[quote="1337math"]If there are 5 library checkout lines and 12 people waiting for books, how many diffrent ways can the 12 people be arranged in the 5 checkout lines?

Plz help?  :?:[/quote]

[hide]err.. I got 792. I usually don't get these kind of problems right though... i just did $\binom{12}{5}$[/hide]

[quote="ch1n353ch3s54a1l"]err.. I got 792. I usually don't get these kind of problems right though... i just did $\binom{12}{5}$[/quote]

That would be more like if the question was asking how many ways could you make one line out of 12 people, if order didn't matter ;) This is more of a subset question, methinks.

Here's a clue. Each person is in either line 1, line 2, line 3, line 4, or line 5. How many combinations are there if we look at the people rather than the lines?

[hide]Each of the twelve people can be in any one of the 5 lines, so we have $5^{12}$. :? [/hide]

[hide]I agree on pianoforte's answer ($5^{12}$), unless the 12 people are indistinguishable or the 5 lines are indistinguishable :roll:[/hide]

[quote="chess64"][hide]The number of solutions to the equation $x_1 + x_2 + x_3 + x_4 + x_5 = 12$ is $\binom{12+4}{4}=\boxed {\binom{16}{4}}$.[/hide][/quote]
That's correct if we interpret that the 12 people are indistinguishable.
But I assume each line has to consist of at least one person, so I would say that the answer is $\binom{7+4}{4}=\binom{11}{4}$ in this case.

hmm, i suppose some lines could be empty, but in any case, does anyone know how we could find the answer if we assumed each person were distinguishable? let's ignore the order of people in the lines for now..

Yeah, in my solution I assumed that all 12 people were clones :lol:

[quote="MCrawford"]This problem is too difficult for the Middle School forums.  I'll move it to a place where you're likely to get a more thorough set of responses.  In general, look around in the forums to make sure you are posting in the right place.  Some of the forums have guidelines to help.[/quote]

xD I'm in middle school though...

The forums aren't set up in terms of grade but rather level.  The middle school forums refer to math that is roughly at that difficulty (it's not for only middle schoolers to post whatever their level).

If each person were indistinguishable, order does not matter: $\binom{5+12-1}{12} = \binom{16}{4}$
If each person is distinguishable, order does not matter: $5^{12}$
If each person is distingushable and order does matter: $12!\binom{5+12-1}{12} = 12!\binom{16}{4}$

[quote="breez"]If each person were indistinguishable, order does not matter: $\binom{5+12-1}{12} = \binom{16}{4}$
If each person is distinguishable, order does not matter: $5^{12}$
If each person is distingushable and order does matter: $5!\binom{5+12-1}{12} = 5!\binom{16}{4}$[/quote]

this may be really obvious, but can you give an explanation for the second two answers, since you seem to be fairly sure of them?

We'll assume all 5 lines are distinguisble, perhaps labeled with $1, 2, 3, 4, 5.$

If people are indistinguishable, and order does not matter, its a matter of placing $x$'s into 5-1 $/$.

Observe:
$xxx/xx/xx/x/xxxx$ would be on possible arrangement.

So we find the number of permutations of $12x$'s and 5-1 $/$'s, which would give us the total number of arrangements of $12$ people in $5$ lines.

Which is: $\binom{5+12-1}{12} = \binom{16}{4}$

This is basically finding combinations with unrestricted repetitions.

If people are distinguishable, order does not matter, then each person can be any of the five lines, so $5$ choices for the 1st, $5$ for the 2nd, etc. This yields $5^{12}$

If each person is distinguishible, and order does matter, its just $\binom{16}{4}$ multiplied by $12!$ as there are 12 unique $x$'s and each permutation is now considered a unique arrangement.

ah okay, that makes a lot of sense. thanks!  :)


A colorless solid $ A$ on strong heating gives a brown gas $ B$ and a colorless gas $ C$ and a residue $ D$. On dissolution of solid $ A$ in $ NH_3$ a complex $ E$ is formed which can oxidize glucose to $ F$. Aqueous solution of $ A$ also gives brick red precipitate with $ K_2CrO_4$.

Identify all with equations
[hide="I think"]

$ A$ can be $ Hg^{2\plus{}}$ salt or $ Ag^\plus{}$ salt.
$ B$ is $ NO_2$
$ C$ is $ O_2$[/hide]

The glucose reaction sort of gives it away. Gives a hint at Tollen's reagent.

So, I think
A=silver nitrate
B=NO2
C=O2
D=metallic silver
E=$ \ce {Ag(NH_3)_2^+}$
F=gluconic acid

Yes these are correct.

I am bad at biochem but I could have guessed Tollen though.  :(


I'm fairily new to this discrete maths thing and I'm stuck on a problem that seems quite simple:

$ Find\ r_{1}\in\mathbb{Z}\ such\ that\ 0\leq r_{1}\leq 40\ and\ 2004^{41}\equiv r_{1}(mod\ 41)$
$ Find\ r_{2}\in\mathbb{Z}\ such\ that\ 0\leq r_{2}\leq 60\ and\ 2004^{6000}\equiv r_{2}(mod\ 61)$

The first problem I think can be solved by Fermat's little theorem. Since 41 is a prime, (2004^41)-2004 would probably be divisible by 41? But r1 should lie between 0 and 40, and I can't figure that part out. 2004 gives a remainder 36 when divided by 41, but then there was the power part.

I'm pretty lost on the second one, but 61 is also a prime, but the power and the divider is not the same so that rules out Fermat?

For the first, Fermat gives $ 2004^{40}\equiv 1\implies 2004^{41}\equiv 2004\equiv \boxed{36}\bmod 41$

In the second, Fermat gives $ 2004^{60}\equiv 1\bmod 61$. How can you use this to rewrite or break down that large exponent?
Hint: Use the rule $ (a^b)^c\equal{}a^{bc}$.

well for the first part grn_trtle is correct.
following from the hint he has given
$ 2004^{60} \equiv 1\mod61$ by fermats theorem.since you have the remainder to be $ 1$ (very useful) you raise the congruent equation to the power $ 100$
$ 2004^{60*100}\equiv 1^{100}\ mod61$
so we have $ r_2 \equal{} 1$
sorry if i have posted something very obvious.


calculus
integration
function
real analysis
complex analysis

For my research i need to solve a complex integral. I'm even not sure if there's a analytical solution for it. The integral has the following form:

$\int^{b}_{a}{\frac{1}{\alpha_{1}-\sqrt{\alpha_{2}-\alpha_{3}e^{-\alpha_{4}x}+\alpha_{5}x}}dx}$

All alpha's are constants, larger than 0. Also both integration limits, a and b, are positive values.
Does somebody have an idea how to solve this equation? Or is trying to find a analytical solution for this integral a waste of time?
Thanks in advance!

Trying to find an analytic solution is a waste of time. The function almost certainly has no elementary antiderivative. Just do it numerically.

By the way, this is a real integral- it belongs in the calculus/real analysis section. There are no complex numbers involved.

That's a pity. Although, it is what I expected.
I was not sure where to post this problem. I assumed that if there was an analytical solution it would be a complex one... that's why I posted it here.

Anyway, thanks for your quick response!!!


Hi again, 3 new questions.

1. Triangle ABC is isosceles with AB=AC and D is the midpoint of BC.  The point K is on the line through C parallel to AB, K ≠ C.  The line KD intersects the sides AB and AC (or the extensions of the sides) at P and Q respectively.  Show that AP • QK = BP • QP.
[img]http://i75.photobucket.com/albums/i285/ngr_sqd/fig1.jpg[/img]

2. PA and PB are tangents to a circle, and a secant passing through P intersects the circle in C and D respectively, D-C-P.  Chords AC, BC, BD, and DA are drawn as illustrated in Fig. 2.  If AC = 9, AD = 12, BD = 10, find BC.
[img]http://i75.photobucket.com/albums/i285/ngr_sqd/fig2.jpg[/img]

3. Side AB of square ABCD is extended to P so that BP = 2AB.  With M the midpoint of DC, BM is drawn meeting AC at Q.  PQ meets BC at R.  Using Menelaus' Theorem find the ratio CR/RB.

[color=red]EDIT: I added Fig. 1 to clear up confusion.[/color]

[quote="jli"]how is D the midpoint of BC.  for the first one, AP = BP so you just need to prove QK = QP.[/quote]

You accidentally combined the first and second questions.  A figure is provided for problem 2.

[quote="GlenthemanN"]how can AP = BP if the line passes through Q K D and P when D is the midpoint of BC?
also QK cannot be equal to QP since they lie on the same line..

unless I drew it out wrong? Q lies ont he extention of AC[/quote]

You are correct, as jli was looking at the figure for the wrong problem.  You can draw this problem out in two ways.

For clarification, here is Fig. 1 for Problem 1:
[color=red]EDIT: I threw Fig. 1 up top.[/color]

[quote="parijat"]1)Triangles BPD and CDK are congruent. so,[b]BP=CK[/b]. Triangles CQK and APQ are similar and hence AP/[b]CK[/b]=QP/QD. REARRANGE TO GET RESULT
2)extend BQ to meet AD extended at M. By midpt. theorem AD=DM.

apply menelaus in ACD and get AQ/QC
apply menelaus in  ABC and get CR/RB
OOPS need to go somewhere urgently (3) remains for someone else to post[/quote]

You actually did #3, not #2.  Leaving #2 open for solution.

[hide] Since we have that $\angle CAP=\angle ADP$ and $\angle CPA=\angle DPA$, we have that $\triangle APD\sim\triangle CPA$.  We thus have that $\frac{AC}{AD}=\frac{AP}{PD}$.  We already know that AC=9 and AD=12, so $\frac{AC}{AD}=\frac{3}{4}$.  Thus, $\frac{3}{4}=\frac{AP}{DP}$.  Now, with similar reasoning, we can say that $\triangle PBD\sim\triangle PCB$.  Thus, $\frac{PB}{PC}=\frac{PD}{PB}$.  We know that AP=PB, so since $\frac{PA}{PD}=\frac{3}{4}$, we know that $\frac{PD}{PB}=\frac{4}{3}$, so $\frac{4}{3}=\frac{BD}{BC}\Rightarrow BC=\frac{3BD}{4}$.  We also know that $BD=10$, so $\boxed{BC=7.5}$ [/hide]


geometry
calculus
derivative
inequalities
quadratics
trigonometry
triangle inequality

So there is a triangle of side lengths 6, 7, and x. What is the maximum area this triangle can be?

Is there any way to do this in just a few minutes without a calculator or any calculus? Because I can do it using derivatives but it's supposed to be an algebra 2 problem...

-G.L.

hello, you can use Heron's fomula $ A\equal{}\sqrt{s(s\minus{}a)(s\minus{}b)(s\minus{}c)}$, let $ a\equal{}6,b\equal{}7$ and $ c\equal{}x$ you have $ A(x)\equal{}\frac{1}{4}\sqrt{(169\minus{}x^2)(x^2\minus{}1)}$.
Sonnhard.

The biggest $ x$ can be by the Triangle Inequality Theorem is $ 6 \plus{} 7 \minus{} 1 \equal{} 12$. Plugging this into what Sonnhard got yields:

$ \frac{1}{4} \times \sqrt{(169 \minus{} 12^2)(12^2\minus{}1)} \implies \frac{5\sqrt{143}}{4}$

Extending this idea, you can then multiply it out and make some kind of substitution such as X^2=a you obtain the square root of a quadratic equation. Since this quadratic equation has a negative leading coefficient, it points down and thus has a maximum at the vertex. Thus, in order to find the maximum, you just take the square root of the value for x whose max is the value of y, and then check to obtain a value as close to this as possible using triangle inequality.

sorry didn't see isabella's post.

[quote="isabella2296"]The biggest $ x$ can be by the Triangle Inequality Theorem is $ 6 \plus{} 7 \minus{} 1 \equal{} 12$. Plugging this into what Sonnhard got yields:

$ \frac {1}{4} \times \sqrt {(169 \minus{} 12^2)(12^2 \minus{} 1)} \implies \frac {5\sqrt {143}}{4}$[/quote]

Only if x has to be an integer.

[quote="isabella2296"]The biggest $ x$ can be by the Triangle Inequality Theorem is $ 6 \plus{} 7 \minus{} 1 \equal{} 12$. Plugging this into what Sonnhard got yields:

$ \frac {1}{4} \times \sqrt {(169 \minus{} 12^2)(12^2 \minus{} 1)} \implies \frac {5\sqrt {143}}{4}$[/quote]

Also, we're trying to maximize the area, not the sidelength. Using greatest possible $ x$ (assuming it doesn't have to be an integer) will actually give you an area of $ 0$.

[u]Two things:[/u]

[b](I)[/b] [i]Regarding Dr. Graubner's approach:[/i]

He gave that. $ A \equal{} \frac {1}{4} \sqrt {(169 \minus{} x^2)(x^2 \minus{} 1)}$. Maximizing this is the same as maximizing $ 4A \equal{} \sqrt {(169 \minus{} x^2)(x^2 \minus{} 1)}$, which is in turn the:[hide="Hint"]Geometric mean of $ 169 \minus{} x^2$ and $ x^2 \minus{} 1$[/hide]

[b](II)[/b] [i]I would suggest this approach:[/i]

$ A \equal{} \frac {1}{2} \cdot 6 \cdot 7 \cdot \sin \theta$, where $ \theta$ is the angle separating the sides of length $ 6$ and $ 7$ (why?)

We can start by multiplying out the constants: $ A \equal{} 21 \sin \theta$. What do we know about $ \sin$ that will help us maximize this expression?

TZF's suggested method is the one that I use most often. Consider the graph of $ \sin\theta$. You should see that the maximum value is 1, which occurs at $ \theta\equal{}\frac{\pi}{2}$. That is why the area of a triangle is at its maximum value when the triangle is right.

Thanks for all the replies.

I found out that all I had to do was make the triangle a right triangle with 6 and 7 as the legs. The hypotenuse would be the square root of 85. I checked this using the Heron's formula method. So the maximum area should have been 21.

I hope this is right :) but that is how my friend explained it and it makes sense, right? =P

Yes, the maximum area of a triangle with two given sides is when it's a right triangle, as said above.

[quote="TZF"]
He gave that. $ A \equal{} \frac {1}{4} \sqrt {(169 \minus{} x^2)(x^2 \minus{} 1)}$. Maximizing this is the same as maximizing $ 4A \equal{} \sqrt {(169 \minus{} x^2)(x^2 \minus{} 1)}$, which is in turn the:[hide="Hint"]Geometric mean of $ 169 \minus{} x^2$ and $ x^2 \minus{} 1$[/hide][/quote]

Is it valid to say, "The geometric mean is at most the arithmetic mean. Thus to maximize it, we set it equal to the arithmetic mean and solve."? It seems dubious, because the arithmetic mean isn't constant.

If this was not the intended meaning of the hint, then what was it?

It was; the arithmetic mean is constant (try it).

OMG WHY AM I SO STUPID  :oops_sign:

I don't deserve to live!!!

Sorry if you had to waste time reading my previous post. And this post. But if you are as stupid as me, you now know that you are not alone.


A standard deck of 52 cards has 13 ranks (Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King) and 4 suits ($ \spadesuit$, $ \heartsuit$, $ \diamondsuit$, and $ \clubsuit$), such that there is exactly one card for any given rank and suit.  Two of the suits ($ \spadesuit$ and $ \clubsuit$) are black and the other two suits ($ \heartsuit$ and $ \diamondsuit$) are red.  The deck is randomly arranged. What is the probability that the top card is red and the second card is black?

There are 26 ways to choose the first card to be red, then 26 ways to choose the second card to be black. There are  ways to choose any two cards.

I would change this to:

There are 26 ways to choose the first card to be red out of the full deck of 52 cards, then 26 ways to choose the second card to be black out of the 51 remaining cards (one red card was removed for the first choice).

26/52 x 26/51 = 1/2 x 26/51 = 26/102 = 13/51

[hide=Solution]
There is $\frac{1}{2}$ chance that the top card is red. There are $51$ cards and $26$ black cards left, so the probability for the second card to be black would be $\frac{26}{51}$. Multiplying both together gives $\frac{1}{2}\times\frac{26}{51}=\boxed{\frac{13}{51}}$.
[/hide]


function
real analysis
real analysis unsolved

Given a subset $E$ of $\mathbb{R}$, is it always possible to find a monotonic function $f$ such that $m^{*}(f(E))>0$ ?

it suffices to do it for $E=(0;1)$ since it is easy to find an increasing function that maps $\mathbb{R}$ in $(0;1)$. Now, if one can find an increasing function that maps $(0;1)$ in a set of Lebesgue measure $0$, it will be done.

I think it suffices to map the real $\overline{0.a_{1}a_{2}a_{2}a_{3}..}$ to $\overline{0.a_{1}0 a_{2}0 a_{2}0 a_{3}..}$. Does it works ?

Well, I actually forgot a piece of information, but this is trivial stuff: in the original problem, $E$ is supposed uncountable. Of course, in the countable case, the answer is no.

Back to the interesting case, i.e, uncountable sets with measure $0$ (for the other ones, the identity works fine). For $E=(0,1)$, the answer is trivially yes. I must admit I have no clue what you tried to prove ...

oh sorry  :blush:  I read it very fast, and I thought you asked a strictly increasing function such that $m^{*}(f(E)) = 0$

Yeah, that's what I understood you read  :P


function
real analysis
number theory
relatively prime

Suppose that $ f(n)$ is a multiplicative function. Define a new function $ g(n)=f(d_{1})+f(d_{2})+\cdots+f(d_{r})$, where $ d_{1},\, d_{2},\ldots ,\, d_{r}$ are the divisors of $ n$. Prove that $ g(n)$ is a multiplicative function.

More generally, suppose that $ f$ and $ g$ are multiplicative functions.  Then show that $ h(n) = \sum_{d \mid n}f(d) g(n/d)$ is multiplicative.  (This is the [Dirichlet] convolution product of $ f$ and $ g$.)  The given result is when $ g(n) = 1$ for all $ n$.  This should also probably be in pre-olympiad.

Bonus: if $ f$ and $ g$ are [i]totally[/i] multiplicative (i.e. $ f(ab) = f(a)f(b)$ for all $ a$ and $ b$, not just relatively prime $ a$ and $ b$), is it true that the convolution of $ f$ and $ g$ must also be totally multiplicative?


I was just wondering how one would easily derive cosines and sines of odd angles such as 18, 36, and 72 degrees algebraically. Any assistance is appreciated.

I don't know about quickly, but if you use trig identities/formulas, let x equal, say, 18, and then

$\sin{5x}=\sin{90}=1=\sin{(2x+3x)}=\sin{2x}\cos{3x}+\sin{3x}\cos{2x}=\dots$
etc, and you could get to a point where you solve for $\sin(18)$.

Also, this may or may not be useful, but since

$[r(\cos{a}+i\sin{a})]^{n}=r^{n}(\cos{(na)}+i\sin{(na)})$, it follows that $(\cos{x}+i\sin{x})^{5}=\cos{5x}+i\sin{5x}=i$ when $x=18$, thus if you determined the value of $\cos{18}$ you might quickly find the value of $\sin{18}$ and such.

Just some ideas.

Take cosine for example.  You have

$\cos 36^{\circ}= \cos 5x =-1$

Use the suggestion in the previous post to evaluate this (take the real part).  You can easily find $\sin 36^{\circ}$ if you have the cosine, so that isn't a problem.  Now you also have the identities

$\cos 2x = \cos^{2}x-\sin^{2}x = 2\cos^{2}x-1$

$\cos\frac{x}{2}= \pm\sqrt{\frac{1+\cos x}{2}}$

so you may find $\cos 18^{\circ}$ and $\cos 72^{\circ}$ after finding $\cos 36^{\circ}$.

This way works, but it isn't a really nice way to do the problem.  There is a nice geometric solution using a pentagon and Ptolemy's Theorem.

For $x=18^{o}$ you have
$\sin 2x = \cos 3x$ which can be reduced to quadratic equation.
[hide] $2 \sin x \cos x = 4 (\cos x)^{3}-3 \cos x$
(since  (cos x is not zero) you can divide by this) and
$2 \sin x = 4 (\cos x)^{2}-3$
if one puts $y=\sin x$ this is nothing but
$2 y = 4(1-y^{2})-3$
or $4y^{2}+2y-1= 0$ and you can solve it.

[hide]$y = \frac{\sqrt 5-1}4$ (You choose the  one of the two values which makes sense) [/hide]
Once you know $sin 18^{o}$ you can know,  other values ( for $36^{o}$ etc)  easily
(By using $sin 2x = 2 sin x cos x$ etc.. [/hide]


inequalities
function
inequalities unsolved

Prove that: $ a^2b^2(a^2 \plus{} b^2)\leq128$ with $ a,b > 0$ and $ a \plus{} b \equal{} 4$

Let $ x \equal{} ab.$ Then $ x\leqslant4$ (by AM–GM) and

[list][list]$ y \equal{} a^2b^2(a^2 \plus{} b^2) \equal{} x^2(16 \minus{} 2x) \equal{} 16x^2 \minus{} 2x^3$[/list]
$ \Rightarrow\ \frac {\mathrm dy}{\mathrm dx} \equal{} 32x \minus{} 6x^2 \equal{} 2x(16 \minus{} 3x) > 0$ for $ x\in(0,\,4]$[/list]
$ \therefore\ y$ is a strictly increasing function of $ x$ for $ 0 < x\leqslant4$ and so $ y_{\text{max}} \equal{} 16(4)^2 \minus{} 2(4)^3 \equal{} 128$

Hello, we have
$ a^2b^2(a^2 \plus{} b^2) < \equal{} 128$ , $ a \plus{} b \equal{} 4$ and $ a,b > 0$.
By using AM-GM on $ a \plus{} b \equal{} 4$ we get $ a^2b^2$<=$ 16$.
And by using $ \frac {a \plus{} b}{2} < \equal{} \sqrt {\frac {a^2 \plus{} b^2}{2}}$
We get $ a^2 \plus{} b^2 > \equal{} 8$
So we now have $ 2(a^2 \plus{} b^2) > \equal{} a^2b^2$

From there we can do:

$ a^2b^2(a^2 \plus{} b^2) < \equal{} 2(a^2 \plus{} b^2)(a^2 \plus{} b^2)$

Now we need to show that $ 2(a^2 \plus{} b^2)(a^2 \plus{} b^2) < \equal{} 128$
From where follows $ a^2 \plus{} b^2 > \equal{} 8$, done.  :)


It is equivalent to
(x-y)^4/2xy+(y-z)^4/2yz+(z-x)^4/2zx>=0

[quote="mathinequs"]For $ x,y,z > 0$ prove :

$ \sum \frac {x^3 \plus{} y^3}{2z} \plus{} 3(xy \plus{} yz \plus{} zx)\ge 4(x^2 \plus{} y^2 \plus{} z^2)$[/quote]
Let $ x,$ $ y$ and $ z$ are real numbers such that $ \sum_{cyc}\frac {x\plus{}y}{z}\geq6.$ Prove that
\[ \sum_{cyc} \frac {x^3 \plus{} y^3}{2z} \plus{} 3(xy \plus{} yz \plus{} zx)\ge 4(x^2 \plus{} y^2 \plus{} z^2)\]

[quote="arqady"]Let $ x,$ $ y$ and $ z$ are real numbers such that $ \sum_{cyc}\frac {x \plus{} y}{z}\geq6.$ Prove that
\[ \sum_{cyc} \frac {x^3 \plus{} y^3}{2z} \plus{} 3(xy \plus{} yz \plus{} zx)\ge 4(x^2 \plus{} y^2 \plus{} z^2)\]
[/quote]
From the given hypothesis, we have
$ (x\plus{}y\plus{}z)\left( \frac{1}{x}\plus{}\frac{1}{y}\plus{}\frac{1}{z}\right) \ge 9.$
Using this inequaliy, we get
$ \begin{aligned}\sum \frac{x^3\plus{}y^3}{z}&\equal{}\left(\sum x^3\right) \left( \sum \frac{1}{x}\right)\minus{}\sum x^2 \\
&\equal{}\left( \sum x^3\minus{}3xyz\right) \left( \sum \frac{1}{x}\right) \minus{}\sum x^2\plus{}3xyz\sum \frac{1}{x}\\
&\equal{}\left( \sum x^2\minus{}\sum xy\right)\left( \sum x\right)\left( \sum \frac{1}{x}\right) \minus{}\sum x^2\plus{}3\sum xy\\
&\ge 9\left( \sum x^2\minus{}\sum xy\right)\minus{}\sum x^2\plus{}3\sum xy\\
&\equal{}8\sum x^2\minus{}6\sum xy.\end{aligned}$
It follows that
$ \sum \frac{x^3\plus{}y^3}{2z} \plus{}3\sum xy \ge 4\sum x^2.$


The sum of several integers (not necessarily distinct) equals $1492$. Decide whether the sum of their seventh powers can equal:

$(a) 1996$
$(b) 1998$

Isn't $1492$ the year Christopher Columbus landed in America?

[hide]Ahh! Remainder modulo 7. :wink:[/hide]

$1492$ is the year the Columbus finally sailed on the trip that landed him in the Caribbean.  Before that, Columbus asked the king and queen, but they were too busy fighting the Moors to finance a sea adventure for him.

[quote="manuel"]what are the are the possible reminders of $k^{7}$ mod 7?[/quote]
The idea is as follows: if $a\equiv k \pmod{7},$ then ...
[hide]$a^{7}\equiv k \pmod{7}.$

I admit I computed this by hand initially (all 7 cases)  :huh:
A more direct way is of course Fermat's Theorem, which only came to my mind later!
[/hide]

[quote="matt276eagles"][quote="Kurt Gödel"]Well, $1996\equiv 1492\mod{7}$, so you probably won't solve (a) with that  :wink: However...
[hide="(b)"]$a^{7}\equiv a \mod{3}$, but $1998 \not\equiv 1492 \mod{3}$, so it's impossible.[/hide][/quote]

You mean $\pmod{7}$, not $\pmod{3}$, right?[/quote]
No. $a^{3}\equiv a \mod{3}$ by FLT, so $a^{7}\equiv a\cdot\left(a^{3}\right)^{2}\equiv a\cdot a^{2}\equiv a^{3}\equiv a\mod{3}$.

EDIT: oh..euhm, yeah, it works modulo 7 as well  :roll:

[hide="(a)"]If we have four 2's and 1484 1's, then the sum of the seventh powers is $4\cdot 2^{7}+1484\cdot 1^{7}=512+1484=1996$, as desired.[/hide]


[quote="mrmbk"]I am a math teacher with 14 years experience, M.Ed.  What steps/suggestions would you give me if I am considering a change to an actuarial or quant career?  Would I necessarily need to return to grad school, or would some companies be willing to hire me as is?[/quote]

For actuarial work, I suggest visiting the various actuarial websites.  You'll probably want to look at knocking off the early tests.

For quant work in the financial world, it really depends on your math and programming background.  If you aren't fairly advanced in either (or both) of those areas, you'll need to return to school.

Pass a couple of actuarial exams first to prove to employers that you really want to be an actuary.

Here is a site of a friend of mine with advice for beginning actuaries:

http://users.aol.com/fcas/advice.html

Good luck,
Joe Orez
Actuary

[quote="rrusczyk"][quote="mrmbk"]I am a math teacher with 14 years experience, M.Ed.  What steps/suggestions would you give me if I am considering a change to an actuarial or quant career?[/quote]

For quant work in the financial world, it really depends on your math and programming background.  If you aren't fairly advanced in either (or both) of those areas, you'll need to return to school.[/quote]

14 years experience means age in the upper 30's or older, and returning to school compounds that.   Quant employers like their novice hirelings to be relatively young (twenties), except for candidates who have technical abilities that demonstrably meet or exceed Wall Street standards (usually ex-academics).   "Math teacher" does not sound like a position that in and of itself demonstrates such capacities, unless one has some specialized skills that have been burnished over the years.

[quote="fleeting_guest"]14 years experience means age in the upper 30's or older, and returning to school compounds that.   Quant employers like their novice hirelings to be relatively young (twenties), except for candidates who have technical abilities that demonstrably meet or exceed Wall Street standards (usually ex-academics).   [/quote]It should be pointed out that this type of hiring practice is illegal under U.S. law.   This basically means that if a person who is 40+ years of age has the same qualifications as a younger person, age cannot be a factor in the employment decision.  See [url]http://www.eeoc.gov/facts/age.html[/url].

Which is not to say that it doesn't happen, because this type of discrimination is [i]very[/i] difficult to prove.  And of course it's not discrimination at all unless you are as equally qualified for the job as the person in their twenties, and the younger person is hired more or less because they're young.

It is rarely if ever the case that all factors other than age are the same.  Companies prefer malleable workers who can learn skills and adapt to the needs of the employers.  There is perceived to be less plasticity in middle age than in one's twenties, particularly when it comes to switching careers, and it is harder for a manager to command a subordinate nearly old enough to be his parent.

Anecdotally I haven't heard of many people successfully making a transition to high-paid quant work in mid career.  I know of cases in actuarial work.

[quote="DPatrick"][quote="fleeting_guest"]It is rarely if ever the case that all factors other than age are the same.  Companies prefer malleable workers who can learn skills and adapt to the needs of the employers.  There is perceived to be less plasticity in middle age than in one's twenties, particularly when it comes to switching careers, and it is harder for a manager to command a subordinate nearly old enough to be his parent.[/quote]I completely agree.  This, however, does not change the fact that it is illegal.  Substituting "malleable" for "young" does not get around the law.[/quote]

No disagreement, but other things being (un)equal, an age-influenced hiring decision can usually be given an unfalsifiable official explanation that does not mention age.   For an age discrimination case to be pursuable (and pursuing it, even successfully, can cause reputational problems of a kind that one would hardly want when starting a second career), the unhired older candidate would need to know and somehow prove that they were unambiguously more qualified than younger, hired applicants.

It would be interesting to hear stories of relatively late career changes into quant or other selective professions.   My impression is that it is rare.  Other than former academics, I know only a couple of cases where this was done.  One guy became a trader in his forties, but it wasn't a quant position where demonstrable previous knowledge was a major consideration.

I feel confident that if you pass a couple actuarial exams and then apply for jobs, someone will hire you as an actuary.

Also, the two jobs you mentioned are totally different, to the extent that it's almost a weird choice to make.

I'm a headhunter for quants, and I know of specific cases of quants starting with household name investment banks past their 40th birthday. >30 is far from unusual. A lot have PhDs, and that helps push the mean age of first job past 27.

I am saddened by the attitude of those who bleat about the law in these circumstances, and I will be frank that anyone who raised it in the pre-screening, would be politely, but firmly told that we would not be forwarding them at all.
To anyone, ever.
Not because we fear being sued, but because they don't have what it takes. Fair enough, it's a specialist activity.

IB quant work is a meritocracy, and harping on about your rights would tell us all we would need to know about their ability to thrive in a competitive environment which values intelligence highly.

We don't do actuaries, maybe it's a good career path, but we see more people wanting to move to quant work from actuarial work to banking than the other way. I'd tell you the ratio, but I get #DIV0 in Excel.

In our Guide to quant careers, we explain at length (143 pages) of stuff you need to know.

A short answer to the original post is that it's a tough road, the first bridge to cross is programming, preferably in C++ [b]not[/b] Java.


analytic geometry
Pascal\u0027s Triangle

does anyone know how to do number 25 on the 2008 grissom math test in alg 2?

(its the one that says in how many ways can you reach the top right corner when starting in the bottom left corner)

Hmmm i got 189, you basically just brut force it: for every point that is the intersection of 2 or more lines, you realize that there are only 2 possible paths to that point: if you came from the point directly to the left of it or from the point below it (and to the left).

So basically what I did for this problem was write down the number of paths you can take (from the starting point) to every point that was an intersection of 2 or more lines (except those at the far right from which there is no path to the ending point)

you can actually ignore all lines that are like \ since the only lines you can "travel" on are / and  _

edit: if there werent holes (that is, if from every point you are have both choices of going up and left/up), then you can like, imagine all the left/up lines as vertical, then put it on coordinates, so that the starting point is at (0,0) and the ending piont is at (m,n), where you need m moves to the right, and n moves up. Then, you know you need a total of m+n moves, and you [b][i]choose[/i][/b] m of them to be moves to the right, so then, the number of paths is $ \dbinom{m\plus{}n}{n}$. (there is a one to one correspondence between the number of ways to  arrange m A's (where an A represents a move to the right) and n B's (which are moves upward). What we are doing is [b][i]modeling[/i][/b]: for example, if m=5, and n=6, one possible model for a path would be AABBBABAABB. Each word you can make with 5 A's and 6B's represents a different path, and each path can be represented by a different word, so there is a one to one correspondence between the number of ways to arrange 5 A's and 6B's with the number of ways to go from (0,0) to (5,6). Understanding this one to one correspondence is key to understanding how to quickly compute the numbers in Pascal's triangle, (which this problem is equivalent to).

however, for this problem, I don't think there's a quick way.


induction
geometry
geometric transformation
reflection

[quote="hpoincare"]B) In a spherical planet, ocens cover a surface greater than the surface covered by earth. Answer if there are two points diametrically opposed which are both in the ocean.[/quote]
[hide] Let's say there are $2n$ points on the planet. We will put each pair of diametrically opposite points into a set, for a total of $n$ sets. There are $n-k$ points of land and $n+k$ points of water, for positive $k$. If we first put each point of land into a set, there will still be $k$ sets which don't have land, and must be filled with 2 points of water. So, there are $k$ pairs of 2 points of water in the same set and therefore, they are diametrically opposite. [/hide]

[quote="hpoincare"]A) an apple is a perfect ball with 3 cm. of radius. A worm has made a hole through a path of at least 6 cm. long, from the entry point E through the exit point O. Show you can find a hemisphere which the worn has not visited.

B) In a spherical planet, ocens cover a surface greater than the surface covered by earth. Answer if there are two points diametrically opposed which are both in the ocean.

Thanx[/quote]

For (a), I'm not sure if you always can find a hemisphere or not, because the worm can dig through all around the apple and you have all hemispheres covered with holess  :D

And for (b), I don't think you always can find diametrically opposite points unless all the ocean must be connected (like lakes can't count as oceans).

[quote="hpoincare"]A) an apple is a perfect ball with 3 cm. of radius. A worm has made a hole through a path of at least 6 cm. long, from the entry point E through the exit point O. Show you can find a hemisphere which the worn has not visited.[/quote]
I think you mean less than 6 cm.

[quote="ZetaX"][quote="lotrgreengrapes7926"]Let's say there are $2n$ points on the planet[/quote]

Why the planet has $2n$ points only¿  :huh:[/quote]

Its an assumption that there are $2n$ points, which is theoritical.

[quote="ZetaX"]And why should this assumption solve the problem¿ I doubt it does.[/quote]

Well, I guess he's just saying that assume there are 2n points and then proving it that way and saying that it should be the same as n reaches infinity (since there are infinite number of points). But I don't know if this does prove the problem. It's sort of like induction the way he did it.

[quote="lotrgreengrapes7926"]There are $n-k$ points of land and $n+k$ points of water, for positive $k$.[/quote]

So, the problem with this proof is the line I've selected here -- it's quite possible that this is not true.  For instance, on the Earth, I might select my first $n$ points in South America, and all of their diametric opposites might be in Asia -- despite the fact that the Earth is 70% water, I could end up with $n$ diametrically opposite pairs and $2n$ points on land.

That said, the basic approach that you're trying here is good.  It fails because you have inexplicably chosen to approach it from a discrete perspective, and that is what ZetaX has been pointing out.

[quote]B) In a spherical planet, ocens cover a surface greater than the surface covered by earth. Answer if there are two points diametrically opposed which are both in the ocean.[/quote]

[hide="simple solution"]if the earth were even 50/50 water and land then each point of water could match up with a point of water but if you have more water than later then there are not enough points  of land to match up with the water so there must be 2 points of water daimetrucally opposite[/hide]

don't get the apple one, if the worm has gone at least 6 inches he could have eaten the whole thing  :huh:

[quote="nodnarbo"]if the earth were even 50/50 water and land then each point of water could match up with a point of water but if you have more water than later then there are not enough points  of land to match up with the water so there must be 2 points of water daimetrucally opposite[/quote]

Same problem as above: prove the matching-up property! (But when speaking of areas, the above one gets correct)
Note: the original property speaks of areas, not of cardinality of sets.

If you say that point A is diametrically opposite to point B then that means that there are no other points that can possibly be diametrically opposite to either of those points, thus all the sets of diametrically opposite points form pairs.

I meant the following part (sorry for not being more clear):
Why are there "not enough" points to match up¿

[quote="nodnarbo"] but if you have more water than later then there are not enough points  of land to match up with the water [/quote]

He wants a better justification of this statement.  It's extremely unrigorous considering the set of points is infinite.

In any case, [hide="The proof is very simple"] Consider the reflection of the water (as a surface) across the center.  If the intersection of this reflection and the water is empty, then the land contains the reflection of the water; contradiction.  [/hide]


induction
combinatorics proposed
combinatorics

Let $A_1$, $A_2$, $\ldots$, $A_n$, $n\geq 2$ be $n$ finite sets with the properties

 i) $|A_i| \geq 2$, for all $1\leq i \leq n$;

ii) $|A_i\cap A_j| \neq 1$, for all $1\leq i<j\leq n$.

Prove that the elements of the set $\displaystyle \bigcup_{i=1}^n A_i$ can be colored with 2 colors, such that all the sets $A_i$ are bi-color, for all $1\leq i \leq n$.

We use induction on $n$.
Base $n=1,n=2$ are clear. Suppose it is true for $n-1$.
Then when we have $n$ substets, color $A_1,..,A_{n-1}$ like we assume by induction.

We look on $A_n$ subset, if it has an element, which doesn't appear in previous $n-1$
subsets we can color it randomly so that $A_n$ will be bi-color.

Suppose all elements are from $A_1,..,A_{n-1}$. And all they are colored in first color.
Take one element $a_x$ from$A_n$ randomly and recolor it in second color.
$A_n$ will satisfy condition, and all $A_i$ too,
because if intersection $A_i$with $A_n$ contains $a_x$, it will have at least two elements in comon and these elements are of two color and it satisfies the condition,
but the rest remained unchange.


integration
inequalities
algebra unsolved
algebra


[hide=" solution"]

A second earlier, that is after $ 59$ seconds.

[/hide]

Since a second earlier the amount of bacteria in the glass is half the amount currently the time when the glass was full [hide="is"]59 seconds[/hide]

[quote="rksmash"]A number of bacteria is placed in a glass. 1 second later each bacterium divides into 2,the next second each of the resulting bacteria divides in two again,etc. After 1 minute the glass is full. When was the glass half-full[/quote]

One second earlier, or 59 seconds.

Guys, please don't just repeat information that has already been states without adding anything new. This is called spamming. And especially, don't revive a resolved topic that is over a week old.


Based off of this [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=225425]this[/url], I am trying to format my document so that I can display some chemical formulas. However, when I'm using the online TeX compiler, the LaTeX is displayed oddly (most likely due to the complexity of the code). Is there a way to format this?

I tried using the \documentclass{article} but then I can't use the fraction code. The \begin{math} code lets me make the fraction, but the text doesn't look as nice.

The formula is

$ \text{Molar mass (solute)} \equal{} \frac{k_{fp} \times \text{g (solute)}}{\text{kg (solvent)} \times \Delta T _{fp}}$

[code]\text{Molar mass (solute)} = \frac{k_{fp} \times \text{g (solute)}}{\text{kg (solvent)} \times \Delta T _{fp}}[/code]

[quote="7h3.D3m0n.117"]However, when I'm using the online TeX compiler, the LaTeX is displayed oddly (most likely due to the complexity of the code). Is there a way to format this? [/quote]
Which online TeX compiler are you using?

Sorry for not specifying. I am using http://dev.baywifi.com/latex/

I suspect you forgot to use amsmath which is required for \text (otherwise you have to use \mbox).
This works fine:
[code]\documentclass{article}
\usepackage{amsmath}
\begin{document}
\[
\text{Molar mass (solute)} = \frac{k_{fp} \times \text{g (solute)}}{\text{kg (solvent)} \times \Delta T _{fp}}
\]
\end{document}[/code]


for 1
$(a+\frac{b}{2})^2(c+\frac{b}{2})\geq 3$ where $a>b>c$ so the minimum value is $3\frac{\sqrt[3]{3}}{\sqrt[3]{4}}$

[quote="hendrata01"]There was another problem that, given $x+y+z=1$ find maximum of $x^2y+y^2z+z^2x$ and turns out the answer is when $x=\frac{2}{3}, y = \frac{1}{3}, z = 0$ (and all its permutations)

Maybe the reverse is true?[/quote]Yes it is, both problems are equivalent. Your solution (can you provide a link?) means $\max_{x,y,z>0}\frac{x^2y+y^2z+z^2x}{(x+y+z)^3}=\frac4{27}$.  From that we would find
$\min f=\sqrt[3]{\min\frac{(x+y+z)^3}{x^2y+y^2z+z^2x}}=\frac3{\sqrt[3]{4}}$.

To Albanian Eagle: That's different from yours. As I don't understand your reasoning anyway, can you explain it please?

EDIT: Albanian Eagle is right, my last line must read instead:
$\min f=\sqrt[3]{\min\frac{(x+y+z)^3}{(x^2y+y^2z+z^2x)/3}} =\frac{3\sqrt[3]{3}}{\sqrt[3]{4}}$. The denominator must be normalized to 1 which I forgot  :blush:

For spanferkel, the link is here: http://www.mathlinks.ro/Forum/post-768.html

I found both (2) and (3) very interesting, and maybe the trick, like (1), is to deal with the converse instead (although I found it difficult  :mad: )

For (b)

OK, let's say $a^2+b^2+c^2=1$, we'll prove that $a^2b+b^2c+c^2a\leq \frac{1}{\sqrt{3}}$ (of course equality is attained when $a=b=c=\frac{1}{\sqrt{3}}$

So we'll prove $(a^2b+b^2c+c^2a)^2\leq \frac{1}{3}(a^2+b^2+c^2)^3$

But that is equivalent to $6(a^2b^3c+a^3bc^2+ab^2c^3)\leq a^6+b^6+c^6+3(a^4c^2+c^4b^2+b^4a^2)+6a^2b^2c^2$

But that follows from $a^6+2a^4c^2+c^4b^2+2a^2b^2c^2\geq 6a^3bc^2$ (AM-GM) and its permutations.

For (c)

Well, from $a^2b+b^2c+c^2a=3$, we take $c=0$, take $a$ big, and $b=\frac{1}{a^2}$, then $ab+bc+ca= \frac{1}{a}$ will approach $0$.

That completes all the problems, I hope.  :)

sorry for not explaining before

I ment this for (a):
let $a\geq b \geq c$ then
$\left( \frac{2a+2b+2c}{3} \right) ^3 \geq (a+\frac{c}{2})^2 (2b+c)\geq 2(a^2 b+b^2 c+c^2 a)=6$
so $a+b+c\geq \frac{3\sqrt[3]{6}}{2}=3 \frac{\sqrt[3]{3}}{\sqrt[3]{4}}$
what's wrong here?
for $a=2b , b=\sqrt[3]{\frac{3}{4}} , c=0$ it is achieved  ;)

What about a "nicer" solution to (b)?

And would it be able to apply to if $a^2b+b^2c+c^2a$ is replaced by $a^3b^2+b^3c^2+c^3a^2$?

[quote="spanferkel"]Yes you are right, I've found an error and edited.
Thank you Albanian Eagle![/quote]

THat's OK :D

[quote="indybar"]Would it be able to apply to if $a^2b+b^2c+c^2a$ is replaced by $a^3b^2+b^3c^2+c^3a^2$?[/quote]Yes, Albanian Eagle's idea works for the condition $a^{n+k}b^n+b^{n+k}c^n+c^{n+k}a^n=3$ with natural $n\ge2,k\ge1$.
Assuming $a\ge b\ge c$ we have by AM-GM
\[ \left((n+k) \frac{a+b+c}{3} \right)^{2n+k} \ge \left( a+\frac{k}{n+k} c \right)^{n+k}\left(\frac{n+k}nb+ c \right)^n  \]
$= \left(\frac{n+k}n \right)^n \left( a+\frac{k}{n+k}c \right)^{n+k}\left(b+\frac{n}{n+k}c \right)^{n}$

$\ge \left(\frac{n+k}n \right)^n(a^{n+k}b^n+b^{n+k}c^n+c^{n+k}a^n)$ by looking at what the two terms with the highest powers of $a$ and $b$ respectively contribute.
From there we find $a+b+c\ge3\cdot\sqrt[2n+k]{\frac3{n^n(n+k)^{n+k}}}$, with equality for $c=0, a=\frac{n+k}nb$. :D
[hide]Hope this is right. (Note that for smaller and non-integer values like $n=k=\frac12$, equivalent to part (b), it is not.) :)  [/hide]

Yes, I also solved the problem $a^3b^2+b^3c^2+c^3a^2=3 \longrightarrow$ find min $a+b+c$  :)

But I was asking for $a^3b^2+b^3c^2+c^3a^2=3 \longrightarrow$ find min $a^2+b^2+c^2$, which is also discussed (and unsolved) on the other thread, I think.

[quote="indybar"]But I was asking for $a^3b^2+b^3c^2+c^3a^2=3 \longrightarrow$ find min $a^2+b^2+c^2$, which is also discussed (and unsolved) on the other thread, I think.[/quote]I think, for that one the equality (=minimal) case is still only $a=b=c$.
See http://www.mathlinks.ro/Forum/viewtopic.php?p=485228#p485228
and http://www.mathlinks.ro/Forum/viewtopic.php?t=2549. In the latter one, Harazi (post #5) proves $x^n y + y^n z + z^n x \leq \frac{n^n}{(n+1)^{n+1}}$ if $x+y+z=1$, and I think his proof, essentially the same as above, is valid for all real $n\ge2$. But not for smaller $n$.


[quote="Thaliwilya"]Let $ p$ be an odd prime. Prove that $ \sum_{k \equal{} 1}^{p \minus{} 1}k^{2p \minus{} 1} \equiv \frac {p(p \plus{} 1)}{2}\ \ (\mod p^2)$[/quote]

First,

$ 2\cdot\sum_{k \equal{} 1}^{p \minus{} 1}k^{p} \equal{} \sum_{k \equal{} 1}^{p \minus{} 1}k^{p} \plus{} \sum_{k \equal{} 1}^{p \minus{} 1}k^{p} \equal{} \sum_{k \equal{} 1}^{p \minus{} 1}k^{p} \plus{} \sum_{k \equal{} 1}^{p \minus{} 1}\left( p \minus{} k\right) ^{p}$
(here we substituted $ p \minus{} k$ for $ k$ in the second sum)
$ \equal{} \sum_{k \equal{} 1}^{p \minus{} 1}\left( k^{p} \plus{} \left( p \minus{} k\right) ^{p}\right)$.

But for every $ k\in\left\{ 1,2,...,p \minus{} 1\right\}$, the binomial formula yields
$ \left( p \minus{} k\right) ^{p} \equal{} \sum_{i \equal{} 0}^{p}\binom{p}{i}p^{i}\left( \minus{} k\right)^{p \minus{} i}$
$ \equal{} \underbrace{\binom{p}{0}}_{ \equal{} 1}\underbrace{p^{0}}_{ \equal{} 1}\underbrace{\left( \minus{} k\right) ^{p \minus{} 0}}_{ \equal{} \left( \minus{} k\right) ^{p}} \plus{} \underbrace{\binom{p}{1}}_{ \equal{} p}\underbrace{p^{1}}_{ \equal{} p}\left( \minus{} k\right) ^{p \minus{} 1}$
$ \plus{} \sum_{i \equal{} 2}^{p}\binom{p}{i}\underbrace{p^{i}}_{ \equal{} p^{2}p^{i \minus{} 2}}\left( \minus{} k\right) ^{p \minus{} i}$
$ \equal{} 1\cdot1\cdot\left( \minus{} k\right) ^{p} \plus{} \underbrace{p\cdot p\cdot\left( \minus{} k\right) ^{p \minus{} 1}}_{ \equal{} p^{2}\left( \minus{} k\right) ^{p \minus{} 1}\equiv0\mod p^{2}} \plus{} \underbrace{p^{2}\cdot\sum_{i \equal{} 2}^{p}\binom{p}{i}p^{i \minus{} 2}\left( \minus{} k\right) ^{p \minus{} i}}_{\equiv 0\mod p^{2}}$
$ \equiv1\cdot1\cdot\left( \minus{} k\right) ^{p} \plus{} 0 \plus{} 0 \equal{} \left( \minus{} k\right) ^{p} \equal{} \minus{} k^{p}\mod p^{2}$

(where $ \left( \minus{} k\right) ^{p} \equal{} \minus{} k^{p}$ is because $ p$ is odd), so that

$ 2\cdot\sum_{k \equal{} 1}^{p \minus{} 1}k^{p} \equal{} \sum_{k \equal{} 1}^{p \minus{} 1}\left( k^{p} \plus{} \left( p \minus{} k\right)^{p}\right)$
$ \equiv\sum_{k \equal{} 1}^{p \minus{} 1}\underbrace{\left( k^{p} \plus{} \left( \minus{} k^{p}\right)\right) }_{ \equal{} 0} \equal{} 0\mod p^{2}$.

Since $ 2$ is coprime to $ p^{2}$ (since $ p$ is odd), this yields

[color=green][b](1)[/b][/color] $ \sum_{k \equal{} 1}^{p \minus{} 1}k^{p}\equiv 0\mod p^{2}$.

Now, for every $ k\in\left\{ 1,2,...,p \minus{} 1\right\}$, Fermat's Little Theorem yields $ p\mid k^{p \minus{} 1} \minus{} 1$, so that $ p^{2}\mid\left( k^{p \minus{} 1} \minus{} 1\right) ^{2}$, or, equivalently, $ 0\equiv\left( k^{p \minus{} 1} \minus{} 1\right) ^{2}\mod p^{2}$. Thus,

$ 0 \equal{} \sum_{k \equal{} 1}^{p \minus{} 1}k\cdot0\equiv\sum_{k \equal{} 1}^{p \minus{} 1}k\cdot\left( k^{p \minus{} 1} \minus{} 1\right) ^{2} \equal{} \sum_{k \equal{} 1}^{p \minus{} 1}k\cdot\left( \left( k^{p \minus{} 1}\right)^{2} \minus{} 2k^{p \minus{} 1} \plus{} 1\right)$
$ \equal{} \sum_{k \equal{} 1}^{p \minus{} 1}k\cdot\left( k^{2p \minus{} 2} \minus{} 2k^{p \minus{} 1} \plus{} 1\right) \equal{} \sum_{k \equal{} 1}^{p \minus{} 1}\left( k^{2p \minus{} 1} \minus{} 2k^{p} \plus{} k\right)$
$ \equal{} \sum_{k \equal{} 1}^{p \minus{} 1}k^{2p \minus{} 1} \minus{} 2\cdot\underbrace{\sum_{k \equal{} 1}^{p \minus{} 1}k^{p}}_{\equiv0\mod p^{2}\text{ by \textbf{(1)}}} \plus{} \underbrace{\sum_{k \equal{} 1}^{p \minus{} 1}k}_{ \equal{} \left( p \minus{} 1\right) p/2}$
$ \equiv\sum_{k \equal{} 1}^{p \minus{} 1}k^{2p \minus{} 1} \minus{} 2\cdot0 \plus{} \frac {\left( p \minus{} 1\right) p}{2} \equal{} \sum_{k \equal{} 1}^{p \minus{} 1}k^{2p \minus{} 1} \plus{} \frac {\left( p \minus{} 1\right) p}{2}$
$ \equal{} \sum_{k \equal{} 1}^{p \minus{} 1}k^{2p \minus{} 1} \plus{} \left( p^{2} \minus{} \frac {p\left( p \plus{} 1\right) }{2}\right) \equal{} \sum_{k \equal{} 1}^{p \minus{} 1}k^{2p \minus{} 1} \minus{} \frac {p\left( p \plus{} 1\right) }{2} \plus{} p^{2}$
$ \equiv\sum_{k \equal{} 1}^{p \minus{} 1}k^{2p \minus{} 1} \minus{} \frac {p\left( p \plus{} 1\right) }{2}\mod p^{2}$,

so that $ \sum_{k \equal{} 1}^{p \minus{} 1}k^{2p \minus{} 1}\equiv\frac {p\left( p \plus{} 1\right) }{2}\mod p^{2}$, qed.

Probably there are shorter proofs.

[b]EDIT:[/b] ZetaX just notified me that $ \left( p \minus{} k\right) ^{p}\equiv \minus{}k^p\mod p^2$ follows from $ p\minus{}k\equiv \minus{}k\mod p$ by Hensel's lemma.

  darij

[quote="darij grinberg"]Probably there are shorter proofs.[/quote]
Maybe :wink:
$ \sum_{k \equal{} 1}^{p \minus{} 1}k^{2p \minus{} 1} \equal{} \sum_{i \equal{} 1}^{\frac {p \minus{} 1}{2}}(i^{2k \minus{} 1} \plus{} (p \minus{} i)^{2k \minus{} 1}) \equal{}$ $ p[(p \minus{} 1)^{2p \minus{} 2} \minus{} (p \minus{} 1)^{2p \minus{} 3} \plus{} ... \plus{} 1] \plus{} p[(p \minus{} 2)^{2p \minus{} 2} \minus{} (p \minus{} 2)^{2p \minus{} 3}2 \plus{} ... \minus{} (p \minus{} 2)2^{2p \minus{} 3}]$ $ \plus{} ... \plus{} p[\left(\frac {p \plus{} 1}{2}\right)^{2p \minus{} 2} \minus{}$ $ \left(\frac {p \minus{} 1}{2}\right)\left(\frac {p \plus{} 1}{2}\right)^{2p \minus{} 3} \plus{} ... \plus{} \left(\frac {p \minus{} 1}{2}\right)^{2p \minus{} 2}]$ $ \equiv p(2p \minus{} 1)$ $ \plus{} p(2p \minus{} 1) \plus{} ...p(2p \minus{} 1)\equiv \minus{} p\left(\frac {p \minus{} 1}{2} \right)\equiv \frac {p(p \plus{} 1)}{2} (mod p^2)$

[i]Clearly solution , Thanks Darij :lol: [/i]

  [i]You seem to be interested in problem involved in Binomial[/i]


In triangle $ ABC$, let $ M$ be the midpoint of $ BC$. The circle with diameter $ AM$ cuts $ AB,AC$ at $ D,E$ respectively. The tangents to the circle at $ D,E$ cut at $ P$. Prove that $ PM$ is perpendicular to $ BC$.

if you draw the perpendicular BC and it meets the circle in F we should prove A(DEMF)=-1 because if it's true then if we suppose pm meets the circle in F' we know (DEMF')=-1 so F and F' are the same point
^AFM=90  &  ^BMF=90 =>  AF || BC    &      M: the midpoint of BC => A(FCMB)=-1 => A(FMDE)=-1
and problem is solved! :idea:


hi, im new in studying physics...
im reading fabric of the cosmos,

about the bucket experiment,

I have two questions,

1) What does it mean on pg 26/27, when they say that there is realtive motion between the bucket and water at first when they let the cord unwind? I don't understand what they mean by [b]relative motion.[/b]  Does it mean that if there [b]"is relative motion"[/b] then the bucket and water are spinning at the same rate?

2) When it [b]"does not have relative motion"[/b] what does that mean?

Thanks!

In physics, the movement of an object can be seen from different points of reference. Think of it this way: If you are on a car going 50km/h and a train is moving alongside you at a rate of 50km/h, from [i]your reference point[/i] it appears that the train is not moving at all. From an observer standing still, however, it will appear that the train is moving at 50km/h.

In this example, the train has no motion [i]relative[/i] to you in the car. However, relative to the observer, the train is moving at 50km/h.

Relative motion is all about the speed of one thing relative to the speed of another thing. If there is no relative motion it does not mean that the objects are still, it just means they're moving at the same rate.

In relative motion,motion of another body is described in terms of another body,whicvh ma be moving or in rest.

If two cars A and B move with 10 and 5 units spped respectively,then relative speed of A with respect to B would be 5 units because B would see A moving with 5 units,not 10 units.

Isn't that book written by Brian Greene? I just recently started reading the other book, "The elegant universe". Some pretty interesting stuff indeed  :lol:

yes it is written by Brian Greene...but i think elegant universe i heard is harder to read without previous physics knowledge and the fabric of the cosmos is easier to read without equations and such..yes it is very interesting!

Yeah i got both books several months ago but i just recently started reading them. I'm reading the elegant universe first just due to chronological order. It was written first.


topology
advanced fields
advanced fields unsolved

Prove that
-Sequentially compactness implies Countably compactness (=if every countable open cover has a finite subcover).
-Countably compactness implies limit-point compactness.

Honestly, I've lost track of all the different kinds of compactness and I don't know very well how to prove this.

Sequentially compact implies countably compact:  I think something along the following lines will work.

Let $ \{U_n\}$ be a countable open cover of $ X$.  If there does not exist a finite subcover, let $ x_n \notin \cup_{i \leq n} U_i$.  Consider $ \{x_n\}$.  By sequentiall compactness, there exists subsequence $ \{y_k\}$ that converges to some $ y \in X$.  Now, this $ y \in U_n$ for some $ n$.  By construction, $ x_i \notin U_n$ for infinitely many n and hence the same is true
for $ y_k$.  Thus, $ \{y_k\}$ cannot possibly converge to $ y$.

As for the other one, Im not sure off the top of my head.

Suppose $ X$ is sequentially compact. Now we will do this by showing that the intersection of any countable family of closed sets with the FIP is nonempty (I think this is a common enough form of countably compact).

Let $ \{U_i\}_{i\equal{}1}^\infty$ be a countable family of closed sets with the FIP. Let $ x_n\in\cap_{i\equal{}1}^n U_i$. So $ x_n$ is in the intersection of the first $ n$. We now have a sequence, so there is a convergent subsequence $ x_{n_k}\to x$. Let $ O$ be an open set containing $ x$. Then $ O$ contains every element of the subsequence after some point.

Thus, $ O$ intersects every set in the family (if there is one that it doesn't intersect then there would have to be elements of the subsequence arbitrarily far along that aren't in $ O$). But this holds now for any open set containing $ x$, thus $ x$ is in the closure of every set in the family. The family consists of only closed sets, so $ x$ is in the intersection of them all.

Hence, $ X$ is countably compact.

Edit: Ah, beat me to it, but mine is a different approach.


I believe this indeed does work.  However, I wouldn't suggest going everywhere and using it in sprint rounds because simply thinking about the problem is usually faster.  Nevertheless, this is a great way to check your answer.

I think it can be derived by calculating in turn the angle from 0 degrees that the hour hand is, and the angle of the minute, then subtracting.  (or something like that).

Memorizing formulas is usually not the way to go. Like kyyuanmathcount said, just thinking will most likely to be faster. it will probably be cool to derive this formula though, try it.

[quote="ragnarok23"]Someone told me to find the degree measure between the two hands of a clock, you use the formula $|\frac{11m-60h}{2}|$ where m=The minute and h=the hour. I tried it and it worked, however, I am not sure about this, or why it might work. Could someone confirm this?[/quote]
I was the person to get the formula. :D

[hide="Here It Is"]$180-|180-30h+\frac{11m}{2}+\frac{11s}{120}|$[/hide]

Well, it probably has been discovered multiple times.  I heard it before you had joined AoPS from somewhere last year.

Could you enlighten us on the terms in the derivation :?:  :)

[quote="kyyuanmathcount"]Well, it probably has been discovered multiple times. I heard it before you had joined AoPS from somewhere last year.[/quote]
 :(

$180-|180-30h+\frac{11m}{2}+\frac{11s}{120}|$

$h$ is hours, $m$ is minutes, $s$ is seconds. Plug them in, and you have the degree between the minute and hour hands. Should I post the explanation?

I encourage you to do so if you have time :).

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=110138[/url]

My original formula had 2 mistakes. The last post has the correct one.


Let $ (x_n)$ be a sequence such that $ x_0$ and $ x_{n \plus{} 1} \equal{} \frac {2x_n \plus{} 1}{x_n \plus{} 2}$ forall $ n \equal{} 0,1.2...$

determine $ [\sum_{k \equal{} 1}^n x_k]$ where $ [x]$ denote the lagest integer not exceeding $ x$.

This is T9/275[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=1256922#1256922[/url]

Should be locked.

Japanese Communities Moderator

kunny


I was wondering if anyone could propose a way of teaching "advanced"  SAT problems for I and II (not that any SAT math problems are very advanced) from the AoPS books. Some stuff I was wondering about was a suggested order of how to do chapters, what parts of the book would actually be relevant for the SAT, and etc. Thanks ahead of time.

I dunno... I really don't think AoPS has a strong correlation with SAT II or SAT I math.  I've never had trouble with any of it, but even some of the AoPS problems are difficult to solve.

As people say, practice, practice, practice.... Just do a lot of SAT math problems, and I think you will see a faster improvement.  IMO.

I think that the set of SAT questions is a set of lower bounds for the set of AoPS questions, when all questions are ordered by difficulty.

[quote="Ragingg"]As people say, practice, practice, practice.... Just do a lot of SAT math problems, and I think you will see a faster improvement.  IMO.[/quote]

Hmm...well it's for a group of kids I'm trying to help out for SAT. SAT books and problems aren't giving  them much success since they can generally score in the 720 - 770 range already. It's just a matter of improving 50 points or so for the perfect score. I've personally never actually practiced SAT problems so I'm not even quite sure if the stuff they're working on is relevant.

If they're already scoring in the 700s, a bit of math team work might be exactly what brings the over the hump to an 800.  Are these students who take the AMCs?  Perhaps working on some harder problems would help them see math with a broader eye enabling them to understand problems more quickly, catch more mistakes, and developing a deeper understanding of the material.

[quote="rrusczyk"]If a student is getting 720 to 770 on the SAT, they should be studying harder stuff than SAT math.  If they do so, the 800 will come naturally.[/quote]

Yes, that's exactly what I was thinking. Perhaps my question in the first post didn't come out right. I was wondering what part of the AoPS Books you guys would recommend them starting?

And yes, I believe they all take AMCs at their schools. I believe one of them even qual'ed for AIME last year. Would you recommend going over some of the easier problems on old AMCs?

Easier problems on old AMCs - have them do first half of the problems at the end of each chapters of AoPS Vol 1.


AMC
AMC 12
probability
number theory
prime numbers

What's the solutions for these 2 questions on the 2005 AMC12:
14 	On a standard die one of the dots is removed at random with each dot equally likely to be chosen. The die is then rolled. What is the probability that the top face has an odd number of dots?

18 	Call a number "prime-looking" if it is composite but not divisible by 2, 3, or 5. The three smallest prime-looking numbers are 49, 77, and 91. There are 168 prime numbers less than 1000. How many prime-looking numbers are there less than 1000?


Five 6-sided dice are rolled. What is the probability that exactly two of the dice show a 1 or a 2?

[quote=mathapple101]i was being stupid and i put 6C2 instad of 5C2[/quote]

I did that too. I solved the problem thinking that there were 6 dice instead of 5. Miraculously I got the [i]exact same answer[/i]. Turns out that $\frac{2}{3}\binom62 = \binom52$. But that's just coincidence.

[quote=cw357]oh wow

i read it as 6 dice on my first try too

[s]i'm seriously considering reporting this problem for that[/s][/quote]

Why? It isn't the problem's fault you read it wrong

please reconsider the $\text{[s]}$

also this is seriously just post farming

[quote=math154]The probability that any one die shows a $ 1$ or a $ 2$ is $ \frac26 \equal{} \frac13$, so the probability otherwise is $ 1 \minus{} \frac13 \equal{} \frac23$. Then the desired probability is

$ \binom52\left(\frac13\right)^2\left(\frac23\right)^3 \equal{} \frac {10\cdot8}{243} \equal{} \boxed{\frac {80}{243}}$.[/quote]

Use [code][hide][/hide][/code] tags

@LuckyG lol the post was from over 10 years ago they're not gonna see

@Possible We certainly shouldn't report the problem for being unclear, because it is clear. though it might be worthwhile to report it for the fact that [b]you can read the problem blatantly wrong and still get the right answer.[/b] for some reason that's happened to me several times on alcumus (solving the wrong problem but getting the right answer) but could be just me

[quote=hunter117]What about the case where there are two 1s and two 2s? Are you not supposed not count them? Or is it already counted under this solution?[/quote]

They are already counted under this solution.

You can interpret the solution in the following way.

Among all the 6^5 possible outcomes after 5 dices are rolled,

there are 5!/3!2! possible ways to pick which 2 dice among the 5 dices in which the 4(2×2) dice permutations consisted of (1,1),(1,2),(2,1),(2,2) will appear. The rest of the 3 dices cannot be numbers 1 and 2, so there are 4×4×4 permutations of dice roll outcomes consisted of numbers 3,4,5 & 6.

So the ratio of all the possible requested permutations among all possible permutations is [(5!/3!2!)×2^2×4^3] : 6^5.

[quote=fishy15]They are not counted, and those aren't supposed to be counted.[/quote]
I don' t think so, please take a look at my argument at  floor #13, thank you.


How many ways can you put six eggs numbered 1-6 in six baskets numbered 1-6 if no egg can be in a basket that has the same number and there can be at most three eggs in basket?

[hide="hint"]ooh...looks like a derangements problem.  Think about inclusion-exclusion principle.[/hide]

[hide]If there is no limit to the number of eggs in each basket, than the answer is , since each egg has five choices of where to go.

If only one egg can go into each basket, than it's a simple derangement problem, and the answer is .[/hide]

i may be totally (extremely) off on this, but if there's only one object per box...is the answer $\boxed{365}$?

i rem. working a similar problem before and writing a formula for it...if this is right i may post it.

Sorry.  It's supposed to be you can have at most 3 eggs in a basket.  I'll change that.  :?


Clara computed the product of the first $n$ positive integers and Valerid computed the product of the first $m$ even positive integers, where $m \ge 2$. They got the same answer. Prove that one of them had made a mistake.

This is a complicated way of asking to solve $ n!\equal{}2^{m}m!$ for $ m\ge2$. Looking at the power of $ 3$ in both sides, we see that $ n<m\plus{}3$ (otherwise $ m!$ has a factor 3 more than the $ n!$), so either $ n\equal{}m\plus{}1$ (then $ m\plus{}1\equal{}2^{m}$ implies $ m\equal{}1$) or $ n\equal{}m\plus{}2$ (then $ (m\plus{}1)(m\plus{}2)\equal{}2^{m}$ implies $ m\equal{}0$). So no solutions.

Firstly $ n>m$, sof $ n\equal{}m$ or $ n\equal{}m\plus{}1$ and no more. That's $ m\plus{}1\equal{}2^{m}$ which is simply impossible.

The same method, we could solve the equation: $ n!\cdot k!\equal{}p^{m}m!$ where $ p$ is prime.


Hi, i am a F.5 student from Hong Kong... :)
Can anyone suggest some really good books about elementary number theory (in English) for me?

I'm afraid I don't know the order of the grades in Hong Kong. What does F.5 mean?

The best basic introduction to number theory for young students (ages 10 to 12) is a classic book called [u]The Higher Arithmatic[/u] by H. Davenport. It is written in a style easy to read for beginners.

Niven and  Zuckerman's [u]An Introduction to the Theory of Numbers[/u] - http://www.amazon.com/exec/obidos/tg/detail/-/0471625469/qid=1078067108/sr=1-1/ref=sr_1_1/002-1562397-5831210?v=glance&s=books is a good introduction for slightly more sophisticated readers (ages 12 to 15 maybe?). It has some good problems.

I also really like [u]A Course in Number Theory[/u] by H.E. Rose - http://www.amazon.com/exec/obidos/tg/detail/-/0198523769/ref=lpr_g_1/002-1562397-5831210?v=glance&s=books It's chapters come in pairs, where the first covers the basics and the second gets into the more advanced topics. I like it because it can be used for a long time. As your knowledge develops you can go from the first chapters to the second. It has many good problems, ranging from basic to hard.

The most famous book in number theory is [u]An Introduction to the Theory of Numbers[/u] by Hardy and Wright - but I would recommend either Davenport, Niven and Zuckerman, or Rose to begin with. Come back to Hardy and Wright later.

For more advanced books I recommend [u]A Classical Introduction to Modern Number Theory[/u] by Ireland and Rosen or [u]Number Theory[/u] by Borevich and Shafarevich. Both of these are University level for most people.

Thanks for your details...
By the way, in HK, generally F.5 is the grade for 16 years old students.  :)

[quote="shobber"]are there nice basic number theory books online??[/quote]

hardy's book. I remember i saw one online before.

[quote="siuhochung"]hardy's book. I remember i saw one online before.[/quote]

But Hardy is difficult enough to read [i]on paper[/i], much less to say online. ;)

[quote="siuhochung"]
hardy's book. I remember i saw one online before.[/quote]

Does anyone have a URL?

I don't have a url for hardy's book, but I suggest the following book
http://modular.fas.harvard.edu/ent/ent.pdf
which is an introduction to the modern elementary number theory and
http://modular.fas.harvard.edu/edu/Spring2004/129/ant/ant.pdf
for an introduction to algebraic number theory from the global and adelic viewpoint.

These are not olympiad oriented, but experience has taught me that theory can help a lot at the olympiad, even if the theory itself is not olympiad oriented (e.g., problem 6, IMO 2001 and number fields of class number 1)

Andrei


Consider a matrix of size $8 \times 8$, containing positive integers only. One may repeatedly transform the entries of the matrix according to the following rules:

-Multiply all entries in some row by 2.
-Subtract 1 from all entries in some column.

Prove that one can transform the given matrix into the zero matrix.

It's sufficent to show that given a column, and the operations that we may multiply any square in the column by 2, and subtract 1 from the whole column, we can get to the zero column.

Suppose there are x ones and 8-x non ones.  The operations "multiply all the 'ones' by 2, then subtract 1 from every square in the column"  reduces the sum of the column if there is atleast one 'not-one', and guarantees that all elements of the column are >= 1.  Because a monotonic bounded function must reach its minimum, so we must reach (1,1,1,1,1,1,1,1).  Result.


you have a postcard with vertical and horizontal dimensions y and x, which you want to stick to the wall. You are going to use two bits of blue tack. where is the best places to put them if you want to make sure the postcard will not fall down?

Does this question revolve around the center of mass of the index card?

If it does, then I don't know how to respond to having two tacks.

My guess (not serious): Put the tacks anywhere on the index card, then knock the wall down flat to the earth so the card won't fall.
Alternate solution:  Tack the index card in any way to the wall, then put the wall in space.

Are you concerned with wind or anything, because it almost seems that the position of tacks won't matter, because in any physical scenario the index card will not fall (assuming only gravitational forces and that the force required to pull the tacks out at some angle is less than the static friction force keeping the tacks in the wall)?

ah, no not tacks, blue tack. its like a sticky thing, like playdough? Im not sure about a solution. say you're just doing it with a square, but if you put one bit right at the bottom, then it's more likely to fall down than one at the top.

similar to glue really

Do you care about sideways forces (like someone pushing it from the side?)

nope, just gravity

I dont know that there is a good answer, I just thought it was quite interesting.


Let $ w$ be circle and $ ABCD$ is quadrilateral with $ AB\equal{}BC\equal{}CD\equal{}DA$(''kvadrat'' in russian) inscribed in $ w$.Prove that if $ X$ is any point of  $ "w"$, then the sum $ XA \plus{} XB \plus{} XC \plus{} XD$ can't be rational.

what is the radius of the circle ? It's important ?

It is not given. Why it is important, justify your answer please!

Note that if $AB = a$ and point $X$ be the intersection point of perpendicular bisector of $AD$ such that, $XA < XB$. Then we can easily get the sum is:
\begin{align*} XA+XB+XC+XD = 2a \sqrt{1-\frac{\sqrt{2}}{2}}(2+\sqrt{2}) \end{align*}.
But we can take $a = \sqrt{2(1+\frac{\sqrt{2}}{2})}(2-\sqrt{2})$ Then sum is $2$. But $2$ is rational. So problem is wrong for arbitrary square or circle.


quadratics
number theory
prime numbers
prime factorization

Let's call an integer lucky if it can be written as the sum of two squares.

a) Prove that the product of two lucky integers is again lucky.
b) Prove that there does not exist a lucky integer a such that 7a is lucky.
c) Find all prime numbers p such that if a is lucky, then pa is "unlucky", for all a.
d) Find all prime numbers p such that if a is lucky, then pa is lucky too, for all a.

just the core ideas

$(a^2 + b^2)(c^2 + d^2) = (ac+bd)^2 + (ab-cd)^2$ (*)

It's a famous theorem that n is lucky iff n has a prime factorization with every prime of the form 4k-3 containing an even exponent.

It's pretty hard to prove, however.  I don't know the full proof.

[hide="a"]$ (a^2\plus{}b^2)(c^2\plus{}d^2)\equal{}(ab\minus{}cd)^2\plus{}(ac\plus{}bd)^2$.[/hide]
[hide="b"]Assume the contrary. Let $ a^2\plus{}b^2$ be a lucky number, such that $ 7(a^2\plus{}b^2)\equal{}c^2\plus{}d^2$ is lucky. Considering mod 4, $ a^2\plus{}b^2$ and $ c^2\plus{}d^2$ must be divisible by 4. By infinite descend, we get contradiction. Q.E.D.[/hide]
I have not solve part c and d. Can anyone give me a hint?

[quote="Johan Gunardi"][hide="a"]$ (a^2 \plus{} b^2)(c^2 \plus{} d^2) \equal{} (ab \minus{} cd)^2 \plus{} (ac \plus{} bd)^2$.[/hide]
[hide="b"]Assume the contrary. Let $ a^2 \plus{} b^2$ be a lucky number, such that $ 7(a^2 \plus{} b^2) \equal{} c^2 \plus{} d^2$ is lucky. Considering mod 4, $ a^2 \plus{} b^2$ and $ c^2 \plus{} d^2$ must be divisible by 4. By infinite descend, we get contradiction. Q.E.D.[/hide]
I have not solve part c and d. Can anyone give me a hint?[/quote]

Why must they be both even? They can all be odd, which , mod 4, gives 7(2)=14 is congruent to 2 mod 4, which is true.

[hide="b"]Consider mod 7. quadratic residues modulo 7 are 0, 1, 4, and 2. c^2 +d^2 must be divisible by 7, but the only combination of quadratic residues that works are 0 and 0, so c and d are multiples of 7. We substitue c=7c' and d=7d' to get   an infinite descent.[/hide]

[hide="hint for d"]A multiple of two lucky numbers is lucky. By a well known result, all primes that are congruent to 1, mod 4, are lucky[/hide]

[hide="c solution"]2 and all primes of the form 4k+1 are lucky, so pa must also be lucky. Assume that $ a^2 \plus{} b^2$ is congruent to 0 modulo p, where p is of the form 4k+3. That implies a^2 is congruent to c, and b^2 is congruent to -c. Since a^2 is congruent to c, a^(4k+2) is congruent to c^(2k+1), and since 2k+1 is odd, b^(4k+2) is congruent to -c^(2k+1). Since $ a^2 \plus{} b^2$ is congruent to 0 modulo p, each both a and b are multiples of p or both are not multiples of p. Assume that they are not multiples of p. Then, by Fermat's Little Thereom, c^(2k+1) and -c^(2k+1) are both congruent to 1. This implies that the sum of c^(2k+1) and -c^(2k+1) is congruent to 0, so 2 must be a multiple of p, a contradiction since 2 is not of the form 4k+3. Therefore, both a and b are multiples of p. Assume that a must be lucky  and equal to $ m^2 \plus{} n^2$. Then m and n are multiples of p. But appllying this continuisly, we get an infinite descent.[/hide]

[hide="d solution"]2 and 4k+1 are lucky, so all primes other than those in the form (4k+3) work[/hide]

This stuff is actually pretty cool when you generalize it to sums of higher numbers of squares. There exist simple bilinear formulas for products of sums of 1, 2, 4, and 8 squares, but it has been proven that the pattern does not continue.


[quote="siuhochung"][quote="cezar lupu"]Let a,b,c>0 such that a^2+b^2+c^2=1. Prove that

     a^2.b^2+b^2.c^2+c^2.a^2+3(2sqrt(3)-3).a^2.b^2.c^2>=2abc.[/quote]
http://www.mathlinks.ro/Forum/viewtopic.php?t=39290
are they the same?[/quote]
you are right!
set $x=\frac{1}{a^2}$, etc and we have the above ineq


function
inequalities
calculus
calculus computations

Find the range of $ k$ such that $ x^3 \minus{} 2\geq 3k(x^2 \minus{} 2)$ for $ \forall x\geq 0$.

Let $ f(x) \equal{} x^3 \minus{} 3kx^2 \plus{} 6k \minus{} 2$.
From $ f(0) \geq 0$ and $ f(2) \geq 0$ we have $ \frac {1}{3}\leq k \leq 1$.
$ f'(x) \equal{} 3x^2 \minus{} 6kx \equal{} 3x(x \minus{} 2k)$.

So, $ f(2k) \geq 0$.

$ f(2k) \equal{} 8k^3 \minus{} 12k^3 \plus{} 6k \minus{} 2 \equal{} \minus{} 4k^3 \plus{} 6k \minus{} 2 \equal{} 2(1 \minus{} k)(2k^2 \plus{} 2k \minus{} 1)\geq 0$.

From $ \frac {1}{3}\leq k \leq 1$ and $ (1 \minus{} k)(2k^2 \plus{} 2k \minus{} 1)\geq 0$ we have $ \frac {\sqrt {3} \minus{} 1}{2} \leq k \leq 1$

[quote="Kirill"]Let $ f(x) \equal{} x^3 \minus{} 3kx^2 \plus{} 6k \minus{} 2$.
From $ f(0) \geq 0$ and $ f(2) \geq 0$ we have $ \frac {1}{3}\leq k \leq 1$. [/quote]

Why do you think them?

[quote="kunny"]
Why do you think them?[/quote]
We have $ f(x)\geq 0 \minus{} \forall x\geq 0$. And $ f(0) \equal{} 6k \minus{} 2\geq 0$, $ f(2) \equal{} 6 \minus{} 6k\geq 0$.
So, $ \frac {1}{3}\leq k \leq 1$.

Sorry, why do you coin the condition $ f(2) \geq 0$?

Because $ f(x)\geq 0$ FOR ALL $ x\geq 0$. And for x=2 $ f(2)\geq 0$ too.

I see. :lol: Anyway,　your solution was right. Thank you!


group theory
abstract algebra
superior algebra
superior algebra unsolved

1. Show that if $ \theta$ is a congruence on G and $ M \equal{} \frac {1}{\theta}$, then the cosets of $ \theta$ are {$ {gM: g\in G}$} = {$ {Mg: g\in G}$}.

If you can translate for me exactly what it is we are after here, then I will be fine. I showed M to be normal. What does a coset of $ \theta$, a relation on G, look like?

2. Let $ \theta$  be a congruence on a group G and N a normal subgroup of G.
Define $ \theta_{N}: {(x,y) \in G^2: xy^ \minus{} 1 \in N}$
Show that the mappings $ \theta \to \frac {1}{\theta}$ and $ N \to \theta_{N}$ define an inclusion-preserving bijection between the set of congruences on G and the set of normal subgroups of G.

I have shown $ \theta_{N}$ to be a congruence. But my question is, how do I define these mappings?

Here is what I did for number 1:
Choose $ g\in G$. Then, for $ s\in G$,
$ s$ is in the congruence class of $ g$ modulo $ M$ if and only if we have:
$ g \equiv s$ mod $ M$
$ \iff g^{ - 1}s \in M$
$ \iff g^{ - 1}s = m$ (for some $ m\M$)
$ \iff s = gm$ (for some $ m\in M$)
$ \iff s\in gM$.

Similarly,
$ s \in Mg$.
Hence $ gM = Mg$.


limit
logarithms
function
real analysis
abstract algebra
real analysis unsolved

Let a sequence $ a_n$ with $ 0<a_1<1$ and $ a_{n\plus{}1}\equal{}a_n(1\minus{}a_n)$ . Prove that $ \lim_{n\to\infty}\frac{n(1\minus{}na_n)}{\ln n}\equal{}1$

Iasu silouan .
since $ a_{1} < 1$ hence we easly see that $ a_{n} > a_{n \plus{} 1}$ and $ a_{n} \to 0$ as $ n \to \infty$
let's define function $ f(x)$ such that $ f(n) \equal{} a_{n}$ and on $ [n,n \plus{} 1]$ define as
\[ f(x) \equal{} (x \minus{} n \minus{} 1)f(n) \plus{} (x \minus{} n)f(n \plus{} 2)
\]
thus we define function on $ [1,\infty[$
after we can approximate via smooth function (for example take convolution of $ f$  with some good kernel)
thus we have smooth function $ g$ for which: $ g(n) \approx a_{n}$
hence we can say that $ g(n \plus{} 1) \minus{} g(n) \equal{} g'(\sigma)$ where $ \sigma \in [n,n \plus{} 1]$ but since $ g$ is monotonic decreasing and tends to $ 0$ as $ x \to \infty$  hence we can say that $ g(x \plus{} 1) \minus{} g(x) \approx g'(x)$  for big $ x$
from your equation follows that :
$ g'(n) \approx g(n \plus{} 1) \minus{} g(n) \equal{} \minus{} (g(n))^{2}$ hence $ g(x) \approx 1/n$ and hence $ a_{n} \approx 1/n$ for big $ n$
let $ a_{n} \equal{} \frac {1}{n} \plus{} \alpha_{n}$ where $ \alpha_{n} \equal{} o(1/n)$ then construct same process for sequance $ 1/n \plus{} \alpha_{n}$
we will have that  (from your equation)
$ 1/n \plus{} \alpha_{n} \minus{} (1/n \plus{} \alpha_{n})^{2} \equal{} \frac {1}{n \plus{} 1} \plus{} \alpha_{n \plus{} 1}$
since $ \alpha_{n} \equal{} o(1/n)$ we have $ (1/n \plus{} \alpha_{n})^{2} \approx 1/n^{2} \plus{} 2\alpha_{n}/n$
and also note that $ 1/(n \plus{} 1) \minus{} 1/n \approx \minus{} 1/n^{2} \plus{} 1/n^{3}$
thus we have
$ \alpha'_{n} \approx \alpha_{n \plus{} 1} \minus{} \alpha_{n} \equal{} \minus{} 2\alpha_{n}/n \plus{} 1/n^{3}$ by solving this differential equation we get :
$ \alpha_{n} \approx \frac {\ln n }{n^{2}}$
finally we get
\[ a_{n} \equal{} \frac {1}{n} \plus{} \frac {\ln n}{n^{2}} \plus{} o\left(\frac {\ln n}{n^{2}} \right)
\]
therefore :
$ \lim_{n\to\infty}\frac {n(1 \minus{} na_n)}{\ln n} \equal{} 1$

It is easy to prove that $ a_n$ is decreasing and $ \lim_{n\to\infty} a_n \equal{} 0$.Now, using  [url=http://en.wikipedia.org/wiki/Stolz-Ces%C3%A0ro_theorem]Stolz's lemma[/url], I will compute:

$ \lim_{n\to\infty}na_n \equal{} \lim_{n\to\infty} \frac {n}{\frac {1}{a_n}} \equal{} \lim_{n\to\infty} \frac {1}{\frac {1}{a_{n \plus{} 1}} \minus{} \frac {1}{a_n}} \equal{} \lim_{n\to\infty}\frac {a_n(1 \minus{} a_n)}{a_n} \equal{} 1$.

$ \lim_{n\to\infty} \frac {n\left(1 \minus{} na_n \right)}{\ln n} \equal{} \lim_{n\to\infty}\frac {na_n\left( \frac {1}{a_n} \minus{} n\right)}{\ln n } \equal{} \lim_{n\to\infty} \frac {\frac {1}{a_n} \minus{} n}{\ln n}$

Again, using Stolz's lemma we obtain:

$ \lim_{n\to\infty} \frac {\frac {1}{a_n} \minus{} n}{\ln n} \equal{} \lim_{n\to \infty}\frac {\frac {1}{a_{n \plus{} 1}} \minus{} \frac {1}{a_n} \minus{} 1}{\ln \left(1 \plus{} \frac {1}{n} \right)} \equal{} \lim_{n\to\infty}\frac {n\left(\frac {1}{a_{n \plus{} 1}} \minus{} \frac {1}{a_n} \minus{} 1\right)}{\ln \left(1 \plus{} \frac {1}{n} \right)^n} \equal{}$

$ \equal{} \lim_{n\to\infty} n\left( \frac {1}{a_{n \plus{} 1}} \minus{} \frac {1}{a_n} \minus{} 1 \right) \equal{} \lim_{n\to\infty} \frac {na_n \minus{} na_n(1 \minus{} a_n)}{a_n(1 \minus{} a_n)} \equal{} \lim_{n\to\infty} \frac {na_n}{1 \minus{} a_n} \equal{} 1$


limit
integration
real analysis
real analysis unsolved

Let us compute the limit when $n$ is even(odd case is the same).
${\Sigma}_{n}=\frac{\sqrt{2n}-\sqrt{2n-1}+\cdots+\sqrt{2}-\sqrt{1}}{\sqrt{2n}}=\frac{1}{\sqrt{2n}}(\frac{1}{\sqrt{2n}+\sqrt{2n-1}}+\cdots+\frac{1}{\sqrt{2}+\sqrt{1}})$.
Then
${\Sigma}_{n}>\frac{1}{\sqrt{2n}}(\frac{1}{2\sqrt{2n}}+\cdots+\frac{1}{2\sqrt{2}})=\frac{1}{4\sqrt{n}}(\frac{1}{\sqrt{n}}+\cdots+\frac{1}{\sqrt{1}})=\frac{1}{4n}(\frac{1}{\sqrt{\frac{n}{n}}}+\cdots+\frac{1}{\sqrt{\frac{1}{n}}})={\Omega}_{n}$,
but ${\Omega}_{n}{\rightarrow}\frac{1}{4}\int_{0}^{1}{\frac{dx}{\sqrt{x}}}=\frac{1}{2}$.Besides it is easy to verify that ${\Sigma}_{n}-{\Omega}_{n}$ tends to zero,so $\lim_{n{\rightarrow}\infty}{\Sigma_{n}}=\frac{1}{2}$.Hope it is correct.


geometry
analytic geometry
3D geometry
tetrahedron
geometry proposed

Given are $\triangle ABC,I(\alpha,\beta,\gamma)$(in barycentric coordinate)
with any $d$ through $I$ and cuts lines $BC,CA,AB$ at $A',B',C'$ prove that
$\frac{\alpha}{\overline{IA'}}+\frac{\beta}{\overline{IB'}}+\frac{\gamma}{\overline{IC'}}=0$
here $\overline{XY}$ shows directed segment. :D

[url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=revisited&t=108719]Here[/url] is some special cases  :)

I think it isn't too ugly, is anybody interested in?
I saw a similar problem for tetrahedron, too :|


Log with the base log 2 with an argument of x is 20. what is x?

If I am reading the original question correctly...

[hide]

(log 2)^20 ?

[/hide]

If I'm interpreting the problem correctly, just writing it in exponential form give the answer.

Maybe there's supposed to be further simplification?

log_2 (x) = 20
x = 2^20 = 1048576

Am I missing something?

I think he meant:

[tex]\mbox{Solve for x:}[/tex]

[tex]\displaystyle log_{~log 2}~x = 20[/tex]

Where I assume log = ln

It doesn't matter what that log is.  Also, you can get your logs to stand up straight by writing \log.


with analitic it's easy, anyway doing it geometrically( and trigo)? thanks.

[hide]
Let $M$ be the midpoint of $AB$. Then $XM=\frac{\sqrt{3}}{2}\cdot AB \Rightarrow  XY = \sqrt{3}\cdot AB$

Theorem of medians in $\triangle CXY: \ \ CX^{2}+CY^{2}= 2CM^{2}+\frac{XY^{2}}{2}$

Theorem of medians in $\triangle CAB: \ \ CA^{2}+CB^{2}= 2CM^{2}+\frac{AB^{2}}{2}$

$(CX^{2}+CY^{2})-(CA^{2}+CB^{2}) = \frac{XY^{2}}{2}-\frac{AB^{2}}{2}$

$= \frac{(\sqrt{3}AB)^{2}}{2}-\frac{AB^{2}}{2}$

$= \frac{3AB^{2}-AB^{2}}{2}$

$= AB^{2}$[/hide]


superior algebra
superior algebra unsolved

First off, just doing these annoying tables for symmetric groups I saw the relationship between the centralizer and conjugate.  The centralizer*conjugate=number of elements in S.  I'm not even sure how to start a proof for this.  We have started the topic of cosets, so I'm assuming that is why this problem was assigned.  I also know that if the left coset=right coset, gxg^{-1}=x which is similar to gx=xg.  Could someone elaborate on this?

If you just started cosets, you probably won't have learned this yet, but that follows from the orbit-stabilizer theorem. See http://en.wikipedia.org/wiki/Group_action#Orbits_and_stabilizers . If you don't know about group actions, read the top of the page first.

The action to use to prove this is G acting on itself by conjugation.


geometry
circumcircle
trigonometry
IMO Shortlist
httpbit.ly3vjUcdg

The tangents at $B$ and $A$ to the circumcircle of an acute angled triangle $ABC$ meet the tangent at $C$ at $T$ and $U$ respectively.  $AT$ meets $BC$ at $P$, and $Q$ is the midpoint of $AP$; $BU$ meets $CA$ at $R$, and $S$ is the midpoint of $BR$.  Prove that $\angle ABQ=\angle BAS$.  Determine, in terms of ratios of side lengths, the triangles for which this angle is a maximum.

[color=red][Moderator edit: The below solution is copied from http://www.kalva.demon.co.uk/short/soln/sh00g5.html .][/color]

We use the triangle $ABA'$ to find $x$. Let $O$ be the center of the circle and $R$ its radius. Then angle $OBA' = 90o$ and ∠A'OB = 1/2 ∠BOC = ∠A, so BA' = R tan A. Also ∠AOB = 2∠C, so AB = 2R sin C. ∠A'BA = ∠A, so ∠A'BA = ∠A + ∠B. Hence ∠AA'B = 180o - ∠A - ∠B - x = ∠C - x. Now from the sine rule, AB/sin AA'B = A'B/sin x. Hence sin(∠C - x) = 2 sin ∠C sin x cot ∠A. So sin ∠C cos x - cos ∠C sin x = 2 sin ∠C sin x cot ∠A. Hence cot x = cot ∠C + 2 cot ∠A.

Applying the sine rule to BMA and BMD gives sin y/sin x = MA/MB = MD/MB = sin MBD/sin MDB = sin(B - y)/sin(B + x), or sin y sin(B + x) = sin x sin(B - y). Expanding and dividing through by sin x sin y sin B gives cot x + cot B = cot y - cot B, or cot y = 2 cot B + cot x = 2 cot A + 2 cot B + cot C. This is symmetric between A and B, so cot ABM = cot BAN. But cot is strictly decreasing over 0o to 180o, so ∠ABM = ∠BAN.

We want to maximise ABM or minimise cot ABM and hence minimise 2 cot A + 2 cot B + cot C. But 2 cot A + 2 cot B = 2 sin(A+B)/(sin A sin B) = 4 sin C/(cos(A - B) - cos(A + B) ) = 4 sin C/(cos(A - B) + cos C) ≥ 4 sin C/(1 + cos C) with equality iff A = B. Thus we must take A = B. We then want to minimise 4 sin C/(1 + cos C) + cot C = 4 (2 sin k cos k)/(2 cos2k) + (1 - tan2k)/(tan k + tan k), where k = C/2, and hence to minimise 7/2 tan k + 1/(2 tan k). That is achieved by taking tan k = 1/√7 (eg complete the square), and hence sin C = sin 2k = 2 tan k/(1 + tan2k) = (√7)/4. Since A = B, we have AC = BC and 1 = AB = 2 AC sin k = 2 AC 1/√8. Hence AC = √2.

The < ABM = < BAN part cries for a synthetic proof. And indeed, there is one, but it is quite hard. I have just uploaded it on [url=http://www.cip.ifi.lmu.de/~grinberg/]my website[/url]: see the paper "Three properties of the symmedian point", Theorem 6 [b]a)[/b].

  Darij

I don't know if the part [b][size=100](a)[/size][/b] of the problem ( $ \angle ABM \equal{} \angle BAM$ ) cries for a synthetic proof yet, but an elementary one, it has already been posted in the topic [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=207440]Angle with tangential circle[/url], of [b][size=100]discredit[/size][/b].

Kostas Vittas.

[hide = No trig]
Let $L,M$ be the midpoints of $BC,CA$ respectively, and let $AL,BM$ meet the circumcircle again at $X,Y$. Let $G$ be the centroid. Since $AP$ is a symmedian, $\angle BAP=\angle XAC$ and clearly $\angle ABP=\angle ABC=\angle AXC$, so $\triangle ABP\sim \triangle AXC$. therefore $\triangle ABQ\sim \triangle AXM$ and $\angle ABQ=\angle AXM=\angle LXM$. Similarly $\angle BAS=\angle MYL$, so it suffices to show $LXYM$ cyclic. But $GA\cdot GX=GB\cdot GY\Rightarrow \frac{GA}{2}\cdot GX=\frac{GB}{2}\cdot GY \Rightarrow GL\cdot GX=GM\cdot GY$, so this quadrilateral must be cyclic and we have the desired angle equality.

To maximize this angle, note that $AX$ must be tangent to the circle with diameter $OC$; otherwise moving $X$ along arc $BC$ away from $C$ will increase the angle. Similarly $BY$ must be also tangent to this circle. Since $GL$ and $GM$ are both tangents, $\triangle GLM$ is isosceles and it follows that $ABLM$ is an isosceles trapezoid $\Rightarrow AC=BC$. Now by PoP we find $AL=AM\sqrt{2}$. Then since $\angle ALB=\angle CML = \angle CLM = \angle LBA$, $\triangle BAL$ is isosceles and $AB=AM\sqrt{2}$ as well. Therefore the ratio $a:b:c = \sqrt{2}:\sqrt{2}:1$.
[/hide]

Here's a quick solution with lengths:

Let $A_1,B_1,C_1$ be the midpoints of $BC,CA,AB$. Note that $Q$ is the foot of the  $A$-symmedian in $\triangle AB_1C_1$. Now let $D=BQ\cap A_1B_1$; similarly define $E=AS\cap A_1B_1$. It follows that $B_1D:BC_1=B_1Q:C_1Q=b^2:c^2\implies B_1D=\dfrac{b^2}{2c}$. It follows from SAS that $\triangle DB_1A\sim \triangle CAB$. Similarly we get $A_1E=\dfrac{a^2}{2c}$ and $\triangle EA_1B\sim \triangle CBA$, so $\angle D=\angle E = \angle C$, so $ABED$ is an isosceles trapezoid, implying $\angle DBA=\angle EAB\implies \angle QBA=\angle SAB$ as desired.

For the second part, let $BD,AE$ meet at $X$ and let $N$ be the projection from $C_1$ onto $A_1B_1$. Then $C_1X:XN=AB:DE=\dfrac{2c^2}{a^2+b^2+c^2}$ using the formulas for $B_1D,A_1E$ from earlier. Now let's fix $A,B$ and vary $C$ on a line parallel to $AB$ so $N$ is fixed. Then to maximize $\angle XBA$ clearly we want to maximize the ratio $C_1X:XN$, which amounts to minimizing $a^2+b^2$. As $C$ varies clearly this happens when $CA=CB$, implying $a=b$. Now let $h_c$ be the length of the altitude from $C$ so that $C_1X=\dfrac{MX}{MX+XN}\cdot \dfrac{h_c}{2}=\dfrac{2c^2}{a^2+b^2+3c^2}\cdot \dfrac{h_c}{2}=\dfrac{c^2\sqrt{4a^2-c^2}}{4a^2+6c^2}$. Now it follows that $\tan \angle XBA=\dfrac{\sqrt{c^2(4a^2-c^2)}}{2a^2+3c^2}$; to maximize this angle we should maximize the tangent, and by AM-GM we have $\dfrac{\sqrt{7c^2(4a^2-c^2)}}{\sqrt{7}(2a^2+3c^2)}\le \dfrac{2a^2+3c^2}{\sqrt{7}(2a^2+3c^2)}=\dfrac{1}{\sqrt{7}}$ with equality when $7c^2=4a^2-c^2\implies a:b:c=\sqrt{2}:\sqrt{2}:1$.

I'll prove the first part of the problem. (Very similar to #5)

Let $M,N$ - midpoints of $BC,AC$. And $AM,BN$ meet $(ABC)$ at $X,Y$.

[b][color=#960000]1)[/color][/b] $ABPQ\sim AXCN$ and hence $\angle ABQ=\angle AXN$. Similarly $\angle BAS=\angle BYN$.

[b][color=#960000]2)[/color][/b] Note that centroid of $ABC$ is incimilicenter of $(ABC)$ and $(MNC)$ and so it lays on their radical axis. Hence by Radical Axis Theorem $MNXY$ is cyclic.

Combine [b][color=#960000]1)[/color][/b] and [b][color=#960000]2)[/color][/b] for the final result.

Let $BQ$ intersect $AC$ at $D$, $AS$ intersect $AB$ at $E$. Let $F$ be on $AC$ and $G$ be on $AB$ such that $PF\parallel BD$ and $RG\parallel AE$. By the properties of a symmedian, we have $BP:PC=AB^2:AC^2$ and $AR:RC=BA^2:BC^2$. Also note that $AD=DF$ and $BE=EG$. Thus, we have
\[\frac{AD}{DC}=\frac{AD}{DF+FC}=\frac{AD}{DF+DF\frac{CP}{BP}}=\frac{AD}{AD\left(1+\frac{AC^2}{AB^2}\right)}=\frac{AB^2}{AB^2+AC^2}\]
and similarly, $\tfrac{BE}{EC}=\frac{AB^2}{AB^2+BC^2}$.

Let $a=BC$, $b=CA$, $c=AB$. Let $\alpha=\angle BAC$, $\beta=\angle ABC$, and $\gamma=\angle CAB$. By the Law of Sines,
\[\frac{\sin(\angle BAE)}{\sin(\alpha)\cos(\angle BAE)-\cos(\alpha)\sin(\angle BAE)}=\frac{\sin(\angle BAE)}{\sin(\angle CAE)}=\frac{BE}{CE}\cdot \frac{AC}{AB}=\frac{bc}{c^2+a^2}\]
When we take the reciprocal, we get
\[\sin(\alpha)\cot(\angle BAE)-\cos(\alpha)=\frac{c}{b}+\frac{a^2}{bc}\]
However, we know that $\cos(\alpha)=\tfrac{b^2+c^2-a^2}{2bc}$ so
\[\cot(\angle BAE)=\frac{3c^2+a^2+b^2}{2bc\sin(\alpha)}\]
Similarly,
\[\cot(\angle ABD)=\frac{3c^2+a^2+b^2}{2ac\sin(\beta)}\]
and since $b\sin(\alpha)=a\sin(\beta)$ it is clear that $\cot(\angle BAE)=\cot(\angle ABD)$. Since $\cot$ is injective in $(0,\pi)$, $\angle BAE=\angle ABD$ as desired. Now, to maximum this number, we need to minimize
\[\cot(\angle BAE)=\frac{3c^2+a^2+b^2}{4[ABC]}=\frac{3c^2+a^2+b^2}{\sqrt{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}}\]
We claim that the minimum is $\sqrt{7}$. Indeed, note that $\left(1,1,\tfrac{\sqrt{2}}{2}\right)$ attains that minimum. It remains to show that $3c^2+a^2+b^2\ge \sqrt{7(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}$. We can square both sides to get
\[9c^4+a^4+b^4+6a^2c^2+6b^2c^2+2a^2b^2\ge -7a^4+14a^2b^2+14a^2c^2-7b^4+14b^2c^2-7c^4\]
Rearranged and divided by $4$, this is
\[4c^4+2a^4+2b^4-2a^2c^2-2b^2c^2-3a^2b^2\ge 0\]
Which is true because
\[\frac{1}{2}(2c^2-a^2)^2+\frac{1}{2}(2c^2-b^2)^2+\frac{3}{2}(a^2-b^2)^2\ge 0\]
And the equality condition is that $a:b:c=2:2:\sqrt{2}$.

Here's a quick and easy proof of part (a) with barycentric coordinates:

Let $\omega$ be the circumcircle of $\triangle ABC$. Suppose $W = BQ \cap AS$. We will show that $W$ lies on the perpendicular bisector of $AB$.

It is well known (and easy to show) that $T= (-a^2:b^2:c^2), \ U = (a^2:-b^2:c^2)$. Therefore $P = (0:b^2:c^2), \ R = (a^2:0:c^2)$.

Thus, the midpoint of $AP$ is given by $Q = (A+P)/2 = (b^2+c^2:b^2:c^2)$, and similarly $S = (a^2: a^2+c^2: c^2)$.

Finally, $W = BQ \cap AS$ is immediately seen to be of the form $(b^2+c^2: * : c^2)$ as well as $(*: a^2+c^2: c^2)$. Hence, $W = (b^2+c^2: a^2+c^2: c^2)$.

It remains to show that $W$ lies on the perpendicular bisector of $AB$. The equation of the perpendicular bisector of $AB$ is well known to be $(a^2 - b^2)z + c^2(x - y) = 0$. And this can clearly be seen to be satisfied by $W$. $\square$


geometry
power of a point
radical axis
geometry unsolved

Prove: The locus of a point such that is the difference of the squares of the distances from two given points is a constant is a line perpendicular to the line containing the two given points.

Prove: If the four tangents from a point on the radical axis of two non-intersecting circles are drawn, then the lines containing the chords formed by the points of tangency of each circle intersect on the radical axis of the two circles.

[quote="amylopectin"]Prove: The locus of a point such that is the difference of the squares of the distances from two given points is a constant is a line perpendicular to the line containing the two given points.
[/quote]
If $MA^{2}-MB^{2}=k$ then $(MA-MB)(MA+MB)=k$, so $2\cdot BA\cdot MI=k$, where $I$ is midpoint of $AB$. Then $2\cdot \overline{BA}\cdot\overline{HI}=k$, where $H\in AM$ and $MH\perp AB$. So $H$ is constant point.
[quote]
Prove: If the four tangents from a point on the radical axis of two non-intersecting circles are drawn, then the lines containing the chords formed by the points of tangency of each circle intersect on the radical axis of the two circles.[/quote]
Assume $MAB,MCD$ are two chord. Then $MA\cdot MB=MC\cdot MD$, done.