calculus
integration
trigonometry
function
logarithms
algebra
rational function

Why can't be calculated integral like this:
$ \int_{0}^{\infty}\frac {\sin x}{x^{2} \plus{} a^{2}}$ ?

Why not? The residue theorem is like an armour-penetrating missile launcher and integrals like that are like tanks without reactive armour...

If you want something a little more mathematical, I'd say it's because it's easy to evaluate the integral of a rational function and sine is analytic.

Actually, the residue theorem isn't so great in this case- it can do the integral from $ \minus{}\infty$ to $ \infty$ easily, but that's just zero here.

[quote="jmerry"]Actually, the residue theorem isn't so great in this case- it can do the integral from $ \minus{} \infty$ to $ \infty$ easily, but that's just zero here.[/quote]
What about the quadrant $ \pi/4 \le \theta \le 3\pi/4$?

Integral of what function? All I see are integrals of functions that decay exponentially.

\[ I(a) \equal{} \int_{0}^{\infty}{\frac {\sin x}{x^2 \plus{} a^2}dx}
\]
Let $ x \equal{} at$, then $ dx \equal{} adt$
\[ I(a) \equal{} \frac {1}{a}\int_{0}^{\infty}{\frac {\sin at}{t^2 \plus{} 1}dt}
\]

\[ G(a) \equal{} \int_{0}^{\infty}{\frac {\sin at}{t^2 \plus{} 1}dt}
\]

\[ G'(a) \equal{} a\int_{0}^{\infty}{\frac {\cos at}{t^2 \plus{} 1}dt}
\]

\[ G''(a) \equal{} \int_{0}^{\infty}{\frac {\cos at}{t^2 \plus{} 1}dt} \minus{} a^2\int_{0}^{\infty}{\frac {\sin at}{t^2 \plus{} 1}dt} \equal{} \frac {1}{a}G'(a) \minus{} a^2 G(a)
\]
We get the equation:
\[ G''(a) \minus{} \frac {1}{a}G'(a) \plus{} a^2 G(a) \equal{} 0
\]
We have:
\[ G(a) \equal{} C_1\sin\left(\frac {a^2}{2}\right) \plus{} C_2\cos\left(\frac {a^2}{2}\right)
\]
We should find constants $ C_1$ and $ C_2$. I try to think about it... Then we get:
\[ \boxed{I(a) \equal{} \frac {1}{a}\left(C_1\sin\left(\frac {a^2}{2}\right) \plus{} C_2\cos\left(\frac {a^2}{2}\right)\right)}
\]

If $ a\equal{}0$ our integral does not exist. Because $ \frac{\sin x}{x^2}\equal{}\frac{1}{x}\plus{}O(x)$ near $ x\equal{}0$.

Sorry, I can't understand what you say, it might be beyond my ability. :(

Near $ 0$, $ \sin x$ looks like $ x$, so $ \frac {\sin x}{x^2}$ looks like $ \frac {1}{x}$, and $ \int_{0}^k \frac {dx}{x}$ does not exist for any $ k > 0$.

[quote="Yuriy Solovyov"]
\[ G(a) \equal{} \int_{0}^{\infty}{\frac {\sin at}{t^2 \plus{} 1}dt}
\]

\[ G'(a) \equal{} a\int_{0}^{\infty}{\frac {\cos at}{t^2 \plus{} 1}dt}
\]
[/quote]
 :blush:  this is not true. In this case you cant diferentiate
( у тебя нет равномерной сходимости)

also:
If you want to calculate integral :
$ \int_{0}^{\infty}\frac {\cos x}{x^{2} \plus{} a^{2}}dx$ it's can be easly calculated by complex analysis . :wink:
becuase you can use method about calculation :$ \int_{ \minus{} \infty}^{\infty}..$
but for $ \sin$ this integral is equal to $ 0$ :
[quote="jmerry"]Actually, the residue theorem isn't so great in this case- it can do the integral from $ \minus{} \infty$ to $ \infty$ easily, but that's just zero here.[/quote]

\[ \frac {1}{t^2 + 1} = \frac {1}{2i}(\frac {1}{t - i} - \frac {1}{t + i})
\]
Now:
\[ \int_{0}^{\infty}{\frac {\sin at}{t + b}dt} = (t = y - b) = \int_{b}^{\infty}{\frac {\sin (ay - ab)}{y}dy} =
\]

\[ = \int_{b}^{\infty}{\frac {\sin (ay)\cos (ab) - \sin (ab)\cos (ay)}{y}dy} =
\]

\[ = \cos (ab)\int_{b}^{\infty}{\frac {\sin (ay)}{y}dy} - \sin (ab)\int_{b}^{\infty}{\frac {\cos (ay)}{y}dy}
\]
Let $ \operatorname{Si}(x) = \int_{0}^{x}{\frac {\sin t}{t}dt}$ and $ \operatorname{Ci}(x) = - \int_{x}^{\infty}{\frac {\cos t}{t}dt}$
\[ \int_{0}^{\infty}{\frac {\sin at}{t + b}dt} = \cos (ab)(\frac {\pi}{2} - \operatorname{Si}(ab)) + \sin (ab)\operatorname{Ci}(ab)
\]
We should put here first $ b = - i$ and then $ b = i$. Will get:
\[ \frac {1}{2i}(\int_{0}^{\infty}{\frac {\sin at}{t - i}dt} - \int_{0}^{\infty}{\frac {\sin at}{t + i}dt}) =
\]

\[ = \frac {1}{2i}(\cos (ai)(\frac {\pi}{2} + \operatorname{Si}(ai)) - \sin (ai)\operatorname{Ci}( - ai) - \cos (ai)(\frac {\pi}{2} - \operatorname{Si}(ai)) - \sin (ai)\operatorname{Ci}(ai)) =
\]

\[ = \frac {1}{2i}(2\cos (ai)\operatorname{Si}(ai) - \sin (ai)\operatorname{Ci}( - ai) - \sin (ai)\operatorname{Ci}(ai)) =
\]

\[ = \frac {1}{2i}(2\cos (ai)\operatorname{Si}(ai) - \sin (ai)(\operatorname{Ci}( - ai) + \operatorname{Ci}(ai)))
\]
We have: $ \operatorname{Ci}( - ai) = \operatorname{Ci}(ai) - \pi i$, so we will have:
\[ I(a) = \frac {1}{2ia}(2\cos (ai)\operatorname{Si}(ai) - \sin (ai)(2\operatorname{Ci}(ai) - \pi i))
\]

\[ \cos (ai) = \cosh a
\]

\[ \sin (ai) = i\sinh a
\]

\[ I(a) = \frac {\cosh a}{a}\cdot\frac {\operatorname{Si}(ai)}{i} - \frac {\sinh a}{a}\cdot (\operatorname{Ci}(ai) - \frac {\pi i}{2}))
\]
Now, easy to show that:
\[ \operatorname{Si}(ai) = i\cdot \sum_{k = 1}^{\infty}{\frac {a^{2k - 1}}{(2k - 1)\cdot (2k - 1)!}}
\]

\[ \operatorname{Ci}(ai) = \gamma + \ln a + \frac {\pi i}{2} + \sum_{k = 1}^{\infty}{\frac {a^{2k}}{(2k)\cdot (2k)!}}
\]
Final answer will be:
\[ I(a) = \frac {\cosh a}{a}\cdot\sum_{k = 1}^{\infty}{\frac {a^{2k - 1}}{(2k - 1)\cdot (2k - 1)!}} - \frac {\sinh a}{a}\cdot (\gamma + \ln a + \sum_{k = 1}^{\infty}{\frac {a^{2k}}{(2k)\cdot (2k)!}}))
\]
:cool:  :D

maple11 gave me answer:
$ ( \minus{} 2*Shi(a)*\cosh(a) \plus{} Ci( \minus{} i*a)*\sinh(a) \plus{} Ci(i*a)*\sinh(a))/a$
it's like your answer. hence it's true. :)
but I'm interesting why we can't to calculate this integral, or I mean Why it' can't be expressd by elementary function or wellknown constant in math. by serie i think can be expressed every function

We ony can give name to some serie and say that it is well known, i found that:
\[ \operatorname{Shi} a \equal{} \sum_{k \equal{} 1}^{\infty}{\frac {a^{2k \minus{} 1}}{(2k \minus{} 1)\cdot (2k \minus{} 1)!}}
\]

\[ \operatorname{Chi} a \equal{} \gamma \plus{} \ln a \plus{} \sum_{k \equal{} 1}^{\infty}{\frac {a^{2k}}{(2k)\cdot (2k)!}}
\]
They have names: first is The Hyperbolic Sine Integral and the second is The Hyperbolic Cosine Integral.
\[ \operatorname{Shi} x \equal{} \int_{0}^{x}{\frac {\sinh t}{t}dt}
\]

\[ \operatorname{Chi} x \equal{} \gamma \plus{} \ln x \plus{} \int_{0}^{x}{\frac {\cosh t \minus{} 1}{t}dt}
\]
and we can give the answer in form:
\[ I(a) \equal{} \frac {\operatorname{Shi} a\cosh a \minus{} \operatorname{Chi} a \sinh a}{a}
\]
Looks like not so ugly... I can give the same question: "why $ \zeta (2)$ could be represented throw well-known constants and $ \zeta (3)$ no?" I think it's the same. We may continue looking for a way of representing or we could give the name to it and use it as a new constant or function...


geometry
circumcircle
trigonometry
geometry proposed

Let $\triangle ABC$ be given with circumradius $R$. Let $K_1$ and $K_2$ be the circles which pass through $C$ and are tangent to $AB$ at $A$ and $B$ respectively. Let $K$ be the circle of radius $r$ which is tangent to $K_1,K_2$ and $AB$.
(a) Express $r$ in terms of $a,b,c$ and $R$.
(b) Suppose $r=\frac{R}{4}$, find $\angle C$.

[i]Dear Mecrazywong[/i], the point (b) is [u]nicely[/u].

[b]I note[/b] $K_1=C(O_1,R_1),\ K_2=C(O_2,R_2),\ K=C(S,\rho)$.

$\blacksquare \ (a).\ R_1=\frac{bR}{a},\ R_2=\frac{aR}{b},\ c=2\left(\sqrt{R_1\rho}+\sqrt{R_2\rho}\right)\Longrightarrow$

$\rho=\frac{c^2}{4\left (\frac{bR}{a}+\frac{aR}{b}+2R\right) } \Longrightarrow \rho =\frac{abc^2}{4R(a+b)^2}.$

$\blacksquare \ (b).\ \rho=\frac{R}{4}\Longleftrightarrow abc^2=R^2(a+b)^2\Longleftrightarrow 16S^2=ab(a+b)^2\Longleftrightarrow$

$4ab\sin^2 C=(a+b)^2\Longleftrightarrow 4ab(1-\cos^2 C)=(a+b)^2\Longleftrightarrow$

$(a-b)^2=-4ab\cos^2 C\le 0\Longrightarrow a=b,\ C=90^{\circ}.$


MATHCOUNTS
LaTeX
trigonometry
geometry
geometric transformation

its a team round, so calculator allowed...technically

Hint:[hide] I think you should use scientific notations...[/hide]

I don't see how scientific notation could help...I may be wrong.

I have no clue how to do this, can anyone who can show an answer and a solution?

[quote="Barnacle"]look at 789^2 and 345^.

6+2+2+5+2+1+1+1+9+0+2+5 = 36.

is this right?[/quote]
Yes, Im pretty sure.

Took me a few seconds of thinking, though.

[quote="Syntax Error"]find the sum of the digits in the decimal representation of (789,000,003,450,000):^2:[/quote]

the answer to this problem is 54

The actual answer is 54 (I remember doing this problem, but I'm not sure my solution is correct).

The number (just in case you were wondering...) is
[hide="number"]622521005444100011902500000000[/hide]

You have to find the sum of the digits in $789^2, 345^2$ and $345\cdot789$.  Those are:
622,521 = 6+2+2+5+2+1 = 18
119,025 = 1+1+9+0+2+5 = 18
272,205 = 2+7+2+2+0+5 = 18

$18\times3=54$

Like I said, I don't know if that is how you are supposed to do it because I forgot how, but remembered the answer.

know what??

The answer, because we had the answer key when we did it (practice???)?

Anyway, I used my computer to get the actual number and verify it.

This is my solution.

[hide]You know that there are enough zero's so they wont overlap.

Technically, this is in scientific notation.  Just take the first part squared, then the second part, then take 2ab, and add up the digits, i dont have time to do it ><[/hide]

Right, this has basically been posted, but this is what you do.

[hide="Click for the solution"]First, you split 789,000,003,450,000 into

$(789*10^{12} + 345*10^4)^2$

So that would equal $(789*10^{12})^2 + 2(789*345*10^4*10^{12})+ (345*10^4)^2$

And then you realize all the 10^x's are irrelevant, so in reality it's:

$(789^2) + 2(789*345) + (345^2)$

So you do those out, find it equals:

$622521 + 544410 + 119025$

And there are enough 0's to seperate any of them from actually adding.

So $6+2+2+5+2+1+5+4+4+4+1+0+1+1+9+0+2+5=54$[/hide]

P.S. Someone needs to teach me how to do some of the codes often used here, it just looks ugly without it.

EDIT: Editted with corrections and made it look better >_>

[quote="JesusFreak197"][url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuideLay.php]LaTeX guide[/url]

Essentially, put dollar signs ($\$$) around your mathematical stuff, and use the Symbols and Commands parts of the guide to know how to represent various things like fractions, $\sin$ and $\cos$, etc.

[b][color=#00f000]J[/color][color=#0ce80c]e[/color][color=#18e018]s[/color][color=#24d824]u[/color][color=#30d030]s[/color][color=#3cc83c]F[/color][color=#48c048]r[/color][color=#54b854]e[/color][color=#60b060]a[/color][color=#6ca86c]k[/color][color=#78a078]1[/color][color=#849884]9[/color][color=#909090]7[/color][/b][/quote]

Thanks, I'll use it later.

[quote="Zenith Xan"]Right, this has basically been posted, but this is what you do.

[hide="Hidden text"]First, you split 789,000,003,450,000 into

(789*10^12 + 345*10^4)^2

So that would equal (789*10^12)^2 + (789*345*10^4*10^12)+ (345*10^4)^2

And then you realize all the 10^x's are irrelevant, so in reality it's:

(789^2) + (789*345) + (345^2)

So you do those out, find it equals:

622521 + 272205 + 119025

And there are enough 0's to seperate any of them from actually adding.

So 6+2+2+5+2+1+2+7+2+2+5+1+1+9+2+5 = 54[/hide]

P.S. Someone needs to teach me how to do some of the codes often used here, it just looks ugly without it.[/quote]

Hm, where'd the 2 in 2ab go?

Whoops, I messed up.

Well 272205*2 = 544410, which still adds up to 18.

[quote="JesusFreak197"]Yeah, I noticed that too. You got lucky. :P

[b][color=#00f000]J[/color][color=#0ce80c]e[/color][color=#18e018]s[/color][color=#24d824]u[/color][color=#30d030]s[/color][color=#3cc83c]F[/color][color=#48c048]r[/color][color=#54b854]e[/color][color=#60b060]a[/color][color=#6ca86c]k[/color][color=#78a078]1[/color][color=#849884]9[/color][color=#909090]7[/color][/b][/quote]

Right, I realized if it wasn't coincidental that the sum of the digits of its double is equal to the sum of its digits, I would have gotten it wrong.

Oh well, dumb luck usually does work to my favor  ;)


Let $ a,b,c\in \mathbb{R}, a,b,c > 0$. Prove that $ \dfrac{a \plus{} b}{c} \plus{} \dfrac{b \plus{} c}{a} \plus{} \dfrac{c \plus{} a}{b}\geqslant 6$. help me. thank!

[hide]
I am assuming that is a greater than or equal to sign:

\[ \frac{a \plus{} b}{c} \plus{} \frac{b \plus{} c}{a} \plus{} \frac{a \plus{} c}{b} \equal{} frac ac \plus{} \frac bc \plus{} \frac ba \plus{} \frac ca \plus{} \frac ab \plus{} \frac bc \ge 6\]

by 6 variable AM-GM since the product of the LHS is one[/hide]

This is not the appropriate forum, but here's the solution.

[hide="My Solution"]
Break up all the fractions and apply AM-GM.

$ \frac {a \plus{} b}{c} \plus{} \frac {b \plus{} c}{a} \plus{} \frac {c \plus{} a}{b} \equal{} \frac {a}{c} \plus{} \frac {b}{c} \plus{} \frac {b}{a} \plus{} \frac {c}{a} \plus{} \frac {c}{b} \plus{} {a}{b} \ge 6\sqrt [6]{\frac {a}{c} \cdot \frac {b}{c} \cdot \frac {b}{a} \cdot \frac {c}{a} \cdot \frac {c}{b} \cdot \frac{a}{b}} \equal{} 6$

Q.E.D
[/hide]


geometry
conics
ellipse
parallelogram
rhombus
rectangle
analytic geometry

Let $ \frac{x^2}{a^2}\plus{}\frac{y^2}{b^2}\minus{}1\equal{}0$ be an equation that describes an ellipse.Find the maximum area of a convex quadrilateral inscribed in this ellipse.

Assume wlog $ a\geq b$ and compress the ellipse along the $ x$-axis until it becomes a circle.  Clearly, the inscribed quadrilateral with area $ S$ transforms into a quadrilateral with area $ S'$ inscribed in a circle with radius $ b$, where $ \frac {S'}{S} \equal{} \frac {b}{a}$.  But the largest area of a quadrilateral inscribed in a circle corresponds to a square, whose area is therefore $ 2b^2$ (half of the diagonal of the square is the radius $ b$ of the circle).  The area is then $ 2ab$.  Note that the square after the transformation does not need to have its sides parallel to the original axes, ie, the maximum area is obtained for an infinitude of parallelograms (for example, the rhombus whose vertices correspond to the intersections of the ellipse with its axes, or the rectangle whose vertices have coordinates $ \left(\pm\frac {a}{\sqrt {2}},\pm\frac {b}{\sqrt {2}}\right)$.  In fact, the general solution (all quadrilaterals with maximum area) corresponds to the quadrilateral with vertices $ \left(a\cos\alpha_i,b\sin\alpha_i\right)$, where $ i\equal{}1,2,3,4$, $ \alpha_i\equal{}\alpha_{i\minus{}1}\plus{}\frac{\pi}{2}$, and $ \alpha_1$ takes any value...


If pens can be purchased for 35 cents each or three for $ \$$1.00, how
many dollars are saved if 63 pens are purchased in sets of three
instead of individually? Express your answer to the nearest
hundredth.

If you buy the pens individually, for every 3 pens you buy, you spend $ (35\times 3)\minus{}100\equal{}5$ cents more than usual. Since you can buy 21 sets of 3 pens, you save $ 0.05\times 21\equal{}1.05$ by buying pens in sets of 3.

[quote="qballmath"]1.05 is the answer.[/quote]

Instead of posting just answers, please post the solutions as well.

Thanks. :wink:

I think it would be best if instead we private messaged the person, rather than post it here. Thanks, everyone.  :)

Locked.


geometry
inradius
circumcircle
incenter
perpendicular bisector

Construct an isosceles triangle ABC given the length of the inradius, and the fact that the orthocenter is on the incircle.

For an isosceles triangle, the circumcenter O, the incenter I and the orthocenter H are collinear. Label the isosceles triangle $\triangle ABC$ so that AB = AC. Then the perpendicular bisector of the side BC, the internal bisector of the angle $\angle A$ and the altitude from the vertex A are coincident. Since the points O, I, H all lie on this line, they are collinear. According to the reply #11 in the thread [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=43819]Geometry[/url], the orthocenter H is one of the 2 intersections $H, H'$ of the circle (M) centered at the distance MO = 2 IO = 2d from the circumcenter O and with radius $r_M = R - 2r$, where R, r are the circumradius and inradius of the isosceles triangle $\triangle ABC$. Consequently,

$r = IH = IM \pm r_M = d \pm (R - 2r)$

On the other hand, Euler's relation between the R, the inradius r, and the inradius to circumradius distance d = IO of any triangle is

$\frac 1 r = \frac{1}{R - d} + \frac{1}{R + d},\ \ R^2 - d^2 = 2rR$

We have 2 possibilites: r = d - (R - 2r) or r = d + (R - 2r).  First, consider the possibility r = d - (R - 2r), d = R - r. Substituting this to Euler's relation, we get

$R^2 - (R - r)^2 = 2rR,\ \ R^2 - R^2 + 2rR - r^2 = 2rR,\ \ r^2 = 0,\ \ r = 0$

This is not possible, so it must be r = d + (R - 2r), d = 3r - R. Substituting this to Euler's relation, we get

$R^2 - (3r - R)^2 = 2rR,\ \ R^2 - 9r^2 + 6rR - R^2 = 2rR,\ \ 9r^2 = 4rR,\ \ R = \frac{9r}{4}$

$d = IO = 3r - R = 3r - \frac{9r}{4} = \frac{3r}{4}$

Thus we can easily construct the circumcenter O at the distance $IO = \frac{3r}{4} = \frac{3 IH}{4}$ from the incenter I opposite to the orthocenter H. The circumcircle (O) intersects the diacentral line $IO \equiv IH$ at a point P opposite to the intenter I with respect to the circumcenter O, hence,

$IP = IO + OP = d + R = 3r - R + R = 3r$

Hence, the point P is easy to find and the circumcircle (O) is then the circle centered at the constructed circumcenter O and with the known radius R = OP. The circumcircle (O) interests the diacentral line IO at one other point Q different form the point P. One vertex of the isosceles triangle $\triangle ABC$, say the vertex A, lies on the diacentral line IO, hence, it is identical either with the point P or with the point Q. But the point P cannot be the desired vertex, because then the point H would be the tangency point of the incircle (I) with the side BC, i.e., it could not be the orthocenter of the isosceles triangle $\triangle ABC$. Hence, the vertex A is identical with the other intersection Q. It remains to construct the vertices B, C. For example, the tangents to the circle (I) from the point $A \equiv Q$ intersect the circumcircle (O) at the remaining 2 vertices B, C. The tangency points E, F of the sides AB, AC are the intersections of a circle on the diameter AI with the incircle (I), hence, they are easily found. Even simpler, the vertices B, C are the intersections of the circumcircle (O) with a normal to the diacentral line IO erected at the point D, the other intersection of the incircle (I) with the diacentral line IO, different from the orthocenter H.


quadratics
algebra
polynomial
number theory solved
number theory

Let $(a_i)_{i\geq1}$, and $(b_i)_{i\geq1}$ have all the terms $\in \mathbb{Z}$. Show that $a_i+b_i$, and $a_ib_i$ cannot be in the same time arithmetical, respectively geometrical progressions, if the $a_i,b_i$ are not constant.

This is untrue in the arithmetical case :
Let $a_i = i$ if $i$ is even and $a_i=1$ if $i$ is odd.
Let $b_i = 1$ if $i$ is even and $b_i =i$ if $i$ is odd.
Then, for all $i$, we have : $a_i+b_i = i+1$ and $a_ib_i=i$ so that both are arithmetical sequences but none of the two initial sequences is constant.

Pierre.

In another hand I can prove the following :
If $(a_i), (b_i)$ are two infinite sequences of integers such that both $(a_i+b_i)$ and $(a_ib_i)$ are arithmetical sequences, then there exists a constant $M$ such that for all $i$, we have $a_i = M$ or $b_i = M$.

For convenience, I assume that the sequences start at $a_0,b_0$.
Assume that $a_i+b_i = \alpha + ip$ and $a_ib_i = \beta + iq$ for all $i$ where $\alpha = a_0 +b_0$ and $\beta = a_0b_0$.
Since the $a_0,b_0,a_1,b_1$ are integers, we deduce that $p,q$ are integers.

Now, $a_i,bi$ are the integer roots of the quadratic : $X^2 - (\alpha + ip)X + (\beta + iq)$, which is a monic polynomial with integer coefficients. Since its roots are integers, it follows that its discriminant is a perfect square.
The discriminant is :
$\Delta_i = (\alpha + ip)^2 - 4(\beta + iq) = (pi+c)^2 + d$,
where $c= \alpha p - 2q$ and $d = -(\alpha p - 2q)^2 + \alpha^2 - 4\beta$ are integers.

If $d > 0$ :
We assume that $p>0$ (the other case can be treated similarly). Then for sufficientely large $i$ we have $(pi+c)^2 < \Delta_i < (pi+c+1)^2$, which contradicts that $\Delta_i$ is a perfect square.
We have a similar reasoning for $d<0$, so that $d=0$.
In that case one of the roots is $M = \frac {\alpha - c} 2$, which is independant of $i$, so that for all $i$,  $a_i$ or $b_i$ equal to $M$, as claimed.

Pierre.

If the problem is that $(a_i+b_i)$ has to be an arithmetical sequence and $(a_ib_i)$ a geometrical one, then if you allow the 0 sequence, my first counter-exemple works again by replacing 1 by 0.

So, Ikap, could you please check the statement. This problem seems interesting, but the statement is unclear or I'm totally off.

Pierre.


[quote="N.T.TUAN"][quote="kcbu"]$m_{a}^{2}+m_{b}^{2}+m_{c}^{2}\leq\frac{9R}{2}$[/quote]
You are wrong![/quote]
Um, sr
$m_{a}+m_{b}+m_{c}\leq\frac{9R}{2}$
 :oops:  :blush:

[quote="kcbu"]Let $\Delta ABC$. Prove that
$a)m_{a}+m_{b}+m_{c}\geq\frac{a^{2}+b^{2}+c^{2}}{2R}$[/quote]
I need a nice solution for \[\Uparrow \]


The book I'm using say solve using a graphic calculator. And I don't have one. So I try to work on their example by using algebra rules and its worked. And I solve some questions also which all begin with  but why I use the same thing with something begin with  my problems start. I think I have do a major problem in dividing because I don't divide usually in inequalities  :(  .
$ 2x^2  \minus{} 5x \plus{} 1 \ge 1 \\
 x^2  \minus{} \frac{5}{2}x \plus{} \frac{1}{2} \ge \frac{1}{2} \\
 x^2  \minus{} \frac{5}{2}x \ge 0 \\
 x^2  \minus{} \frac{5}{2}x \plus{} (\frac{{\frac{{ \minus{} 5}}{2}}}{2})^2  \ge 0 \plus{} (\frac{{\frac{{ \minus{} 5}}{2}}}{2})^2  \\
 x^2  \minus{} \frac{5}{2}x \plus{} \frac{5}{4} \ge \frac{5}{4} \\
 (x \minus{} \frac{5}{4})^2  \ge \frac{5}{4} \\
 x \minus{} \frac{5}{4} \ge  \pm \sqrt {\frac{5}{4}}  $
My final answers:  :(
$ x \minus{} \frac{5}{4} \ge \sqrt {\frac{5}{4}}  \\
 x \ge \frac{{\sqrt 5 }}{2} \plus{} \frac{5}{4} \\
 x \minus{} \frac{5}{4} \le  \minus{} \sqrt {\frac{5}{4}}  \\
 x \le \frac{{ \minus{} \sqrt 5 }}{2} \plus{} \frac{5}{4} $

You made a minor mistake: you forgot to square $ \minus{}\frac{5}{4}$.

[quote="sunehra"]You made a minor mistake: you forgot to square $ \minus{} \frac {5}{4}$.[/quote]
Do I need to why?
See a similar inequality I solve.
$ x^2  \minus{} 6x \plus{} 8 > 3 \\
 x^2  \minus{} 6x >  \minus{} 5 \\
 x^2  \minus{} 6x \plus{} ( \minus{} 3)^2  >  \minus{} 5 \plus{} 9 \\
 (x \minus{} 3)^2  > 4 \\
 x \minus{} 3 > 2 $
My answers:
$ x \minus{} 3 > 5 \\
 x > 5 \\
 (5,\infty ) \\
 x \minus{} 3 <  \minus{} 2 \\
 x < 1 \\
 ( \minus{} \infty ,1) $
And its correct with squaring the other side  :wink: or I need to square if I have $ 2x^2$  :?:

hello, we have $ 2x^2\minus{}5x\plus{}1\geq1 \Leftrightarrow 2x^2\minus{}5x\geq0$
$ x(2x\minus{}5)\geq0$
This gives $ x\geq\frac{5}{2}$ or $ x\le 0$
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, we have $ 2x^2 \minus{} 5x \plus{} 1\geq1 \Leftrightarrow 2x^2 \minus{} 5x\geq0$
$ x(2x \minus{} 5)\geq0$
This gives $ x\geq\frac {5}{2}$ or $ x\le 0$
Sonnhard.[/quote]
WOW  :rotfl:
Thanks a lot "Dr Sonnhard Graubner" that's an amazing answer instead of my long letter  :blush: .


A polynomial of degree four with leading coefficient 1 and integer coefficients has two zeros, both of which are integers. Which of the following can also be a zero of the polynomial?

\[\text{(A)}\ \frac{1+i \sqrt{11}}{2}\qquad \text{(B)}\ \frac{1+i}{2}\qquad \text{(C)}\ \frac{1}{2}+i \qquad \text{(D)}\ 1+\frac{i}{2}\qquad \text{(E)}\ \frac{1+i \sqrt{13}}{2}\]

[hide="Quick Soln."]
From Vieta's formulae we get that $z_{1}+z_{2}\in \mathbb{Z}\wedge z_{1}z_{2}\in \mathbb{Z}$, so this eliminates choices $B$, $C$, and $D$. Then, we see that a factor with root (and conjugate root) of $A$ gives a monic quadratic with integer coefficients and that that with root (and conjugate of) $E$ does not; thus the answer is $\boxed{A}$.[/hide]

[quote="t0rajir0u"]I always thought this problem was poorly worded.  I think it means "two [b]real[/b] zeroes."[/quote]

Nah, who needs the Fundamental Theorem of Algebra? Its not like its important or anything.

But usually when I've seen roots called "zeroes" they have been real, i.e. when the polynomial is graphed on the xy-plane. (At least that has been my experience.  :) )

[quote="t0rajir0u"][quote="cincodemayo5590"][quote="t0rajir0u"]I always thought this problem was poorly worded.  I think it means "two [b]real[/b] zeroes."[/quote]

Nah, who needs the Fundamental Theorem of Algebra? Its not like its important or anything.[/quote]

It's got more to do with multiplicity than that.  You could have $P(x) = (x-p)^{2}(x-q)^{2}$ which wouldn't have any complex roots.[/quote]

can $\ne$ always


Please help with a problem from a contest(in Romania)
I solve it and i don't know where i am wrong

loga (bc)+  logb   (ca)+ logc (ab)>=4( logab (c)+ logbc  (a)+ logca  (b))

I wrote that:
loga  (bc)+ logb  (ca)+ logc (ab)- 4( logab (c)+ logbc  (a)+ logca  (b))>=0

and then I proved that: loga  (bc)+ logb  (ca)+ logc (ab)>=6     (*)
                                  4( logab (c)+ logbc  (a)+ logca  (b))>=6( **)
and then with difference we obtain the inequality
Relation(*)and(**) I'm sure that I proved well.
Can tell me someone where I wrong ( if I wrong somewhere)

$log_{a}(bc)+log_{b}(ca)+log_{c}(ab) \ge 4( log_{ab}(c)+log_{bc}(a)+log_{ca}(b))$
if x=ln(a) , y=ln(b),z=ln(c) then we have to prove
 $\frac{x+y}{z}+\frac{x+z}{y}+\frac{y+z}{x}\ge 4(\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y})$ - it is easy because
$\frac{x}{z}+\frac{x}{y}\ge \frac{4x}{y+z}$
Few words about your proving.Put '0' or something else where is '6' in (*) and (**) and try again. :wink:
Even wrong inequality can be proved like that.

Yes , unfortunately you are wrong.Because if you have $a\geq b$ and $c \geq d$ it not necessary that $a-c\geq b-d$ holds.But this inequality was very easy I managed to solve it in the competition just using means realations.I'll give you a hint.Make the following substitution:\[x=log_{a}b\] \[y=log_{b}c\] \[z=log_{c}a\] and it's very easy


$ \sum(a\plus{}b)(ac\plus{}9)(bc\plus{}9) \geq 3(ac\plus{}9)(bc\plus{}9)(ab\plus{}9)$
$ 10abc\sum ab\plus{}45\sum(a^2b\plus{}b^2a)\plus{}189abc\plus{}243\geq 243\sum ab\plus{}3a^2b^2c^2$
$ 10abc\sum ab\plus{}45\sum(a^2b\plus{}b^2a)\plus{}189abc\plus{}243\geq 81(a\plus{}b\plus{}c)\sum ab\plus{}3a^2b^2c^2$
$ 10abc\sum ab\plus{}243 \geq 36\sum(a^2b\plus{}b^2a)\plus{}54abc\plus{}3a^2b^2c^2$
$ 10abc\sum ab\plus{}243 \geq 36\sum(a^2b\plus{}b^2a)\plus{}6(a\plus{}b\plus{}c)^2abc\plus{}3a^2b^2c^2$
$ 243 \geq 36\sum(a^2b\plus{}b^2a)\plus{}6abc(a^2\plus{}b^2\plus{}c^2)\plus{}2abc(ab\plus{}bc\plus{}ca)\plus{}3a^2b^2c^2$
$ (\sum a)^5 \geq 4(\sum a)^2\sum(a^2b\plus{}b^2a)\plus{}6abc(a^2\plus{}b^2\plus{}c^2)\plus{}2abc(ab\plus{}bc\plus{}ca)\plus{}3a^2b^2c^2$
$ \sum a^5\plus{}\sum (a^4b\plus{}b^4a)\plus{}4abc\sum ab \geq 2\sum(a^3b^2\plus{}b^3a^2)\plus{}2abc\sum a^2\plus{}3a^2b^2c^2$

$ \sum a^5 \sum a\plus{}\sum (a^4b\plus{}b^4a)\sum a\plus{}4abc\sum ab \sum a \geq 2\sum(a^3b^2\plus{}b^3a^2)\sum a\plus{}2abc\sum a^2\sum a\plus{}9a^2b^2c^2$
$ \Rightarrow \sum a^6 \plus{}2 \sum(a^5b\plus{}b^5a)\plus{}3a^2b^2c^2 \geq 4 \sum a^3b^3\plus{}\sum (a^4b^2\plus{}b^4a^2)$
$ \sum (a^6\minus{}a^4b^2\plus{}b^4a^2\plus{}a^2b^2c^2) \plus{}2 \sum (a^5b\plus{}b^5a\minus{}2a^3b^3) \geq 0$
$ \sum a^2(a^2\minus{}b^2)(a^2\minus{}c^2)\plus{} 2\sum ab(a^2\minus{}b^2)^2 \geq 0$
Equality occurs with $ (1,1,1)$ or $ (\frac{3}{2},\frac {3}{2},0)$

Nice solution, cefer!

Other solution is to write the inequality as
$ 10abc\sum ab\plus{}54abc \plus{}243 \ge 108\sum ab\plus{}3a^2b^2c^2$.
Since $ abc\sum ab \ge 3a^2b^2c^2$, it suffices to show that
$ 6abc \plus{}27 \ge (12\minus{}abc)\sum ab$.
By EV-Theorem, it suffices to consider $ b\equal{}c$. We get
$ a(a\minus{}1)^2(a^2\plus{}4a\plus{}1)\ge 0$,
which is clearly true.

See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3359448#p3359448

[quote="arqady"]Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ a \plus{} b \plus{} c \equal{} 3.$ Prove that
\[ \frac {a \plus{} b}{ab \plus{} 9} \plus{} \frac {a \plus{} c}{ac \plus{} 9} \plus{} \frac {b \plus{} c}{bc \plus{} 9}\geq\frac {3}{5}
\][/quote]
Dear friend,do this:

Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ a \plus{} b \plus{} c \equal{} 3.$ Prove that

\[\frac{b+c}{81+49b^2c^2}+\frac{c+a}{81+49c^2a^2}+\frac{a+b}{81+49a^2b^2} \geq  \frac{3}{65}\]

 BQ

and dear you,do this:

$a,b,c>0,a+b+c=3$,prove that:

\[\frac{(a+b)(c+a)}{3+bc}+\frac{(b+c)(a+b)}{3+ac}+\frac{(c+a)(b+c)}{3+ab} \geq  3\]

BQ

[quote="arqady"]Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ a \plus{} b \plus{} c \equal{} 3.$ Prove that
\[ \frac {a \plus{} b}{ab \plus{} 9} \plus{} \frac {a \plus{} c}{ac \plus{} 9} \plus{} \frac {b \plus{} c}{bc \plus{} 9}\geq\frac {3}{5}
\][/quote]
And you do this:

Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ a \plus{} b \plus{} c \equal{} 3.$ Prove that

\[\frac{2+bc}{(b+c)^2}+\frac{2+ac}{(c+a)^2}+\frac{2+ab}{(a+b)^2} \geq  \frac{9}{4}\]

[quote="arqady"]Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ a \plus{} b \plus{} c \equal{} 3.$ Prove that
\[ \frac {a \plus{} b}{ab \plus{} 9} \plus{} \frac {a \plus{} c}{ac \plus{} 9} \plus{} \frac {b \plus{} c}{bc \plus{} 9}\geq\frac {3}{5}
\][/quote]

And no nice,But no easy?

Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ a \plus{} b \plus{} c \equal{} 3.$ Prove that

\[\frac{b+c}{81+7bc+b^2+c^2}+\frac{c+a}{81+7ac+c^2+a^2}+\frac{a+b}{81+7ab+a^2+b^2} \le \frac{1}{15}\]

[quote="arqady"]Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ a \plus{} b \plus{} c \equal{} 3.$ Prove that
\[ \frac {a \plus{} b}{ab \plus{} 9} \plus{} \frac {a \plus{} c}{ac \plus{} 9} \plus{} \frac {b \plus{} c}{bc \plus{} 9}\geq\frac {3}{5}
\][/quote]

And do this:

Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ a \plus{} b \plus{} c \equal{} 3.$ Prove that

\[\frac{b+c+bc}{2+abc}+\frac{c+a+ac}{2+abc}+\frac{b+a+ab}{2+abc} \geq  3\]

\[\frac{b+c+bc}{11+abc}+\frac{c+a+ac}{11+abc}+\frac{b+a+ab}{11+abc} \le  \frac{3}{4}\]

BQ

[quote="xzlbq"][quote="arqady"]Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ a \plus{} b \plus{} c \equal{} 3.$ Prove that
\[ \frac {a \plus{} b}{ab \plus{} 9} \plus{} \frac {a \plus{} c}{ac \plus{} 9} \plus{} \frac {b \plus{} c}{bc \plus{} 9}\geq\frac {3}{5}
\][/quote]

And do this:

Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ a \plus{} b \plus{} c \equal{} 3.$ Prove that

1-\[\frac{b+c+bc}{2+abc}+\frac{c+a+ac}{2+abc}+\frac{b+a+ab}{2+abc} \geq  3\]

2-\[\frac{b+c+bc}{11+abc}+\frac{c+a+ac}{11+abc}+\frac{b+a+ab}{11+abc} \le  \frac{3}{4}\]

BQ[/quote]

$1)\Leftrightarrow \sum ab\geq 3abc \\\\ or \  \sum ab\geq 3\sqrt[3]{a^2b^2c^2}\geq 3abc \ because \ abc\leq \left ( \frac{a+b+c}{3} \right )^3=1$

$2)\Leftrightarrow 4\sum ab\leq 9+9abc $

by Schur we have $ (a+b+c)^3+9abc\ge 4(a+b+c)(ab+ac+bc)\Rightarrow  9+9abc \ge 9+3abc\ge  4\sum ab $

[quote="xzlbq"]
Let $a,$ $b$ and $c$ be nonnegative real numbers such that $a+b+c=3$. Prove that

\[\frac{2+bc}{(b+c)^2}+\frac{2+ac}{(c+a)^2}+\frac{2+ab}{(a+b)^2} \geq  \frac{9}{4}\][/quote]

After homogenizing and brute-force expanding the given inequality, it suffices to show that
\[\begin{aligned}
8\sum_{cyc} a^6+32\sum_{cyc} ab(a^4+b^4)+18abc\sum_{cyc} a^3 + 78a^2b^2c^2 \geq 17\sum_{cyc} a^2b^2(a^2+b^2)+46\sum_{cyc}a^3b^3+18abc\sum_{cyc}ab(a+b)
\end{aligned}\]
which is true according to the (sum of the) following inequalities:

1. $18abc\sum_{cyc} a^3 + 54a^2b^2c^2 \geq 18abc\sum_{cyc}ab(a+b)$ (Schur)
2. $8\sum_{cyc} a^6 +24a^2b^2c^2 \geq 8\sum_{cyc} a^2b^2(a^2+b^2)$ (Schur)
3. $9\sum_{cyc} ab(a^4+b^4) \geq 9\sum_{cyc}\frac{ab(a^2+b^2)^2}{2}\geq 9\sum_{cyc}a^2b^2(a^2+b^2)$ (C-S + AM-GM)
4.  $23\sum_{cyc} ab(a^4+b^4) \geq 23\sum_{cyc}\frac{ab(a^2+b^2)^2}{2}\geq 23\sum_{cyc}\frac{ab(2ab)^2}{2}= 46\sum_{cyc}a^3b^3$ (C-S + AM-GM)

Therefore, the given inequality is true. Equality holds iff $(a,b,c) = (1,1,1)$ and $(a,b,c) = \left(\frac{3}{2}, \frac{3}{2}, 0\right)$ and its permutations.


function
quadratics
inequalities
inequalities open

[quote="conceive"]Prove or disprove
$a,b>0$

$3a^2+3b^2+7ab+2a^2b+2ab^2+1\ge9a+9b$[/quote]

Rewrite this one as function in term of a we have:
$f(a)=(3+2b)a^2+(7b+2b^2-9)a+(3b^2+1-9b)$
Since $3+2b\geq 0$ we only need to prove $\delta=(7b+2b^2-9)^2-4(3+2b)(3b^2+1-9b) \geq 0$
or $4b^4+4b^3+49b^2-26b+69 \geq 0$ but this one is true according to its graph.

Remark the simmetry of $a,b,$ we can set $a+b=u,\ ab=v.$

[quote="arqady"][quote="Gibbenergy"]
Rewrite this one as function in term of a we have:
$f(a)=(3+2b)a^{2}+(7b+2b^{2}-9)a+(3b^{2}+1-9b)$
Since $3+2b\geq 0$ we only need to prove $\delta=(7b+2b^{2}-9)^{2}-4(3+2b)(3b^{2}+1-9b) \geq 0$
or $4b^{4}+4b^{3}+49b^{2}-26b+69 \geq 0$ but this one is true according to its graph.[/quote]
Id est $3a^{2}+3b^{2}+7ab+2a^{2}b+2ab^{2}+1\ge9a+9b$ is wrong. :wink:[/quote]

Oh... I found my mistake. In the quadratic form, $\delta$ should be less than zero in oder to make function positive for all of $a,b \geq 0$. But as my proof, $\delta$ is always $\geq 0$. So there is always exist the interval of $b$ that would definitely change the side of the function.

   Thank you Arqady!

[quote="Gibbenergy"]So there is always exist the interval of $b$ that would definitely change the side of the function.[/quote]

Indeed, $a=b=\frac12$

$3a^{2}+3b^{2}+7ab+2a^{2}b+2ab^{2}+1\ge9a+9b$
$\Longleftrightarrow 3(a+b)^{2}+ab+2ab(a+b)+1\geq 9(a+b).$

Let $a+b=s,\ ab=t,$ we have $a>0\ and\ b>0\Longleftrightarrow s>0\ and\ t>0\ and\ s^{2}-4t\geq 0$

$\Longleftrightarrow s>0\ and\ 0<t\leq \frac{s^{2}}{4}\ \cdots [*].$
Under this condition, we are to prove that $3s^{2}+t+2st+1\geq9s\Longleftrightarrow (2s+1)t+3s^{2}-9s+1\geq 0.$
Now on $t-y$ plane, consider the linear function $y=(2s+1)t+3s^{2}-9s+1$
By $[*],$ this function is an increasing function, so $y_{t=0}=3s^{2}-9s+1$
$=3\left(s-\frac{3}{2\right)^{2}-\frac{23}{4}}\geq-\frac{23}{4}.$
Thus the $y-$ intercept will become positive, zero and negative.So we can prove for some case.

If $0<s\leq \frac{9-\sqrt{69}}{6},\ \frac{9+\sqrt{69}}{6}\leq s,$ then the given inequality holds.


I found the 2007 AMC10 harder than the 2009 AMC10.... Did anyone else?

Did you use a calculator for the 2007 test? If not, it may have seemed harder, because pre-2008 AMC's allowed calculators.

Even with a calculator, the last few questions seemed harder than the 2009's....

I only scored a 100 on the 2007 AMC10, but I scored like a 120+ on the 2009 AMC10... Hmmm...

uh....if you're talking about the 2009 10A, that was probably one of the easiest AMC 10s ever.


Let $ p$ be any primes and $ m$ be any positive integers. If you select some positive integers nicely, then prove that there exists the part of the series of $ m$ consecutive zeros in the decimal representation of $ p^n$.

Very nice!
Solution:
Let $ p$ be an arbitrary prime number,such that $ p\neq 2,5$.
Then it is easy to see,that for every $ m\in N$ there exist a naural number $ n$,such that $ p^{n}\equiv 1 (mod 10^m)$(Just use Euler's theorem).
Hence the last $ m$ digits in decimal representation of $ n$ will be $ \overline{\dots000\dots 1}$.So now we just need to consider 2 cases $ p\equal{}2$ and $ p\equal{}5$.
If $ p\equal{}2$,then for all $ l\in\mathbb{N}$ there exist $ k\in\mathbb{N}$,such that $ 2^k\equiv 1(mod 5^{l})$,hence
$ 2^{k\plus{}l}\equal{}2^{l}\plus{}10^{l}k$.
So if we take enough large $ l$,then the diffrence between the number of digits in decimal representation of $ 10^{l}$ and the number of digits of $ 2^{l}$,can be every natual number,suppose $ m$,but then first digits of $ 2^{k\plus{}l}$ will be $ \overline{\overline{k}00\dots0\overline{2^{l}}}$,and we are done.
The last case $ p\equal{}5$ is totally similar to the case $ p\equal{}2$.

[quote="ZetaX"]Alternatively, you get it to start with $ 100...000...$. In fact that you can archieve any finite starting sequence (proven before, a direct application that $ \alpha m \mod 1$ is dense in $ [0,1]$ for irrational $ \alpha$).[/quote]
If i am not mistaken then the following theorem is true too:
Let $ p\neq 10^n$,be a natural number(not necessary prime),then for any number $ k>1$,there exist a nautal number $ m$,such that
the first $ k$ digits in decimal representation of $ p^{m}$ are exactly the number $ k$.


geometry
Euler
function
3D geometry
tetrahedron
area of a triangle
Heron\u0027s formula

Does there exist a triangle with integer square sides and integer area?
It cannot be a right triangle as a^4 + b^4 = c^4 is impossible with integers (FLT).
Can it be an isosceles triangle? I don't think so (then 4*a^4-b^4=x^2 must have an integer solution).
Can it be a scalene triangle? Then
a^2 + b^2 -c^2 = 2*x^2  and a^2 - b^2+c^2 = 2*y^2 and -a^2 + b^2 + c^2 = 2*z^2
and
a^4 + b^4 + c^4 = 2*w^2
with 1<a<b<c must have an integer solution. I don't know.

kweenie

PS:I posted it in the Combinatorics section, but I think it's more suited for Number Theory.

[quote="kweenie"]Can it be an isosceles triangle? I don't think so (then 4*a^4-b^4=x^2 must have an integer solution).[/quote]
This equation indeed does not have non-zero solutions. It corresponds to the hyper-elliptic curve $ v^2 \equal{} 4 u^4 \minus{} 1$ (where $ u \equal{} \frac {a}{b}$ and $ v \equal{} \frac {x}{b^2}$), which is equivalent to the elliptic curve $ Y^2 \equal{} X^3 \plus{} 16X$ of zero rank (under a certain transformation, according to Maple).
Perhaps, there exists also an elementary proof based on the factorization $ (2a^2 \minus{} b^2)(2a^2 \plus{} b^2) \equal{} x^2$ or the formulas for Pythagorean triples.

Waw, that's already something, but unfortunately the rest remains open!

No wonder, as in fact the Heronian triangle problem with square sides is equivalent to the famous perfect Euler brick problem:
"Is there a rectangular brick so that all 7 distances between the vertices (included the space diagonal) are integers"

To see the relation with a perfect Euler brick, express the product of its volume times its space diagonal as a function of the surface diagonals. You get Heron's formula for the area of a triangle with sides equal to the square of the surface diagonals.

kweenie

PS: Another equivalent formulation for the perfect Euler brick problem is perhaps the following:
Does there exist a rectangular tetrahedron (all face angles at one vertex are rectangular) so that all sides and all areas are integers? I'm not so sure about this last equivalence. Is there some proof for it?


Prove that $ f(x)\equal{}\sqrt[3]{x}(x\in[0,\plus{}\infty))$ is uniformly continuous.

Can you place an upper bound on the expression $ \left|\sqrt[3]{x}\minus{}\sqrt[3]{y}\right|$, in terms of $ \left|x\minus{}y\right|$?  Once you do this, finding a $ \delta$ won't be hard.


Find the smallest perfect square greater than one for which the sum of its digits is its square root.

The sum of the digits must be a perfect square, and the number is one too. So work up through the perfect squares. 1, 4, 9, 16, 25, 36, 49, 64, 81! AHA! $ \boxed{81}$ is obviously the smallest number that fits our criteria.

when you think about sum of digits, 9 should usually come to your mind.
$ 9^2\equal{}81$
checking shows it works, just guess 81.

Is there a more algebraic way of solving this problem other than guess and check?


limit
logarithms
calculus
calculus computations

[quote="Kibu"]$ \lim_{x\to\infty}((x \plus{} a)^{1 \plus{} \frac {1}{x}} \minus{} x^{1 \plus{} \frac {1}{x \plus{} a}}),\qquad a\in\mathbb{R}$[/quote]

Is the answer $ 1$? :D

How about a proof?  Wolfram|Alpha uses some ridiculous substitutions and L'Hopitale twice to get $ a$.  But I would like to see a reasonable solution that a human could do.

[quote] But I would like to see a reasonable solution that a human could do.
[/quote]

And I would like to see any solution...

$ ((x \plus{} a)^{1 \plus{} \frac {1}{x}} \minus{} x^{1 \plus{} \frac {1}{x \plus{} a}}) \equal{} (x \plus{} a) e^{\frac {1}{x} \ln (x \plus{} a) } \minus{} x e^{\frac {1}{x \plus{} a}\ln x}).$
Now use $ e^y \equal{} 1 \plus{} y \plus{} O(y^2)$ as $ y\to0$ ($ \ln x/(x \plus{} a)$ and $ \ln(x \plus{} a)/x$ replace $ y$)

$ (x \plus{} a)\Bigl( 1 \plus{} {\ln (x\plus{}a)\over x }^* \plus{} O({\ln^2 x\over x^2}) \Bigr) \minus{} x \Bigl( 1 \plus{} {\ln x\over x \plus{} a} \plus{} O({\ln^2\over x^2}) \Bigr) \equal{} x \plus{} a \plus{} \ln (x\plus{}a) \plus{} O({\ln x\over x}) \minus{} x \minus{} {x\ln x\over x \plus{} a} \plus{} O({\ln^2 x\over x})  \equal{}
a \plus{} \ln x \plus{} \ln(1\plus{}{a\over x}) \plus{} O({\ln x\over x}) \minus{} {x\ln x\over x \plus{} a} \plus{} O({\ln^2 x\over x})$

whose limit is evident.

*: [b] I wrote $ {\ln x\over x \plus{} a}$ instead of $ {\ln (x\plus{}a)\over x }$ as it must be. [/b]


Hi everyone,

I'm new to this forum and to the AMC/AIME/USAMTS.  Next year i'll be a sophomore in high school, and I want to start competing.    :)

I have a few questions that i haven't been able to find the answer to on the website, if anyone can help me that would be great, so... here goes...

First, what's the difference between the AMC and AIME?  Where can I take them if my school doesn't participate? Where can I get practice material, and are there any mock contests I can do?

Thanks a lot,  :lol:

Vinay

[url=http://www.unl.edu/amc]The AMC Website[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=15513]Mock Contests[/url]
[url=http://www.artofproblemsolving.com/Forum/search.php]Search Feature[/url] :D

For practice, just frequent this forum and try all the problems you find. :)

If your school doesn't participate, you can contact the AMC writers and they'll let you take it at another school.  This is what I'm going to do.  :lol:

The AMC is the American Mathematics Competition; it is the first level test which is open to anyone who wants to attempt it (provided you are in the correct grade range).  The AIME is the American Invitational Mathematics Examination, just as you might conclude from the name, those who do well on the AMC (100 or better AMC 12, 120 or better AMC 10) are invited to take the AIME.


Can anyone guide me how to prepare for mathcounts? Does anyone know of any resources for past competitions besides mathcounts.org?

Heres for starters
http://www.artofproblemsolving.com/Forum

The best way to learn is to do past problems. I would recommend starting with those, and if you feel there is a certain subject that you can't answer many questions about, try learning about that subject.

Mathcounts.org has 2009 tests only.  Is there any other website that I can get past competitions from?  Thanks everyone.

Search "Mathcounts Problems" on Google.
I'm sure it will come up with a lot.

Try reading these links (and searching, this question has been asked many times before):

http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1912104443&t=313618
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1912104443&t=316818
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=286559
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1912104443&t=297329
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1315377612&t=258459

:arrow: My coach recommended [url=http://mathcounts.saab.org/mc.cgi]this site.[/url]

Actual past years' problems aren't usually posted, but MathCounts does sell them. You could also try trading. For a start, PM me what you want and I could give some.


Chad's score on a math test was half of Milton's score. Julia's score was three times Chad's score. The sum of the three scores was 186. What is Julia's score?

you could make an equation:

c=1/2m  or m=2c
3c=j
m+c+j=186

by substitution:

2c+c+3c=6c=186
c=31

so Julia scored a 93


Following a request of Valentin, I will make the following "announcement": All the problems with no source indicated by me are created by me and I post them since I haven't seem them anywhere. So I beg you to announce me when you find that a certain "original" problem was given in a contest or appeared in a mathematical contest. Thank you very much.

I would like to make you writing down the soure in [i]Italic[/i] as follows:

Whenever you don't know where the problem comes from you write below the problem text [i]unknown source [/i] or [i]created by harazi[/i] if you invented the problem.

In the case you know the source you may write something like that below the problem:

[i]German Mathematical Olympiad, grade 12, round 4, problem 3, 2002[/i]

[i]Crux Mathematoricum, 1995, volume 5, page 54, proposed by Walter Janous[/i]

Maybe it is possible for and others to edit your old postings in the case you remember the source,... I think it is easy for you to detect the problems created by you and indicate them as such problems. Thanks.

The purpose of providing sources it that it will be easier navigating in the forum. Moreover it helps to keep the forum structure non-haphazard. In due course there will be more and more problems and it will occur more and more often that a problem (mabye slightly modified), known or unknown by the the mathlinks members, will be reposted. But having very many postings in the solved sections (several pages), it will become difficult to find an old problem. So it might help to remember the source and use the mathlinks search engine. There also should be a better substructure in the solved sections, e.g. maybe the combinatorics solved section should be subdivided in subsections as graph theory etc. But I don't think it is a good idea as it is very subjective and heavily depends on the main idea in the solution. People would have different ideas about where to put it. Another subdivision might be based on the olympiad subsections. Unfortunately that is not easy too as some problems emerge at many olympiads. What will you do ? Having the same problem in many subsections on the forum or putting it into that olympiad section where it was first published. But as you discover an earlier source giving the same problem you would have to move the problem to another section. In the first case it would require another redundant copy at the forum.

I am interested in your ideas about the topic before ! BTW actually my posting should appear in the section "about mathlinks"...

BTW I use to say one creates a problem. So one might argue about the fact whether the problem always has been there and you just did not find it earlier. The other point of view is that you created the problem and the problem did not exist before. What do you think ?

I ussually/never do not take problems from different sources and say they are mine. Anyway, it is very probable that some of the problems that I said there were created by me have been discovered earlier. What should the others do? It's simple: tell the one who said he created the problem that it was created by someone else earlier. This shouldn't offense anyone, since noone is capable to remember all the problems created till a certain moment. That is why I allways say that a certain problem is created by me until somebody else finds it somewhere.

Please use the following to edit the sources field:

- if the problem was given in an olympiad, state it like this
COUNTRY CONTEST YEAR - Author's Name (if any, or it is known).

example: Romanian Olympiad 2003 - Valentin Vornicu or
              Romanian Team Selection Test  1994 - Titu Andreescu ( this can also be written as Romanian TST 1994 - Titu Andreescu) or
              German Olympiad 1998

- if it is YOUR problem just post Created by USERNAME
(example: Source: Created by harazi)

- if there is a well known shortcut for the contest use it. The following are to be used:

USAMO = [b]USA M[/b]athematical [b]O[/b]lympiad
TST = [b]T[/b]eam [b]S[/b]election [b]T[/b]est (for IMO)
IMO = [b]I[/b]nternational [b]M[/b]athematical [b]O[/b]lympiad
ISL = [b]I[/b]nternational_mathematical_olympiad [b]S[/b]hort[b]L[/b]isted problems
ILL =[b]I[/b]nternational_mathematical_olympiad [b]L[/b]ong[b]L[/b]isted problems
APMO = [b]A[/b]sian [b]P[/b]acific [b]M[/b]athematical [b]O[/b]lympiad
BMO = [b]B[/b]alkan [b]M[/b]athematical [b]O[/b]lympiad
(used to be BkMO = [b]B[/b]al[b]k[/b]an [b]M[/b]athematical [b]O[/b]lympiad - which is now deprecated)
CMO = [b]C[/b]hinese [b]M[/b]athematical [b]O[/b]lympiad

- if the problem source is not known, then write Unknown, and nothing else.

I urge all moderators to edit the posts that do not respect these requirements for the source field, and replace the existing (or non-existing text) where possible (for now only in Advanced Section) as soon as possible.

Mr.vornicu
What if the problem is from a book(not a well know to consider it),shall we say
from a book?

Like lomos_lupin says,if its from a local book which is of no use at all,do we still have to write the source?

[quote="José"]Please harazi or you orl explain me what you're trying to say because I don't understand; If I don't know the surce I have to put MADE BY HARAZI?????????[/quote]

No jose,if don't know he source write "Unknown source" and if u created the problem,write "created by Jose"


What do you think? Did Fermat has a proof for his last theorem????????

[hide="My Opinion"] I think he had one, maybe he had one........not a formal proof, but maybe a really nice idea of how to prove it...... [/hide]

According to one source which I forget (but I'm 90% sure that it was AoPS II), Fermat published/circulated proofs for $n=3$ and $n=4$, where $n$ is the exponent, which makes it seem unlikely that he had a proof.  Of course, one [i]could[/i] argue that, given most mathematicians' paranoid attitude in those days (note how Newton didn't publish [i]Principia[/i] ever, yet attacked Leibniz viciously, accusing him of plagiarism, when he published [i]his[/i] methods of calculus; of course one could also argue that Newton was unusually paranoid about it (which he was); but then, Newton was from a later era), but I still find it unlikely that he (Fermat) would have published those proofs and never told anybody his general proof.

i think since it took like, forever, for somebody to actually prove it, and the proof was like a few hundred pages long and used really really advanced stuff, if there really was a nice little "magnificient" proof, somebody would've found it by now.  so he probably didn't actually have a proof.  but there's really no way to know for sure.

Bear in mind that not all of the things that Fermat claimed panned out.

A good marker for the beginnings of modern number theory would be Euler's encounter with Fermat's posthumously circulated annotations of his copy of Diophantus's [i]Arithmetica[/i].

Fermat claimed that if $p$ is prime, $p$ divides $a^p-a.$ Fermat didn't leave a particularly nice proof of this. Euler did provide both a nice proof and an extremely useful generaliztion. The original theorem bears Fermat's name to this day, but Euler was responsible for making the mathematical community aware of it.

Fermat claimed that $2^{2^5}+1$ is prime. Euler (using, in part, the previously mentioned theoerm) proved that this number is divisible by 641 and hence that Fermat was wrong in this instance.

Basically Euler worked through all of Fermat's leftover claims, proving and disproving - with one execption. And that plants a great big warning flag on the problem: the great Leonhard Euler fully considered this claim and could neither prove nor disprove it. Attempt it yourself at your own risk, knowing what you're getting yourself into.

(No, of course Fermat did not have a full proof, at least not beyond the $n=3$ and $n=4$ cases.)

I would add that Fermat was probably not all that paranoid about priority and credit. He did not hold an academic position that depended on his reputation as a scholar. In fact, he was a judge (which included some administrative responsibilities) in an area of France that wasn't particularly close to Paris. His mathematics was something he did on the side.

Also: there weren't scholarly journals at the time, for the most part. Some people had books published, often at their own expense, but not Fermat. We know about him primarily through his correspondence with other mathematicians, either directly, or through intermediaries such as Mersenne.


Check the results here: http://www.iarcs.org.in/inoi/2008/zio2008/results_zio2008.php

Out of the 6700+, 240 got selected. Congrats to all those who cleared this round (including me:D)

~ Shreyas

Me too selected   :P
Sorry for the late reply  :D

:) even i am Selected!!!!!!!!!!!!!!!!!!!! So howz Preparation??

[quote="chemrock"]When did you shreyas join

RAKESH[/quote]

Much before all of you and even madness. If I'm right, I invited madness here;)


superior algebra
superior algebra solved

3.Let A \in Mn(R), n>=2, invertible. Proove that (A*)* = (detA)n-2A.(Where A* is the adjunct of A).

Well you have A.A*=det(A).I_n  => A* = A^(-1).det(A) =>                  det(A*)=det(A)^(n-1)
=> A*.(A*)*=det(A*).I_n  <=> A^(-1).det(A).(A*)*=det(A)^(n-1).I_n
so (A*)*=det(A)^(n-2).A
But this is a known property and what is it doing in Analysis section anyway?


calculus
integration
LaTeX
function
floor function
logarithms
limit

$ \int _0^{\infty}[ne^{\minus{}x}]\ dx$
here  $ [.]$  denotes greatest integer function.

Notational comment and $ \text{\LaTeX}$ tip: while the notation is by no means universal, the convention of calling the greatest integer function the "floor" function is growing, and I would encourage its use. The $ \text{\LaTeX}$ codes for the appropriate symbols are \lfloor and \rfloor.

\lfloor ne^{-x} \rfloor becomes $ \lfloor ne^{ - x} \rfloor.$

\left\lfloor \frac{n-1}{2} \right\rfloor becomes $ \left\lfloor \frac {n - 1}{2} \right\rfloor.$

As for the substance of the problem: for $ x > 0,\ 0 < ne^{ - x} < n,$ so $ \lfloor ne^{ - x}\rfloor < n.$

$ \lfloor ne^{ - x}\rfloor = k$ if $ k\le ne^{ - x} < k + 1,$ which we can rearrange to $ \ln n - \ln k\ge x > \ln n - \ln(k + 1).$

Hence $ \int_{\ln n - \ln(k + 1)}^{\ln n - \ln k}\lfloor ne^{ - x}\rfloor\,dx = k((\ln n - \ln k) - (\ln n - \ln(k + 1))) = k\ln\left(\frac {k + 1}{k}\right)$

That's for $ k > 0.$ For $ k = 0,$ the appropriate integral is zero.

Our overall integral is $ \sum_{k = 1}^{n - 1}k\ln\left(\frac {k + 1}{k}\right).$

I don't immediately see how to render that in simpler form, but since $ \lim_{k\to\infty}k\ln\left(\frac {k + 1}{k}\right) = 1,$ the sum is asymptotic to $ n$ as $ n\to\infty.$ This is consistent with $ \int_0^{\infty}ne^{ - x} = n.$

$ \sum_{k \equal{} 1}^{n \minus{} 1}k\ln\left(\frac {k \plus{} 1}{k}\right) \equal{} \sum_{k \equal{} 1}^{n \minus{} 1} [(k\plus{}1) \ln (k\plus{}1) \minus{} k \ln k\minus{}\ln (k\plus{}1)]$

$ \equal{} n \ln n \minus{} \sum_{k \equal{} 1}^{n \minus{} 1} \ln (k\plus{}1) \equal{} n \ln n \minus{} \ln n! \equal{} \ln \frac{n^n}{n!}$


http://www.freerice.com

Pretty sweet idea, and it's fun. What's your average level? I'm around 35.

[quote="Nerd_of_the_Ages"]Hmm, that's pretty cool. My vocab lvl sucks (probably will get better in time). I'm at only like 20.[/quote]

19 so far, but I've donated 300 grains.

My average level is 30 and i have already donated over 2000

I wasn't concentrating on like the first 400 grains, so I'm at 24, donated 600

39 after about 200 grains.  I missed the third or fourth answer, putting me at 35.  After that I missed maybe one or two.

There are much better people out there.  Some of the words I didn't actually know, but made educated guesses on based on parts of speech and roots.

Very addictive... :D

Level: 36
Grains: 1100

My grandma and me together donated about a pound of rice yesterday.

More than 20 tons of rice has been donated so far.

Here's a post by me in another forum about this:

http://support.invisionfree.com/index.php?showtopic=444008&st=15#entry10892316

32 donating 260 only (I'm short on time, and I gravely doubt that perusal means reading)

31 was my highest, I only did for one night so only donated 2220 grains exactly..

It's updated! Now if you get a word right, it shows that it is right, the previous word, and the definition you picked.

What's it scored out of?  My father got 45.

My score was a 31 and 4200 grains of rice.   :lol:

Okay, I'll stop arguing with D3m0n Shad0w over there.

Sometimes roots help too, but yeah, sometimes you need to guess at a random and obscure word.

Do you think that Free Rice gets those Vocab questions from SATs?

Maybe. It said something about it on Wikipedia. Certainly not the lvl 40+ words, though.

Predictably, there's going to be [url=http://www.thepcspy.com/read/is_freericecom_making_150k_each_day_in_profits]bashers[/url]of FreeRice. They make good points, though. I still want to argue with D3m0n...

Hmm... level 42 is my average... still...
45 is a regular max for me

I'm at the same level I was 4 months ago
gah

I have currently donated 10020 grains on my dad's computer.

[quote="Klebian"]www.freerice.com

Pretty sweet idea, and it's fun. What's your average level? I'm around 35.[/quote]

you do better than my english calss combined

EDIT: basically, everybody does. our class average is 30
 theres a rumor going around that nobody can beat lv 50

It's false, I saw someone spend some words in l55 today without a dictionary, and I personally have gotten up into l51.

[quote="i_like_pie"]Just out of curiosity, what words and choices are given on level 55?[/quote]

Mission: Find what words are in level 55
Tools: Google, Wikipedia, Dictionary.com
Tactics: Use the references!

lvl 55 words

   1. fraise means:
   2. fresh cream
   3. chair
   4. barrier
   5. fruit dish

   1. geeldikkop means:
   2. sheep disease
   3. South African pig
   4. old Dutch headdress
   5. money jar

   1. killick means:
   2. beach bird
   3. small hill
   4. field bird
   5. small anchor

[quote="Ignite168"][quote="sailingdog"]A lot of the time I just guess and get it right. For example, if the word ends in "ing" and starts with "post" or something, and there are 2 nouns and 2 verbs, then it's really easy.[/quote]

HOW CAN YOU TELL IF ITS A GERUND OR INFINITIVE OR PARTICIPLE WITHOUT CONTEXT? PLZ TCH ME HOW[/quote]

I have not encountered that word in the 30's.

Typing in all caps is |\|()()8 behavior.

lol I got to like 43 without dictionary, and 55 with online dictionary... that was kind of incomplete, though.
Anyway, 2800 grains donated...
like 10000 last time I did it though.


Practice for AMC 12:  About 1 hour of cramming before the test plus random intermittent problems.
Result: 130

Practice for AIME:  Irregular, but about six 3-hour tests plus 4 hours per week after school since late January I think (until test).
Result: 11

Practice for USAMO:  About 5 hours of cramming.
Result: 10

Practice for MathCounts:  About 2 hours.
Result:  5th written, booted at countdown, and still not believing I missed 2nd place by an answer in incorrect form.

[quote="probability1.01"]Practice for USAMO:  About 5 hours of cramming.
Result: 10[/quote]

You got a 10 on the USAMO while still of an age to be eligible for Mathcounts? I'm impressed. Do you get any other practice writing proofs for your curricular math classes, or does that all come from hanging out here on AoPS?

I have no formal training in proofs whatsoever, or for that matter, in any math for elementary school < math < college.  I mostly learned from this site how to write proofs.

MathCounts- Monday, Wednesday, Friday mornings from 7:15-815 AM starting at the beginning of the school year (mid-August)...then, after Christmas break we began afterschools as well. We practiced in the mornings like we had before, but began practicing from 3:30-5:00 PM til state. So, about 3 hrs a week for 4 months. Then, 10.5 hrs a week for 3 months...
Result- 1 at chapter, 15th at state. State was a disappointment.

lol i didnt practice and i got 4th at chapter and 11th at state in california

[quote="captcha1000"]well it depends on where you live..i mean 15th in california is a lot better than many people at nats so...im not really sure how hard it is to get in from "not here..."[/quote]

Indiana...

This is kind of odd, but I didn't even know I was going to participate in the chapter competion until the day before. And I amazingly made it to states.

[quote=yif man12]Hi everyone. I'm just wondering how much everyone practiced for Nationals or States, because i'm a little curious to see if practicing is worth it for high school math contests. so could you tell how long you practiced each week, and what your result was? thanks.[/quote]

Practicing is definitely worth it; I would suggest doing past Mathcounts test for practice. Good luck!

Here's a tip! Don't bump 16-year old threads.

Please look at the date before trying to bump threads and use common sense. Have a nice day!

[quote=alexdillonchen]Here's a tip:
If your parents force you to do math, practice for an hour and have fun for an hour! Being unhappy decreases productivity, which means that if you study for a whole day, it means you should've studied for 4 hours with breaks instead..[/quote]

How did you find this thread anyways?

Use  the advanced search function. And searched "oldest"

I feel like people do this intentionally.


1. Convert the rectangular equation $ y \equal{} x^2$ to a polar equation.

2. Express the polar equation $ r \equal{} \csc\theta$ as a rectangular equation.

[hide="1"]

$ r\sin\theta \equal{} (r\cos\theta)^2$

$ r \equal{} \frac {\sin\theta}{\cos^2\theta}$[/hide]

[hide="2"]

$ r\sin\theta\equal{}1$

$ y\equal{}1$[/hide]

maybe this should be moved to intermidiate although the process is simple the concepts are advanced  :(

Well, it is high school level...we learn it in our Algebra 2/Trigonometry class, which is mainly for freshmen and sophomores.


to prove:

if $x\equiv a\pmod m$ and  $x \equiv a \pmod n$ then  $x\equiv a \pmod{mn}$

It's only valid if $m$ and $n$ are coprime.
If $x\equiv a \pmod m$, and $x\equiv a \pmod n$, then $(x-a)|m$, $(x-a)|n$. It then follows that $(x-a)|mn$ as well, because if $(x-a)$ is divisible by two distinct coprime numbers, then its factors must at least include the product of the two numbers. So $x\equiv a \pmod{mn}$ as well.

I think you mean $m$ and $n$
not $a$ and $b$


geometry
trapezoid
perpendicular bisector
geometry proposed

let trapezoid ABCD such AB is parallel to CD.THE perpendicular bisector of AB intersect CD at P.the circumscribed of the triangale PAD and  PBC intersect at Q.prove that BD,AC,PQ are concur.

nobody here cant solve this easy geometry :mad:


limit
floor function
logarithms
integration
real analysis
real analysis unsolved

Let $ R(n) \equal{} 1 \plus{} \frac {n}{n \plus{} 1} \plus{} \frac {n}{n \plus{} 1}\frac {n}{n \plus{} 2} \plus{} ... \equal{} \sum_{k \equal{} 0}^{\infty}{\frac {n!n^k}{(n \plus{} k)!}}$. Prove that
\[ \lim_{n\to\infty}{\left(R(n) \minus{} \sqrt {\frac {\pi n}{2}}\right)} \equal{} \frac {1}{3}
\]

For technical reasons I'll rewrite $ R(n)$ in the form
$ R(n)\equal{}\frac{n!e^n}{n^n}\minus{}\sum_{k\equal{}1}^n\frac{n!n^{\minus{}k}}{(n\minus{}k)!}.$
Denote $ K\equal{}\lfloor n^{5/9}\rfloor$. For $ 1\leqslant k\leqslant K$ one has
\[ \log\frac{n!n^{\minus{}k}}{(n\minus{}k)!}\equal{}\left(n\plus{}\frac12\right)\log n\minus{}n\plus{}\log\sqrt{2\pi}\minus{}k\log n\minus{}\]
\[ \minus{}\left(n\minus{}k\plus{}\frac12\right)\left(\log n\minus{}\frac kn\minus{}\frac{k^2}{2n^2}\minus{}\frac{k^3}{3n^3}\plus{}O(n^{\minus{}16/9})\right)\plus{}\]
\[ \plus{}n\minus{}k\minus{}\log\sqrt{2\pi}\plus{}O(1/n)\equal{}\minus{}\frac{k(k\minus{}1)}{2n}\minus{}\frac{k^3}{6n^2}\plus{}O(n^{\minus{}7/9}),\]
hence
\[ \frac{n!n^{\minus{}k}}{(n\minus{}k)!}\equal{}\left(1\minus{}\frac{k^3}{6n^2}\right)e^{\minus{}\frac{k(k\minus{}1)}{2n}}(1\plus{}O(n^{\minus{}2/3})).\]
Particularly, for large $ n$
\[ \frac{n!n^{\minus{}K}}{(n\minus{}K)!}<e^{\minus{}\frac{n^{1/9}}3}.\]
Since $ a_k\equal{}\frac{n!n^{\minus{}k}}{(n\minus{}k)!}$ is decreasing for $ 1\leqslant k\leqslant n$ one has
\[ \sum_{k\equal{}K\plus{}1}^n\frac{n!n^{\minus{}k}}{(n\minus{}k)!}<n\cdot\frac{n!n^{\minus{}K}}{(n\minus{}K)!}<ne^{\minus{}\frac{n^{1/9}}3}\equal{}o(1),\]
hence
\[ \sum_{k\equal{}1}^n\frac{n!n^{\minus{}k}}{(n\minus{}k)!}\equal{}(1\plus{}O(n^{\minus{}2/3}))\sum_{k\equal{}1}^K\left(1\minus{}\frac{k^3}{6n^2}\right)e^{\minus{}\frac{k(k\minus{}1)}{2n}}\plus{}o(1).\]
In order to estimate the sum in the RHS apply the formula
\[ \sum_{a<k\leqslant b}f(k)\equal{}\int\limits_a^bf(x)\,dx\plus{}\rho(x)f(x)\Bigr|_{x\equal{}a}^b\minus{}\sigma(x)f'(x)\Bigr|_{x\equal{}a}^b\plus{}\int\limits_a^b\sigma(x)f''(x)\,dx,\]
where $ \rho(x)\equal{}\frac12\minus{}\{x\}$, $ \sigma(x)\equal{}\int\limits_0^x\rho(x)\,dx$.
Put $ a\equal{}1/2$, $ b\equal{}K\plus{}1/2$, $ f(x)\equal{}\left(1\minus{}\frac{x^3}{6n^2}\right)e^{\minus{}\frac{x(x\minus{}1)}{2n}}$. Then
$ \rho(x)f(x)\Bigr|_{x\equal{}a}^b\equal{}0;$
$ \sigma(x)f'(x)\Bigr|_{x\equal{}a}^b\equal{}O\left(\frac1n\right);$
$ \left|\int\limits_a^b\sigma(x)f''(x)\,dx\right|\leqslant\frac18\int\limits_0^\infty|f''(x\plus{}1/2)|\,dx\equal{}$
$ \equal{}\frac1{8\sqrt n}\int\limits_0^\infty\left|\frac{d^2}{dy^2}f(y\sqrt n\plus{}1/2)\right|\,dy\equal{}O\left(\frac1{\sqrt n}\right)$;
$ \int\limits_a^bf(x)\,dx\equal{}\int\limits_0^K\left(1\minus{}\frac{(x\plus{}1/2)^3}{6n^2}\right)e^{\minus{}\frac{x^2\minus{}1/4}{2n}}\,dx$.
For fixed $ \alpha>\minus{}1$
$ \int\limits_0^Kx^\alpha e^{\minus{}\frac{x^2}{2n}}dx\equal{}\int\limits_0^\infty x^\alpha e^{\minus{}\frac{x^2}{2n}}dx\plus{}o(1)\equal{}\frac12(2n)^{\frac{\alpha\plus{}1}2}\Gamma\left(\frac{\alpha\plus{}1}2\right)\plus{}o(1)$,
hence
$ \int\limits_a^bf(x)\,dx\equal{}\sqrt{\frac{\pi n}2}\minus{}\frac13\plus{}o(1)$.
Summarizing:
\[ R(n)\equal{}\frac{n!e^n}{n^n}\minus{}\sqrt{\frac{\pi n}2}\plus{}\frac13\plus{}o(1)\equal{}\sqrt{\frac{\pi n}2}\plus{}\frac13\plus{}o(1).\]

RIP used Euler-MacLaurin formula
[url= https://encyclopediaofmath.org/wiki/Euler-MacLaurin_formula] see here [/url]

[quote=ostriches88][quote=Moubinool]Nice
....[/quote]

pls dont bump threads that are 13 yrs old[/quote]

His/her account is 18 years old...

To mathboy I am a man.
In 2008 I did not see this problem on the forum. I saw it recently .


Let be $ f: [0,1]\rightarrow \mathbb{R}$ such that $ f(1)\equal{}1 \ , \ f(x)\ge 0 \ , \ (\forall)x\in [0,1]$ and if $ x,y,x\plus{}y\in [0,1]$ then $ f(x\plus{}y)\ge f(x)\plus{}f(y)$ . Prove that : $ f(x)\le 2x\ , \ (\forall)x\in [0,1]$

[hide="Hint"]By induction $ f(x)\geq2^nf(\frac{x}{2^n})$[/hide]


Twenty-one rectangles of size $ 3 \times 1$ are placed on an $ 8 \times 8$ chessboard, leaving only one free unit square. What position can the free square lie at?

[hide]
Number the sqaures of the chessboard as shown

1 2 3 1 2 3 1 2
2 3 1 2 3 1 2 3
3 1 [b]2[/b] 3 1 [b]2[/b] 3 1
1 2 3 1 2 3 1 2
2 3 1 2 3 1 2 3
3 1 [b]2[/b] 3 1 [b]2[/b] 3 1
1 2 3 1 2 3 1 2
2 3 1 2 3 1 2 3

Every tile covers exactly one 1, one 2, and one 3.  There are 21 1's, 21 3's, and 22 2's.  Therefore, the uncovered square must be a 2.  Now one might as why some 2's are in bold.  This is why.

2 1 3 2 1 3 2 1
3 2 1 3 2 1 3 2
1 3 [b]2[/b] 1 3 [b]2[/b] 1 3
2 1 3 2 1 3 2 1
3 2 1 3 2 1 3 2
1 3 [b]2[/b] 1 3 [b]2[/b] 1 3
2 1 3 2 1 3 2 1
3 2 1 3 2 1 3 2

Now if we number the squares this way (it's the same as the previous one, just each row is reversed), we similar have 21 1's, 21 3's, and 22 2's.  So the uncovered square must be a 2.  Therefore, the uncovered square must be a 2 on both numbered boards.  The only squares which are 2's on both boards are the ones with bold 2's on them.  Then these are the only possibilities for uncovered squares.  If one of them is possible, then all four are, and we can use a simple example to show one of them is possible. (the -'s and |'s connecting three X's in a row indicate a domino.  the O indicates the open square)

EDIT: the diagram I made isn't coming out well, but I'll just state here it's straightforward to find an example.
[/hide]

The following approach should work.  :wink:

[hide="Solution"]Label the squares as $ (1, 1), (1, 2), ..., (8, 1), ..., (8, 8)$.
Then associate to square $ (k, j)$ the number $ \varepsilon_{1}^{k}\varepsilon_{2}^{j}$ where both $ \varepsilon_{1}$ and $ \varepsilon_{2}$ are third roots of unity, but we consider one independent to the other.
Clearly, the sum of the numbers in each $ 3 \times 1$ or $ 1 \times 3$ is zero.

On the other hand we have
\[ \underset{1 \leq k,j \leq 8}{\sum}\varepsilon_{1}^{k}\varepsilon_{2}^{j}= \sum^{8}_{k = 1}\varepsilon_{1}^{k}\sum^{8}_{j = 1}\varepsilon_{2}^{j}= (-1)(-1) = 1\]

So we have to find those squares such that $ \varepsilon_{1}^{k}\varepsilon_{2}^{j}= 1$. These are $ (3, 3), (3, 6), (6, 3), (6, 6)$.[/hide]

WOW awesome solutions. I would not think of that. I suck at these types of problems...

@ Sepp You showed that: If a square is not covered, it must be one of those squares. But you did not show that: if it is one of those squares, a tiling exists.

Comment: we can just as well do a $ kp\pm 1$ by $ kp\pm 1$ grid [signs correspond] with $ p$ by 1 blocks where $ p$ is a prime; $ k$ is a positive integer.

[quote="Altheman"]
@ Sepp You showed that: If a square is not covered, it must be one of those squares. But you did not show that: if it is one of those squares, a tiling exists.
[/quote]
Indeed, I forgot to write it down. Thank you.  :)


If $\sum_{i=1}^{5}x_{i}=1$ and $x_{i}$ are non-negative numbers, then
$x_{1}x_{2}x_{3}+x_{2}x_{3}x_{4}+x_{3}x_{4}x_{5}+x_{4}x_{5}x_{1}+x_{5}x_{1}x_{2}\leq\frac{1}{25}.$
See here too:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=75703

It's also in my book (but in fact, I don't know that Walther Janous is the author of this problem, because I only know it from a high-school test some years ago). I solved it by mixing variable, but I don't really like this solution.

[quote="hungkhtn"]I solved it by mixing variable, but I don't really like this solution.[/quote]
There is a very nice proof of this inequality.(AM-GM only!)

W.Janous's inequality:
If $x_{1}+x_{2}+...+x_{n}=1$ where $x_{i}$ are non-negative real numbers
and $2\leq k< n$ then
$x_{1}x_{2}...x_{k}+x_{2}x_{3}...x_{k+1}+...+x_{n}x_{1}...x_{k-1}\leq max\{\frac{1}{k^{k}},\frac{1}{n^{k-1}}\}.$
If I don't wrong it's open.
In our case $n=5$ and $k=3.$ :wink:

[quote="ehku"]Can you post solution ?
It's a open question (for me).
Thanks ![/quote]

I don't think such a problem is open. If I'm not wrong, I had once seen its solution (or it must be false, of course  :lol: ).

How to prove it by AM-GM, Arqady? Could you show it?

Arquady I think it's time to post your perfect solution

Problem 1. If $x_{1}+x_{2}+...+x_{n}=1$ where $x_{i}$ are non-negative real numbers  and $2\leq k< n$ then \[x_{1}x_{2}...x_{k}+x_{2}x_{3}...x_{k+1}+...+x_{n}x_{1}...x_{k-1}\leq \max\left\{\frac{1}{k^{k}},\frac{1}{n^{k-1}}\right\}.\] Can anyone let me know the source, or reference of this inequality?[/b]

[quote="pvthuan"]Problem 1. If $x_{1}+x_{2}+...+x_{n}=1$ where $x_{i}$ are non-negative real numbers  and $2\leq k< n$ then
\[x_{1}x_{2}...x_{k}+x_{2}x_{3}...x_{k+1}+...+x_{n}x_{1}...x_{k-1}\leq \max\left\{\frac{1}{k^{k}},\frac{1}{n^{k-1}}\right\}. \]
Can anyone let me know the source, or reference of this inequality?[/quote]
It's Crux 1754.

[quote="pvthuan"]Is it still OPEN?[/quote]
I know that it's true for
 $(k,n)\in\{(2,n),(3,n),(4,n),(k,km),(k,km+1)\},$
where $n,k,m$ are natural numbers such that $2\leq k< n.$
See here:
http://kvant.mccme.ru/1996/02/
in Russian.

Does anybody know if this problem is still open except cases $(k,n)$: $(2,n)$; $(3,n)$; $(4,n)$, $n\ne7$; $(k,k+1)$, $k\ge5$; $(k,km)$, $(k,km+1)$, $k\ge5$, $m\ge2$?
If no then what cases is it proved for?
I asked because 15 years passed since W.Janous stated his problem in "Crux Mathematicorum" and 11 years passed since L.Kurliandchik published his article in "Kvant".


Ask a weird Q

and the next person answers it, and asks another weird Q

Can one get drunk on pure orange juice?

No.

In honor of President's day, I'd like to ask: who was the hottest U.S. president?  (At any point in time, preferably his hottest years.  Do the judging solely on hotness; disregard his achievements, if any.)

Abraham Lincoln

Is this thread supposed to be serious, because the previous question definetly wasn't.

No

Why when you go to sleep and wake up on the same day you think its already the next day?

Because I just do.

Is orange juice made out of oranges?

If it's Florida Orange Juice!

why do you sometimes wake up at a hotel and not know where you are and get all panicky?

because you have no clue how you got there

How do you know that chess isnt transitive?

because your alter ego that is sitting on your right shoulder, is arguing with your other alter ego that is sitting on your left shoulder, and since they cant come to an agreement, you feel guilty.

can you figure out if you snore while you sleep without recording yourself or having someone tell you ?


A rocket is rising vertically from a point on the ground 100 meters from an observer at ground level. The observer notes that the angle of elevation is increasing at a rate of 12 degrees per second when the angle of elevation is 60 degrees.

Find the speed of the rocket at that instant.

i got 83.7758 m/s m/sec.

is this correct?


[quote="Jan"]Solve the following system, when $x,y,z \in \mathbb{R}$:
\[\{ \begin{array}{c}x^{2}+2y^{2}+3z^{2}=36\\ 3x^{2}+2y^{2}+z^{2}=84\\ xy+yz+xz =-7 \end{array}\]
Note: The question was posted in Pre-Olympiad, but it remained unanswered so far.[/quote]
1),2) give $x^{2}+y^{2}+z^{2}=30$ therefore $(x+y+z)^{2}=30-2*7=16$.
If $x,y,z$ is solution, then $-x,-y,-z$ solution too. Therefore we can consider only case $x+y+z=4$. From 2)-1) we get $x^{2}-z^{2}=24$.
From $x^{2}+z^{2}+(4-x-z)^{2}=30$ we get $uv=24,u^{2}+v^{2}-8u=14,u=x+z,v=x-z.$ It give $u^{4}-8u^{3}-14u^{2}+576=0.$

here's a hint:
Summing up (1) and (2) we obtain that $x^{2}+y^{2}+z^{2}=30$ and using (3) we have $x^{2}+y^{2}+z^{2}+2(xy+yz+zx)=16$ $\leftrightarrow$ $|x+y+z|=4$

Rust: Ok, but the system is intended to be solved without a calculator. I don't think we can easiley solve that quartic equation. A more clever solution is needed.  :wink:
Besides, the quartic equation you give is not correct, you must have made an error somewhere.

Svejk: Indeed, that's easy to figure out... Now finish the solution if you can please!  :wink:

Please don't spend any more time on this problem. As you suggested, it seems that there are some ugly solutions apart from the integer ones.

I contacted the author of the booklet where I found this problem to ask for the official solution, because I am absolutely sure that a calculator is not intended to be used. I'll keep you informed...


For which natural numbers $ n$ do there exist rational numbers $ a$ and $ b$ which are not integers such that both $ a \plus{} b$ and $ a^n \plus{} b^n$ are integers?

I have posted in here .
http://www.mathlinks.ro/viewtopic.php?t=200222


In Chapter 1, it proves that $ a \plus{} x \equal{} a$ has only one solution ($ x \equal{} 0$) as follows:

$ a \plus{} x \equal{} a \\
( \minus{} a) \plus{} (a \plus{} x) \equal{} \minus{} a \plus{} a \equal{} 0 \\
( \minus{} a \plus{} a) \plus{} x \equal{} 0 \\
0 \plus{} x \equal{} 0 \\
x \equal{} 0.$

Now one thing bugs me, he mentioned all properties (such as additive inverse and etc.) but he forgot to mention that if $ a \equal{} b$, then $ a \plus{} c \equal{} b \plus{} c$. So is the validity of step 2 implied because he assumed that you could perform the same operation on each side of an equation given that it is equal?

EDIT1: He did the same for multiplying both sides of an equation by some real $ c$, again is it implied?

[quote="TachyonPulse"]but he forgot to mention that if $ a \equal{} b$, then $ a \plus{} c \equal{} b \plus{} c$. So is the validity of step 2 implied because he assumed that you could perform the same operation on each side of an equation given that it is equal?[/quote]  This is essentially the definition of "=" -- the meaning of $ a \equal{} b$ is that if you ever have an expression with $ a$ in it, you can replace that with $ b$ without changing the value.  They're the same.


Dear xdragon, I think the problem should be

Let $ x,y,z $ be positive number which satisfy :
$ (x + y + z)^3 = 32xyz $
Find the max and min of the expression $ \frac{x^4+y^4+z^4}{(x+y+z)^4} $

This one is 5th problem of VMO 2004. I remember my teacher,Mr Tran Nam Dung, posted the solution for this problem in March 2004.The problem is beautiful with such an important idea : assume $ xyz=4 $

This idea can also be useful to solve your original problem.


$ x\equal{}y\equal{}1$ and $ z<0$ will give $ \minus{}\infty$ when $ z$ is close to $ 0$.

okay, but what if $ x,y,z$ are positive reals?

[quote="orl"]Let $ x, y, z$ be real numbers such that $ x \plus{} y \plus{} z \geq xyz$.  Find the smallest possible value of $ \frac {x^2 \plus{} y^2 \plus{} z^2}{xyz}$.[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=17142

hello, if the variables are positive the searched minimum is $ \sqrt{3}$.
Sonnhard.

Let $ x \plus{} y \plus{} z \equal{} kxyz$

Then

$ \sqrt {(\frac {x^2 \plus{} y^2 \plus{} z^2}{xyz})^2} \equal{} \sqrt {\frac {x^4 \plus{} y^4 \plus{} z^4 \plus{} 2x^2y^2 \plus{} 2y^2z^2 \plus{} 2z^2x^2}{x^2y^2z^2}} \equal{} \sqrt {\frac {x^4 \plus{} y^4 \plus{} z^4 \plus{} 2x^2y^2 \plus{} 2y^2z^2 \plus{} 2z^2x^2}{(xyz)(\frac {x \plus{} y \plus{} z}{k})}}$

Hence, we need to minimize

$ \frac {x^4 \plus{} y^4 \plus{} z^4 \plus{} 2x^2y^2 \plus{} 2y^2z^2 \plus{} 2z^2x^2}{(xyz)(\frac {x \plus{} y \plus{} z}{k})}$

I claim that $ x^4 \plus{} y^4 \plus{} z^4 \plus{} 2x^2y^2 \plus{} 2y^2z^2 \plus{} 2z^2x^2 \ge 3xyz(x \plus{} y \plus{} z)$

Which is true by 2 trivial applications of muirheads inequality.

Hence, since $ 3xyz(x \plus{} y \plus{} z) \ge 3xyz(\frac {x \plus{} y \plus{} z}{k})$, we have:

$ \frac {x^4 \plus{} y^4 \plus{} z^4 \plus{} 2x^2y^2 \plus{} 2y^2z^2 \plus{} 2z^2x^2}{(xyz)(\frac {x \plus{} y \plus{} z}{k})}\ge 3$

equality will hold when $ x \equal{} y \equal{} z \equal{} \sqrt {3}$. The final minimum is the square root of this answer, which is $ \sqrt {3}$.

Actually to show that $x^2+y^2+z^2\ge\sqrt{3}xyz$ appeared on the Balkan 2001 paper http://www.artofproblemsolving.com/Forum/resources.php?c=1&cid=19&year=2001&sid=5c22a3c467b0fb6290e11cdb1f19768c (this problem is Chinese Olympiad 2001).


function
vector
calculus
derivative
trigonometry
real analysis
real analysis unsolved

Let $ U$ be an open subset of $ \mathbb R^n$. For any nonegative integer $ m$, let $ C^m(U)$ be the vector space of all functions $ f$ which, together its derivatives $ D^\alpha f$ are continuous on $ U$, for every multi-index  $ |\alpha|\leq m$. Let $ C^m(\overline{U}\,)$ be the vector space of all functions $ f$ which, together its derivatives $ D^\alpha f$ are bounded and continuous on $ U$  for every multi-index  $ |\alpha|\leq m$.

(a) Show that $ C^m(\overline{\mathbb R^n}\,)\neq C^m(\mathbb R^n)$

(b) Give an open and bounded subset $ U$ so that $ C^m(U)\neq C^m(\overline{U}\,)$.

$ f(x)\equal{}e^{x}$ is continuous on all of $ \mathbb{R}$ (this is an open subset of $ \mathbb{R}$, namely itself, which is perfectly legal), but it is not bounded.  In fact, it's infinitely differentiable so $ f\in C^{\infty}(\mathbb{R})$ but $ f\notin C^{\infty}(\bar{\mathbb{R}})$.

I'm not sure I interpreted your notation correctly, though, so ignore my response if it isn't what you're looking for.

[quote="JRav"]I'm not sure I interpreted your notation correctly, though, so ignore my response if it isn't what you're looking for.[/quote]

Yes you are right, thanks  :)

Alright then, since I've apparently got your notation correct, for the second one, try $ U\equal{}(\minus{}\pi/2,\pi/2)$ and $ f(x)\equal{}\tan{x}$.  This function is everywhere continuous and differentiable on that interval, but is not bounded.


American Regions Math League is a 4-part math competition that takes place on four campuses nationwide, the closest to us NJ-ers being Penn State. Of the four rounds, there's a Team Round (the entire team gets a set of 10 questions to work on, 20 mins), a Power Round (10 questions, involving a certain concept, 1 hour), an Individual Round (8, or next year 10, questions, with 10 minutes per pair of questions), and a Relay Round (3 or 6 minutes for groups of 3 people, where the 2nd person's question will be based on the answer to the 1st, etc).

NJ sends teams based on location though; one's from Bergen, then AAST, and one from Central Jersey (which picks its members exclusively based upon a regional contest-CJML). And we failed pretty badly ... our CJ A team got 20th in the B division, and our individual scores were pretty low ...

thanks
i don't know my math competitions :lol:

Wait isn't Bergen and AAST the same area? What's the distinction?

I'm not sure; I found out [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=468309#468309]here[/url].


vector
arithmetic sequence
combinatorics unsolved
combinatorics

Xi      1=<i=<n are n real numbers so that |Xi|=<1.prove  that the number of the subsets of the set {Xi|1=<i=<n}which the sum of their members is in the [0,1]  is less than the choose of [n/2] from n.

I am not sure that it's true:

consider $x_{i}=\frac{1}{n}$ so all of the subset of $\{x_{1},x_{2},...,x_{n}\}$ has that property. and number of this subset is $2^{n-1}$ whcih is bigger

than $n \choose [\frac{n}{2}]$

it means that your problem is wrong.
heidari daram sooti midam? :|  :huh:

The statement is false : just consider $n$ real positive numbers which add up to $\frac{1}{2}$. In that case, each subset is ok.

I think the condition is $|x_{i}| \geq 1.$
And the result follows from Sperner's theorem about antichains.

Pierre.

excuse me because saying the problem in a wrong way.the corect form of the problem is that when |Xi|>=1.
thank you.   :P

This question is known as the "Littlewood-Offord" problem, and there are many fascinating extensions/generalizations of it.

For example, Daniel Kleitman proved that the result still held true if you replaced the $n$ real numbers by $n$ vectors each with length at least 1.

Gabor Halasz used Fourier Analysis to prove several extensions including:
-If $|x_{i}-x_{j}|>1$ for each $i$ and $j$, then the maximum number of sums in the interval is bounded by $C*2^{n}*n^{-3/2}$ for some $C$ (the bound in the original case looks like $C*2^{n}*n^{-1/2}$)

-If the $x_{i}$ are in a sense "essentially" $d$-dimensional (located in $R^{d}$, and not too concentrated near any $d-1$-dimensional subspace), then the number drops to $C*2^{n}*n^{-d/2}$.

More recently, Terence Tao and Van Vu have shown sort of an "inverse" version of this theorem: If a large number of the subset sums (say at least $2^{n}* n^{-d}$ for some fixed $d$) are equal, then the $x_{i}$ must be (mostly) part of a structure called a generalized arithmetic progression.

What this means is that there are real numbers $a_{0}, a_{1}, \dots a_{d}$ such that almost all of the $x_{i}$ look like $a_{0}+y_{1}a_{1}+y_{2}a_{2}+\dots y_{d}a_{d}$ with the $y_{i}$ being not too large integers.  $d=1$ corresponds to your standard intuition as to what an arithmetic progression looks like.

hi
thank you for your useful information,but you didnt put a solution for that.
Does it mean that its solution is difficault or something?if it is not could you please send the exact solution for me.
Thank you very much again. :wink:

As a note, I think the question as stated may not be true.  Are you sure it's the closed interval and not the open one?  Assuming you mean the open (or half open) interval, here's a hint:

Suppose there were two subsets $S$ and $T$ of the $x_{i}$ satisfying the following conditions:

1. Every negative number in $S$ is also in $T$.
2. Every positive number in $T$ is also in $S$.
3. The sum of the numbers in $S$ is in [0,1).

What can you say about the sum of the numbers in $T$?

hi kevin.
I think you right.I realized the same mistake in the problem after I send the problem.But maybe the interval should be open of cuorse other wise the problem may not be true.my teacher gave me the problem and as a matter of fact the interval is closed in it.but there can be something else too.may be  the problem just wants different sums.but this is a bit ridicules.I know.any way I should understand the corect form of the problem.

If $x_{1}=-1, x_{2}=1$ then $x_{2}=1, x_{1}+x_{2}=0$ and $2$ is not less than $\binom{2}{1}$. So I prove this problem:

$x_{i}$ are $n$ real numbers so that $|x_{i}|\geq 1$. Prove that the number of the subsets of the set $\{ x_{i}|1\leq i\leq n\}$ which the sum of their members is in the $[0,1)$ is not greater than the choose of $[n/2]$ from $n$.

Also Spener's lemma.

We notice that if $S_{1}=x_{11}+x_{12}+\cdots+x_{1i}, S_{2}=x_{21}+x_{22}+\cdots+x_{2j}$ with $S_{1}, S_{2}\in [0,1)$, we will find that $\{ x_{11}, x_{12}, \cdots ,\ x_{1i}\}$ and $\{ x_{21}, x_{22}, \cdots ,\ x_{2j}\}$ do not include each other as $x_{k}\geq 1$. Then it is clear.

You need to be a little more careful than that.

If for example the numbers are 1.4, -1.3, 1.2, -1.1, then the sets $\{1.4,-1.3\}$ and $\{1.4,-1.3, 1.2,-1.1\}$ both sum in $[0,1)$, and one set contains the other.

IF you're still looking at trying to finish this one off, I'd recommend starting with Libra_Gold's solution, but picking different sets to apply Sperner's Lemma to.

Let $S$ be the subset you're summing over.  The problem with applying Sperner's Lemma directly to $S$ is that adding an element to $S$ will have a different effect on the sum depending on whether the element is positive or negative: Adding positive elements increases the sum, but adding negative elements decreases it.

However, adding positive elements to $S$ has the same effect as [b]removing[/b] negative elements.  If $T$ is whatever set we apply Sperner's Lemma to, therefore, we want adding a positive number to $S$ and removing a negative number from $S$ to have a similar effect on $T$.


Let $ X$ be a normally distributed random variable.
Does there exist a nonlinear transformation $ t$ on $ X$, such that the random variable $ t(X)$ will be normally distributed?

Thanks.

I think, yes.

[hide="idea"]Just make $ Y\equal{}|X|\cdot S(X)$[/hide]

[hide="details"] S(X) is equal to plus one if ($ X>1$ or $ X\in (\minus{}1;0)$) and to minus one if ($ X\in(0;1)$ or $ X<\minus{}1$)[/hide]


En un triangulo $ABC$, $CA \neq CB$, los puntos $A_{1}$ y $B_{1}$ son los puntos de tangencia de las circunferencias exinscritas relativas a los lados $CB$ y $CA$, respectivamente, e $I$ es el incentro.  La recta $CI$ intersecta a la circunferencia circunscrita al triangulo $ABC$ en el punto $P$. La recta que pasa por $P$ y es perpendicular a $CP$ intersecta a la recta $AB$ en el punto $Q$. Pruebe que las rectas $QI$ y $A_{1}B_{1}$ son paralelas.

[quote]En un triangulo $ ABC,$ $ AB \neq AC,$ los puntos $ B_{1}$ y $ C_{1}$ son los puntos de tangencia de las circunferencias exinscritas relativas a los lados $ AC$ y $ AB,$ respectivamente e $ I$ es el incentro.  La recta $ AI$ intersecta a la circunferencia circunscrita al triángulo $ ABC$ en el punto $ P.$ La recta que pasa por $ P$ y es perpendicular a $ AP$ intersecta a la recta $ BC$ en el punto $ Q.$ Pruebe que las rectas $ IQ$ y $ B_{1}C_{1}$ son paralelas.[/quote]
Usemos coordenadas baricentricas con respecto a $ \triangle ABC.$ Entonces:

$ B_1 \equal{} (p \minus{} a: 0: p \minus{} c) \ , \ C_1 \equal{} (p \minus{} a: p \minus{} b: 0).$

Luego la recta $ B_1C_1$ tiene por punto del infinito

$ T_{\infty} \equal{} ((p \minus{} a)(b \minus{} c): b(p \minus{} b): \minus{} c(p \minus{} c))$

Siendo $ P$ es punto medio del arco $ BC$ del circuncirculo, este tiene coordenadas

$ P \equal{} ( \minus{} a^2: b(b \plus{} c): c(b \plus{} c))$

La recta $ \mathcal{L}_a$ que pasa por $ P$ perpendicular a $ AP$ es paralela a la bisectriz exterior de $ \angle BAC$ por ello tienen el mismo punto del infinito $ (c \minus{} b: b: \minus{} c).$ Su ecuación es entonces:

$ \mathcal{L}_a \equiv bc(b \plus{} c)x \minus{} cS_Cy \minus{} bS_Bz \equal{} 0.$ Esta corta a $ BC$ en $ Q \equal{} (0: bS_B: \minus{} cS_C)$

$ \Longrightarrow \ IQ \equiv abcx \minus{} cS_Cy \minus{} bS_Bz \equal{} 0.$ Las coordenadas de su punto del infinito son:

$ R_{\infty} \equal{} (cS_C \minus{} bS_B: b(S_B \plus{} ac): \minus{} c(S_C \plus{} ab)).$ Esta la podemos reescribir teniendo presente las identidades:

$ cS_C \minus{} bS_B \equal{} 2p(p \minus{} a)(b \minus{} c) \ , \ S_B \plus{} ac \equal{} 2p(p \minus{} b) \ , \ S_C \plus{} ab \equal{} 2p(p \minus{} c)$

$\ R_{\infty} \equal{} ((p \minus{} a)(b \minus{} c): b(p \minus{} b): \minus{} c(p \minus{} c)) \Longrightarrow \ T_{\infty} \equiv R_{\infty} \Longrightarrow \ IQ \parallel B_1C_1$


let $ \alpha \neq \beta \plus{} 2k\pi$ and $ \beta \neq \pi\plus{}2 k\pi$
let  $ cos(\alpha)\plus{}cos(\beta)\equal{}1$  and $ sin(\alpha)\plus{}sin(\beta)\equal{}1$
prove that  $ \tan(\frac{\alpha}{2})\plus{}\tan(\frac{\beta}{2})\equal{}1$

We have:

$ 1 \equal{} \frac {\sin(A) \plus{} \sin(B)}{\cos(A) \plus{} \cos(B)} \equal{} \frac {2\sin\frac {A \plus{} B}{2}\cos\frac {A \minus{} B}{2}}{2\cos\frac {A \plus{} B}{2}\cos\frac {A \minus{} B}{2}} \equal{} \tan(\frac {A \plus{} B}{2})$

We have:

$ \tan(\frac {A}{2}) \plus{} \tan(\frac {B}{2}) \equal{} \frac {\sin(\frac {A \plus{} B}{2})}{\cos\frac {A}{2}\cos\frac {B}{2}}$ so combining with above we should prove that:

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

so we should prove that $ \sin\frac {A}{2}\sin\frac {B}{2} \equal{} 0$

I am afraid that this one does not always hold.
Namely after squaring and summing two conditions (equations) we come to conclusion that:
$ \cos(A)cos(B) \equal{} \minus{} \sin(A)sin(B) \equal{} \minus{} 4\sin\frac {A}{2}\sin\frac {B}{2}\cos\frac {A}{2}\cos\frac {B}{2} \equal{} 0$

so one of cosines of $ A$ and $ B$ is 0 so second one is 1. Now it is easy to find that also one $ \sin$ is 0 and the other is 1 which leads to just two special cases and it means that these are only cases in which desired equality holds.

Hope that I have not made some stupid mistake :(


analytic geometry
number theory proposed
number theory

Let $ a,b,c,d$ be an integers,such that $ ad\minus{}bc\equal{}k>0$, $ (a, c)\equal{}1$, $ (b, d)\equal{}1$.Prove that there exist exaclty $ k$ pairs $ (x_1,x_2)$ of real numbers,such that $ 0\leq x_1\leq x_2<1$,such that  numbers $ ax_1\plus{}bx_2$ and $ cx_1 \plus{}dx_2$ are both integers.

[color=green]Hint: Consider the Descartes's coordinate and the integer point in this one.[/color]

See this post:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=124726[/url]


Hello people,

I have an inquiry to the administrators. What topics should be discussed in the forum? It says in the presentation

[b]For serious discussion of other topics.[/b]

Nevertheless, I proposed a topic, may be one that could give a lot of controversy, and I receive this responce

[quote="Valentin Vornicu"]
Yes you are allowed to discuss serious stuff, but not that kind of serious stuff. Topic closed!
[/quote]

I first felt unconfortable, because I really think that everything is up to discussion, specially in a forum like this that have very smart and educated members.

After a while I reallize that I gave for a fact a point that may not be. Everything is up to discussion between two or more parties, if and only if they are capable to have a civilized conversation, without lossing the respect for the others' point of view, even when it is totally opposite to theirs, and that atmosphere is hard (almost impossible) to keep in an open forum like this, specially given the differences of background and age between the members.

Therefore, I agree with Valentin that there have to be rules in what you can post and what you should find another place (probably out of this forum) to post, or just keep it by yourself, and I agree that may be the topic I posted would be in the second of third cathegory I gave.

Now, do such rules exist? I mean, I would not like to be censurated again (it is somehow embarassing), and I am completely willing to follow rules of posting, if they are clear.

Best regards,

I suggest reading section 5 of our terms of use:
[url]http://www.artofproblemsolving.com/Legal/AoPS_L_Comm_Terms.php[/url]

Hello,

[quote="peter"]
Maybe the announcement "Usage of this forum" will help, although it isn't very clear on what topics can and can't be discussed.
[/quote]

Peter, could you please let me know where is it, because I have looked for it for at least twenty minutes, and either for my options it is not displayed, or I am in very uncareful even when I try to be careful, and I don't see the link anywhere (Not directly in mathlinks, but I haven't looked in AoPS site).

DPatrick, I carefully read the point five of the terms of use of AoPS, and it gives a few (quite obvios) limitations according to the nature of the topics, the content itself, but not about the kind of topics itself. Any way, it could be nice to discuss about it (Yes, I am a person who likes to discuss about anything, as I said, trying on every moment to mantain a respectful atmosphere!)

Best regards,

[quote="djimenez"] I proposed a topic, may be one that could give a lot of controversy, and I receive this responce

[quote="Valentin Vornicu"]
Yes you are allowed to discuss serious stuff, but not that kind of serious stuff. Topic closed!
[/quote] [/quote]

What was the topic?
I am curious to know what qualifies as the wrong kind of serious stuff here.

I see that, for example, rrusczyk mentioned Larry Summers' apparently politically incorrect comments on women in his blog, and that there are threads on religion versus science.

Well fleeting_guest, the topic was already deleted, at least I can't find it right now. Nevertheless, at least from my perspective I didn't use at all any politically incorrect comment, I just posted a topic that here in the US is generating controversy.

Nevertheless, I do consider there should be an explicit set of what is considered postable and what is not. I would say that the administrators should be the ones having the last word on this aspect, but I also consider that we could discuss a little about it, or at least I think the site is far from been totalitaristic and usually cares about the opinion of the users.

Best regards,

Sorry,

I feel so stupid. I really spent a lot of time looking for the "Usage of this forum" that fleeting_guest mentined, and I didn't find it, and it was just in front of my nose. It is a little frustrating, therefore, nevermind about what I said before!

Best,

I don't remember exactly what that topic was about, but I think it was someting R-rated as people say in movies. Please be aware that there are numerours kids (and by kids I mean ages less than 14-15) on the site reading these topics.

We will think about a solution for discussing any kind of topics, but only in an appropriate context, and then share it here.

Yeah - it's best to make topics understandable and appropriate for all people who use the forum.


function
calculus
derivative
calculus computations

If you are given the points on the first andsecond derivative function and they ask you for the original function, how do you find it?
Example: If f''(x)=6
f'(2)=7
f(2)=9
f(0)=?

9
7
6
2
0

do you mean $ f''(2)\equal{}6$ instead of $ f''(x)\equal{}6$?


[quote="conceive"]Prove or disprove:

$a,b,c>0$

$\frac{a^2+2b^2}{2a+b}+\frac{b^2+2c^2}{2b+c}+\frac{c^2+2a^2}{2c+a}\ge a+b+c$[/quote]
It's too easy. We have: $\sum(\frac{a^2+2b^2}{2a+b}-b)=\sum\frac{(a-b)^2}{2a+b}\ge 0$


function
integration
geometry
geometric transformation
reflection
calculus
complex analysis

Today i had my final exam in Complex Analysis, and i couldnt answer two questions.   I hope anyone can help me :

1) Let $U$ be a connected open set that contains a line segment $\gamma$ of the real axis.   Let $f$ be analytic over $U$ and suppose $f$ can be extended continuously to $\gamma$, and this extension is zero over $\gamma$.   Show $f=0$ over $U$. ( I was trying to prove that $f$ was analytic over $\gamma$, but i couldnt do it, is that possible to prove? )

2) Let $U$ be a connected open set and $F$ a family of holormorphic functions defined on $U$ such that \[\int \int_{U}|f(z)|dxdy < 1\] .  Is $F$ normal??  (I could prove itis true when $U$ is the open disc centered at 0 and radius 1, but the general case i couldnt)

1) I think you meant that the [b]boundary[/b] of $U$ contains $\gamma$. Use the Schwarz reflection to extend $f$ across $\gamma$ so that $\gamma$ falls into the interior of the extended domain.

2) Yes, because $F$ is uniformly bounded on compact subsets of $U$. If $K$ is such a subset, then its distance to $\partial U$ is some positive number $d>0$. For any $z_{0}\in K$ consider the circle $C_{r}$ with center $z_{0}$ and radius $r\in (0,d)$. By the Cauchy formula $|f(z_{0})|\le \frac{1}{2\pi}\int_{0}^{2\pi}|f(z_{0}+re^{it})|dt$. Multiply both sides by $r$, integrate in $r$, and use Fubini's theorem. The double integral on the right will be bounded by $1$.

when you take $r \in (0,d)$ what is the problem taking $r=d$? I don't see whether there's a problem or not, so can i take that value for $r$ ?.

Anyways, thanks for ur help

Forget my previous post, i dont know what i was thinking, i could see i was wrong


You have two urns, each of which has both black and white balls. The probability of drawing a white ball from the first and second urns are respectively $\frac{1}{3}$ and $\frac{1}{2}$. You will now choose balls from these urns.

Now, with probability $\frac{1}{4}$ you will initially choose a ball from the first urn and with probability $\frac{3}{4}$ you will initially choose a ball from the second urn. If you choose a white ball at any turn, you also choose the next ball from that urn. But if you choose a black ball at any turn, then you choose the next ball at the other urn.

Find the probability that you choose the 5th ball from the first urn.

Maybe I'm missing something, but don't we need to know something about the total numbers of balls?

Not if the choices are with replacement.


If the Smith family drinks milk daily, a carton of milk will last them for 12 days. If the Smiths want milk at home at least $ \frac 45$ of the time, they must purchase milk every how many days?

Just divide $ 12$ by $ \frac45$ for an answer of $ \boxed{15}$.


algebra
function
domain
superior algebra
superior algebra unsolved

Let R be a ring with more than one element.If ax=b has a solution for all non zero a and b then show that R is a division ring.

[quote="rahul1990gupta"]
Let R be a ring with more than one element. If ax=b has a solution for all non zero a and b then show that R is a division ring.[/quote]

$ R$ is a domain: suppose $ ab\equal{}0,$ for some non-zero elements $ a,b \in R.$ We have $ ax\equal{}b, \ by\equal{}x,$ for some $ x,y \in R.$ Thus $ b\equal{}ax\equal{}aby\equal{}0.$ Contradiction!

$ R$ has $ 1$: let $ 0 \neq a \in R$ and choose $ x_0 \in R$ such that $ ax_0\equal{}a.$ Then $ a(x_0a\minus{}a)\equal{}0$ and so $ x_0a\equal{}a.$ Thus for any $ b \in R: \ a(x_0b\minus{}b)\equal{}(bx_0\minus{}b)a\equal{}0.$

Hence $ bx_0\equal{}x_0b\equal{}b$ and so $ x_0\equal{}1_R.$

Now for any $ 0 \neq c \in R,$ there exists $ x \in R,$ such that $ cx\equal{}1,$ and so $ R$ is a division ring.