geometry
trigonometry
number theory
least common multiple
circumcircle
angle bisector
cyclic quadrilateral

we have:

$ \triangle AKL\equal{}\triangle AML$

$ \Rightarrow \angle ALM\equal{}\angle ALK\equal{}\alpha$

so in quadrilateral $ AKLM$,the diagonal $ AL$ bisects the angles $ \angle A,\angle L$ of this quadrilateral,so $ KM\perp AL$ hence $ KM\perp AN$.

now let $ H$ be the intersection point of $ KM,AN$ now $ \triangle LMK$ is isosceles hence $ HM\equal{}HK$ now we have:

$ S_{AKNM}\equal{}S_{AKN}\plus{}S_{AMN}\equal{}\frac 12.KH.AN\plus{}\frac 12.MH.AN\equal{}\frac 12.KM.AN$

so:

$ \boxed{S_{AKMN}\equal{}\frac 12.KM.AN} (I)$

now we know that $ LM\equal{}LK\equal{}h$ so in the right-angled triangle $ \triangle LMH$:

$ \sin \alpha\equal{}\frac{HM}{LM}\equal{}\frac{KM}{2h}$

$ \Rightarrow h\equal{}\frac{KM}{2\sin \alpha}$

now note that:

$ S_{ABC}\equal{}S_{AKLM}\plus{}S_{BKL}\plus{}S_{LMC}$

$ \equal{}\frac 12.KM.AL\plus{}\frac 12.h.BK\plus{}\frac 12.h.MC$

$ \equal{}\frac 12.KM.AL\plus{}\frac 12.h.(BK\plus{}MC)$

$ \equal{}\frac 12.KM.AL\plus{}\frac 12.\frac{KM(BK\plus{}MC)}{2\sin \alpha}$

$ \Rightarrow\boxed{S_{ABC}\equal{}\frac 12.KM.AL\plus{}\frac 12.\frac{KM(BK\plus{}MC)}{2\sin \alpha}} (II)$

now according to (I),(II) in order to complete the proof it's sufficient to show that:

$ \frac 12.AN.KM\equal{}\frac 12.KM.AL\plus{}\frac 12.\frac{KM(BK\plus{}MC)}{2\sin \alpha}$

$ \iff LN\equal{}\frac{BK\plus{}MC}{2\sin \alpha}$ (*)

now in right-angled triangle $ \triangle BKL$ we have:

$ \cos B\equal{}\frac{BK}{LB}\Rightarrow BK\equal{}LB.\cos B$

but $ AL$ is the angle bisector of $ \angle BAC$ so we have:

$ BL\equal{}\frac{ac}{b\plus{}c}$

where $ a\equal{}BC,b\equal{}CA,c\equal{}AB$,now we have:

$ BK\equal{}LB.\cos B\equal{}\frac{ac.\cos B}{b\plus{}c}$

now in the other hand by $ \cos$ law in triangle $ \triangle ACB$ we have:

$ ac.\cos B\equal{}\frac{a^2\plus{}c^2\minus{}b^2}2$

hence:

$ BK\equal{}\frac{a^2\plus{}c^2\minus{}b^2}{2(b\plus{}c)}$

with a similar way we can show that $ MC\equal{}\frac{a^2\plus{}b^2\minus{}c^2}{2(b\plus{}c)}$

so (*) would be equivalent to:

$ (*)\iff LN\equal{}\frac{\frac{a^2\plus{}c^2\minus{}b^2}{2(b\plus{}c)}\plus{}\frac{a^2\plus{}b^2\minus{}c^2}{2(b\plus{}c)}}{2\sin \alpha}$

$ \iff LN\equal{}\frac{a^2}{2(b\plus{}c)\sin \alpha}$

so the problem reduces to show that:

$ LN\equal{}\frac{a^2}{2(b\plus{}c)\sin \alpha}$

now first of all note that in triangle $ \triangle LMA$ we have $ \alpha\equal{}90^\circ\minus{}\frac A2$ hence $ \sin \alpha\equal{}\cos \frac{A}2$.

now in triangle $ \triangle BLN$ we have:

$ \frac{\sin \angle LBN}{LN}\equal{}\frac{\sin \angle LNB}{LB}$

$ \Rightarrow \frac{\sin \frac A2}{LN}\equal{}\frac{\sin C}{\frac{ac}{b\plus{}c}}$

$ \Rightarrow LN\equal{}\frac{\sin \frac A2.ac}{(b\plus{}c).\sin C}$

so we have to show that:

$ LN\equal{}\frac{a^2}{2(b\plus{}c)\sin \alpha}$

$ \iff \frac{\sin \frac A2.ac}{(b\plus{}c).\sin C}\equal{}\frac{a^2}{2(b\plus{}c)\sin \alpha}$

$ \iff \frac{\sin A}{a}\equal{}\frac{\sin C}{c}$

which is true because of the $ \sin$ law in $ \triangle ABC$

QED

Because of $ AKNM$ and $ \bigtriangleup ABC,$ have the quadrilateral $ AKLM,$ as their common area, it is enough to prove that $ (LKN) \plus{} (LMN) \equal{} (LBK) \plus{} (LCM)$ $ ,(1)$

We denote as $ D,$ $ E,$ the orthogonal projections of $ N,$ on the sidelines $ AB,$ $ AC,$ respectively and it is easy to show that $ (LKN) \plus{} (LMN) \equal{} (LKD) \plus{} (LME)$ $ ,(2)$

So, it is enough to prove that $ (LBD) \equal{} (LCE)$ $ ,(3)$

But, because $ NB \equal{} NC$ and $ ND \equal{} NE,$ we conclude that the right triangles $ \bigtriangleup DBN,$ $ \bigtriangleup ECN$ are congruent and so, we have $ BD \equal{} CE$ $ ,(4)$

From $ (4)$ and because of $ KL \equal{} LM,$ as the altitudes of the triangles $ \bigtriangleup LBD$ and $ \bigtriangleup LEC$ respectively, we conclude that $ (LBD) \equal{} (LCE)$ and the proof is completed.

Kostas Vittas.

For a generalize, if $ L$ is arbitrary point on BC, let $ BL\equal{}x,CL\equal{}y$, keep all others assumtion, then
\[ \frac{S_{AMNK}}{S_{ABC}}\equal{}\frac{(yc^2\plus{}xb^2)[(b^2\plus{}c^2\minus{}a^2)(\frac{y^2}{b^2}\plus{}\frac{x^2}{c^2})\plus{}4xy]}{2(x\plus{}y)[xy(b^2\plus{}c^2\minus{}a^2)\plus{}y^2c^2\plus{}x^2b^2]}\]

Now if $ L$ is intersection of bisector, $ x\equal{}\frac{ac}{b\plus{}c},y\equal{}\frac{ab}{b\plus{}c}$ we get this problem.

[quote="kim77"]Such a geometry[/quote]
This is an IMO 1987 number 2 problem.  Here is the link:
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=366535#36653

Here is another solution that has not been posted in the posts above.
[hide="Solution"]
We draw $ KX$ parallel to $ BN$ such that $ X$ lies on $ BC$.  Now, notice that $ \angle KXB \equal{} \angle CBN \equal{} \angle NAC \equal{} \angle BAN \equal{} \angle LMK$ the last one because $ AMLK$ is cyclic.  Hence, $ \angle KXL \equal{} 180 \minus{} \angle KXB \equal{} 180 \minus{} \angle LMK$, which gives us cyclic quadrilateral $ LXKM$.  Therefore, we have that $ \angle LKM \equal{} \angle LXM$ and since $ AL$ is an angle bisector, we have that $ LK \equal{} LM$.  Furthermore, we have that $ BN \equal{} NC$, so
\[ \angle CXM \equal{} \angle LXM \equal{} \angle LKM \equal{} \angle LMK \equal{} \angle KXB \equal{} \angle NBC \equal{} \angle NCB\implies XM\parallel NC
\]
Hence, we have that $ KX\parallel BN$ and $ XM\parallel NC$.  Let $ KN$ and $ XB$ intersect at $ Y$ and let $ XC$ and $ NM$ meet at $ Z$.  Therefore, we have that $ [KYB] \equal{} [XYN]$ and $ [XZN] \equal{} [MZC]$, which give us that
\[ [MZC] \plus{} [KYB] \equal{} [XYN] \plus{} [XZN] \equal{} [YZN]\implies [ABC] \equal{} [AMZYK] \plus{} [MZC] \plus{} [KYB] \equal{} [AMZYK] \plus{} [YZN] \equal{} [AKNM]
\]
[/hide]

[quote="blackmouse"]For a generalize, if $ L$ is arbitrary point on BC, let $ BL \equal{} x,CL \equal{} y$, keep all others assumtion, then
\[ \frac {S_{AMNK}}{S_{ABC}} \equal{} \frac {(yc^2 \plus{} xb^2)[(b^2 \plus{} c^2 \minus{} a^2)(\frac {y^2}{b^2} \plus{} \frac {x^2}{c^2}) \plus{} 4xy]}{2(x \plus{} y)[xy(b^2 \plus{} c^2 \minus{} a^2) \plus{} y^2c^2 \plus{} x^2b^2]}
\]
[/quote]
Who can help me to prove, I don't know solution  :?:

[quote="blackmouse"] [color=darkred][u][b]Generalization.[/u][/b] If $ L$ is arbitrary point on BC, let $ BL \equal{} x,CL \equal{} y$, keep all others assumtion,

then $ \frac {S_{AMNK}}{S_{ABC}} \equal{} \frac {(yc^2 \plus{} xb^2)[(b^2 \plus{} c^2 \minus{} a^2)(\frac {y^2}{b^2} \plus{} \frac {x^2}{c^2}) \plus{} 4xy]}{2(x \plus{} y)[xy(b^2 \plus{} c^2 \minus{} a^2) \plus{} y^2c^2 \plus{} x^2b^2]}$ .

[u][b]Particular case.[/u][/b]  $ \frac {LB}{LC}\equal{}\frac xy$ $ \implies$ $ x \equal{} \frac {ac}{b \plus{} c},y \equal{} \frac {ab}{b \plus{} c}$ we get this problem.[/color] [/quote]
[color=darkblue][b][u]Another extension.[/u][/b] Denote the circumradius $ R$ of $ \triangle ABC$ . Then

$ \left\{\begin{array}{c}
m(\widehat {BAN})\equal{}\alpha\\\\
m(\widehat {CAN})\equal{}\beta\\\\
\frac {LB}{LC}\equal{}k\end{array}\right\|$ $ \implies$ $ [AKNM]\equal{}\frac {a(c^2\plus{}kb^2)\cos (\alpha \minus{}\beta)}{4(k\plus{}1)R}$ .

[b][u]Particular case.[/u][/b] $ \alpha \equal{}\beta$ $ \implies$ $ k\equal{}\frac cb$ $ \implies$ $ [AKNM]\equal{}\frac {a\left(c^2\plus{}\frac cb\cdot b^2\right)}{4(\frac cb\plus{}1)R}$ $ \implies$

$ [AKNM]\equal{}\frac {abc}{4R}$ $ \implies$ $ [AKNM]\equal{}S$ , where $ S\equal{}[ABC]$ .[/color]


calculus
derivative
limit
real analysis
real analysis unsolved

Show that if $ f(x)$ is derivative in $ (a,\infty)$ and $ \lim_{x\to\infty}f'(x)\equal{}0$, that $ \lim_{x\to\infty}\frac{f(x)}{x}\equal{}0$

Apply L'Hospital's Rule.

$ \lim_{x\to\infty}\frac {f(x)}{x} \equal{} \lim_{x\to\infty}\frac {f'(x)}{1} \equal{} 0$

This is only legitimate if also $ \lim_{x \to \infty} f(x) \equal{} \infty$.

No. Check: http://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule
or e.g.: Rudin's Principle of Mathematical Analysis.

[color=green]Moderator says: Never quote the post directly preceeding yours in full; you are replying to it by default.[/color]


For people who don't want to download the file: We want to minimize $ 9x \plus{} 64y \plus{} \frac {1}{x^2y}$ with $ x,y > 0$.

My instinct is to go for AM-GM, seeing the potential $ x$s and $ y$s canceling out. However, you have to split $ 9x$ up into something like $ ax \plus{} bx$ where $ a \plus{} b \equal{} 9$ so that when you apply AM-GM, the $ x$s and the $ x^2$ will cancel out.

Or you can be cheap and use partial derivatives.

OK.
[hide="Cheating"]
$ z\equal{}9x\plus{}64y\plus{}\frac{1}{x^2y}$.

$ \frac{\partial z}{\partial x}\equal{}9\minus{}\frac{2}{yx^3}\equal{}0\implies 9x^3y\equal{}2$.

$ \frac{\partial z}{\partial y}\equal{}64\minus{}\frac{1}{x^2y^2}\equal{}0\implies 64x^2y^2\equal{}1$.

From the first, $ y\equal{}\frac{2}{9x^3}$. Substituting, $ \frac{256}{81x^4}\equal{}1\implies x\equal{}\frac{4}{3}$. (Taking the positive, real fourth root)

Then $ y\equal{}\frac{3}{32}$.

Answer: $ \boxed{24}$.
[/hide]

[hide="Having done that..."]
Let's find an AM-GM solution now.
$ 4\sqrt[4]{64ab}$ needs to be $ 24$ and $ a\plus{}b\equal{}9$. Solving finds $ a\equal{}b\equal{}9/2$. Ah, now wasn't that obvious?

So, your real solution is:

Rewrite as $ \frac{9}{2}x\plus{}\frac{9}{2}x\plus{}64y\plus{}\frac{1}{x^2y}$, apply AM-GM, and find that $ \boxed{24}$ is the minimum. We've already shown it's attainable, so there's your answer.
[/hide]

After you break it up and apply AM-GM, you have to show that the minimum is attainable. According to the equality condition of AM-GM,

$ ax \equal{} bx \equal{} 64y \equal{} \frac{1}{x^2y}$.

According to the first part of the chain, $ x \equal{} 0$. But when $ x \equal{} 0$, the expression becomes undefined. So how would you solve this problem?

Restating the problem in Latex for convenience:

What's the minimum value of $ A\equal{}9x\plus{}64y\plus{}{1\over x^2y}$ if $ x>0, y>0$?

[quote="sunehra"]After you break it up and apply AM-GM, you have to show that the minimum is attainable. According to the equality condition of AM-GM,

$ ax \equal{} bx \equal{} 64y \equal{} \frac {1}{x^2y}$.

According to the first part of the chain, $ x \equal{} 0$. But when $ x \equal{} 0$, the expression becomes undefined. So how would you solve this problem?[/quote]

Quite right. However, you overlooked the possibility that $ a\equal{}b$, which would have gotten my result without the calculus ;)


number theory
prime numbers
real analysis
real analysis unsolved

Sorry, but I'm not sure what the pattern is.  The denominators appear to be products of primes, but what is the pattern in the numerator?

i too could not understand the pattern....thats why i decided to consult with experts.

wat i can observe immediately is that
the n-th term of the series will be
1.2.4....p_(n-1) / 2.3.5....p_(n)
where p_(n) is the n-th prime.

now please try ....

thank you.... :)

That's wrong (you've mis-used parentheses).  The general term appears to be
\[ \frac{\prod_{i \equal{} 1}^{n \minus{} 1}p_i \minus{} 1}{\prod_{i \equal{} 1}^n p_i}
\]
where $ p_i$ is the $ i$th prime, $ p_1 \equal{} 2$.

The partial sums can be written in the form
\[ 1 \minus{} \frac{\prod_{i \equal{} 1}^n p_i \minus{} 1}{\prod_{i \equal{} 1}^n p_i},
\]
and that rephrasing should make your task clearer.

sorry sir ..... i cannot understand how u wrote the numerator in this way....

thank you....

\[ \frac{{\prod\nolimits_{k \equal{} 1}^{n \minus{} 1} {\left( {p_k  \minus{} 1} \right)} }}
{{\prod\nolimits_{k \equal{} 1}^n {p_k } }}
\]
where
\[ \begin{gathered}
  p_1  \equal{} 2 \hfill \\
  p_2  \equal{} 3 \hfill \\
  p_3  \equal{} 5 \hfill \\
  ... \hfill \\
\end{gathered}
\]

well, i see you guys have already found the pattern... :|
the problem with these types of series is that i can
almost always sum them from [b]right to left[/b], but never
from [b]left to right[/b]... i always get stumped! what i mean
is had you not given the answer, it would have probably
taken me more than a few minutes to figure out the sum.

you will see what i mean. here is the [b]right-to-left[/b] proof
of the series in question. we are given the sum is $ 1$. we first
associate $ 1$ with the reciprocal of $ 2$, then we associate
the remainder with the reciprocal of $ 3$, then we associate
the remainder with the reciprocal of $ 5$, then we associate
the remainder with the reciprocal of $ 7$, then we associate
the remainder with the reciprocal of $ 11$, and so forth (that is
take into account all of the [i]prime numbers[/i]).

$ S=1$
$ =1(\boxed{\frac{1}{2}}+\frac{1}{2})$
$ =1(\frac{1}{2}+\frac{1}{2}(\boxed{\frac{1}{3}}+\frac{2}{3}))$
$ =1(\frac{1}{2}+\frac{1}{2}(\frac{1}{3}+\frac{2}{3}(\boxed{\frac{1}{5}}+\frac{4}{5})))$
$ =1(\frac{1}{2}+\frac{1}{2}(\frac{1}{3}+\frac{2}{3}(\frac{1}{5}+\frac{4}{5}(\boxed{\frac{1}{7}}+\frac{6}{7}))))$
$ =1(\frac{1}{2}+\frac{1}{2}(\frac{1}{3}+\frac{2}{3}(\frac{1}{5}+\frac{4}{5}(\frac{1}{7}+\frac{6}{7}(\boxed{\frac{1}{11}}+\frac{10}{11})))))$
$ =1(\frac{1}{2}+\frac{1}{2}(\frac{1}{3}+\frac{2}{3}(\frac{1}{5}+\frac{4}{5}(\frac{1}{7}+\frac{6}{7}(\frac{1}{11}+\frac{10}{11}(\boxed{\frac{1}{13}}+\frac{12}{13})))))))$
$ =1(\frac{1}{2}+\frac{1}{2}(\frac{1}{3}+\frac{2}{3}(\frac{1}{5}+\frac{4}{5}(\frac{1}{7}+\frac{6}{7}(\frac{1}{11}+\frac{10}{11}(\frac{1}{13}+\frac{12}{13}(\boxed{\frac{1}{17}}+\frac{16}{17}))))))))$
$ \ldots\;\;\;\ldots$

$ =\frac{1}{2}+\frac{1}{2\cdot 3}+\frac{1\cdot 2}{2\cdot 3\cdot 5}+\frac{1\cdot 2\cdot 4}{2\cdot 3\cdot 5\cdot 7}+\frac{1\cdot 2\cdot 4\cdot 6}{2\cdot 3\cdot 5\cdot 7\cdot 11}+\frac{1\cdot 2\cdot 4\cdot 6\cdot 10}{2\cdot 3\cdot 5\cdot 7\cdot 11\cdot 13}+\;\;\;\ldots\;\;\;\ldots$

this is what was asked for originally in the question, i.e. to find
$ S=\sum_{n=1}^\infty\;\{\frac{1}{p_n-1}\prod_{k=1}^n\;(1-\frac{1}{p_k})\}$

the thing is anyone can basically start from $ 1$ and go on
to concoct any type of series and then ask for it's sum.... :D


[quote="solafidefarms"]Recently I had some fun when the well that I obtain my water from broke. Actually, the well didn't break :D, but that's beside the point. There are 7 people who get water from that well: 2 in the 2-5 range, 1 in the 5-10 range, two in the 10-15 range, and two in the 30-40 range. Imagine that, for every one year of age, you use one gallon more water per day, and that if someone is between the age of x and y, they are (x+y)/2 years old. How many gallons of water are required if the water is out between 8:30 Tuesday morning and 3:55 Saturday afternoon?

Target problem.[/quote]

Everybody in your family is between the ages of -100 and 100 years old, therefore everybody uses $\frac{-100+100}{2}$, or 0, gallons of water per day. 7 times 0 gallons = [b]0 gallons[/b].

I think he means to find the number of gallons of water needed by the people, assuming each of their ages is the average of the range.  The number of gallons they need is equal to their age (per day).

[quote="bpms"][quote="solafidefarms"]Recently I had some fun when the well that I obtain my water from broke. Actually, the well didn't break :D, but that's beside the point. There are 7 people who get water from that well: 2 in the 2-5 range, 1 in the 5-10 range, two in the 10-15 range, and two in the 30-40 range. Imagine that, for every one year of age, you use one gallon more water per day, and that if someone is between the age of x and y, they are (x+y)/2 years old. How many gallons of water are required if the water is out between 8:30 Tuesday morning and 3:55 Saturday afternoon?

Target problem.[/quote]
I don't understand the question...[/quote]
neither do i...can someone please clarify?

[quote="solafidefarms"]Recently I had some fun when the well that I obtain my water from broke. Actually, the well didn't break :D, but that's beside the point. There are 7 people who get water from that well: 2 in the 2-5 range, 1 in the 5-10 range, two in the 10-15 range, and two in the 30-40 range. Imagine that, for every one year of age, you use one gallon more water per day, and that if someone is between the age of x and y, they are (x+y)/2 years old. How many gallons of water are required if the water is out between 8:30 Tuesday morning and 3:55 Saturday afternoon?

Target problem.[/quote]
Assuming you mean "for every year of age, you use one gallon of water per day," [hide] The 2 people in the 2-5 range have an age of $\frac{2+5}{2}=3.5$. So they use a total of 7 gallons per day. The person in the 5-10 range uses 7.5 gallons, the 2 people in the 10-15 range use a total of 25 gallons, and the 2 people in the 30-40 range use a total of 70 gallons. Therefore, they use $7+7.5+25+70=109.5$ gallons per day. Between 8:30 Tuesday morning and 3:55 Saturday afternoon, there are 4 days, 7 hours and 25 minutes. This is equal to $4+\frac{7}{24}+\frac{25}{1440}=4+\frac{89}{288}\approx4.31$ days. $(4+\frac{89}{288})\cdot 109.5=471+\frac{161}{192}\approx471.84$  :maybe: [/hide]


trigonometry calculus integration function algorithm derivative search

For the first part
$\int\frac{x}{1-x^2}\,dx$ no trig sub is needed!! just use $u=1-x^2$

I'm reading your first problem as this. If this isn't what you meant, add parentheses to clarify:

$\int\frac x{1-x^2}+\sqrt{1-x^2}\,dx.$

Break that into two integrals. The first one, $\int\frac x{1-x^2}\,dx,$ falls to the "natural" substitution $u=1-x^2.$

(Hint: trigonometric substitutions are not your first resort. If a "natural" substitution works, it's much to be preferred.)

The second part is where you will do the trig substitution. I find it useful to draw a triangle. In this case, we'll have a right triangle with legs $\sqrt{1-x^2}$ and $x$ and hypotenuse $1.$ One of the angles is $\theta$, and the substitution works this way:

$x=\sin\theta;\, dx=\cos\theta\,d\theta;\,\sqrt{1-x^2}=\cos\theta.$

That last identity, $\sqrt{1-x^2}=\cos\theta$ - that comes from the Pythgorean trigonometric identity, but the drawing of the triangle is a way to help you remember it.

The basic motivation of trig subsutitions is this: using the various forms of the Pythagorean trig identity, you can always write a sum or difference of two squares as a square, thus simplyfing the radical. If the radical doesn't simplify, then you've picked the wrong substitution.

With that substitution, we get:

$\int\sqrt{1-x^2}dx=\int\cos^2\theta\,d\theta.$

Now you can go consult that section in your calculus books on integrating powers of trigonometric functions.

The thing about integrating powers of trigonometric functions, for instance, $\int\cos^m\theta\sin^n\theta\,d\theta,$ is that there are cases involving whether the integer exponents are odd or even. Three fourths of the possible cases are "easy", but if you actually needed to do that trigonometric subsititution in the first place, then you're going to land in that one fourth of the cases that aren't easy. (There as some exceptions involving reciprocals and turning secants into cosines or vice versa.)

2nd is a little harder

Try $u=\exp(x)$ then $t^2=u-1$

As for $\int\frac{dx}{\sqrt{e^x+1}}$ :

There are many things we could try. There's no automatic algorithm, no machine - you just have to try things on the chance that they might work. Here's one approach.

Let $u=\sqrt{e^x+1}.$ This is going to have to be a "forced" or "backwards" substitution, so we'll have to solve for $x$ in terms of $u.$

$u^2=e^x+1$

$e^x=u^2-1$

$x=\ln(u^2-1)$

$dx=\frac{2u\,du}{u^2-1}$

$\int\frac{dx}{\sqrt{e^x+1}}=\int\frac{2u\,du}{(u^2-1)u}=\int\frac{2du}{u^2-1}= \int\frac{2du}{(u-1)(u+1)}.$

That's a rational function. It's time to use partial fractions.

[quote="fedja"][quote="Kent Merryfield"]
There are many things we could try. There's no automatic algorithm, no machine [/quote]
As a matter of fact, there is an algorithm that, given an elementary function, either produces an elementary antiderivative or tells that no such antiderivative exists. So one can teach a computer to integrate whatever is integrable. This raises an interesting question whether we should teach our students fancy integration skills and what exactly are the benefits one gets from learning various substitutions, but it is a question for another forum. Anyway, that algorithm is not something you would try to follow with pen and paper, so, Kent's words remain partially true.[/quote]

The [url=http://mathworld.wolfram.com/RischAlgorithm.html]Risch Algorithm[/url]

There's another sense I had in mind when I said that.

There is a straightforward algorithm for differentiating elementary functions. The pieces of this algorithm are the specific formulas for the derivatives of each "building block" of the elementary functions, and the sum rule, product rule, and chain rule. As a consequence, there is a theorem that every elementary function has an elementary derivative. The proof is the algorithm.

There cannot be that sort of guaranteed algorithm for antidifferentiation, since there are elementary functions that do not have elementary antiderivatives. Hence the algorithm that fedja and dheardie refer to must be allowed to return "not an elementary function" as an output. By far the trickiest part of designing such an algorithm is knowing when to return that output. (The human pencil-and-paper approximation: if you're pretty good at it and you don't see anything that has a chance of working, that it's a fairly safe assumption that it's not elementary.)

As for fedja's implied criticism of teaching methods: yes, the search for elementary antiderivatives of elementary functions is a limited and limiting game. It cannot be allowed to swallow up the bulk of a calculus course. When I teach that part of the calculus sequence, I hold this part of the course within fairly strict time limits to make sure that I have enough time to wallow in power series.

But being able to manage a change of variables or an integration by parts has uses far beyond the limited game of elementary antidifferentiation. The problems in the "regular" problem sections ("Proposed and Own", "Unsolved", "Open") of this Calculus section of AoPS/Mathlinks give new evidence for that every day.

[quote="ncal"]I have a sheet of practice problems on Trig substitution and using the integral table. Most of it I seem to understand but two problems confuse me.

The first problem is to evaluate the integral x/ 1-x^2 + squareroot (1-x^2) dx

I realize that whenever you have something in the form sqrt (a^2-x^2) it follows the sin format, but Im not sure what to do with the 1-x^2 not under the radical.

The problem states 'Evaluate the integral dx/ squareroot(e^x +1)

For this one, Im not sure where to even begin : ( I appreciate any help you can offer. Thanks![/quote]

Generally speaking, trig-substitution is the method of choice whenever there is a square root in the denominator of the integrand.

Hint: First, sketch the triangle. Second, label the parts of the triangle that you know.

If you're looking at the sketch, things usually become quickly obvious.


linear algebra
matrix
analytic geometry
trigonometry
linear algebra unsolved

A well known fact is that every $ n\times m$ matrix $ {G=\left[\begin{array}{c}G_{1}\\ G_{2}\end{array}\right]}$ (where $ m\leq n$)
can be transformed into its normal form (e.g. $ {\left[\begin{array}{c}{\tilde G}_{1}\\ 0\end{array}\right] }$, where $ {\tilde G}_{1}=1_{m}$ is the identity matrix) by means of an elementary transformation $ Z \in{\mathbb R}^{n\times n}$

Can we actually show that $ Z$ can be choosen as a orthogonal matrix if we do not require that $ {\tilde G}_{1}$ is an identity matrix?  :(

[color=red][Changed the "\Zeta" into a "Z." -mathmanman][/color]

\Zeta is not a code. I'm not actually sure how you would get a capital zeta.

[color=red][Isn't the capital zeta just a "Z?" -mathmanman][/color]

The answer is $ Z^{T}=Z^{-1}$ being the transformation matrix of spherical polar coordinate system, for example

$ n=2~~: ~~{\left[\begin{array}{cc}\sin\theta & \cos \theta\\ \cos\theta &-\sin\theta\end{array}\right] }$

$ n=3~~: ~~{\left[\begin{array}{ccc}\sin\theta\sin\varphi & \sin\theta\cos\varphi & \cos\theta\\ \cos\theta\sin\varphi & \cos\theta\cos\varphi &-\sin\theta \\-\cos\varphi & \sin\varphi & 0 \end{array}\right] }$  :D


For $ n\in N$, let $ d(n)$ be the number of positive divisors of $ n$. Suppose $ r,s$ are positive integers such that $ d(ks)\geq d(kr)$ for all $ k\in N$. Show that $ r$ divides $ s$.


Prove that all positive real number a,b,c

  $ \left(\frac{a}{b\plus{}c}\right)^2\plus{}\left(\frac{b}{c\plus{}a}\right)^2\plus{}\left(\frac{c}{a\plus{}b}\right)^2\plus{}10\cdot\frac{abc}{(a\plus{}b)(b\plus{}c)(c\plus{}a)}\geq2$

[quote="Marius Mainea"]Prove that all positive real number a,b,c

  $ \left(\frac {a}{b \plus{} c}\right)^2 \plus{} \left(\frac {b}{c \plus{} a}\right)^2 \plus{} \left(\frac {c}{a \plus{} b}\right)^2 \plus{} 10\cdot\frac {abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}\geq2$[/quote]
It's equivalent to $ \sum_{cyc}(a^6\plus{}2a^5b\plus{}2a^5c\minus{}a^4b^2\minus{}a^4c^2\plus{}a^4bc\minus{}4a^3b^3)\geq0,$ which obviously.

Well the inequality is quite funny if we don't expand.Put $ \displaystyle \frac{a}{b\plus{}c}\equal{}x$ and the similar ones.We have that $ \displaystyle \sum_{cyc} xy\plus{}2xyz\equal{}1$ and we have to prove that $ \displaystyle \sum_{cyc} x^{2}\plus{}10xyz\geq{2}$.Now we write Schur as $ \displaystyle 9xyz \geq(\sum_{cyc} x)(2\sum_{cyc} xy\minus{}\sum_{cyc} x^{2})$.Now we have $ \displaystyle \sum_{cyc} x \geq \frac{3}{2}$ so we deduce that $ 6xyz \geq 2\sum_{cyc} xy\minus{}\sum_{cyc} x^{2}$.To wrapp things up just add $ 4xyz$ to the last inequality. :)

[quote="Marius Mainea"][quote="arqady"]$ \sum_{cyc}(a^6 \plus{} 2a^5b \plus{} 2a^5c \minus{} a^4b^2 \minus{} a^4c^2 \plus{} a^4bc \minus{} 4a^3b^3)\geq0,$ which obviously.[/quote]

  Why is obviously?[/quote]
For example, $ \sum_{cyc}(a^6 \plus{} 2a^5b \plus{} 2a^5c \minus{} a^4b^2 \minus{} a^4c^2 \plus{} a^4bc \minus{} 4a^3b^3)\equal{}$
$ \equal{}\sum_{cyc}(a^6\minus{}a^5b\minus{}a^5c\plus{}a^4bc)\plus{}\sum_{sym}(a^5b\minus{}a^4b^2)\plus{}2\sum_{sym}(a^5b\minus{}a^3b^3)\geq0,$
which is true by Schur and Muirhead.

[quote="Marius Mainea"]OK. Thanks arqady!

arqady,who is Michael Rozenberg?[/quote]
This is my real name.

excuse me, i have something to ask arqady.
i know that after expansion,using muirhead is quite easy..
but is there any way that we can expand the thing faster?
during my expansion, i have to expand ((a+b)(b+c)(c+a))^2 or a^2(a+c)^2(a+b)2..

i am now breaking out and grouping each terms together which is really slow..like.
((a+b)(b+c)(c+a))^2 =(2abc+a^2b+b^2a+b^2c+c^2b+c^2a+ac^2(^2..

[quote="Marius Mainea"]Prove that all positive real number a,b,c

  $ \left(\frac {a}{b \plus{} c}\right)^2 \plus{} \left(\frac {b}{c \plus{} a}\right)^2 \plus{} \left(\frac {c}{a \plus{} b}\right)^2 \plus{} 10\cdot\frac {abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}\geq2$[/quote]

It here http://www.mathlinks.ro/viewtopic.php?p=1072758#1072758


[quote="Marius Mainea"]If $ a,b,c\in\mathbb{R}$, then $ \frac {|a \plus{} b \plus{} c|}{(a^2 \plus{} 1)(b^2 \plus{} 1)(c^2 \plus{} 1)}\le \frac {25\sqrt {5}}{72}$[/quote]Maybe you mean
$ \frac {5\sqrt {5}}{72}$ :?:

[quote="dima ukraine"]Maybe you mean
$ \frac {5\sqrt {5}}{72}$ :?:[/quote] Nope, try a=1 b=1 c=1


hey
these are probably the easiest problems you will ever see in your life. :lol:

1) The lenght of one base of a trapezoid is 7mm less than twice the length of the base. The height of the trapezoid is 12 mm. The area of the trapezoid is 48mmsquared. Find the lenght in mm of the long base.

2) What is the diameter of a circle whose circumference is numerically equal to its area?

3) What is one-half of the perimeter of a square when five-eights of its area is 40?

HINT: USE UR BRAIN  :lol:

and ya dont answer them if u no how to do them. let others try. PLZ!

Nvr did these before, but they were pretty easy

[hide]1. 5mm
2. 4
3. 16[/hide]

That's y i said these are the EASIEST problems u will EVER see in your lyf.
sry if it sounds mean but i m not :P
i m just obsessed with capital writtings i dont no y? :D
and ya i m going to say the answers after more peeps try coz till now 7 viewed them and only 1 replied.

Hey u mite wanna try the problem called numbers by mathnerd89. I posted that question. Its fairly easy! and ya that is not my SN(screen name-mathnerd) its my bro's.


http://science.slashdot.org/article.pl?sid=06/08/22/1751225

Terence Tao did indeed win one, as did Grigory Perelman (who refused it), Wendelin Werner, and Andrei Okounkov.

How many people had never heard of the Fields, besides me?

[quote="José"]Refusing it is like an insult, I don't see the point in refusing a prize[/quote]

Perelman doesn't think too much of the mathematical community, it seems.

[quote="José"][quote="Elemennop"]What's wrong with an insult? People are too kind nowadays.[/quote]

What I mean is that I don't see the point in refusing a prize[/quote]

To insult the math community?

Sometimes someone refuses a prize if he thinks that it is not enough for his work, that it is a low importance prize.

From what I read, it is either because he's pissed off by somethign to do with his former employers in Russia (he is currently unemployed afaik) or because he just wants to hide and be reclusive and he thinks all this recognition will make that much harder for him.

[quote="tim1234133"]I heard that he is not claiming the $\$$1million prize since he worries that it would disrupt him from doing maths... I imagine he thinks the same about the Fields Medal.[/quote]

He has made no decision on the Clay Prize because it hasn't been offered to him yet.  (See the article rookworm linked to in the thread "Russian math genius disappears into the woods.")

From what I've read, Perelman just doesn't care about prizes and all that.  I don't think it's a knock on the mathematical community, although he's done that before.

[quote="lotrgreengrapes7926"][quote="jclarkemathL314159"]Didn't Terrence Tao win a gold medal at IMO when he was 12?[/quote]
He was 13.  :wink:[/quote]
Wrote the test when he was 12,and received the medal as a 13-year old.
Celebrating his 13th birthday during the competition.:wink:

From a long article I read at http://www.newyorker.com ([url]http://www.newyorker.com/fact/content/articles/060828fa_fact2[/url]) I got the impression that Perelman just isn't interested in winning  prizes or awards. He basically says, "If my work is true, then I have done my job. That is my prize." And, I don't see anything wrong or weird about that. Maybe we are weird in thinking that he is weird :)

Kudos to him. And, Kudos to Terence Tao, who won the prize at the age of 31, making him the youngest Fields Medal winner. Wow!


analytic geometry
invariant
combinatorics unsolved
combinatorics

Determine (with proof) if there exists a polygon on the coordinate plane such that
i) The coordinates of each vertex are rational,
ii) The length of its every edge is equal to one,
iii) The polygon has an odd number of vertices.

[hide="hint"]Let X(P) be the numerator of the x-coordinate of P when expressed as a fully reduced fraction, and let Y(P) be the numerator of the y-coordinate. Find an invariant with this.[/hide]


Take light in glass. What could slow it down. I know that stuff is asorbed and then reemmited, but what about imperfections in the glass its self?? Say you have a piece glass that has impurities(little defects), or the desity of glass(not optical), the actual wave length of light, or for giggles the how bright the light is??? Would any of these affect the speed of light or would light simply travel around them???

I don't understand your question. Please rephrase it.

I don't think light would travel around anything unless there were something to refract from. And personally, I don't think light slows down because it is absorbed and reemitted. Please correct me if I'm wrong.


[quote="SAPOSTO"]On the board it is written a natural number. Then we add a digit different from 9 on the right of the first number. Prove that after finite number of repetitions of this operation we will obtain a composite number.
This was the hardest problem for 9th grade in the Winter Mathematical competition in Bulgaria this year which was held a week ago. Unfortunately I didn't manage to solve it on the competition  :( . However I know a very nice solution  :) .[/quote]

Assume that we can add infinitely many number such that the new number won't be prime.

Let the initial number be $ p > 5$. It must be prime.

If the last added number is divisible by $ 2$ or $ 5$ then the new number will be composite. So we can add only $ 1, 3, 7$.

$ p$ may be $ \equiv 1$ or $ 2 \mod 3$. Take 2 cases:

First case: $ p\equiv 1 \mod p$ then if in next $ p$ digits added to right will be $ 1$ or $ 7$ then take this number as $ p$ and we get the second case. If the next  $ p$ digits added to the right will be only $ 3$ then we will prove that $ n \equal{} 333\cdots3$ if divisible by $ p$. $ n \equal{} \frac13 (10^{p \minus{} 1} \minus{} 1)$ because $ \gcd(p;30) \equal{} 1$ then by Fermat $ n$ divides $ p$

Second case: $ p\equiv 2 \mod p$ then adding $ 1$ or $ 7$ to right will make the new number divisible by $ 3$. So we must add only $ 3$. But after we add $ p \minus{} 1$ digits to right we will get that the new number is divisible by $ p$ as I proofed in the 1st case.

it remains to proof for $ p \equal{} 2,3,5$.

If you found some mistakes, please write.

And if it was left, add a digit to the left so that the number is divisible by 3 (you always can).

[quote="Mathias_DK"][quote="SAPOSTO"]On the board it is written a natural number. Then we add a digit different from 9 on the right of the first number. Prove that after finite number of repetitions of this operation we will obtain a composite number.
This was the hardest problem for 9th grade in the Winter Mathematical competition in Bulgaria this year which was held a week ago. Unfortunately I didn't manage to solve it on the competition  :( . However I know a very nice solution  :) .[/quote]
Well.. If the number is e.g. $ 13$ we can just put on a zero and we have $ 130$.... And it always works since $ 10n$ is composite.... Should it not be on the left site of the number?[/quote]

it is meant that you can get a composite number not depending on what digit you add, isn't it?

Great Work [b]sandu2508[/b]. That's the solution I have too. That's the task of the problem as well.


Let S be a 1990-element set and P be a set of 100-ary sequences $(a_1,a_2,...,a_{100})$ ,where $a_i's$ are distinct elements of S.An ordered pair (x,y) of elements of S is said to [i]appear[/i] in $(a_1,a_2,...,a_{100})$ if $x=a_i$ and $y=a_j$ for some i,j with $1\leq i<j\leq 100$.Assume that every ordered pair (x,y) of elements of S appears in at most one member in P.Show that $|P|\leq 800$.

[quote="Hong Quy"]Any 100-aray sequence $(a_1;a_2;...;a_{100})$ has exactly ${100\choose 2}$ pair $(a_i;a_j)$ with $1\le i<j\le 100$

But any two 100-ary sequences $(a_1;a_2;...;a_{100}) ;(b_1;...;b_{100})$ we have $(a_i;a_j)\neq (b_k;b_l)$ $\forall 1\le i<j\le 100; 1\le k,l\le 100.$ ( because any pair $(x;y)$ appears in at most one member in S).

Therefore: let m be the number of members of P, we have:
$m{100\choose 2}\le 2.{1990\choose 2}$. $\Rightarrow$ $m \frac{100.99}{2}\le 2\frac{1990.1989}{2}$.Thus, $m\le 800$.[/quote]

You are true :D

Chao chu lehoan ,he he


geometry
3D geometry
sphere
function
advanced fields
advanced fields unsolved

Anyone knows how to prove the following statement?

Let P be the real projective plane. Show that any continous map f : P -> P has a fixed point.

I guess that Lefschetz fixed point theorem should be used but dont know how to apply it.

Many thanks,

Rosahara

The Lefschetz number is calculated as $ L(f)\equal{}\sum_{n\ge 0}(\minus{}1)^n tr (H_n(f)\otimes id\mathbb{Q})$ where $ id\mathbb{Q}$ is the identity map $ \mathbb{Q}\to \mathbb{Q}$, and the theorem you use
is that $ L(f)\neq 0$ implies that $ f$ has a fixed point. Now, the homology groups of $ \mathbb{RP}^2$ have torsion for $ n\ge 1$, hence $ H_n(f)\otimes id\mathbb{Q}\equal{}0$ for  $ n\ge 1$.
 In degree 0 we have $ H_0(f)\otimes id\mathbb{Q}\equal{}id\mathbb{Q}$. So $ L(f)\equal{}1\neq 0$, meaning that $ f$ has a fixed point.

In fact this follows without the use of Lefschetz fixed point theorem. Note first that since $ \pi_1(\mathbb{RP}^2) \equal{} 0$, a map $ f: \mathbb{RP}^2\to \mathbb{RP}^2$ lifts to a map $ \tilde{f}: \mathbb{RP}^2\to S^2$ such that $ f \equal{} \pi \circ \tilde{f}$, where $ \pi$ is the canonical projection $ S^2\to \mathbb{RP}^2$. T
his is because $ S^2$ is a covering space of $ \mathbb{RP}^2$. Taking the composition $ \phi : (\tilde{f} \circ \tilde{f}): S^2\to S^2$.
Now use the fixed point theorem for the sphere asserts that any continuous function mapping the sphere into itself either has a fixed point or maps some point to its antipodal point.

Hmm, but $ \pi_1(\mathbb{RP}^2)\equal{}\mathbb{Z}_2$ :)

Yes, you are of course right.. I was thinking about $ \pi_1(S^2)\equal{}0$. :)

And what do you mean by $ \tilde{f} \circ \tilde{f}$? I was thinking, that composing $ \tilde{f} \circ \pi$ we'll got map $ S^2 \to S^2$ that doesn't bring point on sphere to itself or antipodical point. Indeed, if $ \pi\tilde{f}(x) \equal{} x$, then $ f\pi(x)\equal{}\pi\tilde{f}\pi(x) \equal{} \pi(x)$, the same with $ \pi\tilde{f}(x) \equal{} \minus{}x$. That's what you mean?

Yes. $ \tilde{f} \circ \tilde{f}$ should be $ \pi \circ \tilde{f}$, which is a selfmap of the sphere. Sorry.


If

$x-by=1$

$y-ax=1$

$bx+ay=1$

Find the value of $a^2+b^2+ab+a+b$

easy one
We have $ay-a^2x=a$ and $bx-b^2y=b$. Add this two equalities together and get $a+b=ay+bx-a^2x-b^2y$. Thus $a^2+b^2+ab+a+b=a^2+b^2+ab+1-a^2x-b^2y$ since $ay+bx=1$. Eventually, $a^2+b^2+ab+a+b=a^2(1-x)+b^2(1-y)+ab+1=-bya^2-b^2xa+ab+1=ab(1-ay-bx)+1=1$.


[quote="tranthanhnam"]with a,b,c  is povisite  and satisfying abc=1
prove
1/(a^3+b^3+1) + 1/(b^3 +c^3 +1 ) + 1/(c^3 +a^3 +1 ) <=1[/quote]

It is equivalent to

$\sum \frac{1}{a^3+b^3+abc}\leq \sum\frac{1}{ab(a+b+c)}=\sum\frac{c}{a+b+c}=1$

 :)

See http://www.mathlinks.ro/Forum/viewtopic.php?t=20171 .

  darij


Tamika could not remember her scores from five mathematics tests. She did remember that the mean (average) was exactly 80, the median was 81, and the mode was 88. If all her scores were integers with 100 the highest score possible and 0 the lowest score possible, what was the lowest score she could have received on any one test?           <I don't know the repeating times of the grade 88...>  Thanks for help.

ur right, Aznwithabrain.  I got that too

[hide]If the mean was exactly $80$, and she took $5$ tests, then the total score for all 5 of those tests would be $400$.  If she got the score $88$ twice, it would show up enough times for it to be the mode.  If $81$ was the median, then there need to be two answers lower than $81$ because the two $88$ scores are higher.  Then, by adding up $81+88+88$ and then subtracting it from $400$, you get $143$.  Now to get the lowest possible score, $80$ has to be a test score, so then it is lower than $81$, and it doesn't affect the mode.  By subtracting $143-80$ you get $63$, which is the lowest score possible.[/hide]


2007-2008 ARML Power Contest Final results are released.  Here is the summary of top 10 teams:

Place	Team	                              State	R1 Score	R2 Score	Total
1.	Georgia ARML 	                GA	35	34	69
2.T	Colorado Math Circle  	CO	37	31	68
2.T	Thomas Jefferson High School 	VA	36	32	68
4.T	Long Beach Math Circle 	CA	30	37	67
4.T	Lynbrook High School  	CA	28	39	67
6.	AAST #1	                                NJ	36	30.5	66.5
7.T	Utah ARML	                                UT	32	33	65
7.T	Western Washington	                WA	27	38	65
9.	Belmont High School	                MA	23	36	59
10.T	Stuyvesant High School	NY	26	30.5	56.5
10.T	Western Massachusetts	MA	26.5	30	56.5

As you can see, 1st place through 7th place team are separated only by 4 points (65 to 69), while 8th place is 6 points below 7th (59 vs 65).   Western WA team has a fantastic Round 2 with 38 points, the 2nd highest score in Round 2, only losing 2 points in one of the problems while receiving full credits in all others.  A huge "thank you" to all who participated in the Power Contest, especially to your hard work in the weeks prior to R2 to pull such an impressive raise in round 2 score.  Organizer's letter to coaches cited “Lynnbrook High School almost had a come from behind upset with an nearly perfect second round”.  Most of WA ARML Power team members participated ARML Power Contest for the first time this year.
Like Lynnbrook, Western Washington team also registered an 11 point improvement.  A job well done.

You may have realized that 5 of the top 8 teams will meet in Las Vegas this year again at 2008 ARML contest.  Congratulations to all the winners from today’s (WA) State Mu Alpha Theta competition.  As Mr. Norris mentioned in award ceremony, many of you will receive invitations soon to join Washington State ARML team.  Any talented, motivated student who is interested in representing WA in ARML should have your coach / teacher send in a recommendation to Mr. Norris, if you missed today's State Mu Alpha Theta.

In case you have not noticed, ARML is adding a new local contest this year, scheduled on April 5th, 2008 (http://www.artofproblemsolving.com/Forum/viewtopic.php?t=192736).  If you don’t have pre-scheduled commitment (Spring Break travel, State Science Olympiad, etc.),  please consider participating ARML local contest, it will be a good practice for ARML 2008 in May.

Go team Washington!

This message is for members of 2008 Washington ARML team.  If you did not receive my previous email about practice / mini-ARML (Intramural) and competition outlook, send me a PM.

We are going to have a practice this coming Sunday 4/27 1:00pm at Redmond Library.  Please let me know by EOD Saturday if you plan to join.

To all WA ARML 2008 team members:

The date for Washington Oregon Intramural ARML is set on Sunday May 18th.  There are a few students not on my email distribution (from Moses Lake,  LWHS, and a few from TJ), if you plan to go, please drop me a note so that I can send you details.  This also helps to plan (1 team or 2 teams).

Also, our next practice is Saturday May 10th 3:30pm at Microsoft Building 117.  Hope you can join us (especially gold team members shuold practice together before going to Las Vegas).

Here is the summary of the result.  Both teams did very well.

                        OR            WA
---------------------------------------
Team:               24             40
Power:              28             26
Individual:         81             92
Relay:               18             16
---------------------------------------
Total:               151           174

Thanks everyone for your participation.

We will meet Monday (Memorial Day) May 26 9:30am at Microsoft 117.  See you there.


function
quadratics
algebra unsolved
algebra

Given non-zero reals $ a$, $ b$, find all functions $ f: \mathbb{R} \longmapsto \mathbb{R}$, such that for every $ x, y \in \mathbb{R}$, $ y \neq 0$, $ f(2x) \equal{} af(x) \plus{} bx$ and $ \displaystyle f(x)f(y) \equal{} f(xy) \plus{} f \left( \frac {x}{y} \right)$.

[hide]$ f(x)^2 \equal{} f(2x) \plus{} f\left(\frac {x}{2}\right) \equal{} af(x) \plus{} bx \plus{} \frac {f(x) \minus{} \frac {bx}{2}}{a}$, which is quadratic in $ f(x)$.  One can check that the discriminant is not positive for all $ x$ except if $ a \equal{} \frac {1}{2}$, so there does not in general exist such a function well-defined on all of $ \mathbb{R}$.[/hide]

[quote="JoeBlow"][hide]$ f(x)^2 \equal{} f(2x) \plus{} f\left(\frac {x}{2}\right) \equal{} af(x) \plus{} bx \plus{} \frac {f(x) \minus{} \frac {bx}{2}}{a}$, which is quadratic in $ f(x)$.  After solving it, one finds that the discriminant is not positive for all $ x$ for any nonzero $ a,b$, so there does not exist such a function well-defined on all of $ \mathbb{R}$.[/hide][/quote]
Oh! There was a typo. It should be $ f(x)f(y)\equal{}f(xy)\plus{}f \left( \frac{x}{y} \right)$.
Sorry for the trouble...

Trivial.

1. $ f(x)f(y)\equal{}f(xy)\plus{}f\left(\frac{x}{y}\right)\equal{}f(xy)\plus{}f\left(\frac{y}{x}\right)$ $ \Rightarrow f\left(\frac{x}{y}\right)\equal{}f\left(\frac{y}{x}\right)$.

2. $ x\equal{}y\equal{}1$ gives $ f(1)\equal{}2$ or $ f(1)\equal{}0$. But if $ f(1)\equal{}0$, we get $ f(x)\equal{}0$, which doesn't give us a solution.
$ x\equal{}y$ gives $ f^2(x)\equal{}f(x^2)\plus{}2$. $ \Rightarrow |f(x)|>\sqrt{2}$, which give us $ |f(x)|>\sqrt{2\plus{}\sqrt{2\plus{}...\plus{}\sqrt{2}}}$.
$ \Rightarrow |f(x)|\ge 2$.

3. $ f(x^2)\plus{}f(y^2)\equal{}f(xy)f\left(\frac{x}{y}\right)$, since $ x^2\equal{}xy\frac{x}{y}$ and $ y^2\equal{}\frac{xy}{\frac{x}{y}}$.
$ f^2(x)\plus{}f^2(y)\minus{}4\equal{}f(xy)f\left(\frac{x}{y}\right)\equal{}f(xy)\left(f(x)f(y)\minus{}f(xy)\right)$.
$ \Leftrightarrow f^2(x)\plus{}f^2(y)\plus{}f^2(xy)\minus{}f(x)f(y)f(xy)\equal{}4$.

There is a lemma which tells us that if $ x^2\plus{}y^2\plus{}z^2\minus{}xyz\equal{}4$, then there exist $ a,b,c$ such that $ x\equal{}a\plus{}\frac{1}{a}$ and so on.
Therefore + the fact $ |f(x)|\ge 2$ for every real $ x$, we get the following - $ f(x)\equal{}g(x)\plus{}\frac{1}{g(x)}$ for some function $ g$.

4. From $ f^2(x)\equal{}f(x^2)\plus{}2$ we get $ g(x^2)\equal{}g^2(x)$ or $ g(x^2)g^2(x)\equal{}1$ - which gives us nothing.
$ \Rightarrow g^2(x)\equal{}g(x^2)$. bla-bla $ \Rightarrow g(x)\equal{}x^c$.

and we just have to find $ c$ depending on $ a$ and $ b$.

[quote="BG Yoda"]
$ x \equal{} y$ gives $ f^2(x) \equal{} f(x^2) \plus{} 2$. $ \Rightarrow |f(x)| > \sqrt {2}$[/quote]
I might be missing something trivial, but how did you conclude this? We don't know that sign of $ f$.

Use both of the conditions.

By the way, only from $ f(x) \equal{} f\left(\frac {1}{x}\right)$ and $ f(x) \equal{} af\left(\frac {x}{2}\right) \plus{} \frac {b}{2}x$, we can conclude that $ a \equal{} \frac {1}{2}$, which I didn't find yesterday so directly.  :)

$ f\left(\frac {2}{x}\right) \equal{} a.f\left(\frac {1}{x}\right) \plus{} \frac {b}{x} \equal{} a.f(x) \plus{} \frac {b}{x} \equal{} f\left(\frac {x}{2}\right) \equal{} \frac {f(x) \minus{} \frac {b}{2}x}{a}$
$ \Rightarrow f(x) \equal{} \frac { \minus{} b}{a \minus{} \frac {1}{a}}\left(\frac {x}{2a} \plus{} \frac {1}{x}\right)$

But since $ \frac { \minus{} b}{a \minus{} \frac {1}{a}}\left(\frac {x}{2a} \plus{} \frac {1}{x}\right) \equal{} \frac { \minus{} b}{a \minus{} \frac {1}{a}}\left(\frac {1}{2ax} \plus{} x\right)$ $ \Leftrightarrow f(x) \equal{} f\left(\frac {1}{x}\right)$, we find $ a \equal{} \frac {1}{2}$.


For $ a,b,c$ positive reals

$ \frac{\sum a^2}{\sum ab}\plus{}\sum \frac{a(b\plus{}c)}{b^2\plus{}c^2} \geq 4$

[quote="manlio"]For $ a,b,c$ positive reals

$ \frac {\sum a^2}{\sum ab} \plus{} \sum \frac {a(b \plus{} c)}{b^2 \plus{} c^2} \geq 4$[/quote]
Let $ a,$ $ b$ and $ c$ are non-negative such that $ ab \plus{} ac \plus{} bc\neq0.$
The original inequality is equivalent to $ (u^2 \minus{} v^2)w^6 \plus{} P(u,v^2)w^3 \plus{} Q(u,v^2)\leq0,$
where $ a \plus{} b \plus{} c \equal{} 3u,$ $ ab \plus{} ac \plus{} bc \equal{} 3v^2$ and $ abc \equal{} w^3.$
But $ f(w^3) \equal{} (u^2 \minus{} v^2)w^6 \plus{} P(u,v^2)w^3 \plus{} Q(u,v^2)$ is a convex function.
Hence, $ f$ gets a maximal value when $ w^3$ gets an extremal value.
Id est, we need to check two cases:
1) $ w^3 \equal{} 0$;
2) $ b \equal{} c,$ for which it's wrong. Try $ a\equal{}1$ and $ b\equal{}c\equal{}4.$ :wink:


geometry
perimeter
trigonometry
area of a triangle
Heron\u0027s formula
Pythagorean Theorem

[quote="sharkman"]The perimeter of an equilateral triangle is 3S.  In terms of S, find the area of this triangle.[/quote]

I am pretty sure this is how to do it  :maybe:

[hide]1/2S^2sin60[/hide]

That's right. The answer is $ \frac{\sqrt{3}}{4}S^2$.

So if the perimeter is 3S, the side length is S. Plug it into the forumula and you get S^2 x sqrt(3)/4 as the area.

Also consider Heron's Formula (I will show), or solving with the Pythagorean Theorem (divide triangle into 2 right triangles, then use $ \frac{bh}{2}$).

Heron's Formula Solution:

$ s\equal{}\frac{3S}{2}$

Each side is S...

$ A\equal{}\sqrt{\frac{3S}{2}\left(\frac{3S}{2}\minus{}S\right)\left(\frac{3S}{2}\minus{}S\right)\left(\frac{3S}{2}\minus{}S\right)}$

$ A\equal{}\sqrt{\frac{3S}{2}\left(\frac{S}{2}\right)^3}$

$ A\equal{}\sqrt{\frac{3S^4}{16}}$

$ A\equal{}\frac{\sqrt{3}S^2}{4}\equal{}\frac{\sqrt{3}}{4}S^2$

Pythagorean Theorem Solution:

Each side is length S. Draw the line that is perpendicular to one of the edges and that connects to a vertex. Now we have 2 30-60-90 triangles. The hypotenuse is S, so we know the short base is S/2. By Pythagorean Theorem, the taller base =  This is the height of the equilateral triangle, and S is the base. Thus, the area of an equilateral triangle is

$ \frac{S^2\sqrt{3}}{4}$

s^2 * sqrt{3}
___________

       4

sry that my notation is a little wierd :blush:

[quote="blue1421"][quote="sharkman"]The perimeter of an equilateral triangle is 3S.  In terms of S, find the area of this triangle.[/quote]1/2S^2sin60[/quote]

You probably used $ [ABC] \equal{} \frac {1}{2}AB\sin \angle C$; this method is correct, and if you note that $ \sin 60^o \equal{} \frac {\sqrt {3}}{2}$, you get the same answer.

That definetly wasnt High school :lol:

Ohk just to add to the spoils.
h= (3^1/2 )/2 * S

Area = H * S /2

       = (3^1/2)/4*S

[quote]perimeter of triangle is 3S,
then as it ia an equilateral lenght of triangle side will b 'S'

we know that area of triangle with sides a, b, c is equal to
let (a+b+c) = 2s then area A = [s(s-a)(s-b)(s-c)]^(1/2)
hence area of given triangle is
A = [(3S/2)((3S/2)-S)((3S/2)-S)((3S/2)-S)]^(1/2)
= [(3S/2)(S/2)(S/2)(S/2)]^(1/2)
= (3^1.2)[(S/2)^2]
= [(3^1/2)/4](S^2)
[/quote]This is same as fishythefish' solution by Heron's formula.


abstract algebra
group theory
superior algebra
superior algebra solved

Let $G$ be an abelian group and suppose that $G$ has elements of orders $m$ and $n$, respectively. Prove that $G$ has an element whose order is the least common mutiple of $m$ and $n$.

In 'Problems from Herstein', a proof is presented (but not correct)
[quote]
Let $o(a) = m$ and $o(b) = n$. If $(m, n) = 1$, then let $c = ab$.  Clearly, $c^{[m,n]} = c^{mn} = a^{mn} b^{mn} = e$. If $o(c) = r$, we have $a^r = b^(-r)$. Thus, $e = a^{mr} = b^{-mr}$. But then $n|mr$. Since $(m, n) = 1$, $n|r$. Similarly, $m|r$. This implies that $[m, n] = mn|r$ and thus $mn = r$.

If $(m, n) = d > 1$, then $o(a^d) = m/d$ and $o(b^d) = n/d$ and $(m/d, n/d) =
1$. Thus $(ab)^d$ has order $mn/d^2$ by previous case and thus $ab$ has order  $[m, n] = mn/d$.
[/quote]
The last step is not true. In fact, let $G$ be the cyclic group with 30 elements and let $m=6,n=10$.  It is easy to see that the product of elements of orders $6$ and $10$ will have order $15$.

Can anyone provide with a correct proof?

A realy quick proof relying on heavy machinery :).

The structure theorem for finite abelian groups says that the group generated by our two elements is a direct product of cyclic groups. It's obvious that in a direct product of finitely many finite cyclic groups which ahs order $k$, we can find a subgroup of order $d$ for any $d|k$. Since $[m,n]$ divides the order of the group generated by the two elements, we get the conclusion.

Anyway, the problem has been discussed before, and I remember jmerry posting a nice (not long or nasty at all) which did not use the structure theorem. I didn't want to go through the trouble of finding such a solution. Try searching for it; I believe it was in a thread started by the game ("the game" is a username :)).


modular arithmetic
Euler
number theory
prime factorization
number theory proposed

Denote by $ \n u(n)$ the exponent of 2 in the prime factorization of n!. Show that for arbitrary positive integers a and m there exists an integer n>1 for which $ \n u(n) \equiv a \mod(m)$.

P/S: Sorry for my carelessness in the solution posted before.

First we refer an identity of Legendre's function. For prime number $ p$, let $ u_p(n): \equal{} k$ such that $ p^k \parallel n!$ then $ u_p(n) \equal{} \frac {n \minus{} S_p(n)}{p \minus{} 1}$ where $ S_p(n)$ is the sum of digits of $ n$ in the $ p$-base form of $ n$.

In particular, $ u(n) \equal{} u_2(n) \equal{} n \minus{} S_2(n)$ where $ S_2(n)$ is the number of $ 1$ in the binary form of $ n$. Therefore, $ \{u(n)|n\in\mathbb{N}\} \equal{} \{\sum_{k \equal{} 1}^{\infty} \alpha_k (2^k \minus{} 1)|\alpha_k\in\{0;1\}\}$. Let $ m \equal{} 2^st$ where $ t$ is odd. Now there exists $ k$ such that $ 2^k \minus{} 1\equiv q \pmod{m}$ and $ (q,m)\equal{}1$ (indeed, since $ 2^{l\varphi(t)\plus{}1}\minus{}1\equiv 1\pmod{t}$ then $ (2^{l\varphi(t)\plus{}1}\minus{}1,t)\equal{}1$, on the other hand, it is clearly that $ 2^{\varphi(t)\plus{}1}\minus{}1$ is odd so $ k\equal{}l\varphi(t) \plus{} 1,\forall l\in\mathbb{N}$ satisfies). Take $ k>s$, also take $ k'\equal{}k\plus{}r\varphi(t)$ then we have $ m\equal{}2^st|2^k(2^{k'\minus{}k}\minus{}1)$ or $ 2^{k'} \minus{} 1\equiv 2^k \minus{} 1\equiv q \pmod{m}$. In this way, we generated infinitely many $ k$ such that $ 2^k \minus{} 1 \equiv q \pmod{m}$ and $ (q,m) \equal{} 1$. Then, let only those $ \alpha_k\neq 0$ then $ u(n)$ contains a sequence of the form $ qr$ where $ r\in\mathbb{N}^*$ and $ (q,m) \equal{} 1$. Hence, Chinese remainder's theorem yields what desired.

[i]Mashimaru , What is $ q$ in your solution ? :maybe: [/i]

[i]Here is the [b]official solution[/b] :

  Fristly , we will need to use the well -known lemma :

     With $ n \ \in \ \mathbb{N}$ then : $ v_2 (n!) \ \equal{} \sum_{k \equal{} 1}^{\infty} \left[ \frac {n}{2^k} \right] \ (*)$

   Assume that the presentation of $ n$ in $ 2$ base is : $ \overline{d_l d_{l \minus{} 1} ....d_0} \quad |_{2}$

   With $ d_l ; d_{l \minus{} 1} ; ...; d_0 \ \in \ \{ 0 ; 1 \} \ ; \ d_l \equal{} 1$

  Due to $ (*)$ , we have : $ v_2 (n!) \ \equal{} \sum_{k \equal{} 1}^{l} \left[ \frac {\overline{d_l d_{l \minus{} 1} ....d_0} \quad |_{2} }{2^k} \right]$[/i] $ \equal{} \sum_{k \equal{} 1}^{l} \overline{d_l d_{l \minus{} 1} ....d_k} \quad |_{2}$

$ \equal{} \sum_{k \equal{} 1}^{l} \sum_{i \equal{} k}^{l}d_i \cdot 2^{i \minus{} k} \equal{} \sum_{i \equal{} 1}^{l} d_i \left( \sum_{k \equal{} 1}^{i} 2^{i \minus{} k} \right) \ \equal{} \sum_{i \equal{} 1}^{l} d_i (2^i \minus{} 1) \ (**)$

[i]  Now we will construct the infinite increasing sequence $ (i_k)_{k \ge 1 }$ such that $ i_{n} \in \mathbb{N} \ \forall n \in \mathbb{N}$

    $ 2^{i_1} \minus{} 1 \equiv 2^{i_2} \minus{} 1 \equiv ... 2^{i_n} \minus{} 1 \equiv....\equiv r (mod \ \ m)$ , $ r$ is a positive integer such  that $ (r ; m ) \ \equal{} 1$

  Assume that $ m \ \equal{} \ 2^t u$ with $ u \ \in \mathbb{N} \ ; \ 2 \not | u$

   Choose the numbers $ i$ such that  $ i \ge t \ ; \ \varphi (u) | i \minus{} 1$

  Due to [b]Euler 's theorem[/b] , we have  : $ u | 2^{\varphi (u) } \minus{} 1 | 2^{i \minus{} 1} \minus{} 1$

  So , $ 2^{i} \minus{} 1 \equal{} 2( 2^{i \minus{} 1} \minus{} 1) \plus{} 1 \equiv 1 ( mod \ \ u) \ (1)$

and $ 2^{i} \minus{} 1 \equiv \minus{} 1 ( mod \ \ 2^t) \ (2)$ ( $ i \ge t$)

   From $ (1) ; (2)$ we can deduce that : $ (2^{i} \minus{} 1 ;m) \ \equal{} \ 1$
 For example , we can choose $ i_k \ \equal{} \ (t \plus{} k)\varphi (u) \plus{} 1 \ \forall k \ \in \mathbb{N}$ , $ t$ is a sufficient large positive integer  :lol:

    $ r$ is coprime with $ m$ , so there exists a positive integer $ u$ such that $ ur \equiv a (mod \ \ m)$

    Consider the number $ n_1 \ \equal{} \ 2^{i_1} \plus{} 2^{i_2} \plus{} ... \plus{} 2^{i_{u}}$

  We have $ v_2 (n!) \ \equal{} \ v_2 ( (2^{i_1} \plus{} 2^{i_2} \plus{} ... \plus{} 2^{i_{u}}) ! ) \ \equal{} \ \sum_{k \equal{} 1}^{u} ( 2^{i_k} \minus{} 1 )$ [/i]( [i]Due to[/i] $ (**)$ )

[i]   $ \equiv r \plus{} r \plus{} ... \plus{} r \equal{} ur \equiv a (mod \ \ m)$

  Similarly , we can easily see that the sequence $ (u_k)$ , which is defined by $ u_k \ \equal{} \ 2^{i_k} \plus{} 2^{i_{k \plus{} 1}} \plus{} ... \plus{} 2^{i_{u \plus{} k}}$

  have all elements satisfies the condition of the problem :D

  On the other hand , we can see easily that $ n_1 < n_2 < ...... < n_k < .......$ so there are infinite numbers $ n$

  such that $ v_2(n!) \equiv a \ \ (mod \ \ m )$ , QED[/i]


It depends. It certainly seems like a skill when you have good luck.  :)

i think if luck was a skill, luck wouldn't exist.
You are "lucky" if you win a coin toss.
If you were skilled at coin tosses, that would defeat the purpose of luck and the coin toss.

Luck is more of a factor than a skill. A skill is something you can change. Luck is why the two books [u]Introduction to Counting and Probability[/u] and [u]Intermediate Counting and Probability[/u] exist. If you could predict a dice roll, that would be called "hacking". :D

Well by the everyday definition of luck, then yeah it partly is. Somethings that are actually considered "luck" really aren't it at all and have to do with your skill. True "luck" however is truly up to chance.

The whole point of the word "luck" is to describe situations beyond the control of the individual:

[color=darkblue]1 a : a force that brings good fortune or adversity b : the events or circumstances that operate for or against an individual
2 : favoring chance[/color]

However, I believe "luck" is a skill that can be developed in some RPG's, if that's what you meant (of course, since it's forced, it's really a measire of consistency, and not one of luck).

[quote="fedja"]There is a saying that luck is when opportunity meets preparedness. :)[/quote]
The version of that quote that I'm familiar with is attributed to baseball executive Branch Rickey: "Luck is the residue of design."

Well, I'm going to be a sourpuss and note that gauss' question is the only point of interest here: the words "luck" and "skill" both are poorly defined, and as soon as you choose precise definitions for them the answer to the question will be obvious.

(Without thinking too hard, I would choose "the ability to perform a given task, particularly when this ability can be improved by practice" as my personal definition of "skill" and "the component of the consequences of one's actions that is not subject to one's choices and/or which is unforseeable in advance" as my personal definition of "luck," from which it becomes immediately clear that the two are independent of each other.)

In my opinion, there is no such thing as luck in this world, and it is one of the most overused terms. Sorry for being such an extreme, but that's just how I look at it.

i think that if you win a lottery you are lucky and that luck exists to some extent but no, I don't think skill exists in luck.  If you can't practice luck how can you get skilled

(otherwise i already would be  :) )

People often make the mistake of calling skill "luck" so this is harder to define.

i think luck is...luck

these are some vague terms we're using

[quote="Mewto55555"]This being a math website though, one might interpret the question as math-related. In that sense, it would be something like "Is guessing a skill?" which it clearly is. Many times what you guess is determined on a gut reaction, or your subconscious solving the problem or similar, so in that sense, "luck" is a skill.[/quote]

Yes, but guessing in math is often educated guessing.


Prove that $\frac{27n^4 + 10n^3 + 21n^2 + 14n}{12}$ is always an integer for n is an integer greater than 0

cool problem!
[hide]
Note that the numerator must be 0 mod 12.  Then we can subtract the coefficients by multiples of 12 to get $3n^4 - 2n^3 - 3n^2 + 2n \equiv 0 \pmod{12}$.  This factors to $3n^2(n^2 - 1) - 2n(n^2 - 1) = n(3n - 2)(n+1)(n-1) \equiv 0 \pmod{12}$
If we can prove that the LHS is always 0 mod 3 and 0 mod 4, then it will always be 0 mod 12.
One of $n-1, n, n+1$ must be 0 mod 3, so we have that.
The only way none of the previous three are 0 mod 4 is if $n \equiv 2 \pmod 4$.  However, if that's the case, $3n-2 \equiv 4 \equiv 0 \pmod 4$.
So the LHS is always 0 mod 3 and 0 mod 4, thus always 0 mod 12.
QED
[/hide]


[quote="burleyone"]V=(A+a/2)L

Solve for A

The A+a is over 2, I don't know how to type it that way here.

And the other one:

John scored 5 points higher on his midterm, and 13 points higher on his final, than on his first exam.  What did he score on his first exam if his average was 90?

Thanks in advance[/quote]
[hide="1"]$L\left(\frac{A+a}{2}\right)=V$

$\frac{A+a}{2}=\frac{V}{L}$

$A+a=\frac{2V}{L}$

$A=\frac{2V}{L}-a$[/hide]

[hide="2"]$\frac{x+(x+5)+(x+13)}{3}=90$

$3x+18=270$

$3x=252$

$x=84$[/hide]

Thanks for your help and your quick response!


search
number theory proposed
number theory

Find all $ (p,x,y)$ such that $ p$ is prime and $ z,y$ is positive interger such that $ \gcd (z,3) \equal{} \gcd (z,p) \equal{} 1$ and $ x^3 \minus{} p^3 \equal{} z^2$

[quote="aadil"]where is y in the question ?[/quote]
Man, its to find all $ p,x,z$


Is it possible to define the three edges of a triangle when the three distances of the vertexes from the orthocenter are given?

I think it is possible, but I cannot prove it...

The problem [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=27049]"triangle edges and orthocenter"[/url] was also posted and answered in Geometry, Olympiad section. It leads to soving cubic equations, for example, for $x = \frac a b,\ y = \frac b c$, where a, b, c are the triangle sides. Therefore, the problem cannot be solved with compass and straightedge.

This has been posted several times before, in different forms. One nice version of the cubic equation: if the distances from the othocenter are $x,y,z$, the circumradius $R$ satisfies $(2R)^3-(x^2+y^2+z^2)(2R)-2xyz=0$. The unique positive root is associated with the acute triangle that this satisfies, the negative root with larger absolute value is associated with an obtuse triangle, and the other negative root doesn't match a triangle at all.


X1,X2...Xn belong to R
X1^2 + X2^2+ .... + Xn^2 = 1

prove that:

X1/(1 + X1^2) + X2/(1 + X1^2 + X2^2) + ..... + Xn/(1+X1^2+....+Xn^2) < (n/2)^0.5

i think i solved it but i would like to see how others solve it.

[quote="yookon88"][quote="Mathsniper"]
 $\sum{x_{i}} \le \sqrt{2(\sum{x_{i}^2})}$, [/quote]
Why is this part true?[/quote]
sorry...... something I typed wrong here.
because $\frac{\sum{x^2}}{n} \ge (\frac{\sum{x}}{n})^2$. thus $\sum{x} \le \sqrt{n\sum{x^2}} \le \sqrt{n}$. The upperbound is larger than $\sqrt{n/2}$. I'm wrong. Try again. :blush:


function
real analysis
real analysis unsolved

Let $M$ be the set of nonintersecting circles on $Oxy$, all the circles touch the axis $Ox$.Prove:
a).The set of points of contact(where these circles touch $Ox$) can include $Q$.
b).The set of points of contact(where these circles touch $Ox$) can't include all irrational points of $\mathbb{R}$

This set of circles is clearly countable (select a rational point inside each circle), so (2) is obvious.

For the first part, we need to prove that we can assign a positive real $r_i$ to each rational $\alpha_i$ ($(\alpha_n)_n$ is an enumeration of the rationals) s.t. $2\sqrt{r_ir_j}<|\alpha_i-\alpha_j|,\ \forall i\ne j\ (*)$. We can do this inductively: we take $r_1$ to be anything we want, then we take $r_2$ to satisfy $(*)$ for $(i,j)=(1,2)$, then we take $r_3$ to satisfy $(*)$ for $(i,j)=(1,3),(2,3)$, and so on.


The dots in the diagram are equally spaced 2 units apart vertically and horizontally. What is the number of square units in the area of the triangle?

[asy]for(int i = 0; i < 3; ++i){
for(int k = 0; k < 4; ++k){
dot((i,k));}
}
draw((0,0)--(2,0)--(0,3)--cycle);[/asy]

The key here is that it says the dots are equally spaced $ 2$ units apart, not just $ 1$.  Therefore, the left leg has a length of $ 3\cdot2\equal{}6$ and the bottom leg has a length of $ 2\cdot2\equal{}4$.  So our area is $ \frac{4\cdot6}{2}\equal{}\boxed{12}$

You could have also noticed that the triangle is exactly half of the rectangle formed by connecting the dots. The sides of the rectangle have lengths $ 3 \cdot 2  \equal{} 6$ and $ 4 \cdot 2 \equal{} 8$, so it has area $ 6 \times 8 \equal{} 48$.

The triangle is half that, or $ \boxed{12}$.


limit
number theory unsolved
number theory

We have given irrational numbers $a,b,c,d$ such that $a+b=1$.Prove that $c+d=1$ if and only if $[na]+[nb]=[nc]+[nd]$ for all natural n.($[x]$ means the biggest integer not bigger then $x$).


At a particular school, all students participate in at least one out of three activities: debate, music, and community service. If 3 students do all three activities, 21 students do two or more activities, half of all students participate in music, and the ratio of participants in debate, music, and community service, respectively, is $ 5: 4: 3$, what is the total number of students at the school?

First let's define some variables.

Let $ a_1$ be the total number of kids(which is the number of kids doing at least one activity.)

Let $ a_2$ be the number of kids doing at least two activities.

Let $ a_3$ be the number of kids doing all three activities.

We know that the number of kids doing music is $ \frac{a_1}{2}$ and that the ratio of kids doing debate to kids doing music is 5 to 4, so the number of kids doing debate is $ \frac{a_1}{2} \cdot \frac{5}{4} \equal{} \frac{5a_1}{8}$. The ratio of the number of kids doing community service to the number of kids doing music is 3 to 4, so we can say that the number of kids doing community service is $ \frac{a_1}{2} \cdot \frac{3a_1}{8}$.

Now we can say:

$ \frac{a_1}{2} \plus{} \frac{5a_1}{8} \plus{} \frac{3a_1}{8} \equal{} a_1 \plus{} a_2 \plus{} a_3$

This becomes:

$ \frac{a_1}{2} \equal{} a_2 \plus{} a_3$

We know what $ a_2$ and $ a_3$ are, so substituting in those values and solving for $ a_1$ gives us $ \boxed{48}$.

[quote="Residual5000"][quote="isabella2296"]
Now we can say:

$ \frac {a_1}{2} \plus{} \frac {5a_1}{8} \plus{} \frac {3a_1}{8} \equal{} a_1 \plus{} a_2 \plus{} a_3$

This becomes:

$ \frac {a_1}{2} \equal{} a_2 \plus{} a_3$

[/quote]

Can you elaborate on that step?[/quote]

$ \frac {a_1}{2} \plus{} \frac {5a_1}{8} \plus{} \frac {3a_1}{8} \equal{} a_1 \plus{} a_2 \plus{} a_3$

$ \frac {4a_1}{8} \plus{} \frac {5a_1}{8} \plus{} \frac {3a_1}{8} \equal{} a_1 \plus{} a_2 \plus{} a_3$

$ \frac{12a_1}{8}\equal{} a_1\plus{}a_2\plus{}a_3$

$ \frac{3a_1}{2}\equal{}a_1\plus{}a_2\plus{}a_3$

$ \frac {a_1}{2} \equal{} a_2 \plus{} a_3$

[quote="NuncChaos"]Sorry to revive this, but why couldn't you use PIE; 3/2 S - 21 + 3 = S -> S=36 ?? :([/quote]

Answer to question: You made a mistake, it should be $\frac{3}{2}S - 21-3 = S \Rightarrow \frac{S}{2} -24 = 0 \Rightarrow \frac{S}{2} = 24 \Rightarrow S=\boxed{48}$. You should subtract $3$ not add $3$..... If you draw a Venn Diagram, this will be clearer.

Could anyone solve this problem using Venn diagram?
I still can't understand how you get 48.

The number of students who participated in 2 or more activities (21) includes the number of students who participated in all 3 activities (3).

Am I right?

[hide="Solution"]Let $A$ be the amount of students in the school. Then, the number of music students is $A/2$.

Since the ratio of debate, music, and community service students is $5:4:3$, and the "$4$" in the ratio is the number of music students, which is equal to $A/2$, we can say the number of debate students is equal to $5/4 \times A/2 = 5A/8$, and the number of community service students is equal to $3/4 \times A/2 =3A/8$.

Now if we add up the number of students in each club, we get: \[5A/8 + A/2 + 3A/8\]\[\to5A/8 + 4A/8 +3A/8\]\[\to12A/8\]\[\to1.5A\] But remember, we are overcounting! We must correct for the overcounting of the $21$ students who participated in two or more activities and the $3$ super-students that participated in all three.

Notice we counted the $21$ duo-participators twice, when we should have counted them once. So we subtract $21$ from $1.5a$. Then, we counted the $3$ super-students thrice, and we already subtracted them once because they're included in $21$ students who participated in two or MORE activities. So we just subtract $3$ one more time to make sure we counted the super-students once and only once.

Therefore, the total number of students in the school is equal to $3A/2 - 21-3 = 3A/2 -24$. Recall that $A$ is the number of students in the school as well. So: \[3A/2-24 = A\]\[\to3A/2 = A+24\]\[\to3A = 2A+48\]\[\to A=48\] And there we arrive at our answer, $A = \boxed{48}$. [/hide]

What is wrong with this solution: [hide]I marked the total number of people as 8x. This means 4x do music. This means 5x do debate and 3x do community service. When we total these we get 12x. Using the 3 ven diagram forumula we will have 12x-21+3 = 8x. This way 4x=18 so the total is 36. [/hide]

@ awesomemaths
If you want to set the total number of people in music to be 4x then since debate is 5/4 of music and community service is 3/4 of music then the total number of people at the school is 5x+4x+3x = 12x.

[hide=Solution]After using variables, we notice we have overcounted $\frac{n}{2}$, where $n$ is the total number of students. We have overcounted each of the two-subject areas once and counted the three-subject area twice. Through the fact that $21$ students do two or more activities, the two-subject areas sum to $21-3=18$. We subtract $3$ because of the overcounting of the three-subject area. Therefore, $\frac{n}{2}=18+2\times3$
$\frac{n}{2}=24$
$n=48$
There are $48$ students in the school.[/hide]

[hide=another sol] $5x+4x+3x$ counts all the people, but the $21$ people doing $2$ activities are counted twice and the $3$ doing all three are counted three times (two times they are included in the $21$). So the total number of people is $8x$ and we have $$12x-21-3=8x.$$ and then we have $8x=48$

I understand the solution, but why is mine wrong?

Let $m$ be the number of musicians. There are a total of $2m$ students in the School, so by the PIE principle we obtain
$$2m = m + \cfrac{5}{4}m+ \cfrac{3}{4}m-21 + 3,$$
which gives $m=18$.
There are therefore $36$ students in the school.

In the answers above, it is suggested that $3$ should be removed, not added. Why is this?

Thanks


Prove that a^b > b^a where a and b are integer and a>b>2 :)  :)  :)  :)  :starwars:

[quote="samath"]This is not true.
For example, let $a=5$ and $b=4$.

$5^4=625\not> 4^5=1024$
I think you may have written the problem  wrong, or I just missed something.[/quote]

You're correct. Another obvious case is when $a = 4$ and $b = 3$. Obviously, this is false since 64 < 81.

Topic is locked.


Determine all polynomials satisfying $xP(x-1) = (x-23)P(x)$ $\forall x$.

Lemma : If $P(x)$ is a polynomial such that $P(x)=P(x-1)$ for all real number $x$, then $P$ is constant.

Proof.
Assume that such a polynomial $P$ is nonconstant. Let $ \alpha$ be one of its (complex) root. It is easy to verify that $\alpha +1 $ is also a root of $P$. Thus, considering the sequence $U_0 = \alpha$ and $U_{n+1} = U_n + 1$, we deduce that $P$ has an infinite number of root, that is $P = 0$, a contradiction.

Now, let $P_0=P$ where $P$ is a polynomial satisfying the statement of the problem.
Clearly, $P(0)=P(22)=0$, which gives that $P(x)=x(x-22)P_1(x)$ where $P_1$ satisfies
$(x-1)P_1(x-1)=(x-22)P_1(x)$ for all $x \neq 0$ and $x \neq 22$ which means that this equality is in fact satisfied for all $x$.

As above $P_1(1)=P_1(21)=0$ so that $P_1(x)=(x-1)(x-21)P_2(x)$ and $(x-2)P_2(x-1)=(x-21)P_2(x)$.

And so on until we reach $(x-11)P_{11}(x-1)=(x-12)P_{11}(x)$ for all $x$, from which we deduce that $P_{11}(x)=(x-11)P_{12}(x)$ and $P_{12}(x-1) = P_{12}(x)$ for all $x$. From the lemma, we deduce that $P_{12}$ is constant.

It follows that $P(x)=kx(x-1)(x-2)...(x-22)$, for some sonctant $k$.

Conversely, it is easy to verify that these polynomials are solutions of the problem.

Pierre.

Aaargh...I didn't realize it was the beginner's corner....sorry...

Pierre.


trigonometry
calculus
calculus computations

Let $ f: [ \minus{} 1,1]\to\mathbb R$ be given by

$ f(x) \equal{} \begin{cases} {\sin\frac1x,} & {x\neq 0} \\
{0,} & {x \equal{} 0.} \end{cases}$

Show that $ f$ is integrable on $ [ \minus{} 1,1].$

You'd want to prove that for every $ \varepsilon>0$, you can find a partition $ P$ such that $ U(f,P)\minus{}L(f,P)<\varepsilon$.  [url=http://answers.yahoo.com/question/index?qid=20080626174057AAJidT9]This[/url] site provides a proof that $ f$ is integrable on $ [0,1]$ but you should be able to extend this to the other side.  Alternatively, the only discontinuity is at $ x\equal{}0$, which has measure 0, so you can integrate.


Let p(n)  be the greatest prime divisor of n. Show that there exists infinitely many positive integers n such that p(n)<p(n+1)<p(n+2).

Just have a look at :

http://www.mathlinks.ro/viewtopic.php?t=2005&highlight=brazil

Pierre.


Let $ a,b,c\geq0$ s.t. $ a^2\plus{}b^2\plus{}c^2\equal{}3$.Prove that:
$ 12\plus{}9abc\geq7(ab\plus{}bc\plus{}ca)$

Use schur , we need to prove :
$ 2p^3\minus{}6p \ge 7p^2\minus{}45$
<=>$ (2p\plus{}5)(p\minus{}3)^2 \ge 0$
Done

[quote="Inequalities Master"]Let $ a,b,c\geq0$ s.t. $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$.Prove that:
$ 12 \plus{} 9abc\geq7(ab \plus{} bc \plus{} ca)$[/quote]
See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=62537

Hey what kind of Shur did you used?
what is P please tell me about that?


hi people,

I'm arranging a math-contest online, made of some rounds. All begins from italian-oliforum..(the classification is unique!)
The first starts on 12:00AM of 20 september and it ends on 12:00 PM of 21 seprember, so saturday and sunday!
It is made of only 3 problems (each round), and i will give points 0 to 7: 6 for all correct solutions, 7 for only elegants and correct solutions  :lol:
It is tought for high-school student, (and also for undergraduate student, but they cannot use non-elementary-theory!).

The answers will be send to my e-mail, in a unique private message in latex model, without attachments (the questions are quite easy).
If you're interested in it, I need your acknowledgement, since I need some time to traslate the texts..Does someone want to partecipate?

Paolo Leonetti

Hmm... this needs to be moved, but I don't know where...

yes, i've thought it, but i don't know where post it..

however noone is interested in?

I think that this must be on the Italian forum or in the National Communities Round Table.

This is not a game (or a discussion of a game), so this is in the wrong forum.

ok, but italian forum i've already proposed this initiative..
If noone, then the mod can cancel this post..


$ ABCD$ is a convex quadrilateral with $ AB$ not parallel to $ CD$. A circle passing through $ A,B$ is tangent to $ CD$ at $ P$, and a circle passing through $ C,D$ is tangent to $ AB$ at $ Q$. Prove that the common chord of these two circles bisects $ PQ$ if and only if $ AD$ is parallel to $ BC$.


[quote="Ar3a51"]I usually have trouble with simple stuff (so I have been told).

If 3^33+3^33+3^33 = 3^x, what is the value of x.

For some reason, this sole problem has been nagging me for a long time.

Gracias in advance.[/quote]

[hide]Notice that this is just 3*3^33, so this is just 3^34, so x=34[/hide]

aahh... precious one...

[hide="solution"]3*3^33=3^34, x=34[/hide]

[quote="Ar3a51"]Thanks guys, I feel completely retarded now.   :D[/quote]

Don't, that's what we're here for. Those problems always irked me too, and they come up a lot.

I think when you have to solve this type of problems, observing and analyzing it visually does most of the job.

[hide]
$3^{33}+3^{33}+3^{33}= 3^{x}$

Oops, what can you see? You have 3 times $3^{33}$! What's next? You write that sum as $3\cdot 3^{33}$ and that means $3^{34}=3^{x}$ so the solution is $x=34$[/hide]


Σ=1/9n^2+3n-2

n=1-->Infinity

Does it converge? Calculate the sum of Σ

First factorize the denominator and use telescoping method.

Irrelevant, ypsonas.  The partial sums can be explicitly solved for, and they converge.  (Don't post calculus in the high school forums.  That's what the calculus forums are for.)

Also: we do encourage using $\LaTeX$ - that would increase the likelyhood of you getting a reply. But even if you don't, if you write in-line plain text algebra, then you [b]must[/b] use parentheses where needed - and I'd say your first post needs parentheses.


There was once a king who wanted to determine who the smartest person in the land was ). After a bunch of tests, it was down to two men (P and S), who claimed to be "perfect" logicians (meaning there was never any flaw in their logic).

The king then told them their final test:

"I am thinking of two different numbers greater than 1, whose sum is less than 100. I will tell P the product of my two numbers, and I will tell S the sum of the two numbers," the kind stated.

The king whispers in P's ear the product of his two numbers, and whispers to S the sum of the two numbers...The following is the conversation that the logicians had:

P: I don't know the two numbers...
S: I know you don't
P: Ah...now I know them...
S: So do I

Together, both men then stated the two numbers to the king, and he stared in disbelief...

What were the two numbers?

[quote="amanara"]There was once a king who wanted to determine who the smartest person in the land was ). After a bunch of tests, it was down to two men (P and S), who claimed to be "perfect" logicians (meaning there was never any flaw in their logic).

The king then told them their final test:

"I am thinking of two different numbers greater than 1, whose sum is less than 100. I will tell P the product of my two numbers, and I will tell S the sum of the two numbers," the kind stated.

The king whispers in P's ear the product of his two numbers, and whispers to S the sum of the two numbers...The following is the conversation that the logicians had:

P: I don't know the two numbers...
S: I know you don't
P: Ah...now I know them...
S: So do I

$ 2$ and $ 6$

Together, both men then stated the two numbers to the king, and he stared in disbelief...

What were the two numbers?[/quote]

[hide="Mathematic solution"]

Let the numbers be a and b.

Obviously, $ a\cdot b$ isn't prime or its cube (or P could find them in an instant).

From the S's first sentence, we see that the sum isn't sum of 2 primes

Because P knows the numbers now, $ a\cdot b$ has only one factorization in which $ a\plus{}b$ cannot be sum of 2 primes

Now because S knows the numbers, from all possible S that remained, we look for the one that can be only in 1 way split up to 2 numbers which product can't be factored in more than 1 way such that  a+b isn't sum of 2 primes (blah, blah you know what I mean)

Now it's done!

I have found 4,13.
[/hide]


inequalities
function
calculus
derivative
inequalities proposed

That's pretty obvious too. $f(x) = \frac{x}{\sqrt{1 - x}}$ is a convex function, hence \[\frac{1}{n} \sum_{i = 1}^n f\left(x_i\right) \geq f\left(\frac{1}{n} \sum_{i = 1}^n x_i\right) = f\left(\frac{1}{n}\right) = \frac{1}{\sqrt{n}\sqrt{n - 1}}\] and hence \[\sum_{i = 1}^n\frac{x_i}{\sqrt{1 - x_i}} \geq \sqrt{\frac{n}{n - 1}} = \sqrt{\frac{n}{n - 1}} \sqrt{\sum_{i = 1}^n x_i} \geq \frac{\displaystyle \sum_{i = 1}^n \sqrt{x_i}}{\sqrt{n - 1}}\] where the last inequality is just Cauchy-Schwarz.

Jensen's on the LHS with $f(x)=\frac{1}{\sqrt{1-x}}$ simplifies to
\[\displaystyle \frac{1}{\sqrt{1-(x_1^2+...+x_n^2)}}\geq \frac{\sum_{k=1}^{n}\sqrt x_{k}}{\sqrt{n-1}}\]
By Power Mean we know
\[\displaystyle \frac{1}{\sqrt{1-(x_1^2+...+x_n^2)}}\geq \frac{\sqrt{n}}{\sqrt{n-1}}\]
So it suffices to prove
\[\sqrt{n}\geq \sum_{k=1}^{n}\sqrt x_{k}\]
However this follows immediately from Jensen's again with $f(x)=\sqrt{x}$

how can be proven that $f(x) = \frac{x}{\sqrt{1 - x}}$ is a convex function??

Thanks :)

Check the sign of the second derivative:
$f''(x)=\frac{4-x}{4(1-x)^{\frac{5}{2}}}$
which is obviously non-negative for $x$ belonging to interval $(0; 1)$.


AMC
USA(J)MO
USAMO
geometry unsolved
geometry

ABC is an triangle. P is a point inside the triangle that PAB(angle)=10, PBA(angle)=20, PCA(angle)=30, PAC(angle)=40.
prove that ABC is an  isosceles triangle.

It is USAMO 1996 problem. The problem is posted before. As I know it is the first problem discussed on mathlinks.
Take a look at this:
http://www.mathlinks.ro/Forum/viewtopic.php?p=2#2


Is it possible that we can draw uncountable circles such that any circle doesn't intersect another circle and any circle doesn't contain another circle inside of this?

Seems no, because each circle will contain a rational point.


linear algebra
matrix
combinatorics unsolved
combinatorics

Let $G$ be a graph such that the set of neighbours of any two distinct vertexes of it aren't equal.Prove that there exists a vertex of $G$ such as $v$ , such that in $G-v$ also there isn't any two vertexes such that the set of their neighbours are equal.

Apply [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=Kurschak&t=34747]this[/url] to the adjacency matrix of the graph. I think it should work.


inequalities
trigonometry
geometry
inradius
function
inequalities proposed

ABC is  an acute triangle . Show  that  [tex] \sqrt{3}\cdot max(h_a,h_b,h_c)\geq s [/tex]
[tex] h_a , h_b , h_c [/tex]  are  altitudes , and  [tex] s[/tex] is semiperimeter . Thank  you!

Trigo?
Let $h=h_c=\max(h_a,h_b,h_c)$. Then $c=\min(a,b,c)$. We can move $A$ along the line $AB$ increasing $c$ until $c=a$ or $c=b$. We have increased $s$ and remained $\max(h_a,h_b,h_c)$ unchanged. Thus we obtain the isosceles triangle $ABC$. Suppose wlog $AB=AC$. Then $\angle BAC\geq \pi/3$ ($AB$ is the minimal side!!!).
Assume now that $AB=1$. The inequality $ \sqrt{3}\cdot \max(h_a,h_b,h_c)\geq s$ reduces to $\sqrt 3\sin \alpha\geq 1+\sin\alpha/2$ for $\alpha\geq \pi/3$. You can apply various methods to justify it. :)

At first I rewrite the problem to contain the definitions of the involved notations (I have explained in another thread why this is important and useful, so please do it!):

[color=blue][b]Initial problem:[/b] Let ABC be an acute triangle with altitudes $h_a$, $h_b$, $h_c$ and semiperimeter $s=\frac{a+b+c}{2}$. Prove that $\sqrt3\cdot\max\left\{h_a;\;h_b;\;h_c\right\}\geq s$.[/color]
_______________________________________________________________

Does this problem have a nice solution at all? Here is a solution by Peter Scholze and me. I have given up trying to make it halfways elegant.

[b]Lemma 1.[/b] If x, y, z are three reals such that x > 1, y > 1, z > 1, $y+z>2\sqrt3$, $z+x>2\sqrt3$ and $x+y>2\sqrt3$, then xyz > x + y + z.

I put the proof of Lemma 1 into a hider for a good reason - be warned! (And, please, look for something less brutal...)

[hide="Proof of Lemma 1"][i]Proof of Lemma 1.[/i] WLOG assume that $x\geq y\geq z$. Then, we will consider two cases: $y\leq 2\sqrt3-1$ and $y\geq 2\sqrt3-1$.

In the case $y\leq 2\sqrt3-1$, we can use the identity

$xyz-\left(x+y+z\right)=\left(xy-1\right)\left(y+z-2\sqrt3\right)+\left(x-\sqrt3\right)\left(y\sqrt3-y^2+2\right)+\sqrt3\left(x-y\right)\left(y-\sqrt3\right)$.

But

xy - 1 > 0 since x > 1 and y > 1 imply xy > 1;
$y+z-2\sqrt3>0$ since $y+z>2\sqrt3$;
$x-\sqrt3>0$ since $x-\sqrt3=\frac12\left(x-y\right)+\frac12\left(x+y-2\sqrt3\right)$ (but $x-y\geq 0$ and $x+y-2\sqrt3> 0$);
$y\sqrt3-y^2+2>0$ since $y\sqrt3-y^2+2=\left(3\sqrt3-5\right)+\left(\left(2\sqrt3-1\right)-y\right)\left(y+\left(\sqrt3-1\right)\right)$ (but $3\sqrt3-5>0$, $\left(2\sqrt3-1\right)-y\geq 0$ and $y+\left(\sqrt3-1\right)>0$);
$x-y\geq 0$ since $x\geq y$;
$y-\sqrt3>0$ since $y-\sqrt3=\frac12\left(y-z\right)+\frac12\left(y+z-2\sqrt3\right)$ (but $y-z\geq 0$ and $y+z-2\sqrt3>0$).

Thus, we get xyz - (x + y + z) > 0, so that xyz > x + y + z, and the assertion of Lemma 1 is proven in the case $y\leq 2\sqrt3-1$.

The case $y\geq 2\sqrt3-1$ is simpler: We use the identity

$xyz-\left(x+y+z\right)=\left(xy-1\right)\left(z-1\right)+\left(y-\left(2\sqrt3-1\right)\right)\left(x-1\right)$
$+2\left(x+1-2\sqrt3\right)\left(\sqrt3-1\right)+2\left(7-4\sqrt3\right)$.

But

xy - 1 > 0 since x > 1 and y > 1 imply xy > 1;
z - 1 > 0 since z > 1;
$y-\left(2\sqrt3-1\right)\geq 0$ since $y\geq 2\sqrt3-1$;
x - 1 > 0 since x > 1;
$x+1-2\sqrt3\geq 0$ since $x+1-2\sqrt3=\left(x-y\right)+\left(y-\left(2\sqrt3-1\right)\right)$ (but $x-y\geq 0$ and $y-\left(2\sqrt3-1\right)\geq 0$);
$\sqrt3-1>0$;
$7-4\sqrt3>0$.

Thus, xyz - (x + y + z) > 0, so that xyz > x + y + z. Hence, the assertion of Lemma 1 is also proven for the case $y\geq 2\sqrt3-1$. This completes the proof of Lemma 1.[/hide]

Now the solution of the problem will be quite simple:

We have to prove that $\sqrt3\cdot\max\left\{h_a;\;h_b;\;h_c\right\}\geq s$. Assume the contrary, i. e. assume that $\sqrt3\cdot\max\left\{h_a;\;h_b;\;h_c\right\}<s$. Then, since the altitude $h_a$ is trivially $\leq\max\left\{h_a;\;h_b;\;h_c\right\}$, we have $\sqrt3\cdot h_a<s$. Thus, $\sqrt3<\frac{s}{h_a}$.

But the area $\Delta$ of triangle ABC can be written in two ways: On the one hand, $\Delta=\frac12ah_a$, while, on the other hand, $\Delta=sr$, where r is the inradius of triangle ABC. Thus, $\frac12ah_a=sr$. Consequently, $\frac{s}{h_a}=\frac{a}{2r}$. Hence, the inequality $\sqrt3<\frac{s}{h_a}$ becomes $\sqrt3<\frac{a}{2r}$, so that $2\sqrt3<\frac{a}{r}$.

Now, denote $x=\cot\frac{A}{2}$, $y=\cot\frac{B}{2}$, $z=\cot\frac{C}{2}$. Since the triangle ABC is acute, we have A < 90°, so that $\frac{A}{2}<45^{\circ}$, and thus (since the function $f:t\to\cot t$ is monotonically decreasing on the interval ]0°; 90°]), we have $\cot\frac{A}{2}>\cot 45^{\circ}$. In other words, x > 1. Similarly, y > 1 and z > 1.

By the half-angle formulas, we have $\tan\frac{B}{2}=\frac{r}{s-b}$, so that

$y=\cot\frac{B}{2}=1:\tan\frac{B}{2}=1:\frac{r}{s-b}=\frac{s-b}{r}$.

Similarly, $z=\frac{s-c}{r}$. Thus,

$y+z=\frac{s-b}{r}+\frac{s-c}{r}=\frac{2s-\left(b+c\right)}{r}=\frac{\left(a+b+c\right)-\left(b+c\right)}{r}=\frac{a}{r}$.

Since $2\sqrt3<\frac{a}{r}$, we thus have $2\sqrt3<y+z$, so that $y+z>2\sqrt3$. Similarly, $z+x>2\sqrt3$ and $x+y>2\sqrt3$.

So we have obtained x > 1, y > 1, z > 1, $y+z>2\sqrt3$, $z+x>2\sqrt3$ and $x+y>2\sqrt3$. Thus, by Lemma 1, it follows that xyz > x + y + z. In other words,

$\cot\frac{A}{2}\cot\frac{B}{2}\cot\frac{C}{2}>\cot\frac{A}{2}+\cot\frac{B}{2}+\cot\frac{C}{2}$.

But this is impossible because of a well-known trigonometric identity for the angles A, B, C of a triangle:

$\cot\frac{A}{2}\cot\frac{B}{2}\cot\frac{C}{2}=\cot\frac{A}{2}+\cot\frac{B}{2}+\cot\frac{C}{2}$.

Thus, we get a contradiction; consequently, our assumption $\sqrt3\cdot\max\left\{h_a;\;h_b;\;h_c\right\}<s$ must be wrong. Hence, the inequality $\sqrt3\cdot\max\left\{h_a;\;h_b;\;h_c\right\}\geq s$ is proven. $\blacksquare$

My hope is that somebody finds a nice approach to Lemma 1...

  darij

Well, I apologize for the tone, since I misunderstood your solution and thought it would be rather difficult. But it would still interest me whether there is a nice way ("nice" meaning mainly without case distinction) to prove Lemma 1 algebraically...

  Darij


Gauss
linear algebra
matrix
absolute value
linear algebra unsolved

Suppose we have  a matrix of this form:

$A = \left( \begin{array}{cc}
S & M \\
M & S \\
\end{array}
\right) $
S is diagonal, non-singular

What can we say about the convergence of Ax=b with one of the above methods?

Thanx

[quote="amfulger"]The problem text is clearly copied from a book. Most of us don't have that book so please explain the terms:
1. convergence of $Ax=b$;
2. one of the above methods[/quote]
It is not from a book...

I want to solve Ax = b  by using an iteration method.
That it choose $x_0$
and compute $x_{i+1} = f(A,b,x_i)$ until I get a sufficient precision.

The methods are the Jacobi and Gauss-Seidel. Since I wrote it in the subject.. that what I mean by "above"
(More about them one can find in wolfram.math)

Now we know that the GS method converges if the matrix is idd, and they both converge if the matrix is sdd (strictly diagonaly dominant).

We define
A = M-N where M is a diagonal non singular matrix.
Then $C = M^{-1}N$
And if $\rho(C)$ is lower then 1 then the methods converge..
($\rho(C)$ is the radial spectrum, that is the biggest eigenvalue )

I also need to know when one method converges better than the other.

Here is a bit more for Jacobi:
$M = \left( \begin{array}{cc}
S & 0 \\
0 & S \\
\end{array} \right) $
$N = M-A = \left (\begin{array}{cc}
0 & M \\
M & 0 \\
\end{array} \right) $
$C = M^{-1}N = \left( \begin{array}{cc}
0 & S^{-1}M \\
S^{-1}M & 0 \\
\end{array} \right) $
I need to find the biggest (absolute value) eigenvalue of C... if it is lower than 1 then the method converges.
The same for Gauss-Seidel but
$ M = \left(\begin{array}{cc}
S & 0 \\
M & S \\
\end{array} \right)
$


How many ways are there to arrange 5 people in 20 chairs so that none sit next to another?

We can imagine there are 16 chairs and choose 5 for the people, and then place one of the remaining four chairs between each pair, for 16C5 combinations.

But we can also place a chair between each pair, so that 5 people and 4 intermediate chairs --> 9 chairs taken, and then insert the remaining 11 chairs anywhere in the ten slots that the used 9 chairs make. That is 9+11 C 9 = 20C9.

I know the first approach and answer are correct, but I cannot find a mistake in the second approach... :maybe:

I think the second one overcounts the number of cases.
\[| * | * | * | *| \]
Let the $*'s$ represent the intermediate chairs, and the $|'s$ represent the occupied chairs in the diagram above. The second method would distinguish between placing a chair between the first $|*$ and the first $*|$ (reading from left to right), when the resulting setup is the same and so undistinguishable: two unoccupied chairs between the first and second occupied chairs.


geometry
circumcircle
power of a point
radical axis
geometry proposed

$ AB,AC$ are tangent to circle $ (O)$. $ M,N$ are the midpionts of $ AB,AC$. $ X,Y$ lie on $ MN$ such that $ YAX\equal{}ANM$. prove that  the circumcircle of $ AYX$ si tangent to $ (O)$

I shall re-word the problem to match my solution, the new problem being equivalent to the original one:

[b][color=darkblue][i][size=117]Problem:[/size][/i]
Let $ O$ and $ O'$ two externally tangent circles, $ T$ being their common point. From a point $ A$ belonging to $ O'$, $ A \neq T$, one draws the tangents $ AB$ and $ AC$ to the circle $ O$, $ B$ and $ C$ belonging to the circle $ O$. $ BT$ and $ CT$ intersect the circle $ O'$ at $ \ { T, D\ }$ and $ \ { T, E\ }$ respectively.
Prove that, $ D$ and $ E$ belong to the midline of $ \triangle ABC$[/color][/b]

Let $ \ {M' \ } \in DE \cap AB$ and $ \ {N' \ } \in DE \cap AC$. $ AB$ being tangent to the circle $ O$, we get $ \widehat {ABC} = \widehat {BXC}$, $ X$ being any point on the arc $ BC$ not containing $ T$ of the circle $ O$, so $ \widehat {ABC} = \widehat {ETB} = \widehat {EM'B}$ ( 1 ) ( $ DE \parallel{} BC$, from the homothety). From (1) we get $ BEM'T$ being cyclic and $ \widehat {M'BT} = \widehat {M'ET}$, but $ \angle M'ET = \angle DET = \angle DAT$, consequently $ \angle DAT = \angle M'BT$ ( 2 ), i.e. $ AD$ is tangent to the circle $ (BAT)$, hence $ AD^2 = DT \cdot DB$ ( 3 ), similarly finding $ AE^2 = ET \cdot EC$ ( 4 ), the relations (3) and (4) showing that $ D$ and $ E$ belong to the radical axis of the "circles" $ O$ and $ A$, which is the midline of $ \triangle ABC$, thence $ M' \equiv M$ and $ N' \equiv N$, thus the original claim is proved.

Best regards,
sunken rock

thank you.my solution is very near to you.i solved the main problem.

Thank mathlinker for nice problem!!  :) [size=200] [color=red]Happy new year[/color][/size]


A piston weighing $ 3kg$ has the form of a circular disc of radius $ R \equal{} 4cm$. The disc has a hole into which a thin walled tube of radius $ 1cm$ is inserted. Initially the piston is at the bottom of the cylinder. Now, $ 700gm$ water is poured into the pipe.

Now, answer the following questions:
$ \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{}$

$ \color{red} 1)$ The height to which the piston will rise is:

$ A)$ $ 5cm$

$ B)$ $ 10cm$

$ C)$ $ 15cm$

$ D)$ $ 20cm$

$ \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{}$

$ \color{red} 2)$ The length of the liquid column in the pipe is:

$ A)$ $ 50cm$

$ B)$ $ 54cm$

$ C)$ $ 60cm$

$ D)$ $ 64cm$

$ 700 \equal{} 1 x 3.14 x 16 x h$
$ 15cm \equal{} h$

Now pressure just below the cylinder = $ P_o \plus{} \frac {3g}{\pi((4cm)^2 \minus{} (1cm)^2)}$= 6370pa

This pressure is compensated in the hole by the rise in liquid.
=> $ 1000gh \equal{} 6370$
$ h \equal{} 64cm$

[quote="Conjurer"]$ 700 \equal{} 1 x 3.14 x 16 x h$
$ 15cm \equal{} h$

[/quote]
can you explain this part,i got the 2nd part but in the solution,it is given that the piston will rise by 10cm

It might, because I have neglected the rise in the capillary, so actually answer has to be less than 15 so 10. :D

[quote="Conjurer"]
Now pressure just below the [b]cylinder[/b] = $ P_o \plus{} \frac {3g}{\pi((4cm)^2 \minus{} (1cm)^2)}$= 6370pa

[/quote]
[b]
Cylinder[/b]? I suppose you mean piston, don't you? Thanks, I got it! Nice thinking! Initially I couldn't understand why the water wouldn't pour out of the pipe...now I get it!

[quote="Conjurer"]$ 700 \equal{} 1 x 3.14 x 16 x h$
$ 15cm \equal{} h$
[/quote]

vivek_n, he means to say the rise would have been $ 15cm$ without the hole and the pipe.

[quote="Rajiv"]

vivek_n, he means to say the rise would have been $ 15cm$ without the hole and the pipe.[/quote]

got it

Rajiv, do you have any idea when the results for "part test 3" will be out

[quote="vivek_n"]
Rajiv, do you have any idea when the results for "part test 3" will be out
[/quote]
No idea!

But, seems the problem refers to the rise of the piston with the hole and the pipe, that are part of the problem !


Hi beginners. Solve this one please. Don't leave it to rot.  :D  :P  :D

If it sits still this long in Getting Started, it probably needs a promotion to Intermediate.

Hint:   ( Millisecond answer)   if the number is odd   so its square  root is also odd... and square of  an odd number is always 1 modulo  8. (why?)

 See if  you  can get something like:  [hide] x^2(2x+1)=4*k(k+1) [/hide]  and see if you can  solve it.  Remember that there are no common factor between k and k+1  (or x and 2x+1)) ...

Hello, DPatrick!

[quote]In what base(s) (less than or equal to 12) is 2101 a perfect square?[/quote]
$2101_3 \;= \;64 \;= \;8^2$
$2101_8  \;= \;1089 \;= \;33^2$