function
limit
calculus
derivative
real analysis
real analysis unsolved

Let $ f$ be a real function on $ (0, \plus{} \infty]$ , differentiable .Prove that :
 If $ \lim_{x \to \plus{} \infty}{(f(x) \minus{} 2\sqrt {x}f^{(1)}(x))} \equal{} 0$ then $ \lim_{x \to \plus{} \infty}{f(x)} \equal{} 0$

[quote="thefuturebird"]$ \lim_{x \to \plus{} \infty}{(f(x) \minus{} 2\sqrt {x}f^{(1)}(x))} \equal{} 0$
$ \lim_{x \to \plus{} \infty}{\left(f(x) \minus{} 2\sqrt {x}\left( \frac {f(x \plus{} 1/x) \minus{} f(x)}{1/x} \right)\right)} \equal{} 0$
$ \lim_{x \to \plus{} \infty}{(f(x) \minus{} 2x\sqrt {x}\left( f(x \plus{} 1/x) \minus{} f(x) \right))} \equal{} 0$
$ \lim_{x \to \plus{} \infty}{(f(x) \minus{} 2x\sqrt {x}\left( f(x) \minus{} f(x) \right))} \equal{} 0$
$ \lim_{x \to \plus{} \infty}{(f(x) \minus{} 2x\sqrt {x}\left( 0 \right))} \equal{} 0$
$ \lim_{x \to \plus{} \infty} f(x) \equal{} 0$[/quote]

how did you make first step??? and why we can write $ f(x)$ instead of $ f(x\plus{}\frac{1}{x})$ ??

1st step... That's the difference quotient.

Is there something wrong with writing f(x) instead of $ f(x\plus{}\frac{1}{x})$ ?

 :o

[quote="thefuturebird"]1st step... That's the difference quotient.

Is there something wrong with writing f(x) instead of $ f(x \plus{} \frac {1}{x})$ ?

 :o[/quote]

 We can write $ f'(x) \equal{} \lim_{y\to\infty} \frac {f(x \plus{} 1/y) \minus{} f(x)}{1/y}$ but $ f'(x)\neq \frac {f(x \plus{} 1/x) \minus{} f(x)}{1/x}$.
and are you sure that  $ \lim_{x\to\infty} f'(x) \equal{} \lim_{x\to\infty} \frac {f(x \plus{} 1/x) \minus{} f(x)}{1/x}$ ?

I'm not certian but but the function is differentiable so... the direction of the derivative won't matter... right? This is path independent... right? I'm not really certian at all anymore.

The claim is false to begin with.  $ f(x) \equal{} ce^{\sqrt {x}}$ with any $ c \neq 0$ is a counterexample.  It satisfies $ f \minus{} 2\sqrt {x}f'(x)\equiv 0$, but $ \lim_{x\to \infty}|f(x)| \equal{} \infty$.

As to why your argument fails, the first problem is indeed here:
[quote="anik_andrew"]are you sure that $ \lim_{x\to\infty} f'(x) \equal{} \lim_{x\to\infty} \frac {f(x \plus{} 1/x) \minus{} f(x)}{1/x}$ ?[/quote]Try, for example, $ f(x) \equal{} x^3$.  What is $ \lim_{x\to \infty}\left[ f'(x) \minus{} x\left(f\left(x \plus{} \frac {1}{x}\right) \minus{} f(x)\right)\right]$?

But even if you could justify that step (e.g. if somehow you knew a priori that $ \lim_{x\to \infty}f'(x)$ exists and is finite), this step:
[quote="thefuturebird"]$ \lim_{x \to \plus{} \infty}{(f(x) \minus{} 2x\sqrt {x}\left( f(x \plus{} 1/x) \minus{} f(x) \right))} \equal{} 0 \\
\lim_{x \to \plus{} \infty}{(f(x) \minus{} 2x\sqrt {x}\left( f(x) \minus{} f(x) \right))} \equal{} 0$[/quote]is also unjustifiable unless, again, you know a priori that $ \lim_{x\to \infty}f(x)$ exists and is finite.

I guess Quybac had one more condition $ e^{ \minus{} \sqrt{x}}f(x) \rightarrow 0$ as $ x \rightarrow \infty$. Then we may consider $ f(x) \equal{} \frac {e^{ \minus{} \sqrt {x}} f(x)}{e^{ \minus{} \sqrt {x}}}$ and use L'hospital rule.


let 51 subsets A1,A2,........,A51 of a finite X set such that the element numbers of each set > X/3
prove that: we can find a subset B of X ,the element numbers of B<=7 and B have least 1 element common for A1,A2,..........,A50

I thing the true is $|B| \le 8$
 If it is true, this is my idea:
 We have $51$ subsets, each of them have more than $\frac{X}{3}$ elements, so all of them have more than $\frac{51X}{3}= 17X$ elements, so there are $a_{1}\in X$ such that there are at least $18$ subsets such that $a_{1}$ is an common elements of them.
 We remove them, and use the same idea, we have:
 There are at least $[\frac{(51-18)X}{3}]+1=[\frac{33X}{3}]+1=12$ different subsets have $a_{2}$ as an elements.
 There are at least $[\frac{(33-12)X}{3}]+1=[\frac{21X}{3}]+1=8$ different subsets have $a_{3}$ as an elements.
There are at least $[\frac{(21-8)X}{3}]+1=[\frac{13X}{3}]+1=5$ different subsets have $a_{4}$ as an elements.
There are at least $[\frac{(13-5)X}{3}]+1=[\frac{8X}{3}]+1=3$ different subsets have $a_{5}$ as an elements.
There are at least $[\frac{(8-3)X}{3}]+1=[\frac{5X}{3}]+1=2$ different subsets have $a_{6}$ as an elements.
There are at least $[\frac{(5-2)X}{3}]+1=[\frac{3X}{3}]+1=2$ different subsets have $a_{7}$ as an elements.
 And we have only an subset, assume that $a_{8}$ in it.
 So we get $B=\{ a_{i}|i=1,2,..,8 \}$. This set have 8 elements and satisfied.

 Sorry for my English.


inequalities
function
logarithms
real analysis
real analysis unsolved

let f a function in $ C^2(R,R)$ such that for all $ x,y \in R$ we have:
$ f(x\plus{}y)f(x\minus{}y)<\equal{}f(x)^2$
prove that $ ff''<\equal{}(f')^2$

Taking logs of both sides, $ \log f(x \plus{} y) \plus{} \log f(x \minus{} y) \leq 2 \log f(x)$, so $ \log f(x)$ is convex (or is that concave?) and $ \frac{d^2}{dx^2}\log f(x) \leq 0$.  Equivalently, $ \frac{f''(x)f(x) \minus{} f'(x)^2}{f(x)^2} \leq 0$ and so $ f''(x)f(x) \leq f'(x)^2$.

Ok. What if $ f$ is always negative or zero ? How do you deal with taking logs. More care is needed here.


Nancy generates a two-digit integer by rolling a six-sided die twice. The result of her first roll is the tens digit, and the result of her second roll is the ones digit. What is the probability that the resulting integer is divisible by 8? Express your answer as a common fraction.

The solution was far more complicated than it needed to be. Just list the multiples of 8 between 0 and 66, then eliminate those that have digits of 7,8,9,and 0. There are 5 of these, then put it over 36

1. If it is divisible by 8 and atwo digit positive number, there aren't that many: I can list them all  16, ..., 64. So,
I counted and entered 7/36.

2. Wrong! I was told. So I look at the list again in disbelief: how could it be anything else. Well, 48 does not count!
So, I entered 1/6, which is a nice answer.

3. Wrong again! Why hadn'tI just looked at the list more carefully, 40 is impossible. Too late.

[hide=This is what Alcumus wrote]The largest number Nancy can generate is 66. The positive two-digit multiples of 8 less than 66 are 16, 24, 32, 40, 48, 56, and 64. 40 and 48 cannot be produced from the numbers on a 6-sided die, but the other 5 multiples of 8 in the list can. Therefore, there are 5 equally likely ways to produce a multiple of 8 out of the $6\cdot6=36$ total combinations for the two dice, so the desired probability is $\boxed{\dfrac{5}{36}}$.

(Special thanks to 5849206328x for this solution.)[/hide]

(@ above )

[size=75]Don't just post the Alcumus solution, even if you gave credit. [/size]

:thumbup:  :omighty: Ok thanks. I understand

Lol, why did I think that 0 was a number on a dice :/ ...

[hide=Solution]
Each ten you can produce $6$ numbers, one ending in $1\text-6$. There are $6$ tens, so there are $6\times6=36$ total possibilities. The ones that are divisible by $8$ are $16, 24, 32, 56, 64$. There are $5$ successful outcomes, so the probability that the resulting integer is divisible by $8$ is $\boxed{\frac{5}{36}}$.
[/hide]

Who is 5849206328x? Also, here is a[hide=hint]Remember that a six sided dice cannot display numbers less than one or more than six, so not every two digit multiple of eight is possible.[/hide].

[quote=ElectricFox]Who is 5849206328x?[/quote]

[url]https://artofproblemsolving.com/community/user/42575[/url]


Call a positive integer $n$ [i]good[/i] if there are $n$ integers, positive or negative, and not necessarily distinct, such that their sum and products are both equal to $n$. Show that the integers of the form $4k+1$ and $4l$ are good.

[quote=Pikachu12]hey, 4k+1=(4k+1)+1-1+1-1....4k times
        and product also 4k+1

but 4l=2l+2+1+1+1....(2l-2) times...+1-1+1-1.....2l times
        product is also 4l.

doesnt this work? we dont need cases.[/quote]

@Pikachu12 no.. if $l$ is odd you have odd many $-1$ s which will make the product $-4l$


geometry
circumcircle
inradius
conics
ellipse
incenter
geometric transformation

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14374[/img]

 :(

Let $ \mathbf P = \left\{\triangle ABC\ |\ (O), (I) = \text{const}\right\}$ be a set of triangles with a common circumcircle $ (O)$ and incircle $ (I).$ $ R$ is their circumradius, $ r$ the inradius, and $ d = OI.$ When $ AI \parallel OI,$ the $ \triangle ABC$ is isosceles and $ IT_a = r$ is minimum. On the other hand, when $ AI \perp OI,$ $ AI$ is maximum. $ S$ is the triangle area and $ s$ is the semiperimeter. In this case

$ 2rR = R^2 - OI^2 = AI^2 = \frac {4bcs(s - a)}{4s^2} = \frac {4RS(s - a)}{as} = \frac {4rR(s - a)}{a}$

$ \Longrightarrow\ b + c = 2a$ (which obviously means that $ GI \parallel BC,$ where $ G$ is the triangle centroid, see [url=http://www.mathlinks.ro/viewtopic.php?t=210342]Relation of lines OI and GI[/url]). It also means that maximum value of $ IT_a$ for $ AI \perp OI$ is $ IT_a = \frac {AI}{2} = \frac {\sqrt {2rR}}{2}.$ From this we can conjecture that the locus of $ T_a$ for $ \triangle ABC \in \mathbf P$ is an ellipse with center $ I,$ semiminor axis $ r$ parallel to $ OI,$ and semimajor axis $ \frac {\sqrt {2rR}}{2}$ perpendicular to $ OI.$ But a proof by direct calculation is hopeless.

Let $ D, E, F$ be the incircle $ (I)$ tangency points with $ BC, CA, AB,$ respectively. Invert the figure in the incircle. The triangle vertices $ A, B, C$ go to the midpoints $ A', B', C'$ of $ EF, FD, DE,$ so that the inverted $ \triangle A'B'C'$ becomes medial triangle of the intouch $ \triangle DEF.$ The circumcircle $ (O)$ of the $ \triangle ABC$ goes to the circumcircle $ (O')$ of the inverted $ \triangle A'B'C',$ i.e., into the 9-point circle of the intouch $ \triangle DEF.$ Radius of $ (O')$ is therefore $ \frac {r}{2}.$ The incenter $ I$ becomes the orthocenter of $ \triangle A'B'C'$ The sidelines $ a \equiv BC, b \equiv CA, c \equiv AB$ of the original $ \triangle ABC$ go to circles $ (O_a), (O_b), (O_c)$ with diameters $ ID = IE = IF = r,$ all congruent to $ (O'),$ intersecting in the inversion center $ I,$ the orthocenter of the inverted $ \triangle A'B'C'.$ The lines $ AI, BI, CI$ through the inversion center $ I$ stay in place and $ T_a$ goes to the intersection $ T_a'$ of the A-altitude $ A'I$ of the inverted triangle with the circle $ (O_a)$ other than $ I.$ Thus $ T_a'$ is a reflection of the vertex $ A'$ in the sideline $ B'C'$ of the $ \triangle A'B'C'.$

With respect to the intouch $ \triangle DEF,$ the point $ T_a'$ becomes the orthogonal projection of its vertex $ D$ to the perpendicular bisector $ IA'$ of its side $ EF.$ Suppose that the original $ \triangle ABC$ is rotated according to Poncelet's porism. The circumcenter $ I$ and the 9-point circle center $ O' \in OI$ of the intouch $ \triangle DEF$ are fixed, $ O'I \equiv OI$ is its Euler line, hence its orthocenter $ K \in O'I$ is also fixed. Let its altitude $ DK$ cut its circumcircxle $ (I)$ again at $ U$ and let the perpendicular to $ DK$ at $ D$ cut $ (I)$ agein at $ V.$ $ UV$ is a diameter of $ (I),$ $ T_a'$ is the midpoint of $ DV$ and $ IT_a'$ is midline of the right $ \triangle DUV,$ $ IT_a' \parallel UD,$ $ IT_a' = \frac {UD}{2}.$ Denote $ k = KI,$ $ \varrho' = IT_a'$ and let $ \theta$ be the angle of altitude $ UD$ with Euler line line $ KI \equiv OI.$ By Pythagorean theorem for half of the isosceles triangle $ \triangle DUI,$

$ \varrho'^2 = IT_a'^2 = \frac {UD^2}{4} = r^2 - k^2 \sin^2 \theta.$

This is polar equation of the locus of $ T_a'.$ (BTW, this is polar equation of elliptic lemniscate with center $ I.$) Now we invert the locus equation back:

$ \varrho^2 = IT_a^2 = \frac {r^4}{IT_a'^2} = \frac {r^4}{\varrho'^2} = \frac {r^2}{1 - (k^2/r^2) \sin^2 \theta}$

But polar equation of the ellipse with center $ I$ and semiminor / semimajor axes $ r, \frac {\sqrt {2rR}}{2}$ is

${ \frac {\varrho^2 \cos^2 \theta}{r^2} + \frac {2\varrho^2 \sin^2 \theta}{rR} = 1\ \Longrightarrow\ \varrho^2 = \frac {r^2R}{R - (R - 2r) \sin^2 \theta} = \frac {r^2}{1 - \epsilon^2 \sin^2 \theta}}$

where $ \epsilon$ is the ellipse excentricity, $ \epsilon^2 = \frac {R - 2r}{R}.$ This proves that the locus of $ T_a$ is an ellipse, the axes of which are already known.

Substituting the Euler formula for $ r = \frac {R^2 - d^2}{2R}$ into the ellipse equation, we get a quadratic equation for square of the circumradius $ R^2$:

$ 4R^2 \varrho^2 \cos^2 \theta + 4(R^2 - d^2) \varrho^2 \sin^2 \theta = (R^2 - d^2)^2$

$ R^4 - 2(d^2 + 2 \varrho^2) R^2 + d^2(d^2 + 4 \varrho^2 \sin^2 \theta) = 0$

with positive roots

$ R^2 = d^2 + 2 \varrho^2 \pm 2 \varrho \sqrt {\varrho^2 + d^2 \cos^2 \theta}$

where $ d = OI, \varrho = IT_a$ and $ \theta = \angle OIT_a$ are given. When $ d \rightarrow 0$ (approaching an equilateral triangle), $ R^2 \rightarrow 4 \varrho^2 \rightarrow 4r^2$ for the positive sign and $ R^2 \rightarrow 0$ for the negative sign $ \Longrightarrow$ only the positive sign is acceptable. This acceptable root can be constructed with compass and straightedge and I am certainly not going to describe it here. After constructing $ R,$ the inradius $ r$ can be constructed using Euler formula. Having both the circumcircle $ (O)$ and incircle $ (I),$ the ray $ T_aI$ cuts $ (O)$ at $ A$ and tangents to $ (I)$ from $ A$ cut $ (O)$ again at $ B, C.$

Is this all you say for getting solution of a difficult problem ? Did you ever hear about feet of the angle bisectors being on an ellipse about the incenter, defined by $ r, R$ only ? Talk to me, man.

$ R^2 \equal{} IO^2 \plus{} 2\ IT^2 \plus{} 2\ IT \sqrt {IT^2 \plus{} IO^2 \cos^2 \widehat{OIT}}$

From this, $ R$ is extremely simple segment to construct and I wonder, why do you endlessly post problems about solving quadratic equations, if you do not apply it.

1. Draw a circle $ \mathcal U$ with center $ I$ and radius $ IT.$

2. Drop perpendicular from $ O$ to the line $ IT,$ let $ F$ be the foot. $ IF \equal{} IO \cos \widehat{OIT}.$

3. Perpendicular to $ IT$ at $ I$ cuts the circle $ \mathcal U$ at $ D.$ Construct rectangle $ IDEF$ with sides $ ID, IF$ and take its diagonal

$ IE \equal{} \sqrt {ID^2 \plus{} IF^2} \equal{} \sqrt {IT^2 \plus{} IO^2 \cos^2 \widehat{OIT}}.$

4. Double the segment $ IE$ into $ IG,$ so that $ IG \equal{} 2\ IE.$ Line $ IG$ cuts the circle $ \mathcal U$ at $ H,$ so that $ IG, IH$ are opposite rays. Draw circle with diameter $ GH.$ Perpendicular to $ GH$ at $ I$ cuts this circle at $ K,$

$ IK \equal{} \sqrt {IH \cdot IG} \equal{} \sqrt {2\ IT \sqrt {IT^2 \plus{} IO^2 \cos^2 \widehat{OIT}}}.$

5. Construct square $ ITMD$ and take its diagonal $ IM \equal{} IT \sqrt 2.$

6. Transfer $ IM$ on the perpendicular to $ IO$ at $ I$ by a circle with center $ I$ and radius $ IN \equal{} IM.$ Construct rectangle $ IOPN$ with sides $ IO, IN$ and take its diagonal $ IP$:

$ IP \equal{} \sqrt {IO^2 \plus{} IN^2} \equal{} \sqrt {IO^2 \plus{} 2\ IT^2}.$

7. Transfer $ IK$ on the perpendicular to $ IP$ at $ I$ by a circle with center $ I$ and radius $ IS \equal{} IK.$ Construct rectangle $ IPQS$ with sides $ IP, IS$ and take its diagonal $ IQ$:

$ IQ \equal{} \sqrt {IP^2 \plus{} IS^2} \equal{} \sqrt {IO^2 \plus{} 2\ IT^2 \plus{} 2\ IT \sqrt {IT^2 \plus{} IO^2 \cos^2 \widehat{OIT}}} \equal{} R$

Translate the segment $ IQ$ to $ OZ \equal{} IQ$ and draw the circumcircle $ (O)$ with radius $ OZ \equal{} R.$

8. The simplest way to construct the inradius $ r$ is by Euler formula. (There are other ways.) $ (O)$ cuts the line $ IO$ at $ X.$ Invert $ X$ into $ Y$ in a circle with center $ O$ and radius $ OI.$ $ XY \equal{} OX \minus{} OY \equal{} R \minus{} \frac {OI^2}{R} \equal{} 2r.$

ok i make fourth proportional to solve
\[ R \minus{} \frac {{OI^2 }}{R} \equal{} 2r
\]
to calculate OI^2
\[ \frac {1}{{OI}} \equal{} \frac {{OI}}{{OI^2 }}
\]
but R construction is not clear with semantics :

"Draw circle with diameter GH Perpendicular to GH at I cuts this circle at K"
???

 :(

[quote="jrrbc"]"Draw circle with diameter GH Perpendicular to GH at I cuts this circle at K"
???[/quote]

This is standard way to construct square root of segment length product, or their geometric mean, or their mean proportional. I could have said just that: Construct the geometric mean (mean proportional) of IG, IH. If you are familiar with the fourth proportional and its construction, you are familiar with the mean proportional and its construction. The perpendicular to GH at I cuts the circle with diameter GH at 2 points K, K', but IK = IK' and only the segment length with one endpoint I is needed.

As for the inradius, $ \mathcal O$ is circle with center O and radius $ R$ (the circumcircle), $ \mathcal D$ is circle with center O and radius OI. Again, the circle $ \mathcal O$ cuts the line OI at 2 points X, X', but the segments OX = OX' are equal and only one is needed. Circle with diameter OX cuts the circle $ \mathcal D$ at 2 points U, V and perpendicular UV to OI cuts the line OI at Y. Y is reflection of X in the circle $ \mathcal D.$ Then $ XY \equal{} OX \minus{} OY \equal{} OX \minus{}\frac{ OI^2}{OX} \equal{} R \minus{} \frac{OI^2}{R} \equal{} 2r.$

with  R=IQ  and  XY=2r   my construction is not correct with the Ta point
i saw  problem with square ITMD (with the D point defined it's not a square)

 :(

I did the construction when I put down the recipe a month ago and now I did it again, following my own recipe exactly. There is not a single typo in it. If you understand what's being done, you should be able to follow the recipe even with a typo, but there is none.

A test after the construction is finished: Having circles (O), (I) with given centers O, I and constructed radii R, r, the line IT cuts (O) at A, so that A, I, T follow on the line in this order. Tangents to (I) from A cut (O) again at B, C. Line BC has to be tangent to (I) and points B, T, C have to be collinear - angle <BTC straight. My own construction passed the test.

As far as ITMD being a square: $ \mathcal U$ is a circle with center I and radius IT. Perpendicular to IT at I cuts $ \mathcal U$ at D. ID = IT and angle <DIT is right. What's preventing you to construct a square ITMD ?

ok i think the problem is with my construction about the square IMTD (is square i see now), i make other correct construction today to see if it is all ok.

 :wink:

See a step by step construction in the attachment. Howgh.

I got a reasonable construction that in my opinion falls into the category CWW.

Still, if anobody needs some words I have some.

 Happy New Year!

 M.T.

Let H be foot of $ O$ on $ IT_a$ and let  $ X$ be a point on $ IT_a$ that $ XT_a.(2XH) \equal{} XI^2$ we can find $ X$ Because:
Let $ XH \equal{} x$ we have :$ (x \minus{} HT_a).(2x) \equal{} (x \plus{} HI)^2$ Since we have $ HT_a$ and $ HI$ hence $ x$ is constructable , notice that $ X$ is on circumcircle of $ \triangle ABC$ , and the relations is equivalent to $ XT_a.XA \equal{} XI^2$ hence We have triangle $ ABC$ .

Happy new year


trigonometry
inequalities unsolved
inequalities

[quote="sylvect"]Let $ x\in \left(0,\frac {\pi}{2}\right]$. Prove that

$ \frac {1}{\sin^2x} \minus{} \frac {1}{x^2}\leq 1 \minus{} \frac {4}{\pi^2}.$[/quote]$ f(x)\equal{}\frac {1}{\sin^2x} \minus{} \frac {1}{x^2},$

$ f'(x)\equal{}\frac{2}{x^3}\minus{}\frac{2\cos{x}}{\sin^3{x}}.$

$ g(x)\equal{}\sin^2{x}\tan{x}\minus{}x^3,\left(0\leq x<\frac{\pi}{2}\right).$

$ g'(0)\equal{}g''(0)\equal{}0;g'''(x)\equal{}2\tan^4{x}(3\plus{}4\cos^2{x})>0,\left(0<x<\frac {\pi}{2}\right)$

$ \Longrightarrow g(x)> g(0)\equal{}0 \Longleftrightarrow f'(x)>0 \left(0< x<\frac {\pi}{2}\right)$

$ \Longrightarrow f(x)\leq f\left(\frac {\pi}{2}\right)\equal{}1 \minus{} \frac {4}{\pi^2}.$


conics
parabola
quadratics
linear algebra
matrix
algebra
system of equations

I have a parabola that crosses through the origin (0,0). The parabola also crosses through (60,10) and (24,d).
I need help in finding the equation of the graph in terms of d.

Is anyone able to help me please. Try and show the basic steps of how you came to solve it.

Thanks

[hide="Solution"]
Set up a system of equations.
All parabolas can be represented as an equation of the form $y=ax^{2}+bx+c$
If the parabola crosses the origin, then we know immediately that $c = 0$ (why?)
So now we have a system of equations for A and B.

We have
$10 = a(60)^{2}+b(60)$

and

$d = a(24)^{2}+b(24)$

Let's solve the first one for $b$.
$b = \frac{10-3600a}{60}= \frac{1-360a}{6}$

Now sub that back into the second one.

$d = a(24)^{2}+24 \cdot \left(\frac{1-360a}{6}\right)$
$d = 576a+4(1-360a)$
$d = 576a+4-1440a$
$d = 4-864a$
$864a = 4-d$
$a = \frac{4-d}{864}$

Now we do the same thing for $b$:
Solve the first one for $a$:
$10 = a(60)^{2}+b(60)$
$3600a = 10-60b$
$360a = 1-6b$
$a = \frac{1-6b}{360}$

Now sub that in for $a$ in the second equation.
$d = \frac{1-6b}{360}\cdot(24)^{2}+b(24)$
$d = 1.6(1-6b)+24b$
$d = 1.6-9.6b+24b$
$d = 1.6+14.4b$
$14.4b = d-1.6$
$b = \frac{d-1.6}{14.4}$

So now we have our two values.
We can put these into the quadratic:
$f(x) = \frac{4-d}{864}\cdot x^{2}+\frac{d-1.6}{14.4}\cdot x$

I think that's right.

Might be some arithmetic mistakes, but that's the general solution.
[/hide]

what if we set the determinant of this matrix equal to 0:

$0=\begin{tabular}{|cccc|}x^{2}&x&1&y\\ 0&0&1&0\\ 3600&60&1&10\\ 576&24&1&d\\ \end{tabular}$

[b]DiscreetFourierTransform[/b] you are a legend. Thanks heaps and I hope you are able to help with my new problem that I have just posted. Your working out was the greatest!
[/b]

Wow!  :blush:
That's very nice of you.

I'll see what I can do.


AMC
AIME
function
geometry
calculus
number theory
prime factorization

What are the AMC's rules regarding graphing calculator applications and programs?  I find a lot of program's useful (such as prime factorization), so I would like to know.  Do they clear your memory before you start the test?

Maybe the SAT has a rule about this that the AMC follows, but I don't have anything to do with the SAT as of yet.

Also, is it legal to store information (such as formula's etc.) in something like NoteFolio.  Could you go as far as to store former problems and solutions on the calc.

Yes, it's legal. But in reality, it will not help you all that much. The test writers designed the test with the fact in mind that TI 89s are permitted.

I would beg to differ.  The calculators can be very useful indeed.  It takes the skill (and fun) out of finding the elegant solution to a brute force problem.

The rules are very lax: there's no memory-clearing, and probably for good reason, too...it would get ugly if calculators got messed up and AMC was blamed, not to mention the problems associated with teaching teachers how to clear the memory.  And not all teachers would do it, so your score could depend on how sharp your teacher was...

Theoretically, you could store anything you want on a calculator.  However, there are certain moral issues at stake: will you feel good about your AMC score if you know you couldn't have gotten it without consulting your notes? And if your calculator is your only hope for qualifying for AIME, why bother cheating, as you'll be knocked out at that stage for sure?

Still, I don't think having little programs (as opposed to gigantic note files) is a big deal, especially if you wrote them yourself and know how to use the formulas anyway.

I use the distance formula obsessively during the test.  Me and my analytic geometry.

The Teachers' Manual ( Section IV, page 5, rule 4.) says: "...Students
may use scratch paper, graph paper, ruler, compass, protractor, and
four-function, scientific or graphing caclulators.  No problem on the
test will require the use of a calculator.  However, any
non-typewriter keyboard calculator accepted for use on the SAT may be
used during the contest.  See
http://www.collegeboard.com/counselors/hs/sat/registration/calc.html Students
may not share a calculator."

Practically speaking, the TI-89 is allowed.  The non-typewriter (or
QWTERTY) keyboard rule disallows some other equivalent calculators and
PDAs.  We understand that the boundary between all electronic devices
is blurring, and we will revise the rule at some time in the future.
The exsiting rule seems to be an adequate compromise between an
enforceable rule and the spectrum of choices running from allowing no
calculator to allowing laptops running computer algebra systems.

Disallowing all calculators for the AMC 10 and AMC 12 contests is
probably not a practical step anymore, because calculators are
recognized as a viable teaching and learning tool in mathematics
classes.  The outcry would be too great.

The committee  composing the AMC  10 and AMC  12 is well aware  of the
capabilities  of   the  TI-89   and  takes  these   capabilities  into
consideration when creating the  contests.  The committee strives hard
to  create "calculator-neutral"  contests.   The committee  recognizes
when a "brute-force", graphical, or computational solution would solve
a  problem   as  well  as  a  "paper-and-pencil"   solution.   If  the
computation is easier or quicker, the problem is usually rejected, and
some good  problems have been omitted  for that reason.   On the other
hand, in a few cases the  committe has decided that if a student knows
how  to use a  calculator intelligently  to find  a solution,  and the
method seems  to offer no  significant speed advantages,  then solving
with a calculator is another form of valid mathematical reasoning, and
the problem is kept.

In my own personal experience in the editing of the AMC 10 and and AMC
12 contests, I frequently use both a calculator and a computer algebra
system to check problems, check cases, or check ideas.  I have not
found a case where either of them offers a significant advantage over
using my mind well!

Steve Dunbar
AMC Director

[quote="AMCDirector"]The Teachers' Manual ( Section IV, page 5, rule 4.) says: "...Students
may use scratch paper, graph paper, [b]ruler, compass, protractor[/b], and
four-function, scientific or graphing caclulators.  No problem on the
test will require the use of a calculator.  However, any
non-typewriter keyboard calculator accepted for use on the SAT may be
used during the contest.  See
http://www.collegeboard.com/counselors/hs/sat/registration/calc.html Students
may not share a calculator."[/quote]

How would a ruler, compass or protracter really help you in any way?  I've tried thinking of a way, but I can't seem to find one.

They help you draw better pictures.  It is often easier to do a problem if what you look at is nice and neat.  They re not supposed to make any problems trivial or else they probably wouldn't be allowed.

you could attempt measuring an angle if you draw a sufficiently good figure, or measuriing lenghts. its even possible off the diagrams on the actual test.

I actually have a friend (a junior in college, now) who tried to solve a geometry problem on the AIME (or possibly the AMC -- not positive) by creating a massive to-scale diagram and using his protractor, but he was unable to decide if the angle he was measuring was 72 or 73 degrees.

Anyhow, geometry is vastly easier when you can quickly make to-scale diagrams, and that makes having a compass and straight-edge very useful.  (I don't think I ever really used a protractor, myself, but I could see how someone might.)

[quote="AMCDirector"]In my own personal experience in the editing of the AMC 10 and and AMC 12 contests, I frequently use both a calculator and a computer algebra system to check problems, check cases, or check ideas.  I have not found a case where either of them offers a significant advantage over using my mind well![/quote]

I asked a group of parents of mathematically gifted kids once online what they thought about the frequently made statement that the TI-89 is the secret weapon for scoring high on the AP Calculus BC exam. They thought that that was silly, based on what their kids had reported about taking that test. I know FOR SURE that tests at the level of the SAT I math section, the AMC 8, and the AMC 10 are amenable to being taken with no calculator at hand at all--after all, I took my SAT I that way back in the Stone Age, and when I take AMC 8s for self-practice as a math coach, I can get the answers without a calculator. The AMC 10 begins to expose the gaps in my mathematics education  :blush:  and the AMC 12 is HARD for me--as it is for most teenagers--but I don't think the hardness comes from being unfamiliar with a calculator, because I am pretty familiar with calculators these days. I sometimes attempted to check my son's curricular math homework, after he had finished each assignment, with Mathematica to spot careless errors, but I soon found that it took me as long to get through the homework assignments for spot-checking as it did for him to do the whole assignment for a grade, showing his work on paper. Good mathematical thinking ability is definitely the most important issue for any well-designed mathematics test for the twenty-first century: the problems that involve brute-force calculation are mostly left to the machines, and the human brains are challenged by problems that can't easily be automated.

We can't wait to get into this year's AMC season. Keep up the good work.


A fair, six-sided die is rolled, and the # obtained spinners are spun. The spinners have the #s 1-3 on them. What is the probability that the product of the #s spun is even?

[hide]1522/2187[/hide]
wow that took soo long.. chances you get the 6 numbers, then multiply by 1-chances you get odd

there is a pretty quick way to do it, though. hint:geometric series


Let $n\geq3$ be any real number. Assume $a,b,c>0$ with $abc=1$. Prove that
\[
\frac{1}{a^n(b+c)}+\frac{1}{b^n(c+a)}+\frac{1}{c^n(a+b)}
\geq\frac{3}{2}.
\]

You are very unlucky, since your generalization is already known...

http://www.mathlinks.ro/Forum/viewtopic.php?t=16352

  Darij


Prove that $\forall x,y,z>0: xy+yz+zx+xyz=4$ we have
$\frac{x+y+z}{xy+yz+zx}\le 1+\frac{(x-y)^2+(y-z)^2+(z-x)^2}{48}$
 :)  ;)

[quote="pvthuan"]It was my problem.[/quote]

What about your solution   ??
 :)

[quote="pvthuan"]It was my problem.[/quote]
But I sure that it is my problem ,too. :)

[quote="pvthuan"]I sent this inequality to your teacher, MDNghi. The solution will appear in my book.[/quote]
But I gave my teacher this pro,too.But I think this is no problem,because it is not important


I'm sort of confused, is an only child mean that a couple only has one child, or does it mean that only one child exists in an entire family's (aunts, uncles, etc) generation?

I've heard they're usually spoiled  :P I understand that in China, you're only allowed one child per household.

I believe an only child is when a family has only one child.
They're USUALLY spoiled because they recieve all the attention and all things are bought for them (not split between a sibling). They also also TEND to be selfish since they're not used to sharing (if you have a sibling, you know what I'm talking about)

Notice how I put usually and tend in capital letters. Not all only childs are spoiled brats. I know a  bunch of only children who are basically the opposite of a spoiled brat.

About the thing about 1 child being allowed in China. If you give birth to twins, triplets... etc, I believe you get to keep all the kids. So the basic law is you can only give birth once. If you give birth twice, you must either pay a fine (I can't remember if it was a large sum, or a certain amount every month or year) or give the child to the government. That's why you can adopt so many chinese kids.

I never heard of giving a child to the government.
Of course you should all know the population situation in China. They you should understand why it has a one-child policy. But if you give birth twice, then you will have to pay some money. And then it is ok.

and then i know some non-only child kids that are WAY super spoiled.

Ahhhh!! GoBraves is [i]spoiled[/i]

just kidding  :P

So there are such families in China with two children?

[quote="GoBraves"]Im an only Child (rare for an Indian  :P )[/quote]

Well, actually I think that isn't that rare anymore now, at least for the Indian families I know.  I can't think of any Indian family with more than 2 children that are around my age.

Off topic, but this is my 1,000th post!

[quote="Drunken_Math"]About the thing about 1 child being allowed in China. If you give birth to twins, triplets... etc, I believe you get to keep all the kids. So the basic law is you can only give birth once. If you give birth twice, you must either pay a fine (I can't remember if it was a large sum, or a certain amount every month or year) or give the child to the government. That's why you can adopt so many chinese kids.[/quote]

That's most definitely not the whole truth. I will write my rant another day, though. I don't have enough hours in a day to do all the math I need as it is. :P

[quote="Buffalo"][quote="236factorial"]So there are such families in China with two children?[/quote]
It seems you've got too much "information" from rumor. Try to find the answer by yourself.[/quote]

But I can't. I don't live in China.

[quote="236factorial"]I'm sort of confused, is an only child mean that a couple only has one child, or does it mean that only one child exists in an entire family's (aunts, uncles, etc) generation? [/quote]

Only child means that you're the only child your parents had.  If you include youre aunts and uncle's children, and you're still the only one, you're the only [i]grand[/i]child, and so on :roll:

is anyone here an only grandchild as well?

extra points if you CAN'T "get anything you want."

[quote="Buffalo"]Man, it's 21th century now, google "china population policy", you can find tons of materials.[/quote]

I just don't see why i have to  ;)  I did though anyway.

[quote="star99"]and what did you find?

I have heard that chinese parents come here to have their baby, and then they go back, so its not against the law.  But that doesn't seem true, because why would they come here and do all that hassle.  why not just come to live here?[/quote]
As we all know, parents have to pay some money if they born two childs. The fee is much lower the plane tickets.

[quote="shobber"][quote="236factorial"]But I've never seen a family with two kids my whole trip back to China  :?[/quote]
That is because you went to the cities. There are less two-kids-family in the cities than in the countryside.[/quote]

Yes, in fact, in rural areas, it is very difficult to find one-child families.

Also, minorities are not restricted by the one-child laws.

[quote="mel"]
Yes, in fact, in rural areas, it is very difficult to find one-child families.

Also, minorities are not restricted by the one-child laws.[/quote]

I saw no children in the rural areas.  ;)

Are minorities like americans and indians?

The dismaying demographic fact about China, a result of the official population policies interacting with traditional culture and economic reality, is the severe shortage of women in the generation now growing up.

http://www.washingtonpost.com/ac2/wp-dyn/A77925-2001May25

http://www.guardian.co.uk/international/story/0,3604,714412,00.html

http://app1.chinadaily.com.cn/star/2002/1024/fo5-1.html


[quote="mmrproblem"]There are 12 coins on a desk and half of them are A face and other half are B face.You can not see and touch that coins.Can you separate A face coins two groups that each group have equal A face coins?How?[/quote]

i don't understand what you are asking for

I think he means:
There are 12 coins on a desk. They all look the same on one side (the one displayed), but half of them have an A face on the reverse side, and the other half have a B face.
You can't see or touch the coins.
Is it possible to seperate the coins into two groups, such that they have the same number of A face coins?

---

I would say no. Unless you have superpowers, it's impossible to do.

But I might have misinterpreted it. Maybe you could clarify?

You interpreted the question correctly, but it is possible. :)

This is a very famous problem/puzzle.

Wait, isn't it that each coin has an A face and a B face? So you can flip them to change the faces... I think that's it, I've heard the problem before like that only with red and green faces.

This is very confusing... Can somebody please post the corrected question??? the way Hexaditidom phrased it, I'm quite sure it's impossible...

If you don't know anything about the coins and you can't do anything to the coins, it's pretty obvious that it's impossible ;)

Try this one then: You have $2n$ coins, each with a red face and a green face. $n$ of them are red face up and the other $n$ are green face up. Without looking, separate the coins into two piles such that the number of green face up coins in each pile is equal. You may move and flip coins.

Ah, I see it...

[hide="Solution"]Just take any $n$ coins into one of the piles, and flip all of them.  The other $n$ coins remain unflipped, and constitute the other pile.[/hide]

Seperate coins into n pairs, and put one element of the pair in pile A, the other element of the pair in pile B.   Then flip all coins in B.

If a pair has different colors face up, then after flipping one element in it, it becomes the same, and doesnt change the difference in green-face-up.

If a pair is red-red, it changes to red green, and B has one more green up.
If a pair is green-green, it changes to green-red, and A has one more green up.

But the number of pairs that are red-red and green-green are equal, so the number of green-up's are equal.

---

So in short:  split it into two equal piles, and flip all in one pile.


ABCD is a cyclic quadrilateral.prove that <AOB + <COD = 180degree

Hm...Is this always true?

Let the intersection of AC and BD be P.

$ \angle AOB \equal{} \angle ADB \plus{} \angle ACB$, since $ \angle AOB \equal{} 2\angle ADB \equal{} 2\angle ACB$

Similarly,

$ \angle COD \equal{} \angle CAD \plus{} \angle CBD$

If the statement was true,

$ \angle AOB \plus{} \angle COD \equal{} \angle ADB \plus{} \angle ACB \plus{} \angle CAD \plus{} \angle CBD \equal{} 180$

This implies that $ \angle APD \plus{} \angle BPC \equal{} 180$, which implies that $ \angle APD \equal{} \angle BPC \equal{} 90$, which isn't always true...

You could also consider an extreme case, when A and D are very close, and B and C are very close. The sum of the angles are obviously above 180.

Use the fact that for $ \triangle ABC$ with circumcenter $ O$, $ \angle ABC \equal{} \frac{1}{2} \angle AOC$. In other words, if the vertex of the angle in a circle is itself on the circle, then its measure is half that of the intercepted arc.

In a quadrilateral, the intercepted arcs of diagonally opposite angles add up to the full circle, or 360 degrees, so the measures of the two angles add up to half of that, or 180 degrees.

It doesn't seem like this is usually the case. Consider a circle with center O and A and B lying very close to each other (making <AOB close to 0) and C and D on the opposite side of A and B respectively.

yes i also thought this and got confused.......so i gave this in this forum


calculus
derivative
integration
logarithms
calculus computations

Write y=a+bx+cx^2+dx^3+...
Then y'=b+2cx+3dx^2+4ex^3+...
Thus b+2cx+3dx^2+4ex^3+...=a+(b+1)x+cx^2+dx^3+...
Equating terms, we get b=a, c=(b+1)/2, d=c/3, etc.
Putting that all together, we get y=k+kx+(k+1)/2*x^2+(k+1)/6*x^3+..., which looks an awful lot like (k+1)e^x except for a few terms at the beginning. In the end we get y=(k+1)e^x-x-1, and we can replace k+1 by k since it's an arbitrary constant, so the solution is y=ke^x-x-1.

[color=cyan]Ooh, I just thought of another way, in case you don't know Taylor series methods:

Take two derivatives of the whole expression.  Then
[b](1)[/b] y'' = y' + 1
[b](2)[/b] y''' = y''
So that means y'' = r*e^x (from [b](2)[/b])
So that means y' = r*e^x - 1 (from [b](1)[/b])
So that means y = r*e^x - x + c

Then solve the initial equation to get rid of a constant we don't need:
y' = y + x
r*e^x - 1 = r*e^x - x + c + x
-1 = c
y = r*e^x - x - 1, which is luckily exactly what was already gotten.

I do like Taylor Series, though.[/color]

That's a nice method. Another way to solve this problem, if you know the standard theory of differential equations, is to write it as y'-y=x, multiply the entire expression by e^x to get y'*e^x-y*e^x=xe^x, or d/dx(e^x*y)=xe^x and integrate with respect to x.

The question is how you know to multiply by e^x. The idea is that if we have dy/dx + P(x) y = Q(x), then we multiply the entire expression by e^integral of P(x) dx, and then the left hand side becomes factorable. That's the standard technique, but I usually prefer to look for more creative ways of doing problems if I think they won't be too difficult. That's why I looked for the Taylor series method before thinking about putting the differential equation into the assembly line and solving it quickly.

[color=cyan]You mean e^(-x).  I wish I remembered all my diffeq.  That was fun stuff.[/color]

Ah yes, sorry about that. I worked it out a few days ago and then tried to do it from memory. That wasn't such a good idea.

whoa that was a superior method

I will remember this alternate method of solving DiffEqs :)

An other approach

$\frac{dy}{dx}=x+y$

Let $x+y=t\implies 1+\frac{dy}{dx}=\frac{dt}{dx}$

$\implies \frac{dt}{dx}=1+t\implies \ln (1+t)=x+c\implies 1+t=e^{x+c}\implies y=e^{x+c}-x-1$

 :)


number theory
greatest common divisor
modular arithmetic
number theory proposed

Let $ m,n$ be a positif integers such that $ \text{gcd}(m,n)\equal{}1$ calculate: $ \text{gcd}(5^m\plus{}7^m,5^n\plus{}7^n)$

[hide="Alternately"]
Consider a prime $ p$ which divides both; $ 5^m\equiv \minus{} 7^m\pmod{p}$, $ 5^n\equiv \minus{} 7^n\pmod{p}$. We can take inverses and multiply as many times as we wish to get $ 5^{am \plus{} bn} \equal{} ( \minus{} 1)^{a \plus{} b}\cdot7^{am \plus{} bn}\pmod{p}$; since $ (m,n) \equal{} 1$, by Bezout's we get $ p|12$ and $ p|2$ respectively when both are odd and one is even by considering parity of the exponent of the $ \minus{} 1$ term. Hence when $ m,n$ are odd, we have gcd=12, otherwise, gcd=2.
[/hide]


A gambler plays the following coin-tossing game. He can bet an arbitrary positive amount of money. Then a fair coin is tossed, and the gambler wins or loses the amount he bet depending on the outcome. Our gambler, who starts playing with  $ x$ forints, where $ 0<x<2C$, uses the following strategy: if at a given time his capital is $ y<C$, he risks all of it; and if he has $ y>C$, he only bets $ 2C\minus{}y$. If he has exactly $ 2C$ forints, he stops playing. Let $ f(x)$ be the probability that he reaches $ 2C$ (before going bankrupt). Determine the value of $ f(x)$.

Fair game, should be [hide]$ f(x)\equal{}\frac{x}{2C}$[/hide]

Formal proof: [hide]Fortune, $ F_{t}$ is a martingale, winning (or loosing) time $ T$ satisfies optionall stopping time theorem, so $ F_{0}\equal{}E(F_{T})\equal{}p2C\plus{}(1\minus{}p)0$, so $ p\equal{}\frac{F_{0}}{2C}$ [/hide]


Find all natural a so that there exists a natural number n>1 such that:
a^n +1 is divisble by n^2.
Try to solv

Ok, here's a first guess (this isn't a solution):

I think all numbers NOT of the form 2^k-1 work. If a isn't of this particular form then there exists an odd number t>1 s.t. t|a+1. This means that t|S=a^(t-1)-a^(t-2)+a^(t-3)-..-a+1 because each of the terms is 1 (mod t) and there are t terms. It follows that t^2|S*(a+1)=a^t+1, so we have found our n for this specific a. I don't know if numbers of the form 2^k-1 aren't good; I'm still thinking.

I think I've got the last part of the soln:

Let's assume there is an a of the form 2^k-1 s.t. we can find n with the property n^2|a^n+1. It's obvious that n is odd, because 2|n => 4 | n^2 | a^n+1 = a^(2x)+1 = 2 (mod 4), contradiction. n^2 | (2^k-1)^n+1 | (2^(2k)-2^(k+1)+1)^n-1 (I just multiplied the right term with              (2^k-1)^n-1). Let p be the smallest prime which divides n. We have p|(2^(2k)-2^(k+1)+1)^n-1 and p|(2^(2k)-2^(k+1)+1)^(p-1)-1 (the last one from Fermat). From these 2 and the fact that the sequence                  (2^(2k)-2^(k+1)+1)^n-1 is a Mersenne sequence we get the fact that p|2^(2k)-2^(k+1)+1-1=2^(k+1)*(2^(k-1)-1), so p|2^(k-1)-1. This means that a=2^k-1=1(mod p), so a^n+1=2 (mod p), so p | n^2 | a^n+1=Mp+2 is a contradiction because p is odd (we know that n is odd).

I hope it's clear enough...


There are $L$ given empty cells and $N$ boxes. Each box is $K$ cells length.
And, $N \times K \le L$

I want to arrange the boxes in the cells so they don't overlap. For example: $L = 50, N = 2, K = 10$

I can arrange the first box in cells $1-10$, and the second in $15-24$. but I can't put the second in $10-19$ because it will overlap the first one.

How many different ways are there to arrange the boxes? All the boxes are same in shape, size, colour, et al.

[hide]Ok so there are N boxes, and L-(N x K) open cells when all boxes are placed, so the ways to rearrange them is  (L-NK+N)! / N!(L-NK)![/hide]


Divisibility
induction
prime
number theory

this is easy of course but cute :
we know that if $d|n$ then $d+1|n+1$.prove that n is prime :)

We will first show that we cannot have [tex]1 < d < n[/tex]. (I assume [tex]d[/tex] is not [tex]1[/tex].)
Suppose to the contrary that [tex]1 < d < n[/tex]. Saying that [tex]d|n[/tex] is equivalent to [tex]n = kd[/tex] for some integer [tex]k > 1[/tex] (because [tex]1 < d < n[/tex]). Thus [tex]d|kd[/tex] and by hypothesis [tex]d|kd \Rightarrow d + 1| kd + 1 \Rightarrow \ldots \Rightarrow 2 d| ( k + 1 ) d[/tex], in general [tex]md| ( k + m - 1 ) d[/tex] for any [tex]m \geqslant 1[/tex]. Choose an [tex]m[/tex] such that [tex]( m, d ) = 1[/tex], then [tex]m|k + m - 1[/tex] which imply that [tex]m|k - 1[/tex] (note that [tex]k - 1 > 0[/tex] which is important elsewhere if [tex]k - 1 = 0[/tex] then for any [tex]m[/tex] we would have [tex]m|k - 1[/tex], since [tex]m| 0[/tex]). But since [tex]m[/tex] can be arbitrary large and [tex]k[/tex] is fixed, we can choose an [tex]m[/tex] such that [tex]m > k - 1 > 0[/tex] and [tex]( m, d ) = 1[/tex], arriving at a contradiction.
Since we cannot have [tex]1 < d < n[/tex], there is no divisor of [tex]n[/tex] being smaller than [tex]n[/tex], and clearly [tex]n[/tex] is a prime number.

I am not tu sure about your solution, but anyway, suposse $n=ab$, then we would have:

$a+1|ab+1$ and $b+1|ab+1$ it is easy to see $(a+1,b+1)|2$ so we have $(a+1)(b+1)|2ab+2$ and hence $ab+a+b+1|2a+2b$. this implies $a+b$ is at least $ab+1$, and equality only holds for $a=1$ or $b=1$, so $n$ is prime.

[quote="Pascual2005"]I am not tu sure about your solution[/quote]

I'm also unsure about it  :( . I would be really happy if anyone could certify or refute it ...

you stated that iff $d|n$ then $d+i|n+i$ for all $i$ and thats not what the problem says! ;)

Right ! ;)  We have [tex]d|n \Rightarrow d+1|n+1[/tex] but not necessarly [tex]d + 1| n + 1 \Rightarrow d + 2| n + 2[/tex] :oops:

The problem statement maybe somehow midleading.
For example, d = 2, n = 8, then 2 | 8, 2 + 1 | 8 + 1, but n is not prime.

So I think the condition might be just like induction, that is:
d + i | n + i for i = 0, 1, 2, ...,

Man! the problem says for all divisors of $n$, if you put $n=8$, then we get $d$ can be $1,2,4,8$ but taking $d=4$ clearly you get a contradiction!

Now what you are trying to solve is a problem from colombia last round 2003, that was something like: prove that for every $m$ and $d$ we can find $m$ numbers $D+1, D+2,...,D+m$ such that $d+i|D+i$. But we cannot find infinitely many ( i mean $m$ cannot be replaced by infinitely). This is certainly a very easy problem


inequalities
induction
inequalities unsolved

For $x_1,x_2,...,x_n$ positive real numbers whose product is less than 1, prove the old forum ineq of Harazi

$\frac{1}{n-1+x_1}+\frac{1}{n-1+x_2}+...+\frac{1}{n-1+x_n} \leq \frac{n}{n-1+G}$

where $G^n=(x_1x_2\cdot\cdot\cdot x_n)$

Do you mean there is still no solution for this inequality (except my unposted one :) )?

I don't remember any solution for it, but I can be in error.

In my notes I didn't find any solution for it.

It is strange, because I think I saw solution, but I am not sure  :?

I solve the same problem:
If $x_1,x_2,...,x_n>0$ and $x_1x_2...x_n=1$ then for all $k \ge n-1$ we have:
$\sum {\frac{1}{x_i+k}} \le \frac{n}{1+k} (*)$

We use induction by n:
For n+1:
$x_1 \le x_2 \le ... \le x_n \le x_{n+1}$
Take $c$ st:
$c^n.x_1x_2....x_n=1$ then $c \ge 1$
We have:
$\sum{\frac{1}{ck+cx_i}} \le \frac{n}{ck+1}$
Then
$\frac{1}{k+x_1}+\frac{1}{k+x_2}+...+\frac{1}{k+x_n}+\frac{1}{k+x_{n+1}} \le \frac{nc}{ck+1} + \frac{1}{k+x_{n+1}}(**)$

Because $x_1x_2...x_nx_{n+1}=c^n.x_1x_2...x_n $
then $x_{n+1}=c^{-n}$
Take $x_{n+1}$ in (**) then it's not difficult to have result.

[quote="harazi"]I do not revendicate this inequality![/quote]
What is "revendicate"?

[quote="harazi"]Oh, sorry, after starting to learn french, I started also to mix languages: romanian, english, french.[/quote]

It doesn't look similar to anything I know of in English or French...I wonder exactly what you mixed together, harazi. :?

perhaps it is romenian? I guess he mustn't have spoken that for a while :P

Dear Hungkhtn,

I tried to complete your nice solution by induction of

$\displaystyle \frac{1}{k+x_1}+\frac{1}{k+x_2}+...+\frac{1}{k+x_n}+\frac{1}{k+x_{n+1}} \le \frac{nc}{ck+1} + \frac{1}{k+x_{n+1}}$

but I obtained

$kc^{n+1}(n+1) +cn +c^n \leq n+1+ck+nkc^n $

which is not true.

Can you please to detail better your proof.

Thank you very much. :lol:

[quote="Kuba"]actually "revendicate"(=demand the return of property, reclaim property) is an English word  :D[/quote]

:blush: Apparently, my English dictionary isn't thick enough. :wacko:

[quote] Apparently, my English dictionary isn't thick enough. [/quote]

Apparently, neither is google's.  :lol:

[quote="manlio"]I tried to complete your nice solution by induction of
$\displaystyle \frac{1}{k+x_1}+\frac{1}{k+x_2}+...+\frac{1}{k+x_n}+\frac{1}{k+x_{n+1}} \le \frac{nc}{ck+1} + \frac{1}{k+x_{n+1}}$
but I obtained
$kc^{n+1}(n+1) +cn +c^n \leq n+1+ck+nkc^n $
which is not true.
Can you please to detail better your proof.[/quote]
Hungkhtn was wrong when he wrote $x_{n+1}=c^{-n}$. Maybe it is the core of your problems, Manlio?
Actually, $x_{n+1}=c^n$, $c\geq 1$ and we are to prove the following inequality:
\[(k-n)c^{n+1}+(n+1)c^n-k(n+1)c+1\geq 0,\]
which is true, since $LHS'_c=(n+1)(k-n)c^n+n(n+1)c^{n-1}-k(n+1)\geq 0$ for $c\geq 1$ and $LHS|_{c=1}=0$.

[quote="manlio"]This solution by induction works only for $k\geq n$ and what happens for $ n- 1 \leq k \leq n$  :?  :? ??????????????????[/quote]
But we are proving induction step, so $k\geq (n+1)-1=n$ !!!!!!!!!!!!!!!!

:blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush: :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:

I would make better to choose another hobby more suited for my stupid mind  :(

[quote="harazi"]I do not revendicate this inequality![/quote]
[quote="Myth"] What is "revendicate"?[/quote]
[quote="harazi"]Oh, sorry, after starting to learn french, I started also to mix languages: romanian, english, french. Sorry, for me "' revendicate'"' means to consider that it is your creation and to claim that. I said this because Manlio said one of Harazi's inequalities. I do not remember anything about this one except the fact that I have a solution.[/quote]
Meanwhile, I "revendicate" this inequality, because it is the core of my paper
[i]Extensions of some cyclic inequalities[/i],  G.M. 10, 1991.
But if someone can indicate a previous reference about this inequality, then
I waive a claim.

:) When 2 argue, the 3rd one comes and wins :D

[quote="nickolas"][quote="hardsoul"]I think I found a proof without using induction.
Put $x_i=a_i^n$  then adding  $\frac{1}{n-1+x_i}=\frac{a_1a_2...a_n}{(n-1)a_1a_2...a_n+a_i^n}\leq \frac{1}{n-1}-\frac{a_i^{n-1}}{(n-1)(a_1^{n-1}+a_2^{n-1}+...+a_n^{n-1})}$ we get the desired result.
The last inequality is an easy application of AM-GM inequality.
 :D[/quote]

That  solotion works  if we had [tex]x_1x_2...x_n=1[/tex] , But we have  [tex]x_1x_2...x_n\leq1[/tex][/quote]
We  have $x_1x_2...x_n\ge 1$ if not just take all of x_i's  aproaching to zero .


a function f is such that f(x)>0 and f`(x) is continous at every real x.If f`(t)>=square root of f(t) then show that (f(x))^1/2>=(f(1))^1/2+1/2(x-1).....for all x>=1 :)

SORRY FOR NOT USING LATEX...ACTUALLY I AM NOT GOOD AT IT.

OHHHH!
I also dont know good latex........

here are my hints f`(t)/(f(t))^1/2>=1
Integrate both the sides taking limit from 0 to x........... :)  :)  :wow:


[quote="mathking123"][quote="leroyjenkens"]Here's an example I ran across that I don't quite understand.
[img]http://www.wtamu.edu/academic/anns/mps/math/mathlab/beg_algebra/beg_alg_tut11ex5b.gif[/img]

Why didn't the 15 become negative?[/quote]

if you are referring to the last line, isnt it already negative??[/quote]
I don't think so. That's the minus sign between the terms, which was already there.

Notice the -3 turns the 4 into 12, but he just turns the + in to a - instead of making the 12 a negative. But when he multiplies times the 5, the 15 didn't change at all.

[quote="leroyjenkens"]Notice the -3 turns the 4 into 12, but he just turns the + in to a - instead of making the 12 a negative. [/quote]

Turning the $ \plus{}$ into a $ \minus{}$ does make the $ 12$ negative...

[quote="leroyjenkens"] But when he multiplies times the 5, the 15 didn't change at all.[/quote]

Are you sure you understand that subtracting a positive number and adding a negative number are the same thing?  I want to rewrite this example without any subtraction, using negative signs instead:

$ 2(6x \plus{} (\minus{}5)) \plus{} (\minus{}3)(5x \plus{} 4) \equal{}$
$ 2(6x) \plus{} 2(\minus{}5) \plus{} (\minus{}3)(5x) \plus{} (\minus{}3)(4) \equal{}$
$ 12x \plus{} (\minus{}10) \plus{} (\minus{}15x) \plus{} (\minus{}12)$

Is this clear now?

[quote="hunter34"]A bit more steps.

\begin{align*} 2(6x - 5) - 3(5x + 4) & = 2(6x - 5) + ( - 3)(5x + 4) \\
& = ((2\cdot6x) - (2\cdot5)) + ((( - 3)(5x)) + (( - 3)(4))) \\
& = 12x - 10 - 15x - 12 \\
& = - 3x - 22 \end{align*}[/quote]
Ok thanks that helped. I think I understand now.
[quote]Turning the  into a  does make the  negative...
[/quote]
What I meant was literally turning the 12 into a negative. For example 15x + (-12). I know the result is the same thing, I just mean writing it that way.
[quote]Is this clear now?[/quote]
Yeah, thanks guys. It's just a little confusing because it goes against what you initially learn as a kid.


[quote="asdfG"]2. The eigen values of matrix A are the values of z for which det(zI - A) = 0. Find the difference of the largest and smallest eigenvalues of the matrix
| 2 -5 4 |
| 0 1 8 |
| 0 0 -3

Thanks for any help.|[/quote]

This is a do-it-in-your-head question, asdfG.

$\begin{bmatrix}2&-5&4\\0&1&8\\0&0&-3\end{bmatrix}$ is an upper triangular matrix.

The eigenvalues of an upper triangular (or lower triangular) matrix are the elements on the main diagonal, hence the eigenvalues of this one are $2, 1,$ and $-3.$ The difference between the largest and smallest is $5.$

Thanks. I didn't know what eigenvalues were before. I get it now.


Hi mathlinkers.

$ a^{2} \plus{} 1 \equal{} b(a \plus{} b)$
$ a,b\in{N}$

Find $ (a,b)$

There are a few missing.  Those ordered coordinates seem to be in a pattern, don't they?  :P

[hide="other answers"]
along with -1,-1, they're ascending fibonacci. eg. (21,13)
[/hide]

this is a known fact-
for any 4 consecutive fibonacci numbers - a,b,c,d in that order
we have $ ab\minus{}cd \equal{} \plus{}\minus{} 1$
now reduce that to a,b, and u will have this equation.
so infinite
for the property u can use fibonacci formula or use induction

how can you say that a,b have to be only fibonacci numbers

i dont think he is laying a condition that $ (a,b)$ are fibo no.s , he is only finding infinite solutions for $ (a,b)$ which ofcourse doesnt solve the problem as u have to find all $ (a,b)$ if [b]finite[/b] all prove it to be infinite for all cases..

The equation writes $ b^2 \plus{} ab \minus{} a^2 \minus{} 1 \equal{} 0$, with natural solution $ b \equal{} \frac { \minus{} a \plus{} \sqrt {5a^2 \plus{} 4}} {2}$ when $ 5a^2 \plus{} 4 \equal{} c^2$, i.e. $ c^2 \minus{} 5a^2 \equal{} 4$.

This is a Pell equations with seed $ (c, a) \equal{} (3,1)$ for odd elements $ c, a$; while for $ c \equal{} 2c', a \equal{} 2a'$ it comes to $ c'^2 \minus{} 5a'^2 \equal{} 1$, with seed $ (c', a') \equal{} (9, 4)$. Therefore all solutions can be found.

Could it be possible that when he asked for $ (a,b)$ he meant $ \gcd(a,b)$? For all the solutions that have been shown the $ \gcd(a,b)$ has been $ 1$. Just putting an idea out there, could be wrong.


Hi, I am posting using my son's account.

I suggest you create several forums for the parents of math students to share ideas and experiences from a parental point of view on how to help their kids advance in math skills, competitions, college choices, etc.

James

We would be happy to make such a forum if we have someone willing to moderate it.  If we created a parents forum, we would want to make it private, meaning that only parents could access it, since otherwise parents might be considerably less likely to use the forum.

Do we have any volunteers among our parents to act as a moderator of the forum?  The primary duty the moderator would have is being the gatekeeper to the forum, taking requests from parents to have access to the forum, etc.

How many moderates do you need?  Count me as one.  I can get few more to join if you need more.

James

[quote="zmli"]How many moderates do you need?[/quote]One suffices.

Hello,

I don't know if it is just me or if it is the rule, but the Parents forum says that

"Parents only. Private message the moderator or the site administrators to join."

and I try to see the forum, but can't get in (I click in the link, but it stays there forever, doing nothing).

I am not really a parent (not yet, and my wife wants to keep it that way for now  :D ), but I have been math student almost all my life, I have been Math teacher in middle and high school, Math Instructor in Colleges in Costa Rica and the US and even Math Olympiad Trainer for some national and international competitions. Therefore, I suppose that I would find nice to be able to read from the forum, even if I am not allowed to give an opinion right there.

Best regards,

[quote="djimenez"]
I don't know if it is just me or if it is the rule, but the Parents forum says that

"Parents only. Private message the moderator or the site administrators to join."

[/quote]

Just write them a PM and ask for permission to join. I am pretty sure they will let you in.  ;)


Theresa's parents have agreed to buy her tickets to see her favorite band if she spens an average of 10 hours per week helping around the house for 6 weeks. For the first 5 weeks she helps around the house for 8,11,7,12 and 10 hours. How many hours must she work during the final week to earn the tickets?

$ \textbf{(A)}\ 9 \qquad
\textbf{(B)}\ 10 \qquad
\textbf{(C)}\ 11 \qquad
\textbf{(D)}\ 12 \qquad
\textbf{(E)}\ 13$

[quote="Dojo"]Theresa's parents have agreed to buy her tickets to see her favorite band if she spends an average of 10 hours per week helping around the house for 6 weeks. For the first 5 weeks she helps around the house for 8,11,7,12 and 10 hours. How many hours must she work during the final week to earn the tickets?

$ \textbf{(A)}\ 9 \qquad \textbf{(B)}\ 10 \qquad \textbf{(C)}\ 11 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 13$[/quote]

[hide]Total, she must spend $ 10 \times 6 \equal{} 60$ hours of working.  She has already spent $ 8 \plus{} 11 \plus{} 7 \plus{} 12 \plus{} 10 \equal{} 48$, so she needs to spend $ 60 \minus{} 48 \equal{} \boxed{\textbf{(D)}\ 12}$[/hide]

Dojo: I'm not sure, but I think you should use the format described [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1056197#1056197]here[/url].

[hide="Solution"]$ (D)$ The first 5 weeks Theresa works a total of $ 8\plus{}11\plus{}7\plus{}12\plus{}10\equal{}48$ hours. She has promised to work $ 6\cdot 10\equal{}60$ hours. She must work $ 60\minus{}48\equal{}12$ hours during the final week.[/hide]

@Dojo:

May the title something to do with the problem and the sub-title "AMC 8 2007 Problem 1"

Or for a faster way to solve this problem, use:
[hide]Theresa needs to spend an average of 10 hours a week cleaning, she has already spent: 8,11,7,12,10 hours. so we can "smooth out the averages" so take 2 hours from the 12 hour week and add it on to the week with 8 hours. take one hour from 11 hour week and add it to the 7 hour week, so we have:10,10,8,10,10, she needs to clean 10 hours as a requirement so add two more hours from the 8 hour week to get 12 hours. Its basically the same method as described above but just a little faster and you don't have to deal with all of the business regarding multiplying hours and subtracting hours. just even out the averages. [/hide]

D! 6*10=60
she has worked 48 so far
60-48=12
D

[quote="pisqrt314"]Is this a Q or Challenge?
[hide]D[/hide][/quote]

The problem has been already answered--multiple times.  If you had a different way to solve it, you should have posted it, but since you just provided the answer, your post is useless.  BTW: This is a problem that is going to be added to the resources section, so you HAVE to post solutions.


trigonometry
ratio
Pythagorean Theorem
geometry
trig identities
Law of Sines
Law of Cosines

[quote="jimi"]This is probably so obvious to lots of you but not to me. How do we know that sin 30 deg is exactly 1/2?[/quote]
In an equilateral triangle $ ABC$, draw altitude $ AD$ from $ A$  to $ BC$. You can now see that the altitude bisects $ \angle A$ into $ 30^{\circ}$s. If the side length of the equilateral triangle is $ 2a$ then $ BD \equal{} a$. So $ \sin 30^{\circ} \equal{} \frac{BD}{AB} \equal{} \frac{1}{2}$.

Another reason is that the ratio of the side lengths in a 30-60-90 triangle is

$ 1: \sqrt{3}: 2$

Using this, it can easily be shown that sin 30 is 1/2.

However, if you want a proof when you don't know the ratio of the side lengths, then mthiyaga's proof would be fine.

thanks mthiyaga. So simple I didn't think of it.
fishy - wouldn't you need to know trig before you knew the ratio of the sides?

No...you would only need to know pythagorean theorem, since you know the hypotenuse is twice one of the legs...

[quote="Brut3Forc3"]But how would you know that the hypotenuse is double the length of the leg without knowing the value of $ sin 60^{\circ}$? I think it's circular reasoning. :P[/quote]

No circular reasoning: just Pythagoras and the fact that when you bisect a segment you make two segments that are half as long as the original one.
Out of an equilateral triangle, you can make two congruent 30-60-90 triangles by drawing the altitude as mthiyaga described. Two of the sides of the equilateral triangle becomes the hypotenuses in the right triangles. The other (equally long) side is split into two segments of equal length that become the short legs of the right triangles. Hence they are half as long as the hypotenuses.

And if this was circular reasoning, how was the ratio of the sides discovered in the first place? :maybe:

Because of halving an equilateral triangle and using the Pythagorean Theorem, as these guys have already said.

[quote="jimi"]This is probably so obvious to lots of you but not to me. How do we know that sin 30 deg is exactly 1/2?[/quote]

You could use the law of sines for a 30-60-90 triangle:
\[ \frac{\sin 30^{\circ}}{1} \equal{} \frac{\sin 90^{\circ}}{2}\]

It is self-evident, looking at the unit circle, that $ \sin 90^{\circ}\equal{}1$; therefore, the above reduces to:
\[ \sin 30^{\circ} \equal{} {1\over{2}}\]

[quote="Dr. No"][quote="jimi"]This is probably so obvious to lots of you but not to me. How do we know that sin 30 deg is exactly 1/2?[/quote]

You could use the law of sines for a 30-60-90 triangle:
\[ \frac {\sin 30^{\circ}}{1} \equal{} \frac {\sin 90^{\circ}}{2}
\]
It is self-evident, looking at the unit circle, that $ \sin 90^{\circ} \equal{} 1$; therefore, the above reduces to:
\[ \sin 30^{\circ} \equal{} {1\over{2}}
\]
[/quote]

I'm kind of confused.  Wouldn't you need to use mthiyaga's solution to get the side lengths?  And then, why would you bother with the law of sines, since you already had the ratio?

[quote="Brut3Forc3"]Well, I meant just "randomly" drawing a 30-60-90 triangle, and citing that its special properties make it have a $ 1: \sqrt3: 2$ ratio.[/quote]
The 30-60-90 triangle is far from "random."  It's constructed by cutting an equilateral triangle in half.

At this level, trigonometry is just a shorthand for using the properties of similar triangles.  The two are equivalent for practical purposes; trigonometry is just faster.

[quote="b-flat"]I'm kind of confused.  Wouldn't you need to use mthiyaga's solution to get the side lengths?  And then, why would you bother with the law of sines, since you already had the ratio?[/quote]

Only to the extent that I had noticed the property t0rajir0u explained in his post and wrote down the lengths by inspection. For a 30-60-90 triangle the hypotenuse is always twice as long as the short leg. There is another solution that is less direct but no less elegant.  Consider an equilateral triangle having sides of length $ a$. By the law of cosines,
\[ a^2 \equal{} a^2 \plus{} a^2 \minus{}2a^2\cos 60^{\circ} \\ \\ \Rightarrow a^2 \equal{} 2a^2\left(1 \minus{}\cos 60^{\circ}\right) \\ \\\Rightarrow {1\over{2}} \equal{} 1\minus{}\cos 60^{\circ} \\\\ \Rightarrow \cos 60^{\circ} \equal{} {1\over{2}}\]

Now using the identity that $ \cos (90^{\circ} \minus{}\theta) \equal{} \sin\theta$ we have,
\[ \cos 60^{\circ} \equal{}\cos(90^{\circ} \minus{}30^{\circ}) \equal{}\sin 30^{\circ} \\ \\ \therefore \sin 30^{\circ} \equal{}{1\over{2}}\]


Minimize $ x \plus{} \frac {9}{x}$ where $ x$ is a real number greater than 0.

wow I'm impressed by the simplicity there.  We've been doing these (and more complicated minimization/maximization) problems with derivatives in my Calc class.  Like this:

$ f(x) \equal{} x\plus{}9/x$
$ f'(x) \equal{} 1\minus{}\frac{9}{x^2} \equal{} \frac{x^2\minus{}9}{x^2}$
$ f'(x)\equal{}0 \iff x^2\minus{}9\equal{}0 \iff x\equal{}\pm3$

$ f(3) \equal{} 6$ is the minimum

[quote="sunehra"][hide="My Solution"]
We can simply use AM-GM because $ x$ is greater than 0.
$ \frac {x \plus{} \frac {9}{x}}{2}\ge \sqrt {x\cdot\frac {9}{x}} \\
x \plus{} \frac {9}{x} \ge 2(3)$

Thus, the minimum value is $ \fbox{6}$.

[/hide][/quote]

You will need to show that 6 is attainable:

$ x\plus{}\frac{9}{x}\equal{}6 \implies x^2\plus{}9\equal{}6x \implies x^2\plus{}9\minus{}6x\equal{}(x\minus{}3)^2\equal{}0 \implies x\equal{}3$.

Thus there exists a $ x>0$ such that $ x\plus{}\frac{9}{x}\equal{}6$.

ThetaPi, it is always possible to attain the value that you find with AM-GM.  It is only a matter of showing that it occurs when $ x>0$.  The equality in AM-GM only holds if all the numbers are equal:  $ x\equal{}9/x \iff x^2\equal{}9 \iff x\equal{}\pm 3$


ratio
geometry
ARML
topology
perimeter
percent
arithmetic sequence

RULES:

- I will post three* problems about twice a week. The variable t.1 refers to the answer to problem 1, t.2 to problem 2, etc.
- After every checkpoint, submit all of the answers from the start of the checkpoint to the end.
- This is an open contest, you may join in whenever you want.

* Some times this may change.

Please do not disclose answers to problems until I say so.

GUTS ROUND - CYCLIC COMPUTATION

1. If 2,a,b,c,(t.3) is an arithmetic progression, find a+b+c.
2. What is the sum of the last two digits of (t.1)^43?
3. Triangle ABC has cevians AD, BE, CF. If D is the midpoint of BC, AE=4, CE=6, and BF=(t.2), find AF.

Note: There may be multiple solutions, for which either triple will be accepted.

DEADLINE = May 22, 2009

So where do we post the solutions? Do we pm them to you? Can we start with the first three already?

I like the idea though.

Lol; this competition involves more guess-and-check then the other competitions I'm used to.  :wink:

Nice idea though.  :)

I'll do it (excuse me, I mean TRY to do it).

PM submissions, forget about that "checkpoint" idea, that was a really bad idea.

Leaderboard coming up 5/22

Due to the awesome number of submissions... [b]three[/b]...  :roll:

The answer was (15,12,8). Everyone who submitted got it correct.

QUESTIONS CORRECT LEADERBOARD
[hide]yaofan 1
Twin Prime Conjecture 1
ytrewq 1[/hide]

TIME RANKINGS LEADERBOARD
[hide]yaofan 3
Twin Prime Conjecture 2
ytrewq 1[/hide]

RELAY ROUND - STRAIGHTFORWARD

4) Given that the ratio of a regular pentagon's diagonals to its sides is A, and the ratio of the area of a square constructed with the diagonal of a regular pentagon as the side to the area of a square with the side of the regular pentagon as the diagonal of the square is B, find $ \frac{A^{2}}{B}$
5) Determine the number of sides of a regular polygon with an exterior angle of (t.4)+1.5 degrees.
6) Find the sum of the real zeroes of the following equation: $ x^4\plus{}42x^3\minus{}(t.5)x^2\plus{}42x\minus{}(t.5)\equal{}\minus{}1$

Do we submit the answers to problems 4, 5, and 6, or just problem 6?

[quote="AIME15"]For 4), what do you define as the "ratio of a regular pentagon's diagonals to its sides"?  Average length?  Number of them?[/quote]

TPC: All of them - however for scoring purposes only the last one counts.

AIME15 - Well, there is only one diagonal length for the regular pentagon.

Yes, length.

END DATE - MEMORIAL DAY (Monday)

It says "END DATE - MEMORIAL DAY" which was yesterday.

Ok then.

Two days late... but:

QUESTIONS CORRECT LEADERBOARD
ytrewq 1.5
yaofan 1
Twin Prime Conjecture 1

TIME RANKINGS LEADERBOARD
yaofan 3
ytrewq 2.5
Twin Prime Conjecture 2

ytrewq submitted incorrect solution for #4, #5, but #6 correct. As in ARML, last question correct is OK, but here it merits half points.

SUBMISSIONS: 1  :(

Answer was: (0.5,180,-42)

SUPER RELAY - PART ONE
7) What is the least number of edges needed to be removed from a complete K6 graph to result in a planar graph?

Hint for 7: A finite graph is planar if and only if it does not contain a subgraph that is homeomorphic to K5 or K3,3.
lol, will that help?

8) How many times does the digit t.7 occur in the first (t.7)*100 numbers written in base 10?

9) Express t.8 as a^b, where a is an integer as small as possible and b is an integer. What is a+b?

10) What is the length of the repeating decimal block for 1/(t.9)?

11) How many digits does (t.10)^(t.10) have?

12) Determine the perimeter of a regular polygon with sidelength (t.11) with (t.11) sides.

13) What percent of the first (t.12) primes have a reciprocal larger than 0.38?

Only 7 is hard-ish, the rest is easy.

ARML-like practice?

[quote="pianoforte"]uh, why is this in Classroom Math?[/quote]
No other place is appropriate for this "tournament".


There are given $ (n\plus{}1)$ positive integers $ \le 2n$. Prove that there are two numbers among them so that one of them is divisible by the other.

[quote="moldovan"]There are given $ (n \plus{} 1)$ positive integers $ \le 2n$. Prove that there are two numbers among them so that one of them is divisible by the other.[/quote]
[hide="Hint"]
[color=blue]
Write all $ \ (n \plus{} 1)$ numbers in the form of $ \ 2^{k_i}r_i$ where $ \ k_i$  is a whole number and $ \ r_i$ is an odd natural number.
[/color][/hide]
[color=blue]
[hide="Complete Solution"]
Now there can be at the most $ \ n$ odd numbers in $ \ n \plus{} 1$ numbers as all are less than or equal to $ \ 2n$

$ \implies$ there exist at least two numbers say $ \ a$ and $ \ b$ among $ \ n \plus{} 1$ numbers in which $ \ r_i$ is same.

If $ \ a > b\implies b|a$ or if $ \ b > a\implies a|b$
[/color]
[/hide]


Find all function $f: \mathbb{R^+} \rightarrow \mathbb{R}$ s.t. $\forall x,y\in \mathbb{R^+} , f(x)+f(y)=f(f(x).f(y))$

Choose $a$ arbitrarily and let $f(a)=c$.

For $x=y=a$ we have:
\[ f(c^2)=2c\quad (1) \]

For $x=y=c^2$, (1) gives:
\[ f(4c^2)=4c\quad (2) \]

For $x=a$ and $y=4c^2$ we have:
\[ f(4c^2)=5c\quad(3) \]

From (2) and (3) follows $c=0$, and hence the only solution is $f(x)\equiv 0$.


Given a triangle $ABC$ with angle $A$ obtuse and with altitudes of length $h$ and $k$ as shown in the diagram, prove that $a+h\ge b+k$. Find under what conditions $a+h=b+k$.
[asy]
size(6cm);

pair A = dir(105), C = dir(170), B = dir(10), D = foot(B, A, C), E = foot(A, B, C);

draw(A--B--C--cycle);
draw(B--D--A--E);

dot(A); dot(B); dot(C); dot(D); dot(E);

label("$A$", A, dir(110));
label("$B$", B, B);
label("$C$", C, C);
label("$D$", D, D);
label("$E$", E, dir(45));

label("$h$", A--E, dir(0));
label("$k$", B--D, dir(45));
[/asy]

you can transform in into a trig inequality...

$h=b*\sin C$
$k=a*\sin C$

then it becomes:

$a+b\sin C\ge b+a\sin C$
$(a-b)(1-\sin C)\ge 0$

which is true because of the obtuse angle is the largesta angle, so the opposite side is longest...

equality? i am not sure...

[hide="Just Getting it Engrained"]Note that $h=b\sin C$ and $k=a\sin C,$ so we can rewrite our expression as $a+b\sin C\ge b+a\sin C.$ Factoring by grouping, we see that this is clearly true because $(1-sin C)(a-b)\ge 0.$ ($\sin C\le 1$ and $a>b$ because $\angle A$ is the largest in the triangle.)

Equality occurs when $(1-\sin C)(a-b)=0,$ or when $\sin C=1$ or $a=b.$ Both occur if and only if the triangle is degenerate.  :D    [/hide]

Alternate solution: let $AD=d$. By Pythagoras, $(b+d)^2 + k^2 = a^2$, so it follows that $b^2+k^2 <= a^2$, with the equality case being when d = 0.
The area of ABC can be computed as $\frac{ah}{2} = \frac{bk}{2}$, so
$ah=bk$
$2ah=2bk$
Combine this with the first result derived to get
$a^2 + 2ah >= b^2+2bk+k^2$
Add k^2 to the LHS of the inequality to get squares of a+h and b+k. Equality hence holds if d = 0 and k = 0, implying that the triangle is degenerate.


1.prove that the lenght of the latus rectum is 1/a.
2.consider three cylindrical drums of equal radii,if a rope is wound round them what is the lenght of the rope given the radius is 2
3.a man cycles to his destination with the wind at a speed of 8m/s and back to his starting point at 4m/s. Calculate his average speed.

2) If the drums are lined next to each other, then the length is
$ \frac{2\pi (2)}{2}\plus{}\frac{2\pi (2)}{2}\plus{}2(8)\equal{}4\pi\plus{}16$

all the cylindrical drums touch each other :oops:


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let H be a subgroup of G with[G:H]=n
Show that there exits a normal subgroup N of G if finite index with N subgroup og H.
Give explicitly a number m (dependent on m)such that g^n is in H for all g in G

The first part is posted before, the second one is incorrectly stated (probably $g^{m}$ shall be in $H$; then take $m=n!$).

Yes You are right there is a miswritten.
But why n!. I could not understand the reason. Thanks


Let $ a, b, c \geq0$. Prove that

$ \prod(a^4\plus{}7a^2\plus{}10)\geq216(a\plus{}b\plus{}c)^3$

----------------
 :wink:  Panagiote LIGOURAS

[quote="Ligouras"]Let $ a, b, c \geq0$. Prove that

$ \prod(a^4 \plus{} 7a^2 \plus{} 10)\geq216(a \plus{} b \plus{} c)^3$

----------------
 :wink:  Panagiote LIGOURAS[/quote]

We have :

$ \prod(a^4 \plus{} 7a^2 \plus{} 10) \geq 729(a^2\plus{}1)(b^2\plus{}1)(c^2\plus{}1) \geq 216(a \plus{} b \plus{} c)^3$

Done ! :)

[quote="Materazzi"][quote="Ligouras"]Let $ a, b, c \geq0$. Prove that

$ \prod(a^4 \plus{} 7a^2 \plus{} 10)\geq216(a \plus{} b \plus{} c)^3$

----------------
 :wink:  Panagiote LIGOURAS[/quote]

We have :

$ \prod(a^4 \plus{} 7a^2 \plus{} 10) \geq 729(a^2 \plus{} 1)(b^2 \plus{} 1)(c^2 \plus{} 1) \geq 216(a \plus{} b \plus{} c)^3$

Done ! :)[/quote]
your solution is not true! :P
We have $ (a^2\plus{}2)(b^2\plus{}2)(c^2\plus{}2) \geq 3(a\plus{}b\plus{}c)^2$
and my ineq:$ (a^2\plus{}5)(b^2\plus{}5)(c^2\plus{}5) \geq 72(a\plus{}b\plus{}c)$
=>dpcm! :lol:

$ lemma$           $ a^4\plus{}7a^2\plus{}10\geq6(a^3\plus{}2)$
$ (a^3\plus{}2)(b^3\plus{}2)(c^3\plus{}2)\geq(a\plus{}b\plus{}c)^3$  It's    very      easy    caushy-shvarz :rotfl:

[quote="2424"][quote="Materazzi"][quote="Ligouras"]Let $ a, b, c \geq0$. Prove that

$ \prod(a^4 \plus{} 7a^2 \plus{} 10)\geq216(a \plus{} b \plus{} c)^3$

----------------
 :wink:  Panagiote LIGOURAS[/quote]

We have :

$ \prod(a^4 \plus{} 7a^2 \plus{} 10) \geq 729(a^2 \plus{} 1)(b^2 \plus{} 1)(c^2 \plus{} 1) \geq 216(a \plus{} b \plus{} c)^3$

Done ! :)[/quote]
your solution is not true! :P
We have $ (a^2 \plus{} 2)(b^2 \plus{} 2)(c^2 \plus{} 2) \geq 3(a \plus{} b \plus{} c)^2$
and my ineq:$ (a^2 \plus{} 5)(b^2 \plus{} 5)(c^2 \plus{} 5) \geq 72(a \plus{} b \plus{} c)$
=>dpcm! :lol:[/quote]

Sorry :D My mistakes :P


Hi guys!!!!
how r u?
I wish u good luck & success!!!!!!!

Kazakhstan team on BMO2007:
1.Baev Alen  :lol:
2.Daliev Asset  :lol:
3.Kartabaev Sayat :wink:
4.Nursultanov Medet :)
5.Kassenov Iskander :)
6.Stasij Igor'   :)

I'm looking forward to rezults!!!!
hope u could post them here.

Alen & Asset r great, nice persons from my school.
hope you'll take good medals!!!!

of course all of them r nice and smart!!!

especially Sayat is generous and kind, isn't he?
(nadeius' ty pomniw' 4to mne obewal)

a ewe ia rada 4to Igor popal v komandu !!!

Zhdem ot vas vestej!!!!

P.S. Alen, zhelaiu tebe 4toby priwla geometria tvoej me4ty!!!!

:blush:
We are very glad to find this topic, but i think we aren't as good as you write...
Anybody haven't done more than 2 problems...

It's about 1-2 hours before we will have known our rezults. As soon as we get it, I write it here...

[quote]
My english is as good as my rezults :oops:
[/quote]

the bmo 2007 is discussed here
http://www.mathlinks.ro/Forum/viewtopic.php?p=827991#827991

[quote="A_L_E_N"]Kaz's rezults:
Bayev Alen (KAZ1) 10 4 9 1 [b]24[/b]
Stassiy Igor' (KAZ2) 10 3 1 1 [b]15[/b]
Kassenov Eskendir (KAZ3) 10 2 1 1 [b]14[/b]
Daliyev Asset (KAZ4) 10 4 0 9 [b]23[/b]
Nursultanov Medet (KAZ5) 10 1 0 0 [b]11[/b]
Kartabayev Sayat (KAZ6) 10 2 0 0 [b]12[/b][/quote]

Three medals and three HMs-sounds like a good result... Congratulations!

Congratulations!!!!!

especially for Alen and Asset!

FIZ-MAT  :clap2:

Привет
как дела? как учеба? как жизнь молодая? Айзана пришли пожалуйста фото которую сняли в аэропорту Любляны.

пока...


Dear Mathlinkers,
let ABC be a triangle, 0 the circumcircle of ABC, I the center of the incircle of ABC, A' the second intersection of AI with 0, A" the antipole of A' wrt 0, 1 the A-mixtilinear incircle of ABC, R the point of contact of 1 with 0, U the symmetric of I wrt R and Ia the A-excenter of ABC.
Prove that AA", BC and IaU are concurrent.
Sincerely
Jean-Louis

Incircle $ (I)$ with radius $ r$ is tangent to $ BC, CA, AB$ at $ D, E, F.$ Inversion in $ (I)$ takes the vertex $ A$ into the midpoint $ A^*$ of $ EF$. The sidelines $ AB, AC$ go into circles with diameters $ IF, IE$, the circumcircle $ (O)$ goes to the 9-point circle $ (O^*)$ of the $ \triangle DEF$ and the A-mixtilimnear incircle $ (1)$ into a circle $ (1^*)$ tangent to these 3 congruent circles with radii $ \frac{_r}{^2}$ $ \Longrightarrow$ it is centered at $ A^*$ with radius $ r$ (see [url]http://www.mathlinks.ro/viewtopic.php?t=186117[/url] for details) $ \Longrightarrow$ the image $ R^* \in (O^*)$ of $ R \in (O)$ is diametrically opposite to $ A^* \in (O^*).$ $ A'^* \in (O^*)$ is the image of $ A' \in (O).$ The angle $ \angle A^*A'^*R^* \equal{} \angle IA'^*R^*$ is right $ \Longrightarrow$ the angle $ \angle IRA'$ is also right. (Numerous other ways to prove this, all more tedious.)

Since $ A'A''$ is diameter of $ (O)$ and the angle $ \angle IRA'$ is right, $ U, R, I, A''$ are collinear. The angle $ \angle A''AI_a \equal{} \angle A''AA'$ is right. $ II_a$ is diameter of a circle $ (A')$ with center $ A'$ and radius $ A'B \equal{} A'C.$ Since $ A'U \perp IU$ at $ R,$ the midpoint of $ IU,$ it follows that $ U \in (A')$ and the angle $ \angle A''UI_a \equal{} \angle IUI_a$ is also right. Therefore, $ AA''I_aU$ is cyclic with a circumcircle $ (P).$ Pairwise radical axes $ A''A, BC, I_aU$ of the circles $ (O), (A'), (P)$ are concurrent at their radical center.


There is already some vote on AMC12A 2008. If you took AMC10A this year, please vote and THANKS!

um this topic has already been posted but i suppose that ur poll is a little more straightforward so i guess its okay.


group theory
abstract algebra
Galois Theory
superior algebra
superior algebra unsolved

Prove
Suppose that $ B_1, B_2$ are Galois extensions of F with respective Galois groups $ G_1, G_2$, and that $ E\equal{}B_1 \otimes B_2$ is a field. Then it is Galois over F with Galois group isomorphic to $ G_1 \times G_2$.

#2
Suppose that $ E$ is a Galois extension of $ F$ with Galois group $ G$ and that $ G$ contains subgroups $ H$ and $ N$ with $ N$ normal in $ G$, $ H \cap N\equal{}${1} and $ HN\equal{}G$. Let $ B_N$ be the fixed field of N (so $ B_N$ is Galois over $ F$) and $ B_H$ be the fixed field of $ H$. Prove that $ E$ is isomorphic to $ B_N \otimes B_H$. (If $ H$ is also normal in G then $ G\cong H \times N$ giving a converse to the preceding)

For 1), recall that one equivalent condition for an extension to be Galois is that the number of automorphisms which fix the base field equals the degree of the extension. It is clear (why?) that G_1 times G_2 is contained in the group of automorphisms of B_1 tensor B_2 which fix F. Now argue that any such automorphism is necessarily in G_1 times G_2. Now check orders to confirm that the extension is Galois.

For 2) read the paragraph above the statement of the problem on the midterm: it gives a condition for when the tensor product of two fields is actually equal to the product of the fields (when the fields are linearly disjoint). Verify that this holds here. Then show that this product has to be all of E. Make sure you use each of the hypotheses (they're all needed).

In the parenthetical remark Murray is suggestion the following "well-known" result from group theory: if H,K are two normal subgroups of a group G such that they intersect trivially and HK=G, then G is isomorphic to H times K. So he wants you to use this to show that if E=B_N tensor B_H with H and K satisfying the above, then G equals H times K. This is pretty easy (I can't tell if Murray wanted you to actually write something here or was just making a comment).

Best,
- Colin


function
Support
real analysis
real analysis unsolved

Let $ \Omega \subset \mathbb{R}^n$ be open and $ \phi \in C_c^{\infty}(\Omega)$. Prove that there exist nonnegative functions $ \phi_1$ and $ \phi_2$, both in $ C_c^{\infty}(\Omega)$, such that $ \phi\equal{}\phi_1\minus{}\phi_2$.

Can you not just say that $ \phi$ is bounded below, since it is compactly supported? If $ \phi$ is bounded below by $ m < 0$, then $ \phi_2(x) \equal{} \minus{}m$ and $ \phi_1(x) \equal{} \phi(x)\minus{}m$ are both nonnegative functions with difference equal to $ \phi(x)$.

(Excuse me if I'm being silly or something.)

Ah yes, thank you :). Knew something was wrong there.

Here's a construction:

First, let $ F$ be $ C^{\infty}$, nonnegative, compactly supported in $ \Omega$, and identically $ 1$ on the compact support $ K$ of $ \phi$; a bump function.

Now, let $ \phi_1\equal{}F\cdot\frac{\phi^2 \plus{} 1}{2}$. Since $ F$ is supported in $ \Omega$, $ \phi_1$ is as well. Since it's constructed by multiplying and adding $ C^{\infty}$ functions, it's $ C^{\infty}$
Since $ \frac{a^2\plus{}1}{2}\ge |a|$, $ |\phi_1(x)|\ge |\phi(x)|$ for all $ x\in Ka$. Outside $ K$, $ \phi_1(x)\ge 0\equal{}\phi(x)$, so $ \phi_2(x)\equal{}\phi_1(x)\minus{}\phi(x)\ge 0$ everywhere.


Determine all possible values of

S = a/(a+b+d) + b/(a+b+c) + c/(b+c+d) + d/(a+c+d)

for a,b,c,d arbitrary positive numbers.

it is very clear to see that S is in (1,2)...
For any 1+[i]t[/i] \in (1,2), ([i]t[/i]\in (0,1)), let [i]a[/i]=[i]c[/i]=1, S=1+t,
we need to find out [i]b[/i] and [i]d[/i] such that

2/(1+b+d) - 2/(b+2) + d/(d+2) = t.                              (1)

Take [i]b[/i] small enough such that 2/((b+1)(b+2)) > t > b/(b+2), thus the quadratic function

f(d) = (1+b+d)(d+2) (2/(1+b+d) - 2/(b+2) + d/(d+2) - t)

is positive at d = 0, and it is negative as [i]d[/i] tends to \infty. So there must exist [i]d[/i] such that f([i]d[/i]) = 0, i.e, there exist [i]b[/i] and [i]d[/i] such that (1) holds.

This is an IMO problem?It's not too hard!Well, it's not too hard after i saw ,your solution!!!!

some of the first imo questions we have nowadays at the Baccalaureate exam in Romania, so IMO question doesn't necesarily imply very hard problem :D

no , they are not very hard.
some are taken as our school exam.


quadratics
algebra
polynomial
real analysis
real analysis unsolved

I want to minimize  x/f(x), where f(x) is a quadratic polynomial
subject to certain constraint g(x) < k

Is there any method to do it

Yes, you use Lagrange multipliers, a handy technique from multivariable calculus.


I have problems with the following problem:

Sequence {x_n} s.t. x_(n+1) --- x_n --->0. Prove that the set of it's partial limits is
                           ---
interval [lim x_n ; lim x_n ]  , where the interval bounds are upper and lower partial
             ---
limits of {x_n}

Thanks for you attention
[/img]

Let $x_{u_n},\ x_{v_n}$ be subsequences tending to $u,v\ne \pm\infty$ (a similar argument works for infinite limits) respectively, and let $\ell=\lambda u+(1-\lambda)v$ for some $\lambda\in[0,1]$. We need to show that there is a subsequence $x_{\ell_n}\to\ell$.

We may assume WLOG that $u\le v$ and $x_{u_n}\le x_{v_n},\ \forall n$. We clearly have $x_{v_n}-x_{u_n}\to v-u$. For each $k\in\mathbb N^*$ there is an $n_k$ s.t. each of the intervals $(x_{u_n},x_{u_n}+\frac{x_{v_n}-x_{u_n}}k),(x_{u_n}+\frac{x_{v_n}-x_{u_n}}k,x_{u_n}+2\frac{x_{v_n}-x_{u_n}}k),$...$,(x_{u_n}+(k-1)\frac{x_{v_n}-x_{u_n}}k,x_{v_n})$ contains an element of the sequence, $\forall n\ge n_k$. This means that for $n\ge n_k$, there is an element of the sequence (call it $x_n(k)$) within $\frac 1k$ of $\lambda x_{u_n}+(1-\lambda) x_{v_n}$.

We can now select a strictly increasing sequence indexed by $k$, $t_k\ge n_k$. We have $x_{t_k}(k)\to \ell$ as $k\to\infty$.

I hope it's Ok.

Thanks, i almost understand you solution, except that moment when ypu say that
there is an element of the sequence x_n(k) within \frac 1k  of \lambda x_{u_n}+(1-\lambda) x_{v_n}  ?

Yes, that's because if we divide the interval $(x_{u_n},x_{v_n})$ in $k$ equal subintervals, then in each of those subintervals there is a term of the sequence, and, at the same time, $\lambda x_{u_n}+(1-\lambda) x_{v_n}$ must be in one of those intervals.


inequalities
trigonometry
inequalities unsolved

If $ \alpha \plus{} \beta \plus{} \gamma \equal{} 2\pi$ radian then, what is max. value of   $ \sin^2 \alpha \plus{} \sin^2 \beta \plus{} \sin^2 \gamma$ ?

$ \sin ^ 2 \alpha \plus{} \sin ^ 2 \beta \plus{} \sin ^ 2 \gamma$

$ \equal{} \frac {1 \minus{} \cos 2\alpha}{2} \plus{} \frac {1 \minus{} \cos 2\beta}{2} \plus{} \frac {1 \minus{} \cos 2\gamma}{2}$

$ \equal{} \frac {3}{2} \minus{} \frac {1}{2}(\cos 2\alpha \plus{} \cos 2\beta \plus{} \cos \gamma)$

$ \equal{} \frac {3}{2} \minus{} \frac {1}{2}\{2\cos (\alpha \plus{} \beta)\cos (\alpha \minus{} \beta) \plus{} \cos 2\gamma\}$

$ \equal{} \frac {3}{2} \minus{} \frac {1}{2}\{2\cos (2\pi \minus{} \gamma)\cos (\alpha \minus{} \beta) \plus{} \cos 2\gamma\}$

$ \equal{} \frac {3}{2} \minus{} \frac {1}{2}\{2\cos \gamma \cos (\alpha \minus{} \beta) \plus{} \cos 2\gamma\}$

$ \leq \frac {3}{2} \minus{} \frac {1}{2}( \minus{} 2\cos \gamma \plus{} \cos 2\gamma)$ the equality: $ \cos (\alpha \minus{} \beta) \equal{} \minus{} 1$

$ \equal{} \frac {3}{2} \minus{} \frac {1}{2}(2\cos ^ 2 \gamma \minus{} 2\cos \gamma \minus{} 1)$

$ \equal{} \frac {9}{4} \minus{} \left(\cos \gamma \minus{} \frac {1}{2}\right)^2$ the equality: $ \cos \gamma \equal{} \frac {1}{2}$

$ \leq \frac {9}{4}$

[quote="kunny"]

... $ \leq \frac {3}{2} \minus{} \frac {1}{2}( \minus{} 2\cos \gamma \plus{} \cos 2\gamma)$ the equality: $ \cos (\alpha \minus{} \beta) \equal{} \minus{} 1$

... the equality: $ \cos \gamma \equal{} \frac {1}{2}$
[/quote]

tnx kunny for solution. But I have a question. If equality hold for $ \cos (\alpha \minus{} \beta) \equal{} \minus{} 1$ and $ \cos \gamma \equal{} \frac {1}{2}$ then, for example $ \alpha\equal{}195^o, \beta\equal{}105^o, \gamma \equal{} 60^o$. But this values of $ \alpha, \beta, \gamma$ are not satisfy the equality.

May be $ \cos \gamma <0$ and $ \alpha\equal{}120^o, \beta\equal{}120^o, \gamma \equal{} 120^o$. I have no solution.

hello, your solution is wrong, the problem is here
$ \minus{}\frac{1}{2}\left(2\cos^2(\gamma)\minus{}2\cos(\gamma)\minus{}1\right)\equal{}\minus{}\cos^2(\gamma)\plus{}\cos(\gamma)\plus{}\frac{1}{2}\ne\minus{}\left(\cos(\gamma)\minus{}\frac{1}{2}\right)^2$.
Sonnhard.

hello kunny, the maximum is indeed $ \frac{9}{4}$ but your proof is wrong.
Sonnhard.

Dear Dr Sonnhard Graubner, have you got a proof? May you explain your solution please? tnx.

[quote="kunny"]$ \sin ^ 2 \alpha \plus{} \sin ^ 2 \beta \plus{} \sin ^ 2 \gamma$

$ \equal{} \frac {1 \minus{} \cos 2\alpha}{2} \plus{} \frac {1 \minus{} \cos 2\beta}{2} \plus{} \frac {1 \minus{} \cos 2\gamma}{2}$

$ \equal{} \frac {3}{2} \minus{} \frac {1}{2}(\cos 2\alpha \plus{} \cos 2\beta \plus{} \cos \gamma)$

$ \equal{} \frac {3}{2} \minus{} \frac {1}{2}\{2\cos (\alpha \plus{} \beta)\cos (\alpha \minus{} \beta) \plus{} \cos 2\gamma\}$

$ \equal{} \frac {3}{2} \minus{} \frac {1}{2}\{2\cos (2\pi \minus{} \gamma)\cos (\alpha \minus{} \beta) \plus{} \cos 2\gamma\}$

$ \equal{} \frac {3}{2} \minus{} \frac {1}{2}\{2\cos \gamma \cos (\alpha \minus{} \beta) \plus{} \cos 2\gamma\}$

$ \leq \frac {3}{2} \minus{} \frac {1}{2}( \minus{} 2\cos \gamma \plus{} \cos 2\gamma)$ the equality hold: $ \cos (\alpha \minus{} \beta) \equal{} \minus{} 1$

$ \equal{} \frac {3}{2} \minus{} \frac {1}{2}(2\cos ^ 2 \gamma \minus{} 2\cos \gamma \minus{} 1)$

$ \equal{} \frac {9}{4} \minus{} \left(\cos \gamma \minus{} \frac {1}{2}\right)^2$ the equality: $ \cos \gamma \equal{} \frac {1}{2}$

$ \leq \frac {9}{4}$[/quote]

$ \equal{} \frac {3}{2} \minus{} \frac {1}{2}\{2\cos \gamma \cos (\alpha \minus{} \beta) \plus{} \cos 2\gamma\}$

$ \leq \frac {3}{2} \minus{} \frac {1}{2}( \minus{} 2\left|{\cos \gamma}\right| \plus{} \cos 2\gamma)$ the equality: $ \cos (\alpha \minus{} \beta) \equal{} \minus{} 1$

$ \equal{} \frac {3}{2} \minus{} \frac {1}{2}(2\cos ^ 2 \gamma \minus{} 2\left|{\cos \gamma}\right| \minus{} 1)$

$ \equal{} \frac {9}{4} \minus{} \left(|{\cos \gamma}| \minus{} \frac {1}{2}\right)^2$ the equality: $ |{\cos \gamma}| \equal{} \frac {1}{2}$

$ \leq \frac {9}{4}$


group theory
abstract algebra
function
superior algebra
superior algebra unsolved

Prove that if H is a subgroup of the group G, then $C_{G}(H) \lhd N_{G}(H)$ and $N_{G}(H)/C_{G}(H)$ is isomorphic to a subgroup of Aut(H).

First of all, it's obvious that $C_{G}(H) \leq N_{G}(H)$. Next, we prove that $x^{-1}y x \in C_{G}(H)$, where $x \in N_{G}(H)$ and $y \in C_{G}(H)$. In other words, we have to prove that $\left( x^{-1}y x \right) z = z \left( x^{-1}y x \right), \, \forall z \in H$, which is equivalent to $y \left( x z x^{-1}\right) = \left( x z x^{-1}\right) y$. By the definition of $N_{G}(H)$ we have that $x z x^{-1}\in H$; $y$, being in $C_{G}(H)$, commutes with it, so we're done.

Finally, we assign to each coset $a C_{G}(H)$ the automorphism $x \mapsto a x a^{-1}$. It remains to see that the function is well-defined, is a morphism and is bijective, which is kind of straightforward.

PS: This should be in any textbook, shouldn't it?

Actually, it is much simpler: by definition of $N_{G}(H)$, conjugation defines a map $N_{G}(H)\to\mathrm{Aut}\,H$. It is a homomorphism (well-known) whose kernel is $C_{G}(H)$, again just by definition. owk


The addition problem below has a unique solution. Each of the
letters A, B and C represents a different nonzero digit. What is
the product of A, B and C?
\begin{tabular}{cccc}
& A & B & B \\
+ & B & B & A \\ \hline
C & A & B & C \\
\end{tabular}

[quote="GameBot"]The addition problem below has a unique solution. Each of the
letters A, B and C represents a different nonzero digit. What is
the product of A, B and C?
$ \begin{tabular}{cccc}
& A & B & B \\
\plus{} & B & B & A \\ \hline
C & A & B & C \\
\end{tabular}$[/quote]
Clearly, C=1. Now from the last column, we know that A+B=11, since neither one can be 0. Since this makes a 1 carry over to the next column, we know that B=9. That also gives us A=2. The answer is 2*9*1=18.


pigeonhole principle
number theory unsolved
number theory

Prove that in any infinite subset of $N^{\ast}$,we can find two elements in it such that their sum has a prime factor that's larger than $10^6$.

Assume the opposite.  Call the largest prime factor of a sum of any two members of the set $p_k < 10^6$, and take the set $\bmod p_n > 10^6$.  If any of the remainders add up to $p_n$, we are done.

Else, take the set $\bmod p_{n+1} > p_n$.  If any of the remainders add up to $p_{n+1}$, ...

We must now prove that, if we do this an infinite number of times, any set that does not have remainders that add up to $p_n, ... p_{\infty}$ has a finite number of elements.  This is simple.  If it does not contain remainders that add up to $p_{n+k}$ when taken $\bmod p_{n+k}$ then the set, taken as a sequence, can be defined by $s_i = a_i + p_{n+k} b_i$, where $a_i$ is a finite subset of the integers $0 < a_i < p_{n+k}$ and $b_i$ is another sequence of non-negative integers.

Repeating this process for all $p_{n+k}$ we find that the set's members are defined by $s_i = a_i + \left( \prod_{i=0}^{\infty} p_{n+i} \right) b_i$, where $a_i$ is a finite subset of the integers $0 < a_i < p_n$.  Thus, the set contains a finite number of elements - a contradiction.

(I'm not actually sure if this proof works or not  :blush: )

It doesn't. You can easily avoid some infinite collections of primes. If our set is $1,4,16,64,256,\dots$, no prime which is 3 mod 4 divides any of the sums. You never used that we have to avoid all of the large primes.

Here's an argument:

Suppose that $S$ is an infinite set of positive integers, and that $p_1,p_2,\dots,p_n$ are primes. We will show that there exists some $q$ not equal to any $p_k$ such that some element of $S+S$ is divisible by $q$.

Step 1: Construct infinite sets $S_k$ so that $S_0=S$, $S_k\subset S_{k-1}$, and either
(a) For some $j$, no element of $S_k+S_k$ is divisible by $p_k^j$
or (b) For each $j$, all but finitely many elements of $S_k$ are divisible by $p_k^j$.

Proof/Construction:
Case 1: $p_k>2$. If $S_{k-1}$ already satisfies condition (b), let $S_k=S_{k-1}$.
Otherwise, choose $j$ such that infinitely many elements of $S_{k-1}$ are not divisible by $p_k^j$. By the Pigeonhole Principle, there is some $r\not\equiv 0$ such that infinitely many elements of $S_{k-1}$ are congruent to $r$ mod $p_k^j$. Let $S_k=\{x: x\in S_{k-1},x\equiv r\mod p_k^j\}$. Since $2r\not\equiv 0\mod p_k^j$, no element of $S_k+S_k$ is divisible by $p_k^j$ and condition (a) is satisfied.
Case 2: $p_k=2$. If $S_{k-1}$ already satisfies condition (b), let $S_k=S_{k-1}$.
Otherwise, choose $j$ such that infinitely many elements of $S_{k-1}$ are not divisible by $2^j$. By the Pigeonhole Principle, there is some $r\not\equiv 0,2^j$ such that infinitely many elements of $S_{k-1}$ are congruent to $r$ mod $2^{j+1}$. Let $S_k=\{x: x\in S_{k-1},x\equiv r\mod2^{j+1}\}$. Since $2r\not\equiv 0\mod 2^{j+1}$, no element of $S_k+S_k$ is divisible by $2^{j+1}$ and condition (a) is satisfied.

Now, let $a$ be a particular element of $S_n$. We can find integers $j_1,j_2,\dots$ such that for all but finitely many members $b$ of $S_n$, $p_k^{j_k}$ does not divide $a+b$ for each $k$. Since $S_n\subset S_k$, it satisfies either (a) or (b) for each prime $p_k$. If no element of $S_n+S_n$ is divisible by $p_k^t$, let $j_k=t$. Otherwise, it satisfies (b) for $p_k$. In this case, choose $j_k$ such that $p_k^{j_k}$ does not divide $a$; since it divides all but finitely many $b\in S$, it does not divide $a+b$ for all but finitely many such values.
Now we have infinitely many $b\in S_n$ such that $p_k^{j_k}$ does not divide $a+b$ for each $k$. For each such $b$, either $a+b| p_1^{j_1-1}p_2^{j_2-2}\cdots p_n^{j_n-1}$ or $a+b$ has some prime factor other than $p_1,p_2,\dots,p_n$. Choosing $b> p_1^{j_1-1}p_2^{j_2-2}\cdots p_n^{j_n-1}$, we have the second, and our desired result is proved.


LaTeX
inequalities unsolved
inequalities

Let a+b=m, x_1, x_2, ..., x_n >0, a, b>=0. Show that B (see file attached) is true.

We have $\frac{x_{i}^{m}}{\sum_{j\not =i}x_{j}^{a}}+x_{i}^{b}=x_{i}^{b}\cdot\frac{S}{S-x_{i}^{a}}$, where $S=\sum x_{i}^{a}$. Therefore ineq is equivalent to $\sum \frac{S}{S-x_{i}^{a}}\geq \frac{n}{n-1}$ or $(\sum (S-x_{i}^{a}))(\sum\frac{1}{S-x_{i}^{a}})\geq n^{2}$. This is trivial!

P.S: You can learn use $\text{LaTex}$ here http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php Very easy!  :wink:


$ a\ge b\ge c>0 \,\,and\,\, b\plus{}c>a,a\plus{}b>c,c\plus{}a>b$
this inequality:
$ \frac{1}{ac}\plus{}\frac{1}{bc}>\frac{(a\minus{}b)^2}{b^2c^2}$

is true?

or how to prove it?

We need to prove that

$ \frac{a\plus{}b}{abc} \ge \frac{(a\minus{}b)^2}{(bc)^2}$

it is equivalent to the following

$ bc(a\plus{}b) \ge a(a\minus{}b)^2$

However, we have

$ abc \ge ac^2 \ge a(a\minus{}b)^2$

Therefore, we complete our proof.