probability
logarithms
integration
expected value
conditional probability
probability and stats

Fix $ c > 0$.  We have $ n$ boxes of which $ n \minus{} 1$ are empty, while the last box holds $ n$ balls.  $ cn^2$ other balls are distributed independently and uniformly among the $ n$ boxes.  Compute the probability that after this process is complete, the $ n$th box still holds more balls than any other box.

is there a closed form expression of this probability or you are expecting to describe the asymptotic behavior when $ c$ or $ n$ goes to infinity ?

Sorry for being unclear -- probably there is no nice closed form.  (At least, I personally don't expect one.)  $ c$ is a constant.  We're interested in what happens as $ n \to \infty$ (so, as you say, asymptotics of some sort).

With such conditions, the probability that the box originally containing $ n$ balls is not the one with the maximal number of balls in the end is exponentially small, i.e., about $ e^{\minus{}\beta(c)n}$ with some $ \beta$ that can, probably, be explicitly computed. Are you sure that the head start is $ n$ and not $ \sqrt n$? The latter would make much more sense because then you would have a decent chance to beat the original box.

No, I'm not sure ;).  Amanda says that either $ \sqrt n$ or $ \log n$ is most interesting to her.

Well, in that range, you may be pretty sure that for the head start $ H>0$, the quantity $ P(H,n)\equal{}\Phi(s_n\minus{}a)$ where $ \Phi(t)\equal{}\frac{1}{\sqrt{2\pi}}\int_t^\infty e^{\minus{}s^2/2}\,ds$ is the tail probability of the normal distribution, $ s_n>0$ is determined by $ \Phi(s_n)\equal{}\frac 1n$, and $ a\equal{}\frac H{\sqrt{cn}}$ is a fairly good approximation to the answer in the sense that the true probability is $ P(H,n)e^{O(1)}$ when $ P(H,n)$ is small and $ P(H,n)\plus{}o(1)$ when $ P(H,n)$ is comparable to $ 1$ as $ n\to\infty$. If this answer is to your liking and this precision is enough for your purposes, I can try to verify the details of my back of the envelope computation and post the full argument. If not, tell what exactly you want.

Hi, this is Amanda, the original poser of the problem.  Yes, the answer is to my liking!  It'd be very helpful to see the full argument and computations, if you don't mind.  Thanks so much.

Hi, Amanda! Welcome to the Forum! ;)

OK, let's go step by step though I'll skip some proofs (if you have difficulty recovering them, let me know).

1) Consider $ n$ independent copies of the Poisson process with intensity $ \lambda\equal{}cn$ for each box (i.e., the probability to have $ k$ balls in the box is $ \frac{\lambda^k}{k!}e^{\minus{}\lambda}$. The process you are interested in is exactly this process conditioned to the total number of balls being $ N\equal{}cn^2$. My first claim is that this conditioning is fairly irrelevant. Indeed, the expected number of balls is $ N$ and the probability to deviate from it by $ n^{1\plus{}\delta}$ (with $ \delta\equal{}0.01$, say) is small (well below $ 1/n$). Now, you may view the full Poisson process as following: you distribute a ball and then decide whether to stop or to continue with some probability depending only on the number of distributed balls. Suppose that your box with the head start $ H$ was leading after $ N\minus{}n^{1\plus{}\delta}$ steps that you almost certainly made. Then the conditional probability that it will remain leading (by the $ N$-th step, or in the entire process) if you change $ H$ to $ H\plus{}n^{2\delta}$ is very close to $ 1$ because it is very unlikely that any box will get at least $ n^{2\delta}$ balls to it during the last $ 2n^{1\plus{}\delta}$ steps (after which you almost certainly will stop). Arguing further along these lines, you reach the conclusion that changing $ H$ by $ n^{2\delta}$ has more influence on the outcome than conditioning on the exact number of balls.

2) Now we have a much easier problem because we have to compare one random variable with the maximum of $ n\minus{}1$ other independent random variables with the same distribution. The next step is to replace the Poisson law by the normal law using the CLT and assume that we have just Gaussians with variance $ \lambda$ one of which has the expectation $ H$ and the other ones expectation $ 0$. (You need to make some computations to show that the CLT holds in a wide enough range but it is a piece of cake for Poisson; just remember that you can always neglect the events of probability much less than $ 1/n$ because your final answer is at least $ 1/n$). Renormalizing, you get the formula $ \int_\mathbb R [1\minus{}\Phi(t\plus{}a)]^{n\minus{}1}\frac{1}{\sqrt{2\pi}}e^{\minus{}t^2/2}\,dt$ in which $ a\equal{}H/\sqrt\lambda$ as a good approximation to your probability in the modified problem.

3) Now comes some asymptotic analysis. Note that the first factor changes much faster than the second one when $ \Phi(t\plus{}a)$ is well above $ 1/n$. Wnen you go down from $ s_n$, it takes not more time to change the value of $ \Phi(t\plus{}a)$ twice than to change the value of $ e^{\minus{}t^2/2}$ twice but the first change results in far stronger decay of the first factor than just twofold. Thus, you have a sharp cutoff at $ s_n\minus{}a$ and, thereby, the answer is $ \Phi(s_n\minus{}a)$ with precision mentioned in my last post (this argument explains why it is true up to a bounded factor but when $ t$ is of constant size, $ e^{\minus{}t^2/2}$ hardly changes at all when $ [1\minus{}\Phi(t\plus{}a)]^{n\minus{}1}$ rises from almost $ 0$ to almost $ 1$, so you get an $ o(1)$ additive error.

4) Now, it remains to note that changing $ H$ by $ n^{2\delta}$ (or, equivalently, changing $ a$ by $ n^{2\delta\minus{}\frac 12}$ has no effect on the final answer within the claimed precision, so we can switch back to the original problem for free.

Feel free to ask as many questions as you want :).

Thanks!  I'm currently going over it and seeing how I can apply it to another problem I'm working on.  I'll let you know if I come up with any questions.


superior algebra
superior algebra unsolved

Let G be the set of all mappings \[ T_{ab}: \Re\rightarrow\Re\] defined by \[ T_{ab}(r) = ar + b\] for any real number r, where a, b are real numbers and \[ a \neq 0\]
Let H be a subset of G and H is defined by \[ H = T_{ab} \in G \] where a is rational and b is real
Show that \[ T_{ab}*V*(T_{ab})^{-1}\in H \] if  \[ V \in H\]

I can find these two:
\[ T_{ab}*T_{cd} = T_{ac, ad+b}\]

\[ (T_{ab})^{-1} = T_{a^{-1}}, _{-a^{-1}b}\]

But do not understand the steps to get

\[ T_{ab}*T_{cd}*T_{ab}^{-1} = T_{ac, ad+b} * T _{a^{-1}}, _{-a^{-1}}_{b}=\]

is equal to

 \[ T_{a, ad-bc+d}\]

Any ideas?

Shouldn't there be "is equal to $ T_{a, ad\plus{}b\minus{}bc}$?

According to the student manual the answer is with a "d "
	\[ T_{a,ad-bc+d}\]
	Could the student (solution) manual be wrong?
	I worked the problem to
	\[ T_{aca^{-1}, -aca^{-1}b+ad+b
}\]

but wasn't sure how to reduce further, etc.

Any help would be greatly appreciated.

Yes, it could be wrong.

We have

\[ T_{ab} * T_{cd} * (T_{ab})^{\minus{}1} \equal{} T_{ab} * (T_{ca^{\minus{}1},\minus{}a^{\minus{}1}bc\plus{}d}) \equal{} T_{aca^{\minus{}1}, a\cdot(\minus{}a^{\minus{}1}bc\plus{}d) \plus{} b} \equal{} T_{c, \minus{}bc\plus{}ad \plus{} b} \in V,\]

since $ c \in \mathbb{Q}$.


inequalities
triangle inequality
calculus
calculus computations

Suppose you want to find maximum and minimum modulli of $ f(z) \equal{} 3iz^2\plus{}6$ on $ |z| \leq 1$. First of all what is $ |f(z)|$? I rewrote it as $ f(z) \equal{} 3e^{i \pi /2}e^{i \theta}\plus{}6$. But isn't $ |f(z)| \equal{} 9$?

By the triangle inequality, $ |f(z)| \leq |3i z^2| \plus{} |6| \equal{} 3|z|^2 \plus{} 6 \leq 9$.  Thus, the value of for $ |f(z)|$ on this region cannot be larger than 9.  It remains to show that it is actually possible for $ |f(z)|$ to attain the value 9 somewhere on this region.  (Note that $ |f(z)|$ can certainly be less than 9 on this region -- consider, for example, $ z \equal{} 0$ or $ z \equal{} 1$.)

For $ z \equal{} e^{i{\pi\over 4}} \equal{} {\sqrt{2}\over 2}(1\plus{}i) \equal{}> |f(z)| \equal{} 3 \;$ minimum.

For $ z \equal{} e^{i{{3\pi}\over 4}} \equal{} {\sqrt{2}\over 2}(\minus{}1\plus{}i) \equal{}> |f(z)| \equal{} 9\;$ maximum.


abstract algebra
superior algebra
superior algebra unsolved

How to find the order of :

$ |Aut(\overbrace{\mathbb Z_p \times \mathbb Z_{p^2 } \cdots \mathbb Z_{p^r}}^{r\minus{}Z})|$ where $ p$ is odd prime .

Is there a good way to count it ?? if yes ? what is it?

Also how to find :

$ |Aut(\overbrace{\mathbb Z_p \times \mathbb Z_{p} \cdots \mathbb Z_{p}}^{r\minus{}Z})|$ where $ p$ is odd prime .

I'm interested to know how to calculate them . :)

You might find the article "Automorphisms of Finite Abelian Groups" from Christopher J. Hillar and Darren L. Rhea in AMM Dec 2007, useful for this question.


Give at least three ways a barometer can be used to determine the height of a tall building.

1.
[hide]go to the top and measure the height using the barometer.  when you reach the bottom, you will have the height of the building in however many barometer lengths.[/hide]

2.
[hide]go to the top, drop it, and time how long it takes to reach the ground.  then plug into kinematics equation for distance from top to ground (the height).[/hide]

3.
[hide]step back and bring the barometer up close to your eye so that its entire length exactly blocks out the height of hte building.  measure the distance from your eye to the barometer, and the distance from your eye to the building.  that is the same ratio as the height of the barometer to the height of the building, assuming zero curvature of the earth.[/hide]

4.
[hide]threaten to attack someone with the barometer if he/she doesnt tell you the height of the building[/hide]

5.
[hide]offer someone a free barometer in exchange for the information [/hide]

LOL!  That's a great solution!  And in all of those 5 cases, you could replace "barometer" with any other random object!

The solution that the book gave which I liked the most went something like:

[hide]  Go to the top of the building and hang the barometer using a string so that it is right at the ground.  Now swing it like a pendulum and calculate the length of the string by measuring the frequency of the pendulum first  :P [/hide]

[quote][hide]Go to the top of the building and hang the barometer using a string so that it is right at the ground. Now swing it like a pendulum and calculate the length of the string by measuring the frequency of the pendulum first [/hide][/quote]
Or you could just measure the length of the string...

[hide]As an improvement on alan's #1, measure your height and the length of your shadow with the barometer. Divide the first by the second and call the result r. Measure the length of the building's shadow and multiply by r. This is better because you don't have to crawl down the side of the building to do it.[/hide]

[quote="LynnelleYe"]
Or you could just measure the length of the string...
[/quote]

Well, yeah, they gave that answer also, but I liked the pendulum better  :P .


calculus
derivative
function
geometry
3D geometry
calculus computations

1. The purpose of this question is just to be ugly and see if you remember what the derivative of $\arcsin$ is.  Also, "x^x^e" is ambiguous.

2. This is also ambiguous; is the burnt shape always supposed to be a rectangular box, with the direction of burning a constant value in a direction normal to the 5 faces?  (This is not how things actually burn.)

[quote="Semigroups"]1,Find Dy/Dx where  y=(x^x^e)*arcsin(1+lnx)

2,A cube of wood sits in a fire. It burns at the same rate on five sides,but remains untouched on the bottom side. Determine what proportion of the cube has burnt away when the height of the remaining block is twice the width.[/quote]
in first: $x^{x^{e}}=e^{ln(x)*x^{e}}$ after is simpl


This example explains it:
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=70016[/url]
(you get one extra www, after which the link doesn't work).

Yeah I knew about that one I'll fix it in a couple of hours.


If $ a,b,c,d$ are real numbers, then
$ \frac{3}{2}\sqrt{(a^2\plus{}b^2)(c^2\plus{}d^2)}\plus{}\sqrt{(a^2\plus{}c^2)(b^2\plus{}d^2)\minus{}3abcd} \ge (a\plus{}d)(b\plus{}c).$


function
Discrete intermediate value theorem

There are 30 shoes in a row, 15 left ones and 15 right ones. Show that for each permuation of the shoes there 10 successive shoes such that there are five left and five right shoes each.

suppose it were not. consider for i=1 to 21 the sequence ($i,i+1,i+2,...,i+10$). By increasing i by one, the amount of shoes can only change by one. if there is no 5-5 division, then WLOG all 20 sequences got at least 6 left shoes. That's impossible with 15 of each though.

Indeed, I had solved this problem by the same way. (I, however, got in troubles with writing down my solution.)

[quote="Peter VDD"]By increasing i by one, the amount of shoes can only change by one.[/quote]

I really like this conclusion: a kind of discrete shade of the continuity method. The underlying theorem is almost trivial:

[i]If $Z$ denotes the set of all integers, and we have an integer function $f : Z\to Z$ such that for any integer k, we have $\left| f\left(k+1\right)-f\left(k\right) \right| \leq 1$, then for any two integers a and b and for any integer h lying between f(a) and f(b), there exists another integer g lying between a and b such that f(g) = h.[/i]

(Hereby, we say that a number z lies between two other numbers x and y if either $x \leq z \leq y$ or $x \geq z \geq y$.)

  Darij

[quote="darij grinberg"]The underlying theorem is almost trivial:
[i]If $Z$ denotes the set of all integers, and we have an integer function $f : Z\to Z$ such that for any integer k, we have $\left| f\left(k+1\right)-f\left(k\right) \right| \leq 1$, then for any two integers a and b and for any integer h lying between f(a) and f(b), there exists another integer g lying between a and b such that f(g) = h.[/i]
(Hereby, we say that a number z lies between two other numbers x and y if either $x \leq z \leq y$ or $x \geq z \geq y$.)[/quote]

Haha :D

Yes darij, but somewhat simpler than your function (didn't use math):
when moving $1$, 9 out of 10 shoes considered are the same. So only $1$ is new. :)

Yeah, clear! All what I did is Bourbakizing this trivial fact...

  darij

[quote="isotomion"]Yeah, clear! All what I did is Bourbakizing this trivial fact...

  [b]darij[/b][/quote]

heh

are you imposting Darij? or is that a second account of you? :D

Yes, it's a second account of myself. Actually, I'm at the summer academy for school students in Metten now, and we have good (and free  :D ) internet access here, but the firewall is very agressive, hence I have big problems with logging in on MathLinks (the cookies are really mad here). And it appears helpful to have two identities - each time, at least one of them works!

  darij

[quote="isotomion"]each time, at least one of them works![/quote]

Now give a mathematical proof of that and you're the man! :D

What about a confirmation from the first Darij Grinberg?

And yeah, comparing IP addresses could give some clues, too  :D

  dg = darij = darij grinberg = isotomion = (all my other nicks on all the other fores)

heh :)

actually I meant proving there'll be always at least one working ;)

Ah well, it doesn't bother me as long as it works...

  dg

Consider any $30$ bit string $S$ that has exactly $15$ ones, where a $1$ represents a right shoe and a $0$ represents a left shoe. We want to show that there are $10$ consecutive bits in $S$ such that exactly $5$ of them are $1$'s. For $1 \leq i \leq 21$, let $S_i$ be the number of ones in the $i$th from the right to the $(i + 9)$th bit from the right. First, assume that we had $S_1 > 5$. If $S_{11} > 5$ and $S_{21} > 5$, then $S$ would have more than $15$ ones, which is a contradiction. Also, if any of those two were equal to $5$, we would be done, so we can assume that at least one of them is less than $5$. First, say that $S_{11} < 5$. Consider the change from $S_i$ to $S_{i+1}$. We note that $S_i$ can decrease or increase by at most $1$. Thus, for some $i$ in between $1$ and $11$, we must have that $S_i = 5$. Likewise, if $S_{21} < 5$, the for some $i$ in between $1$ and $21$, we must have $S_i = 5$. So, we are done when $S_1 > 5$. If $S_1 = 5$, then we would be done, so now assume that $S_1 < 5$. Assume that $S_{11} < 5$ and that $S_{21} < 5$. Then we would have less than $15$ ones in $S$, which is a contradiction, so we may assume that at least one of $S_{11}$ and $S_{21}$ are greater than $5$, because if any of them were equal to $5$, then we would be done. If $S_{11}$ were less than $5$, then for some $i$ in between $1$ and $11$, we must have $S_i = 5$. If $S_{21}$ were less than $5$, the for some $i$ in between $1$ and $21$, we would have that $S_i = 5$, so we are done. $\blacksquare$


function
limit
Euler
complex analysis
complex analysis unsolved

$ a,b,c,d\in\mathbb{C}\setminus\{0,\minus{}1,\minus{}2,\dots\}$ satisfy $ a\plus{}b\equal{}c\plus{}d$
find
\[ \lim_{m\to\infty}\frac{\Gamma(a\plus{}m)\Gamma(b\plus{}m)}{\Gamma(c\plus{}m)\Gamma(d\plus{}m)}\]

[quote="jens18"]$ a,b,c,d\in\mathbb{C}\setminus\{0, - 1, - 2,\dots\}$ satisfy $ a + b = c + d$
find
\[ \lim_{m\to\infty}\frac {\Gamma(a + m)\Gamma(b + m)}{\Gamma(c + m)\Gamma(d + m)}\]
[/quote]

Use the definition of the Gamma function used by Euler, i.e.,
\[ \Gamma (z): = \lim_{n\to\infty}\frac {n!n^z}{\prod_{s = 0}^{n}(z + s)}.\]
Hence,
\begin{align*} L & : = \lim_{m\to\infty}\frac {\Gamma(a + m)\Gamma(b + m)}{\Gamma(c + m)\Gamma(d + m)} = \\
& = \lim_{m\to\infty}\frac {\Gamma (a)\cdot\Gamma (b)}{\Gamma (c)\cdot\Gamma (d)}\cdot\prod_{r = 1}^{m}\frac {(m + a - r)(m + b - r)}{(m + c - r)(m + d - r)} = \\
& = \lim_{m\to\infty}\frac {\Gamma (a)\cdot\Gamma (b)}{\Gamma (c)\cdot\Gamma (d)}\cdot\frac {m!\cdot m^c\cdot m!\cdot m^d}{m!\cdot m^a\cdot m!\cdot m^b}\cdot\frac {(m + c)(m + d)}{(m + a)(m + b)}\cdot\prod_{r = 0}^{m}\frac {(m + a - r)(m + b - r)}{(m + c - r)(m + d - r)} = \\
& = \frac {\Gamma (a)\cdot\Gamma (b)}{\Gamma (c)\cdot\Gamma (d)}\cdot\lim_{m\to\infty}\frac {(m + c)(m + d)}{(m + a)(m + b)}\cdot\frac {\frac {m!\cdot m^c}{\prod_{r = 0}^{m}(m + c - r)}\cdot\frac {m!\cdot m^d}{\prod_{r = 0}^{m}(m + d - r)}}{\frac {m!\cdot m^a}{\prod_{r = 0}^{m}(m + a - r)}\cdot\frac {m!\cdot m^b}{\prod_{r = 0}^{m}(m + b - r)}} = \\
& = \frac {\Gamma (a)\cdot\Gamma (b)}{\Gamma (c)\cdot\Gamma (d)}\cdot\frac {\Gamma (c)\cdot\Gamma (d)}{\Gamma (a)\cdot\Gamma (b)} = 1. \end{align*}

Note that $ m^a\cdot m^b=m^{a+b}=m^{c+d}=m^c\cdot m^d$.


Hey, its Alex from the Waterloo math seminar.  Does anyone know if and when the Toronto math circle thing Larry Rice mentioned is starting.  I'd love to go to it.
Thx.


Panowie i Panie- co jest... na tym forum nastąpił jakiś zastój... trzeba się troche postarać aby było więcej tematów, postów i w ogóle :P

Zgadzam sie ! Ale o czym mielibysmy pisac ?

mhmmm.... matematyka - to brzmi ciekawie - to nazwa jakiejs potrawy? bo tak slyszalem cos o tym kilka razy, ale nie jestem pewien

a temat jest: "nowy rzad a poziom matematyki w Polsce"  :D


$ (1)$ $ 16^{2^{x}} \equal{} 2^{(4)(2^{x})}$.  Notice that we now have two equal bases.  Thus, we can equate exponents:
$ 16^{x} \equal{} (4)(2^{x}) \longmapsto 2^{4x} \equal{} 2^{x \plus{} 2}$.
We can once again equate exponents, to get $ 4x \equal{} x \plus{} 2 \longmapsto x \equal{} \boxed {\frac {2}{3}}$.

$ (2)$ We know that $ (2x)^{x} \equal{} 36 \bullet 6 \equal{} 6^{3}$.
Quickly, we see that $ \boxed {x \equal{} 3}$ fits the requirements for this problem.


[quote="janieluvsmusic"]How many ways are there to give 28 identical apples to 4 people if each person must have at least one?[/quote]

[hide]Give an apple to each person first. Then the question asks "How many solutions in non-negative integers are there in the equation $a+b+c+d=24$, which I think has $\boxed{2925}$ solutions.[/hide]

[hide="Hmmm"]Give each of them an apple.  Then we have 24 apples left to distribute to 4 people with no restrictions.  Let o be an apple, and let each person be represented by two lines, and let that person have all apples between those 2 lines:

ooooooooooooooooo|ooo|oo|oo

So we have 27 spots, and want to choose any 3 to be lines, so the answer is $27 \choose 3$ = $\boxed{2925}$.[/hide]


search
number theory unsolved
number theory

Let A(n) be the number of divisors of natural number n and B(n) be the sum of the divisors of n.

Prove that

1) $ A(1)\plus{}A(2)\plus{}...\plus{}A(n)\equal{}[\frac{n}{1}]\plus{}[\frac{n}{2}]\plus{}...\plus{}[\frac{n}{n}]$

2) $ B(1)\plus{}B(2)\plus{}...\plus{}B(n)\equal{}[\frac{n}{1}]\plus{}2[\frac{n}{2}]\plus{}...\plus{}n[\frac{n}{n}]$

the first one posted before [url=http://www.mathlinks.ro/viewtopic.php?search_id=1263994292&t=146482]here[/url] ...
the second one can be solved using the same method and also considering what represents the fraction $ [\frac{n}{i}]$...


geometry
circumcircle
geometric transformation
homothety
ratio
trigonometry
angle bisector

An isosceles triangle $ \bigtriangleup ABC$ with $ AB \equal{} AC$ is given and let $ AA'$ be, the diameter of its circumcircle $ (O).$ Through vertex $ B,$ we draw two isogonal lines with respect to the angle $ \angle B$ and we denote as $ D,$ $ E,$ their intersection points from $ AA'$ $ ($ $ E,$ between $ A,$ $ D$ $ ).$ Through vertex $ C,$ we draw two lines symmetric to $ BC$ with respect to $ CD$ and $ CE,$ which intersect the line segments $ BD,$ $ BE,$ at points $ M,$ $ N$ respectively. Prove that the line segment $ MN,$ passes through the point $ A'.$

My best wishes to all for the next year, Kostas Vittas.

just some hints.
first prove A is Ex-homothecy point of circumcircles of $ \triangle BEC$ and $ \triangle BFC$ and $ A'$ is their In-homethecy center.
then use pascal or polar.

Denote by $ x$ the magnitude of the angles $ \angle{ABE} \equal{} \angle{DBC} \equal{} \angle{BCD} \equal{} \angle{ECA}$. Now, we will express the ratios $ NB/NE$, $ A^{\prime}E/A^{\prime}D$, and $ MD/MB$ in terms of the side lenght $ BC \equal{} a$, of the magnitudes of the angles $ \angle{B} \equal{} \angle{C} \equal{} B$, and of $ x$.

(1) By the angle bisector theorem, $ BE/NE \equal{} a/CN$. Since $ \frac {CN}{CE} \equal{} \frac {\sin{(2B \minus{} 2x)}}{\sin{(3B \minus{} 3x)}}$, and $ CE \equal{} \frac {a}{2\cos{(B \minus{} x)}}$, it follows that $ \frac {BN}{NE} \equal{} \frac {BE}{NE} \plus{} 1 \equal{} 4\cos^{2}{(B \minus{} x)}$.

(2) By the Law of Sines, $ \frac {ED}{A^{\prime}D} \equal{} \frac {CE}{CA^{\prime}} \cdot \frac {\sin{(B \minus{} 2x)}}{\cos{(B \minus{} x)}} \equal{} \frac {\sin{B}\sin{(B \minus{} 2x)}}{\cos^{2}{(B \minus{} x)}}$, thus
\[ \frac {EA^{\prime}}{A^{\prime}D} \equal{} \frac {ED}{A^{\prime}D} \plus{} 1 \equal{} \frac {\sin{B}\sin{(B \minus{} 2x) \plus{} \cos^{2}{(B \minus{} x)}}}{\cos^{2}{(B \minus{} x)}} \equal{} \frac {\cos^{2}{x}}{\cos^{2}{(B \minus{} x)}}.
\]
(3) By the angle bisector theorem, $ DM/DB \equal{} CM/a$. Since $ \frac {CM}{CD} \equal{} \frac {\sin{2x}}{\sin{3x}}$, and $ CD \equal{} \frac {a}{\cos{x}}$, it follows that $ \frac {DM}{MB} \equal{} \frac {1}{4\cos^{2}{x}}$.

Hence, by (1), (2) and (3),
\[ \frac {NB}{NE}\cdot\frac {A^{\prime}E}{A^{\prime}D}\cdot\frac {MD}{MB}\equal{} 4\cos^{2}{(B \minus{} x)} \cdot \frac {\cos^{2}{x}}{\cos^{2}{(B \minus{} x)}} \cdot \frac {1}{4\cos^{2}{x}} \equal{} 1,\]
thus, by Menelaus' theorem, applied in triangle $ EBD$, the points $ N$, $ M$, $ A^{\prime}$ are collinear.

[quote="Amir.S"]
 $ \triangle BEC$ and $ \triangle BFC$
[/quote]
$ \triangle BFC$ should be $ \triangle BDC$

Thank you dear friends for your interest.

Another approach, is to consider the points $ D',$ $ E',$ on $ CM,$ $ CN$ respectively, such that $ DD'\parallel EE'\parallel BC$ and then it is easy to show that the triangles $ \bigtriangleup MDD',$ $ \bigtriangleup NEE',$ are perspective and so, the line segments $ MN,$ $ DE,$ $ D'E',$ are concurrent at one point.

$ ($ this result is also true, in general case of the lines through vertices $ B,$ $ C,$ as arbitrary ones, with only needed condition, to be $ DD'\parallel EE'\parallel BC$ $ ).$

So, it is enough here to prove that the line segment $ D'E',$ passes through the fixed point $ A'.$

I know a not difficult way to prove the collinearity of $ A',$ $ D',$ $ E',$ based on harmonic conjugation arguments and kept in mind the remarks mentioned by [b][size=100]Amir.S[/size][/b], about the points $ A,$ $ A',$ as the external and the internal homothetic centers respectively, of the circumcircles of the triangles $ \bigtriangleup DBC,$ $ \bigtriangleup EBC$ $ ($ also not difficult to prove $ ).$

As I owe to you dear [b][size=100]Cosmin[/size][/b], I will send you as soon is possible, the complete solution I have in mind, before its posting in this forum. Give me please a little time.

Best regards, Kostas Vittas.

PS. I am looking forward to more details of [b][size=100]Amir.S’s[/size][/b] approach, based on the polar theory.


If $ a,b,c,>0$, then the following inequality holds

$ \frac{x^2(x^2\plus{}y^2)}{(y^2\plus{}xz)^2}\plus{}\frac{y^2(y^2\plus{}z^2)}{(z^2\plus{}xy)^2}\plus{}\frac{z^2(z^2\plus{}x^2)}{(x^2\plus{}yz)^2}\geq \frac{3}{2}$

[quote="marin.bancos"]If $ a,b,c, > 0$, then the following inequality holds

$ \frac {x^2(x^2 \plus{} y^2)}{(y^2 \plus{} xz)^2} \plus{} \frac {y^2(y^2 \plus{} z^2)}{(z^2 \plus{} xy)^2} \plus{} \frac {z^2(z^2 \plus{} x^2)}{(x^2 \plus{} yz)^2}\geq \frac {3}{2}$[/quote]

It's true by CS and Nesbit  :)

[quote="new_member"]That's the most [b]ANNOYING[/b] thing in the world for me.
If you don't have time to write a solution for the problem PLEASE don't just say we will use this and this and
then we are done.Thanks[/quote]
Mhm, you're right...

However, what [b]caubetoanhoc94[/b] meant may be that by Cauchy-Schwartz, we have
$ (y^2\plus{}xz)^2\leq(y^2\plus{}x^2)(y^2\plus{}z^2)$,
plugging this into the original inequality, we get
$ \sum_{cyc} \frac{x^{2}(x^{2}\plus{}y^{2})}{(y^{2}\plus{}xz)^{2}}\geq\sum_{cyc}\frac{x^2}{y^2\plus{}z^2}$
which is clearly greater than or equal to $ \frac{3}{2}$ by Nesbit.  :)

[quote="Martin N."][quote="new_member"]That's the most [b]ANNOYING[/b] thing in the world for me.
If you don't have time to write a solution for the problem PLEASE don't just say we will use this and this and
then we are done.Thanks[/quote]
Mhm, you're right...

However, what [b]caubetoanhoc94[/b] meant may be that by Cauchy-Schwartz, we have
$ (y^2 \plus{} xz)^2\leq(y^2 \plus{} x^2)(y^2 \plus{} z^2)$,
plugging this into the original inequality, we get
$ \sum_{cyc} \frac {x^{2}(x^{2} \plus{} y^{2})}{(y^{2} \plus{} xz)^{2}}\geq\sum_{cyc}\frac {x^2}{y^2 \plus{} z^2}$
which is clearly greater than or equal to $ \frac {3}{2}$ by Nesbit.  :)[/quote]

Okay!.It is my solution  :)


superior algebra
superior algebra solved

problem: Show that $U(8)$ and $U(12)$ are isomorphic.

I know that these groups are non-cyclic, and their non-identity elements have the same order.  But I dont know if thats enough to show they are isomorphic.   Also, if i define a map $\phi:U(8)\rightarrow U(12)$,  what would $\phi(x)$ be?

Hmm. both of them are isomorphic to Klein four group., isn't it clear?

There are only two groups of order 4: $Z_4$ and $K_4$ - up to isomorphism. You know that $U(8)$, and $U(12)$ ARE groups on 4 elements and that they are not cyclic groups.


How do you do the cartesian plane on geogebra?

Thanks.

All such questions should be asked on Asymptote forum. The support forum is reserved for "something doesn't work right" issues, not for "I don't know how" questions.

Just right click on the background and check axes and/or grid.

When I get a window showing ''Click to view applet''- when I click it, it says that I don't have special  components for viewing the picture. What is it? Internet browser-Google Chrome, OS: Windows XP, Board Style-Mathlinks.

Can anyone help?

You know that Chrome is still in Beta phase, right¿ I've heard about such bugs form other people trying it, so it may not be the forums fault.

This is not my fault. You need a correct Java setup on your Chrome browser. Google on how to do that, but as Zetax pointed out Chrome is a really really beta browser, I don't recommend using it.

The "Properties" often do not work properly when using the "minimal" edit window, especially when you try to change the attributes of dependent objects (so if you right click on the object and then choose "properties" from the menu, nothing happens). The "full" window doesn't seem to have this problem, so it may be something AoPS GeoGebra configuration related rather than GeoGebra related (unless it is my browser fault). It would be nice to look into it if the idea is to have the "minimal" edit window as the main option :)

I'm using Firefox 3.0.1 under windows XP.


Suppose there are $2006$ natural numbers from $1$ to $2006$ . How many way can we choose $10$ numbers from the $2006$ numbers such that no two numbers are adjacent to each other?

Between each of the $10$ numbers, we know we need at least $1$, making a total of $9$. Of the remaining $2006-9-10 = 1987$ numbers, we want to distribute them among the $11$ spots ($9$ between the numbers, $2$ at the ends). Using the standard argument of XXOO...XOX with $10$ X's and $1987$ O's, the answer should be

$\binom{1997}{10}$.


search
geometry
geometric transformation
reflection
rotation
Asymptote
parallelogram

draw a triangle such that lower part members of triangle  are   evident           1)B-C,a,b                                                   2)draw a triangle such that AB, AC and the point of lomovan  are evident

(1) Draw a triangle, given the following elements: $a, b, \angle B - \angle C$.
(2) Draw a triangle, given AB, AC and the Lomovan point.

As for the 2nd problem, I have never heard of the Lomovan point and the internet search yields nothing :?.

[color=blue][b](1a)[/b][/color] As for the 1st problem, it would be trivial, if the sides b = CA, c = AB and the angular difference $\angle B - \angle C$ of the opposite angles $\angle B, \angle C$ were given. Assuming that the problem has been solved, reflect the triangle $\triangle ABC$ in the altitude AD from the vertex A to the side a = BC into an oppositely congruent triangle $\triangle A'B'C'$. The triangles $\triangle ABB', \triangle ACC'$ are both isosceles with AB = AB' = c and AC = AC' = b and the angles $\angle CAB' = \angle C'AB$ are both equal to the angular difference $\angle B - \angle C$. For example, the angle $\angle B = \angle AB'C'$ is the external angle of the triangle $\triangle AB'C$, equal to the sum of the 2 opposite internal angles:

$\angle B = \angle AB'C' = \angle B'CA + \angle CAB' = \angle C + \angle CAB'$

$\angle CAB' = \angle B - \angle C$

Consequently, it is sufficient to construct the triangle $\triangle AB'C$ with the known sides $CA = b, AB' = c$ and the known angle $\angle CAB' = \angle B - \angle C$ and then the isosceles triangle $\triangle ABB'$ with the the known sides $AB = AB' = c$ and the known base angles $\angle ABB' = \angle AB'B = \angle B$.

[color=blue][b](1b)[/b][/color] However, since the sides a = BC, b = CA and the angular difference $\angle B - \angle C$ are given, the problem cannot be solved with compass and straightedge, because it leads to finding the intersection of a known circle with a conchoid of Nicomedes. Assuming that the problem has been solved, reflect the triangle $\triangle ABC$ in the altitude AD from the vertex A to the side a = BC into an oppositely congruent triangle $\triangle A'B'C'$, as before. We can construct the segment CA = b and the angle $\angle CAB' = \angle B - \angle C$, but we do not know the position of the point B'. Let (C) be a circle centered at the vertex C and with radius a. The remaining vertex B of the triangle $\triangle ABC$ lies on this circle. Let $B_1$ be a point on this circle and let the ray $CB_1$ meet the ray AB' (the point B' still being unknown) at a point $B_1'$. Now we have to rotate the point $B_1$ around the vertex C, i.e., move it on the circle (C), until the triangle $\triangle AB_1B_1'$ becomes isosceles.

Since the angle $\angle CDA = 90^\circ$ is right, the altitude foot D lies on a circle (N) with the diameter CA, centered at the midpoint N of the side CA. Let M be the midpoint of the segment CB'. Since MN is a midline of the triangle $\triangle AB'C$, the lines $AB' \parallel MN$ are parallel and the line MN can be constructed, with the position of the point M unknown. Since D is the midpoint of segment BB', the segment length

$DM = DB' + B'M = \frac{BB'}{2} + \frac{B'C}{2} = \frac{BC}{2} = \frac a 2$

is known. Let $M_1$ be the intersection of the ray $CB_1$ with the line MN and let $D_1$ be a point on the ray $CB_1$, between the points $M_1, B_1$, such that $M_1D_1 = MD = \frac a 2$. As the point $B_1$ moves on the circle (C) and the ray $CB_1$ is rotated around the vertex C, the point $M_1$ moves on the line MN, while the lenght $M_1D_1 = \frac a 2$ remains constant. When the triangle $\triangle AB_1B_1'$ becomes isosceles, the points $B_1 \equiv B, B_1' \equiv B', M_1 \equiv M, D_1 \equiv D$ coincide. Thus the desired point D is the intersection of the circle (N) with the locus of the points $D_1$. By definition, this is a [url=http://mathworld.wolfram.com/ConchoidofNicomedes.html]conchoid of Nicomedes[/url] with the focus C and asymptote MN (the locus of points with a constant distance from a given line measured along lines from a given point). It is well known that finding the intersection of a circle with a conchoid of Nicomedes requires solving at least a quartic equation, hence, it is not possible to find it with compass and straightedge.

Yetti

Yetti, thanks a lot for handling the first problem (I tried it for some time, now I see it was rather useless).

As for the second problem, the only interpretation that seems plausible to me is to understand "Lomovan" as a special Iranian spelling of "Lemoine" ;) . In this case, the problem is to reconstruct a triangle ABC, given the sidelines CA and AB and the symmedian point K of triangle ABC. But this problem is really easy:

The point A is immediately located as the point of intersection of the lines CA and AB. Let the parallel to the line AB through the point K meet the line CA at Y, and let the parallel to the line CA through the point K meet the line AB at Z. Then, the quadrilateral AYKZ is a parallelogram, so that its diagonal AK bisects its diagonal YZ. But K is the symmedian point of triangle ABC, and thus the line AK is the A-symmedian of triangle ABC. Hence, the segment YZ, being bisected by this A-symmedian, must be antiparallel to the side BC of triangle ABC (since symmedians bisect antiparallels and only antiparallels). Hence, < AYZ = < ABC and < AZY = < ACB; thus, the triangle AYZ is similar to the triangle ABC. So now we have constructed a triangle similar to the required triangle ABC; the rest is a trivial application of the method of similitude.

  Darij


What is the area of quadrilateral with vertices at (1,1),(5,2),(4,4) and (2,3)


Let xk, 1 \leq k \leq n, be real numbers with sum equal to 0.
Prove that there exists an index i such that
xi + xi+1 + ... + xj \geq 0 for all i \leq j < i + n,
where indices are considered modulo n.

Consider m=maxa,k(xa+xa+1+...+xa+k) where a=1..n, k=0..n-1. Let (a,k) be extremum pair. It is fact m \geq 0.

Suppose xa+xa+1+...+xa+l < 0 where l<k ==> xa+l+1+xa+l+2+...+xa+k > m !!!

Suppose xa+xa+1+...+xa+l < 0 where k < l < n ==> xa+l+1+xa+l+2+...+xa-1 > 0 ==> xa+l+1+xa+l+2+...+xa+k >m !!!


Prove the identity

${a\choose b}= \sum_{k=0}^{c}{c \choose k}{a-c \choose b-k}$

for all $a,b,c \in \mathbb{N}$ if $a>b\geq c$

[hide]
$\binom{a}{b}$ is clearly the number of ways to choose $b$ things from $a$ of them.

Now split the $a$ things into two groups: $c$ things (Pile 1) and $a-c$ things (Pile 2). We want to choose $b$ things from these: we can do this in a number of ways.

$0$ from Pile 1 and $b$ from Pile 2 - $\binom{c}{0}\binom{a-c}{b}$

$1$ from Pile 1 and $b-1$ from Pile 2 - $\binom{c}{1}\binom{a-c}{b-1}$

...

$c$ from Pile 1 and $b-c$ from Pile 2 - $\binom{c}{c}\binom{a-c}{b-c}$

Summing these all up, we have indeed

$\binom{a}{b}= \binom{c}{0}\binom{a-c}{b}+\binom{c}{1}\binom{a-c}{b-1}+\cdots+\binom{c}{c}\binom{a-c}{b-c}= \sum_{k=0}^{c}\binom{c}{k}\binom{a-c}{b-k}$.[/hide]

[hide]If we were to block walk from ${0\choose0}$ to ${a\choose b}$ on Pascal's Triangle, we must pass through row c, though any of ${c\choose k}$, for $0\le k \le b$. From that entry, we need to make an additional a-c moves, and b-k of those be right turns. Summing all of these, $\sum_{k=0}^{c}{c\choose k}{a-c \choose b-k}={a\choose b}$.[/hide]


number theory
relatively prime
greatest common divisor
algebra solved
algebra

Let $a,b$ are integers and $(a,b)=2$, solve the following system equation :
 $\left\{\begin{array}{l}(a+b)^2=25a\\18a^4+b^4=41ab^2\end{array}\right.$

Good luck ! ;)

[quote="quantum leap"][hide]Guess and check's a fine method when the algebra can get ugly :P
From the first equation, $b=5\sqrt{a}-a$, where $a$ and $b$ are integers.  Hence, $a\in\{4,16,36,64,144,\cdots\}$ [i]somewhere[/i] in that set of even squares.  Testing the first in the set, with $a=4$ we get $b=6$.  Plugging these values into the second equation, we see that $(4,6)$ indeed works.
(I'm not guaranteeing this is the only solution, just the most convenient and less algebraic to find.)[/hide][/quote]
Thank you, you so good ! ;)

Suppose $x,y,z$ are integers, solve the equation : $\frac{(y+z+1)^2}{10}=\frac{6x+5}{3\sqrt{y-z+1}+ \frac{1}{x}+\frac{1}{y}+\frac{1}{z}}$ ;)

Am I too late ??? Hopefully not...
Maybe I have a simpler solution :

Because $(a,b) = 2$, let $a = 2k, b = 2l$
Plugging this to both equations implies
$2k^{2}+4kl+2l^{2}= 25k (1)$
$36k^{4}+2l^{4}= 41kl^{2}(2)$

Obviously $(k,l) = 1$, then
From (1), $k|2$ ( also from (2) )
From (2), $l|36$

By checking $k =-2,-1,1,2$ to $(1)$, we get
$(k,l) = (2,3)$ as the only solution.

Hence, $(a,b) = (4,6)$

Raymond.

Let a=2x and b=2y where x and y are relatively prime then for equation (1) we have that (x+y)^2=25x/2 then x=2z and (2z+y)^2=25z.Let y be odd then z becomes odd if y be even z becomes 4r. For first we have ...
Thanx.

(4,6) is my answer. I can't really post well.

[quote="Differ"]What does $ (a, b) \equal{} k$ mean?[/quote]
The greatest common divisor of $ a$ and $ b$ equals $ k$. And it mean than $ k\mid a$ and $ k\mid b$ and no $ k^\prime >k$ satisfies $ k^\prime\mid a$ and $ k^\prime \mid b$.

[quote="puuhikki"][quote="Differ"]What does $ (a, b) \equal{} k$ mean?[/quote]
The greatest common divisor of $ a$ and $ b$ equals $ k$. And it mean than $ k\mid a$ and $ k\mid b$ and no $ k^\prime > k$ satisfies $ k^\prime\mid a$ and $ k^\prime \mid b$.[/quote]

Thanks.


linear algebra
matrix
inequalities
advanced fields
advanced fields unsolved

[quote="djimenez"][quote="alexrait1"]Reminder:
Frobenious norm is defined as $||A||_F=\sqrt {\sum_{i,j}{|A_{i,j}|^2 }}$

I am trying to solve directly, but I can't work it out.[/quote]

No, it is not. Easy way to think about it, take a nonzero nilpotent matrix and you have a nice counterexample. That is, for example, take $A=\left(\begin{array}{cc}0&1\\0&0\end{array}\right)=B$. Obiusly, $\|A\|\cdot\|B\|=1$, but $\|AB\|=0$.

I hope it helps!  ;)[/quote]

You are wrong here... since that is not the definition of a multiplicative norm.
The definition says: $\phi(xy)\le\phi(x)\phi(y)$
And your counter example does not work here.

[quote="alexrait1"]
You are wrong here... since that is not the definition of a multiplicative norm.
The definition says: $\phi(xy)\le\phi(x)\phi(y)$
And your counter example does not work here.[/quote]

Upss  :blush:  :oops: . I am sorry. It is just that I thought by multiplicative it should mean an equality (because though, all norms are "aditive", because they hold with the triangular inequality) Nevertheless, of course,if you are taking norms in a ring, chances are you'll have zero divisors and then the equality I thought has no sence  :blush: .

Does anyone has an example of a norm in a ring that is not multiplicative?


The sum of the squares of $5$ real numbers $a_1, a_2, \ldots, a_5$ is $1.$ Show that $\min(a_i - a_j)^2$ when $1 \leq i < j \leq 5$ is smaller or equal to $\frac{1}{10}.$

Use case-bash.
First WLOG let a_1>=...>=a_5 and after that do a case bash on whether a_5 > or < 0.

Bomb

Assume $\min_{i,j \in \{1,2,3,4,5\}}(a_i-a_j)^2 >\frac{1}{10}$, thus $\min_{i,j \in \{1,2,3,4,5\}}|a_i-a_j| > \frac{1}{\sqrt{10}}$.

It is clear that we can assume $a_i\ge0$, because involving different signs can only decrease the minimal mutual distance for equal sum of squares. For the same reason we can assume $a_1=0$.

So we have $a_1=0, a_2 > \frac{1}{\sqrt{10}}, a_3 > \frac{2}{\sqrt{10}}, a_4 > \frac{3}{\sqrt{10}}, a_5 > \frac{4}{\sqrt{10}}$. But this means $a_1^2+a_2^2+a_3^2+a_4^2+a_5^2>\frac{1+2+3+4}{10}=1$, contradiction.

[hide="comment to bomb"]I don't know why you need casebashing, but I think my reasoning can perfectly be generalised.[/hide]


Prove that

$\displaystyle {n\choose k}=\sum_{i=k}^n{{i-1}\choose {k-1}}$

[hide]
Repeated application of Pascal's Identity:

\begin{eqnarray*} {n\choose k} &=& {n-1\choose k} + {n-1\choose k-1}\\ &=& {n-2\choose k} + {n-2\choose k-1} + {n-1\choose k-1}\\ &=& {n-3\choose k} + {n-3\choose k-1} + {n-2\choose k-1} + {n-1\choose k-1} \end{eqnarray*}

Doing this repeatedly eventually gets the original identity.
[/hide]

You could also easily prove this by induction:
the number in pasal's triangle is equal to the sume of two numbers above it and one of those numbers is equal to $\sum_{i=k}^{n-1}{{i-1}\choose {k-1}}$ and the other -- to ${n-1}\choose{k-1}$
The result follows. :)


Anyone here ever play MMBN games (Megaman battle network)
It's an pretty good game. Post here to talk stuff about it.

I've played Megaman Battle Network. I like the other Megaman games a whole lot more especially the Megaman X series.  Battle Network is still pretty fun.  I've only played the first one.

cool, did you beat it yet? I've beaten all of them, well the main storyline.
The Megaman X games are pretty cool.


[quote]assume that $ \frac{(x\minus{}y)^{2}\times 3}{4}$ is max

$ 4x^{2}\plus{}4y^{2}\plus{}4z^{2} \geq 4xy\plus{}4yz\plus{}4xz\plus{}3x^{2} \minus{} 6xy\plus{}3y^{2}$

$ x^{2}\plus{}2xy\plus{}y^{2} \plus{} z^{2} \geq 4yz\plus{}4xz$

$ (x\plus{}y\minus{}2z)^{2}\geq 0$[/quote]

I jut made small typo to this very nice and simple solution Thanks [b]vg23nov[/b].

assume that $ \frac{3}{4}(x\minus{}y)^2$ is the maximum
rewrite the LHS as :
$ \frac{1}{2}(x\minus{}y)^2\plus{}\frac{1}{2}(y\minus{}z)^2\plus{}\frac{1}{2}(x\minus{}z)^2$
so we need to prove that
$ (y\minus{}z)^2\plus{}(x\minus{}z)^2 \geq \frac{1}{2}(x\minus{}y)^2$
it is equivalent to $ a^2\plus{}b^2 \geq \frac{1}{2}(a\plus{}b)^2$ which is true , if we let $ a\equal{}y\minus{}z ; b\equal{}z\minus{}x$

[quote=delegat]For arbitrary reals $ x$, $ y$ and $ z$ prove the following inequality:

$ x^{2} \plus{} y^{2} \plus{} z^{2} \minus{} xy \minus{} yz \minus{} zx \geq \max \{\frac {3(x \minus{} y)^{2}}{4} , \frac {3(y \minus{} z)^{2}}{4} , \frac {3(y \minus{} z)^{2}}{4} \}$[/quote]

very inteligent problem


Could someone please explain to me a question please. The question is as follows:

Prove that if p is a prime number and p | a^5 and a^4 | b^3 then p^2 | b.

I'm actually not really sure how to go about these kinds of problem, and what tips and/or techniques to look at. Another question is something like

Prove that if a | b + c and a | b^2 then a | c^2.

Again, this question is of similar nature as the previous, but i just don't how to go about it. Is there something i should be looking for? In terms of powers, patterns etc ....BTW, sorry if i'm posting this in the wrong section.

The first problem concerns products and powers only - notice that there are no additions.  In that case, the smartest thing to do is to reason about the prime factorizations of $a, b$, which will lead to a solution quickly.

The second involves addition, so some algebra will be necessary.  Specifically, how can you transform the given divisibility so that it is phrased in terms of $b^{2}, c^{2}$ rather than $b, c$?

[hide="A useful lemma"] If $x | y$ and $x | z$, then $x | (ay+bz)$ for arbitrary integers $a, b$.  (Positive or negative!) [/hide] [hide="A useful corollary"] $x | y, x | y \pm z \implies x | z$ [/hide] [hide="A final very large hint"] $b^{2}-c^{2}= (b+c)(b-c)$ [/hide]

I'm still struggling of this. Could someone show me the method of doing these please? Thanks.

[hide="1"] If $p | a^{5}$ and $p$ is a prime, then $a^{5}$ must contain $p$ in its prime factorization.  But since any prime in the factorization of $a^{5}$ must also be in the factorization of $a$, it follows that $p | a$.

If $a^{4}| b^{3}$ then $p^{4}| b^{3}$, so there must be at least four $p$s in the prime factorization of $b^{3}$.

[b]Case:[/b]  $b$ has less than two $p$s in its prime factorization.  But then $b^{3}$ can only have up to three $p$s in its prime factorization, and it needs four.

[b]Case:[/b]  $b$ has at least two $p$s in its prime factorization.  Then $p^{2}| b$ and we are done. [/hide]
 [hide="2"] If $a | b+c$ then $a | (b+c)(b-c) = b^{2}-c^{2}$.  But since $a | b^{2}$ it follows that $a | b^{2}-(b^{2}-c^{2}) = c^{2}$. [/hide]


I'm not sure if this is what you're referring to, but some marine organisms possess an open circulatory system--that is, instead of a complex system of arteries, veins, and capillaries as seen in our [i]closed[/i] circulatory system, they instead have blood that is 'free-flowing' throughout their body.

There is plenty of literature online about this, of course, but I do believe the starfish is one of the typical examples of a creature with this kind of circulatory system.

[quote="Sanjar"]What is the diferent betwen  circulatory system of  fish and circulatory system of human?[/quote]

Fish have single circulations whereas humans have double circulation...

This means blood passes twice to the heart in humans for a complete circuit around the body...

In fish, blood only passes once to the heart in a complete circuit around the body...[/i]


In this maze you may move only in a downward direction.  Starting with 0.4 at the top, you must continue to add each new number you land on to the sum of the previous numbers you've been on.  Your trail is complete when you land on and add the last value, which is 0.4 at the bottom.  What is the least possible sum for a complete trail through the maze?  Express your answer as a decimal to the nearest tenth.
[metapost=fdc6fa965bee6a90d3470fe4f7cf2eb4]
 save u,i;

 % Set the value of u to scale the entire image
 u:=1cm;

 % Horizontal row of numbers
 label(btex $ 0.8$ etex,-3u*right);
 label(btex $ 0.9$ etex,-2u*right);
 label(btex $ 0.7$ etex,-1u*right);
 label(btex $ 0.6$ etex,origin);
 label(btex $ 0.5$ etex,1u*right);
 label(btex $ 0.6$ etex,2u*right);
 label(btex $ 0.3$ etex,3u*right);

 % Vertical row
 label(btex $ 0.4$ etex,1.6u*up);
 label(btex $ 0.3$ etex,.8u*up);
 label(btex $ 0.7$ etex,-.8u*up);
 label(btex $ 0.4$ etex,-1.6u*up);

 % Others
 label(btex $ 0.1$ etex,(-2u,0.8u));
 label(btex $ 0.7$ etex,(2u,0.8u));
 label(btex $ 0.3$ etex,(2u,-0.8u));
 label(btex $ 0.6$ etex,(-2u,-0.8u));

 % Arrows
 for i=-1 upto 1:
 drawarrow (i*2u,.6u)--(i*2u,.2u);
 drawarrow (i*2u-.3u,.6u)--(i*2u-u+.1u,.2u);
 drawarrow (i*2u+.3u,.6u)--(i*2u+u-.1u,.2u);

 drawarrow (i*2u,-.2u)--(i*2u,-.6u);
 drawarrow (i*2u-u+.1u,-.2u)--(i*2u-.3u,-.6u);
 drawarrow (i*2u+u-.1u,-.2u)--(i*2u+.3u,-.6u);
 endfor;

 drawarrow (0,1.4u)--(0,1u);
 drawarrow (-.3u,1.4u)--(-1.7u,1u);
 drawarrow (.3u,1.4u)--(1.7u,1u);

 drawarrow (0,-u)--(0,-1.4u);
 drawarrow (-1.8u,-1u)--(-.4u,-1.4u);
 drawarrow (1.8u,-1u)--(.4u,-1.4u);
[/metapost]

[geogebra]16ab0d270fea1b89e7fcf4c9e41e38bb37b12e2e[/geogebra]

This is the diagram

I got 1.9;
solution to come


Let $ a,b,c,d\ge 0$ such that $ a^2\plus{}b^2\plus{}c^2\plus{}d^2\equal{}2$. Then \[ a^4 \plus{} b^4 \plus{} c^4 \plus{} d^4
  \plus{}1 \ge
 2 (a^3 b \plus{} b^3 c \plus{} c^3 d \plus{} a d^3)\]

$ 1)$ $ a^4 \plus{} b^4 \plus{} c^4 \plus{} d^4 \plus{} a^2b^2 \plus{} b^2c^2 \plus{} c^2d^2 \plus{} d^2a^2\geq 2 (a^3 b \plus{} b^3 c \plus{} c^3 d \plus{} a d^3)$ (because $ a^4 \plus{} a^2b^2\geq 2a^3b$) (AM-GM) but $ a^2b^2 \plus{} b^2c^2 \plus{} c^2d^2 \plus{} d^2a^2 \equal{} (a^2 \plus{} c^2)(b^2 \plus{} d^2)\leq (\frac {a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2}{2})^2 \equal{} 1$ adding :

$ a^4 \plus{} b^4 \plus{} c^4 \plus{} d^4 \plus{} 1\geq 2 (a^3 b \plus{} b^3 c \plus{} c^3 d \plus{} a d^3)$ we get the result

Sorry I didn't notice I have the same solution as SplashD


Cheryl, Don, Oleg, Shandra and Hans each have exactly one pet.  Their five pets are a ferret, parrot, dog, cat and fish, but not necessarily in that order.  In how many different ways can the five friends be paired with the five animals?

Not sure: $ 5!\equal{}120$. I think that first one has 5 choices, second 4, ... fifth has 1 choice.

Yes, you are correct.  For these type of problems, the answer should be $ n!$.

Just a sidenote: these are called permutation problems, and are covered in Introduction to Counting and Probability(and also my blog!). Bugi explained why you do $ n!$(pronounced "n factorial"), so make sure you understand why we do so, rather than blindly memorizing.


If 0<a<b<c<1 prove that

 3(a+b+c) \geq (1/(1+a) + 1/(1+b) + 1/(1+c))(a+b+c +3abc)

It can be generalized to n variables.

It is easy to prove that sum 1/(1+a)<=3/(1+(abc)^1/3) (just use the concavity of f(x)=1/(1+e^x). So we must prove that 3/(1+P^1/3)(3S/P+1)<=3, where S=sum a and P=abc. This one can be written P^1/3>=3P/S and follows from AM-GM and the fact that P<1. The generalization can be deduced from these arguments.


inequalities
logarithms
inequalities unsolved

note that:

$ x\equal{}\log_{a}^{bc}\equal{}\frac{\log b\plus{}\log c}{\log a}$ and etc.

let $ A\equal{}\log a,B\equal{}\log b,C\equal{}\log c$ we have:

$ x\equal{}\frac{B\plus{}C}{A}$ and etc.

so our inequality is equivalent to:

$ \sum\frac{1}{\frac{B\plus{}C}{A}\plus{}2}\leq\frac 34$

$ \iff\sum\frac{C}{A\plus{}B\plus{}2C}\leq\frac 34$ (*)

now note that $ 0<a,b,c<1$ thus $ A,B,C<0$ let $ A\equal{}\minus{}X,B\equal{}\minus{}Y,C\equal{}\minus{}Z$ where $ X,Y,Z\in\mathbb{R}^\plus{}$ then (*) would be equivalent to:

$ \sum\frac{Z}{X\plus{}Y\plus{}2Z}\leq\frac 34$

$ \iff\sum\left(1\minus{}\frac{X\plus{}Y\plus{}Z}{X\plus{}Y\plus{}2Z}\right)\leq\frac 34$

$ \iff\sum\frac{X\plus{}Y\plus{}Z}{X\plus{}Y\plus{}2Z}\geq\frac 94$

now note that $ X,Y,Z$ are positive numbers,thus by Cauchy-Schwartz inequality we have:

$ \left(\sum\frac{X\plus{}Y\plus{}2Z}{X\plus{}Y\plus{}Z}\right)\left(\sum\frac{X\plus{}Y\plus{}Z}{X\plus{}Y\plus{}2Z}\right)\geq 9$

$ \Rightarrow\sum\frac{X\plus{}Y\plus{}Z}{X\plus{}Y\plus{}2Z}\geq\frac 94$

QED


A ball is dropped from a height of $ 3$ meters.  On its first bounce it rises to a height of $ 2$ meters.  It keeps falling and bouncing to $ \frac23$ of the height it reached in the previous bounce.  On which bounce will it not rise to a height of $ 0.5$ meters?

$ \textbf{(A)}\ 3\qquad
\textbf{(B)}\ 4\qquad
\textbf{(C)}\ 5\qquad
\textbf{(D)}\ 6\qquad
\textbf{(E)}\ 7$

After the first bounce the ball will be two meters high, and after the second bounce, it will be 4/3 meters high. After the third bounce, it will be 8/9 of a meter high. After the fourth bounce it will be 16/27 of a meter high. After the fifth bounce it will be 32/81 of a meter high. So the answer is 5, but if it started at a very high height, then it would be best to use an inequality to solve the problem.

Isn't that kind of vague, though? I mean, on the 6th bounce, it won't reach 0.5 meters either, and same goes for the 7th bounce.

So technically those are correct, too...?  :ninja:

They probably meant, which is the [b]first[/b] bounce in which the ball does not reach 0.5 meters or something.  :huh:

Wasn't this the problem they said that they were taking any answer for, because they realized it was a mistake?  :)

Yeah, the official answer key on their site says C, D, or E.


$ g(x)$ is a function which was  defined.
$ f(x)$ is a function such that: $ f(x)\equal{}f(g(x))$
find $ f(x)$?
I found it has many applied!Any one can sovle it? :wink:

Define an equivalence relation $ \sim$ on $ \mathbb{R}$:
\[ x\sim y\Leftrightarrow \exists m,n\in \mathbb{N}: g^m(x) \equal{} g^n(y)
\]
where $ g^1(a) \equal{} g(a)$ and $ g^k(a) \equal{} g\left(g^{k \minus{} 1}(a)\right)$ for $ k\geq 2$.  The relation $ \sim$ induces a set of equivalence classes, which we denote by $ [x] \equal{} \{y\in \mathbb{R}: x\sim y \}$.

Claim: Given any function $ h: \left( \mathbb{R}/\sim \right) \to \mathbb{R}$, the function $ f(x) \equal{} h([x])$ satisfies $ f(x) \equal{} f(g(x))$ for all $ x\in \mathbb{R}$.


Remove a corner of an $8\times 8$ chessboard.
Can the remaining board be covered with $3\times 1$ (or $1\times 3$) dominoes?

Why is this in superior algebra¿

Let the missing corner be the upper left one and let the squares canonically be $(1,1);(1,2);...;(1,7);(2,1);(2,2);...;(2;8);(3;1);...;(8,8)$.
Now colour the square $(a,b)$ black iff $3|a-b$ and white otherwise. Then every stone occupies exactly one black square, and there are $22$ of them, so they totally occupy $66$ squares which is impossible.

[quote="ZetaX"]Why is this in superior algebra¿ [/quote]

Because of this:

Assign to the square $(i,j)$ the monomial $X^{i+j}$ for $i,j=\overline{0,7}$.
The sum of all the monomials on the board from which the corner $(0,0)$ was removed is $S(X)=(1+X+\ldots+X^7)^2-1$.
If we could cover the board with $3\times 1$ dominoes, then it is easy to see that $1+X+X^2|S$ in $\mathbb{Z}[X]$.
Take $\varepsilon$ be a primitive root unity of order 3. Then $1+\varepsilon+\varepsilon^2=0$, so $S(\varepsilon)=0$. But you can contradict this by direct computation.

Your solution is shorter, nicer, simpler, but my proof is the first I found because it was inspired by another problem.


algebra
polynomial
induction
vector
advanced fields
advanced fields unsolved

Let $ k$ be a field, $ F\in k[X_1,\ldots,X_n]$, $ a_1, \ldots, a_n\in k$.

(a) Show that \[ F\equal{}\sum{\lambda_{(i)}(X_1\minus{}a_1)^{i_1}\cdots(X_n\minus{}a_n)^{i_n}},\quad\lambda_{(i)}\in k\]

(b) If $ F(a_1,\ldots,a_n)\equal{}0$, show that $ F\equal{}\sum_{i\equal{}1}^n{(X_i\minus{}a_i)G_i}$ for some (not unique) $ G_i$ in $ k[X_1,\ldots,X_n]$.

a) induction on the total degree of the polynomial $ F$. A monomial $ c_iX_1^{i_1}...X_{n}^{i_n}$ of maximal degree $ N$ in the representation of $ F$ rewrite it as $ c_i(X_1\minus{}a_1)^{i_1}...(X_n\minus{}a_n)^{i_n}\plus{}R$ where $ R$ means a polynomial of total degree less than $ N$. You do this for each monomial of maximal degree and now you have $ F \equal{} \sum_{i, |i|\equal{}N}c_iX_1^{i_1}....X_n^{i_n} \plus{}G$ where $ G$ is a polynomial of total degree $ \le N\minus{}1$ and now apply induction.

b) It follows from a): you group together the terms which are divisible by $ (X_1\minus{}a_1)$ (if any), then from the rest of the sum you group together the terms which are divisible by $ X_2\minus{}a_2$ (if any), and so on...
Since F(a_1,....,a_n)=0 it follows that in the representation of a) none of the terms in the representation from a) is a constant, hence the conclusion.

Note that a) ist just the Taylor expansion around a point.
Also note that a) holds for any commutative ring with unity. Then a) is a direct corollary from it's subcase $ n\equal{}1$.
b) now follows from a) as the condition says that the coefficient at exponents $ (0,0,0,...0)$ is $ 0$, and for every other vector os exponents, there is at least one entry $ \geq 1$ (say that at position $ i$), thus that summand is divisible by $ (X_i\minus{}a_i)$.


analytic geometry
algebra proposed
algebra

Let us consider $a,b$ two integers. Prove that there exists and it is unique a pair of integers $(x,y)$ such that: \[(x+2y-a)^{2}+(2x-y-b)^{2}\leq 1.\]

see the point (a,b) in the cordinates plane, and the circle w, with center in (a,b) and ray 1.
with easy computations we see that is impossible have 2 points of form (x+2y, 2x-y) inside the circle.  :)

$x=\frac{c+2d+(a+2b)}{5}$
$y=\frac{2c-d+(a-2b)}{5}$
and
$c+2d\equiv-(a+2b)[5]$
$2c-d\equiv 2b-a[5]$ (it have unique solution $mod\ 5$)
so $c$ and $d$ are unique because with $\{c,d\}\in \{-1,0,1\}$
then $(x,y)$ is unique


Let be a regular $ 12$ -gons $ A_{1}A_{2}...A_{1}2$ .Denote $ A_{1}2$ number $ -1$ and other vertices ,denote $ 1$.In each moves select a isocceles triangle with $ 3$ vertices of polygon and changes sign of $ 3$ numbers of $ 3$ vertices .After finite moves,can we get all number in $ 12$ vertices are $ -1$

Sure: do triangles $ A_{k}A_{12-k}A_{6}$ for $ 1 \leq k \leq 5$.  Each of the vertices $ A_{i}$ with $ 6 \not | i$ gets its sign changed once, vertex $ A_{12}$ gets its sign changed 0 times and $ A_{6}$ gets its sign changed 5 times.


function
linear algebra
matrix
linear algebra unsolved

For two real valued functions $ f,g$ on R. Let $ L,A$ be Hermitian, show that $ tr[f(A)Lg(A)L]\le \frac{1}{2}tr[(f(A)L)^2 \plus{} (f(B)L)^2]$

[quote="miwalin"]For two real valued functions $ f,g$ on R. Let $ L,A$ be Hermitian, show that $ tr[f(A)Lg(A)L]\le \frac {1}{2}tr[(f(A)L)^2 \plus{} (f(B)L)^2]$[/quote]What's $ B$? Should you replace $ f(B)$ by $ g(A)$?

Sorry, the $ B$ should be replaced by $ A$. Thank you, JC_Math

$ tr[f(A)Lg(A)L]\le\frac{1}{2}tr[(f(A)L)^{2}\plus{}(g(A)L)^{2}]$ is false.
$ f(A),g(A)$ are hermitian matrices.
$ tr(((f(A)\minus{}g(A))L)^2)\equal{}tr((f(A)L)^2)\plus{}tr((g(A)L)^2)\minus{}2tr(f(A)Lg(A)L)$.
It remains to prove that $ tr(((f(A)\minus{}g(A))L)^2)\geq{0}$.
Let $ U\equal{}f(A)\minus{}g(A)$ ; it's  a hermitian matrix without any property (see the case $ f\equal{}id,g\equal{}0$).
Unfortunately $ tr((UL)^2)$ may be negative  as follows from this counterexample:
$ U\equal{}\begin{pmatrix}52&\minus{}61\\\minus{}61&64\end{pmatrix},L\equal{}\begin{pmatrix}160&59\\59&\minus{}70\end{pmatrix},tr((UL)^2)\equal{}\minus{}263102$


ceiling function
floor function
number theory unsolved
number theory

I'm unsure how to solve the following question.  Can someone assist me?

Find the smallest integer greater than $ (\sqrt{6} + \sqrt{5})^6$ or $ \lceil (\sqrt{6} + \sqrt{5})^6 \rceil = ?$ without using a calculator.

The equation eventually reduces to:

$ (\sqrt{6} + \sqrt{5})^6 \& = 5291 + 966\sqrt{30}$

Using a calculator, the smallest integer of the term $ 966\sqrt{30}$ turns out to be the first term $ 5291$. So,

$ \lceil (\sqrt{6} + \sqrt{5})^6 \rceil \& = 2 * 5291$

    or

$ \lceil 966\sqrt{30} \rceil = 5291$

This appear to be true for all binomial equations of the form $ (\sqrt{a} + \sqrt{b})^n$  which can be reduced to $ x + y\sqrt{ab}$ . I.e,

$ (\sqrt{a} + \sqrt{b})^n = x + y\sqrt{ab}$

$ \lceil  (\sqrt{a} + \sqrt{b})^n \rceil = 2x$

$ \lceil y\sqrt{ab} \rceil = x$

where $ n$ is a even number, $ a, b, x$ & $ y$ are positive integers.

Does anyone know how to prove this?

Let $ a,b,n\in\mathbb{N}$, then $ S(a,b,n) \equal{} (\sqrt {a} \plus{} \sqrt {b})^{2n} \plus{} (\sqrt {a} \minus{} \sqrt {b})^{2n} \equal{} \sum_{k \equal{} 0}^n\left({2n\atop k}\right)a^{\frac {k}{2}}b^{\frac {2n \minus{} k}{2}}[1 \plus{} ( \minus{} 1)^{2n \minus{} k}]$ is an integer, since the square root terms only occur for odd $ k$ in the summation, which are then cancelled by the factor $ [1 \plus{} ( \minus{} 1)^{2n \minus{} k}]$. Now, if $ 0 < \sqrt {a} \minus{} \sqrt {b} < 1$ holds, we conclude $ \left\lfloor(\sqrt {a} \plus{} \sqrt {b})^{2n}\right\rfloor \equal{} \left\lfloor S(a,b,n) \minus{} (\sqrt {a} \minus{} \sqrt {b})^{2n}\right\rfloor \equal{} S(a,b,n) \minus{} 1$ and similarly, $ \left\lceil(\sqrt {a} \plus{} \sqrt {b})^{2n}\right\rceil \equal{} S(a,b,n)$

For instance for $ m > 1,a \equal{} m \plus{} 1,b \equal{} m$ the condition $ 0 < \sqrt {a} \minus{} \sqrt {b} < 1$ is fulfilled. This covers your case with $ a \equal{} 6,b \equal{} 5,n \equal{} 3$


We are given $n$ planets in space, where $n$ is a positive integer. Each planet is a perfect sphere and all planets have the same radius $R$. Call a point on the surface of a planet private if it cannot be seen from any other planet. What is the total area of the private area?

[hide="a way to visualize it"]
Take a big rubber balloon-like thing, and stretch it around the planets (similar to stretching a rubber band around a bunch of circular pegs in a plane to make the convex boundary polygon).  It is not difficult to show the resulting figure is a convex polyhedron, and it is not difficult to show, from that, that a point is private iff the balloon touches it.  From there, we can show that the balloon never touches two points (excepting those on tangency areas) at the same "angle (ie with the same normal vector in the same direction).  This is by no means a proof, but it is a way to see the answer, and a framework for one. [/hide]

[hide="A way to simplify it first"]Try doing it with circles, it might be easier to visualize, or at least, it was for me.[/hide]

[hide]This is similar to K81o7's solution: If nothing can see you, then you can't see anything. For every given "direction", consider the corresponding point on the surface and draw planes tangent to that point for each planet. Exactly one of these planes will have nothing on one of the sides ("there's nothing in the sky you can see"), and this is the "private" one. Since each infinitesimal direction has precisely one planet which is private, the total private area is the surface area of one planet, 4piR².[/hide]


logarithms
function
calculus
derivative
limit
algebra unsolved
algebra

[hide="1"]The first one has no real solution.

For all $ t > 0$ we have: $ 12^t > 2.7^t$ and $ 2^t > 0$.

Thus, $ 2^t \plus{} 12^t > 2.7^t$ for all $ t > 0$

For all $ t < 0$ we have: $ 2^t > 2.7^t$ and $ 12^t > 0$.

Thus, $ 2^t \plus{} 12^t > 2.7^t$ for all $ t < 0$

Finally, for $ t \equal{} 0$ we have $ 1 \plus{} 1 \equal{} 1$ which is false.

Thus, there are no real solutions. [/hide]

EDIT: If by $ 2.7^t$ you meant $ 2\cdot7^t$ then ignore the previous "solution". The code for the dot is "\cdot" without the quotes.

I'm not sure that there are any algebraic methods to solving $ 2^t \plus{} 12^t \equal{} 2\cdot7^t$, but you can find the two solutions by guess and check or "inspection", and then prove that there are no more solutions through derivatives.

For 1:By graphing the function, I see it have $ \boxed {t\equal{}0 , t\equal{}1}$ is the roots. So it have real solution.But I don't know how to solve it by algebra.

hello, probably he meant $ 2^t\plus{}12^t\equal{}2\cdot7^t$.
Sonnhard.

[b][quote="Dr Sonnhard Graubner"]hello, probably he meant $ 2^t \plus{} 12^t \equal{} 2\cdot7^t$.
Sonnhard.[/quote]
[color=blue]That's right :lol: [/b][/color]

[color=blue][b][quote="doabli"]For 1:By graphing the function, I see it have $ \boxed {t \equal{} 0 , t \equal{} 1}$ is the roots. So it have real solution.But I don't know how to solve it by algebra.[/quote]
Your solution is not persuasive. Why don't you apply Rolle theorem (meant to use derivative) to guess the root of the equation? Do you know this?[/b][/color]

hello, rewriting your equation in the form $ \left(\frac{2}{7}\right)^t+\left(\frac{12}{7}\right)^t-2=0$ and defining the function $ f(t)=\left(\frac{2}{7}\right)^t+\left(\frac{12}{7}\right)^t-2$ so we have
$ \lim_{t \to -\infty}f(t)=+\infty$, $ \lim_{t \to +\infty}f(t)=+\infty$ further we have
$ f^'(t)=\left(\frac{2}{7}\right)^t\ln\left(\frac{2}{7}\right)^t+\left(\frac{12}{7}\right)^t\ln\left(\frac{12}{7}\right)$ and we get from here $ f^'(t)=0 \Leftrightarrow t^{\star}=\frac{\ln\left(-\frac{\ln\left(\frac{2}{7}\right)}{\ln\left(\frac{12}{7}\right)\right)}}{-\ln(6)}$ and we get $ f(t^{\star})<0$ as a global minimum. So our equation $ f(t)=0$ hase exactly two solutions $ t=0$ or $ t=1$.
Sonnhard.

hello, to 1) obviously is $ x\equal{}1$ a solution, i have not found other real solutions.
Sonnhard.

[quote="doabli"][quote="quocbao153"][color=blue][b]
Your solution is not persuasive. Why don't you apply Rolle theorem (meant to use derivative) to guess the root of the equation? Do you know this?[/b][/color][/quote]
Could you tell me your solution by using Rolle theorem please, thank you.[/quote]
[color=blue][b]Thanks ever so much for Sonhard' solution. I am very sorry for borrowing Sonhard's solution to explain my ideas.
hello, rewriting your equation in the form $ \left(\frac {2}{7}\right)^t + \left(\frac {12}{7}\right)^t - 2 = 0$ and defining the function $ f(t) = \left(\frac {2}{7}\right)^t + \left(\frac {12}{7}\right)^t - 2$ so we have
$ \lim_{t \to - \infty}f(t) = + \infty$, $ \lim_{t \to + \infty}f(t) = + \infty$ further we have
$ f^'(t) = \left(\frac {2}{7}\right)^t\ln\left(\frac {2}{7}\right)^t + \left(\frac {12}{7}\right)^t\ln\left(\frac {12}{7}\right)$
$ f^{''}(t)>0$, so $ f^'(t) =0$ has at most 1 root then $ f(t) =0$ has at most 2 roots. Hence we will guess and find out that given equation has 2 only roots which are $ t=0,t=1$
That's all I am using the Rolle theorem to guess the root of the equation via using derivative. Is it OK?[/b][/color]

[color=blue][b]and how is the 2nd problem?[/b][/color]


If $a,b,c>0$ and $abc \leq 1$, prove that
$\frac{1+ab}{2a+ab}+\frac{1+bc}{2b+bc}+\frac{1+ca}{2c+ca}\geq \frac{6(abc+2)}{4abc+5}$.

It seems dirty but there are some nice lemmas like $\frac{p^{2}r}{q}+\frac{q}{r}\geq 2p$ in your solution. Making the lemmas should be not so easy as you should consider both degree and equality conditions. Nice job!

Actually there is another solution using a substitution $a=t\frac{p}{q}, b=t\frac{q}{r}, c=t\frac{r}{p}$ where $t^{3}=abc$.


first place written = david yang.
2nd written = max schindler
3rd =Bobby shen
4th = calvin deng

after cd...
bobby shen = 1st
david yang = 2nd
max schindler + alan zhou = semis

me...
im pretty sure that i got 37, and rumors have it that the person that got 12th also got 37...
so i got tied with the person that got 12th, but he beat me in the tie breaker thing....
i fail

Gah Massachusetts got 7th. What happened? Congratulations on making top 10 still though :P but Massachusetts definitely should've gotten in the top 3 given that I'm not on the team anymore :P

I hope I made top 20... Bleh well let's see the full top 57 results tomorrow.

Reproducing part of my posts from the other thread:

Top ten teams:
1. Texas
2. Missouri
3. Washington
4. California
5. Maryland
6. Indiana
7. Massachusetts
8. North Carolina
9. New Jersey
10. Virginia

Top twelve individuals, written:
1. David Yang, CA
2. Max Schindler, MO
3. Bobby Shen, TX
4. Calvin Deng, NC
5. Alan Zhou, MA
6. Anya Katsevich, FL
7. Victor Wong, MO
8. John Zhang, WA
9. Mark Sellke, IN
10. Ray Li, PA
11. Steven Chen, TX
12. Alexander Clifton, NJ

Hi guys, hopefully this hasn't already happened...

Many of the kids in the countdown are active on AoPS/FTW. Please don't tie their real names/place to the user names. If they wish to reveal who they are, let them choose to do that, don't choose for them. I know some are already out there, but please don't state the rest or repeat names already given.

BTW - Best countdown ever, IMHO.

[quote="echen"]Will they release the 2009 written test anytime soon? I want to take it and see what I get. Also, does anyone know the scores of anyone.[/quote]

Unless their policy changes this year, nationals tests are not publicly released.  To obtain a copy, try contacting some local coaches or the State Coordinator.  I know the state coach gets a copy of the test each year, and my state has them for many years back for use as practice tests.
Alternatively, you may ask someone on AoPS who gets their hands on a copy (likely from their coaches)

[quote="veezbo"]uhm, where did darryl wu go?[/quote]

everyone was rather confused; he was clearly a fluke last year. i'd say that the sprint was a bit harder than usual, and that target was a bit harder. michigan failed; we should've gotten top ten (AT LEAST).

blah. i guess it's fine. mathcounts is dumb anyways; AMCs are better.

(excuses. excuses, excuses, excuses.)

Darryl is a two-time USAMO qualifier, like David Yang, Calvin Deng, and five other middle school students that didn't make the countdown (some of which weren't in the contest at all). If making it last year was a fluke, it's a fluke when anybody does it.

Actually, all twelve of the countdown competitors qualified for the USAMO this year.

I am waiting to see what our state (North Dakota) got. My teacher was the coach since we got first place team. I would have made it if the countdown counted (in my state), but it doesn't.  :(

Scores:

45,44,43,42,42,and then some...

I epic phailed countdown... don't watch the web cast  :(

epic epic epic epic epic epic epic epic epic epic epic epic epic epic epic epic epic epic epic epic

Congrats to Bobby Shen, 2009 MathCounts Champion.

If you made countdown, that is pretty good. I could never do that good.

I didn't watch the web cast yet (because I was at school).

[quote="Textangle"]

So yes, Hawaii got an 8 on the team round, 25th place team =).
[/quote]

Hawaii got 8 right?  Kansas got 8 too and got 13th.

Well we got 8 too but only 14th place...

That yet really lies on the strengths of the individuals...Like most people say, even a 10/10 alone on the team round won't guarantee you top five, cuz your individuals need to do decently enough, which, unfortunately, didn't happen for my team :(

[quote="modularmarc101"]I was hoping Puerto Rico(my state) wouldn't get last :([/quote]

You guys got last?  I thought it was the Virgin Islands that got last almost every year... (at least that's what the coach told me last year)

hahaha... we got a 9 on the team round, but our individual scores were TERRIBLE!!!! We made so many stupid mistakes it's not even funny, so we got 21st.

[quote="modularmarc101"]Yeah we completely failed this year.. One of my team mates told some kid that asked him that VI got last lol maybe that was you[/quote]

Eh, Quickster94 is out of MATHCOUNTS now.

[quote="AIME15"]
Eh, Quickster94 is out of MATHCOUNTS now.[/quote]
This is true, thought I'd confirm it to be safe :D

oh ok well i am too this was my last year :(

How did u do? team place and total score? Individual score and place?

[quote="modularmarc101"]really bad...
[hide][hide][hide]i did terrible. last place team, 2problems. indvidual score : 8/4 = 12 (i messed up 3 of the target problems), ranking: #203[/hide][/hide][/hide][/quote]
ouch... oh well, everyone has bad days. our entire team had a bad day, including pinkmuskrat. we just pulled it together during the team round, and got 9. but we still epicly phailed. We could've gotten top 10 if we didn't do so badly...

[quote="levans"]Many of the kids in the countdown are active on AoPS/FTW. Please don't tie their real names/place to the user names. If they wish to reveal who they are, let them choose to do that, don't choose for them. I know some are already out there, but please don't state the rest or repeat names already given.[/quote]

That post was unnecessary. You could have pm'ed dnkywin to find out who he is, instead of making a forum post inquiring, when levans specifically said not to.

228 I think. 57 States X 4 People per team.

Also, if you want to know who dnkywin is, read his post and you can probably guess.


pigeonhole principle
combinatorics unsolved
combinatorics

If you take 12 different numbers from the group $10\leq n \leq 99$

Prove that amongst those 12 numbers will always be 2 numbers, that the difference between them is a double digit number, and it can be divided by 11

This is a very simple use of the pigeonhole principle...


Find the first three positive values for which y = 5

$y = 1+5sin\frac{\pi}{2}(x+0.7)$

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=105078

If you plug in 2.7097 or 6.7097 into the original equation, you get $-3$ instead of $5$.


Find all continuous functions $ f: (0; \infty ) \to \mathbb R$ such that
\[f(f(x)) = x f(x)\]
for all reals $x>0$.

[quote="aktyw19"]Find all $ f: (0; \infty ) \rightarrow R$
$ f(f(x)) \equal{} x f(x)$[/quote]

Besides the two trivial solutions $ x^{\frac{1\plus{}\sqrt 5}2}$ and $ x^{\frac{1\minus{}\sqrt 5}2}$, I'm pretty sure there are infinitely many solutions.

Are you sure there is no other constraint (for example continuity) in this problem ?.
Thanks for checking ...

[quote="aktyw19"]Find all $ f: (0; \infty ) \rightarrow R$
$ f(f(x)) \equal{} x f(x)$[/quote]

Plus, since aktyw19 is so kind with us (I wonder what was exactly the problem he got in his class / contest / book ) :

[quote="aktyw19"]ok let f continuity[/quote]

$ f(x) > 0$, else $ f(f(x))$ would not be defined for some $ x$
$ f(a) \equal{} f(b)$ $ \implies$ $ f(f(a)) \equal{} f(f(b))$ $ \implies$ $ af(a) \equal{} bf(b)$ $ \implies$ $ a \equal{} b$ (since $ f(a) \equal{} f(b) > 0$) and so $ f(x)$ is injective.
If $ \lim_{x\to \plus{} \infty}f(x) \equal{} c > 0$ we have a contradiction when using $ x\to \plus{} \infty$ in $ f(f(x)) \equal{} xf(x)$ ($ LHS\to f(c)$ while $ RHS\to \plus{} \infty$)
If $ \lim_{x\to 0 \plus{} }f(x) \equal{} c > 0$ we have a contradiction when using $ x\to \plus{} \infty$ in $ f(f(x)) \equal{} xf(x)$ ($ LHS\to f(c)$ while $ RHS\to 0$)

$ f(x)$, as a continuous injective function is monotonous.
Considering the previous lines about limits, it's immediate to establish then that $ f((0, \plus{} \infty)) \equal{} (0, \plus{} \infty))$ and that $ f(x)$ is a bijection.

Then, let $ f^{[ \minus{} 1]}(a) \equal{} b$ : $ f(b) \equal{} a$ and $ f(f(b)) \equal{} bf(b)$ so $ b \equal{} \frac {f(f(b))}{f(b)} \equal{} \frac {f(a)}a$ and we get $ f(\frac {f(x)}x) \equal{} x$

Now let $ a\ne 1 > 0$ and $ b \equal{} \frac {\ln(f(a))}{\ln(a)}$ (such that $ f(a) \equal{} a^b$)

Using $ f(f(x)) \equal{} xf(x)$, it's easy to establish that $ f(a^{u_n}) \equal{} a^{u_{n \plus{} 1}}$ $ \forall n\ge 0$ where $ u_n$ is the fibonacci sequence starting with $ u_0 \equal{} 1$ and $ u_1 \equal{} b$

Using $ f(\frac {f(x)}x) \equal{} x$, it's easy to establish that $ f(a^{u_n}) \equal{} a^{u_{n \plus{} 1}}$ $ \forall n\le 0$ where $ u_n$ is the fibonacci sequence (extended to negative index) with $ u_0 \equal{} 1$ and $ u_1 \equal{} b$

Let then $ u \equal{} \frac {1 \plus{} \sqrt 5}2$ and $ v \equal{} \frac {1 \minus{} \sqrt 5}2$, we have $ u_n \equal{} \frac {(b \minus{} v)u^n \minus{} (b \minus{} u)v^n}{u \minus{} v}$ $ \forall n\in\mathbb Z$

Since $ f(x)$ is strictly monotonous, $ \frac {f(x) \minus{} f(y)}{x \minus{} y}$, when $ x\ne y$, has a constant sign ($ \plus{}$ is increasing, $ \minus{}$ is decreasing).

So $ A_n \equal{} \frac {f(a^{u_{n \plus{} 1}}) \minus{} f(a^{u_n})}{a^{u_{n \plus{} 1}} \minus{} a^{u_n}}$ has a constant sign.

$ A_n \equal{} \frac {a^{u_{n \plus{} 2}} \minus{} a^{u_{n \plus{} 1}}}{a^{u_{n \plus{} 1}} \minus{} a^{u_n}}$ $ \equal{} a^{u_{n \minus{} 1}}\frac {a^{u_n} \minus{} 1}{a^{u_{n \minus{} 1}} \minus{} 1}$ and so sign of $ A_n$ is sign of $ \frac {u_n}{u_{n \minus{} 1}}$

So sign of $ A_n$ is sign of $ B_n \equal{} \frac {(b \minus{} v)u^n \minus{} (b \minus{} u)v^n}{(b \minus{} v)u^{n \minus{} 1} \minus{} (b \minus{} u)v^{n \minus{} 1}}$

If $ b\ne v$ $ \lim_{n\to \plus{} \infty}B_n \equal{} u > 0$ and so the only decreasing solution may be obtained when $ b \equal{} v$ $ \forall a\ne 1$ and so $ f(x) \equal{} x^v$, which indeed is a solution.
(notice that we obtained $ f(x) \equal{} x^v$ only for $ x\ne 1$ but we extend this at $ f(1)$ thru continuity)

If $ b\ne u$ $ \lim_{n\to \minus{} \infty}B_n \equal{} v < 0$ and so the only increasing solution may be obtained when $ b \equal{} u$ $ \forall a\ne 1$ and so $ f(x) \equal{} x^u$, which indeed is a solution.
(notice that we obtained $ f(x) \equal{} x^u$ only for $ x\ne 1$ but we extend this at $ f(1)$ thru continuity)

Hence the only two continuous solutions :

$ f(x) \equal{} x^{\frac {1 \plus{} \sqrt 5}2}$ (the unique increasing solution)

$ f(x) \equal{} x^{\frac {1 \minus{} \sqrt 5}2}$ (the unique decreasing solution)


geometry
circumcircle
incenter
ratio
inradius
Euler
geometric transformation

$ ABC$ is a triangle and $ X,Y,Z$ are the points of tangency of the incircle of $ \triangle ABC$ with the sides $ BC,CA,AB$ respectively. Let $ X',Y',Z'$ be the projections of $ X,Y,Z$ on $ YZ,ZX,XY$ respectively.

1. Show that $ AX',BY',CZ'$ concur at some point $ K$.

2. Show that $ K,I,O$ are collinear, where $ O$ and $ I$ are respectively the circumcenter and the incenter of triangle $ ABC$.

3. Find the ratio $ KI/KO$ in terms of $ r$ and $ R$, where $ r$ and $ R$ are respectively the inradius and the circumradius of triangle $ ABC$.

I think it is easy with barycentric coordinates. :)

1. It's true also a more generally fact: Let ABC a triangle, P and Q two point, and A'B'C' the cevian triangle of P in ABC (with A' on BC a.s.o.), A''B''C'' the cevian triangle of Q on A'B'C' (with A'' on B'C' a.s.o.). So AA'',BB'',CC'' are concurrent.

In this case is also siple the particolar case because the orthic triangle and the tangential triangle are homotetic so K is the center of homothety.

2. As well know OI is the euler line of the triangle XYZ so it pass also through the nine point center of XYZ that is the circumcircle of X'Y'Z' that is on the line OK for the homothety.

3. It's equal to find the ratio of the homothety that is equal to the ratio berwen the circumradius of X'Y'Z' that is $ \frac{r}{2}$ and the circumradius of ABC. So it's $ \frac{r}{2R}$

[quote="¬[ƒ(Gabriel)³²¹º]¼"]3. It's equal to find the ratio of the homothety that is equal to the ratio berwen the circumradius of X'Y'Z' that is $ \frac {r}{2}$ and the circumradius of ABC. So it's $ \frac {r}{2R}$[/quote]

I think this is wrong.

Actually I was looking for a basic and detailed solution..

[quote="nayel"]I think this is wrong.

Actually I was looking for a basic and detailed solution..[/quote]

oops, I mistud the preblem sorry you are right..

$ I_a, I_b, I_c$ are excenters of the $ \triangle ABC.$ The the intouch $ \triangle XYZ$ and the excentral $ \triangle I_aI_bI_c$ are centrally similar, having parallel sides, for example, $ YZ \parallel I_bI_c$ (both $ \perp AI$). $ K \equiv I_aX \cap I_bY \cap I_cZ$ is their similarity center. $ X', Y', Z'$ are altitude feet of the $ \triangle XYZ,$ $ \triangle X'Y'Z'$ is its orthic triangle. $ A, B, C$ are altitude feet of the $ \triangle I_aI_bI_c,$ $ \triangle ABC$ is its orthic triangle. The 2 orthic triangles are centrally similar with the same similarity center $ K \equiv AX' \cap BY' \cap CZ'$ and so are their circumcircles $ (O'), (O)$ with radii $ \frac{_r}{^2}, R.$ The similarity coefficient $ \frac{\overline {KO'}}{\overline{KO}} \equal{} \frac{r}{2R} > 0$ of the $ \triangle XYZ \sim \triangle I_aI_bI_c, \triangle A'B'C' \sim \triangle ABC$ is positive, because the sides $ YZ \parallel I_bI_c$ are equally directed, therefore $ K$ is the external similarity center of $ (O'), (O).$

Inversion in the incircle $ (I)$ takes $ A, B, C$ into the midpoints $ A', B', C'$ of $ YZ, ZX, XY$ and the $ \triangle ABC$ into the medial $ \triangle A'B'C'$ of the $ \triangle XYZ.$ Image of the circumcircle $ (O)$ of the $ \triangle ABC$ is the circumcircle $ (O')$ of the $ \triangle A'B'C'.$ Therefore, $ (O')$ is the 9-point circle of the $ \triangle XYZ$ and and also circumcircle of its orthic $ \triangle X'Y'Z'.$ The inversion center $ I$ is the similarity center of the circle $ (O)$ and its image $ (O'),$ different from $ K,$ $ I$ is the internal similarity center of the circles $ (O'), (O).$ The similarity coefficient $ \frac{\overline{IO'}}{\overline{IO}} \equal{} \minus{}\frac{r}{2R} < 0$ is then negative. The circle centers $ O', O$ are collinear with their similarity centers $ K, I.$

$ \overline{O'O} \equal{} \overline{KO} \minus{} \overline{KO'} \equal{} \overline{O'I} \plus{} \overline{IO}$

$ \frac{\overline{KI}}{\overline {KO}} \equal{} 1 \minus{} \frac{\overline{IO}}{\overline {KO}} \equal{} 1 \minus{} \frac{\overline{KO} \minus{} \overline{KO'}}{\overline{KO}} \cdot \frac{\overline{IO}}{\overline{O'I} \plus{} \overline {IO}} \equal{} 1 \minus{} \frac{2R \minus{} r}{2R} \cdot \frac{2R}{2R \plus{} r} \equal{} \frac{2r}{2R \plus{} r}$


function
geometry
geometric transformation
homothety
topology
vector
real analysis

$H$ a hilbert space.
$f: [0;1]\to H$ is called a "weird" curve if:
1/$f$ is continuous injective, $f(0)=0$
2/ for any $0 \leq a \leq b \leq c \leq d \leq 1$, $(f(b)-f(a) | f(d)-f(c))=0$

Suppose that $f_1$ and $f_2$ are two weird curves on $H$.
$H_1$ the vectorial space generated by $\{f_1(x) | x \in [0;1]\}$
$H_2$ the vectorial space generated by $\{f_2(x) | x \in [0;1]\}$
Prove that there exists a bijective isometry from $Adh(H_1)$ to $Adh(H_2)$, an homothety $h$ on $H$, an homeomorphism $g$ of $[0;1]$ into itself, such that:
$f_1(x)=h(u(f_2(g(x)))$

PS: give an example of such weird function.

I didn't give it that much thought, but I think this should work, basically.

In order to get that $g$ out of the way, let's observe that the statement is equivalent to finding $u,h$ such that the sets $f_1([0,1]),(h\circ u\circ f_2)([0,1])$ coincide. It will become clear later on, I hope, that we can also forget about $h$ by assuming that $\|f(1)-f(0)\|=\|g(1)-g(0)\|\ (*)$. All we need now is the bijective isometry $u$ from $\overline H_1$ to $\overline H_2$ which will ensure that $f_1([0,1])=(u\circ f)([0,1])$ when we have the additional assumption $(*)$.

There are unique points $a_1,b_1\in[0,1]$ for which $\|f(a_1)-f(0)\|=\|f(1)-f(a_1)\|$ and $\|g(b_1)-g(0)\|=\|g(1)-g(b_1)\|$. This, together with $(*)$, actually implies $\|f(a_1)-f(0)\|=\|f(1)-f(a_1)\|=\|g(b_1)-g(0)\|=\|g(1)-g(b_1)\|$. We make $u$ map $f(a_1)-f(0),f(1)-f(a_1)$ onto $g(b_1)-g(0),g(1)-g(b_1)$, respectively. After that, we do the same for the intervals $[0,a_1],[a_1,1]$ and $[0,b_1],[b_1,1]$, finding $a_2,a_3$ in the first two respectively and $b_2,b_3$ in the last two respectively such that all the norms $\|f(a_i)-f(a_j)\|,\|f(1)-f(a_i)\|,\|f(a_j)-f(0)\|$ are equal to the corresponding norms $\|g(b_i)-g(b_j)\|,\|g(1)-g(b_i)\|,\|g(b_j)-g(0)\|$. We make $u$ map $f(a_i)-f(a_j),f(1)-f(a_i),f(a_j)-f(0)$ onto $g(b_i)-g(b_j),g(1)-g(b_i),g(b_j)-g(0)$ respectively. I think it's clear how to proceed. The construction will yield the sequences $(a_n)_n,(b_n)_n$ which will be dense in $[0,1]$, and it will define an isometry $u$ on the set consisting of vectors of the form $f(a_i)-f(a_j),f(1)-f(a_i),f(a_j)-f(0)$, which we will be able to extend by continuity to a map from $\overline H_1$ to $\overline H_2$.

i think this idea works .(although is it totally clear that $(a_n)$ is dense ?)
And now we have prove a kind of unicity, to conclude it suffices to give such a curve  :)

In order to see that $(a_n)_n$ is indeed dense, we could do this, for example:

Consider the function $\varphi: [0,1]\to[0,\|f(1)\|]$ defined by $\varphi(x)=\|f(x)\|$. It's clear that $\varphi$ is a strictly increasing continuous surjection, so it must be a homeomorphism. It should also be clear from the construction of $(a_n)_n$ that $(\varphi(a_n))_n$ is dense in $[0,\|f(1)\|]$, so $(a_n)_n=\varphi^{-1}((\varphi(a_n))_n)$ must also be dense in $[0,1]$.

As for the construction of such a function, I don't really want to get into all the details, but we can take $f$ to be the limit of a sequence of functions. We take define $f(1)$ arbitrarily and take $f_1$ to be $f_1(x)=xf(1),\ \forall x\in[0,1]$. Then we take some $f(\frac 12)$ such that $f(\frac 12)-f(0),f(1)-f(\frac 12)$ are orthogonal and with equal norms (which will be $\frac{\|f(1)\|}{\sqrt 2}$), and define $f_2(x)=\{\begin{array}{c}2xf(\frac 12),\ x\in[0,\frac 12]\\(2-2x)f(\frac 12)+(2x-1)f(1),\ x\in[\frac 12,1]\end{array}$. After that we find $f(\frac 14),f(\frac 34)$ such that the vectors $f(\frac 14)-f(0),f(\frac 12)-f(\frac 14),f(\frac 34)-f(\frac 12),f(1)-f(\frac 34)$ are mutually orthogonal and have equal norms (equal to $\frac{\|f(1)\|}2$), and define $f_3$, and so on. I think you get the idea. This sequence of functions will be uniformly convergent to our continuous function $f$.

yep, I got the idea  :)
I although thought to give such a construction with a sequence of functions but I finally found a simpler one:
Just take $f: [0;1]\to L^2([0;1])$, $f(x)$ defined by:
$f(x)(y)=1$ if $y \in [0;x[$, $0$ otherwise.


geometry
perimeter
inequalities
triangle inequality

Find the number of distinct scalene integer-sided triangles that exist such that the length of one side is prime, the length of a second side is of the form $2^{k}$ where $k\in\mathbb{N}$, the length of a third side is the square of an integer, and such that the perimeter of the triangle is less than 50.

[hide="Solution.. :maybe: "]

let $a,b,c$ denote the sides of $\triangle{ABC}$

We have,

$a=p$
$b=2^{k}$, where $k\in\mathbb{N}$
$c=n^{2}$, where $n\in\mathbb{N}$

We also know,

$a\ge2$
$b\ge2$
$c\le49$

$a+b+c<50$

if $c=49$ then $a+b=0$ which cant happen, therefore $c\le36$
if $c=36$ then $0<a+b<13$, however $a+b>36$, therefore $c\le25$
if $c=25$ then $0<a+b<24$, however $a+b>25$, therefore $c\le16$

So if $c=16$ then

From triangle inequality,

$a+b>16$
$b+16>a$
$a+16>b$

Also by definiton

$a+b<33$, so

$16<a+b<33$

We see that $b<16$ as $b=16=c$ which would be isosceles thus
$b=8$, $c=16$, and $a\in(11,13,17,19,23)$
$b=4$, $c=16$, and $a\in(13,17,19)$
$b=2$, $c=16$, and $a=17$ is the only solution

so if $c=9$

$9<a+b<41$
$b+9>a$
$a+9>b$

so for:
$b=16$, $c=9$, and $a\in(11,13,17,19,23)$
$b=8$, $c=9$, and $a\in(11,13)$
$b=4$, $c=9$, and $a=11$ is the only solution

so if $c=4$

$4<a+b<46$
$4+a>b$
$4+b>a$

so for:
$b=16$, $c=4$, we can see above since its the same as $c=16$, $b=4$
$b=8$, $c=4$, and $a\in(5,7,11)$
$b=2$, $c=4$, and $a=5$ is the only solution

so if $c=1$

$1<a+b<49$
$1+a>b$
$1+b>a$

this can only happen if $a=b$ but since its a scalene triangle there are no possible solutions for this case..

So counting them up there are a total of $S_{Total}=\boxed{21}$

[/hide]hmm htats my solution please let me know if i missed some or did something incredibly stupid

[quote="SM4RT"][hide="Solution.. :maybe: "]

let $a,b,c$ denote the sides of $\triangle{ABC}$

We have,

$a=p$
$b=2^{k}$, where $k\in\mathbb{N}$
$c=n^{2}$, where $n\in\mathbb{N}$

We also know,

$a\ge2$
$b\ge2$
$c\le49$

$a+b+c<50$

if $c=49$ then $a+b=0$ which cant happen, therefore $c\le36$
if $c=36$ then $0<a+b<13$, however $a+b>36$, therefore $c\le25$
if $c=25$ then $0<a+b<24$, however $a+b>25$, therefore $c\le16$

So if $c=16$ then

From triangle inequality,

$a+b>16$
$b+16>a$
$a+16>b$

Also by definiton

$a+b<33$, so

$16<a+b<33$

We see that $b<16$ as $b=16=c$ which would be isosceles thus
$b=8$, $c=16$, and $a\in(11,13,17,19,23)$
$b=4$, $c=16$, and $a\in(13,17,19)$
$b=2$, $c=16$, and $a=17$ is the only solution

so if $c=9$

$9<a+b<41$
$b+9>a$
$a+9>b$

so for:
$b=16$, $c=9$, and $a\in(11,13,17,19,23)$
$b=8$, $c=9$, and $a\in(11,13)$
$b=4$, $c=9$, and $a=11$ is the only solution

so if $c=4$

$4<a+b<46$
$4+a>b$
$4+b>a$

so for:
$b=16$, $c=4$, we can see above since its the same as $c=16$, $b=4$
$b=8$, $c=4$, and $a\in(5,7,11)$
$b=2$, $c=4$, and $a=5$ is the only solution

so if $c=1$

$1<a+b<49$
$1+a>b$
$1+b>a$

this can only happen if $a=b$ but since its a scalene triangle there are no possible solutions for this case..

So counting them up there are a total of $S_{Total}=boxed{21}$

[/hide]hmm htats my solution please let me know if i missed some or did something incredibly stupid[/quote]

Hey, nice solution, I think its correct.


You'll have to buy the tests (or just ask a friend or teacher), other then that, just do sample problems. Trust me, I've tried to obtain a lot of AMC-12 tests from online but with no luck, just ask your teacher.

There are a few sites that have past AMC tests.  For copyright purposes, I'm not posting the links (I think I forget the links anyways  :oops: ) but they are out there.  As far as AMC 10 goes, I don't think there are too many of them out there.  The AMC 10 is fairly new anyways, so there aren't many past tests.  I recommend just using past AHSME's/AMC 12's to study for AMC 10.  Just ignore the trig stuff.

Btw, most of the past AMC's you will find online are not that great quality.   For this reason it would be better to buy past tests from AMC.

[quote="ben12345_5"]Does anybody know of a website where I can find previous AMC 10 problems?  :?:[/quote]

You can always study the AMC Level problems.

[quote="Iversonfan2005"][quote="Drunken_Math"]You can search up "AMC 10 problems" on google.com and you'll get some pretty good results![/quote]

Hmmm..I haven't really found any sites that give you links to previous tests. However, by searching AIME problems, I found this site that gives $all$ of the previous AIME tests with solutions :D  :D  :D[/quote]

[b]A lot[/b] of people on this site know about Kalva.


AMC USA(J)MO USAMO geometry trigonometry

In $ \bigtriangleup ABC$, $ < BAC \equal{} 50^{\circ}$.  Given that $ < DAB \equal{} 10^{\circ}, < DBA \equal{} 20^{\circ}, < DCA \equal{} 30^{\circ}$, and $ D$ lies within the triangle, find $ < DBC$.

Although this problem has been posted before, most of the solutions are trig.  There is a very nice geometry one.

is the answer [hide]60?[/hide] if so ill post my solution

You should post your solution anyways... Posts like those are discouraged...

Well I can see why all the solutions are trig because this is extremely straightforward with Ceva in trigonometric form.

[quote="mathcrazed"]However, I have a very nice geometry proof as well.  It requires only basic knowledge, including the properties of cyclic quadrilaterals.  The only tricky part is the construction.[/quote]
Then post it!

[quote="mathcrazed"]Ermmm. Fine: I was hoping someone else would find it as well though.

[hide="Solution"]Thought process: Since $ 40^\circ \equal{} (30 \plus{} 10)^\circ$, it seems that we should split $ \angle DAC$.

Remark: This solution uses a very important method of constructing an equilateral triangle.  Many problems use this method, and it should be remembered.

Construct segment $ AE$ ($ E$ is on the line $ BC$, but that can be proved later, although it is unnecessary) such that $ AE \equal{} AB$, and $ \angle DAE \equal{} \angle DAB \equal{} 10^\circ$

Let the point at which $ AE$ intersects $ DC$ be $ F$.
Draw the segment $ DE$.  Notice that $ \bigtriangleup DEA$ is congruent to $ \bigtriangleup DBA$.

Thus, $ \angle BDE \equal{} 60^\circ$, which implies that $ \bigtriangleup BDE$ is equilateral.
Thus, $ \angle FEB \equal{} 60 \plus{} 20 \equal{} 80^\circ$

We also have that $ \angle ADC \equal{} 110^\circ$ because $ \angle DAC$ and $ \angle DCA$ are given.
This implies that $ \angle BDC \equal{} 100^\circ$ and that quadrilateral $ BDFE$ is cyclic due to opposite angles being supplementary.
If we connect $ B$ to $ F$, we see that $ \angle DBF \equal{} 20^\circ$, also due to the quadrilateral being cyclic.

Also notice that when we first constructed $ AE$, we created another isosceles triangle, namely $ \bigtriangleup AFC$.

Since $ AF \equal{} FC$, $ \bigtriangleup CFB$ is congruent to $ \bigtriangleup AFB$ by side-angle-side.

Thus, we have that $ \angle FBC \equal{} \angle FBA \equal{} 40^\circ\rightarrow\angle DBC \equal{} 60^\circ$
This also proves the $ E$ lies on $ BC$, and that the triangle is isosceles.

Cognos599: you were correct  :)
[/hide][/quote]
That is pretty crazy, nice job ;) I used trig since I ran out of angles to find and I suck at geo.

[quote="mathcrazed"]Ermmm. Fine: I was hoping someone else would find it as well though.

[hide="Solution"]Thought process: Since $ 40^\circ \equal{} (30 \plus{} 10)^\circ$, it seems that we should split $ \angle DAC$.

Remark: This solution uses a very important method of constructing an equilateral triangle.  Many problems use this method, and it should be remembered.

Construct segment $ AE$ ($ E$ is on the line $ BC$, but that can be proved later, although it is unnecessary) such that $ AE \equal{} AB$, and $ \angle DAE \equal{} \angle DAB \equal{} 10^\circ$

Let the point at which $ AE$ intersects $ DC$ be $ F$.
Draw the segment $ DE$.  Notice that $ \bigtriangleup DEA$ is congruent to $ \bigtriangleup DBA$.

Thus, $ \angle BDE \equal{} 60^\circ$, which implies that $ \bigtriangleup BDE$ is equilateral.
Thus, $ \angle FEB \equal{} 60 \plus{} 20 \equal{} 80^\circ$

We also have that $ \angle ADC \equal{} 110^\circ$ because $ \angle DAC$ and $ \angle DCA$ are given.
This implies that $ \angle BDC \equal{} 100^\circ$ and that quadrilateral $ BDFE$ is cyclic due to opposite angles being supplementary.
If we connect $ B$ to $ F$, we see that $ \angle DBF \equal{} 20^\circ$, also due to the quadrilateral being cyclic.

Also notice that when we first constructed $ AE$, we created another isosceles triangle, namely $ \bigtriangleup AFC$.

Since $ AF \equal{} FC$, $ \bigtriangleup CFB$ is congruent to $ \bigtriangleup AFB$ by side-angle-side.

Thus, we have that $ \angle FBC \equal{} \angle FBA \equal{} 40^\circ\rightarrow\angle DBC \equal{} 60^\circ$
This also proves the $ E$ lies on $ BC$, and that the triangle is isosceles.

Cognos599: you were correct  :)
[/hide][/quote]
Woah, that's absurd. I can't usually find the crazy construction solutions, though maybe given 4.5 hours I could. :)


modular arithmetic
search
number theory unsolved
number theory

Show that if $ p$ is prime then $ \dbinom{p\minus{}1}{k} \equiv (\minus{}1)^k \pmod p$ for $ 0 \le k \le p\minus{}1$.

Ah, my TA just explained it to me the exact same way :) Thanks for the help though!
[hide="Same proof but more formal"]
Before we begin, we first show the trivial case where $ p = 2$ and see that $ \dbinom{2}{k}$ will equal $ 1$ for both $ k = 0$ and $ k = 1$. By inspection, $ 1 \equiv ( - 1)^k \pmod 2$ for any $ k$. Thus we will assume $ p$ is odd.

$ \dbinom{p - 1}{k} \& = \frac {(p - 1)!}{k!(p - k - 1)!}$, which gives us $ \frac {(p - 1)!}{k!(p - k - 1)!} \equiv ( - 1)^k \pmod p$. Since $ {k!(p - k - 1)!}$ is an integer, we can multiply the congruence by it to get $ (p - 1)! \equiv k!(p - k - 1)!( - 1)^k \pmod p$.

Note that
$ \begin {align*}p - (k - 1) \& \equiv - k - 1 \pmod p \\
p - (k - 2) \& \equiv - k - 2 \pmod p \\
. \\
. \\
. \\
1 \& \equiv - p + 1 \pmod p \end{align*}$

Thus,
$ \begin {align*}(p - k - 1)! \& \equiv ( - k - 1)( - k - 2)...( - p + 1) \pmod p \\
\& \equiv ( - 1)^{p - k - 1} (k + 1)(k + 2)(k + 3)...(k + (p - k - 1)) \pmod p \\
\end {align*}$.

Plugging this back into our original congruence equation, we get
\begin{align*} (p - 1)! & \equiv k!(p - k - 1)!( - 1)^k \pmod p \\
& \equiv k!(k + 1)(k + 2)(k + 3)...(k + (p - k - 1))( - 1)^k( - 1)^{p - k - 1} \pmod p \\
& \equiv (p - 1)!( - 1)^{p - 1} \pmod p \end{align*}

Now, using the theorem that $ x \equiv y \pmod {\frac {m}{(a,m)}}$ if $ ax \equiv ay \pmod m$, we divide by $ (p - 1)!$ to get $ 1 \equiv ( - 1)^{p - 1} \pmod {\frac {p}{(p, (p - 1)!)}}$. However, $ (p, (p - 1)!)$ is just $ 1$ since $ p$ is a prime and the only factor in $ (p - 1)!$ that divides $ p$ is $ 1$.

So the congruence is equal to ${ 1 \equiv ( - 1)^{p - 1} \pmod p}$.
Since we assumed $ p$ is odd, this congruence is equivalent to $ 1 \equiv 1 \pmod p$. This is clearly true for all values of $ p$.

Note that this process of derivation is reversible because the other way would  involve multiplication, which works for all congruences, and division by $ k!(p - k - 1)!$, which works because $ (p, k!(p - k - 1)!) = 1$ (due to $ 0 \le k \le p - 1$) and
$ \frac {(p - 1)!}{k!(p - k - 1)!} = \dbinom {p - 1}{k}$ is an integer (It can generally be interpreted that choosing $ k$ sets out of $ p - 1$, where $ k \le p - 1$, will always hold an integer. Formal proof of this can be seen at http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1474740710&t=113106).

Thus, $ \dbinom{p - 1}{k} \equiv ( - 1)^k \pmod p$ for $ 0 \le k \le p - 1$.
[/hide]

$ (x \plus{} 1)^p \equiv x^p \plus{} 1 \bmod p \implies$
$ (x \plus{} 1)^{p \minus{} 1} \equiv \frac {x^p \plus{} 1}{x \plus{} 1} \equal{} x^{p \minus{} 1} \minus{} x^{p \minus{} 2} \pm ... \plus{} 1 \bmod p$.

[quote="kevinatcausa"]Question:  Is the following conjecture true, and, if so, is there a combinatorial proof?

[b]Conjecture:[/b] The number of $ k$ element subsets of $ \{1, \dots p \minus{} 1\}$ with sum congruent to $ 1$ modulo $ p$ is equal to $ \frac {\binom{p \minus{} 1}{k} \minus{} ( \minus{} 1)^k}{p}$.[/quote]

Yes it is true.

It follows by this fact:

(*) the number of $ k$ element subsets of $ \mathbb{Z}/p\mathbb{Z}$ with sum congruent to $ 1$ modulo $ p$ is equal to $ \frac {\binom{p}{k}}{p}$ (with $ 1\le k\le p\minus{}1$)

Infact called $ p_k$ the number of these sets and called $ x_k$ the number of sets you are looking for you have that $ x_1\equal{}p_1\equal{}1$ and $ x_k\plus{}x_{k\minus{}1}\equal{}p_k$ Now the thesis follows from an easy calculation with sums of binomials.

Now let's prove (*):

definition: two $ k$ element subsets of $ \mathbb{Z}/p\mathbb{Z}$ that we call $ A\equal{}\{a_1,\dots ,a_k\}$ and $ B\equal{}\{a_1,\dots ,a_k\}$ are equivalent if there exists an integer $ j$ such that $ B\equal{}\{a_1\plus{}j,\dots ,a_k\plus{}j\}$ (the equality is always meant modulo $ p$)

Now A equivalent to B is a relationship of equivalence. Every class of equivalence has exactly p elements of which exactly one has some of the elements equal to 1 (no two distinct equivalent subset can have the same sum of the elements because the difference of the sums is jk which is not a multimple of p). The thesis follows.

As you probably have already noticed, to prove the problem in a combinatorial way it is enough to say that \[ \binom{p\minus{}1}{k} \equal{}\sum_{i\equal{}0}^k \binom{p}{i}(\minus{}1)^{k\minus{}i}\equiv (\minus{}1)^k \pmod p\]


Given a real number $ A$ and an integer $ n$ with $ 2 \le n \le 19,$ find all polynomials $ P(x)$ with real coefficients such that $ P(P(P(x)))\equal{}Ax^n\plus{}19x\plus{}99.$

If $ P(x)\equal{}ax^k\plus{}...$, then $ P(P(P(x)))\equal{}a^{k(k\plus{}1)\plus{}1}x^{k^3}\plus{}...$, therefore if $ A\not \equal{}0$ $ k\equal{}2,n\equal{}8,a\equal{}A^{1/7}$.
But from $ P(x)\equal{}ax^2\plus{}bx\plus{}c\to P(P(P(x)))\equal{}a^7x^8\plus{}4a^6bx^7\plus{}...\to b\equal{}0.$. If $ P(x)\equal{}ax^2\plus{}c$, Then $ P(P(P(x))\not \equal{}Ax^8\plus{}19x\plus{}..$
Therefore $ A\equal{}0, P(x)\equal{}ax\plus{}b,P(P(x))\equal{}a^2x\plus{}(a\plus{}1)b,P(P(P(x)))\equal{}a^3x\plus{}(a^2\plus{}a\plus{}1)b$ and $ a\equal{}\sqrt[3]{19}, b\equal{}\frac{11(\sqrt[3]{19}\minus{}1)}{2}.$