function
algebra
domain
logarithms
real analysis
real analysis unsolved

1) Find all continuos functions $ f: R_ \plus{} \to R$ s.t $ f(a^x) \plus{} f(a^{2x}) \equal{} x$, $ x\in {R}$,where a is a real pozitive and a is not $ 1$
2) Find all continuos functions s.t $ f(xy) \equal{} f(x^2) \plus{} f(y^2) \minus{} ln(xy)$ for every pozitive $ x$,$ y$ and $ f(1) \equal{} 0$

[quote="posabogdan"]1) Find all continuos functions s.t $ f(a^x) \plus{} f(a^{2x}) \equal{} x$, $ x\in {R}$
2) Find all continuos functions s.t $ f(xy) \equal{} f(x^2) \plus{} f(y^2) \minus{} ln(xy)$ for every pozitive $ x$,$ y$ and $ f(1) \equal{} 0$[/quote]
1) Let $ x$ be $ log_ak$ so $ f(k) \plus{} f(k^{2}) \equal{} log_ak$ so $ k$ is in the domain of the function $ f$ but $ k\in R_\plus{}$ so $ f: R_\plus{}\to R$ but it shows the function is not continuos anymore.
2) i have no any idea on it.

Set $ y \equal{} x$ to get $ f(x^2) \equal{} \log x^2$.  Since $ x \mapsto x^2$ is a bijection on $ {\bf R}_{> 0}$, $ f(x) \equal{} \log x$ is the unique solution.  (We didn't need either continuity or the given value.)


real analysis
function
Support
integration
geometry
real analysis unsolved

Show, that the Graph of $ \Gamma_f : \equal{} \{(x, f(x)) \vert \ x \in A \} \subset \mathbb{R}^n$ of a continuous function $ f: A \rightarrow \mathbb{R}$ with compact $ A \subset \mathbb{R}^{n\minus{}1}$ is a lebesgue null set.

Here's a more general result: if $ f$ is a nonnegative measurable function on some measurable set $ E$, then $ m(\Gamma(f,E))\equal{}0$ (note that a continuous function with compact support is measurable, and a function can be written as a difference of two nonnegative functions).

Given $ \epsilon>0$ and $ k\equal{}0,1,\ldots$, let $ E_{k}\equal{}\{k\epsilon\leq f<(k\plus{}1)\epsilon\}$.  The $ E_{k}$ are disjoint and measurable, and their union is a subset of $ E$ where $ f$ is finite.  Since $ m^*(\Gamma(f,E_{k}))\leq\epsilon m(\Gamma(f,E))$, we obtain $ m(\Gamma(f,E))\leq\sum_{k} m^*(\Gamma(f,E_{k}))\leq\epsilon\sum_{k}m(E_{k})\leq\epsilon m(E)$.  If $ m(E)<\infty$, this implies that $ m(\Gamma(f,E))\equal{}0$.  Otherwise, write $ E$ as the countable disjoint union of measurable sets of finite measure.  Then $ \Gamma(f,E)$ is the countable union of sets of zero measure.

Could you explain, what is meant by $ \Gamma_f : \equal{} \{(x, f(x)) \vert \ x \in A \} \subset \mathbb{R}^n$?

That bracket notation means "the set of all ... such that ...". It's simply the precise definition of the set; in this case, the graph of the function.

This is trivial. By Fubini's theorem, $ m\left(\Gamma\left(f,E\right)\right)\equal{}\int_{E}m\left(\left\{ f\left(x\right)\right\} \right)dx\equal{}\int_{E}0\ dx\equal{}0$.

So does it mean the set of all $ \Gamma_f(x, f(x)) \equal{} \int_{f(x)}^\infty t^{x\minus{}1} e^{\minus{}t} dt$ ?

No. This has absolutely nothing to do with the gamma function. This question is about the measure of a set that happens to be named $ \Gamma$.
If the graph of a function is measurable, that measure must be zero, because we can stack uncountably many translates. The issue at hand is to prove that the graph of an arbitrary continuous function is a measurable set. Note that this makes CommandoGuard's argument insufficient; Fubini's theorem includes measurability as a hypothesis, which we don't have until we prove it.

Ok, thanks.

Can I do it like this?

We know that $ f$ is continuous,  this means $ f^{ \minus{} 1}(O)$ ist open for open sets $ O$ in the image set. $ f^{ \minus{} 1}(\infty) \equal{} \{\}$ and $ f^{ \minus{} 1}( \minus{} \infty) \equal{} \{\}$.

Open sets are measurable because they are unions of countably many open (disjunct) intervals and open intervals are measurable by definition as well as countably many unions of disjunct measurable sets.

(Can you explain the word "translates" for me pls?)

Thanks

Drawing pictures to find motivation (try with $ n \equal{} 2$) for this sort of problem is very useful.  If $ n \equal{} 2$, the problem is essentially asking you to show that the graph of a continuous function of one variable (a one-dimensional curve in a two-dimensional plane) has $ 0$ 'area'.  This is intuitively clear because, if it were any larger than $ 0$, we could stack infinitely many disjoint copies of the graph very close together and wind up with something that fits in a box but has infinite area (clearly a contradiction; I think this is something like the uncountable 'translates' jmerry was getting at).

One of the difficulties in transitioning from intuition to rigorous argument in this case is that, as jmerry already said, it isn't clear that the area of the graph is a well-defined notion for any arbitrary continuous function, i.e. we don't know if $ \Gamma_f$ is measurable.  So first thing you need to do is show that $ \Gamma_f$ satisfies one of the definitions you have for what a measurable set is.  This depends entirely on what theorems you have already proven in your course that characterize measurable sets.

Here's one strategy:  If you've already proven that open (and therefore closed) sets are Lebesgue measurable, then to show $ \Gamma_f$ is $ \mathbb{R}^n$ Lebesgue measurable it suffices to show that $ \Gamma_f$ is closed.  You should be able to do this directly from the definition (for intuition go back to the $ n \equal{} 2$ case and think about why you can always find an open ball in the complement of the graph that is disjoint from it).  Once you've found that it is measurable it follows that it has measure $ 0$ by any of the arguments in the previous posts.

Note that $ \Gamma\left(f,E\right)\equal{}\left(E\times\mathbb{R}\right)\cap\left\{ \left(x,y\right)\in\mathbb{R}^{n}\mid y\equal{}f\left(x\right)\right\}$. The former set is measurable as a product of measurable sets and the latter set is measurable since it is the preimage of $ \{ 0 \}$ under the measurable function $ \left(x,y\right)\mapsto f\left(x\right)\minus{}y$.


Is entropy channge ${\Delta}{S_{0}}$ with respect to both  the system and the surroundings for an adiabatic process always zero?
What is the entropy change for an irreversible isothermal process when the external pressure is $P$ with respect to both system and surroundings?

the entropy change for the system in an adiabatic process
 is irrespective of whether it is reversible or otherwise, but not
 so for the surroundings. To find entropy change for
 surroundings, it is imperative that you calculate $\int \frac{dq}{t}$. For [b]reversible[/b] processes, ,
$\Delta S_{0,total}$ is $0$, whereas for [b]irreversible[/b]
it is $+$ve.

AS far as irreversible isothermals go it is the same. If
 i am correct, an irreversible isotherm is where the initial and final states have same temperature. so if given the trajectory , then you can find entropy change as above.

If we have a spontaneous endothermic process ,can we always say that the entropy change is positive?
(By the way I didn't understand ur point ,It'd be better if u could xplain in terms of some equations.

[quote="ashwinkrishnan"]If we have a spontaneous endothermic process ,can we always say that the entropy change is positive?
[/quote]

Yes, as can be seen by the Gibbs Free Energy equation $\Delta G = \Delta H-T\Delta S$.  If $\Delta G < 0$ and $\Delta H > 0$, then $\Delta S > 0$.  Of course, the process will only be spontaneous at high enough temperatures.


In a right triangle with integral side lengths, the hypotenuse has length 39 units. How many units is the length of the shorter leg?

[hide] We recognize that this triangle has side lengths which are simply multiples of the $ 5$-$ 12$-$ 13$ triangle. Hence, the shorter leg has length $ 5(3)\equal{}\boxed{15}$.[/hide]


topology
real analysis
real analysis unsolved

$X$ is a Banach space. $p$ is map from $X$ to $\mathbb{R}$, which satisfies
(1) $p(x) \geq 0$ for all $x$ in $X$
(2) $p(ax)=ap(x)$ for all $a>0$ and $x$ in $X$
(3) $p(x+y)\leq p(x)+p(y)$ for all $x,y$ in $X$
(4) If $x_n \to x$ then $\varliminf_{n\to\infty} p(x_n) \geq p(x)$.
Show that there exists $M$, such that $p(x) \leq M\|x\|$ for all $x$ in $X$.

$p(0)=0$.
By continuity of $p$, there exists some $\epsilon$ such that $|p(x)|\le1$ for all $x$ with $\|x\|\le\epsilon$. Let $M=\frac1\epsilon$.

How do you obtain the continuity of $p$?

I was misreading the $\ge$ in (4) as $=$.

Also, I'm not familiar with that compressed notation for $\liminf$; I read it as an ordinary limit.

Are you sure thisis how you found it though, liyi? It works, but $\limsup$ suffices, so I don't see the reason to put $\liminf$ there. It actually lead me astray at first :).

Let $A_n$ be the set of elements of $X$ with $p(x)\le n$ (for $n\ge 1$). It's clear that $X=\bigcup_{n\ge 1}A_n$. Also, each $A_n$ is closed, since $x_k\to x$ and $p(x_k)\le n,\ \forall k$ implies $p(x)\le\limsup_k p(x_k)\le n$.

$X$ is a complete metric space, so it cannot be the union of countably many closed sets with void interior, meaning that there is an $n$  s.t. $A_n$ has non-empty interior. Let $B(x_0,r)$ be a ball contained in $A_n$. We have, for all $x\in B(0,r),\ p(x)=p(x+x_0-x_0)\le p(-x_0)+p(x+x_0)\le p(-x_0)+n\le N$ for some $N$, and now we can take $M=\frac Nr$.


linear algebra
matrix
linear algebra unsolved

$M$ a real matrix. Is it true that there always exists a matrix $C$ such that:
$MC=CM$ and $det(C)<0$

I would say no?
consider $M=\left(\begin{array}{cc} 0&1\\0&0\end{array}\right)$


Prove that if $f: N\to N_{0}$ satisfies

$f(mn)=f(m)+f(n)+3f(m)f(n)$ for all natural numbers $m,n$, then $f$ is not bijective.

Let g(x)=3f(x)+1. Using the given equation (multiply by 3 and add 1 on each side) we get
g(mn)=g(m)g(n),
so g is determined by the values of g(p) where p is a prime. Function f is bijective iff g is bijective mapping of $N$ to
$\{3k+1,k\in N_{0}\}$. For contradiction suppose f is bijective.
Obviously g(x)=1 iff x=1.
We must have primes p,q,r such that g(p)=25, g(q)=4, g(r)=10 (if the x, for which f(x)=t, $t\in\{4,10,25\}$  was composite x=mn, we'll have t=g(m)g(n) but t is not product of two numbers >1 that are 1 mod3).
And voila:
$g(pq)=100=g(r^{2})$, contradiction and QED


#1 Find all integer solutions $ x,y,z$ of the equation $ x^2\plus{}5y^2\plus{}10z^2\equal{}4xy\plus{}6yz\plus{}2z\minus{}1$.

#2 Show that $ \sum_{k\equal{}1}^{n}a_k^2 \geq a_1a_2\plus{}a_2a_3\plus{}\cdots\plus{}a_{n\minus{}1}a_n\plus{}a_na_1$.

I think you can prove #2 by rearrangement but can you prove it without it?

#2 definitely is rearrangement, cut and dried. To prove it without it, search it in Wikipedia, copy the proof of the Rearrangement theorem, and paste it here, done.

Rearrangement basically states $ \sum_{k \equal{} 0}^{n}x_ky_k\geq\sum_{k \equal{} 0}^{n}x_ky_{n \minus{} k}$ if $ x_0\leq\cdots\leq x_n$ and $ y_0\leq\cdots\leq y_n$, where each $ x_i,y_i\in\mathbb{R}$.

Edit: Oh, you want it proved without rearrangement. Well, see http://en.wikipedia.org/wiki/Rearrangement_inequality

[quote="mathwizarddude"]
#2 Show that $ \sum_{k \equal{} 1}^{n}a_k^2 \geq a_1a_2 \plus{} a_2a_3 \plus{} \cdots \plus{} a_{n \minus{} 1}a_n \plus{} a_na_1$.

I think you can prove #2 by rearrangement but can you prove it without it?[/quote]

#2 is completing squares. Multiplying both sides by $ 2$, #2 becomes:

$ \sum_{i \equal{} 1}^{n} (x_{i\plus{}1}\minus{}x_i)^2 \geq 0$, where $ x_{n\plus{}1}\equal{}x_1$, which is obviously true.

I guess stephencheng means "#1 is completing squares":

$ (x\minus{}2y)^{2} \plus{} (y\minus{}3z)^{2} \plus{} (z\minus{}1)^{2} \equal{} 0$

$ z \equal{} 1, y \equal{} 3, x \equal{} 6$

[quote="AndrewTom"]I guess stephencheng means "#1 is completing squares":
[/quote]

Both of them are.

Sorry: I presumed you were answering both parts of the question.


modular arithmetic
algebra unsolved
algebra

[quote="orl"]A magician asked a spectator to think of a three-digit number $ \overline{abc}$ and then to tell him the sum of numbers $ \overline{acb}$, $ \overline{bac}$, $ \overline{bca}$, $ \overline{cab}$, and $ \overline{cba}$. He claims that when he knows this sum he can determine the original number. Is that so?[/quote]

The sum is $ 122a\plus{}212b\plus{}221c$ and the question is to know if $ 122a\plus{}212b\plus{}221c\equal{}122a'\plus{}212b'\plus{}221c'$ with $ a,b,c,a',b',c'\in\{0,1,2,3,4,5,6,7,8,9\}$ implies $ a\equal{}a'$, $ b\equal{}b'$ and $ c\equal{}c'$

So the question is to know if $ 122u\plus{}212v\plus{}221w\equal{}0$ with $ u,v,w\in\{\minus{}9,\minus{}8,..., 8, 9\}$ has a solution different from $ u\equal{}v\equal{}w\equal{}0$

This equation implies $ w\equal{}2t$ even and so $ 61u\plus{}106v\plus{}221t\equal{}0$ with $ u,v\in\{\minus{}9,\minus{}8,..., 8, 9\}$ and $ t\in\{\minus{}4,\minus{}3,\minus{}2,...,2,3,4\}$

Taking this equality mod $ 61$, we get that :
$ v\in\{\minus{}9,\minus{}8,..., 8, 9\}$ $ \implies$ $ 106v\in\{0,3,6,10,13,16,19,22,26,29,32,35,39,42,45,48,51,55,58\}\pmod {61}$
$ t\in\{\minus{}4,\minus{}3,\minus{}2,...,2,3,4\}$ $ \implies$ $ \minus{}221t\in\{0,8,15,23,30,31,38,46,53\}\pmod{61}$

And so, since $ 106v\equal{}\minus{}221t\pmod{61}$, $ v\equal{}t\equal{}0\pmod{61}$ and so $ u\equal{}v\equal{}w\equal{}0$

And so the result : the magician is right.

Denote $ a\plus{}b\plus{}c \equal{} \sigma$. By adding all the numbers the magician gets, he obtains $ 222\sigma \minus{} \overline{abc}$. Assume the same value is obtained as $ 222\sigma' \minus{} \overline{a'b'c'}$ with, let's say, $ \sigma \geq \sigma'$. Then $ \overline{abc} \minus{} \overline{a'b'c'} \equal{} 222(\sigma \minus{} \sigma') \equal{} 222k$.

Now, from $ \sigma \minus{} \sigma' \equal{} k$ one gets $ (c\minus{}c') \equal{} k \minus{} (a\minus{}a') \minus{} (b\minus{}b')$, while $ 221k \equal{} \overline{abc} \minus{} \overline{a'b'c'} \minus{} k \equal{} 100(a\minus{}a') \plus{} 10(b\minus{}b') \plus{} (c\minus{}c') \minus{} k \equal{} 99(a\minus{}a') \plus{} 9(b\minus{}b')$.
This means that $ 9$ divides $ k$, only possible when $ k \equal{} 0$, since  $ 222k \equal{} \overline{abc} \minus{} \overline{a'b'c'} < 1000$.

Therefore $ \overline{abc} \equal{} \overline{a'b'c'}$, and so the number is uniquely determined.


Just pretend there are $ 10$ parts total in each, so there are a total of $ 7$ acid parts and $ 13$ water parts.  The answer is $ \boxed{\frac{7}{13}}$.

[quote="GameBot"]A solution that is $ 20\%$ acid and $ 80\%$ water is mixed with a solution that is $ 50\%$ acid and $ 50\%$ water. If twice as much $ 50\%$ acid solution is used as $ 20\%$ solution, then what is the ratio of acid to water in the mixture of the solutions?[/quote]

It says that the answer is $ 2/3$, and plus you cant assume both solutions have the same amount of solution.

If there are $ 10$ oz. of acid and $ 40$ oz. of water in the original solution, then there are $ 20$ oz. of acid and $ 20$ oz. of water in the $ 50\%$ acid solution. If you combine both, you get a total  of $ 30$ oz of acid, and $ 60$ oz. of water....so what I got for the ratio is $ 1/2$

Help anyone?

There are (5+5+2)/30 or 2/5 acid in the mixture of the solutions. There are (8+5+5)/30 or 3/5 water in the solution. Thus the ratio of acid to water is 2/3. <- The answer is correct.

Edit: Remember there is twice as much of the second solution as there is of the first solution in the final solution. (I under that might be confusing)

Can you tell us what you specifically don't understand? Then we can help you better. :)

if we assume that there are $ 10$ parts in each solution, then the first solution has $ 2$ parts acid and $ 8$ parts water
the second solution has $ 5$ parts acid and $ 5$ parts water

when we mix them together, we use twice as many of the second solution as the first solution
in order to show this, we can simply add in the second solution twice

there are $ 2\plus{}5\plus{}5\equal{}12$ parts acid, and $ 8\plus{}5\plus{}5\equal{}18$ parts water

thus, the acid to water ratio is $ \frac{12}{18}$ or $ \frac{2}{3}$

does that help a little more?


logarithms algebra polynomial Vieta AMC AIME

For some real numbers $ a$ and $ b$, the equation
\[ 8x^3 \plus{} 4ax^2 \plus{} 2bx \plus{} a \equal{} 0
\]
has three distinct positive roots.  If the sum of the base-$ 2$ logarithms of the roots is $ 5$, what is the value of $ a$?

(A) -256
(B) -64
(C) -8
(D) 64
(E) 256

Thanks in advance!

Let the roots be $ r,s,t > 0$ such that $ \log_2r \plus{} \log_2s \plus{} \log_2t \equal{} 5$.

We then have $ \log_2r \plus{} \log_2s \plus{} \log_2t \equal{} \log_2{rst} \equal{} 5\implies rst \equal{} 2^5 \equal{} 32$, and by Vieta,

$ 32 \equal{} rst \equal{} \minus{} \frac a8$, so

$ a \equal{} \minus{} 8\cdot32 \equal{} \minus{} 256\implies\boxed A$.

Edit: *sigh*... :P

[quote="math154"]Let the roots be $ r,s,t > 0$ such that $ \log_2r \plus{} \log_2s \plus{} \log_2t \equal{} 5$.

We then have $ \log_2r \plus{} \log_2\color{red}s\normalcolor \plus{} \log_2t \equal{} \log_2{rst} \equal{} 5\implies rst \equal{} 2^5 \equal{} 32$, and by Vieta,

$ 32 \equal{} rst \equal{} \minus{} \frac a8$, so

$ a \equal{} \minus{} 8\cdot32 \equal{} \minus{} 256\implies\boxed A$.[/quote]

Minor correction :D

wasn't this a problem from the AMC or AIME?


calculus
integration
trigonometry
calculus computations

(i) Show that $ \int_0^1 \frac {\sin^{ \minus{} 1} x}{\sqrt {1 \minus{} x^2}} \, dx \equal{} \frac {\pi^2}{8}$.

(ii) By writing $ \sin^{ \minus{} 1} x \equal{} \int_0^x \frac {dy}{\sqrt {1 \minus{} y^2}}$, show that $ \sin^{ \minus{} 1} x \equal{} \sum_0^{\infty} \frac {(2n)!}{2^{2n} (n!)^2} \frac {x^{2n \plus{} 1}}{2n \plus{} 1}$.

(iii) Let $ I_n \equal{} \int_0^1 \frac {x^{2n \plus{} 1}}{\sqrt {1 \minus{} x^2}} \, dx$. Show that $ I_n \equal{} \frac {2n}{2n \plus{} 1} I_{n \minus{} 1}$ and hence that $ I_n \equal{} \frac {2^{2n} (n!)^2}{(2n \plus{} 1)!}$.

Deduce that $ \sum_0^{\infty} \frac {1}{(2n \plus{} 1)^2} \equal{} \frac {\pi^2}{8}$.

Deduce further that $ \sum_1^{\infty} \frac {1}{n^2} \equal{} \frac {\pi^2}{6}$.


If  $ a\plus{}b\plus{}c\equal{}6.12$ and $ abc\equal{}6.12$
Can you find(guess) $ a,b$ and $ c$ ?  :roll:

There are infinitely many solutions. I am doing a generalization, for $ x$, not $ 6.12$.

$ a\equal{}\frac{b\plus{}c}{bc\minus{}1}$

$ b^2c\plus{}bc^2\equal{}xbc\minus{}x$

$ b^2c\plus{}b(c^2\minus{}xc)\plus{}x\equal{}0$

$ b\equal{}\frac{xc\minus{}c^2\pm\sqrt{c^4\plus{}x^2c^2\minus{}2c^3x\minus{}4xc}}{2c}$ Now when $ b$ is defined, we can express $ a$ as a function of $ c$, and we have a family of solutions.

Example:

$ c\equal{}1,x\equal{}6.12$.

$ b\equal{}\frac{6.12\minus{}1\pm\sqrt{1\plus{}37.4544\minus{}12.24\minus{}24.48}}{2}$

Take the plus sign, and $ b\equal{}3.21848\ldots$

$ a\equal{}\frac{b\plus{}c}{bc\minus{}1}\equal{}1.90151\ldots$

$ abc\equal{}6.12,a\plus{}b\plus{}c\equal{}6.12$

ah , nice problem from "Plus" magazine.

It would be better if you post the whole story about this problem Obel1x

See the story of problem:

A costumer has bought three articles in the shop and the seller told him he has to pay the price $ 6.12$ euros.That day was 6th december $ (6.12)$ and the costumer easily realised that the pay he had to give,  was equivalent to date,that's why he asked the seller to explain him the way he does calculate the pay.
The seller answered$ :$
I multiplied prices of  articles you bought,sir $ !$
You don't have to multiply but to add hem - said the angry costumer$ .$
Don't worry sir,even if u add them, you will get the same price$ .$
The costumer was really surprised and hardly find the door to go out from the shop$ .$
Can you find the prices of these $ 3$ articles $ ?$

at enndb0x: that was a good idea $ !$

I suppose it's safe to assume that prices are in dollars and cents. So if we putt everything into cents we find

$ x \plus{} y \plus{} z \equal{} \frac {xyz}{10^4} \equal{} 612$ with $ x,y,z \in \mathbb{N}$

So $ xyz \equal{} 2^6\cdot 3^2 \cdot 5^4 \cdot 17$ and $ x \plus{} y \plus{} z \equal{} 612$

Now we can reduce this equation a bit with mod arguments like $ x \plus{} y \plus{} z\equiv 2 \mod 5$ and since $ 5^4 > 612$ we have exactly two numbers with factors of $ 5$ and one equivalent $ 2\mod 5$

So wlog $ 25x' \plus{} 5y' \plus{} z \equal{} 612$

anyway, after a while i got $ (2.72, 2.50, 0.90)$

[quote="Bugi"]Ah, that's the story :D
I've done all real numbers  :D[/quote]

Of course, you did great job,you generalized it without knowing the story.Absolutely is well-done :lol: But it was my mistake for not showing the story in the beggining,it would be more interesting and you wouldn't work on it by generalizing, which wasn't the point of this problem.Anyway,thanks in advance :D
Moreover ocha's solution $ (2.72, 2.50, 0.90)$ is correct.  :D


I found this question in a pre-olympiad test paper.  I don't understand the question.

The diagram represents the street plan of part of an American town.  The police are to capture a bank robeer who is making his way from A to B and they must catch him before he reaches B where he has an accomplice waiting in a car.  The police can run faster than the robber who is hampered by foot.  How many policemen must be dispatched from A so that every possible from A to B is covered to make sure that the robber will be caught?  Only consider travel northwards and eastwards.

I think it's just asking you...

What is the minimum number of continuous segments drawn from point A to B that will cover all streets? You may only go upwards and to the right.

sorry.  I forgot about this.  I couldn't post this before.
The road plan is like a 4x3rectangle with B at top right hand corner and A at bottom left hand corner.
I'll try and write this:
 ㅡㅡㅡㅡB
 l  l  l  l  l
 ㅡㅡㅡㅡ
 l  l  l  l  l
 ㅡㅡㅡㅡ
 l  l  l  l  l
Aㅡㅡㅡㅡ

This is a classic problem. Won't spoil the fun by giving away the solution. Hope someone provides a meaningful hint!

[quote="max1546"]I know the answer but I'm still not sure.  Could anyone explain for me please?[/quote]
As required by the problem, you are only allowed to move northward or eastward. This means any shortest path from $A$ to $B$ is exactly $7$ units in length. For example, one path could be traversed by first moving eastward four units and then moving northward three units (7 units in length.) Or, you could first move three units eastward, then 1 unit northward, followed by 1 unit eastward and finally 2 units northward (again 7 units in length.) Now, note that all such paths require you to move 4 units eastward and 3 units northward irrespective of how you reach $B$ from $A.$ Now, think of each unit of path as an object; so we have 7 objects. So, the answer to the problem is the number of ways of choosing 4 objects (corresponding to 4 units of path eastward) from a total of 7 objects. This is equal to $\binom{7}{4}$.

Still, there is another way to look at it -- this is a fairly standard technique employed in such problems. Let us denote moving 1 unit eastward by $1$ and moving 1 unit northward by $0.$ Therefore, the two paths mentioned just above would be denoted by $1111000$ and $1110100,$ respectively. As you can see, any path from $A$ to $B$ can be represented by a string of four $1's$ and three $0's$ and vice versa. So, now we can rephrase our original question: how many ways can we rearrange (permute) a set of string consisting of four $1's$ and three $0's$? Answer: $\frac{7!}{3! 4!},$ which is again equal to $\binom{7}{4}$.

Hope this explanation helps.

As a sidenote, you can extend this problem to the three dimensional case. What if you have a 4x3x5 box and $A$ and $B$ are located at the ends of the long diagonal?

I think there is a trick in the question itself, as the ways of getting from one place to the other is 35ways.  However, the answer is 31policemen.

Maybe a trick in the speeds of the policemen???


limit
inequalities
inequalities unsolved

Obviously we only have to check for k=1 and k=4 since the equation is concave in k when you move everything over to the L.H.S.

k=0 gives $ ab \plus{} ac \plus{} bc \ge 3$ which is true by trivial AM-GM

Then we just have to do k=4

$ 1 \plus{} \frac {12}{a \plus{} b \plus{} c}\geq\frac {15}{ab \plus{} bc \plus{} ca}$

[quote="Marius Mainea"]Let a,b,c ,k positive numbers such that  abc=1and $ k\in[0,4]$.  Prove that

  $ 1 \plus{} \frac {3k}{a \plus{} b \plus{} c}\geq\frac {3(k \plus{} 1)}{ab \plus{} bc \plus{} ca}$[/quote]
setting $ p \equal{} a \plus{} b \plus{} c, q \equal{} ab \plus{} bc \plus{} ca$ and $ r \equal{} abc \equal{} 1$

$ \Longleftrightarrow 1 \plus{} \frac {3k}{p}\ge\frac {3(k \plus{} 1)}{q}$

$ \Longleftrightarrow k\le \frac {p(q \minus{} 1)}{3(p \minus{} q)} \equal{} \frac {pq \minus{} p}{3(p \minus{} q)}$

we have $ pq\le\frac {2p^3 \plus{} 9r}{7} \equal{} \frac {2p^3 \plus{} 9}{7}$

$ p^2\ge3q \Longrightarrow 3p \minus{} q \equal{} 3p \minus{} \frac {p^2}{3}$

thus, $ k\le\frac {6p^3 \minus{} 21p \plus{} 27}{63p \minus{} 7p^2}$
then, find $ k_{max}$ (easy :wink: )

[quote="bunhiacovski"][quote="Marius Mainea"]Let a,b,c ,k positive numbers such that  abc=1and $ k\in[0,4]$.  Prove that

  $ 1 + \frac {3k}{a + b + c}\geq\frac {3(k + 1)}{ab + bc + ca}$[/quote]
setting $ p = a + b + c, q = ab + bc + ca$ and $ r = abc = 1$

$ \Longleftrightarrow 1 + \frac {3k}{p}\ge\frac {3(k + 1)}{q}$

$ \Longleftrightarrow k\le \frac {p(q - 1)}{3(p - q)} = \frac {pq - p}{3(p - q)}$

we have $ pq\le\frac {2p^3 + 9r}{7} = \frac {2p^3 + 9}{7}$

$ p^2\ge3q \Longrightarrow 3p - q = 3p - \frac {p^2}{3}$

thus, $ k\le\frac {6p^3 - 21p + 27}{63p - 7p^2}$
then, find $ k_{max}$ (easy :wink: )[/quote]
Can you detail the last,....find$ k_{max}...$
My proof is still used mixing variables With $ c=max{a,b,c}$ then $ c \geq\1 ,1\geq\ ab$The proof is easier

[quote="rofler"]

Then we just have to do k=4

$ 1 \plus{} \frac {12}{a \plus{} b \plus{} c}\geq\frac {15}{ab \plus{} bc \plus{} ca}$[/quote]
Try $ a \equal{} b \equal{} 0.4$ and $ c \equal{} 6.25.$ :wink:
[quote="bunhiacovski"]
thus, $ k\le\frac {6p^3 \minus{} 21p \plus{} 27}{63p \minus{} 7p^2}$
then, find $ k_{max}$ (easy :wink: )[/quote]
But $ \lim_{p\rightarrow9^\plus{}}\frac {6p^3 \minus{} 21p \plus{} 27}{63p \minus{} 7p^2}\equal{}\minus{}\infty.$ :wink:

[quote="Marius Mainea"][quote="Marius Mainea"]Let a,b,c ,k positive numbers such that  abc=1and $ k\in[0,2]$.  Prove that

  $ 1 \plus{} \frac {3k}{a \plus{} b \plus{} c}\geq\frac {3(k \plus{} 1)}{ab \plus{} bc \plus{} ca}$[/quote][/quote]
The maximal value of $ k$ for which the inequality
\[ 1 \plus{} \frac {3k}{a \plus{} b \plus{} c}\geq\frac {3(k \plus{} 1)}{ab \plus{} ac \plus{} bc}
\]
holds for all $ a,$ $ b$ and $ c$ such that $ abc \equal{} 1$ is $ k \equal{} \min_{(0,1)}\frac {2x^4 \plus{} 4x^3 \plus{} x \plus{} 2}{3(x \minus{} x^3)} \equal{} 2.67747357...$ :wink:


factorial
modular arithmetic
binomial coefficients

Find the sum of the last 2 digits of 9^103.

What other ways to do it besides Euler's theorem and finding the pattern of it?

if you use factorials for (a-b)^103 you can write the equation back to front and add it together to get twice the value but cancelling out the intermediate factorials then divde the equation by half to get the number itself then just add the last two digits of it.

Someone explain it?

Yes.
Why does it become 9 in the end?
I used the chinese remainder theorem, and I got 9^103==29 (mod 100)
And can you really get the number in the end?
Since it's such a big number

And did you get the equation right?
Since 9^2=(10-1)^2=10^2-10^1-10^0
but it doesn't equal to 81
Or am I missing something here?

No, Johan left out the binomial coefficients, so the last part of the expansion should be \[ \binom{103}{102}\cdot 10^1\cdot 1^{102}\minus{}\binom{103}{103}\cdot 1^{103} \equiv 29 \pmod{100}\]

You still have to answer the question though: $ 2\plus{}9\equal{}\boxed{11}$


Let n be a positive integer not less than 2. Suppose for all integer $k\in[0,\sqrt{\frac{n}{3}}]$, $k^2+k+n$ is a prime. Prove that for all integer $k\in[0,n-2]$, $k^2+k+n$ is a prime.

Posted too many times already.

IMO 1987 Problem 6.

If I remember correctly, this is Myth's all time - favourite - IMO problem by the way :)

[i]Edit: see http://www.mathlinks.ro/Forum/viewtopic.php?t=60767[/i]


Let $ F_n$ denote the $ n^{th}$ Fibonacci number.Prove that if $ p>5$ is a prime,then
$ F_p$ is congruent to ($ \frac{p}{5}$) modulo $ p$
Note: ($ \frac{p}{5}$) denotes the Legendre symbol.


Perhaps I'm just really technologically ignorant, but as far as I know there's nowhere you can look up a piece of music by, say, entering a small part of the melody into a box and clicking "search"... I know they have songs indexed by the first line's melody in some hymnbooks.

Wouldn't this be really useful? Technically, I don't see anything difficult about it.

And I have this melody stuck in my head right now, and I can't tell if I came up with it myself or if I just heard it somewhere.

This has been tried, though I have no citations. :) Someone believes the first few seconds of a song / first few notes are enough to identify a song - at least, that's how we identify songs in our head. (Ever heard Every Breath You Take and not realized it was it in about 3 drumbeats? Ever heard a song and been convinced it was another song, before the words started?)

I think verizon phones can do something similar, and the iPhone?  you hold it up to playing music, then it links you to the song, and of course, gives you the opportunity to buy it.

I have wished many times you could search by sound video or photo.  I'm not sure how much harder it would be, b/c these formats have much more information than sound right?  So they searching would be more difficult?  I'm not sure, I don't really know anything about searching.


[quote="soltixnj"]In each cell of an n x n board there is a real number.  Assume that the sum of the numbers in each [i]cross[/i] (formed by a row and a column) is at least [i]A.[/i]  Find the smallest possible sum of all numbers in the board.

Note: This problem was taken from German Math Competition, 1971/1972.[/quote]
[hide="Solution"]
We add up the sum of all of the crosses.  Clearly, there are $ n^2$ sums and each sum counts $ 2n \minus{} 1$ numbers.  Thus, by symmetry, we can see that each number is counted $ 2n \minus{} 1$ times.  Let the sum of all of the squares be $ S$.  Hence, $ (2n \minus{} 1)S\ge n^2A\implies S\ge \boxed{\frac {An^2}{2n \minus{} 1}}$ since there are $ n^2$ crosses (one for every square) and each cross has a sum of at least $ A$.  This equality is reached when each of the numbers is $ \frac {A}{2n \minus{} 1}$.  [/hide]

[quote="soltixnj"]In each cell of an n x n board there is a real number.  Assume that the sum of the numbers in each [i]cross[/i] (formed by a row and a column) is at least [i]A.[/i]  Find the smallest possible sum of all numbers in the board.

Note: This problem was taken from German Math Competition, 1971/1972.[/quote]

[hide]
each square is in exactly $ 2n\minus{}1$ crosses
and there are $ n^2$ crosses total
so if we add up the minimum values of each cross it's $ an^2$
but each number is used $ 2n\minus{}1$ times in this sum
so the actual minimum is $ \frac{an^2}{2n\minus{}1}$

[/hide]

dang you quatto.


geometry ratio logarithms LaTeX incenter circumcircle inradius

I complied some getting started level challenges.

Please give your solution with explanations.

1. Find the circumference of a circle with area $81 \pi$.

2. For non-zero real numbers, let $f(x) = x^{-1}$. Compute:

$f(f(f(....(f(2))....)))$,

where the above expression contains 20 "$f$"s.

3. Let $f(x) = x^2 - 1$. Then $f(f(x)) = $ what?

4. Let ABC be a triangle, and let D be the midpoint of AC. What is the ratio of area of triangle ABD to the area of the triangle DBC?

5. In a triangle with sides 3,4, and 5, what is the distance between the centers of the inscribed circle and circumscribed circle?

6. Find $x$, if $3^{27^{x}} = 27^{3^{x}}$.

7. Let ABC be an equilateral triangle with side length 1. Construct circles with unit radii centered at A,B, and C. What is the area of the intersection of these three circles?

8. How many different 4-digit integers are there for which one digit is 5, one digit is 6, and two digits are 7's?

9. Let $p(x) = x^2+1$. Then $p(x^2+1) =$

10. Suppose $\log_4 (\log_3 (\log_2 x)) = 5$. Find $\log_2 (\log_3 (\log_2 x))$.

11. Find the smallest positive integer $n$ such that $n!$ is a multiple of $(15!)^2$.

Enjoy! :)

1.[hide]If the area is $81\pi$ then the radius is $9$ and the circumference is $18\pi$.[/hide]
2.[hide]Note that it follows a pattern like this:
$f(x)=\frac{1}{2}$
$f(f(x))=2$
$f(f(f(x)))=\frac{1}{2}$
So the answer is $2$.[/hide]
3.[hide]$(x^{2}-1)^{2}-1=x^{4}-2x^{2}$[/hide]
4. Doesn't this depend on the triangle?
5. Eh, I'm too lazy to do geometry right now.
6.[hide]We can rewrite it like this:
$3^{(27^{x})}=27^{(3^{x})}$
$3^{(27^{x})}=3^{3(3^{x})}$
$3^{(27^{x})}=3^{(3^{x+1})}$
So:
$27^{x}=3^{x+1}$
$3^{3x}=3^{x+1}$
and $x=\frac{1}{2}$[/hide]
I'll do the rest later.

4.
[hide]ABD and DBC have the same base (D is the midpoint of AC) and the same height.
Thus, their areas are equal. 1:1.[/hide]

8.
[hide]Out of 4 digits, choose 2 to put the 7's.. 4C2.
For the other two digits, 5 can come before 6, or the other way around.
The answer is 4C2 x 2 = 12.[/hide]

[quote]
SnowStorm said:

$\displaystyle 4^5 = \log_3(\log_2(x))$
$\displaystyle 3^{4^{5}} = \log_2(x)$
$\displaystyle 2^{3^{4^{5}}} = x$
[/quote]

since you said that: $\displaystyle 4^5 = \log_3(\log_2(x))$
you can cut down the work by saying $4^5 = 2^\left(10 \right)$

$\displaystyle 2^\left(10 \right) = \log_3(\log_2(x))$
$\displaystyle \log_2(\log_3(\log_2(x))) = 10$

that is simpler, isnt it?

EDIT: GOSH I AM TIRED OF LATEXING... i dont want to edit that exponent error anymore.... :(  i need some $\text{\LaTeX}$ practice

[hide]
[quote="Silverfalcon"]1. Find the circumference of a circle with area $81 \pi$[/quote]
Area of a circle is $\pi r^2$ so $r^2=81$ and $r=9$. The circumference of a circle is $2 \pi r = 18 \pi$ which is our answer.

[quote="Silverfalcon"]2. For non-zero real numbers, let $f(x) = x^{-1}$. Compute:

$f(f(f(....(f(2))....)))$,

where the above expression contains 20 "$f$"s. [/quote]
We see that $f(x)=\frac{1}{x}$ and that $f(f(x))=x$ From this, we see that if there are an even number of "$f$"s, the expression will equal $x$. Therefore, since 20 is an even number, the expression with 20 "$f$"s is equal to $x$.

[quote="Silverfalcon"]3. Let $f(x) = x^2 - 1$. Then $f(f(x)) =$  what?[/quote]
We substitute $x^2-1$ for $x$ in $f(f(x))$ to get $f(f(x))=(x^2-1)^2-1$. This is a difference of two squares so it factors to $f(f(x))=(x^2+1-1)(x^2-1-1)=x^2(x^2-2)$.
By the way, can we use eqnarray in forums?

[quote="Silverfalcon"]4. Let ABC be a triangle, and let D be the midpoint of AC. What is the ratio of area of triangle ABD to the area of the triangle DBC?[/quote]
Since D is the midpoint of side AC, we have $AD=DC$. That means the bases of $\triangle ABD$ and $\triangle DBC$ are equivalent. Also, we can clearly see that $h_b$ for both triangles are the same so their areas are the same and therefore in a $1:1$ ratio.

[quote="Silverfalcon"]5. In a triangle with sides 3,4, and 5, what is the distance between the centers of the inscribed circle and circumscribed circle?[/quote]
I had no idea what to do for this one, so I went on MathWorld and what do you know? The distance between the incenter and circumcenter happens to be $\sqrt{R(R-2r)}$ where $R$ is the circumradius and $r$ is the inradius. Now, the problem is a lot simpler. Since the triangle (call it $\triangle ABC$) is a right triangle, the circumcenter is the midpoint of the hypotenuse, and therefore the circumradius is 2.5. For the inradius, we use the fact that $[ABC]=rs$ to find that $r=\frac{6}{6}=1$. Therefore, the distance is $\sqrt{\frac{5}{2}(\frac{5}{2}-2(1))}=\frac{\sqrt{5}}{2}$.

[quote="Silverfalcon"]6. Find $x$, if $3^{27^{x}} = 27^{3^{x}}$.[/quote]
Since $3^3=27$, we can simplify the equation to be $3^{3^{3x}}=3^{3\cdot3^x}=3^{3^{x+1}}$.
Now, we see that $3x=x+1$ and we get $x=\frac{1}{2}$ which is our answer.

[quote="Silverfalcon"]7. Let ABC be an equilateral triangle with side length 1. Construct circles with unit radii centered at A,B, and C. What is the area of the intersection of these three circles?[/quote]
[hide="Faulty answer--someone help me find the error"]
Wow! I can't believe I misread the question so many times!. First, let's label the center of equilateral $\triangle ABC$ as $D$. Now, we know that $AD=BD=CD$ and that each of the three angles in the center are 120^\circ. We can separate the Reuleaux triangle into three sectors so that the area of each sector is $\frac{\pi r^2}{3}$. However, there are three congruent sectors so the area is just the area of the circle, namely $\pi r^2$.

Now we look for $r$. We notice that the height of $\triangle ABC$ is $\frac{\sqrt{3}}{2}$. We also know that the center is the centroid, and therefore cuts off the median (which is also the height in this case) into a 2:1 ratio. We are looking for the longer segment, so we have:

$\frac{\text{longer segment}}{\text{median}}=\frac{r}{\frac{\sqrt{3}}{2}}=\frac{2}{3}$.

Solving for $r$, we find that $r=\frac{\sqrt{3}}{3}$ and plugging it into $\pi r^2$ yields the area to be $\frac{\pi}{3}$.[/hide]

Doing it the right way this time, we find the area of 1/6 the unit circles times 3 minus twice the area of the equilateral triangle, and we end up with $\frac{\pi-\sqrt{3}}{2}$ which better be the correct answer or I'm gonna think I'm slowly going insane lol.

[quote="Silverfalcon"]8. How many different 4-digit integers are there for which one digit is 5, one digit is 6, and two digits are 7's?[/quote]
We see that the amount of different permutations of the four digits is $4!=4\cdot3\cdot2\cdot1=24$. However, since there are 2 sevens, we have counted each number twice (the two sevens can be switched around without changing the number). Therefore, our answer is $\frac{24}{2}=12$.

[quote="Silverfalcon"]9. Let $p(x) = x^2+1$. Then $p(x^2+1) =$[/quote]
Simply substitute $x^2+1$ for $x$ in $p(x)$ and we get $p(x^2+1)=(x^2+1)^2+1=x^2+2x+2$.

[quote="Silverfalcon"]10. Suppose $\log_4 (\log_3 (\log_2 x)) = 5$. Find $\log_2 (\log_3 (\log_2 x))$.[/quote]
This is equivalent to $4^5=\log_3(\log_2 x)$. We are looking for $2^y=\log_3(\log_2 x)$. Because $2^2=4$, the answer we are looking for is just $5\cdot2=10$.

[quote="Silverfalcon"]11. Find the smallest positive integer n such that $n!$ is a multiple of $(15!)^2$.[/quote]
First, we know that the first $15!$ can cancel as long as $n\geq15$. Now, we need to cancel the other $15!$. Prime factorize it into $2^{11}\cdot3^6\cdot5^3\cdot7^2\cdot11\cdot13$. Now, I just went through 16 and up and started subtracting away exponents. By the time I reached 28, $15!$ was completely cancelled so I assume the answer is $28$.[/hide]

[hide=""To All Previous Solvers of this topic---Click here to see if this applies to you""]
[quote="eryaman"]...
4. Doesn't this depend on the triangle?
5. Eh, I'm too lazy to do geometry right now.
...
I'll do the rest later.[/quote]
4. No this does not depend on the triangle. Think what is area of a triangle and what is peculiar about its shape (Hint: same base same height but different shape, what happens to their area?)
5. lazy was the answer? wrong :P

[quote="SnowStorm"][hide]...
3. $(x^2 - 1)^2 - 1$
$x^4 - 2x^2 + 1 - 1$
$x^2(x^2 - 2)$
$0, \sqrt{2}, -\sqrt{2}$
These are roots, but I don't know exactly what you are looking for when you say "= what"
...[/hide][/quote]
Silverfalcon was asking for the expression of the function $f(f(x))$, there's no need to find the roots...

[quote="hiitsme"][hide]
6.
3^3^3^x=3^3^3^x
x=1

7. The cicles have lenth 1? What if the circles are infinitely large?

8.Guess=375

9. p(x^2 + 1)=(x^2+1)^2 +1=x^2+2x+1+1=x^2+2x+2[/hide][/quote]
6. You are confused with the order of operations when there are ladder exponents.
7. Read the problem carefully. It says "circles with unit radii". So it cannot be infinitely large.
8. Sorry but wrong guess.
9. Please work carefully with the exponents when you square things.

[quote="xxreddevilzxx"][quote]
SnowStorm said:

$\displaystyle 4^5 = \log_3(\log_2(x))$
$\displaystyle 3^{4^{5}} = \log_2(x)$
$\displaystyle 2^{3^{4^{5}}} = x$
[/quote]

since you said that: $\displaystyle 4^5 = \log_3(\log_2(x))$
you can cut down the work by saying $4^5 = 2^{10}$

$\displaystyle 2^{10} = \log_3(\log_2(x))$
$\displaystyle \log_2(\log_3(\log_2(x))) = 10$

that is simpler, isnt it?
EDIT: GOSH I AM TIRED OF LATEXING... i dont want to edit that exponent error anymore.... :(  i need some $LaTeX$ practice[/quote]
Thx for pointing out that, I was gonna say it too to SnowStorm. Also I fixed the exponents problem by reposting them here. As Silverfalcon said, $\LaTeX$ not $LaTeX$ :D

[quote="Zyloch"] No need to requote you[/quote]
Please rework on 3, 7, 11[/hide]

Problems 7, 11 are still open for grabs, they are not solved yet :)

Thanks :D

I fixed the stupid error on 3, and redid 7. Now I gotta work on 11 *runs away scared* lol

Btw, I meant $n=30$, not $30!$. Huge typo there lol.

Zyloch, again you still need to work on number 7.
Hint: find the area of the "Rouleaux" triangle
For number 15, the answer is not 30  :o  :what?:  :P
Hint: write out the factors of $(15!)^2$ and give it some thoughts

Wow! I can't believe I misread the question that many times lol. Thanks a lot for pointing that out to me. I think I have the idea right this time--I only hope that I did not make any stupid calculation errors heh ;)

I'll give #11 some thought.

[hide="11."]

reorganizing $(15! )^2$, we have:

$15! * 1 * 2 * ..... * 15$
$ = 15! * 2^(11 ) * 3^6 * 5^3 * 7 * 11 * 13$
$ = 15! * 2^4 * (3^2 *2 ) * (2^2 * 5 ) * (3 * 7  *
(11 * 2 ) * (2^3 * 3 ) * 5^2 *13 * 3^2$
$ = 15! * 16 * 18 * 20 * 21 * 22 * 24 * 25 * 13 * 9$
$ = 16! * ........$

since we cannot go further, the answer is 16. [/hide]

that should be like a number one on AIME. it was easy, but cannot be assumed.

EDIT: UGGGGGGGGGGGGH how do i show an exponent of two digit without having all of the equation screw up LIKE THAT!!!!!!!!!!!!!!!!!!!11 :?

Lol! You would put the exponent in brackets--something like this:

[code]2^{11} \cdot 3^6[/code]

Regarding your answer, I'm pretty sure 16 isn't it because 16 isn't larger than $15!$ and we're looking for multiples.

Close. You left out an extra 7 in the original prime factorization.

oops, so would it be 28 then?

btw, i am REALLY tired right now. CANT THINK. mostly because of ton of hmwk and preparing for AIME. So, Sorry for all the miscalculations and etc. as im writing this, im also booking a plane ticket for my mom's trip next month. you can see how busy and stressed i am.

Don't worry about it lol. I miscalculated at least 4 times on #7, which I definitely should've got right :D I know how you feel with the stress and all heh

maybe next time, or when i take aime , i will be more focused.

[quote="xxreddevilzxx"]

reorganizing $(15! )^2$, we have:

$15! * 1 * 2 * ..... * 15$
$ = 15! * 2^{11} \cdot 3^6 * 5^3 * 7 * 11 * 13$
$ = 15! * 2^4 * (3^2 *2 ) * (2^2 * 5 ) * (3 * 7  *
(11 * 2 ) * (2^3 * 3 ) * 5^2 *13 * 3^2$
$ = 15! * 16 * 18 * 20 * 21 * 22 * 24 * 25 * 13 * 9$
[/quote]

I dont follow what you are doing but there is an odd number of brackets in the 3rd line of your solution. Is the missing right bracket supposed to go after the 7?

yes, you are right about the bracket thats supposed to go after 7. like i said, i was really tired and i was latexing about 10 replies yesterday.

but in the third line, after breaking down the second 15! into primes, i am trying to group them, so that i could start with 16, and go on. there are some missing (primes like 17 and 19) and some that have no other multiples to form a number i want... (13 needs 2 to be 26, and 3^2 needs another three to be 27)

this helped me solve the problem because i was able to find the biggest factorial by lining up the numbers like that.

is there a more sophisticated way of doing number 11 than just cancelling out exponents from 16 onwards cause i dont understand xxreddevilxx's way.  thx

and btw (x^2+1)^2+1= x^4+2x^2+2[/hide]

[quote="sleepwarrior"]
is there a more sophisticated way of doing number 11 than just cancelling out exponents from 16 onwards cause i dont understand xxreddevilxx's way. thx

and btw (x^2+1)^2+1= x^4+2x^2+2
[/quote]

well i am trying to break down the second set of 15! so that i can take the first set of 15! and multiply by 16, 17, 18..... by using the second set of 15!... ultimately theres a point where i cannot get the number to continue, such as prime numbers...

Thanks a lot to all people who have tried out this.

Now, here are the answers:

[hide="Answers"]

1. $18 \pi$
2. $2$
3. $x^4-2x^2$
4. $\frac {1}{1} \Rightarrow 1$
5. $\frac {\sqrt 5}{2}$
6. $\frac {1}{2}$
7. $\frac {\pi - \sqrt 3}{2}$
8. $12$
9. $x^4+2x^2+2$
10. $10$
11.$28$[/hide]

Good job to everyone who tried it! :)

BTW, there is going to be another thread with Getting Started Challenges 2! Check it out! :D


search
function
parameterization
probability and stats

I try to search what is meant by complete statistic and I only know its definition which is

$ s(X)$ is a complete statistic if and only if for every measurable function $ g$,

$ E(g(s(X))) \equal{} 0$ for all $ \theta$ $ \implies$ $ P_\theta(g(s(X)) \equal{} 0) \equal{} 1$ for all $ \theta$

Whereas from wikipedia i found this

"completeness is a property of a statistic where the statistic optimally obtains information about the unknown parameters characterizing the distribution of the underlying data"

How is complete statistic different from sufficient statistic?

I know that a statistic $ T$ is sufficient if once we know $ T$, we know every information of $ \theta$ the unknown parameter. But what about complete statistic?

I just want to know the concept of completeness as I cant imagine it from the definition given.
Thanks

i will try to explain the notion of complete statistics and the difference
between complete and sufficient statistics.

let $ X_1,X_2,...,X_n$ be n observations from a distribution.
sufficient statistic is defined formally as a statistic such that the conditional
distribution of $ X_1,X_2,...,X_n$ given the sufficient statistic is independent of
the population parameters.
what this essentially means is that given the value of the sufficient statistic,
it is possible to simulate observations from the joint distribution of $ X_1,...,X_n$,
without any knowledge of the population parameters.

the notion of a complete statistic is different from that of that of a sufficient statistic.
the definition you have specified and the wikipedia comment are essentially same.
the wikipedia comment emphasizes that if T is the complete statistic, then the expectation (or average)
of a function of T is 0 (or a constant ) (i.e. independent of the population parameters)
essentially implies that it must be trivially a constant function of T
almost everywhere.therefore , a complete statistic essentially obtains the information about the
population parameters of the underlying distribution of the data in an optimal fashion.

both the notion of sufficiency and completeness are essentially concerned with the reduction of data
but their notions are independent.

hope this helps you.
you can check any book on point estimation for a better discussion of these concepts. :)


Benjamin Franklin half-dollars were minted in the United States in 1975. Each coin had a diameter of $ 1\frac3{16}$'' and a thickness of $ \frac1{16}$''. If $ 80\%$ of each coin was silver, how many cubic inches of silver were in one of these half-dollars? Express your answer as a decimal rounded to the nearest hundredth.

The diameter is $ 1\frac{3}{16}$ so the radius is $ \frac{19}{32}$.

The volume of a cylinder with radius $ r$ and height $ h$ is $ \pi r^2h$.  In this case, $ r\equal{}\frac{19}{32}$ and $ h\equal{}\frac{1}{16}$, so the volume is \[ \pi \left(\frac{19}{32}\right) ^2\left(\frac{1}{16}\right) \equal{}\frac{361\pi}{16384}\]

The silver makes up $ 80\% \equal{}.8$ of this, so the amount of silver in one of the half-dollar coins is \[ .8\cdot \frac{361\pi }{16384} \approx \boxed{.06}\]

hmm that answer seems small

Oh my goodness... I put [b]0.06[/b] and it said it was incorrect??? How can that be???

0.06 [i]does[/i], in fact equal .06. I don't see why that's such a surprise..


national olympiad
Kyiv mathematical festival

[b][color=cyan]Here are problems from the 7-th Kyiv mathematical festival (9.05.2007)
Any comments are welcome![/color][/b]

[b][size=200][color=orange]The seventh Kyiv mathematical festival[/color][/size][/b]

[b][size=150][color=indigo]8-th form[/color][/size][/b]

[url=http://www.mathlinks.ro/viewtopic.php?t=204348]1.[/url] Find all positive integers $ k$ for which equation $ n^m \minus{} m^n \equal{} k$ has solution in positive integers.

[url=http://www.mathlinks.ro/viewtopic.php?t=204349]2.[/url] Aladdin has a set of coins with weights $ 1, 2, \ldots, 20$ grams. He can ask Genie about any two coins from the set which one is heavier, but he should pay Genie some other coin from the set before. (So, with every question the set of coins becomes smaller.) Can Aladdin find two coins from the set with total weight at least $ 28$ grams?

[url=http://www.mathlinks.ro/viewtopic.php?t=204350]3.[/url] Prove that among any 7 integers there exist three numbers $ a,b,c$ such that $ a^2 \plus{} b^2 \plus{} c^2 \minus{} ab \minus{} bc \minus{} ac$ is divisible by 7.

[url=http://www.mathlinks.ro/viewtopic.php?t=204351]4.[/url] Let $ K,L,M$ and $ N$ be the midpoints of sides $ AB,$ $ BC,$ $ CD$ and $ AD$ of the convex quadrangle $ ABCD.$ Is it possible that points $ A,B,L,M,D$ lie on the same circle and points $ K,B,C,D,N$ lie on the same circle?

[url=http://www.mathlinks.ro/viewtopic.php?t=204353]5.[/url] Some $ m$ squares on the chessboard are marked. If among four squares at the intersection of some two rows and two columns three squares are marked then it is allowed to mark the fourth square. Find the smallest $ m$ for which it is possible to mark all squares after several such operations.

[b][size=150][color=indigo]9-th form[/color][/size][/b]

[url=http://www.mathlinks.ro/viewtopic.php?t=204355]1.[/url] Prove that there exist infinitely many positive integers $ k$ for which equation $ n^m \minus{} m^n \equal{} k$ has unique solution in positive numbers.

[url=http://www.mathlinks.ro/viewtopic.php?t=204349]2.[/url] Aladdin has a set of coins with weights $ 1, 2, \ldots, 20$ grams. He can ask Genie about any two coins from the set which one is heavier, but he should pay Genie some other coin from the set before. (So, with every question the set of coins becomes smaller.) Can Aladdin find two coins from the set with total weight at least $ 28$ grams?

[url=http://www.mathlinks.ro/viewtopic.php?t=204350]3.[/url] Prove that among any 7 integers there exist three numbers $ a,b,c$ such that $ a^2 \plus{} b^2 \plus{} c^2 \minus{} ab \minus{} bc \minus{} ac$ is divisible by 7.

[url=http://www.mathlinks.ro/viewtopic.php?t=204352]4.[/url] Points $ K,L,M$ and $ N$ lie on sides $ AB,$ $ BC,$ $ CD$ and $ AD$ of the convex quadrangle $ ABCD.$ Let $ ABLMD$ and $ KBCDN$ be inscribed pentagons and $ KN \equal{} LM.$ Prove that $ \angle BAD \equal{} \angle BCD.$

[url=http://www.mathlinks.ro/viewtopic.php?t=204353]5.[/url] Some $ m$ squares on the chessboard are marked. If among four squares at the intersection of some two rows and two columns three squares are marked then it is allowed to mark the fourth square. Find the smallest $ m$ for which it is possible to mark all squares after several such operations.

[b][size=150][color=indigo]10-th form[/color][/size][/b]

[url=http://www.mathlinks.ro/viewtopic.php?t=204355]1.[/url] Prove that there exist infinitely many positive integers $ k$ for which equation $ n^m \minus{} m^n \equal{} k$ has unique solution in positive numbers.

[url=http://www.mathlinks.ro/viewtopic.php?t=204349]2.[/url] Aladdin has a set of coins with weights $ 1, 2, \ldots, 20$ grams. He can ask Genie about any two coins from the set which one is heavier, but he should pay Genie some other coin from the set before. (So, with every question the set of coins becomes smaller.) Can Aladdin find two coins from the set with total weight at least $ 28$ grams?

[url=http://www.mathlinks.ro/viewtopic.php?t=204352]3.[/url] Points $ K,L,M$ and $ N$ lie on sides $ AB,$ $ BC,$ $ CD$ and $ AD$ of the convex quadrangle $ ABCD.$ Let $ ABLMD$ and $ KBCDN$ be inscribed pentagons and $ KN \equal{} LM.$ Prove that $ \angle BAD \equal{} \angle BCD.$

[url=http://www.mathlinks.ro/viewtopic.php?t=204356]4.[/url] The sequences of integers $ \{a_n\}$ and $ \{b_n\}$ are defined by $ a_1 \equal{} 2,$ $ b_1 \equal{} 1$ and $ a_{n \plus{} 1} \equal{} 2a_n^2 \minus{} 1,$ $ b_{n \plus{} 1} \equal{} (a_n \minus{} b_n)(a_n \plus{} 3b_n) \minus{} 1,\, n\ge1$. Prove that for every $ n\ge1$ numbers $ a_n$ and $ b_n$ are relatively prime.

[url=http://www.mathlinks.ro/viewtopic.php?t=204354]5.[/url] Some $ m$ squares on the chessboard are marked. If among four squares at the intersection of some two rows and two columns three squares are marked then it is allowed to mark the fourth square. Find the maximal possible number of marked squares which can appear after several such operations.

[hide="[color=cyan"]authors[/color]]
The problems are [supposed to be] by:
V.Brayman (8.2=9.2=10.2, 8.3=9.3, 8.4), V.Brayman and O.Rudenko (8.5=9.5~10.5, 9.4=10.3), G.Kryukova (8.1~9.1=10.1), O.Rudenko (10.4)[/hide]

The 7-th Kyiv mathematical festival was not in 2007 but in 2008. Sorry for mistake


Alternatively, Let the points be $P_i$($i = 1,2,3,4,5$).
Let $L_{ij}=0$ be the equation of the line passing through $P_i$ and $P_j$.
We may first contruct the  equation of family of  conics through $P_1,P_2,P_3,P_4$ by:
$L_{12}L_{34}+\lambda L_{13}L_{24} = 0$
Then determine the value of $\lambda$ by substituting the coordinates of $P_5$ into this equation.

Can we set the equation of the conic to: \[ Ax^2+By^2+Cxy+Dx+Ey=1, \] and solve a system of five equations?

All of these are good methods, but is there any quick method without Mathematica/Maple/etc. ?


Let $ a,b,c > 0$ such that $ a \plus{} b \plus{} c \equal{} abc$. Prove that:
\[ \frac {a^2}{b^2} \plus{} \frac {b^2}{c^2} \plus{} \frac {c^2}{a^2}\geq \sqrt {a^{2} \plus{} b^{2} \plus{} c^{2}}\]

Cheers  :)

My solution is a bit ugly

[hide]
Homogenise the $ RHS$
$ \frac {a^{2}}{b^{2}} \plus{} \frac {b^{2}}{c^{2}} \plus{} \frac {c^{2}}{a^{2}}\geq\sqrt {\frac {(a \plus{} b \plus{} c)(a^{2} \plus{} b^{2} \plus{} c^{2})}{abc}}$

We have
$ \sum_{cyc} \frac {a^4}{b^4} \plus{} 2\sum_{cyc} \frac {a^2}{c^2} \ge \sum \frac {a^2}{bc} \plus{} \sum_{sym} \frac {a}{b}$

Now by AM-GM
$ 7 \frac {a^4}{b^4} \plus{} 4\frac {b^4}{c^4} \plus{} \frac {c^4}{b^4} \ge 12\sqrt [12]{\frac {a^{24}}{b^{12}c^{12}}}$

Cyclicly summing this gives $ \sum \frac {a^4}{b^4} \ge \sum \frac {a^2}{bc}\qquad(*)$

by AM-GM
$ \sum \frac {a^2}{c^2} \ge \frac {1}{3}\left(\sum\frac {a}{c}\right)^2 \ge \sum \frac {a}{c} \qquad(**)$

By rearrangement
$ \sum \frac {a^2}{c^2} \ge \sum \frac {a}{c}\cdot \frac {c}{b} \equal{} \sum \frac {a}{b}\qquad(***)$

From (*),(**),(***), our result follows
[/hide]

[hide="My solution"]Since \[ \left(\sum_{cyc}\frac{a^2}{b^2}\right)^2\ge 3\left(\sum_{cyc}\frac{a^3}{b^2c}\right)\] it'll be enough to prove that:

\[ \sum_{cyc} \frac {a^{3}}{b^{2}c} \equal{} 3\sum_{cyc} \frac {a^{4}}{abc\cdot b} \equal{} 3\sum_{cyc} \frac {a^{4}}{(a \plus{} b \plus{} c)b}\geq \sum_{cyc} a^{2}\Leftrightarrow\] \[ 3\sum_{cyc} \frac {a^{4}}{b}\geq (a^{2} \plus{} b^{2} \plus{} c^{2})(a \plus{} b \plus{} c)\]

And using Cauchy-Schwartz :

\[ 3\sum_{cyc} \frac {a^{4}}{b}\geq 3\frac {(a^{2} \plus{} b^{2} \plus{} c^{2})^{2}}{a \plus{} b \plus{} c}\geq(a^{2} \plus{} b^{2} \plus{} c^{2})(a \plus{} b \plus{} c)\] \[ \Leftrightarrow 3(a^{2} \plus{} b^{2} \plus{} c^{2})\geq (a \plus{} b \plus{} c)^{2}\][/hide]


It says that problems are coming out this week.
Are there any details as to when this week they are coming out?
Thanks.

Also, does anyone happen to remember the due date of the round. They removed it from the webpage (I guess accidentally).

Round 3: due date January 11, 2010

It's good to know that I'm not the only person refreshing avidly for the new problems.

I have nothing to add to the Round 3 release: they will be released this week.

The solution deadline will be January 11.


geometry
analytic geometry
integration
trigonometry
calculus
calculus computations

Hello, i've got this problem.

I need to compute the area enclosed by the loop of the descartes folium. I know that I can find on many books the measure of this area, but I need to demostrate it, the problem is that I dont know  how to make it because the explicit rectangular eq  is very complicated, but in polar coordinates it is $ f(\theta) = \frac {3asin\theta cos\theta}{sin^3\theta + cos^3\theta}$.

I do not know how to integrate $ \frac {1}{2}\int (f(\theta))^2d\theta$ , any suggestion?

I think $ r(\theta)\equal{}\frac{3a\sin \theta \cos \theta}{\cos ^ 3 \theta \plus{}\sin ^ 3 \theta}$

:blush: sorry it is a typo, thanks for pointing it out  Kunny.

Any ideas to solve this integral? :?: I just need some hints

I have rearshed on Mathworld , and this area measures $ \frac {3}{2}a^2$.

I don't know how to get to this answer, Maybe is not that hard, but i do not seem any progress.

http://mathworld.wolfram.com/FoliumofDescartes.html


Let a,b,c be real numbers such that abc=1. Prove that

   ab^2+bc^2+ca^2 \geq ab+bc+ca

We want to prove b/c + c/a + a/b \geq 1/c + 1/b + 1/a.

We have

b/c + b/c + c/a \geq 3b
c/a + c/a + a/b \geq 3c
a/b + a/b + b/c \geq 3a

by AM - GM. Summing gives the result.


Can anyone explain why four vectors are always linearly independent in $ R^3$? How is it shown mathematically?

You mean dependent?

This is a consequence of the theorem that states that $ n$ linearly independent vectors in $ \mathbb{R}^n$ span it.  This is because the matrix $ \mathbf{T}$ whose columns consist of each of those $ n$ vectors is invertible, hence the transformation $ \mathbf{x} \mapsto \mathbf{Tx}$ is invertible, its range is $ \mathbb{R}^n$, and so its column space is $ \mathbb{R}^n$.

For more details, consult any textbook on linear algebra.

Let $ v_1,v_2,v_3,$ and $ v_4$ be the columns of a $ 3\times 4$ matrix $ A.$ Now ask if we can find a linear combination of these four vectors that equals zero.

That is, we seek to solve $ x_1v_1\plus{}x_2v_2\plus{}x_3v_3\plus{}x_4v_4\equal{}0.$ We can write this as a matrix equation; it is $ Ax\equal{}0.$

Now this has an obvious solution: the trivial solution $ x\equal{}0.$ The question of dependence is the question of whether there is also some nontrivial solution.

We seek to solve $ Ax\equal{}0$ by row reduction. But with only three rows to the matrix, we can find at most 3 pivot elements. With 4 columns of the coefficient matrix, there must be at least one column without a pivot element. Hence there is at least one free variable, and hence more than one solution, including nontrivial solutions. Hence the original set of four vectors must be dependent.


can we have some geometrical inequalities proved using harmonics,projectively and using inversion...of course transformations whic make the proof simpler??

You can prove Ptolemee inequality with a tricky inversion. It's very nice.

yes,thanks i have seen that it reduces to tringle inequality  some other


limit
geometric series
calculus
calculus computations

Hi,
  How can I find out the limit for this problem algebraically?

   lim (x^1/5-1)/(x^1/2-1)
   x->1

   Thanks.

Put $u=x^{\frac 15}$, $v=x^{\frac 12}$

Then $\frac{u}{v}=\frac{(u-1)(v+1)}{u^5-1}$

By geometric series, we know that

$\frac{u-1}{u^5-1}=\frac{1}{1+u+u^2+u^3+u^4}$

So our limit is

$\lim_{x\to 0} \frac{v+1}{1+u+u^2+u^3+u^4}=\frac{2}{5}$

You can see this easily using l'hopital's rule or binomial expansion, but I assume that you don't know those techniques.

Of course Blahblah...solution is the most elementary. You can use L' hospital's rule , if you have been tought it , as well.

Babis


analytic geometry
pigeonhole principle
geometry unsolved
geometry

Do there exist 12 rectangular parallelepipeds  $P_1,\,P_2,\ldots,P_{12}$ with edges parallel to coordinate axes $OX,\,OY,\,OZ$ such that $P_i$ and $P_j$ have a common point iff $i\ne j\pm 1$ modulo 12?

(Based on official solution.)  Suppose that 12 such parallelepipeds exist.  Since the edges of each parallelepiped are parallel to an axis, parallelepipeds $P_i$ and $P_j$ intersect if and only if their projections onto the $x$-axis, their projections onto the $y$-axis, and their projections onto the $z$-axis intersect.

Taking the contrapositive, this says that if parallelepipeds $P_i$ and $P_j$ do not intersect, then there is at least one axis such that the projections of $P_i$ and $P_j$ onto that axis do not intersect.

Consider the four pairs of parallelepipeds $(P_1, P_2)$, $(P_4, P_5)$, $(P_7, P_8)$, and $(P_{10}, P_{11})$.  We know that the parallelepipeds within each pair do not intersect, so from the above, for each pair, there is an axis such that the projections of the parallelepipeds in the pair onto that axis do not intersect.  There are four pairs and three axes, so by the Pigeonhole Principle, some axis appears at least twice.  WOLOG, let this axis be the $x$-axis, and assume that the corresponding pairs are $(P_1, P_2)$ and $(P_4, P_5)$.

Let the projection of parallelepiped $P_i$ onto the $x$-axis be the interval $[a_i, b_i]$.  From the above, intervals $[a_1, b_1]$ and $[a_2, b_2]$ do not intersect, and intervals $[a_4, b_4]$ and $[a_5, b_5]$ do not intersect.  WOLOG, assume that $a_1 < b_1 < a_2 < b_2$, and $a_4 < b_4 < a_5 < b_5$.  Since parallelepipeds $P_1$ and $P_5$ intersect, intervals $[a_1, b_1]$ and $[a_5, b_5]$ intersect, so $a_5 \le b_1$.  Then $b_4 < a_2$, which implies that intervals $[a_2, b_2]$ and $[a_4, b_4]$ do not intersect, so parallelepipeds $P_2$ and $P_4$ do not intersect, contradiction.  Therefore, 12 such parallelepipeds do not exist.


Δίνεται μία σκακιέρα (2n+1)*(2n+1) γεμισμένη με 2 σύμβολα το + και το - (κάθε κουτί έχει 1 σύμβολο).
Μπορείς να κάνεις "αλλαγή" των + και - ως εξής:

Επιλέγεις 2n+1 κουτάκια που ανά 2 δεν βρίσκονται στην ίδια γραμμή και στήλη και αλλάζεις το πρόσημο στο κάθε
κουτί από αυτά που επέλεξες( αν είναι - το κάνεις + και αν είναι + το κάνεις - ). Να δείξεις ότι μπορείς μετά από συγκεκριμμένες "αλλαγές" να αφήσεις στην σκακιέρα το πολύ 2n + και τα υπόλοιπα να είναι - (ανεξάρτητα από την αρχική κατάσταση της σκακιέρας)!

Αφιερωμένη στο φίλο μου το Στέργιο που ελπίζω να κατεβεί Αθήνα!

Exo tin entiposi oti einai pio efkolo na grapseis algorithmo epiliseis gia to problima para austiri mathimatiki apodeiksi.  :P

I geniki idea mou einai na katataseis tis seires se fthinousa seira os pros to plithos ton + pou periexoun. Ean exoume parapano apo 2n , tote mporoume na epileksoume ta tetragona oste o arithmos "megisto plithos + se mia seira" na min auksithei. Afto einai mia morfi analoiotis.
(Arketa boliki analoioti einai to "megisto plithos + se mia seira sin megisto plithos + se mia stili")
Apo ekei kai pera exoume na antimetopisoume merikes tetrimenes periptoseis, opos na exoume akrivos 2n+1 (+) se mia stili mono k.t.l

Tha ta poume stin Athina meta tis 10 tou mina.

Stergios

Γράψε κώδικα για τη λύση σου!

Πρέπει να βρεις μία αναλοίωτη. Αρχικά μετράω τα + και αν είναι περιττά σε αριθμό αλλάζω την κύρια διαγώνιο ώστε να γίνουν άρτια
σε αριθμό(ανα δύο δεν είναι στην ίδια γραμμή και στήλη)! Αν είναι άρτια τα αφήνω!
Αρχική σκέψη είναι ότι μπορείς να αλλάξεις τα σύμβολα των κορυφών ενός οποιουδήποτε ορθογωνίου
χωρίς να αλλάξεις ουσιαστικά κανένα άλλο σύμβολο πάνω στην σκακιέρα ( αλλάζοντας αρχικά δύο από τις απέναντι κορυφές και κάποια άλλα 2n-1 και στη δεύρερη αλλαγή αλλάζοντας τις άλλες δύο  που μένουν και τα ίδια 2n-1 που επέλεξες στην πρώτη αλλαγή)
Επίσης να παρατηρήσεις ότι με αυτο τον τρόπο δεν αλλάζει το πλήθος των + mod 2 .  Με αυτή τη διαδικασία μπορείς να πετύχεις
κάθε φορά  τα τετράγωνα  i_1,i_2,i_3,.....,i_i,i-1_i , i-2_i ,......,1_i  να έχουν το πολύ 1 + και έτσι προχωράς από i=2n+1 μέχρι i=2.
Άρα στο τέλος θα μείνουν το πολύ 2n+1 +. Όμως δεν γίνεται να είναι περιττού αριθμού τα +. άρα θα έχουν μείνει το πολύ 2n +.

Παρατήρηση: Την αρχική ιδέα για το ότι μπορείς να αλλάξεις τις κορυφές του τετραγώνου
χωρίς να αλλάξεις κανένα άλλο σύμβολο ( δηλαδή το 50% του προβληματος )
τη βρήκε ο φίλος μου Θεοδόσης!

ΡΕjj-the-king ΠΟΝΗΡΙΔΗ ΕΠΕΙΔΗ ΣΕ ΞΕΡΩ ΚΑΙ ΜΕ ΞΕΡΕΙΣ ΜΗΠΩΣ ΒΓΑΙΝΕΙ ΜΕ ΤΙΣ "ΧΩΡΙΣ ΓΝΩΣΕΙΣ" ΑΡΧΗ ΤΟΥ DIRICHLET?  :rotfl:  :rotfl:  :rotfl:  ΤΟ ΡΩΤΑΩ ΠΛΗΡΟΦΟΡΙΑΚΑ ΚΑΘΟΤΙ ΔΕΝ ΤΗΝ ΕΧΩ ΔΕΙ ΤΗΝ ΑΣΚΗΣΗ ΟΚ?
ΧΑΙΡΕΤΑΙ


Refer to http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=43&year=2008&p=1044539

Prove that the inradius of a right triangle is equal to the result when the hypotenuse is subtracted from the semiperimeter.
i.e.
Prove that the inradius $ r$ of a right triangle with hypotenuse $ a$ and legs $ b$ and $ c$ is equal to $ r \equal{} \frac {a \plus{} b \plus{} c}{2} \minus{} a$.

[hide]$ \left( \frac {b \plus{} c}{2} \plus{} \frac {a}{2} \right) \left( \frac {b \plus{} c}{2} \minus{} \frac {a}{2} \right) \equal{} \frac {b^2 \plus{} 2bc \plus{} c^2}{4} \minus{} \frac {a^2}{4} \equal{} \frac {bc}{2}$

Substituting, we have: $ s \left( \frac {b \plus{} c}{2} \minus{} \frac {a}{2} \right) \equal{} A$. But we know $ A \equal{} sr$.

$ \therefore r \equal{} \left( \frac {b \plus{} c}{2} \minus{} \frac {a}{2} \right) \equal{} \frac {a \plus{} b \plus{} c}{2} \minus{} a \equal{} s \minus{} a$[/hide]

Let ABC be a right triangle with angle A=90. Let the incircle touch BC, CA, AB at D,E,F, respectively. Let I be the incenter. Observe that quadrilateral IEAF has three right angles and AE=AF, so IEAF is a square with side r. It is easy to show that for ANY triangle that AE=AF=(a+b+c)/2-a, so we are done since AE=AF=r in the right triangle case.

Thanks all!
I don't get what you mean, Zuton Force. Are you using coordinate geometry? I'm pathetic at geometry.
Yea, I can understand your solution Altheman. I notice that you were the one who actually wrote the solution to that problem.


[quote="mdk"]Find the smallest integer n such that 2^n - 1 is a multiple of 47. How about 2^n+1?[/quote]

Isn't the first problem trivialized by
[hide="this"]Fermat's Little Theorem?[/hide]
[hide="Solution to 1"]By Fermat's Little Theorem, the smallest $ n$ such that
$ 2^{n}\equiv 1\pmod{47}$,
is $ \phi(47)$
and since $ 47$ is prime, $ \phi(47) = \boxed{46}$.[/hide]

[quote="mdk"]yes, yes it is, but you can also use a pigeon hole argument.[/quote]

mdk, this is interesting, I wonder how a pigeon hole argument can be used here. :)


Hi everyone! MIT admissions results came out today at 2 PM. Who got in?

I got in. MIT is pretty awesome, but I'm probably more interested in Harvard or Princeton, so I'll wait until those decisions come out.

I got rejected!

First rejection, but it hurt so good. Makes my decision much easier. Now hopefully HMC doesn't reject me, and I'm set. :)

Accepted, after being deferred early action.  I also have a lot of places I'm considering.

Accepted! I'm not going to Caltech's Prefrosh anymore since MIT is (and has long been) my first choice. See you guys at CPW!

My friend,a Vietnamese also got admitted from MIT .Congratulation him. Should I spend time practicing SAT  and apply MIT in the fall this year a or continue learning at current school ??

Waitlisted, so agonizing!

Now, to convince as many people as I can to choose Harvard/Yale/Princeton/UChicago (or some other prestigious school I didn't apply to) over MIT...

Congrats to those that got in and good luck to those waitlisted hoping to get in!  I'm not very active on AoPS that much anymore, but if anyone has any questions about MIT, I can provide a current MIT undergrad's perspective on things.  Feel free to PM me as well.


geometry
incenter
inequalities unsolved
inequalities

1. Let ABC be a triangle with incenter [i]I[/i] such that its centroid [i]G[/i] lies inside [i](I)[/i]. Let [i]a, b, c[/i] be the lengths of the sides [i]BC, AC, AB[/i] respectively. Find the greatest and least value of the following expression.
$ P\equal{}\frac{a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ac}$
2. Let x, y, z be positive numbers satisfying: $ x^2\plus{}y^2\plus{}z^2\equal{}\frac{1\minus{}16xyz}{4}$.
Find the least value of the expression:
$ S\equal{}\frac{x\plus{}y\plus{}z\plus{}4xyz}{1\plus{}4xy\plus{}4yz\plus{}4xz}$

hello, to 2) we have $ S\geq \frac{13}{28}$ and the equal sign holds if $ x\equal{}y\equal{}z\equal{}1/4$
Sonnhard.

For 1) ; I think (without geometry) $ 2>P\geq 1$
Equality holds for max value when $ a\equal{}b\equal{}c$
(As $ a^2\plus{}b^2\plus{}c^2\geq ab\plus{}bc\plus{}ca$ from AM-GM
And;
$ a\plus{}b> c \implies ac\plus{}cb>c^2$
Summing up yields the result)

It's not maximum value. I think so, too but it's not true. When does the equality happen?

[quote="Long2011"]1. Let ABC be a triangle with incenter [i]I[/i] such that its centroid [i]G[/i] lies inside [i](I)[/i]. Let [i]a, b, c[/i] be the lengths of the sides [i]BC, AC, AB[/i] respectively. Find the greatest and least value of the following expression.
$ P \equal{} \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ac}$[/quote]

1.$ P \ge 1$

2.$ IG^2 \equal{} \frac {1}{9}(s^2 \plus{} 5r^2 \minus{} 16rR) \le r^2 \equal{} > s^2 \le 4(r^2 \plus{} 4rR)$

$ P \equal{} 2\frac {s^2 \minus{} (r^2 \plus{} 4rR)}{s^2 \plus{} (r^2 \plus{} 4rR)} \le \frac {6}{5}$

Equality holdes for : $ s^2 \equal{} 4(r^2 \plus{} 4rR) \Leftrightarrow 5(a^2 \plus{} b^2 \plus{} c^2) \equal{} 6(ab \plus{} bc \plus{} ca)$.


Two people pull on opposite sides of a rope. The person on the left pulls with a force of 10 lbs, and the person on the right pulls with a force of 10 lbs. What is the total tension in the rope?

Hmm... according to what we learned in class, there is a tension of 10 N in the rope

Yeah...our teacher also said the answer is 10 lbs. I guess the term "total tension" is a bit misleading.

Everything about our physics class is not rigorous. I truly wish the AP process required more rigor. Oh well

We only had tension defined in the case of a rope. The definition we were given was that it is a measure of how tense the rope is, i.e. the sum of the magnitudes of the forces acting on either end of the rope.

This definition seems like it does not generalize and is rather useless, because I'm sure you can define tension in terms of integrals and stuff for all three dimensional objects.

[quote="Carcul"]The total tension is zero, because due to the force applied by each person there is a corresponding tension in the rope with the same intensity, direction, but opposite sense, and those two tensions sum to zero.[/quote]

I think you have newton's law confused with tension.....the total force exerted on the rope would add to 0 because the forces are in opposite directions but there would still be tension in it

Yes, I was referring to the forces of tension, because that is what is implied by the problem as it was stated (or at least is the usual way as it is stated in my country).


superior algebra
superior algebra unsolved

Let $ p$ be prime and $ GL(n,p)$ be the multiplicative group of invertible $ n \times n$ matrices over $ F_p$. Prove that there is an element in $ GL(n,p)$ of order $ p^n\minus{}1$.


Consider an equilateral triangle. Divide each side of it into $4$ equal segments. If you count all the subdivision points and the three vertices, you find that you have $12$ points. Draw a small circle around each of these $12$ points so that no two circles overlap. The numbers $1, 2, 3, 4, ..., 12$ are arranged, one per circle so the sum of the numbers on each side of the triangle are the same. What is the least sum possible?

[hide]So each side has 5 circles

then one side would have a + b + c + d + e
next would have e + f + g + h + i
last would have i + j + k + l + a

make the overlaps as small as possible

a = 1, e = 2, i = 3

First row has 3
Second row has 5
third row has 4

Now numbers 4 thru 12 are left.

This is a sum of 72.
Divided evenly, you'd have 24 per row. But some of the rows are uneven.
So add 25 for first row
23 for second row
24 for last row.

This gives a sum of 28 for each side of triangle.
[/hide]

[hide]The smallest numbers have to be on the outside, so 1, 2, and 3 are the corners. Then I put 12 and 7 on one side, 11 and 8 on another, and 10 and 9 on the last. So far the sides' sums are 24, 23, and 22. So I put 6 on the side with a sum of 22, 5 on the side with a sum of 23, and 4 on the side with a sum of 24. So the sum of each side is[color=green] 28[/color].[/hide]


Find all functions $ f: R\rightarrow R$ such that:
$ f(x \minus{} f(y)) \equal{} f(f(y)) \plus{} xf(y) \plus{} f(x) \minus{} 1$ for all $ x,y\in R$


My dad just asked me this question and the generalizations...I thought a bit, and found it's really cool (the solution is beautiful).

You have a plane with 100 seats, and 100 passengers.  The passengers walk in one at a time and take their assigned seats.  However, the first guy is blind.  He doesn't want anyone to know, though, so he walks in and takes an arbitrary seat.  After that, each person walks in, and if his seat is vacant, he takes it: otherwise, he picks an arbitrary seat.  Question is, what is the probability that the last person gets his seat?

Generalizations: n seats and n passengers?  k blinnd guys?  They come not at the start, but at random times throughout the boarding process?

I first heard this question years ago... I don't remember the solution, but it was very nice, yes.

[hide]1/2[/hide]
I forgot how to get the answer though.

Here was the basic idea of my solution...

[hide] When the blind guy comes in, let's say he takes the seat of guy A.  Then all goes as planned until guy A comes in.  Guy A then takes an arbitrary seat, so we can pretend that he goes blind when he walks in and sees his seat is taken.  Then, WLOG, his assigned seat is now the blind guy's assigned seat (but he's blind, so he takes an arbitrary seat.  If he takes his "assigned seat", ie the blind guy's assigned seat, all goes well).  In other words, having the blind guy come at any point during the flight is the same as having the blind guy come at any other point in the flight, because the guy whose seat he takes comes later, and that guy becomes like the new blind guy.  Also notice that because of this, the number of seats doesn't matter (because the passengers before the blind guy are irrelevant).  This logic extends to k blind guys.  Thus, the probability in question with k blind guys coming at random times is the same as the probability when we have k+1 seats, and k blind guys come, followed by the last guy.  The probability in question is then $\frac{1}{k+1}$ [/hide]

My dad's solution was a little different (and could probably also be generalized) [hide] Let's say the seat of the blind guy is seat 1, and the last guy's seat is seat n.  For 1 blind guy:
When the blind guy walks in, he could sit in
1) Seat 1.  If this happens, the last guy on the plane will see seat n.
2) Seat n.  Then, the last guy on the plane will see seat n.
3) One of the n-2 other seats.  In that case, because the other people pick arbitrary seats, and by symmetry, the last guy could see seat 1 or seat n with equal probability.

Hence 1/2.[/hide]

I dreamed about this problem last night :o

Cool problem!

[hide="My idea"]Represent an event with sequence of ones who find seat at random, eg 12478...(n-2)n. 1, the blind guy, must be in this sequence.
If n is in the sequence, the last one does not get his seat. There are $2^{n-2}$ sequences that contain n. [Choose YES/NO for 2,3,4,...,n-1 to let them in the sequence or not]
If n is not in the sequence, the last one gets his seat. There are $2^{n-2}$ sequences that does not contain n. [Choose YES/NO for 2,3,4,...,n-1]
Note: Evey event is equally likely to happen.[/hide]

Hmm... when learning about probability and recursion at SIMUW this problem came up... and then the direct solution was exactly like K8107's found by Basham...

This is a very well known problem, nice because it is easy :)


What is the tens digit in the decimal form of $ 11^8$?

We see that the 2 digit end of exponents of 11 is 11,21,31,...
Answer: 81
This can be proved, but FTW is speed and not proof contest.

It only asks for he tens digit so $ \boxed{8}$  :wink:


algebra
binomial theorem
combinatorics unsolved
combinatorics

How do you handle this problem?

$\sum_{k=0}^{n}\sum_{i=0}^{k}(-3)^{i}\frac{n!}{i!(n-k)!(k-i)!}$

all i got was that this thing equals :$\sum_{k=0}^{n}\sum_{i=0}^{k}(-3)^{i}\binom{k}{i}$

what do i do next?!

If the expression you wrote as a simplification were correct, you would then apply the binomial theorem on the inner sum.  However, you lost a term in the course of simplifying, and you should go back and figure out where it went.  After you find it, you should look at the expression you have and consider the various tools you have and which ones you can apply.  In fact, something which looks a lot like the binomial theorem could be helpful.

$\frac{n!}{i!(n-k)!(k-i)!}={n \choose i,n-k,k-i}$. Also ,it would be better if you had done ;
$\frac{n!}{i!(n-k)!(k-i)!}=\frac{n!}{(n-k)!k!}\frac{k!}{i!(k-i)!}={n \choose k}{k \choose i}$ ,and solve each term separately  :lol:.
(Please correct me if you see any problems)

Regards

Come on, now, you aren't trying very hard.

$\frac{n!}{i!(n-k)!(k-i)!}= \frac{(n-k)!k!}{i!(n-k)!(k-i)!}$ ?  Since when is $(a+b)! = a!b!$?

[quote="Lena"]can i do

$\sum_{k=0}^{n}\sum_{i=0}^{k}(-3)^{i}\binom{n}{k}\binom{k}{i}= \sum_{k=0}^{n}\binom{n}{k}\sum_{i=0}{k}(-3)^{i}\binom{k}{i}$  ?[/quote]

There shouldnt be any problems with that.

Edit:Sorry for the typo.  :roll: .

[quote="Lena"]what is the problem that's going to arise from that ?[/quote]

No there isn't. I think he meant "There shouldn't be any problems with that." :)

And it should be $\sum_{k=0}^{n}\sum_{i=0}^{k}{\left((-3)^{i}\binom{n}{k}\binom{k}{i}\right)}= \sum_{k=0}^{n}{\left(\binom{n}{k}\sum_{i=0}^{k}{\left((-3)^{i}\binom{k}{i}\right)}\right)}$


If $ a      \plus{}    b  \equal{}23$ , $ ax    \plus{}  by  \equal{}79$ , $ ax^2\plus{}by^2\equal{}217$ and  $ ax^3\plus{}by^3\equal{}691$

Compute the value of: $ ax^4\plus{}by^4$

[hide="solution"]Note that $ a\plus{}b$, $ ax\plus{}by$, $ ax^2\plus{}by^2$ and $ ax^3\plus{}by^3$ are the first four $ f_k$, if $ f_1\equal{}a\plus{}b$, $ f_2\equal{}ax\plus{}by$ and $ f_k\equal{}(x\plus{}y)f_{k\minus{}1}\minus{}(xy)f_{k\minus{}2}$.  We know this because the general solution to that recurrence is $ ax^k\plus{}by^k$.

Let $ x\plus{}y\equal{}s$ and $ xy\equal{}p$.  Now we need to find $ s$ and $ p$, and to do this, we set $ 217\equal{}79s\plus{}23p$ and $ 691\equal{}217s\plus{}79p$.  We solve to get $ s\equal{}1, p\equal{}6$.  Hence $ f_5\equal{}691\plus{}6\cdot217\equal{}\boxed{1993}$.[/hide]


Let$ a.b.c>0$. Prove $ \frac{a}{\sqrt{a^{2}+2bc}}+\frac{b}{\sqrt{b^{2}+2ca}}+\frac{c}{\sqrt{c^{2}+2ca}}\leq \frac{a+b+c}{\sqrt{ab+bc+ca}}$

It follows Cauchy and the fact that (proved by directly expanding)

\[ \frac{a}{a^{2}+2bc}+\frac{b}{b^{2}+2ca}+\frac{c}{c^{2}+2ab}\le \frac{a+b+c}{ab+bc+ca}.\]

Or by Jensen. After expending it's Schur and Muirhead only.


This is a national contest that not much people attend, but I have...

this is the contest problem from year 2000.

If 5 oranges cost as much as 2 grapefruits, and 1 grapefruit costs as much as 3 apples, then 10 oranges cost as much as ? apples

this is an easy problem, so I won't give you the multiple choices

5 oranges = 2 grapefruits = 6 apples
10 oranges = 12 apples.

[hide="Solution"]10 oranges cost as much as 4 grapefruits, from the first part of the problem.
4 grapefruits cost as much as 12 apples, from the second part of the problem.
From the two above sentences, 10 oranges cost the same as 12 apples.[/hide]

Please take care to post contest problems in The Basics or in MATHCOUNTS rather than Classroom Math -> I've moved this to The Basics.

[quote="shinwoo"]Thank You..( to administer )

I actually thougth it was easy enough for 5th graders to solve.. That's why I put it up there..

well, too bad..[/quote]

i think it is easy enough for 5th graders...but if rruczyk says so...

actually, I learned this in the end of 3rd grade, talk about starting problem solving early in china....  okay, nvm...

what does MLC stand for?

2 shinwoo: what grade r u in? just wondering

I'm in 7th grade..

The following problem is from MLC which stands for Math League Competition
It's a national test

I took one in 6th grade.

It's an interesting test

you have to get certain score to make the school team

If you make it, your score is added up to the  team score

I've made the school team, but most of the kids in my school are stupid, so we were only in the 17th place in Pennsylvania... It stunk..

[quote="KBabe"]i think it is easy enough for 5th graders...but if rruczyk says so...[/quote]

Rusczyk moved it based on the source of the problem, not the difficulty.

Yup he did

Mr.Rusczyk moved it to here because it's a practice problem for MLC


I went to Barnes and Noble(I really did!) and purchased a book of mathematica formulas. It cost 10 dollars, and had a 7% tax. A Barnes and Noble member would have saved 2 dollars after tax, and it just so happens that a non-member could buy a book about mathematical proof with the same amount of money it would take for the Barnes and Noble member to buy the book on mathematical formulas. How much would the tax come to for the book on mathematical proofs the non-member?

[hide]

10.00 x 1.07= 10.70

10.70-2.00 = 8.70

8.70/107 (cost of object + tax)

= a endless number x 7 (to get .07 or tax)

is approximately .56915887854 or rounded up is

57 cents[/hide]


modular arithmetic
calculus
integration
Euler

Hi, Im have trouble with a few questions. Can anybody help me? I not sure if these questions should be posted  in the Getting Started section or the Intermediate section.

1.(a) Is there a perfect square that has exactly 2000 positive divisors?
   (b)  " "          " "          "             "    "        "       1997 positive divisors?

2. If x,y are positive integers, show that (x^1996) + (y^2000) cannot be a perfect square.

Thanks in advance.

[quote="m5obrien"]
1.(a) Is there a perfect square that has exactly 2000 positive divisors?
   (b)  " "          " "          "             "    "        "       1997 positive divisors?

2. If x,y are positive integers, show that (x^1996) + (y^2000) cannot be a perfect square.[/quote]

[hide="#1"](a) No perfect square can have an even number of divisors because if you think of a factor rainbow (???) there is one number that isn't paired with another so then the total number of factors is odd.

(b) For a reason that I cannot explain, yes there is ($2^{1996}$)[/hide]

[quote="m5obrien"]
2. If x,y are positive integers, show that (x^1996) + (y^2000) cannot be a perfect square.
[/quote]

[hide]
Consider x^1996 and y^2000 both mod 4.

x^1996 is either 0 or 1 (mod 4)
y^2000 is either 0 or 1 (mod 4)

A number n, can be expressed as the sum of two positive integer squares if and only if, $n \equiv 3 \pmod 4$, but clearly no combination of two, 0 or 1s can give 3 mod 4.
[/hide]

[quote="lkryptonicl"][quote="José"][quote="shadysaysurspammed"]huh?This isn't a marathon,we can give as many related questions as we want[/quote]

It's better [b]not to[/b][/quote] Says who?[/quote]

I doubt it would be better "not to" have two related questions being posted in the same topic... by the same person. Would you really want someone to create a ridiculous amount of topics just because they have multiple questions on one concept, yet they cannot pull it all together?

Well you are thinking of a particular case when the person posts the same type of questions but one might not neccessarily do so.So,it's better not to go around saying this(but if someone does post the same type of questions,do tell him)  :)

[quote="JavaMan"]
x^1996 is either 0 or 1 (mod 4)
y^2000 is either 0 or 1 (mod 4)
[/quote]
I'm really bad at mods. How do we get this?  :?  :oops:

So am I.Those are the perfect square rules,that if the perfect square number is odd,it is $\equiv1(mod4)$ and if even,$0(mod4)$

it is easy to prove.

Then can you post the proof? I would like to know how.  Thanks.

Oh, and on the subject of posting more than one problem in a topic, [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=23215]clicky![/url]

Posts #4 and 5. :)

[quote="JavaMan"]A number n, can be expressed as the sum of two positive integer squares if and only if, $n \equiv 3 \pmod 4$, but clearly no combination of two, 0 or 1s can give 3 mod 4.[/quote]
I'm so confused.  :huh: $1^2+2^2=5\equiv 1\pmod4$. :?
Isn't $3\pmod4$ the only one that CAN'T be the sum of 2 positive squares?

I thought the perfect square rule was...

If $n$ is a positive integer such that $n\mod{4}$ is $2$ or $3$, then $n$ is not a perfect square.

Therefore, it is easy to prove that if $(x,y)$ are odd integers than there sum cannot be a perfect square (one can prove this without the prior knowledge of $\mod{4}$ though). But if one of them is even...

[quote="robinhe"]Then can you post the proof? I would like to know how.  Thanks.

[/quote]

If the number is even,then it follows...

If it is odd,then,

$(2n+1)^2=4(n^2+n)+1$

to surge--I never meant to STATE any rule or something,I just said that it was one thing related to it.

[quote="shadysaysurspammed"]to lotofgreengrapes--Javaman messed up the concept,it is as follows

A positive integer $n$ can be represented as sum of two perfect squares iff all prime factors $p\equiv3(mod4)$ of $n$ occur an even number of times in the factorisation of $n$. $n$ can be written as sum of squares  iff the previous condition holds and it has at least one prime factor  $\equiv1(mod4)$ or $v_2(n)$ is odd.[/quote]

I see, thanks  :)

[quote="lkryptonicl"][quote="José"][quote="shadysaysurspammed"]huh?This isn't a marathon,we can give as many related questions as we want[/quote]

It's better not to[/quote] Says who?[/quote]

Look in other forums, moderators and administrators say it, not me....

[quote="José"][/quote][quote="lkryptonicl"][quote="José"][quote="shadysaysurspammed"]huh?This isn't a marathon,we can give as many related questions as we want[/quote]

It's better not to[/quote] Says who?[/quote][quote="José"]

Look in other forums, moderators and administrators say it, not me....[/quote]

Thats ridiculous. I've never heard of this.

If someone wanted to post 10 closely related problems they should make 10 different topics?
I would urge someone to put them together, not to spread them out.

[quote="ShoeFactory"][/quote][quote="José"][/quote][quote="lkryptonicl"][quote="José"][quote="shadysaysurspammed"]huh?This isn't a marathon,we can give as many related questions as we want[/quote]

It's better not to[/quote] Says who?[/quote][quote="José"]

Look in other forums, moderators and administrators say it, not me....[/quote][quote="ShoeFactory"]

Thats ridiculous. I've never heard of this.

If someone wanted to post 10 closely related problems they should make 10 different topics?
I would urge someone to put them together, not to spread them out.[/quote]

In the link that robinhe provided, Mr. Crawford even says that the problems should be posted in separate topics.  It is a problem when someone posts related problems, but the difficulty is different, which makes it hard for moderators to move the threads to the correct forums.

However, I think that when the problems could make up a part a, b, c, etc. of a question (assuming none are too difficult for the forum and they are all relate), they should be posted together.

Stop arguing, 9 of the posts have been about this (now 10).

There is nothing wrong with it.

[quote="10000th User"]Let $a=x^{499}$ and $b=y^{500}$. Then, it asks equivalently if there is an integral solution to $a^4+b^4=c^2$. This has been proven many times and in fact I think Euler or Lagrange proved by infinite descent that there are no integral solutions.[/quote]

Do you know any links with a proof?  :)

10000th user is right.This is pretty well known...

Someone please check my work

The equaton is of the form

$a^4+b^4=c^2$

Clearly $a,b$ both cannot be odd.If both are even...

$(2x)^4+(2y)^4=(2z)^2$

$\implies 16x^4+16y^4=4z^2$

$\implies 4x^4+4y^4=z^2$

$\implies z$ is even(we are obviously taking $x,y$ to be odd )

$\implies x^4+y^4=k^2$

Which isn't possible if $x,y$ are both odd.If both $x,y$ are even,this process would go on implying that both a,b cannot be even.

Now third case, if $a$ is odd,other is even

$\implies a^4+16x^4=c^2$

Now again,for this to be possible,$x$ has to be even,so this goes on and on implying that there is no minimal integral slution for $b$ when it is even.Hence combining all three cases,we see there is no solution possible.

Please tell me if i am wrong anywhere

For whatever its worth,     I will reproduce a proof as it is not too hard.

Let   $z >1$   be the smallest  positive integer such that one can write its square as sum of two fourth powers, that is :

$x^4+y^4 = z^2$

Obviously  ($x,y,z$ ) have no common factor (otherwise one can divide  each  and get a smaller value of $z$)  and thus  one of them is odd.
Also  $z$ is odd (This is by taking  mod 4,  if  z  is even,   $z^2$  is 0 mod 4, which is not possible if both x and y are odd)
(in fact z has to be 1 mod 4)

So WLOG let  us say   $y$ is even , $x$ is odd. (And of course  both (x and z are 1 mod 4) )

So  $y^4 = z^2-x^4= (z+x^2)(z-x^2)$

We do know $y^4$ is divisible by 16
  $z+x^2$  is divisible by 2 but NOT 4
Hence  $z-x^2$  must be divisible by 8   and we get (all integers)
$(y/2)^4 = ( \frac{z+x^2} 2 ) (\frac{z - x^2} 8)$

(Since  two factors in RHS have no common factors, each must be a fourth power)
and  Let us have
$\frac{z+x^2} 2 = a^4$
and $\frac{z-x^2} 8 = b^4$
and we get $x^2 = a^4 - 4b^4$
or $b^4 = (\frac{a^2+x} 2 )(\frac{a^2-x} 2 )$
Now same logic as above (since no common factors, each must be a  fourth power so we get
$\frac{a^2+x} 2 = u^4$
$\frac{a^2 - x} 2 = v^4$
and we have $a^2 = u^4+v^4$
(Since  $a < z$  (because  $z = a^4+4b^4$ etc ) we found a $a^2$ which can be written as sum of two fourth powers.

But that is a contradiction as we  started with $z$ being smallest such number.
Hence no solution exist.

[quote="Gyan"][quote]
Now third case, if $b$ is odd,other is even

$\implies a^4+16x^4=c^2$

Now again,for this to be possible,[b]x has to be even,[/b]so this goes on and on implying that there is no minimal integral slution for b when it is odd.Hence combining all three cases,we see there is no solution possible. [/quote]

Not clear, how can you claim, "x has to be even" ( (x could be odd  still  $16x^4$ would be even) . (Besides , it sees that,   you are stating,  "$b$  is odd"  yet claiing $b^4 = 16 x^4$  :) )[/quote]

Yes sorry,I am bad at nt :D Bymistake i neglected $16$ thinking 'its a constant so don't take it'.Bad mistake.My solutiuon to the third case is wrong.


function
number theory proposed
number theory

Can please solve this problem by explaining clear , what is the  problem And really i could not understand the problem ;)
 [color=darkblue]Problem[/color]
Let $D(n)$ be the digital sum of the positive integer $n$.
(a) Does there exist an $n$ such that $n+D(n)$
(b) Prove that at least one of any two succesive positive integers can be represented in the form $E_n=n+D(n)$

[quote="abdurashidjon"]
Let $D(n)$ be the digital sum of the positive integer $n$.
[/quote]
I think $D(n)$ is the sum of all the decimal digits, e.g.
$D(999)=27$ and $D(1005)=6$.

(a) Does there exist an $n$ such that $n+D(n)$

This question is incomplete. Maybe the book is asking if the mapping $n\mapsto n+D(n)$ is injective; in other words
\[ m+D(m)\ne n+D(n)\text{ whenever }m\ne n \]
That is true, but it is ticklish to prove. Say $m=\sum a_k 10^k$ and $n=\sum b_k 10^k$ where $a_k$ and $b_k$ are all in $\{0,1,\ldots,9\}$. Reduce to the case $a_k b_k=0$ for all $k$. Then reduce to the case in which one or the other is nonzero for any $k$ except the highest.

(b) Prove that at least one of any two succesive positive integers can be represented in the form $E_n=n+D(n)$

I have not solved this yet :) Interesting problem.

[quote="Larry Hammick"]
(a) Does there exist an $n$ such that $n+D(n)$
This question is incomplete. Maybe the book is asking if the mapping $n\mapsto n+D(n)$ is injective; in other words
\[ m+D(m)\ne n+D(n)\text{ whenever }m\ne n  \]
That is true, but it is ticklish to prove. Say $m=\sum a_k 10^k$ and $n=\sum b_k 10^k$ where $a_k$ and $b_k$ are all in $\{0,1,\ldots,9\}$. Reduce to the case $a_k b_k=0$ for all $k$. Then reduce to the case in which one or the other is nonzero for any $k$ except the highest.
[/quote]

Actually what you are saying about part [b](a)[/b] is wrong because for example $91+D(91)=100+D(100)=101$
I remember one of the problems of Romanian TST was this :
Prove that for each $n$ there exist $n$ different numbers like $a_1,\ldots,a_n$ such that
\[ a_1+D(a_1)=a_2+D(a_2)=\ldots=a_n+D(a_n)  \]

For part [b](b)[/b] just note that
\[ ((n+1)+D(n+1))-(n+D(n))\leq 2 \]
And that the function $n+D(n)$ takes very large values.

[quote="Nima Ahmadi Pour"]
I remember one of the problems of Romanian TST was this :
Prove that for each $n$ there exist $n$ different numbers like $a_1,\ldots,a_n$ such that
\[ a_1+D(a_1)=a_2+D(a_2)=\ldots=a_n+D(a_n)  \][/quote]

A weaker version of this problem appeared in 1999 Polish Olympiad. See:
http://www.kalva.demon.co.uk/polish/pol99.html

[quote="Nima Ahmadi Pour"]
Actually what you are saying about part [b](a)[/b] is wrong because for example 91+D(91)=100+D(100)=101
[/quote]
Yes, sorry. I was thinking about a restricted case :blush:


Find all solutions to this equation: \[ x^{y}\equal{}y^{x}\]

Let $ y\equal{}px$.
$ x\equal{}p^{\frac1{p\minus{}1}}$
$ y\equal{}p^{\frac p{p\minus{}1}}$

I forgot to say x and y have to be rational numbers.

$ p\equal{}\frac yx$ has to be rational. Let $ p\equal{}\frac mn$.
$ x\equal{}(\frac mn)^{\frac{n}{m\minus{}n}}$ and $ y\equal{}(\frac mn)^{\frac{m}{m\minus{}n}}$
Since $ gcd(m,n)\equal{}1$, we have $ an\plus{}bm\equal{}1$.
$ x^ay^b\equal{}(\frac mn)^{\frac1{m\minus{}n}}$ is rational. Or $ m^{\frac1{m\minus{}n}}$ and $ n^{\frac1{m\minus{}n}}$ are rational.
wlog, assume $ k\equal{}m\minus{}n>0$. $ m\equal{}p^k$ and $ n\equal{}q^k$ for some integer $ p,q$. When $ k>1$, $ p^k\minus{}q^k>k$ conflicts with $ m\minus{}n\equal{}k$. So $ m\minus{}n\equal{}k\equal{}1$.
Conclusion, for any natural $ n$,
$ x\equal{}(1\pm \frac1n)^{\pm n}$
$ y\equal{}(1\pm \frac1n)^{1\pm n}$


inequalities
geometry
geometric transformation
reflection
triangle inequality

Let $ A,B,C,D,E$ be five distinct points, such that no three of them lie on the same line. Prove that
\[ AB\plus{}BC\plus{}CA \plus{} DE < AD \plus{} AE \plus{} BD\plus{}BE \plus{} CD\plus{}CE .\]

Consider points $ E$ and $ D$, two of the points $ (A,B,C)$ lie in the same side of the line $ {ED}$wlog suppose they are $ A\&B$ then, one of the segment pairs $ (AD,BE)$ or $ (AE,BD)$ produces an intersection (Both segments intersect). wlog suppose $ AD\cdot BE=O$ because of triangle inequality: $ AO+OB>AB$ and $ EO+OD>ED$ $ \Rightarrow$ $ AD+BE>ED+AB$. On the other hand, $ AE+EC>AC$ and $ BD+DC>BC$ (Triangle inequality) adding up all inequalities: $ BD+CD+AE+EB+AD+CE>AB+AC+BC+ED$ as we wanted.

[img]http://img98.imageshack.us/img98/2469/14328624xh7.png[/img]

From the picture and from triangle inequality, we get:

$ AE\plus{}EB>AB$

$ BD\plus{}DC>BC$

$ AF\plus{}FC>AC$

$ DF\plus{}FE>DE$

where F is the intersection of line segment $ AD$ and $ CE$

Adding up the inequalities we had, we get:

$ AE\plus{}EB\plus{}BD\plus{}DC\plus{}AF\plus{}FC\plus{}DF\plus{}FE>AB\plus{}BC\plus{}CA\plus{}DE$

$ AE\plus{}BE\plus{}BD\plus{}CD\plus{}(AF\plus{}FD)\plus{}(CF\plus{}FE)>AB\plus{}BC\plus{}CA\plus{}DE$

$ \therefore AE\plus{}BE\plus{}BD\plus{}CD\plus{}AD\plus{}CE>AB\plus{}BC\plus{}CA\plus{}DE$

that is $ AB\plus{}BC\plus{}CA\plus{}DE<AD\plus{}AE\plus{}BD\plus{}BE\plus{}CD\plus{}CE$

If none of points A,B,C,D,E are on the same line than ABCDE is pentagono.We can easily prove that the BC+DE<BD+CE.We can take it from the cuadrangulo BCDE (sum of the diagonals is bigger than a sum of two oposite sides - prove easy).Than we can prove that AD+CD>AC and AE+BE>AB , it's because this sides are in one triangulos each.
AB+BC+CA+DE<AD+AE+BD+BE+CD+CE
Lets first put BC+DE<BD+CE :
BD+CE+AB+AC<AD+AE+BD+BE+CD+CE
AB+AC<AD+CD+AE+BE
Lets put AD+CD>AC :
AB+AC<AC+AE+BE
AB<AE+BE
Thats the right inequality (we had already proved that) so the starting inequality was also right.That means
AB+BC+CA+DE<AD+AE+BD+BE+CD+CE
So we are done.

P.S I'm bad in English so if there are some mistakes in my speech dont mind and try to understand.

Since no three points are collinear, any three points make up a triangle.

Draw a horizontal line segment $ DE$, with D to the left of E without loss of generality. Triangle $ ABC$ will be positioned with either 3 vertices above the line or 2 vertices above and 1 below, and the reflections of both of these cases. In any case 2 vertices are above or 2 are below, so without loss of generality, let A and B always be on the same side of the triangle. Note all possible positions of $ \triangle ABC$ have been considered.

Without loss of generality, let A and B be positioned above the triangle, and C anywhere. There are two possibilities: B is inside triangle ADE or A is inside triangle BDE (1 possibility), or B is outside triangle ADE and vise versa (again, note there are no other possibilities). In the first case, where without loss of generality B is inside $ \triangle ADE$,  $ AD \plus{} DC > AB \plus{} BC$ (or $ AE \plus{} EC > AB \plus{} BC$, depending on whether C is to the right or left of line AB). Then $ AE \plus{} CE > AC$ (or $ AD \plus{} CD > AC$), and $ BD \plus{} BE > DE$, so putting this all together yields the desired result.

For the second possibility, if B is not inside triangle ADE and A is not in triangle BDE, then two of the segments on the greater than side of the inequality must intersect each other. Without loss of generality, let A be to the left of B (or more exactly, there exists a line normal to DE to which A is left of and B right of). Note if neither is to the left of the other then we have the previous possibility occuring. Then AE and BD (and it must be these because of the second sentence of the problem) cross each other at a point we'll creatively call F. So $ AE \equal{} AF \plus{} FE$ and $ BD \equal{} BF \plus{} FD$. Note no restrictions have been placed on C (besides existence); it could be anywhere and the reasoning of this explanation is in no way affected. Then the inequality is a conglomeration of the following triangle inequalities: $ AF \plus{} BF > AB$, $ FD \plus{} FE > DE$, $ CD \plus{} AD > AC$, and $ CE \plus{} BE > BC$.