a subset of $ X$: we have to choose some factors in $ \mathbb{F}_2^n$.


Suppose all roots of the polynomial \[ x^4  \minus{} 4x^3  \plus{} ax^2  \plus{} bx \plus{} 1 \equal{} 0
\]
are positive real numbers.Show that all roots of the polynomial are equal.

First, use Vieta. Then use AM-GM, confirm that the inequality becomes an equality, and then use the equality case.


[quote="AntonioMainenti"][hide="hint"]Consider $x$ in several well-chosen intervals.[/hide][hide="tedious solution"]The absolute value signs around the entire LHS may obviously be disregarded.

We'll look at $x$ in the intervals $(-\infty,-3),(-3,0),(0,3),(3,5),(5,\infty)$.

In the first case we have $-x-3+5-x=7\Longrightarrow{x}=-\frac{5}{2}$. But this is a contradiction, so no solution in this interval.
Second, $3+x+5-x=7\Longrightarrow8=7$. No solutions here.
Third, $3-x+5-x=7\Longrightarrow{x}=\frac{1}{2}$.
Fourth, $x-3+5-x=7\Longrightarrow2=7$. Nope.
Fifth and finally, $x-3+x-5=7\Longrightarrow{x}=\frac{15}{2}$.

The two roots of the equation are thus $\{\frac{1}{2},\frac{15}{2}\}$.[/hide][/quote]

Actually, it's not tedious. Could there possibly any other way (I did it your way too) that shorter?

[quote="236factorial"]Actually, it's not tedious. Could there possibly any other way (I did it your way too) that shorter?[/quote]I doubt it. Absolute value signs generally don't lend themselves to pretty solutions, but casework.

Yeah, I think that's about the best way we can do.


Prove that a list can be made of all the subsets of a finite set in such a way that:

(i) The empty set is first in the list,
(ii) each subset occurs exactly once, and
(iii) each subset in the list is obtained either by adding one element to the preceding subset of by deleting one element of the preceding subset.

Solutions which did not use graph-theoretic principles would be appreciated.

[quote="blahblahblah"]Prove that a list can be made of all the subsets of a finite set in such a way that...[/quote]

What zabelman says is essentially a construction of the binary "Gray Code".  I suggest you search the Internet for more information.

I think you'll find a rich set of proofs, properties, and descriptions of Gray Code that are consistent with what you seek to prove.

Does anyone with a passing familiarity with graph theory know if Ore's Theorem provides a simple solution to this problem if we consider the 2^n subsets of S as vertices of a graph in this manner:

{}
{0}, {1}, ...... {n} --> C(n,1)
{0,1}, {0,2}, ...... {n-1,n} --> C(n,2)

......

{0,1, ..... , n-1, n}

I had been tinkering around with a solution similar to zableman's, but I did not notice how writing it backwards allows us to complete the inductive step.  Thus, this graph theoretic idea was the best I could come up with.

As far as I can tell, the conditions of Ore's theorem are satisfied here, so there should be a Hamiltonian cycle.

[quote="blahblahblah"]As far as I can tell, the conditions of Ore's theorem are satisfied here, so there should be a Hamiltonian cycle.[/quote]

I don't claim to be an expert at Graph Theory.  Every node of your graph has degree exactly equal to n.  Right?  So every pair of non-adjacent nodes would have a combined degree exactly equal to 2n.

Doesn't Ore's Theorem require the combined degree of every pair of non-adjacent nodes to be at least equal to the total number of nodes (2^n)?  I don't believe the condition is satisfied for n>2.

::blinks::

Yeah, I need remedial counting classes.

::blinks::


floor function
inequalities
inequalities proposed

let $a,b,k\in N$ such that
$a+b\ge 2k+1$ prove that
$\lfloor\frac{a}{2}\rfloor+\lfloor\frac{b}{2}\rfloor\ge k$

[quote="Megus"]counterexample- let have equality: $\frac{a+b}{2}-\frac{1}{2}=k$ and let $a=2m+1$, $b=2n+1$ then your inequality equals: $m+n \geq k$ but on the other hand $m+n+\frac{1}{2}=k$ so you obtain opposite sign in your ineq.[/quote]
when we have equality then both of a,b cant be odd ! :?

ah, sorry  :blush: how could I have been so stupid, of course you're right.

is'nt easy problem ?
Do u  think i will lost any points?

yes it's easy (though I don't have for now any idea other than regarding few cases, any short way?)

I think whether you lose points or not depends only on checker now, because it is just a diminutive overlooking.


induction
inequalities
real analysis
real analysis unsolved

Find the limit
$lim_{n \to \infty}\frac{[x]+[2x]+...+[nx]}{n^{2}}=?$

Using sigma notation:

$ \frac{{x}{\sum_{r=1}^{n}{r}}}{n^{2}}$

This can use the formula (which can be proved by induction if needed, but it should acceptable to use it as stated)

$ \frac{x({\frac{1}{2}})({n})({n+1})}{n^{2}}$

Hopefully you can work it out from here, a little neating up and it should become obvious. (I'll post the rest of the solution if you're still stuck)

I assume $[x]$ means the integer part of $x$. Since $x-[x]$ is at most $1$, the difference between $[x]+\dots+[nx]$ and $x+\dots+nx$ is at most $n$. This difference disappears in the limit, as $\frac{n}{n^{2}}\to 0$. Therefore, the limit stays the same if you remove the square brackets. Then you can follow the suggestion of [b]artalyx[/b].

I am still unable to find the limit. I think Sandwich Theorem may work as [x]<= x. But the other inequality I am not getting.

[quote="Soarer"]you need x - 1 <= [x][/quote]

Thanks Soarer, I got it.
I don't know whether your English/German is correct or not but your maths is correct.


geometry
trigonometry
trig identities
Law of Sines

Compute the area of a right triangle with hypotenuse $2$ and the median to the hypotenuse divdes the right angle into $\frac{2}{3}$ and $\frac{1}{3}$.

[hide]I'm assuming that you understand the picture and the variables. If not, ask me what you don't get and I will explain.

(Don't ask me about my mixing up with degrees and radians)

From the law of sines on $\triangle ABD$,

$\frac{x}{\sin(\theta)}=\frac{1}{\sin(60^{\circ})}$

On $\triangle BDC$,

$\frac{\sqrt{2-x^{2}}}{\sin(\pi-\theta)}= \frac{1}{\sin(30^{\circ})}$

But notice that if we cross multiply, we get that two quantities are equal:

$\sin(\theta) = x \cdot \sin(60^{\circ})$

$\sin(\pi-\theta) = \sqrt{2-x^{2}}\cdot \sin(30^{\circ})$

These are equal, so we can solve for $x$:

$\frac{x\sqrt{3}}{2}= \frac{\sqrt{2-x^{2}}}{2}$

$\sqrt{3x^{2}}= \sqrt{2-x^{2}}$

$4x^{2}= 2$

$2x = \sqrt{2}$

$x=\frac{\sqrt{2}}{2}$

$\sqrt{2-x^{2}}= \frac{\sqrt{6}}{2}$

So the area is $\frac{\sqrt{3}}{4}$[/hide]

[hide]
Since the median of the hypotenuse divides the right angle into 1/3 and 2/3 it divides it clearly into 60 and 30 degrees.

Since it is the median we now have atleast two isoseles triangles. This leads us to realize that we have a 30-60-90 triangle.

Since the hypotenuse is 2, the base is $\sqrt{3}$ and the height is 1.

$k = \frac{1}{2}bh = \boxed{\frac{\sqrt{3}}{2}}$
[/hide]


calculus
integration
function
trigonometry
calculus computations

Given a continuous function $ g$ on $ \mathbb{R}$, let $ h(x)\equal{}\frac{1}{2}\int_{0}^{x}\sin2(x\minus{}y) g(y) dy.$ Show that $ h''\plus{}4h\equal{}g$

Expand $ \sin 2(x\minus{}y)\equal{}\sin 2x\cos 2y\minus{}\cos 2x\sin 2y$ in order to separate $ x$ from the integral. You'll then have
\[ h(x)\equal{}\frac{\sin 2x}{2}\int_0^x g(y)\cos 2y\ dy \minus{}\frac{\cos 2x}{2}\int_0^x g(y)\sin 2y\ dy\]
Now it's straightforward enough, since $ \frac{d}{dx} \left[ \int_0^x A(y)\ dy \right] \equal{}A(x)$ is now useable.


Does every polynomial equation (of odd degree) with rational coefficients have at least one rational root?

obviously not! take $ x^3 \minus{} 2 \equal{} 0$

Realy I don't under stand your problem.
You can prvove that for each n there exist an polynomial  with rational conficient and degree n has no rational solution.
For example :
$ f(x)=x^n+a_{n-1}x^{n-1}+...+1$
Where $ a_i\in Z$
If it has a rational solution then this solution only ${ \{-1,1}$
if we chose $ a_i$ for $ f(1),f(-1)$ is dinferent from 0 then it has no rational solution.


If $ x$, $ y$ and $ z$ are real numbers such that \[ x^2 \plus{} y^2 \plus{} z^2 \minus{} xy \minus{} yz \minus{} zx \equal{} 8\] what is the maximum possible difference between any two of $ x$, $ y$ and $ z$?

[hide="Solution"]$ 2x^2 \minus{} 2xy \plus{} 2y^2 \minus{} 2yz \plus{} 2z^2 \minus{} 2zx \equal{} 16 \iff (x \minus{} y)^2 \plus{} (y \minus{} z)^2 \plus{} (z \minus{} x)^2 \equal{} 16$

Let $ a \equal{} x \minus{} y \; \; b \equal{} y \minus{} z \; \; c \equal{} z \minus{} x$. Then, $ a \plus{} b \plus{} c \equal{} 0$ and $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 16$. We want to maximize one of these numbers.

Substituting for $ c$, we have:
$ 16 \equal{} a^2 \plus{} b^2 \plus{} ( \minus{} a \minus{} b)^2 \equal{} 2a^2 \plus{} 2ab \plus{} 2b^2 \equal{} 2 \left( \frac {a}{2} \plus{} b \right)^2 \plus{} \frac {3}{2} a^2$

Thus, $ \frac {3}{2} a^2 \leq 16 \implies \boxed{a \leq \frac {4}{3} \sqrt {6}}$[/hide]


i cnt get this problem. I knw there should be an easy sequence bt i cnt find it. Id appreciate it if u guys could!

A grocer stacks oranges in a pyramid-like stack whose rectangular base is 5 oranges by 8 oranges. Each orange above the first level rests in a pocket formed by four oranges in the level below. The stack is completed by a single row of oranges. How many oranges are in the stack?

(A) 96
(B) 98
(C) 100
(D) 101
(E) 134


just out of curiousity, should there be a $ x_{100}^2 \cdot x_1$ term? that would make things nice and symmetric :P

This is ridiculous. We are posting stronger versions of [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=214705]difficult IMO shortlist problems[/url] (this is an A6) in intermediate topics?

[quote="vinskman"]Suppose $ x_1^2 \plus{} x_2^2 \plus{} x_3^2 \plus{} \cdots \plus{} x_{100}^2 \equal{} 1$

Find the maximum value of

$ x_1^2*x_2 \plus{} x_2^2*x_3 \plus{} x_3^2*x_4 \plus{} \cdots \plus{} x_{99}^2*x_{100}$[/quote]

Sorry , I think I was wrong to post it.
[color=blue]Now, edit[/color]
[color=green]Find the maximum value of[/color]

$ x_1^2*x_2 \plus{} x_2^2*x_3 \plus{} x_3^2*x_4 \plus{} \cdots \plus{} x_{99}^2*x_{100}\plus{}x_{100}*x_1$

 :blush:


trigonometry function calculus integration algebra polynomial

I was wondering, how could one get an exact solution for a trig equation in which the variable is inside and outside of the trig function, such as x+sinx=1, or xsinx=38?

I don't think it's possible. It requires some really advanced math to prove this, though.

I don't think it really requires advanced math.  I once asked a professor why the integral of sin(x^2) can't be expressed in closed form and he said the proof was basically combinatorial and messy: showing that no linear combination of the product of elementary functions (trigonometric, logarithmic, exponential, polynomial) can be the answer.  I think solving equations like x = sinx is the same.  I doubt there is any answer that can be written in closed form.

Actually Polymorphic, the example you gave of x=sinx is the only example of problems like this to have a simple answer, because if x is 0, the equation works.

yup...sorry my bad
I meant x = cosx.  then x is approximately .7390851332 which one can get by evaluating cos^n(x) for large values of n.

You can graph the two functions and find the intersections.

yes, I have been able to find solutions approximately through graphing, but those haven't helped in finding exact solutions.

As I said, I don't think it's possible to find a closed form for the solution to x = cosx using elementary functions.  For example the solution to the equation x^2 =2 we define as  :sqrt: 2.  Similarly I think the best we can do in this case is call the solution to that equation pi, phi, epsilon or some other random name.


trigonometry
geometry
trig identities
Law of Cosines

If the positive reals x,y,z satisfy

x:^2: + xy + y:^2:/3 = 25
y:^2:/3 + z:^2: = 9
z:^2: + zx + x:^2: = 16

Find xy + 2yz + 3zx.

Algebra? I think not... Ok I should say no more. Have fun with this problem. It's a very nice one.

yeah...definitely doesn't look like algebra to me...looks more like [hide]the law of cosines and areas of right triangles![/hide]

Those are some pretty big hints.

Now I've got this quadrilateral, right, and its diagonals, yes, and how do I find the value of that expression?

Call (3^1/2/2)*x=a and x/2=b and y/3^1/2=c.
First we can add the last 2 to get the 3'rd one (9+16=25), so 2z^2+zx=xy iff z/a=c/(z+b) (*) (you'll see why it's written like this).
We can rewrite it (the system) like this:

(c+a)^2+b^2=25

c^2+z^2=9

(z+b)^2+a^2=16.

From (*) it means that we must have equality in Cauchy for the ineq (c^2+z^2)*((z+b)^2+a^2) >= (c*(z+b)+za)^2, so we can replace >= with =. This is where we used (*). This means that 144=9*16=(cz+cb+za)^2=
((zy)/3^1/2+(xy)/2*3^1/2+(xz)*(3^1/2/2))^2 = ((1/2*3^1/2)*(xy+2yz+3zx))^2 = 1/12*(xy+2yz+3zx)^2=144=12^2, so (xy+2yz+3zx)^2=12^3, so xy+2yz+3zx=12^3/2. I might have made some calculation mistakes, but this is definitely the idea.


Given an infinite sequence of positive integers: $a_1< a_2< a_3< \cdots$ . Denote $S = \{ a_i + a_j | i, j \in \mathbb{N} \}$. Prove that there exists an infinite subset of $S$ such that in this subset, no element is a multiple of another element.

This is way too hard of a problem, seems like we have nothing to work with... :(

Anyone?

I am interested in it too.
Maybe I will find time to think about...


If $ f(f(x))$ is a continuous function, is $ f(x)$ also continuous?

If $ f(f(x))$ is differentiable, then is $ f(x)$ also differentiable?

Consider f(x) = 0 for rational numbers, f(x) = 1 for irrational numbers.


geometry
3D geometry
floor function
number theory

A rectangular solid block of wood is 6 cm high, $ M$ cm wide, and $ N$ cm long, where $ M > N > 1$ are positive integers.  The outside surface of the block is painted green and the block is then cut into $ 6MN$ unit cubes.  Exactly half of these cubes have no paint at all on them.  Find the least possible value of $ N$.

Here's what I have:

[hide]
So we have $ 6MN$ total cubes, and $ 4(M\minus{}2)(N\minus{}2)$ with no paint.  Then $ \frac{6MN}{4(M\minus{}2)(N\minus{}2)}\equal{}2$.
[/hide]

I'm stuck after that though.  Help please!

[hide]

For a step-by-step solution:

$ \frac{6MN}{4(M \minus{} 2)(N \minus{} 2)} \equal{} 2 \newline
3MN \equal{} 4(M \minus{} 2)(N \minus{} 2) \newline
3MN \equal{} 4MN \minus{} 8M \minus{} 8N \plus{} 16 \newline
MN \minus{} 8M \minus{} 8N \equal{} \minus{}16 \newline
MN \minus{} 8M \minus{} 8N \plus{} 64 \equal{} 48 \newline
M(N \minus{} 8) \minus{} 8(N \minus{} 8) \equal{} 48 \newline
(M \minus{} 8)(N \minus{} 8) \equal{} 48$

If $ N \minus{} 8$ is a factor of 48, then the least value $ N$ can take is when $ N \minus{} 8 \equal{} 1 \rightarrow \boxed{N \equal{} 9}$
[/hide]

Unless I screwed up somewhere, of course.

Thanks, I shoulda experimented a bit farther.

Another (quite unrelated) number theory problem:

What is the largest prime, $ p$, such that $ p^2$ divides
\[ \frac {300!}{(150!)^2}?
\]

This one follows hand in hand:

How many perfect squares divide
\[ \frac {300!}{(150!)^2?}
\]

$ \left\lfloor\frac {300}{p}\right\rfloor \plus{} \left\lfloor\frac {300}{p^2}\right\rfloor \ge 2 \left(\left\lfloor\frac {150}{p}\right\rfloor \plus{} \left\lfloor\frac {150}{p^2}\right\rfloor\right) \plus{} 2$

It's obvious that $ p^2 \le 300$, therefore $ p \le 17$. And in fact, $ 17^2|\frac {300!}{(150!)^2}$.

I don't get what you did.   Can you please explain a little more?  Pretty please?

Also, could you help me on the related problem too?

For some background -- the number of factors of $ p$ that divide $ n!$ is exactly

$ \lfloor\frac{n}p\rfloor + \lfloor\frac{n}{p^2}\rfloor + \lfloor\frac{n}{p^3}\rfloor + ...$

, where $ \lfloor x \rfloor$ denotes the smallest integer less than or equal to $ x$, and the sum goes until $ p^k > n$.

Now, the number of factors of $ p$ that divide $ 300!$ is

$ \lfloor\frac{300}p\rfloor + \lfloor\frac{300}{p^2}\rfloor + \lfloor\frac{300}{p^3}\rfloor + ...$

And the number of factors of $ p$ that divide $ (150!)^2$ is

$ 2(\lfloor\frac{150}p\rfloor + \lfloor\frac{150}{p^2}\rfloor + \lfloor\frac{150}{p^3}\rfloor + ...)$

Now, we want the biggest value such that it divides $ 300!$ twice more than it divides $ (150!)^2$. If $ p^2 > 300$, then we have

$ \lfloor\frac{300}p\rfloor \ge \lfloor\frac{150}p\rfloor + 2$

But, as $ \lfloor2x\rfloor \le 2\lfloor x \rfloor + 1$ (do you see why?), then this cannot be so. Therefore $ p^2 < 300 \Rightarrow \boxed{p \le 17}$

By inspection, we can verify that 17 divides $ 300!$ exactly $ \lfloor\frac{300}{17}\rfloor + \lfloor\frac{300}{289}\rfloor = 17 + 1 = 18$ times, while it divides $ (150!)^2$ exactly $ 2\lfloor\frac{150}{17}\rfloor = 2 \cdot 8 = 16$ times.

Therefore it divides $ \frac{300!}{(150!)^2}$ exactly $ 18 - 16 = 2$ times, which means $ 17^2|\frac{300!}{(150!)^2}$. But we have proved that no larger $ p$ can satisfy this condition, so 17 is the largest.

The other problem is pretty similar -- in fact, there are few squares that divide $ \frac{300!}{(150!)^2}$...


People are complaining of the scrollbar missing in my blog. When I look (using IE7), it's there, but I haven't tried with other browsers.

You messed up your CSS somehow. You can fix it, just google for a CSS tutorial and see what you did wrong (or alternatively take every command out step by step and see which one is causing the trouble.)


Let $ a,b,c$ and $ p$ be nonnegative real numbers. Find the largest range of $ p$ such that

$ \frac {a(b\plus{}c)}{pa^2 \plus{} b^2 \plus{} c^2} \plus{} \frac {b(c\plus{}a)}{pb^2 \plus{} c^2 \plus{} a^2} \plus{} \frac {c(a\plus{}b)}{pc^2 \plus{} a^2 \plus{} b^2} \le \frac 6{p \plus{} 2}$.

And $ p_{min} \equal{} 5 \minus{} 2\sqrt {6}$.


On a $ 25$-question multiple choice test, Dalene starts with $ 50$ points. For each correct answer, she gains $ 4$ points; for each incorrect answer, she loses $ 2$ points; for each problem left blank, she earns $ 0$ points. Dalene answers $ 16$ questions correctly and scores exactly $ 100$ points. How many questions did she answer incorrectly?

We have \[ 50\plus{}16(4)\minus{}2x\plus{}0\equal{}100 \Rightarrow x\equal{}\boxed{7}\]


limit
real analysis
real analysis unsolved

Find the limit:
$ \lim_{n->{\infty}} {\frac{x+ \sqrt{x}+ \sqrt[3]{x}+ . . . + \sqrt[n]{x} - n}{ln n}} $
$ x>0$

I think, you mean $x>0$ instead of $a>0$, dont you?

Use the following theorem:
If $\lim{x_{n}}=a$,then $\lim{\frac{x_{1}+x_{2}+\ldots+x_{n}}{n}}=a$.

I found the result. It is $ ln x $. Thanks anyway!  :D

As you can see I already posted the solution, but thanks anyway !  :D

[quote="cezar lupu"]nemesis,just use Cesaro-Stolz's theorem and the problem becomes trivial. ;)[/quote]

what does that theorem state?

Just look at the National romanian Forum and try to find a topic named "Teorema Stolz-Cesaro". ;)


[u] [i]  Problem[/i][/u]

 [i] The expansion of  $ f(x) \equal{} (x \plus{} 1) \plus{} 2(x \plus{} 1)^{2} \plus{} 3( x \plus{} 1)^{3} \plus{} .... \plus{} 2008(x \plus{} 1)^{2008}$

  is the polynomial $ f(x) \equal{} a_{0} \plus{} a_{1}x \plus{} a_{2}x^{2} \plus{} ...... \plus{} a_{2008}x^{2008}$

  Calculate  $ a_{100}$

   Nice problem

  Argady , Scorpius119 and mathlinkers   I am  waiting for your nice solutions

 Love An Forever
 Huỳnh Võ Phương An[/i]

[i] You are really intelligent Rust

  Nice solution for nice problem

   Congratulation[/i] :lol:

  [i]Love An forever[/i]
[i]  Huỳnh Võ Phương An[/i]

[i]  But i think your technique can not apply for this problem

    The expansion of  $ P(x) \equal{} f(x) \equal{} (x \plus{} 2) \plus{} 2(x \plus{} 2)^{2} \plus{} 3( x \plus{} 2)^{3} \plus{} .... \plus{} 2008(x \plus{} 2)^{2008}$

  is the polynomial  $ P(x)  \equal{} a_{0} \plus{} a_{1}x \plus{} a_{2}x^{2} \plus{} ...... \plus{} a_{2008}x^{2008}$

  Calculate  $ a_{100}$ :wink: [/i]


How does one prove that $ \left[\frac{x}{n}\right]\equal{}\left[\frac{[x]}{n}\right]$ for $ x\in\mathbb{R}$ and $ n\in\mathbb{Z}_{>1}$?  Obviously, this does not hold if $ n$ can be negative but I'd like to see a proof for the case above.

Write $ x\equal{}nt\plus{}r\plus{}r'$ with $ n,r\in\mathbb{N}_0,r'\in\mathbb{R}$ and $ 0\leq r\plus{}r' <n$ and $ 0\leq r'<1$. This gives $ \frac{x}{n}\equal{}t\plus{}\frac{r\plus{}r'}{n}\Longrightarrow \left[\frac{x}{n}\right]\equal{}t$ and $ \frac{[x]}{n}\equal{}t\plus{}\frac{r}{n}\Longrightarrow \left[\frac{[x]}{n}\right]\equal{}t$


the question is
Find real numbers m and b such that S= \sum 50j=1 j +mj +b
is minimal

correct me if I am wrong, but isn't this sum equal with:
S=1275+1275m+50b  and obviously this sum doesn't have a minimal value for m and b real variables.
There must be something wrong with your problem's statement.

Find real numbers m and b such that \sum(j=1 to 50) (j^2 + mj +b)^2 is minimal(i am sorry abt the wrong statmentof the problem earlier.

there are more computations to make but the general idea still holds:
S=m2 \sum j2+2m (\sum j)(2b +  \sum j) + b2+2b \sum j2+ \sum j4

This is a second degree function of m (or b if you like) and it's minimal value is obtained when:
m=-(\sum j(2b+ \sum j)/(\sum j2)

Plugging this m into the sum, we will obtain an second degree function of b, so we can easily compute the minimal value for b too.

I leave the computation to you.


Balak ko sana magpa math marathon dito sa forum natin ... hmmmm... basta may mag apply na at least 6 na tao lang.... pag merong interesado mag reply lang kayo sa topic na ito. i-take note ko nalang names nyo... tapos later ko nalang state yung rules... Pero kung makita ko wala pa tlgang mag reply sa topic na ito then forfeit nalng.. hayyy...

Mag register na kasi yung mga nakikibasa jan!!

[quote=Math-wiz]I can't understand even a single word lol[/quote]

lol I have Filipino class in school so I understand


A right circular cylinder with its diameter equal to its height is inscribed in a right circular cone. The cone has diameter $ 10$ and altitude $ 12$, and the axes of the cylinder and cone coincide. Find the radius of the cylinder.

$ \textbf{(A)}\ \frac83 \qquad
\textbf{(B)}\ \frac{30}{11} \qquad
\textbf{(C)}\ 3 \qquad
\textbf{(D)}\ \frac{25}{8} \qquad
\textbf{(E)}\ \frac{7}{2}$

[hide] Let the diamter of the cylinder be 2r. Examing the cross section of the cone and cylinder, we find two similar triangles. Hence, $ \frac{12-2r}{12}=\frac{2r}{10}$ which we solve to find $ r=\frac{30}{11}$. Our answer is $ \boxed\text{(B)}\ \frac{30}{11}$.[/hide]

Since the axes coincide, we can take the problem to $2$ dimensions. We have isosceles triangle $ABC$ with $AB=AC$, and a square $DEFG$ inscribed inside, with $G$ between $B$ and $F$ on $BC$. By the Pythagorean Theorem, we have $AB=AC=13$. Set $BG=x$. This means that $DG=GE=10-2x$. However, we have $\triangle BDG \sim \triangle BAO$, where $O$ is the midpoint of $BC$. Solving for ratios, we find that $x=\frac{25}{11}$, which means half the sidelength of the square is $$\frac{1}{2}\cdot\left(10-\frac{50}{11}\right) = \boxed{\frac{30}{11}}.$$

Similar triangle gives $\frac{5-r}{2r}=\frac{5}{12}\to r=\frac{30}{11}$.


Quick question for everyone:

Does anyone have access to PDFs of past USAPHO exams as well as answers.  I'm self-studying AP Physics B & C this summer as well as preparing for the Olympiad in general (using Halliday/Resnick/Walker and Irodov), and I was interested in seeing how I would fare on past exams.  I've literally searched everywhere on the web, and have had no success in locating any exams.  I would greatly appreciate if someone could point me towards a specific website that has some past exams, or send me PDFs or Docs via PM or my e-mail (Tommy2030@gmail.com).  Thanks for your time,

-Tommy

This website has sample exams from 1997:

http://www.compadre.org/PSRC/items/detail.cfm?ID=390

Unfortunately, the rest of the exams from past years are not free.  However, there is a link on that page to where you can purchase a CD with copies of past exams.

Irodov is great, btw.

While we're talking about great books, Halliday/Resnick is great (although I'm not sure if this is similar to your text with the added Walker). I used Knight in Gen Physics I and II, but I always studied for my final from Halliday and Resnick.

Ahh, ok. Well, I checked out the store and unfortunately the product (the CD of past exams) has been discontinued....sooo.....does that mean that there is no way for me to obtain them then?

Whoa, that's surprising.  Well, actually, starting this year, the format of the first two USAPhO screening exams has changed, but you can still look at the 1997 semifinal exam on that page to get a feel for its format, which hasn't changed.  I think the problems in Irodov should be good enough if you want to practice for the USAPhO.

Yeah, I took a look at the 1997 multiple choice.  Has it increased in difficulty over the years at all? Because right now I'm strictly looking at the Mechanics related problems (because that's all I've covered so far), and they seem pretty easy to me.  If that is the normal level of difficulty.....then I should slaughter the exam lmao.

I think the multiple-choice problems have slightly increased in difficulty - in previous years, it seems like the multiple-choice exam was just a test of speed.  Now, there are three rounds: an "F=ma" mechanics multiple-choice exam (25 questions in 75 minutes, like the AMC) that's a bit harder than previous multiple-choice exams (at least near the end), a four-question free-response quarterfinal exam covering only mechanics and electrostatics, and the same old semifinal exam covering everything.  The main thing to study for would be the semifinal exam.

@Jaspher: If you want usapho type problems check out baupc stuff or even AP French's mechanics book

Aptitude Test Problems in Physics by S.S.Krotov is a very good book for Physics Olympiad .Its Just that it is somehow hard to get...But hey!! its cheap Edition is available in India !!!!


LaTeX
logarithms
limit
trigonometry
calculus
calculus computations

lim x-> infinite

x^( (ln9)/(1+lnx) )

lim x-> 0+

(7x+1)^cotx

lim x-> infinite

...................(4x+3)
[  4x-1    ]  ^
[  ------   ]
[  4x+5   ]

the three brackets on top of each other represent one big brackt

Pretty version (in $ \text{\LaTeX}$):

Find the following limits:
[quote="sneaky"]$ \lim_{x\to\infty} x^{\frac{\ln 9}{1+\ln x}}$

$ \lim_{x\to 0^+} (7x+1)^{\cot x}$

$ \lim_{x\to\infty} \left[\frac{4x-1}{4x+5}\right]^{4x+3}$[/quote]Quote me if you want to see how that works.

On the actual solutions, I'll just say this: when dealing with exponential limits, it's usually a good idea to start by taking logarithms of everything.

i now understand how you do the last two questions, but i am still lost on the first one

Answer to the first question that is the limit on ln9 is 9..

$ \lim_{x\to\infty}{x^{\frac {\ ln 9}{1 + \ ln x}} = \lim_{x\to\infty}e^{\ ln{x^{\frac {\ ln 9}{1 + \ ln x}}}} = \lim_{x\to\infty}\left(e^{\ ln 9}\right)^{\frac {\ lnx}{1 + \ ln x}} = \lim_{x\to\infty} 9^{\frac {\ lnx}{1 + \ ln x}} = 9.}$

$ \mbox{Beucase: } a^x=y\Leftrightarrow log_a y=x; \\
\mbox{then: } y=a^x=a^{log_a y}\mbox{ and } x=log_a y=log_a a^x.$

i dont get how the last step equals to 9

Because $ \lim_{x\rightarrow{\infty}}\frac {\ln{x}}{1 \plus{} \ln{x}} \equal{} \lim_{x\rightarrow{\infty}}\frac {1}{\frac {1}{\ln{x}} \plus{} 1} \equal{} 1$.
             $ (\because\lim_{x\rightarrow{\infty}}\frac{1}{\ln{x}}\equal{}0)$


topology
function
limit
real analysis
real analysis unsolved

Does there exist a homeomorphism $f: (0,1)\rightarrow\mathbb{R}$ which takes rationals to rationals and irrationals to irrationals?

An example:
$f(x)=\begin{cases}\frac1{1-x}-2&\frac12\le x<1\\ 2-\frac1x&0<x<\frac12\end{cases}$

thats right. just out of curiosity,how did you think of your example jmerry?

Well, jmerry can speak for himself, but the whole class of linear fractional transformation commends itself to the process:

Let $f(x)=\frac{ax+b}{cx+d}$ where $a,b,c,d$ are rational numbers with $ad-bc\ne0.$ Then $f^{-1}(x)=\frac1{ad-bc}\cdot\frac{dx-b}{-cx+a}$ is another such function. You can see that (with the exception of the point for which the function is undefined), $f$ maps rational numbers to rational numbers, and so does $f^{-1}.$ If you're looking to map the rationals into the rationals, then algebraic functions will do the job - but what algebraic functions have algebraic inverses? That's where the linear fractional tranformations stand out.

Simple linear functions $f(x)=ax+b$ are a subset of this class, but a simple linear function won't map the bounded interval onto an unbounded set.

Of course, you can't do the whole task here with a single linear fractional transformation, so you look to splice two or more together to create what you need. And that line of thought leads either to jmerry's example or to something else in the same spirit.

we can also splice infinitely many linear functions: Let $p_{n}$, $n \in \mathbb{Z}$, be a bi-infinite sequence of rational numbers in (0,1) such that $\lim_{n\to\infty}p_{n}= 1$,$\lim_{n\to-\infty}p_{n}= 0$,$p_{n}< p_{n+1}$. Let $q_{n}$, $n \in \mathbb{Z}$, be similar bi-infinite sequence of rational numbers in (-$\infty$,$\infty$). Take linear map from [$p_{n}$,$p_{n+1}$] onto [$q_{n}$,$q_{n+1}$] and splice them together to make a map from (0,1) to (-$\infty$,$\infty$).

We can replace the rational numbers with any countable dense set from the problem. Let $P$ and $Q$ be countable dense subsets of open intervals $I$ and $J$, respectively. Then there is a homeomorphism $f$ from $I$ to $J$ with $f(P)=Q$. To prove that, it is enough to construct an order-preserving bijection $g$ from $P$ to $Q$. WLOG we may assume $P$ is the set of diadic numbers $k/2^{n}$ in (0,1) and $I$=$J$=(0,1). Let $q_{n}$ be an enumeration of $Q$. We say $n$ is an index of $q_{n}$. Let g(0) = 0, g(1)=1, g(1/2) = $q_{1}$. Let g(1/4) be the number in $Q$ $\cap$ (g(0),g(1/2)) with the smallest index. Let g(3/4) be the number in $Q$ $\cap$ (g(1/2),g(1)) with the smallest index and so on. We defined an order-preserving injection g from $P$ $\cup$ {0,1} to $Q$ $\cup$ {0,1}. Because of "with the smallest index" choice, this map is onto. Now extend g to f.

There is a classical result along this line: all countable metric spaces with no isolated points are homeomorphic to the rationals.


geometry
rhombus
similar triangles
geometry unsolved

Let $ ABCD$ be a rhombus with $ \angle ABC\equal{}60^{\circ}$. A line through $ B$ cuts $ CD,CA$ at $ N,M$ respectively, and $ AN$ cuts $ CM$ at $ E$.

Prove that $ \angle AEC\equal{}60^{\circ}$

Obviously, M is on AD!
From similar triangles ABM and CNB we get AM*CN=AB^2 ( 1 ), but AC=AB, and, with (1), triangles MAC and ACN are similar, hence <NAC=<AMC ( 2 ). From triangle CND we have <CND+<DCN=60 ( 3 ), but <DAN+<NAC=<DAC=60 ( 4 ), from (3) and (4) getting <DAE = <DCE, i.e. DACE cyclic.

Best regards,
sunken rock

Thanks for the correction, sunken rock! And a nice solution!


The very best of Luck to all  IMO contestants especially the IRISH Team of
Fiachra Knox,
Stephen Ryan,
Seamus O'Boyle,
Richard Ellard
Richard Martin
and the very young Galin Ganchev.
Hopefully Valentin will be able to keep all of us at home informed of the progress and problems and now that Geoff S is posing on AOPS we should be able to get his updates also !

Good luck to all contestants!!! :)  :)
I have a question.
When will the contest for the Imo take place?
I think the IMO organisation has already started! :D

The first day of the contest is Wednesday.

[quote="socrates"]I think the IMO organisation has already started! :D[/quote]Yup, I'm already here :D Good luck to y'all.

As long as I am not breaking any news I will be informing you of the progress from down here in Mexico.

[quote="socrates"]Ok. Have a good time in Mexico! :D and keep us informed!! :)  :)  :lol:[/quote]Thanks, I'll try :D

Valentin,
Will you be able to post the 3 problems from today's ( Wednesday 13th ) session later on today or do you have to wait until  Thursday's session is complete before posting them.

Ladies and Gentelmen by the premission of mr. :yoda: may i informe you that ,The contest of  the World Championship Mathematics Competition for High School students  is right now held in merida  :yup:

The announcer in the contest hall :
Girls and Boys , you can now pick up your question sheets.

one of the contestants:
:read: :read:  hmm, What a  :coolspeak:   exam , but the problems seems a little easy .Dont you agree ?

[quote="lomos_lupin"]
one of the contestants:
:read: :read:  hmm, What a  :coolspeak:   exam , but the problems seems a little easy .Dont you agree ?
[/quote]

We must see the problems for agreement or disagreement, yes?


calculus integration Asymptote conics hyperbola

So my Calc BC class is designing class t-shirts and at my school, it's pretty much a competition between the AP math and science classes to see who has the better t-shirts.

I need a quote that's fresh and cool.I was thinking 'Calc BC - Take Math to the Limit" was kinda cool but I saw it on a calc t-shirt website. I'm trying to go for something original. Any ideas?

[quote="PerpetualMotion"]Asymptote: You can look but you can't touch.[/quote]

Or rather, a graph of a hyperbola with "You can look but you can't touch.
The phrase reminds me of something like Xeno's Paradox. Someone might be able to use it.

Integrating our problem solving skills to the limit.

Or you can say,"Limits undefined"?!
Or you can use my signature if you want....I won't feel bad!
Its said that all of analysis is centered in that equation....!


Hey,

How do I create braces such as $ {\Bigl\{}$ and $ {\Bigl\}}$. Around the system of equations:

$ {{x \plus{} y} \equal{} {5}}$
$ {{{x}{y}} \equal{} {6}}$

Thanks,

-APSStudent

[code]$\left\{
\begin{array}{ll}
x+y=5 &\quad  \\
xy=6 &\quad
\end{array}
\right.$[/code]

$ \left\{
\begin{array}{ll}
x\plus{}y\equal{}5 &\quad  \\
xy\equal{}6 &\quad
\end{array}
\right.$

Hey,

[quote="kunny"]
[code]$\{
\begin{array}{ll}
x+y=5 &\quad  \\
xy=6 &\quad
\end{array}
$[/code]

$ \{ \begin{array}{ll} x + y = 5 & \quad \\
xy = 6 & \quad \end{array} $
[/quote]

Thanks for the reply Kunny. However, in the following System of Equations how do I make the second equation's fraction look bigger so it does not look like it is squeezed?

$ {\{}\begin{array}{ll}
{{{S}_{T}} = {{{2}{\pi}{r}}{{(}{{h}+{\sqrt{{{h}^{2}}+{{r}^{2}}}}}{)}}}} \\
{{{V}_{T}} = {\dfrac{{5}{\pi}{h}{{r}^{2}}}{3}}} \\
\end{array}{\}}$

Thanks,

-APSStudent

Sorry, I can't help you, because As a rule, [code]V_T=\displaystyle \frac{5\pi hr^2}{3}[/code] could work, but this site's doesn't work.

kunny

[quote="APSStudent"]However, in the following System of Equations how do I make the second equation's fraction look bigger so it does not look like it is squeezed?

$ {\{}\begin{array}{ll} {{{S}_{T}} = {{{2}{\pi}{r}}{{(}{{h} + {\sqrt {{{h}^{2}} + {{r}^{2}}}}}{)}}}} \\
{{{V}_{T}} = {\dfrac{{5}{\pi}{h}{{r}^{2}}}{3}}} \\
\end{array}{\}}$

Thanks,

-APSStudent[/quote]
Use [b]\cfrac[/b]: $ \{\begin{array}{ll}S_T = 2\pi r(h + \sqrt {h^2 + r^2}) \\
V_T = \cfrac{5\pi hr^2}{3}\end{array}\}$

Lil' bit better I think to make a small vertical space between two lines:

\[ \left\{\begin{array}{ll}S_T \equal{} 2\pi r\left(h \plus{} \sqrt {h^2 \plus{} r^2}\right) \vspace{5pt}\\
    V_T \equal{} \cfrac{5\pi hr^2}{3}\end{array}\right\}\]


calculus
integration
trigonometry
function
calculus computations

Calculate the integral    $ \int_{C}(3y\minus{}e^{\sin x})dx\plus{}(7x\plus{}\sqrt{y^{4}\plus{}1})dy$ using the Green's Theorem, where $ C$ is the circle $ x^{2}\plus{}y^{2}\equal{}9$ taken in counter clockwise direction.

Well, just do exactly what is suggested, i.e., use Green's theorem. You'll get an integral of a constant function over the disk.


abstract algebra
limit
inequalities
real analysis
real analysis solved

I would know  when we have the equivalence between $(u_n)$ is convergent and $(\frac{\sum_1^n u_k}{n})$ is convergent

For the series $\sum_{n=1}^{\infty}u_n$ we define:

The partial sums $S_n=\sum_{k=1}^nu_k$, and

the Cesaro means $\sigma_n=\frac1n\sum_{k=1}^n S_k= \sum_{k=1}^n\left(1-\frac{k-1}{n}\right)u_k.$

The following theorems are true:

The Abelian theorem:

If $\lim_{n\to\infty}S_n=L$ then $\lim_{n\to\infty}\sigma_n=L.$

The simple Tauberian theorem:

If $\lim_{n\to\infty}\sigma_n=L$ and $u_n=o\left(\frac1n\right)$, then $\lim_{n\to\infty}S_n=L.$

The difficult Tauberian theorem (due to Littlewood and/or Hardy):

If $\lim_{n\to\infty}\sigma_n=L$ and $u_n=O\left(\frac1n\right)$, then $\lim_{n\to\infty}S_n=L.$

I'm still looking on my bookshelf for who has the best treatment of this. Summation by parts is the key technique here. One place to look would be the classic Hardy, Littlewood, and Polya [i]Inequalities[/i].

That the Tauberian conditions, or something like them, are needed can be seen by taking $u_n=(-1)^{n-1}$. $S_n$ diverges but $\sigma_n\to\frac12.$


Three coplanar circles intersect as shown. What is the maximum number of points on the circles that a line passing through all three circles can touch?

[asy]import graph;
draw(Circle((-9,9),15));
draw(Circle((0,-9),15));
draw(Circle((9,9),15));[/asy]

Draw a line which touches each circle twice. By not going through intersections of circles, we maximize the number of intersections for the line.

3 circles*2= $ \boxed{6}$

Is there any proof to it???????????[size=200][/size]

The maximum number of points at which a line can intersect 1 circle is 2 distinct points. Thus, for 3 circles, the maximum should be $3 \times 2 = 6$ points at most. If you're going for speed, you should probably guess 6 points at this point with a reasonable degree of certainty. If you have time and want to be certain, you should only check for the existence of a line that intersects the three circles at $\boxed{6}$ distinct points, because it is impossible that a line could intersect the circles at more than 6 points. (There are, in fact, many lines that satisfy the conditions.)

You can have a line from the top starting with intersecting the right circle, then intersect the right circle, and stop intersecting the third part. The go through the rest of the way for 3*2=6


In triangle $ABC$, prove that $\sum \sin A = 4 \prod \cos \frac{A}{2}$ and $\sum \sin 2A = 4 \prod \sin A$

The second one has already been discussed. This is a nice pfoof by towersfreak2006 ;)

[quote="towersfreak2006"][quote="shobber"]\[\sin{2A}+\sin{2B}+\sin{2C}=4\sin{A}\sin{B}\sin{C}\][/quote]

Yeah, that's true for the three angles $A,B,C$ in a triangle:

Want to show: $\sin{2A}+\sin{2B}+\sin{2C}=2\sin(A+B)\cos(A-B)+\sin(2C)$ with $A+B+C=180^\circ$

$\sin{2A}+\sin{2B}+\sin{2C}=2\sin(180^\circ-C)\cos(A-B)+2\sin(C)\cos(C)$ by sum to product and double angle identities.
$\sin{2A}+\sin{2B}+\sin{2C}=2\sin(C)\left(\cos(A-B)+\cos(C)\right)$ since $\sin\theta=\sin(180^\circ-\theta)$.
$\sin{2A}+\sin{2B}+\sin{2C}=2\sin(C)\left(2\cos\left(\frac{A-B+C}{2}\right)\cos\left(\frac{A-B-C)}{2}\right)\right)$ by sum to product.
$\sin{2A}+\sin{2B}+\sin{2C}=4\sin(C)\left(\sin\left(\frac{180^\circ-A+B-C}{2}\right)\sin\left(\frac{180^\circ-A+B+C}{2}\right)\right)$  since $\cos\theta=\sin(90^\circ-\theta)$
$\sin{2A}+\sin{2B}+\sin{2C}=4\sin(C)\left(\sin(B)\sin(A)\right)$ since $\sin\theta=\sin(180^\circ-\theta)$ and $A+B+C=180^\circ$.
$\sin{2A}+\sin{2B}+\sin{2C}=4\sin{A}\sin{B}\sin{C}$[/quote]

I don't know it the firs one has been discused but I might proof it :).

$\sin{A}+\sin{B}+\sin{C}=4\cos{\frac{A}{2}}\cos{\frac{B}{2}}\cos{\frac{C}{2}}$

$\sin{A}+\sin{B}+\sin{C}=4\cdot\frac{1}{2}\left[\cos{\left({\frac{A+B}{2}\right)}+\cos{\left(\frac{A-B}{2}\right)}\right]\cos{\frac{C}{2}}}$ using the product to sum formula.

$\sin{A}+\sin{B}+\sin{C}=2\left[\cos{\left(\frac{180^{0}-C}{2}\right)}+\sin{\left(\frac{180^{0}-A+B}{2}\right)}\right]\cos\frac{C}{2}$
using $\cos{\alpha}=\sin{(90^{0}-\alpha)}$.

$\sin{A}+\sin{B}+\sin{C}=2\left[\cos{\left(90^{0}-\frac{C}{2}\right)}+\sin{\left(\frac{180^{0}-A+B}{2}\right)}\right]\cos\frac{C}{2}$

$\sin{A}+\sin{B}+\sin{C}=2\left[\sin{\frac{C}{2}}+\sin{\left(\frac{180^{0}-A+B}{2}\right)}\right]\cos\frac{C}{2}$

$\sin{A}+\sin{B}+\sin{C}=2\sin{\frac{C}{2}\cos\frac{C}{2}+2\sin{\left(\frac{180^{0}-A+B}{2}\right)\cos\frac{C}{2}\right]}}$

$\sin{A}+\sin{B}+\sin{C}=\sin{C}+2\cdot\frac{1}{2}\left[\sin{\left(\frac{180^{0}-A+B+C}{2}\right)}+\sin{\left(\frac{180^{0}-A+B-C}{2}\right)}\right]$
using the fact that $2\sin{\alpha}\cos{\alpha}=\sin{2\alpha}$ and $\sin{\alpha}\cos{\beta}=\frac{1}{2}[\sin{(\alpha+\beta)+\sin{(\alpha-\beta)}}]$

$\sin{A}+\sin{B}+\sin{C}=\sin{C}+\sin{(B+C)}+\sin{B}$. But $\sin{(B+C)}=\sin{(180^{0}-A)}=\sin{A}$ so that's it :)!


I read in Problem Solving Strategies that the following theorem is valid, but the proof is complex.

The problem states that if you have a lattice plain, and you assign to each lattice point the arithmatic mean of the values of its four neighbours, that is, above, below, right, and left, then all real numbers are equal.

The proof is very simple when all values are integers, but it is not so clear when they are real numbers.  I have posted this I believe in the Pre-olympiad forums, but received no positive results.  Could someone please solve this problem or lead me an article in which will give me some ideas on how to solve this?  Thanks in advance.

Masoud Zargar

This problem is.. "complex", yes, to say the least.. :
http://www.mathlinks.ro/Forum/viewtopic.php?t=5327


Consider a triangle with area 1.

If two of the sides are trisected by lines coming from their respective opposite vertices, what is the area of the "middle quadrilateral" formed by these trisectors?

Here is a similar problem:

http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1539736619&t=146987


geometry
circumcircle
trigonometry
geometric transformation
trig identities
Law of Sines
Law of Cosines

Let $ T \in AD$ such that $ BC\equal{}AB\equal{}AT$. $ ABT$ is an isosceles triangle with $ \angle{A}\equal{}60^{\circ}$, so it is equilateral and $ \angle{ABT}\equal{}60^{\circ}$. $ \angle{B}\equal{}150^{\circ}$, $ \angle{TBC}\equal{}150^{\circ}\minus{}60^{\circ}\equal{}90^{\circ}$. But $ BT\equal{}AB\equal{}BC$, so $ TBC$ - isosceles and $ \angle{BCT}\equal{}45^{\circ}$. $ \angle{ACT}\equal{}45^{\circ}\minus{}15^{\circ}\equal{}30^{\circ}$. $ \angle{TCD}\equal{}\angle{ACD}\minus{}\angle{ACT}\equal{}30^{\circ}\minus{}30^{\circ}\equal{}0^{\circ}$, so $ T\equal{}D$ and $ \angle{DBC}\equal{}90^{\circ}$

[quote="MAT"][color=darkred]Let $ ABCD$ be a quadrilateral with $ \|\begin{array}{c} m(\widehat {CAD}) = 45^{\circ}\ ,\ m(\widehat {ACD}) = 30^{\circ} \\
 \\
m(\widehat {BAC}) = m(\widehat {BCA}) = 15^{\circ}\end{array}$ . Prove that $ BD\perp BC$ .[/color][/quote]

[color=darkblue][u][b]Proof I.[/u][/b] Since the triangle $ ADC$ is obtuse, $ BA = BC$ and $ m(\widehat {ABC}) = 360^{\circ} - 2\cdot m(\widehat {ADC})$ obtain that

the point $ B$ is the circumcenter of the triangle $ ADC$ , i.e. $ BA = BC = BD$ what means $ BD\perp BC$ .

[b][u]Proof II[/u][/b] (similarly with [b]Ahiles[/b]'s). Denote $ D'\in CD$ for which $ BD'\perp BC$ . Observe that $ BD' = BC = BA$ ,

i.e. $ BD' = BA$ and $ m(\widehat {ABD'}) = 60^{\circ}$ . Thus, the triangle $ ABD'$ is equilateral and  $ m(\widehat {CAD'}) = 45^{\circ}$ ,

i.e. $ D'\in AD$ . In conclusion, $ D'\in CD\cap AD$ , i.e. $ D'\equiv D$  and $ BD\perp BC$ .

[b][u]Proof III[/u][/b] (trigonometric). Denote $ m(\widehat {CBD}) = x$ . Then $ \|\begin{array}{c} m(\widehat {BDA}) = x - 30^{\circ} \\
 \\
m(\widehat {CDB}) = 135^{\circ} - x \\
 \\
m(\widehat {ABD}) = 150^{\circ} - x\end{array}$ . Apply the well-known relation :

$ \sin\widehat {ABD}\cdot\sin\widehat {DAC}\cdot\sin\widehat {CDB}\cdot\sin\widehat {BCA} = \sin\widehat {DBC}\cdot\sin\widehat {CAB}\cdot\sin\widehat {BDA}\cdot\sin\widehat {ACD}$ $ \Longleftrightarrow$

$ \boxed {\begin{array}{c} \\
\ \sin (30 + x)\cdot\sin 45\cdot\sin (45 + x)\cdot\underline {\sin 15} = \sin x\cdot\underline {\sin 15}\cdot\sin (x - 30)\cdot\sin 30 \\
 \\
\sin (30 + x)\cdot [\cos x - \cos (90 + x)] = \sin x\cdot [\cos (x - 60) - \cos x] \\
 \\
\sin (30 + x)\cdot\cos x + \sin x\cdot\sin (30 + x) = \sin x\cdot\sin (30 + x) - \sin x\cdot\cos x \\
 \\
\cos x\cdot [\sin x + \sin (30 + x)] = 0 \\
\ \end{array}}$

Since $ 0 < x < 150^{\circ}$ obtain $ \sin x + \sin (30 + x) > 0$ . In conclusion, $ \cos x = 0$ $ \Longleftrightarrow$ $ x = 90^{\circ}$ , i.e. $ BD\perp BC$ .[/color]

[quote="MAT"]Give quadrilateral $ ABCD$ with $ CAD \equal{} 45$, $ ACD \equal{} 30$, $ BAC \equal{} BCA \equal{} 15$, find the value of $ DBC$.

(angles)[/quote]
[hide="Solution"]
Consider point $ P$ such that $ \triangle ABP\sim \triangle ADC$ and let $ PB$ meet $ AC$ at $ X$.  Thus, a spiral similarity maps $ \triangle ACP\sim \triangle ADB$.  Now, that means that $ \angle APB \equal{} \angle ACD \equal{} 30$ and $ \angle BAP \equal{} \angle CAD \equal{} 45$.  Hence, $ \angle CAP \equal{} \angle BAP \plus{} \angle BAC \equal{} 45 \plus{} 15 \equal{} 60$.  It follows that $ \angle AXP \equal{} 180 \minus{} \angle XPA \minus{} \angle XAP \equal{} 180 \minus{} 30 \minus{} 60 \equal{} 90$.  Thus, $ BX\perp AC$ and since $ BA \equal{} BC$, we have that $ BX$ is the perpindicular bisector of $ AC$, so $ PX$ is the perpindicular bisector of $ AC$.  It follows that $ PA \equal{} PC$ and $ \angle PAC \equal{} 60$, so $ \angle PCA \equal{} \angle PAC \equal{} 60$, which means that $ \angle CPA \equal{} 60$, and we see that $ \angle ABD \equal{} \angle APC \equal{} 60$.  Finally, we conclude that $ \angle DBC \equal{} 180 \minus{} \angle BAC \minus{} \angle BCA \minus{} \angle ABD \equal{} 180 \minus{} 15 \minus{} 15 \minus{} 60 \equal{} 90$. [/hide]

This can easily be trig bashed. WLOG, let AB=BC=1.Then, AC=2sin 75. We now use law of sines on triangle ACD, since sin 75=sin 105, this makes it easy to find DC=$ \sqrt{2}$. Now use law of cosines on triangle BCD to find BD=1. So BCD is isoceles, so BCD=90.


1.How many 4 digit numbers divisible by 5 can be formed with the digits 0,1,2,3,4,5,6 & 6. (two sixes) ?

2.from 4 gentlemen and 4 ladies a committee of 5 is to be formed. The committee consists of 1 president, 1 vice-president and 3 secretaries. What will be the number of ways of selecting the committee with atleast 3 women such that atleast 1 women is either the president or the vice-president?

3.out of a pack of 52 cards 1 card is missing. From the remaining, Two cards are drawn and are found to be spades. Find the chance the missing card is a spade?

4.if a times the ath  term of a A.P is equal to b times the bth term, find (a+b)th   term.

Number $4$:
Assuming $a$ and $b$ are different, we have:
$a(t+(a-1)d)=b(t+(b-1)d)$ where $t$ is first term and $d$ is comon ratio.
$t(a-b)+d(a^{2}-b^{2})-d(a-b)=0$
$(a-b)(t+d(a+b-1))=0$
$a-b$ is not $0$ so $t+d(a+b-1)=0$ which is the $(a+b)$th term.

thanks for that....any more anyone... i need help with the other three very badly

[quote="fizsle"]1.How many 4 digit numbers divisible by 5 can be formed with the digits 0,1,2,3,4,5,6 & 6. (two sixes) ?

3.out of a pack of 52 cards 1 card is missing. From the remaining, Two cards are drawn and are found to be spades. Find the chance the missing card is a spade?[/quote]

For 3, I'm assuming the missing card is randomly chosen.
[hide="3"] We are looking for $\frac{P(\text{a spade is missing and you draw 2 spades})}{P(\text{a spade is missing and you draw 2 spades})+P(\text{the missing card is not a spade and you draw 2 spades})}$

$\text{P(a spade is missing and you draw 2 spades)}=\frac{1}{4}\cdot\frac{12}{51}\cdot\frac{11}{50}=\frac{33}{2550}$

$\text{P(the missing card is not a spade and you draw 2 spades)}=\frac{3}{4}\cdot\frac{13}{51}\cdot\frac{12}{50}=\frac{39}{850}$

$\frac{\frac{33}{2550}}{\frac{33}{2550}+\frac{39}{850}}=\frac{\frac{33}{2550}}{\frac{150}{2550}}=\frac{33}{150}=\boxed{\frac{11}{50}}$ [/hide]
[hide="Hint for 1"] There are 2 cases:
the number ends in 0
or
the number ends in 5[/hide]

[hide="Number 1"] Since the number is divisible by 5, the last digit is either 0 or 5.  If the last digit is 0, and the number has less than 2 sixes, there are $6\times 5\times 4=120$ ways to choose the other  3 digits. If there are 2 sixes, there are 5 ways to choose the third digit and 3 ways to arrange them for a total of 15 numbers.  Then we must multiply this total by 2 since there can also by a 5 in the units digit, for a total of $2(120+15)=270$ numbers.[/hide]

[hide="Number 2"] There are $\frac{4!}{3!1!}\times \frac{4!}{2!2!}=4\times 6=24$ ways to choose 3 women and 2 men; in this case out of $\frac{5!}{3!}=20$ ways to arrange the committee, only 2 do not work (when all 3 women are secretaries) for a total of 18.  Then there are a total of $24\times 18=432$ committees with 3 women and 2 men.

There are 4 ways to choose a commitee with 4 women and a man, and all of these must contain a women as president or vice-president.  Since there are 20 ways to arrange each committee, there are a total of 80 committees with 4 women and a man.  Therefore, there are a total of 512 committees that have at least 3 women with a women as president or vice-president.[/hide]

[hide="Number 3"]  If the removed card was not a spade, there would be a probability of $\frac{3}{4}\times \frac{13}{51}\times \frac{12}{50}=\frac{39}{850}$ of drawing two spades.  If the card removed was a spade, there would have been a probability of $\frac{1}{4}\times \frac{12}{51}\times \frac{11}{50}=\frac{11}{850}$ chance of drawing two spades.  Then the probability of a spade being removed is $\frac{\frac{11}{850}}{\frac{39}{850}+\frac{11}{850}}= \frac{11}{50}$[/hide]

[hide="Number 4"]
$t_{1}$ be the first term and d be the common difference.  We have that $at_{a}=bt_{b}$, or $a(t_{1}+ad-d)=b(t_{1}+bd-d)$  We have that $(a-b)t_{1}+(a^{2}-b^{2}-a+b)d=0$  Dividing by (a-b), we get $t_{1}+(a+b-1)d=0$, which means that $t_{a+b}=0$[/hide]

Lotrgreengrapes7926:  The probability of the missing card not being a spade is $\frac{3}{4}$, not $\frac{1}{4}$.  Somehow you got the right answer though.

number 1.... exactly the answer i got 270.... but there is no such answer choice...further the given answer is 588.... something missing???? anybody else any clue??? or is it just that the book is wrong???

[quote="samath"]Lotrgreengrapes7926:  The probability of the missing card not being a spade is $\frac{3}{4}$, not $\frac{1}{4}$.  Somehow you got the right answer though.[/quote]
I probably copy and pasted and forgot to change it to $3$. Thanks for pointing it out. :D

for 3)....  what is the logic behind the method used to solve? why is the chance of the missing card being a spade calculated in this fashion.... please explain in detail....

It's conditional probability.
$P(B|A)=\frac{P(A\cap B)}{P(A)}$ ([url=http://mathworld.wolfram.com/BayesTheorem.html]Bayes' theorem[/url])
In this case, $B$ is the event 'the removed card was a spade' and $A$ is the event 'two spades are drawn from a deck missing a card'. In the numerator, the $P(A\cup B)$ represents 'the probability that two spades are drawn from a deck missing a spade'. In the denominator, $P(A)$ is represented by the sum of two fractions because the event 'two spades are drawn from a deck missing a card' can happen in two, non-intersecting cases: the missing card is a spade, or the missing card is not a spade. Each fraction represents one of these cases.

Note: You can't just multiply the number of ways to make a 4 digit number ending in 0 by 2 to get the answer since you can't have 0 as the first digit.

So, it should be:

[hide="number 1"]
With 0 as the last digit:

We add the number of ways with 0 sixes, 1 six, and 2 sixes.

With 0 sixes:

We choose 3 numbers out of 1,2,3,4,5 and permute them however we like $\binom{5}{3}\cdot6 = 60$

With 1 six:
We choose 2 numbers out of 1,2,3,4,5 and permute them with a six however we like $\binom{5}{2}\cdot6 = 60$

With 2 sixes:
We choose 1 number out of 1,2,3,4,5 and permute it with 2 sixes (3 permutations)
$\binom{5}{1}\cdot3 = 15$

With a 5 as the last digit:

With 0 sixes and no 0:
We choose 3 numbers out of 1,2,3,4 and permute them $\binom43\cdot6=24$
With 0 sixes and one 0:
We choose 2 numbers out of 1,2,3,4 and can permute them in 4 ways.
$\binom42\cdot4=24$
With 1 six and no 0:
We choose 2 numbers out of 1,2,3,4 and permute them
$\binom42\cdot6=36$
With 1 six and one 0:
We choose 1 number out of 1,2,3,4 and permute them
$\binom41\cdot4=16$
With 2 sixes and no 0:
We choose 1 number out of 1,2,3,4 and permute them
$\binom41\cdot3 = 12$
With 2 sixes and a 0:
We permute them
$2$

So we have the total at $60+60+15+24+24+36+16+12+2=249$

[/hide]


Let $a,b,c$ be integers different from zero. It is known that the equation $a \cdot x^2 + b \cdot y^2 + c \cdot z^2 = 0$ has a solution $(x,y,z)$ in integer numbers different from the solutions $x = y = z = 0.$ Prove that the equation \[ a \cdot x^2 + b \cdot y^2 + c \cdot z^2 = 1 \] has a solution in rational numbers.

assume that $(x_0,y_0,z_0)$ satisfying $ax^2+by^2+cz^2=0$ and WLG assume $z_0\neq0$. Then consider triplet $(x,y,z)=(x_0(1-t);y_0(1-t);z_0(1+t))$ where $t$ is rational. Then
$ax^2+by^2+cz^2=ax_0^2(1-t)^2+by_0^2(1-t)^2+cz_0^2(1+t)^2=4tcz_0^2$
Chose $t=\frac{1}{4cz_0^2}$ (cause $c\neq0$) then we get the requirment!


Let $P(x)=(x-1)(x-2)(x-3)$. For how many polynomials $Q(x)$ does there exist a third degree polynomial $R(x)$ such that $P(Q(x))=P(x)R(x)$

This problem was posted before.

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=357250#p357250


Are there positive integers $a, b$ with $b>2$such that $2^{a}+1$ is divisible by$2^{b}-1$?

It is easy to show $b|2a$. It give $b=1, a-all$, or $b=2, a-odd$.


I have a problem :)
I started digging  in the endless stuff we have here at my house,and I found a letter dated 98 :| UNOPENED!!! so I opened it,and it was in pakistani.if someone can help me translate I'd be obliged... the adress it should have been taken to is nearby so I will take it soon. thx
[img]http://66.98.192.31/~jeriko/baber%20mehrad.png[/img]

post your message here:  http://www.artofproblemsolving.com/Forum/index.php?f=363&sid=103d9f6f00236cdf4dde5274f4d49b1e

That's the pakistan forum. ;)


search
graph theory
combinatorics proposed
combinatorics

I was wondering if someone would be so kind and direct me to a site(or something like this) where there is some information about graph theory, because I have nothing regarding graph theory. Or some books that I can find in Romania and where? And google search or other search engines didn't do too good! :(

Thank you very much! :D  :D

i will try to make up a reading list and put it on the site
but thats going to happen when i get back from Japan

up untill then you could try studing I. Tomescu's book - Introduction into combinatorics or the Graph Theory problems

i would also be interesed 2 find some books about graph theory.
Unfortunately i do not have Tomescu's book(an i don't where to get it from).Till there i'll try to study some of my father's old Uviversity books,but as i told you i can't finf to mch there.
So pls could somene give me a sugestion?? :(  :(  :(

try http://www.math.uni-hamburg.de/home/diestel/books/graph.theory/GraphTheoryII.pdf for a graph - theory book.

I believe the best book in Graph Theory is :

Introduction to Graph theory (2nd edition)
by Douglas B. West

I don't have the original book, but I borrow it from a university in Esfahan (Iran) and get a photocopy of it  :lol:

The cover of this book attached :


Galois Theory
superior algebra
superior algebra unsolved

Let $K \subset F$ -Galois, $E$ - intermediate field.Show that there exists a unique minimal field $L$ with $K \subset F \subset L \subset F$, such that $K \subset L$ - Galois.

I'm assuming your stack of fields is $K\subset E\subset L\subset F$ from the wording. Are you allowed to use the fundamental theorem of Galois theory? From this you know that $K \subset L$ is Galois iff $\text{Gal}(F/L) \triangleleft \text{Gal}(F/K)$. I'd start here with the standard techniques of showing uniqueness and minimal (actually I think that if you show existence, then just assume minimal it will work).

Actually now that I'm thinking it through (from writing that), I haven't worked the details so don't hold me to this, but try showing existence, assume minimal, then you should get uniqueness from the 1-1 correspondence between intermediate fields and Galois groups, then the Galois extension from what I wrote above. Also note that if you have that normality then: $\text{Gal}(L/K) \cong \text{Gal}(F/K)/\text{Gal}(F/L)$. Maybe this will work? Just kind of musing though.


How many nonsimilar triangles have angles whose degree measures that are all integers?

[quote="13375P34K43V312"][quote="kyyuanmathcount"][quote="13375P34K43V312"][hide]x+y+z=180 x,y,z >0
equivalent to
a+b+c=177 a,b,c $\ge$ 0

by balls and urns

${179\choose2}$[/hide]

pwn[/quote]

Wait, but does that method imply that each angle of the triangle is different, or am I being a vegetable?[/quote]

but yeah like XX...X|XX...X|XX...X where we have 59 X's in each of the three groups generates a=b=c=59.[/quote]
But he said non-similar, your method counts 30-60-90 and 90-60-30 as the same. For this I think we need case work on isosceles, scalene, equilateral, etc.

Either that, or you could have fun and list out all the partitions of 180 into three positive integers.  But, we only need how many, so we can equate this with how many ways we can partition 180 into parts, the largest of which is 3.  Would that somehow lead to an easier solution?

[hide="PIE"]

Well from what we had before there are ${179\choose2}=15931$ if order matters. Now we will find how many are equilateral, how many are isosceles, and how many are scalene.

Case 1: equilateral

well duh 1

Case 2: isosceles

First choose the vertex angle, then the base angles are forced, however the vertex angle must be even, 2, 4, .., 178, giving 89. However there are 3 ways to order these so we counted 267 isosceles triangles originally.

Then by complimentary counting, we counted 15931-(1+277)=15654 scalene

Now we correct for overcounting...

1/1 = 1
267/3 = 89
15654/6 = 3609

Add those to get a final answer of 3699.[/hide]

I love how perfect628 posts problems that are "simple" but actually turn out to be somewhat hard.

kyyuanmathcount, are you suggesting generating functions somehow?

[quote="13375P34K43V312"][hide="PIE and diophantients idea"]

Well from what we had before there are ${179\choose2}=15931$ if order matters. Now we will find how many are equilateral, how many are isosceles, and how many are scalene.

Case 1: equilateral

well duh 1

Case 2: isosceles

First choose the vertex angle, then the base angles are forced, however the vertex angle must be even, 2, 4, .., 178, giving 89. However there are 3 ways to order these so we counted 267 isosceles triangles originally.

Then by complimentary counting, we counted 15931-(1+277)=15654 scalene

Now we correct for overcounting...

1/1 = 1
267/3 = 89
15654/6 = 3609

Add those to get a final answer of 3699.[/hide]

I love how perfect628 posts problems that are "simple" but actually turn out to be somewhat hard.

kyyuanmathcount, are you suggesting generating functions somehow?[/quote]
You can't put apostrophes in hide tags
Generating functions might work,
$(\sum^{179}_{k=0}x^{k})^{3}$, look for the coefficient of 180

Eh, lets use some elementary methods for this one, or else it might get moved...mwahhaaha....

I like 1337's method.

Well let's see... elementary? count 'em all.

Can we come up with a final answer?  I got a different value than any already posted, but I only thought about it for a couple minutes, so I am very likely off.

Would somebody like to post a solution that we can work off of?

Here's my [hide="solution"]Use the above reasoning to get $x+y+z=177$. To eliminate any overcounting, first allow any nonnegative integers for the variables. WLOG, say $x=0$. There are then 89 triangles with at least one 0. Similar reasoning yields 88 with only 1 and above, 86 2 and above, 85 with 3 and above, and so on, until 1 with 57s. Taking the sum, we get $(1+89)44.5-3(1+29)14.5=\boxed{2700}$[/hide]

I'm not 100% on that, but it doesn't seem flawed. And it's a nicer # than the others.

[hide="The secret of this solution is organization"]$A+B+C=180.$  wlog (without loss of generality), assume $A\le B\le C.$

Consider odd A:
A=1:  min B=1; max B=89; #triangles=89
A=3:  min B=3; max B=88; #triangles=86
A=5:  min B=5; max B=87; #triangles=83
...
A=59:  min B=59; max B=60; #triangles = 2
Note the #triangles forms an arithmetic sequence.

Consider even A:
A=2:  min B=2; max B=89; #triangles = 88
A=4:  min B=4; max B=88; #triangles = 85
A=6:  min B=6; max B=87; #triangles = 82
...
A=60:  min B=60; max B=60; #triangles = 1
Again, the #triangles forms an arithmetic sequence.

For any arithmetic sequence, the sum of the sequence is:
$S=n\times m$, where $n$ is the number of terms, and $m$ is the average term, which can be obtained by averaging the first term and the last term.

Total cases with odd A:  $30\times\frac{89+2}{2}=15\times 91$

Total cases with even A:  $30\times\frac{88+1}{2}=15\times 89$

Total cases:  $15\times 180=2700$[/hide]

Also, leet's method (using balls and urns, with PIE) is fine in priciple, but it contains several minor mistakes.

Good, I did have the right answer originally.

[quote="rcv]
Also, leet's method (using balls and urns, with PIE) is fine in priciple, but it contains several minor mistakes.[/quote]

Elaborate on the mistakes please?

1.You divided incorrectly by 6
2. You overcounted the equilateral case

Without these mistakes, you get 2700.

We have

Isoscles (non-equilateral)*3+equilater+scalene*6=179C2 (from balls and urns)

the isosceles non equilaterals can be arranged in 3 ways and scalens in 6 ways, but when we summed them all up, we counted all of the ways

so isoscels triangles are 2a+b=177 (from same method as getting all the triangles), b must be odd so 2a+2k=176, a+k=88, there are 89 isoscles traignes, since there 1 of those triangles is equilateral, there are 88 isoceles non equilateral trainges.

so 88*3+1+6x=179C2, from here x=2611

so there are 2611 scalene trainges
88 isoscles non equalateral traingesl
1 equalateral triangle

add them and you get 2700.

Is there a way we can put these interesting problems, which seem to have no competition source, along with good and final solutions on the Wiki?

Yeah, AOO's solution is just the corrected version of leet's solution.


Find all integers $ n\geq 3$ such that for any two positive integers $ m, r$, both less than$ n \minus{} 1$, there exist $ m$ distinct elements of the set $ \{1,2,...,n \minus{} 1\}$ whose sum is congruent to $ r$ when considered in modulo $ n$


[color=indigo][size=150]What is the smallest positive root of the equation

[ sin(1-x) ]^1/2  = [ cosx ] ^ 1/2[/size][/color]

[quote="UTTARA"]What is the smallest positive root of the equation $ [ \sin(1\minus{}x) ]^{\frac 12}  \equal{} [ \cos(x) ] ^ {\frac 12}$[/quote]

So $ \sin(1\minus{}x)\equal{}\cos x$ with $ x>0,\sin(1\minus{}x)>0,\cos(x)>0$

So $ x\equal{}\frac 12\minus{}\frac{\pi}4\plus{}k\pi$ with $ x>0,\sin(1\minus{}x)>0,\cos(x)>0$

And the first $ k$ matching the requirements $ x>0,\sin(1\minus{}x)>0,\cos(x)>0$ is $ k\equal{}2$

Hence the solution $ \boxed{x\equal{}\frac{2\plus{}7\pi}4}$

[quote="prester"][quote="pco"]

So $ x \equal{} \frac 12 \minus{} \frac {\pi}4 \plus{} k\pi$

[/quote]

I cannot see that. Dear pco, can you explain where does it come from? Many thanks[/quote]

General property : $ \sin(a)\equal{}\sin(b)$ $ \iff$ $ a\equal{}b\plus{}2k\pi$ or $ a\equal{}\pi\minus{}b\plus{}2k\pi$

Here : $ \sin(1\minus{}x)\equal{}\cos(x)\equal{}\sin(\frac{\pi}2\minus{}x)$ $ \iff$ $ 1\minus{}x\equal{}\frac{\pi}2\minus{}x\plus{}2k\pi$ or $ 1\minus{}x\equal{}\pi\minus{}\frac{\pi}2\plus{}x\plus{}2k\pi$.

First case is impossible (it would mean $ 1\equal{}\frac{\pi}2\plus{}2k\pi$), so : $ \sin(1\minus{}x)\equal{}\cos(x)$ $ \iff$ $ 1\minus{}x\equal{}\pi\minus{}\frac{\pi}2\plus{}x\plus{}2k\pi$ $ \iff$ $ x \equal{} \frac 12 \minus{} \frac {\pi}4 \plus{} k'\pi$

Hence my assertion.


If $ x,y,z$ are distinct real numbers such that
$ (x^2\plus{}y^2\plus{}z^2)^2\equal{}3(x^3y\plus{}y^3z\plus{}z^3x)$,
then
$ 2(x^2\plus{}y^2\plus{}z^2)\equal{}3(xy\plus{}yz\plus{}zx)$.

there is probably some typo, e.g. ,$ x\equal{}0,y\equal{}1$ yields $ (1\plus{}z^2)^2\equal{}3z^3$ and $ 2(1\plus{}z^2)\equal{}3z$ which is not simultaneously solvable

Equation $ (1\plus{}z^2)^2\equal{}3z^3$ hasn't real roots.

Using this following transformation:x=min(x,y,z).Let y=x+a,z=x+b we will prove that (x^2+y^2+z^2)^2>=3(x^3y+y^3z+z^3x)(quadratic fuction)(Vasc'ineq)  :lol:
Equality happens when x=y=z or (x,y,z)=(ksin^2(4pi/7),ksin^2(2pi/7), ksin^2(pi/7))
Then replcaing these 2 cases into the second equation we 'll get the result easily!


calculus
integration
function
real analysis
algebra
polynomial
derivative

Find all functions $f$ in $L^{1}(\mathbb{R})$ that satisfy the integral eq.

\[\int_{\mathbb{R}}f(x-y)e^{-|y|}dy=e^{-x^{2}}\]

I really appreciate some help with the solution and the intermediate steps.

One technical question: where are you used to having the $2\pi$'s in the definition of the Fourier transform? Please write down your definition.

Then, as hints, I'd suggest you write out all of the following:

In detail, the theorem that says something like "the Fourier transformation of a convolution is the pointwise product of the Fourier transforms."

The Fourier transform of $e^{-x^{2}}.$

The Fourier transform of $e^{-|x|}.$

[quote="Kent Merryfield"]One technical question: where are you used to having the $2\pi$'s in the definition of the Fourier transform? Please write down your definition.
[/quote]

By $\int_\mathbb{R}$ i mean $\int_{-\infty}^{\infty}$. Let $f$ be a function on $\mathbb{R}$ s.t. $\int_\mathbb{R}|f(t)|dt<\infty$.
The fourier transform $\hat{f}=\int_\mathbb{R}f(t)e^{-i\omega t}dt$.

[quote="Kent Merryfield"]So you would then have (for sufficiently regular functions) that

$f(x)=\frac1{2\pi}\int_{-\infty}^{\infty}\widehat{f}(\omega)e^{i\omega x}\,d\omega$

Now, what about my three hints? In particular, what is the Fourier transform of the convolution

$f*g(x)=\int_{-\infty}^{\infty}f(x-y)g(y)\,dy\ ?$[/quote]

Yes there is a theorem:  $\widehat{f*g}=\widehat{f}\widehat{g}$  but how to use it?
So basically we hav a convolution eq. $\widehat{f}\widehat{g}=\widehat{h}$, where $\widehat{h}=\sqrt{\pi}e^{-\omega^{2}/4}$ and $\widehat{g}=\frac{2}{1+\omega^{2}}$ but what to do next?

$\widehat{f}=\frac{\widehat{h}}{\widehat{g}}= (1+\omega^{2})\frac{\sqrt{\pi}}{2}e^{-\omega^{2}/4}$
but this does not make it easier since the integral formula for $f$ would be hard to solve. How should I solve this? Please can you give me an more concrete answer, i.e. a solution?

[quote="Kent Merryfield"]Split that into two functions by distributing across the $1+\omega^{2}.$

One term will be $C\cdot e^{-\omega^{2}/4}.$ That's a Gaussian and the Fourier transform of a Gaussian. We recognize it.

Now, what about the term that is $C\omega^{2}e^{-\omega^{2}/4}$ ?

Here, we can use this theorem:

$\widehat{f'}(\omega)=i\omega\widehat{f}(\omega).$

As a corollary, $\widehat{f''}(\omega)=-\omega^{2}\widehat{f}(\omega).$[/quote]

First of all thanks. I came about this far but i still dont know what to do whith any of the terms of the integral, since we still have a $e^{i\omega x}$ times both of the integrands, like this term
\[\frac{\sqrt{\pi}}{2}\frac1{2\pi}\int_{-\infty}^{\infty}e^{-\omega^{2}/4}e^{i\omega x}\]
Sorry for asking a lot but I really appreciate your help with the details.
Thank you Kent Merryfield.

is the solution $\frac{5}{4}e^{-x^{2}}$?

[quote="Kent Merryfield"]The solution to the whole problem will be a (presumably second degree) polynomial times $e^{-x^{2}}.$[/quote]

Could you please post the steps? what about the $e^{i\omega x}$ factor? It is not an $e^{-i\omega x}$ as in the transform, so how should it be solved?

Because when I identified the first integral term with the transform of the gaussian function I got Cexp(-x^2) but I didnt bother the minus sign in the factor mentioned above, which propably is wrong.
(And the second integral became as the first after using $\widehat{f'}(\omega)=i\omega\widehat{f}(\omega).$ so it became also a  Cexp(-x^2). )
Perhaps If I fix the minus sign these will have a polynomial factor as you mentioned, but how?

So I would really appreciate if you could post some solution with some steps, please.

[quote="Kent Merryfield"]Just from the Fourier inversion theorem, as a transform-inverse transform pair,

$\int_{-\infty}^{\infty}e^{-x^{2}}e^{-i\omega x}\,dx=\sqrt{\pi}e^{-\omega^{2}/4}$ and

$\frac1{2\pi}\int_{-\infty}^{\infty}\sqrt{\pi}e^{-\omega^{2}/4}e^{i\omega x}\,d\omega=e^{-x^{2}}.$

You seem to know some part of this result already - that the Fourier transform of a Gaussian is a Gaussian, with some appropriate scaling. So use all of that result, backwards as well as forwards.

And read the post in which I talked about second derivatives carefully.

---

In the end, I'm not going to do all of your work for you. You need to figure this out for yourself.[/quote]

Ok thank you very much. The problems I had (I think I have solved them)
were the minus sign i the factor exp(i omega x) I mentioned an the second derivative formula thing, but at least I have an answer
which i something like $(1/2-2x^{2})e^{-x^{2}}$. Hopefully it is somewhat correct. I got some expressions like
\[\frac{1}{4\sqrt{\pi}}\int_{-\infty}^{\infty}e^{-\omega^{2}/4}e^{i\omega x}+\frac{1}{4\sqrt{\pi}}(-4\omega^{2})\int_{-\infty}^{\infty}e^{-\omega^{2}/4}e^{i\omega x}\]
	\[f(x)=\frac{1}{2}e^{-x^{2}}+(-2x^{2})e^{-x^{2}}=(\frac{1}{2}-2x^{2})e^{-x^{2}}. \]
Sorry for asking trivial questions. I really appreciate your help, thank you.

[quote="color"]I have done something wrong in the solution above.
I think it is the $\omega^{2}$ factor.
can anyone tell me how to cerrect it?
[color=blue]PLEASE [/color]help.[/quote]


Hi,
What is the greatest possible number of points of intersection for 8 distinct lines in a plane?
I've seen this question appear several times, especially MathCounts. I guessed the answer by doing [hide="this"]8 Choose 2=8!/(2!*6!)=[b]28[/b][/hide]
However, I don't know if this is the right answer. Can someone please explain this to me?

This is correct. ans: $ \boxed{28}$.

To maximize the number of intersections, we want all the lines to be able to intersect (no parallel), and we want each intersection point to have only two lines going thorugh. Thru this, we have designed our problem such that each of the 8 lines intersects each of the other lines at unique points. We can choose any 2 lines from 8 to intersect, so C(8,2) is the answer.

alternatively, we can name the lines $ L_1, L_2, ... L_8$. we see that $ L_1$ ideally intersects 7 other lines. we see that $ L_2$ ideally intersects 6 other lines. eventually, we arrive at $ 1\plus{}2\plus{}3\plus{}4\plus{}5\plus{}6\plus{}7\equal{}\frac{7\times8}{2}\equal{}\boxed{28}$, as pinkmuskrat said. :)

And there is this
http://www.artofproblemsolving.com/Alcumus/Videos/Video.php?video_id=36&m=5f0443d5822ef063ee3428c64af62988


Anybody else here camp out for a Wii?  I got to Circuit City at 5 AM.  I was 15th in line, and now I have my long-awaited Wii. :lol:

why'd you want one? JW...I didn't think that people actually camped out for them....(or anything for that matter)

I drove by circuit city when I saw a long line of people waiting in line. Then there were some weird dudes protesting. People get way too into this.

hows your wii 360fan?

Extremely fun, thanks.  I got Red Steel and Legend of Zelda.

Some people waited in tents for days to get a PS3.

[quote="laughinghead505"]That's pretty sad... :roll:[/quote]

It's sad that he waited patiently for something he wanted?

[quote="mysmartmouth"]Um, no.  Not sad at all.  $\$$500 cost, $\$$3000 eBay resell value.  $\$$2500 profit per unit. 4 people in a tent, four units. $\$$10,000.  That's $\frac{1}{3}$ of the average US income.[/quote]
problem: There are 4 people, so it's really $\frac{1}{12}$ of the average US income.

Still pretty good for just a wait.... :roll:  :wink:

[quote="mysmartmouth"]Um, no.  Not sad at all.  $\$$500 cost, $\$$3000 eBay resell value.  $\$$2500 profit per unit. 4 people in a tent, four units. $\$$10,000.  That's $\frac{1}{3}$ of the average US income.[/quote]

The average US income is 40,000.

[quote="Klebian"]another problem: They sell for more like 1500-2000 on ebay, not 3000[/quote]

They were over $\$$3000 when they first came out (PS3, not Wii).

I just realized US average income is significantly higher than SC average income.  :rotfl:  :rotfl:

average family income is like 48 grand
-jorian

its so funny to laugh at my brother because he wasn't early enough to get a wii :rotfl:
-jorian

30,000 is the average? mines 840 gosh i must be poor and therefore that is why i should wait in lines at CC and sell it to people. OK i thought the average income was 100,000 but that is kind of weird because when i think of it no one i know gets the money to the nearest ten thousand....

2 of my friends camped out for it...they thought I was weird for NOT!

lol what games are you gonna get for it?

the only WII games i know are like nintendo games and stuff...nothing actually serious (like Medal of Honor or Halo, etc.)...all the games are all fantasy (and for the games like James Bond you can get for other systems...) related...

have u guys seen the commericals for the Wii
"Wii would like to play" :rotfl:  :rotfl:

[quote="Thu7"]
"Wii would like to play" :rotfl:  :rotfl:[/quote]

 :rotfl:  that is really stupid...probably took 5 seconds to think of that

[quote="juicybooty911"]camp out a little for a thousand bucks, not bad in my opinion...[/quote]

probably why when I stopped by gamestop to get new games today for cheap, I saw like 10 people sleeping outside before it was open waiting in line for the wii. When I came back like half an hour later, they were in, first in line.
same as the ps3, but those sold out in my local gamestop a few days ago

Then you get people, like me, that wait a couple years for the prices to go own.  I still think that the wii is realy inexpensive though!!

It's what they were on when they were thinking of a name.

Wiiiid!

It also gives you a clue about how you're supposed to pronouce it.

[quote="Koda"]What is it anyway? all I know is it is some time of a gaming console.
I have heard of XBox and PlayStation, but what is Wii? :maybe:[/quote]
Google is your friend....

I want a wii but im gonna wait till the prices drop. I'll get Brawl and maybe StarFox if they have a starfox.

wii owns! my friend got it and I played a little.   He is the nephew of one of the guys who helped create the wii so he got it early.      :lol:  :D

nice!....disgusting Wii joke:

Hey I'm going down to my basement to play with my Wii...you wanna join me?

do u have a wii?

Also I might. jk What state do u live in?

maryland..and no i have a 360...not a Wii

[quote="nutz_for2.718281828"]Why in the three infinities would you want one? It's just an electronic box.[/quote]
Why would you want a TV? It also is an electronic box, and I know you watch TV.

Did you know that the Wii holds the world record for Most Dirty Jokes About A Video Game System?

[quote="nutz_for2.718281828"]Why in the three infinities would you want one? It's just an electronic box.[/quote]

why would you want a computer? Its just a electronic box. :D

[quote="Ihatepie"]u have Halo 2?
thats the best game.[/quote]

yea i started playing Halo when it came out(i was 7 or something...not exactly the best lol) and my parents got strict so they sold my original xbox. now i got a 360 with halo and i just bought halo 2 on black friday


inequalities
real analysis
real analysis unsolved

Let $ (X, \mathcal{B}, \mu)$ be a measure space. Use Hölder's inequality to prove that if $ \frac{1}{p}\plus{}\frac{1}{q} \equal{} \frac{1}{r}$ and $ r \geq 1$ then for every $ f \in L^p(X, \mathcal{B}, \mu)$, $ g \in L^q(X, \mathcal{B}, \mu)$, and $ fg \in L^r(X, \mathcal{B}, \mu)$ with $ \parallel{} fg \parallel{}_r \leq \parallel{} f \parallel{}_p \cdot \parallel{} g \parallel{}_q$.

I don't see how to prove this right now. I know that this looks a lot like Young's inequality which follows from Hölder's inequality, so I was thinking to adopt a similar proof of that theorem. Right now, I just need a few hints on how to proceed. Thank you.

Let $ \tilde {f} \equal{} |f|^r$, $ \tilde{g} \equal{} |g|^r$, then $ \tilde{f}\in L^\frac {p}{r}(X,\mathcal {B},\mu)$ and $ \tilde{g}\in L^\frac {q}{r}(X,\mathcal{B},\mu)$.  By Hölder's inequality $ \Vert \tilde{f}\tilde{g}\Vert _1^\frac {1}{r}\leq \Vert \tilde{f}\Vert _\frac {p}{r}^{\frac {1}{r}}\left \Vert \tilde{g}\right\Vert_{\frac {q}{r}}^{\frac {1}{r}}\Leftrightarrow \Vert fg\Vert _r\leq \Vert f\Vert_p \Vert g\Vert_q$


[quote="ShoeFactory"][quote="vishalarul"]Factor $ax^{2}+bx+c$.[/quote]

Are you joking? Stop the spam.[/quote]

??? this problem is perfectly fine,

$ax^{2}+bx+c=a\cdot \left(x-\frac{-b+\sqrt{b^{2}-4ac}}{2a}\right)\cdot\left(x-\frac{-b-\sqrt{b^{2}-4ac}}{2a}\right)$

[quote="Altheman"]$ax^{2}+bx+c=a\cdot \left(x-\frac{-b+\sqrt{b^{2}-4ac}}{2a}\right)\cdot\left(x-\frac{-b-\sqrt{b^{2}-4ac}}{2a}\right)$[/quote]
Are you sure it isn't $\left(x-\frac{-b+\sqrt{b^{2}-4ac}}{2a}\right)\left(x-\frac{-b-\sqrt{b^{2}-4ac}}{2a}\right)$?  :maybe:

No, I believe Altheman is correct. Take $2x^{2}+3x+1$ for example. With Altheman's solution, the factors are $2(x+{1 \over 2})(x+1) = (2x+1)(x+1) = 2x^{2}+3x+1$ whereas yours will generate $x^{2}+{3 \over 2}x+{1 \over 2}$. Although these quadratics have the same roots, they are different.

[u]Have fun with this![/u]

Factor $ax^{3}+bx^{2}+cx+d$.  :rotfl:

[quote="Altheman"][quote="cincodemayo5590"][quote="lingomaniac88"]Factoring $kx^{3}+ax^{2}+bx+c$:
Define $r_{1}$, $r_{2}$, and $r_{3}$ according to this URL:
http://planetmath.org/encyclopedia/CubicFormula.html
The answer is $k(x-r_{1})(x-r_{2})(x-r_{3})$. :rotfl:[/quote]

 :rotfl: Awesome.  How in the world did they derive that??[/quote]

look for cardano...there are a few substitutions that solve it[/quote]

Right, i didnt see the links at the bottom. Its actually not tooo bad. But still...  :rotfl:

Alternately, we can use Galois Theory to derive both the Cubic and the Quartic formulas.

Not the way Ramanujan did it, but...  :P


Let $ a, b, c$ be the reals numbers satisfying $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 2(ab \plus{} bc \plus{} ca)$. Prove that:
\[ \frac {|a \minus{} b|}{\sqrt {2ab \plus{} c^2}} \plus{} \frac {|b \minus{} c|}{\sqrt {2bc \plus{} a^2}} \plus{} \frac {|c \minus{} a|}{\sqrt {2ca \plus{} b^2}} \ge 2\]
PS: When does equality occur? :)

Merry Christmas to nguoivn and other members of inequality box! :)

We have
\[ \sum_{cyc}\frac{|a\minus{}b|}{\sqrt{2ab\plus{}c^2}} \ge \sum_{cyc}\frac{2(a\minus{}b)^2}{a^2\plus{}b^2\plus{}c^2} \equal{}2\]

The equality doesn't occur since we must get $ a^2\plus{}b^2\plus{}c^2\equal{}0$ so $ a\equal{}b\equal{}c\equal{}0$ and this is a contradiction because the denominators must be distinct from 0.

Yes, your solution is right. It's also my hint to creat it.
But this sentence is not trues  :)
[quote="great math"]
The equality doesn't occur since we must get $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 0$ so $ a \equal{} b \equal{} c \equal{} 0$ and this is a contradiction because the denominators must be distinct from 0.[/quote]
Because equality can occur when $ a \equal{} b; c \equal{} 0$ or $ a \equal{} 4b \equal{} 4c$  :lol:

sorry bro, I was a bit careless when considering the cases of equality. Indeed, we have $ |a\minus{}b|\equal{}0$ which implies 2 other cases, namely $ c\equal{}0$ and $ c\equal{}2(a\plus{}b)$

The conclusion follows by nguoivn's post.