MATHCOUNTS
number theory
relatively prime
prime numbers

Every time Albatross does the worm, he gets pelted by thirteen rotten tomatoes. If he does the super-worm, he gets pelted by twenty-nine rotten tomatoes.

What is the most number of rotten tomato hits Albatross [b]cannot[/b] receive?

[hide="Solution"]I just learned (a few minutes ago) that the greatest number that cannot be achieved is ab-a-b. So in this case, the greatest number that cannot be achieved is 29*13-29-13 (wow Albatross must stink at dancing :D ) which equals 335, that is if that formula is correct.[/hide]

That formula only works for relatively prime numbers a and b.
For example, lets say the numbers were 6 and 9, instead of 13 and 29.
The formula would give 54-9-6=39, but that isn't the largest number he can't get(as a matter of fact, he can get 39).

Yeah, when the numbers aren't relatively prime there is no biggest number that you can't get.

As for this problem with 3 numbers, if there is a formula I don't know it.

[hide]this is a trick i learned in mathcounts but it only works for two numbers
ab-a-b
13*29-13-29
377-13-29
377-42
335

YAY!!!! :D [/hide]

[quote="yif man12"][quote="hockeyadi23"]So how would you get it with non-relatively prime integers? Is there a formula?

I just noticed that there is no solution to the question what is the greatest integer that cannot be attained by adding 6 and 9, as you would need a multiple of three, however, is that true in all cases? Like 2 and 26, anything that isn't a multiple of 2 won't work... I think it is... Final Question: How would you do this with three relatively prime integers?[/quote]

You answered your own question but decided to go and ask it anyways. Forgive me for not seeing the logic in this.[/quote]

Lol...sorry... I edited my post when I realized something... and I kind of wrote as I thought... Vacation does things to you... o well does anyone know the answer to my last question.. Is there f formula for three relatively prime integers?

[quote="jasonho2004"]I think the answer's 377[hide] made you look[/hide][/quote]
Howd you get 377? I got 335 by using the formula...

Thx. I never knew that formula, but I always used mods to do those sort of problems.  For 13 and 29, it would work like:

29=3 mod 13 so,

Above 13, only 1,2,3,4,5,6,7,8,9,10,11,12 mod 13 cannot be written

Above 13+29, only 1,2,4,5,6,7,8,9,10,11,12 mod 13 cannot be written

Above 13+29+29, only 1,2,4,5,7,8,9,10,11,12 mod 13 cannot be written

And so on

  If there is no formula for three numbers, this method would work, even if it took a long time.

I don't get the concept of like how its not possible to impossibly get pelted by rotten tomatoes.  Can someone explain the question this problem is asking in a much clearer sense? Thanks

[quote="lightspeed"]I don't get the concept of like how its not possible to impossibly get pelted by rotten tomatoes.[/quote]

It is just a context to put the problem in.  It could be that you are picking a collection of numbers from the set with only 13 and 29.  If you sum up all the numbers in your collection, what is the largest number you cannot get as a sum.

Sorry if that made things worse.

To dasherm:

If thats the case, then wouldn't the answer be like googolplex? Or infinity?  Thanks for your explanation, but i am still confused...sorry!

[quote="Hexahedron"]the largest number you [b]cannot[/b] get
after awhile, you can get every number using only combinations of 13 and 29[/quote]

Wait wait wait are you saying you can get 101 using combinations of 29 and 13?  And can you get googol with combining 13 and 29?  This makes no sense to me!  Sorry if i am slow... :blush:

[quote="Hexahedron"]any number greater then 13*29-29-13 can be made using a combination of 13's and 29's[/quote]

Lol really?  Can you prove it to me?  Thanks!

[quote="Hexahedron"]uh, i don't know the proof[/quote]
LOL! Halarious!  Thanks anyway!  Can you please post that formula on my thread?  Thanks!

[quote="lightspeed"]
Wait wait wait are you saying you can get 101 using combinations of 13 and 29? And can you get googol with combining 13 and 29?[/quote]

You cannot get 101, we haven't said you couldn't.  Since 335 is the [b]largest[/b] number you cannot attain, 101 could still be unattainable.

As for googol:

googol = 3 mod 13

Like I said before, 29 = 3 mod 13

So, 29+13*some large number = googol

It helps to know, in general that (a+b) mod n = (a mod n) + (b mod n) and that (a*b) mod n = (a mod n) * (b mod n)

I think I can prove the formula:

The largest number that CANNOT be written by 13 and 29:
Lets start with 13*29
That can obviously be made by all 13 or all 29
but if you minus a 29, you can't do it by 13s
and if you further minus a 13, you can't do 29s either.
so 13*29-13-29

and there's your proof!

Hold on...
Oh, great...

How am I' supposed to prove that's the LARGEST???
(Help)

OKay so you use the formula and get a number.  Now if you added 2 to that number....can you still get that number with 13 and 29?  This concept is pretty hard to grasp. :blush:

And... this modulus stuff is confusing me.. I mean I get the formula and all, but does three mod 13 mean 3R13 on the older TI calcuators that you used in elementary school? Sorry this stuff is a bit new to me...

[quote="lightspeed"]OKay so you use the formula and get a number.  Now if you added 2 to that number....can you still get that number with 13 and 29?  This concept is pretty hard to grasp.[/quote]

Remember, the number you get for the formula is one you can't get. So, it has no bearing on whether you can add 2 and get it. Besides:

335+2=337

29*4+13*17=337

[quote="dwx314"]No, 3 (mod 13) means "a remainder of 3 when divided by 13", or 13R3
I know what you mean. WHen I first saw mods I had the same thought. "What???" Try looking at the modulus string, or maybe you could ask your math teacher, or maybe take Number Theory/Mathcounts Prob Series[/quote]

Wait.. isn't 13R3 is different from 3(mod 13)? as far as I see it... 13R3 would be the answer to say 53/4, while 3 mod 13 is the answer to say 17/13... pls help me... I'm a bit confused here.... and I just want how to apply modulus to the states (that are a week away!), not really the number theory aspect of it... thanks for your responses..


trigonometry
geometry
inequalities
geometry solved

Prove that for any triangle we have:
$\sum \sqrt{\sin A}\geq 4\sum{\sqrt{\cos \frac A2}\sin\frac B2 \sin \frac C2}$

[color=blue][b]Problem.[/b] Prove that for every triangle ABC, we have

$\sum\sqrt{\sin A}\geq 4\sum\sqrt{\cos\frac{A}{2}}\sin\frac{B}{2}\sin\frac{C}{2}$,

where the $\sum$ sign stands for cyclic sums.[/color]

[i]Solution.[/i] Let $s= \frac{a+b+c}{2}$ be the semiperimeter and $\Delta$ the area of triangle ABC. Then, by a well-known area formula for triangles, $\Delta=\frac12bc\sin A$, so that $\sin A=\frac{2\Delta}{bc}$. Thus,

$\sqrt{\sin A}=\sqrt{\frac{2\Delta}{bc}}=\frac{\sqrt{2\Delta}}{abc}\cdot a\sqrt{bc}$.

On the other hand, by the Heron formula, $\Delta=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$, and the half-angle theorem yields $\cos\frac{A}{2}=\sqrt{\frac{s\left(s-a\right)}{bc}}$. Hence,

$\frac{\cos\frac{A}{2}}{\Delta}=\frac{\sqrt{\frac{s\left(s-a\right)}{bc}}}{\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}}=\frac{1}{\sqrt{bc\left(s-b\right)\left(s-c\right)}}$,

so that, after taking the square root, we get

$\sqrt{\frac{\cos\frac{A}{2}}{\Delta}}=\frac{1}{\sqrt[4]{bc\left(s-b\right)\left(s-c\right)}}=\frac{\sqrt[4]{bc\left(s-b\right)\left(s-c\right)}}{\sqrt{bc\left(s-b\right)\left(s-c\right)}}=\frac{\sqrt{\sqrt{b}\cdot\sqrt{c}}\cdot\sqrt{\sqrt{\left(s-b\right)\left(s-c\right)}}}{\sqrt{bc\left(s-b\right)\left(s-c\right)}}$.

Now, by the GM-AM inequality, $\sqrt{\sqrt{b}\cdot\sqrt{c}}\leq\frac{\sqrt{b}+\sqrt{c}}{2}$ and $\sqrt{\left(s-b\right)\left(s-c\right)}\leq\frac{\left(s-b\right)+\left(s-c\right)}{2}$, what, in view of (s - b) + (s - c) = 2s - (b + c) = (a + b + c) - (b + c) = a, becomes $\sqrt{\left(s-b\right)\left(s-c\right)}\leq\frac{a}{2}$; hence, we obtain

$\sqrt{\frac{\cos\frac{A}{2}}{\Delta}}=\frac{\sqrt{\sqrt{b}\cdot\sqrt{c}}\cdot\sqrt{\sqrt{\left(s-b\right)\left(s-c\right)}}}{\sqrt{bc\left(s-b\right)\left(s-c\right)}}\leq\frac{\frac{\sqrt{b}+\sqrt{c}}{2}\cdot\sqrt{\frac{a}{2}}}{\sqrt{bc\left(s-b\right)\left(s-c\right)}}$.

Consequently, using the half-angle theorems $\sin\frac{B}{2}=\sqrt{\frac{\left(s-c\right)\left(s-a\right)}{ca}}$ and $\sin\frac{C}{2}=\sqrt{\frac{\left(s-a\right)\left(s-b\right)}{ab}}$, we obtain

$4\sum\sqrt{\cos\frac{A}{2}}\sin\frac{B}{2}\sin\frac{C}{2}=\sqrt{2\Delta}\cdot\sum 2\sqrt2\cdot\sqrt{\frac{\cos\frac{A}{2}}{\Delta}}\sin\frac{B}{2}\sin\frac{C}{2}$
$\leq\sqrt{2\Delta}\cdot\sum 2\sqrt2\cdot\frac{\frac{\sqrt{b}+\sqrt{c}}{2}\cdot\sqrt{\frac{a}{2}}}{\sqrt{bc\left(s-b\right)\left(s-c\right)}}\cdot\sqrt{\frac{\left(s-c\right)\left(s-a\right)}{ca}}\cdot\sqrt{\frac{\left(s-a\right)\left(s-b\right)}{ab}}$
$=\sqrt{2\Delta}\cdot\sum\frac{\left(s-a\right)\sqrt{a}\left(\sqrt{b}+\sqrt{c}\right)}{abc}=\sqrt{2\Delta}\cdot\sum\frac{\left(s-a\right)\left(\sqrt{ab}+\sqrt{ca}\right)}{abc}$
$=\frac{\sqrt{2\Delta}}{abc}\cdot\sum\left(s-a\right)\left(\sqrt{ab}+\sqrt{ca}\right)=\frac{\sqrt{2\Delta}}{abc}\cdot\left(\sum\left(s-a\right)\sqrt{ab}+\sum\left(s-a\right)\sqrt{ca}\right)$
$=\frac{\sqrt{2\Delta}}{abc}\cdot\left(\sum\left(s-b\right)\sqrt{bc}+\sum\left(s-c\right)\sqrt{bc}\right)=\frac{\sqrt{2\Delta}}{abc}\cdot\sum\left(\left(s-b\right)+\left(s-c\right)\right)\sqrt{bc}$
$=\frac{\sqrt{2\Delta}}{abc}\cdot\sum a\sqrt{bc}=\sum\frac{\sqrt{2\Delta}}{abc}\cdot a\sqrt{bc}=\sum\sqrt{\sin A}$,

what solves the problem.

  Darij


Let 2 two digit numbers be ab and cd.(e.g. in 23 a=2, b=3)
Find ab and cd such that $(ab+cd)^2=abcd$

is it allowed that one of the $a$ , $b$ ,$c$ or $d$ equal to zero?

[hide]ab+cd is less than 100.
Let's rewrite this as (10a+b+10c+d)^2=1000a+100b+10c+d.
Looking at this in mod 4, d is not 3 or 2 mod 4 since squares are 0 or 1 mod 4.  Let's say d was 1.  Then, we need to make (b+1)^2=1 mod 4 so b is 0 or 2.  Let's say d was 0.  Then, we need to make (b+0)^2=0 and again, b is 0 or 2.  So b is either 0,2,4,6,8 and d is 0,1,4,5,8,9.  Now let's look at this in mod 10.  If d is 0, so is b.  Not only that, c would 0 and that's not possible since 10c+d is two digits.  So d is not 0.  If d is 1, b would be either 0 or 8.  If d is 4, b is either 8 or 4.  If d is 5, b must b 0 and c must 2.  If d is 8, there is no such b that satisfies the solution.  If d is 9, b is 4 or 8.  Notice b is never 2 or 6 and d is not 0 or 5.  So we have b=0,4,8 and d=1,4,5,9.  Now, let's find some cases.  Remember that when d=5, b=0, and c=2.  Let's use that to say (10a+25)^2=1000a+25.  We square to get 100a^2+500a+625=1000a+25 so 100a^2-500a+600.  We factor out 100 to get 100(a^2-5a+6) or 100(a-2)(a-3) so a=2 or 3.  We have (20+25)^2=2025 and (30+25)^2=3025.[/hide]

A little oversight in that proof:
[hide]$d$ can be 2 mod 4 if $c$ is odd.[/hide]

Are there any other answers to this problem?  Or is that it?
Yeah, you're right.  C does matter.  That might increase the possibilities for answers.

Dengmi,
if you got onle 3025 and 2025. Then what about $9801$?

[quote="seamusoboyle"]ab+cd is not nescerrarly less than 100. Less than 200 i think... (or 198)[/quote]

$(ab+cd)^2=abcd$
Since $abcd$ is 4 digit number hence $(ab+cd)^2$ is also 4 digit. Thus ab+cd must be les than 100.

varun, ab and cd must be 2 digit numbers.  So 9801 doesn't work.

(10a+b+10c+d)^2=1000a+100b+10c+d
(10a+b)=x and (10c+d)=y so x^2+y^2+2xy=100x+y or x^2+y^2+2xy-100x-y=0.  We solve quadratically for x.  a=1, b=2y-100, c=y^2-y.  [100-2y plusorminus sqrt(4(y^2-100y+2500)+4y-4y^2)]/2.  This is [100-2y plusorminus sqrt(-396y+10000)]/2 or [100-2y plusorminus 2sqrt(-99y+2500)]/2=50-y plusorminus sqrt(-99y+2500)=x.  Of course, x is an integer.  How many values of y satisfy this equation?  We know that 1 does and that results in the 9801 and a trivial 0000 answer.  We know that 25 is an answer that results in 25 plusorminus 5.  This results in 30 or 20 yielding the 3025 and 2025 answers.  Since imaginaries aren't digits and 2500/99 is 25+25/99, we know that y greater than 25 will fail.  Now, 50^2-99y will only be a square when the -99-97-95-... the squares subtraction equals some multiple of 99.


logarithms
number theory unsolved
number theory

Let $ x\equal{}\left(1\plus{}\frac1n\right)^n, y\equal{}\left(1\plus{}\frac1n\right)^{n\plus{}1}$. Find $ y^x\minus{}x^y$.

A good assumption is: the result is very simple. So, Suppose both numbers are equal.

$ \log \left(y^x\right)\equal{}\log \left(x^y\right)$
$ \left(1\plus{}\frac{1}{n}\right)^n (n\plus{}1)\equal{}\left(1\plus{}\frac{1}{n}\right)^{n\plus{}1} n$

Now is very easy to see that both numbers are equal.


Find parametric equations for the line through the point $(0, 1, 2)$ that is perpendicular to the line $x=1+t,\ y=1-t,\ z=2t$ and intersects this line.

[quote="mlok"]Solution using a bit of calculus: the line we want passes through the point on the given line that is nearest to $(0,1,2)$. Find this point by computing the distance squared: \[(1+t)^{2}+(-t)^{2}+(2t-2)^{2}=6t^{2}-6t+\dots\] has minimum at $t=1/2$. This gives the nearest point and consequently the line.

There should be a calculus-free way of doing this too.[/quote]

This is true. But I need another way without any Calculus.

[quote="mlok"]Write down an equation of the plane that passes through $(0,1,2)$ and has normal vector $\langle 1,-1,2\rangle$. The line you are looking for lies in this plane. It remains to find the point of intersection of the plane with the given line.[/quote]

True. Here is another solution:
Let $L$ be the given line. Then $(1,1,0)$ is the point on $L$ corresponding to $t=0$. $L$  is in the direction of $a=\langle 1,-1,2 \rangle$ and $b=\langle-1,0,2 \rangle$ is the vector joining $(1,1,0)$ and $(0,1,2)$.

Then $b-proj_{a}b = \langle-1,0,2 \rangle-\frac{\langle 1,-1,2 \rangle \cdot \langle-1,0,2 \rangle}{1^{2}+(-1)^{2}+2^{2}}\ \langle 1,-1,2 \rangle = \langle-1,0,2 \rangle-\ \frac{1}{2}\left \langle 1,-1,2 \right \rangle =$
$\left \langle \frac{-3}{2},\frac{1}{2},1 \right \rangle$ is a direction vector for the required line. Thus $2 \left \langle \frac{-3}{2},\frac{1}{2},1\right \rangle$ is also a direction vector.
The parametric equations are: $x=-3t, y=1+t, z=2+2t$


Let there be given n points in the plane, no three on a line and no four on a circle. Is it true that these points must determine at least n distinct distances, if n is large enough ?

Hmmm....are you sure that it is an unsolved problem and not an open question?
As far as I know, the minimum number f(n) of distinct distances determined by a set of n points of the plan in general position (that is no three collinear and no four on a circle) satisfies :
f(n) =  O(n 1 + c/ \sqrt(log(n))  and it not known whether this function is superlinear in n.

Reference :
P.Erdös, Z.Füredi, J.Pach, Z.Ruzsa, "The grid revisited", Discrete Math. 111 (1993), 189-196.

Pierre.


Sorry, I forgot how it was like last year. Will the grading rubric for the Round 1 problems be released at the same time scores are released?

Yes, I believe so.

This is so that you can protest if nessacary.


vector
function
calculus
calculus computations

It is actually a easy question, but I don't know how to give a nice proof of it

The question is a particle is moving under influence of a central force field Q=cr, with c a scalar function of x, y and z
How do I proof on a nice way that the trajectory of it lies in a fixed plane through the origin

What I did was

Take a = F/m = c/m r
integrate and get v and again for the position vector
The particle has a initial position and an initial velocity which I choose to be in the plane in which it is accelerated
But can someone help a little bit, cause I don't know if I am forgetting something  :blush:

well a simple and direct proof would involve physics .
since the force field is central the there is no net torque about the source point .hence the angular momentum of the particle about that point is conserved.now if the particle were to change its' plane of motion then the line along which angular momentum exists is no more the same hence there is atleast change in direction a direct contradiction.

We have an initial position $ \mathbf{r}(0)$ and an initial velocity $ \mathbf{r}'(0).$

The system evolves under the equation $ \mathbf{r}'' \equal{} c(\mathbf{r})\mathbf{r}.$

(I'm ignoring the mass $ m,$ which is the same thing as choosing a system of units in which $ m \equal{} 1.$)

Now, define $ \mathbf{L}\equal{}\mathbf{r}\times\mathbf{r}'.$

$ \frac{d}{dt}\mathbf{L}\equal{}\mathbf{r}'\times\mathbf{r}'\plus{}\mathbf{r}\times\mathbf{r}'' \equal{}\mathbf{0}\plus{}\mathbf{r}\times[c(\mathbf{r})\mathbf{r}] \equal{}\mathbf{0}.$

Hence, $ L$ is constant. The set $ \{\mathbf{r}:\mathbf{L}\cdot\mathbf{r}\equal{} 0\}$ is a fixed plane through the origin. But since $ \mathbf{r}\times\mathbf{r}' \equal{}\mathbf{L}$ at all times, we have $ \mathbf{r}\cdot\mathbf{L}\equal{}\mathbf{r}\cdot(\mathbf{r}\times\mathbf{r}') \equal{} 0.$ Hence the motion remains within this plane.

Of course, although I've phrased this in calculus terminology, this is an essentially physical insight: the conservation of angular momentum. $ \mathbf{L}$ is angular momentum, and it is a vector quantity which determines the plane of motion.


\begin{tabular}{|l|c|c|c|c|}   \hline
& Hamburgers & Cheeseburgers & Fries & Sodas  \\ \hline
Mr. Jones Needs: & 15 & 12 & 25 & 30   \\ \hline
1 Meal Deal includes: & 2 & 2 & 3 & 4  \\  \hline
\end{tabular}
What is the least number of Meal Deals that Mr. Jones must
purchase to get all of the food and beverage that he needs?

[quote="GameBot"]$ \begin{tabular}{|l|c|c|c|c|}   \hline
& Hamburgers & Cheeseburgers & Fries & Sodas  \\ \hline
Mr. Jones Needs: & 15 & 12 & 25 & 30   \\ \hline
1 Meal Deal includes: & 2 & 2 & 3 & 4  \\  \hline
\end{tabular}$
What is the least number of Meal Deals that Mr. Jones must
purchase to get all of the food and beverage that he needs?[/quote]

To get enough hamburgers, he needs 8 meal deals.
To get enough cheeseburgers, he needs 6 meal deals.
To get enough fries, he needs 9 meal deals.
To get enough sodas, he needs 8 meal deals.
Therefore, to get everything he needs, he would need to by 9 meal deals.


function
integration
calculus
real analysis
real analysis unsolved

Let $ (X,M,m)$ be a finite measure space.If $ f$ is a measurable function,let
$ r(f) \equal{} \int \frac {|f|}{1\plus{}|f|} dm$.
Show that a sequence $ (f_n)$ of measurable functions converges in measure to $ f$ if and only if $ r(f_n\minus{}f) \rightarrow 0$.

Since measure space is finite, $ m(X)<\infty$.
1) If $ (f_n)$ converges in measure to $ f$, then for each $ \varepsilon>0$
\[ m(\{x\in X \colon |f_n(x) \minus{} f(x)|>\varepsilon\}) \to 0, \quad n\to \infty\]
For every $ S\subset X$ we have
\[ r(f_n \minus{} f) \equal{} \int_{S} {|f_n\minus{}f|\over 1\plus{}|f_n \minus{}f|}dm \plus{} \int_{X\setminus S} {|f_n\minus{}f|\over 1\plus{}|f_n \minus{}f|}dm\]
For each $ \varepsilon>0$ we can take $ S\equal{}\{x\in X \colon |f_n(x) \minus{} f(x)|> \varepsilon/(2 m(X))\}$.
Then there exists $ N\in \mathbb N$, such that for $ n>N$ the first integral is estimated as $ < m(S) \le \varepsilon/2$.
The second integral is estimated as $ \le \varepsilon/2$, since in $ X\setminus S$ $ |f_n(x) \minus{} f(x)|\le \varepsilon/(2m(X))$.
So, $ r(f_n\minus{}f)<\varepsilon$.

2) If $ r(f_n\minus{}f)\to 0$ but $ f_n \not\to f$ in measure, then, passing to subsequence if necessary, there exists $ \varepsilon>0$ and $ \delta >0$:
\[ m(\{x\in X \colon |f_n(x) \minus{} f(x)|>\varepsilon\}) > \delta, \quad n\in \mathbb N\]
Then
\[ \begin{gathered}
\int \frac{|f_n \minus{} f|}{1\plus{} |f_n \minus{}f|} dm \ge \int_{\{x\in X \colon |f_n(x) \minus{} f(x)|>\varepsilon\}} \frac{|f_n \minus{} f|}{1\plus{} |f_n \minus{}f|} dm >\\
>\int_{\{x\in X \colon |f_n(x) \minus{} f(x)|>\varepsilon\}} \frac{\varepsilon}{1\plus{} \varepsilon} dm
>\frac{\varepsilon \delta}{1\plus{} \varepsilon} \not\to 0, \quad n\to \infty
\end{gathered}\]
which is a contradiction.


analytic geometry
geometry solved
geometry

In triangle $ABC$, $M\in(BC)$, $\frac{BM}{BC}=\alpha$, $N\in(CA)$, $\frac{CN}{CA}=\beta$, $P\in(AB)$, $\frac{AP}{AB}=\gamma$.
Let $AM\cap BN=\{D\}$, $BN\cap CP=\{E\}$, $CP\cap AM=\{F\}$. Prove that
$S_{DEF}=S_{BMD}+S_{CNE}+S_{APF}$ iff $\alpha+\beta+\gamma=1$.

I think it can be done by repeated application of menlaus? But then it sounds too easy for a Moldova tst :?

Hmm... What did they mean? Baricentric calculations?

Is this a joke?
\[(\alpha + \beta + \gamma)S_{ABC} - S_{ABC}
= S_{ABM} + S_{BCN} + S_{CAP} - S_{ABC}
= S_{ADP} + S_{BDM} + S_{CEN} - S_{DEF}\]
and the result follows...

All students except me and Sasha I think solved it like Valiowk did.
Sasha solved it with baricentric coordinates till the end (as I remember),
but I didn't finish my calculations to the final rezult :(  :blush:
(what a shame !).

[quote="pestich"]Moldova should stop borrowing problems from Polish olympiads.

     Sgt.  Pestich[/quote]

You are right, becuase it this way it will decrease the level of Moldavian TST since it was the easiest problem from all 12.


In trapezoid ABCD, DC//AB, and DC:AB = 1:3. Extend AD, BC to intersect at P, then make a line XiYi (i =1,2,...,n), on which P is its midpoint.  Mi, and Ni are the midpoints of AXi and BYi repestively, prove: all the lines MiNi pass through a same point. ( hint: this same point is the intersection of AC and BD.)

[img]http://i227.photobucket.com/albums/dd72/TPFC/codinh.png[/img]
Call $ O \equal{} AC\cap BD,PO\cap AB \equal{} E$. Condition $ AB \equal{} 3CD$ mean $ O$ is midpoint of $ PE$. (Very easy to check that by Thales theorem or Menelaus theorem)
Very well-known that: $ E$ is midpoint of $ AB$ and $ MENP$ is a parallelogram.
It mean $ MN$ pass through midpoint $ O$ of $ PE$.

It is clear that the midpoint of $ MN$ and the midpoint of $ PE$ are the same since they are both the gravity center of $ ABY_iX_i$. But $ P$, $ E$, are fixed so we are done.

Cute problem!

Just a remark: The lline $XY$ does not need to be in the same plan $ABC$!

Best regards,
sunken rock


Points $ A, B, C$ and $ D$ have these coordinates:
$ A(3,2)$, $ B(3,\minus{}2)$, $ C(\minus{}3,\minus{}2)$ and $ D(\minus{}3,0)$.
What is the area of quadrilateral $ ABCD$?

ABCD is a trapezoid with height 6 and base lengths 2 and 4. Thus, the area is $ \frac{2\plus{}4}{2} \times 6 \equal{} 18$.

[size=59][color=red]Fixed LaTeX. -Isabella[/color][/size]


Chose any n+1 element subset from {1,2,...2n}. Show that this subset must contain two integers that are relatively prime.

This is old and easy.

Do the partition $ \{1,2\}$, $ \{3,4\}$,...,$ \{2n\minus{}1,2n\}$.
Then you must tooke two elements from same subset. This two are relatively prime.

Tougher variant: Given $ n\plus{}1$ elements from $ \{1, 2, \ldots, 2n\}$, show that one of them divides another. Also show that this is tight (namely that $ n\plus{}1$ cannot be reduced to $ n$).

Regroup like:

$ (1,2), (2,4), (3,6),...,(n,2n)$

since there are $ n$ pairs and $ n\plus{}1$ numbers to choose, then there are two in the same pair.

For tightness:

take $ n\plus{}1, n\plus{}2, n\plus{}3,...2n$. There are $ n$ numbers in here, but clearly none of them divides each other.

[quote="Pul de Algodoncito"]Regroup like:

$ (1,2), (2,4), (3,6),...,(n,2n)$

since there are $ n$ pairs and $ n \plus{} 1$ numbers to choose, then there are two in the same pair.
[/quote]

This is false, since the pairs are not disjoint. For instance, with $ n\equal{}3$ I can choose the numbers $ 1,4,5,6$, and I don't have two numbers in one of your pairs. As you can see, your proof seems to show that there are always two numbers such that one is twice the other, and this is simply not the case.

a famous problem due to Erdos.
suppose that $ A\subset\{1,2,\ldots ,2n\}$ with $ \left| A\right| = n + 1$,now wirte every number $ a\in A$ in the form $ a = 2^km$ where $ m$ is an odd number between $ 1$ and $ 2n - 1$. since there are $ n + 1$ numbers in $ A$,but only $ n$ different odd parts,there must be two numbers in $ A$ with the same odd part,hence one is a multiple of the other...


function inequalities trigonometry logarithms algebra domain integration

Prove that there is no twice differentiable function $ f$ defined on $ [0, \infty)$ such that
\[ f^{\prime}^{\prime}(x)f(x) +1\leq 0, \forall x\geq 0.
\]

EDIT:

Define $ f(x) : \equal{} \minus{} e^x$.

Then $ f'' \equal{} f' \equal{} f$.

So we have that $ f'' \minus{} 2f' \plus{} 2f \plus{} e^x \equal{} 0 \leq 0, \forall\, x \geq 0$.

(if you want strict inequality, just take something like $ \minus{}2 e^x$).

As for the question you deleted, put $ f(x) \equal{} (A \cos x \plus{} B \sin x \minus{} \gamma) e^x$ and you get your equation to be true for any $ \gamma \geq 1$.

And for the current question, we can simply try to solve the equation:

$ f''(x)f(x)\plus{}1\equal{}0$

$ f''(x)f(x)\equal{}\minus{}1$

$ f''(x)\equal{}\minus{}\frac{1}{f(x)}$

$ f''(x)f'(x)\equal{}\minus{}\frac{f'(x)}{f(x)}$

$ \frac{1}{2}f'(x)^2\equal{}\minus{}\ln f(x)$

I believe that from this we can see that the function we are looking for is complex, hence there is no solution on the domain you need.

Well, can you find an explicit formula for all its solutions?

I can't but Mathematica obviously can. It gives:

$ f_{(1,2)}(x) \equal{} \exp\left(\text{erf}^{\minus{}1}\left(\frac{\pm\sqrt{2}\,x\plus{}i\,C}{\sqrt{\pi}}\right)^2\right)$

But to get something that is defined you should put $ C \equal{} 0$ but even then it would be defined only on $ (\minus{}1,\,1)$.

The last equation of milin's first post is incorrect- it misses solutions.

We have an autonomous second order equation $ y''\equal{}g(y)$. Physical insights are useful for understanding this, especially for qualitative analysis. We have a potential function $ h(y)\equal{}\int \minus{}g(y)\,dy$, and solutions are either periodic near a local minimum of $ h$ or run to an edge of the domain (in both time directions) where $ h$ decreases.
In this case, $ h(y)\equal{}\ln(y)\plus{}C$ on $ (0,\infty)$ (and $ \ln(\minus{}y)\plus{}C'$ on the other side). This has no local minima, so there are no stable periodic solutions. The potential increases toward $ \infty$ and decreases toward zero, so all solutions go to zero eventually both as $ t$ increases and as $ t$ decreases.
Do they get to zero in finite time? Of course they do. If $ f(t_0)\equal{}a>0$ and $ f'(t_0)\equal{}b<0$, $ f$ will reach zero in less than $ \frac{a}{b}$ time; it's not slowing down. This is a finite estimate if we're anywhere after the maximum.

We can also use this to solve the equation, at least to a point. We have $ |y'|\equal{}\sqrt{2E\minus{}2h(y)}$. Separating variables, $ \int\frac1{\sqrt{2E\minus{}2h(y)}}\,dy\equal{}\pm t\plus{}C$. In this case, we specialize to $ \int_{y}^{\exp(E)} \frac1{\sqrt{2E\minus{}2\ln(s)}}\,ds\equal{}t\minus{}t_0$ after the maximum, and $ \int_{y}^{\exp(E)} \frac1{\sqrt{2E\minus{}2\ln(s)}}\,ds\equal{}\minus{}t\plus{}t_0$ before the maximum. Since $ \int_0^{\exp(E)} \frac1{\sqrt{2E\minus{}2\ln(s)}}\,ds$ always converges, all solutions reach zero in finite time.


For a,b,c positive reals prove that

a^3/(a^2+ab+b^2)  + b^3/(b^2+bc+c^2) + c^3/(c^2+ca+a^2)  \geq (a+b+c)/3

We can observe that our sum = b^3/(a^2+ab+b^2)+etc. (you replace a with b in the first fraction, b with c in the second one and c with a in the third one). Then your ineq is
 \sum (a^3+b^3)/(a^2+ab+b^2) >= 2(a+b+c)/3. It's easy to prove 3(a^3+b^3)>=(a+b)(a^2+ab+b^2) and the analogous ones, and then we just add them up.

Sorry, but your proof is logically flawed because you can't replace the variables.

You can't do it usually, I agree, but in this particular case it works because a^3-b^3=(a-b)(a^2+ab+b^2), so  \sum (a^3-b^3)/(a^2+ab+b^2) =  \sum (a-b)=0. That's why I like this solution. I've seen sth like this a LONG time ago, but I've never forgotten it because it's so nice :D  :D

That's exactly what I said, but unfortunately this kind of thing works for just A FEW inequalities... It almost makes me sad :D  :D  :D


I need help with #17 from 1989 MCNat Sprint.

If $ f(x) \equal{} 3x \plus{} 2$, for what value(s) of $ z$ does $ f^{\minus{}1} (z\minus{}1) \equal{} 4?$

First we need to find $ f^{-1}$. That means we have to solve for $ x$ in terms of $ y$, like so: $ 3x + 2 = y \implies x = \frac{y - 2}{3}$, so $ f^{-1)(x) = \frac{x-2}{3}}$. Now we substitute $ z - 1$ for $ x$ and equate it with 4 to get $ \frac{z - 3}{3} = 4 \implies z = \boxed{15}$.

[hide="Solution"]Let $ y\equal{}f(x)$. If we find $ x$ in terms of $ y$, that is the inverse function. We have:

$ y \equal{} 3x \plus{} 2$
$ 3x \equal{} y \minus{} 2$

$ x \equal{} \frac {y \minus{} 2}{3}$

Now, let $ z \minus{} 1 \equal{} y$. We have:

$ \frac {y \minus{} 2}{3} \equal{} 4$
$ y \minus{} 2 \equal{} 12$
$ y \equal{} 14$

And finally, solving for $ z$:

$ z \minus{} 1 \equal{} 14$

$ \boxed{z \equal{} 15}$
[/hide]

EDIT: Beaten.

Thanks so much :)  My Math Coach gave me a CD with all the previous MCNats ever, and I am doing them over the summer.


Daca avem un triunghi echilateral ABC de afixe a, b, c, este adevarata urmatoarea afirmatie : $ a \equal{} \epsilon b$, unde $ \epsilon \equal{} cos\frac {2\pi}{3} \plus{} i sin\frac {2\pi}{3}$? Si daca da, de ce?
Multumesc  :maybe:

Relatia nu este adevarata pentru orice triunghi echilateral. Relatia este adevarata in cazul in care centrul triunghiului echilateral este punctul de coordonate $ (0,0)$ si varfurile sunt in ordine trigonometrica.

In general, insa, este valabila relatia:

$ \boxed {c\minus{}a\equal{}(b\minus{}a)(\cos\alpha\plus{}i \sin\alpha)}$

unde $ c$ se obtine din $ b$ printr-o rotatie cu centrul in $ a$ egala cu $ \alpha$ in sens trigonometric.

Si la mine de ce este adevarata cu a si b, si nu cu diferentele c-a, b-a etc?
Si de ce este $ \frac {2\pi}{3}$ in loc de $ \frac {\pi}{3}$?

pt ca centrul este in (0,0) si nu mai e nevoie de a .
pt a2a intrebare se stie ca o rotatie completa inseamna $ 2\pi$ deci pt ca avem 3 unghiuri (cu aceeasi masura) iar suma unghirilor e $ \pi$ o rotatie cu $ \frac {2\pi}{3}$ inseamna un unghi de $ \frac {\pi}{3}$.parca era si ceva in clasa a7a care spunea ca pe cerc daca luam unghiul pe lungimea cercului iar unghiul este x el determina pe partea cealalta a cercului un arc de lungime 2x (asta daca raza cercului este 1)

Multumesc mult, am inteles. Asa este, in clasa a 7-a am invatat ca masura unui unghi inscris in cerc este egala cu jumatate din masura arcului cuprins intre laturile lui (sau cu masura unghiului la centru)


Show that $ x^2 \plus{} y^2 \equal{} 4z \minus{} 1$ has no integer solution.

For the left side of the equation we show that any integer square may be only $ 0$ or $ 1 \mod{4}$. Then the sum of two such integers is 0,1, or 2. For the right side $ 4z \minus{} 1 (\mod{4}) \equal{} \minus{} 1$.

Why can we conclude from this that the diaphantine eq. has no sln?

Thanks!

Didn't you just conclude that since the rhs not equal to lhs mod4, the statement cannot be true??? :huh:

If two numbers are equal, then they must give the same remainder upon division by $ 4$ (since they are... well, [i]equal[/i]).  But you just showed that that was impossible.

This problem is beyond the level of HSB forum. If unsure, please read [url=http://www.mathlinks.ro/viewtopic.php?t=32837]High School Basics Difficulty - Read BEFORE Posting Problems[/url]


Any idea which campus of Penn State this is at? Or what the address of that campus is?

State College, PA -- the main Penn State campus.

Thankies. May need to be finding my way there by public transportation  :maybe:  because of stupid senior prom blegh.


Prove that for $n, p$ integers, $n \geq 4$ and $p \geq 4$, the proposition $\mathcal{P}(n, p)$
\[\sum_{i=1}^{n}\frac{1}{{x_{i}}^{p}}\geq \sum_{i=1}^{n}{x_{i}}^{p}\quad \textrm{for}\quad x_{i}\in \mathbb{R}, \quad x_{i}> 0 , \quad i=1,\ldots,n \ ,\quad \sum_{i=1}^{n}x_{i}= n,\] is false.

[i]Dan Schwarz[/i]

[hide="Remark"]In the competition, the students were informed (fact that doesn't actually relate to the problem's solution) that the propositions $\mathcal{P}(4, 3)$ are $\mathcal{P}(3, 4)$ true.[/hide]

Just take  for $(4,4)$, $x_{1}=2$, $x_{2}=x_{3}=x_{4}=\frac{2}{3}$, the general case can be obtained by putting $x_{j}=1$, $j > 4$ and for $p>4$, take $x_{1}= 2^{\frac{4}{p}}$, etc.

[quote="pohoatza"] and for $ p > 4$, take $ x_{1} \equal{} 2^{\frac {4}{p}}$, etc.[/quote]

Umm... take $ x_{1} \equal{} 2^{\frac {4}{p}}$, and what about other $ x_i$'s? Sorry if I'm missing something obvious...

And how do you know when substituting that the ineq is true?


1): Let f : A to B and g : B to C be functions. Prove the following implications:

f and g injective implies g of f injective ;
f and g surjective implies g of f surjective ;
g of f injective implies f injective ;
g of f surjective implies g surjective :

Give an example of functions where g of f is injective but g is not, and where
g of f is surjective but f is not.

When I say g of f, I mean g is composed of f etc...

[quote="mdk"]f and g injective implies g of f injective ;
f and g surjective implies g of f surjective ; [/quote]
These two implications are proven previously. Look at them below.
[url]http://www.mathlinks.ro/viewtopic.php?t=183789[/url]
The rest are open for a try.

Problem 1:
1) See above post.
2) See above post.
3) Suppose $ f(a) \equal{} f(b)$. Then $ g(f(a)) \equal{} g(f(b))$. Since $ g(f(x))$ is injective, $ a \equal{} b$. Done.
4) Suppose $ c$ is an element of $ C$. There exists an element $ a \in A$ such that $ g(f(a)) \equal{} c$ since $ g(f(x))$ is surjective. Then consider $ f(a) \in B$. $ g$ of this element is $ c$. So $ g$ is surjective since $ c$ is arbitrary.

Counterexamples:
Let $ A \equal{} \{0\}, B \equal{} \{0,1\}, C \equal{} \{0\}$. $ g(f(x))$ is injective and surjective but $ f$ is not surjective and $ g$ is not injective, no matter how you define $ f$ and $ g$ (total of 2 choices :D).


A point $ P$ is randomly selected from the square region with vertices at $ (\pm 2, \pm 2)$. What is the probability that $ P$ is within one unit of the origin? Express your answer as a common fraction in terms of $ \pi$.

Well, the answer the area of a circle of radius 1 unit divided by the total area of the square.

The area of the square is 16. The area of a unit circle is pi.

Thus, the answer is pi/16

We know that the total area of the square os 16. The area of the possible area is just a circle with radius 1. That means that area of the circle is pi so the answer is pi/16.


geometry
trapezoid
invariant
symmetry
similar triangles

Triangles $ABC$ and $ABD$ are isosceles with $AB =AC = BD$, and $BD$ intersects $AC$ at $E$. If $BD$ is perpendicular to $AC$, then $\angle C + \angle D$ is
[asy]
size(130);
defaultpen(linewidth(0.8) + fontsize(11pt));
pair A, B, C, D, E;
real angle = 70;
B = origin;
A = dir(angle);
D = dir(90-angle);
C = rotate(2*(90-angle), A) * B;
draw(A--B--C--cycle);
draw(B--D--A);
E = extension(B, D, C, A);
draw(rightanglemark(B, E, A, 1.5));

label("$A$", A, dir(90));
label("$B$", B, dir(210));
label("$C$", C, dir(330));
label("$D$", D, dir(0));
label("$E$", E, 1.5*dir(340));
[/asy]
$\textbf{(A)}\ 115^\circ \qquad \textbf{(B)}\ 120^\circ \qquad \textbf{(C)}\ 130^\circ \qquad \textbf{(D)}\ 135^\circ \qquad \textbf{(E)}\ \text{not uniquely determined}$

[hide="Solution"]Let $\angle ABD = x$ and $\angle BAC = y$. Since triangles $ABC$ and $ABD$ are isosceles, $\angle C = \frac{180^\circ - y}2$ and $\angle D = \frac{180^\circ - x}2$.

Noting that $x + y = 90^\circ$ as two of the angles in right triangle $ABE$, $\angle C + \angle D = \frac{360^\circ - (x + y))}2 = 135^\circ$.[/hide]

[hide="solution"]Since the two lines, as you see in the figure, are perpendicular, $DE=CE\implies AE=BE$.  Thuse, traingle $ABE$ is a 45-45-90 traingle.  Thus, $\angle ACB=\frac{180-45}{2}=67.5$ The same thing goes for $\angle D$.  Thuse, $\angle C+\angle D=135$.[/hide]Masoud Zargar

[quote="Iversonfan2005"]Wow, this problem really should not be a number 21 :o[/quote]That's what I was thinking. :D

If I can solve it, it definately should not be a 21 =p

[hide="alternate solution"]
Note that the figure is an isosceles trapezoid by similar triangles and alternating angles if you connect D and C. Then note that $AE = BE$ by symmetry(remember, isoceles trapezoid). So then we find $EBA = 45$ and $BAD = 135/2 = \angle D = \angle C$ so the sum is $\boxed{135}$
[/hide]

We see that $\triangle{AED}$ is isoceles, and is thus a ${45^\circ}-{45^\circ}-{90^\circ}$ triangle. Thus, $\angle{BAD} + \angle{D} + {45^\circ} = 180$
Then:
$\angle{BAD} = \angle{D} = \angle{C} = \frac{135^\circ}{2}$
Then we have: $\frac{135^\circ}{2} + \frac{135^\circ}{2} = {135^\circ} \implies \boxed{D}$

[quote="me@home"]boxedexe, your first statement was DE=CE. How do you know this is true?[/quote]
Well, his method works because this sum is invariant, but as a proof, I don't think he's right... neither are agolsme and LordoftheMorons  :P  It is not given that $\triangle ABC \cong \triangle ABD$. Hence, there isn't symmetry at all. In fact, if $\triangle ABD$ is equilateral and we choose $C$ on the internal bisector of $\angle BAD$ such that $AB = AC$, we'll get that $AC$ is perpendicular to $BD$, as asked. We still got $\angle C+\angle D=135^{\circ}$, but $DE\neq CE$, $AE\neq BE$ and $\triangle AED$ isn't a 45°-45°-90°-triangle.

However, 4everwise gave a correct and nice solution  :)

All of the above solutions assume that points C and D are on the same side of the line AB. If they are on opposite sides of it, then the sum of angles C and D cannot be uniquely determined.

Problems on original pdfs came with a diagram that clarified that C and D were in fact on the same side of AB. Here is the image:
[asy]
size(200);
pair A, B, C, D, E;
real angle = 70;
B = origin;
A = dir(angle);
D = dir(90-angle);
C = rotate(2*(90-angle), A) * B;
draw(A--B--C--cycle);
draw(B--D--A);
E = extension(B, D, C, A);
draw(rightanglemark(B, E, A, 1));

label("$A$", A, dir(90));
label("$B$", B, dir(210));
label("$C$", C, dir(330));
label("$D$", D, dir(0));
label("$E$", E, 1.5*dir(340));
[/asy]

Hey Taher,

I recommend you post your problem by creating ONE thread rather than by creating 3 posts in 3 different threads that are unrelated.

[quote=Taher.mm.oo]I have one amazing problem about  angle chasing :) ABC is triangle. [AC]=10 ,Let D point at [AB] and [DA] =2
 And Let E point at [BC] and [CE]=2
Let (QF) is line , F   is midpiont at [AC] and also Q midpiont at [ED]
Let
<C=55°
 <A=33°

Show <QFA=??[/quote]

You do not need the size of AC or CE, you only need BD=CE; answer 101.

Best regards,
sunken rock


abstract algebra
group theory
vector
linear algebra
matrix
algorithm
search

Suppose we have an abelian group $ G$ of order $ p^{B}$ isomorphic to $ Z_{p^{b_1}}\oplus Z_{p^{b_2}}\oplus ... \oplus Z_{p^{b_n}}$ where $ Z_p$ is the cyclic group of order $ p$

Thinking of $ G$ as the internal direct product of these subgroups; are the subgroups $ Z_{p^{b_1}}, Z_{p^{b_2}}, ... ,Z_{p^{b_n}}$ the unique subgroups that $ G$ 'decomposes' to? or could we sometimes find a different set of subgroups in $ G$ that yield an internal direct product. (this set would have to have cyclic groups of same order up to multiplicity)

No.  Think of a vector space over a finite field.  More concretely, the two rows of any invertible $ 2 \times 2$ matrix with entries in $ \mathbb{F}_p$ generate $ \mathbb{F}_p^2$ as an abelian group (this is the definition of full rank).

I have not studied fields yet... if the decomposition is unique, shouldn't this be a theorem?

The decomposition is not unique.  Consider p=2, and Z/2Z × Z/2Z.  It can be written as an internal direct sum of nonzero cyclic subgroups in 3 distinct ways.

Ah, I keep forgetting to try simple examples...

Will any disjoint subgroups of the appropriate structure be able to provide an internal direct sum of $ G$. If so then generators of these subgroups are looking quite similar to basis vectors...

They [b]are[/b] basis vectors, in the case of a vector space over a finite field.  (Substitute "the ring of integers $ \bmod p$" if you're not sure what this means, or just the cyclic group $ C_p$ equipped with the obvious action of $ \mathbb{Z}$.)

So which abelian groups are 'isomorphic' to finite vector spaces? are those the only ones were we can decompose them into an internal direct product using an algorithm?

Precisely those of the form $ (\mathbb{Z}/p\mathbb{Z})^n$ for a prime $ p$.  I don't understand your second question; every external direct product is an internal direct product, and an algorithm trivially exists because you can search through every possible decomposition of a finite abelian group.

I am trying to answer this question now:

Suppose we have an abelian group G of order $ p^{B}$ isomorphic to $ Z_{p^{b_1}}\oplus Z_{p^{b_2}}\oplus ... \oplus Z_{p^{b_n}}$ where $ Z_p$ is the cyclic group of order $ p$. $ (b_1, ... , b_n)$ add up to $ B$.

now if we pick any subgroup of $ G$ isomorphic to $ Z_{p^{b_x}}$ can this subgroup be used in an internal direct product decomposition of $ G$ ? Is there a 'nice' algorithm to decompose $ G$ into a direct product? like pick $ x \in G$ \{e} and find a special subgroup containing $ x$ call it $ [x]$ (some $ Z_p^{b_x}$) this subgroup is your first factor, now pick arbitrary $ y \in G$ \[x] and find $ [y]$ etc..

[quote="question?"]I am trying to answer this question now:

Suppose we have an abelian group G of order $ p^{B}$ isomorphic to $ Z_{p^{b_1}}\oplus Z_{p^{b_2}}\oplus ... \oplus Z_{p^{b_n}}$ where $ Z_p$ is the cyclic group of order $ p$. $ (b_1, ... , b_n)$ add up to $ B$.

now if we pick any subgroup of $ G$ isomorphic to $ Z_{p^{b_x}}$ can this subgroup be used in an internal direct product decomposition of $ G$ ?[/quote]

No.  A generator x of a direct summand is such that there is no y with py=x (the keyword for this concept in torsion abelian groups is "height").  For instance in the group 4×2 = <a,b:a^4=b^2=1,ba=ab>, the subgroup <a^2> of order 2 is not a direct summand because a^2 is a square.

[quote="question?"]Is there a 'nice' algorithm to decompose $ G$ into a direct product? like pick $ x \in G$ \{e} and find a special subgroup containing $ x$ call it $ [x]$ (some $ Z_p^{b_x}$) this subgroup is your first factor, now pick arbitrary $ y \in G$ \[x] and find $ [y]$ etc..[/quote]

Sort of.  You take elements of height zero and maximal exponent, and they generate a direct summand.  In Kaplansky's Infinite Abelian Groups this is explained reasonably well (Kulikoff's theorem on bounded implies DSC).  A more module theoretic version is given in Lam's Lectures on Modules and Rings, where you observe the ring Z/p^nZ is self-injective, and so submodules isomorphic to Z/p^nZ are direct summands.  After removing all the Z/p^nZ you are left with a module over Z/p^(n-1)Z and you can remove those direct summands.

Now for an actual algorithm this is awful; how do you test the height of an element?  In a blackbox group, there was some recent work on doing some of this a little more efficiently, but basically it is an awful problem very similar to solving discrete log problems.  In a finitely presented group, there is an algorithm that just uses linear algebra to find generators of cyclic direct summands (a so-called "independent set" or "basis").  I think it is described in lots of modern algebra texts, such as Dummit&Foote, since it provides a constructive proof of the "structure theorem for finitely generated modules over PIDs".  For permutation groups there is another quick algorithm, but I don't have the reference handy.  I don't know about matrix groups, but I suspect any such algorithm would need a discrete log oracle in order to decompose a very common type of abelian matrix group called a "torus".  Basically you'd need to decompose the multiplicative group of the field, which is usually either assumed to have an oracle or to be entirely too hard.

Thanks again JackSchmidt,

I was asking these questions in hopes they will give me an opening to proving that an abelian group of order $ P^B$ is indeed isomorphic to some $ Z_{p^{b_1}}\oplus Z_{p^{b_2}}\oplus ... \oplus Z_{p^{b_n}}$. I know there is an inductive solution I would like to avoid it.

In a formal sense it is impossible to prove such statements without induction, so I'm afraid I don't see the point.  Are you familiar with the standard proof of the structure theorem?

oh, you can't? lol, good that I know this. Well I have not looked at the proof yet. I am trying to think of it myself... I may give up soon though.  I am curious as to why it is impossible to prove without induction?

You can't prove any statement about every positive integer (in this case all values of $ B$) without induction.

What if I take an arbitrary abelian group of order $ P^B$ and provide a construction that represents it as $ Z_{p^{b_1}}\oplus Z_{p^{b_2}}\oplus ... \oplus Z_{p^{b_n}}$ am I using induction?

Yes.  Whenever you say "...and this construction terminates," you're using induction.

So if I were to write the induction more formally I would write:

...now that we have proved that $ P^B$ is isomorphic to (direct product...) by the same reasoning $ P^{B\plus{}1}$ is isomorphic to (some other direct product), so once we have verified the truth of $ B\equal{}1$ by induction the statement is true for all $ B \in Z^\plus{}$

So is the construction not valid until $ B\equal{}$ (some integer) is verified to be true?
also this is a little different to what I usually see in induction where we assume truth of statement at value $ B$ and the work is in showing that implies the truth of $ B\plus{}1$. Here we assumed nothing the work would be in the construction.

is this right? I feel a bit confused..

There's a distinction here between the proof-writing method of induction and the formal logic axiom of induction.  There are lots of theorems for which one can avoid invoking induction in the way that you mention, i.e., there are lots of proof-writing techniques that circumvent the explicit invocation of induction.  However, it is frequently the case that this avoidance is superficial in the sense that a formalization of both proofs would still require invoking the axiom of induction.  This is what t0rajir0u is pointing out.  It can still be instructive to express a construction in an explicit, nonrecursive way, in the sense that your construction doesn't require you to construct all (or many) smaller cases in order to deal with a given instance.


[hide]595 breaks down into 17*7*5. 119 and 595 are too big, so he either has to be 17*5 85 or 7*5 35. In both cases he will be dead in 69 years, but in the first case his son will be dead too. So sam is 35[/hide]

[quote="Treething"]Sam's current age is a positive factor of 595, and in 69 years, he will be dead (keep in mind everybody in Daesunland dies at age 100). However, in 69 years, his son still be alive. Given that Sam hasn't engaged in any underage fun, how old is he at the moment?[/quote]
[hide="Answer"]All you do is find a factor of 595 between 31 and 100. Those are 85 and 35. If he is 85, his son is dead too. It is [b]35[/b][/hide]


trigonometry
inequalities
geometry
incenter
quadratics
inequalities proposed

[quote="fuzzylogic"]Let A, B, C be angles of a triangle. Prove that

\[
7 + 3(\cos{A}\cos{B}+ \cos{B}\cos{C}+\cos{C}\cos{A}) \geq 6(\cos{A}+ \cos{B}+\cos{C}) + 2\cos{A}\cos{B}\cos{C}.
\][/quote]

fuzzylogic's inequality is equivalent \[ p^2 + 5r^2 \geq 16Rr \]  (*)

And (*) is equivalent to \[ IG \geq 0 \] (I is the incenter and G is the centroid of triangle ABC).

If you expand (*) with a,b,c you will find that (*) is equivalent to Schur Inequality.

I was going to submit this problem also
It has a more symmetric form in half angles

3(Symmetric sum( SinA/2)^2.(Sin B/2)^2 >=Symmetric sum(SinA/2)^2.(CosA/2)^2

This could be treated directly using reaarangement inequaltity, Cauchy or maybe by consideration of discriminants -The inequality being quadratic in SinA/2)^2, etc

Iam tryng to find a proof using these techniques

MIchael


Who wants a tournament for players rated 1200 or below? I would like as many players as possible (6-32) and well figure out the format, maybe even a team format, please reply if interested

Hm...my rating is somewhere in those numbers.  wth, /me joins.

only two people...lol
and no i DON'T join (my rating's over 1600 anyway)

even though my rating is 1337, i have not played a rated game in a year  :D  so my technical rating is proably somewhere around 1100. I will join, if playing unrated games is an option  :D

Well need a few more to make this complete

can I join? my rating is constantly dropping to below 1200. basically that means every time I play, my rating goes down, like the stock market right now  :D my rating now is like 1168

haha, I haven't played before, so my ranking is 1200.

*cough*

lol, kl2836 has rating like 1600's

[quote="kl2836"]my rating is 950, and im a 3rd grader[/quote]

I doubt that, seeing as how you were at AMSP.

I shall "Try" to get to the tourney, and my current rating is 1038 (hey, Im a fifth grader) and my highest rating is 1200 :oops:

Sorry for double posting, but here's the list.

1. 007math  (rating is apparently "under 1200")
2. 3.141592653589  (1056)
3. jgao1995 (unknown)
4. mz94 (either 1337, or 1100)
5. tcjy (around 1168)
6. davidd2 (1200)
7. kl2836 (950, still being debated)
8. louiethechewy123 (1038)

Everyone agree?

OK. i will try to check this page everyday untill contest end.

[quote="andrewnycpops"]this tourney was originally my idea, but now rating is 1300+ SO I Think im out of this[/quote]
Lol. Well you can't really get 32 people. There aren't that many people with 1200-.

There are a lot of people! 1200 is average, remember? Think!

Did the tournament already happen :huh:
or are we even going to go on with this

[quote="levans"]The rating system doesn't quite work on a certain value is an average. A lot of people with below 1200 averages don't play. They played a few games, realized they weren't winning, got frustrated, and moved on with life. I wish they'd continue to play, but I certainly understand their frustration and lack of desire to play.[/quote]
Sorry for spamming or reviving, but I was kind of like that. I got below like 1100 and I got really frustrated. But I loved cd and I was determined to get better so I played izzy and r15 and lots of good people and now I am like a lot better. Some people just don't understand these concepts! It's not about rating! It's about getting better!  :(


ratio
geometry
cyclic quadrilateral
similar triangles
geometry proposed
power of a point

Let $E$ be the intersection of the diagonals of the cyclic quadrilateral $ABCD$. Let $F$ and $G$ denote the respective midpoints of the sides $AB$ and $CD$. Prove that the line through $G$ which is perpendicular to $AC$, the line through $F$ which is perpendicular to $BD$ and the line through $E$ which is perpendicular to $AD$ are concurrent.

I don't think we really need $F$ and $G$ to be the midpoints of $AB$ and $CD$, I think the fact that $BF/AB = CG/CD$ is sufficient (so that's a more general condition). Am I right? (I'll post my solution later.)

Yes, you are indeed correct. The midpoint condition was unnecessary but it does make the question a little easier.

Hints:

[hide]Show that the prependicular in A on AC, and in D on BD intersect on prepedicular from E to AD, in T. (The ortocenter of AED)
[hide] Show that the prependicular in C on AC, and in B on BD intersect on prepedicular from E to AD, in S. (Angle chasing)
[hide] The point of intersection of the three lines in problem is the midpoint of ST.
[/hide][/hide]

It can be easily generalized.[/hide]

My solution is quite different, I think.

(In my case the midpoint thingie didn't really make things easier. I was confused for a while about the midpoints, in fact.)

So, supposethat $F$ and $G$ are points on $AB$ and $CD$ such that $BF/AB = CG/CD$. Let $H$ be the foot of the perpendicular from $F$ to $BD$ and let $I$ be the foot of the perpendicular from $G$ to $AC$. Let $FH$ and $GI$ intersect at $S$. We only need to show that $SE \perp AD$. Notice that triangles $BHF$ and $CIG$ are similar. Hence we have (using similarity of triangles) that $BH/CI = BF/CG = AB/CD = BE/CE$. This implies that $BC \parallel HI$. Using the fact that $SHEI$ is cyclic, we get $\angle HSE = \angle HIE = \angle BCE = \angle BCA = \angle BDA$. The result follows. $\blacksquare$

Essentially the solution is all about similar triangles along $BD$ and $AC$. So your solutions aren't too different.

Here $X$ is the foot of the perpendicular on $BC$ from $F$ and $Y$ is the foot of the perpendicular on $AD$ from $G$. And $Z$ being the foot of the perpendicular from $E$ to $AD$. $FX$ and $GY$ cross at $P$.

The official solution talked about showing $AXYD$ is cyclic and angle chasing to get $\angle DEZ = \angle XEP$ showing $Z, E, P$ collinear.

I did something a little less elegant and showed the orthocentre $H$ of $\triangle AED$ was collinear to $E$ and $P$ because I did not spot the nice angle chase at the end and left everything to ratios of lengths along $AC$ and $BD$.

Still quite a nice, easy geometry question which a fair number of us got out eventually.

Here's my solution which seems to be a little different from others.

[hide=Solution]
Let $X$ be the foot of the perpendicular from $F$ to $BD$. Let $Y$ be the foot of the perpendicular from $E$ to $AD$. Let $Z$ be the foot of the perpendicular from $G$ to $AC$. We also denote $H$ to be the intersection of $GZ$ with $YE$. We wish to prove that $F, X, H$ are collinear. Notice that it would be enough to prove that $XHDY$ is a cyclic quadrilateral since then $\angle HYD = \frac{\pi}{2} = \angle DXH$ as the both subtend the same arc and therefore $\angle HXE + \angle FXE = \pi$ which implies that $F, X, H$ are collinear.

Notice that $\angle AYH = \angle AZH = \frac{\pi}{2}$ and therefore $AHZY$ is cyclic. Thus, by the Power of a Point theorem we have that
$$
EY \cdot EH = EA \cdot EZ
$$

Further notice that triangles $ABE$ and $DEC$ are similar. Moreover, since $F$ and $G$ are the midpoints of corresponding sides of similar triangles, we know that triangles $FBX$ and $FZC$ are similar. From this we see that $\frac{EB}{EX} = \frac{EC}{EZ} \implies XZ \parallel BC \implies \angle BCA = \angle XZE$ but we also know that $\angle BCA = \angle BDA$ since they subtend the same arc. Therefore, $\angle XZA = \angle XDA$ which shows that $XZDA$ is cyclic. Using the Power of a Point theorem again we have that
$$
EX \cdot ED = EA \cdot EZ
$$
which we combine with the first equation to get
$$
EY \cdot EH = EX \cdot ED
$$
which shows that $XHDY$ is a cyclic quadrilateral by the converse of the Power of a Point theorem.
$\blacksquare$
[/hide]

Someone please check my solution :).

Regards,
ThePingu


I have a question!!! How could I crate a table here? Can anyone give me an example. (a table with 9 rows and 8 columns). Please  :help:  :read:

What kind of table? You can use the following \begin{array}{cc...c} [code] \end{array} like below

$ \left| \begin{array}{ccccc}
 1 & 2 & 3 & 4 & 5 \\
 ab & cd & ef & gh & ij \\
 x & y & z & t & u
\end{array} \right| $

[quote="Slizzel"]A table with borders also :?[/quote]Try the following then
$  \begin{tabular}{|c|c|c|}
  \hline
  a & b & c \\
  \hline
  b & c & a \\
\hline
  c & a & b \\
  \hline
\end{tabular} $

Thanks a lot!! At least 250 posts   :D i got a half of star  :surf:  :trampoline:  :agent:


calculus
integration
inequalities
trigonometry
function
calculus computations

Show

$\int_{0}^{x}\frac{\sin t}{t+1}\, dt \geq 0$ for every $x \geq 0$

Let's try integrating by parts. Let $dv=\sin t\,dt$ and choose the constant so that $v=1-\cos t.$
Then we have $u=\frac1{t+1}$ and $du=-\frac1{(t+1)^{2}}\,dt.$

$\int_{0}^{x}\frac{\sin t}{t+1}\,dt=\left.\frac{1-\cos t}{t+1}\right|_{0}^{x}-\int_{0}^{x}(1-\cos t)\left(-\frac1{(t+1)^{2}}\right)\,dt$

$=\frac{1-\cos x}{x+1}+\int_{0}^{x}\frac{1-\cos t}{(t+1)^{2}}\,dt$

Since $1-\cos x\ge 0$ for all $x,$ both parts of that are clearly nonnegative.

By taking the limit as $x\to\infty,$ we have also demonstrated the convergence of $\int_{0}^{\infty}\frac{\sin t}{t+1}\,dt.$

Let $f(x)=\int_{0}^{x}\frac{\sin t}{t+1}dt$, defined on $[0,\infty)$.

Then $f'(x)=\frac{\sin x}{x+1}$. Critical points are $k\pi$. The functions increases on each interval of the form $[(2k-2)\pi, (2k-1)\pi]$ and decreases on $[(2k-1)\pi, 2k\pi]$, hence it sufices to prove that for all $k\in N$

$\int_{0}^{2k\pi}\frac{\sin t}{t+1}dt\geq 0$.

A calculation shows that

$\int_{0}^{2k\pi}\frac{\sin t}{t+1}dt=\sum_{m=1}^{k}\int_{(2m-2)\pi}^{(2m-1)\pi}+\sum_{m=1}^{k}\int_{(2m-1)\pi}^{2m\pi}$

which after some calculations is equal to

$\int_{0}^{\pi}\sin x \sum_{m=1}^{k}\frac{(y+1)(1+(-1)^{m})+\pi+(2m-2)\pi(1+(-1)^{m})}{((2m-2)\pi+x+1)((2m-1)\pi+y+1)}dx$, which is definitely greater than or equal to 0.
I hope I made no computational mistakes.

Let $I_{2k}=\int_{2k \pi}^{(2k+1) \pi}\frac{\sin t}{t+1}dt$ and $I_{2k+1}=\int_{(2k+1) \pi}^{(2k+2) \pi}\frac{\sin t}{t+1}dt$.  Now clearly $I_{2k}\geq 0$ for all $k \geq 0$.  Additionally, $I_{2k}+I_{2k+1}\geq 0$ for all $k \geq 0$ (one can substitute $t=t+\pi$ and verify this if this is not clear; the analysis will be similar to what is shown below).

Now, by the above, if $x \in I_{2k}$ for some $k$ we must have $I(x) \geq 0$. So let us suppose $x \in I_{2k+1}$ for some $k$.  Now notice that:

$\int_{(2k+1) \pi}^{x=(2k+1) \pi+\delta}\frac{\sin t}{t+1}dt =-\int_{2k \pi}^{2k \pi+\delta}\frac{\sin t}{t+1+\pi}$ where we have substituted $t=t+\pi$ and assumed $0 \leq \delta < \pi$.

Now to show $I(x) \geq 0$ in this case, since $I_{2k}+I_{2k+1}\geq 0$ for all $k$, we need only show $I(x)-I(2k) \geq 0$.  This means:

$I(x)-I(2k)=\int_{2k \pi}^{(2k+1) \pi}\frac{\sin t}{t+1}dt-\int_{2k \pi}^{2k \pi+\delta}\frac{\sin t}{t+1+\pi}dt \geq \int_{2k \pi+\delta}^{(2k+1)\pi}\frac{\sin t}{t+1}dt$
$\geq 0$
which is clearly true.

So $I(x) \geq 0$ for all $x \geq 0$.

P.S.  Kent is a much faster typer than I am, but at least my solution is sufficiently different  :)

Or just integrate by parts using $1-\cos t$ for the antiderivative of $\sin t$ :) This trick leads to the following generalization. Suppose that $\int_{0}^{x}f(t)\,dt\ge 0$ for all $x>0$ and $g$ is a non-increasing non-negative function. Then $\int_{0}^{x}f(t)g(t)\,dt\ge 0$ for all $x>0$.


geometry
geometric transformation
rotation
AMC

How do scores that one gets on the AHSME and AMC 12 compare? I have old AHSME's that I ordered, but they are differented from the AMC 12's in that you have 90 minutes rather than 75 minutes and there are 5 more questions. If I get a 120 or so on the AHSME, is that appx. what I will get on the AMC 12 or is there some conversion formula?

There's no conversion factor, because there's never been any year-to-year stability in scores. The AMC isn't ETS. They can't prescreen questions by giving them to a sample of people taking it, and gradually work questions into their rotation, knowing their difficulty and knowing how they compare to previous question. The can't do that for several reasons, the most obvious being that the only possible test subjects are the very students who want to compete in the contest. That, and all of you would remember a question you've seen before. Scores do bounce around from year to year.

The 90-minute, 30-question AHSME tests were graded with 5 points per right answer, 2 per blank, and 0 per wrong answer. There is the sense that 150's, and possibly also all scores from 130 up, got a little more common after the change from 30 questions to 25. It's just a little easier to survive 25 questions without screwing up than 30. But I would say that 100 is 100, except for that year-to-year variance.

Oh - and if you get your hands on a 1994 AHSME, understand that that one was way out of whack: far easier than any other year that I know of.


THIS IS NOT SPAM

I just figured out that I lost my cell phone. It was a black SAMSUNG T Mobile with camera. Since I can't find it after many times of searching my house, I think it's possible that I lost it in the GCTM testing room. Did anyone, by any chance, have a clue about it? Or who should I contact? Lots of thanks!

was this the one that i called you on? cuz i called you after GCTM for results remember? if you're talking about that phone, you probably lost it afterwards

No. The one you are talking about is my father's.

OK, I just realized I lost a notebook at the Lassiter practice. Did anyone happen to recover it? Thanks!

Could you describe the notebook more specifically, and perhaps say where you might have been when you lost it?  I didn't notice one, but seeing as I go to Lassiter, I might find it in the next week.  (Or Ms. Poss/Mr. Slater might.)

Anyway, Tim, maybe this is a stupid thing to ask, but did you try calling your phone from another phone?


http://portal.acs.org/portal/fileFetch/C/WPCP_012023/pdf/WPCP_012023.pdf
How do you do #3?

Since you are looking at a graph of $ \Delta T$ in a calorimetry experiment, there is inherent heat loss. To correct for this, you have to do some extrapolation.

Take the curve after it begins to dip down from the increase in temperature and draw a best fit line through that. Then find the temperature on that line when the graph just begins to increase in temperature (when the reaction begins).

In this case, the extrapolated line reports a temperature of 36 degrees. So $ \Delta T \equal{} 12 ^\circ C$

When I do what you say, I do not see an increase in temperature after "it begins to dip down from the increase in temperature."

You said "Take the curve after it begins to dip down from the increase in temperature and draw a best fit line through that. " I'm confused. The line of best fit after the increase does not go through 36.

See the picture.

now extend that line backwards until you hit the right spot (see image)

Is there any intuitive reason why this works?

Why does a line of best fit taken from the decreasing part of graph give us a temperature related to delta T?

Also, is there something I should have read in a general chemistry book to know to do this? I have not seen questions like this before.

im actually not really sure on the intuitive reason why its the best fit line... only reason i know this is because i looked this up when i did the problem before.

haven't seen this in any normal general chem txtbook ive gone thru. :/ sry bout that

Thanks for helping me, adchia.

Where did you look up this method? What book was it in, if it was in a book?

http://books.google.com/books?id=qciCdSFpFPkC&printsec=frontcover&source=gbs_summary_r&cad=0

Found it in this book, confirmed with some other engineering books i found on google books.

hope that helps!


function
calculus
derivative
calculus computations

A function is defined by
$f(x)=x^{m}(x-1)^{n}$ where $m$ and $n$ are both positive integers greater than $1$.

Show that the curve $y=f(x)$ has a stationary point in the range $0<x<1$.

Also, find the condition involving $n$ which alters the nature of this stationary point.

$f(0) = f(1) = 0 \Rightarrow$ By Rolle's theorem $\exists c \in (0,1)$ such that $f'(c) = 0.$ $n \le 0$ changes the nature of this stationary point, assuming $m$ is a positive integer. I guess a nicer way of showing this is by actually differentiating the given function.

With $n\le 0$ you no longer have $f(1)=0$. Anyway, both $m$ and $n$ are assumed to be positive integers. The important thing is whether $n$ is even or odd.

A point where the derivative of a function is zero. Reason for the name: the rate of change of the function is zero.

A related concept: a point where the derivative is either zero or does not exist is called a critical point. For example, $f(x)=|x|$ has a critical point at $0$, but I would not call it a stationary point.

ok. i just learned it as a different name (critical point).
i just wanted to make sure it wasn't the point $x_{0}$ where $f(x_{0})=x_{0}$

[quote="maokid7"]the point $x_{0}$ where $f(x_{0})=x_{0}$[/quote]

That would be a fixed point.  (I personally don't like the term "stationary point," partly because it sounds too much like it could mean "fixed point.")

Man! I've been looking for a fixed point... a stationary point $x_{sp}\in(0,1)$ is easy: $x_{sp}=\frac{m}{m+n}$ and $f\left(\frac{m}{m+n}\right) =(-1)^{n}\frac{m^{m}n^{n}}{(m+n)^{m+n}}$ which is a max or a min according as n is even or odd.


I have a question when the equality holds. I have come across many versions of that. Could you give strict answer. Links to some useful materials about that inequality will we very appreciated :)

Though I've got this one thanks.
But on the other website I've found, that this inequality holds if and only if $z_1=y_1$ $z_2=y_2$ ... and so on
where $z$ is the permutation of $y$
Is it true?


inequalities
AMC
USA(J)MO
USAMO
function
inequalities theorems

What is the hardest inequality you have ever seen heard or deal with ? :)  :)

I am sure that the hardest is Iran 1996:
$(xy+yz+zx)(\frac{1}{(x+y)^2}+\frac{1}{(y+z)^2}+\frac{1}{(z+x)^2})\geq\frac{9}{4}$

Here is the hardest inequality as far as I am concerned:

     \[ \sum\frac{1+a^{2}b^{2}}{(a+b)^{2}}\geq\frac{5}{2}. \]

Guys, I'm amazed. What do you say about problems such as the Mongolian Inequality, USAMO 1999 and even some other Mathlinks inequalities? ( ;) ). But of course, these problems interest noone, since they have ideas and no stupid computations and thus they cannot be difficult because noone is interested in them. I said this at least 3 times till now and I will say it one more time: god help those who still think inequalities = bunching and olympiad problems= mathematics.

I think the most difficult inequalities I dealt with were those solvable only (???) by vasc's (n-1)EV... but then again for me all inequalities are difficult  :P

and also usa mo 2000 (i think harazi was talking about this one) is very nice and difficult

Let $\sigma (n)$ be the divisor function and $H_n=\sum_{k=1}^nk^{-1}$. Prove that $\sigma (n)\leq H_n+e^{H_n} \text{ ln } H_n$. It has been proved that this inequality is equivalent to the Riemann hypothesis, but I was unable to prove the inequality or Riemann hypothesis.

Well, it would be quite big news if you was able to :).

$a,b,c$ are positive numbers, such that $\sqrt{abc}=\sqrt{ab}+\sqrt{bc}+\sqrt{ca}$ and $ab+bc+ca=2abc$.
Prove that their product is not smaller than $2$, i.e. $abc\ge 2$.

This problem is mine, but the idea is not mine :).

Beat:

[b]Your hypothesis doesn't work. From the first equation you can square both sides and factor you get:

0 = m.n.p.(m + n + p) where m = a^(1/2) , n = b^(1/2) and p = c^(1/2). Thus you get m = 0 or n = 0 or p = 0
.So a = 0 or b = 0 or c = 0 which contradicts your assumption that a, b, and c > 0.[/b]
m.c.

Sorry the problem is wrong and i cannot find the mistake.
Anyway i think that the hardest problem is Vietnam 95/96, too

[quote="harazi"]Guys, I'm amazed. What do you say about problems such as the Mongolian Inequality, USAMO 1999 and even some other Mathlinks inequalities? ( ;) ). [/quote]

What's the Mongolian Inequality?

And which inequality at USAMO 1999 are you refering to?
At  http://www.kalva.demon.co.uk/usa/usa99.html  there is no tough USAMO 1999 inequality...

That's an impossible question to answer! My three favorite inequalities are those below. They are very hard and still have beautiful solutions. The first one is astonishingly difficult and stumped many people, myself among them.

(Reid) Let $n$ be a positive integer and let $x_1, \dots, x_n$, and $y_1, \dots, y_n$ be positive real numbers. $z_2, \dots, z_{2n}$ is a sequence of positive real numbers such that $z_{i+j}^2 \ge x_iy_j$ for all $1 \le i, j \le n$. Let $M = \textrm{max}\{z_2, \dots, z_{2n}\}$. Prove that
\[ \left(\frac{M + z_2 + \cdots + z_{2n}}{2n}\right)^2 \ge \left(\frac{x_1 + \cdots + x_n}{n}\right)\left(\frac{y_1 + \cdots + y_n}{n}\right)  \]

(Anh-Cuong) For positive reals $a,b, c$,
\[ a^3+b^3+c^3+3abc \ge ab\sqrt{2a^2+2b^2} + bc\sqrt{2b^2+2c^2} + ca\sqrt{2c^2+2a^2}  \]

(Harazi) For positive reals $a, b, c, x, y, z$ with $xy+yz+zx=3$,
\[ \frac{a(y+z)}{b+c} + \frac{b(z+x)}{c+a} + \frac{c(x+y)}{a+b} \ge 3  \]

For the second inequality we can assume: a + b + c = 1. Can some one take it further ?

for the second one, use S.O.S(Sum of Square) which appeared in the forum, then it's not hard.

for $a,b,c$ with the property that $ab+bc+ca=1$ we have
$\sum\frac{1+a^{2}b^{2}}{(a+b)^{2}}\geq\frac{5}{2}$
This inequality is much stronger than the Iran 1998
$(xy+yz+zx)\left(\frac{1}{(x+y)^{2}}+\frac{1}{(y+z)^{2}}+\frac{1}{(z+x)^{2}}\right)\geq\frac{9}{4}$

the hardest i have ever seen atleat i couldn,,t solve that till now is from [b]imo tst 2006 india [/b].see the problem i have posted that in the [b]unsolved problem section  [/b] and please try solving that . any approaches are also appreciated.....


Prove that there are infinitely many positive integers n such that
         $ n | 2^n \plus{} 2$ and $ n \minus{} 1 | 2^n \plus{} 1$.

[hide]
It is easy to see that $ n \equal{} 2$ satisfies the problem. Now prove that if $ n$ satisfies the problem then so does $ 2^n \plus{} 2$.
[/hide]

A much more challenging problem is to find every positive integer $ n$ such that $ n| 2^n \plus{} 2$ and $ n \minus{} 1| 2^n \plus{} 1$. This has been once posted on Mathlinks but no accurate answer received yet.

My idea for that problem is first, we see that sequence defined above satisfies the problem. Then, suppose that there is some $ n$ satisfies the problem but is not in the sequence, we assume the minimum $ n$ and then try to construct a smaller one and then, generate that $ n$ must be in the sequence, which will yields a contradiction. But ideas are always ideas :D.

[quote="Mashimaru"]
It is easy to see that $ n \equal{} 2$ satisfies the problem. Now prove that if $ n$ satisfies the problem then so does $ 2^n \plus{} 2$.
[/quote]
Could you please explain how $ 2^n\plus{}2$ satisfies the problem?

[quote="Mashimaru"][quote="keyree10"]
Could you please explain how $ 2^n \plus{} 2$ satisfies the problem?[/quote]

Suppose that $ n \minus{} 1 | 2^n \plus{} 1$. We have $ 2^{n \minus{} 1} \plus{} 1 | 2^{2^n \plus{} 1} \plus{} 1$ and so $ 2^{n} \plus{} 2 | 2^{2^n \plus{} 2} \plus{} 2$.
Now suppose that $ n | 2^n \plus{} 2$  then $ 2^{n} \plus{} 1 | 2^{2^n \plus{} 2} \plus{} 1$ so $ (2^{n} \plus{} 2) \minus{} 1 | 2^{2^{n} \plus{} 2} \plus{} 1$.
Since then, $ 2^{n} \plus{} 2$ also satisfies the problem.[/quote]


Does anyone know where to find the solutions for Edward Zhu's Mock AIME 2007? I was doing it the other day and wanted to see the solutions and answers. If anyone has it, please post.

I don't know if there's a soln pdf or anything but
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=106177&start=20[/url]
captcha's post has the basic soln idea for all the probs


geometry
geometric transformation
parallelogram
rotation
ratio
dilation

I'm having trouble understanding the solution to this problem.

Given that DXC, BWC, BYA, and DZA are all similar to each other. You need to prove that WXZY is a parallelogram.

[hide="solution in the book"]
The composite mapping

$A_{\alpha}\circ A\left(\frac{|AB|}{|AY|}\right)\circ C\left(\frac{|AY|}{|AB|}\right)\circ C_{-\alpha}$

maps Y to W and Z to X. Since the mapping is a translation, $\overrightarrow{YW}=\overrightarrow{ZX}$ which makes it a parallelogram
[/hide]

[hide="my question"]
What I do not understand is WHY that composite mapping must be a translation.
[/hide]

I did not open the book yet  to see the solution , but you must keep in mind that the composition of two rotations about two points , the first at angle $\phi$ and the second at angle $\rho$ with $\phi+\rho = 0(mod180)$ is a translation. Since $\alpha+(-\alpha ) = 0$ , the compostition of the two rotations about A and C is a translation.

 But I must see it more carefully in the book by Yaglom : geometric Transformations.

 Babis

[quote="silouan"]It seems that I haven't only this problem with the translations.
Mr Babis do you remember that a year ago I made you the same question ?

@mr Babis : The second team will not go to Rodos , will we write from Athens :([/quote]

Yes Silouan ! I remember it very well !

   For Rhodes ,  I  :(  too!

[quote="tjhance"][quote="stergiu"][quote="tjhance"]right, but I'm having more difficulty with the dilations. how do we know the composition of them is a translation as well?[/quote]

This happens, since

    $\frac{AB}{AY}\cdot \frac{AY}{AB}= 1$

that is, the ratios of the two dilations have product equal to $1$.

   Babis[/quote]

Yeah, so its an isometry.
But how do I know the isometry doesn't include a rotation? Why must it be a pure transformation?[/quote]

 It does not include rotation , because the sum of the rotation angles round A and C , that is $\alpha$ and $-\alpha$ ,  is zero !

 babis


is it true?  if $ {a}\geq{b}\geq{c}>0$ then $ a^b\plus{}b^c\plus{}c^{a}\geq{b}^a\plus{}a^c\plus{}c^b$

No. Take $ a \equal{} 4, b \equal{} 2, c \equal{} 1$

There was a discussion about $ a^a\plus{}b^b\plus{}c^c\geq a^b\plus{}b^c\plus{}c^a$... which is true, if i remember correctly :wink:

What is true is the above inequality extended to four variables, i.e. $ a^b\plus{}b^c\plus{}c^d\plus{}d^a\geqslant b^a\plus{}c^b\plus{}d^c\plus{}a^d$ for all $ a,b,c,d \in \mathbb{R}$ with $ a\geqslant b\geqslant c\geqslant d>0$.  :)

[quote="Georg-A."]No. Take $ a \equal{} 4, b \equal{} 2, c \equal{} 1$

There was a discussion about $ a^a \plus{} b^b \plus{} c^c\geq a^b \plus{} b^c \plus{} c^a$... which is true, if i remember correctly :wink:[/quote]That one was posted [url=http://www.mathlinks.ro/viewtopic.php?t=118722]here[/url]. It looks like Vasc has proved it (post #17) but has only given hints, not his full solution with 4 cases.
[quote="Martin N."]What is true is the above inequality extended to four variables, i.e. $ a^b \plus{} b^c \plus{} c^d \plus{} d^a\geqslant b^a \plus{} c^b \plus{} d^c \plus{} a^d$ for all $ a,b,c,d \in \mathbb{R}$ with $ a\geqslant b\geqslant c\geqslant d > 0$.  :)[/quote]Certainly not. Take again $ (4,2,1,1)$  :maybe:

[quote="spanferkel"][quote="Georg-A."]No. Take $ a \equal{} 4, b \equal{} 2, c \equal{} 1$

There was a discussion about $ a^a \plus{} b^b \plus{} c^c\geq a^b \plus{} b^c \plus{} c^a$... which is true, if i remember correctly :wink:[/quote]That one was posted [url=http://www.artofproblemsolving.com/viewtopic.php?t=118722]here[/url]. It looks like Vasc has proved it (post #17) but has only given hints, not his full solution with 4 cases.
[/quote]
See here a full solution (Proposition 4.2).
http://www.emis.de/journals/JIPAM/article1077.html?sid=1077


The positive integers are written consecutively in groups of five so that the first row contains 1, 2, 3, 4, 5; the second row 6, 7, 8, 9, 10; etc. The row which has a sum nearest to the value of 150 is the:

a. 5th row b. 6th row c. 7th row d. 8th row e. 9th row

[hide=" click to reveal hidden content"]

b. 6th row.

$ n$th row is $ 5n\minus{}4, 5n\minus{}3, 5n\minus{}2, 5n\minus{}1, 5n$ with sum $ 25n\minus{}10$

$ 25n\minus{}10 \equal{} 150$ gives $ n\equal{}6.4$, so it's either the $ 6$th or the $ 7$th row, checking, $ n\equal{}6$ gioves $ 140$, while $ n\equal{}7$ gives $ 165$.

Therefore $ n\equal{}6$: the $ 6$th row.

[/hide]


geometry
area of a triangle
Heron\u0027s formula

Let $T = (a,b,c)$ be a triangle with sides $a,b$ and $c$ and area $\triangle$.  Denote by $T' = (a',b',c')$ the triangle whose sides are the altitudes of $T$ (i.e., $a' = h_a, b' = h_b, c' = h_c$) and denote its area by $\triangle '$.  Similarly, let $T'' = (a'',b'',c'')$ be the triangle formed from the altitudes of $T'$, and denote its area by $\triangle ''$. Given that $\triangle ' = 30$ and $\triangle '' = 20$, find $\triangle$.

$\triangle = \frac{h_aa}{2} = \frac{h_bb}{2} = \frac{h_cc}{2}$

$\triangle' = \frac{h_a'h_a}{2} = \frac{h_b'h_b}{2} = \frac{h_c'h_c}{2}$

$\frac{a}{h_a'} = \frac{b}{h_b'} = \frac{c}{h_c'} = \frac{\triangle}{\triangle'}$

Using Heron's formula for the areas $\triangle, \triangle''$ of the triangles $\triangle(a, b, c), \triangle(h_a', h_b', h_c')$:

$\triangle = \frac 1 4 \sqrt{(a + b + c)(b + c - a)(c + a - b)(a + b - c)} =$

$= \frac 1 4 \left(\frac{\triangle}{\triangle'}\right)^2 \sqrt{(h_a' + h_b' + h_c')(h_b' + h_c' - h_a')(h_c' + h_a' - h_b')(h_a' + h_b' - h_c')} =$

$= \left(\frac{\triangle}{\triangle'}\right)^2 \triangle''$

$\triangle = \frac{(\triangle')^2}{\triangle''} =\frac{900}{20} = 45$


[color=blue]Find all integer solution (m.m) to[/color]

$n^{4}+2n^{3}+2n^{2}+2n+1=m^{2}$

I am an amateur solving this kind of problem. The only thing I got is this:

$(n^{2}+n)(n^{2}+n+1)=(m-1)(m+1)$

I don't know if factorization is helpful or how to use it. Any help? Thank you!


calculus
integration
function
calculus computations

Here we go: I had to solve this integral: -4/1+X^2 DX. I'm saying it's -4 arctan X + C and my brother says it's 4 arccot X + C (albeit with another C). Can you help prevent a family drama about an indefinite integral? Because I'm quarelling about this function with my brother.

arctanx+arccotx=pi/2 so you are both right


calculus
integration
calculus computations

hello,

$ \Large \int_{0}^{\frac{\pi}{4}}(cos(2x)^4\plus{}sin(2x)^4)ln(1\plus{}tanx)dx$

Use this formula

$ \int\limits_{a}^{b}f(x)dx\equal{}\int\limits_{a}^{b}f(a\plus{}b\minus{}x)dx$,

and you should get it.

hello
some more help to ur solution::
$ I \equal{} \Large \int_{0}^{\frac {\pi}{4}}(cos(2x)^4 \plus{} sin(2x)^4)ln(1 \plus{} tanx)dx$
$ I \equal{} \large \int_{0}^{\frac {\pi}{4}}(cos(2x)^4 \plus{} sin(2x)^4)ln (1 \plus{} tan(\frac {\pi}{4} \minus{} x))dx$
$ I \equal{} \large \int_{0}^{\frac {\pi}{4}}(cos(2x)^4 \plus{} sin(2x)^4)ln (\frac {2}{1 \plus{} tan(x)})dx$
$ I \equal{} \large \int_{0}^{\frac {\pi}{4}}(cos(2x)^4 \plus{} sin(2x)^4)ln (2)dx \minus{} I$
$ 2I \equal{} \large ln(2) \int_{0}^{\frac {\pi}{4}}(cos(2x)^4 \plus{} sin(2x)^4)dx$
now i feel u can proceed by ur own.
Thank u


Parallel chords in a circle have lengths 12 and 16, and the distance between them is 7. Another chord is midway between them. What is its length?

[hide]
Let the radius=r, and distance from center to chord of length 16 equal to x.

Then,
$(7+x)^2+6^2=r^2$   and  $x^2+8^2=r^2$
Solving yields $r=\frac{\sqrt{265}}{2}$ and $x=-3/2$

So the distance from the center to the third chord is 7/2-3/2, and length of the chord,
$l=2(\sqrt{r^2-2^2})=\frac{249}{2}$
[/hide]

i think there are two possible solutions.

I am not great at geometry, but couldn't you draw lines to connect the parellel chords, thus forming an iscoceles trapezoid. Then you could apply that midsegment rule to conlclude that the middle one is
(16+12)/2=14

shobber--I don't think there are two possible solutions because a quadrilateral has only one circumcircle.  Remember also that x can be positive or negative.

Drunner, that will not work. The midsegment of the trapezoid is not a chord of the circle (draw it and see). However, it is a good thought.

Probability, I think shobber means that the chord of length 16 could be a regular chord of the circle, or it could be a diameter, each yielding a different answer.


geometry
3D geometry
combinatorics unsolved
combinatorics

I suppose you know the rubik's cube. I have a following problem:

Let's have a sequence S of "moves", that is, turns of faces in clockwise direction (you hold the 'center cubelet' static, as in the standard Rubik's cube notation). Let's denote P(S) the least positive integer such that when we repeat the sequence S P(S) times, we return to the orignal state.

What is the maximal possible P(S) (and for what sequence S)?
What are all the possible values of P(S)?

It is easy to see that P(S) $\leq$ 43252003274489856000.
I think this can be lowered hugely, but not sure how to do that.

The answer is 1260.

Here's the idea (call it proof):

There are two kinds of pieces: corner pieces, those that have 3 facelets, and edge pieces, those that have 2 facelets. Note that corner pieces (from now on: corners) cannot replace edge pieces (from now on: edges) so that we can look at them separately.

There are 8 corners and 12 edges. The corners can have 3 orientations and the edges can have 2 orientations.

Now think of the effect of S as some cycles of corners and edges. Pieces can also get twisted in the course of one cycle, so that after we complete the cycle we see that all the pieces in that cycle are twisted. We can then multiply by 2*3=6 (3 corner orientations and 2 edge orientations) so that we can ignore orientations and concentrate on the permutations.

We'll look for some edge cycles which use less than 12 edges (because there are only 12 edges on the cube!) and for which the product of their orders is as big as possible. It's quite easy to see that the case where we have a 5-cycle and a 7-cycle is the answer: we need to repeat the move 35 times in order to get the pieces back to their original positions.

We cannot have 5-cycles or 7-cycles for the corners, because the would have the same order as the edge cycles so that we will still need 35 moves for solving all pieces, not more. A corner 8-cycle would seem the best solution. But an 8-cycle is an odd cycle (because 8 is even) and we cannot have an odd number of odd cycles (both edge cycles are even), so it's not possible to have an 8-cycle. The next best solution is a 6-cycle and a 2-cycle  (6+2=8), so then we need 35*6=210 moves to bring both the edges and the corners to their original positions. Multiply then by 6 to get rid of the orientation, and the result is 1260.

If we had other options for edges than 5+7 it would still be seen that they are worse than this one presented here. It's not a strict proof, but it works because the only way to check for the best solution is to simply take all possible cases.

A move which needs to be repeated 1260 timesin order to return to the solved position again is R U2 D' B D' (R right, U up, D down, B back. The two others are L left and F front. 2 after the letter means you turn that face 180 degrees, and ' means you turn it 90 degrees in anticlockwise direction.)

Now for the second question, the answers are (taken from the [url=http://games.groups.yahoo.com/group/speedsolvingrubikscube/]international speedcubing forum[/url]):

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 14, 15, 16, 18, 20, 21, 22, 24, 28, 30, 33, 35, 36, 40, 42, 44, 45, 48, 55, 56, 60, 63, 66, 70, 72, 77, 80, 84, 90, 99, 105, 110, 112, 120, 126, 132, 140, 144, 154, 165, 168, 180, 198, 210, 231, 240, 252, 280, 315, 330, 360, 420, 462, 495, 504, 630, 720, 840, 990, 1260.

[quote="portoseb"]The answer is 1260.[/quote]
Wow :o . Much lower than what I expected.

Whoa, i thought this up a year ago (and solved it), to find it is now on this website. Very weird. I started wondering about it when i started doing a weird combo over and over, and wondering how long it would take to come back.... cool!


Consider the following increasing sequence of four numbers: 1,3,n,120. What is the value of n if the result of adding 1 to the product of any two of them is always a square?

[hide=answer? or hint?]n+1 is a square, 3n+1 is a square, 120n+1 is a square. As far as I can tell guess-and-check is our best method. Take squares, subtract 1. 8 works.


topology
real analysis
real analysis unsolved

Consider $ f: A \subset \mathbb{R} \rightarrow \mathbb{R}$, if we know that for each closed $ B \subset \mathbb{R}$, $ f^{ \minus{} 1}(B)$ is relatively closed in $ A$, proof then that $ f$ is continuous.

What is your viewpoint here? This is almost the definition of topological continuity - i.e. that for each open $ E \subset \mathbb{R}$, $ f^{ \minus{} 1}(E)$ is open in $ A$. Can you show that the metric continuity definition is equivalent to the latter statement?

I dont understand you, we are working with closed sets here.

The statements for open and closed sets are equivalent.

We know that f is continuous if for all open $ V \subset \mathbb{R}$, $ f^{\minus{}1}(V)$ is relatively open in $ A$.

Suppose now that V is an open set such that $ f^{\minus{}1}(V)$ is not relatively open.
$ V^c$ is a closed set, so $ f^{\minus{}1}(V^c)\equal{}f^{\minus{}1}(V)^c$ (easy to proof) is relatively closed in A. This means that $ f^{\minus{}1}(V)$ is relatively open in A. Contradiction.


inequalities
trigonometry
geometry proposed
geometry

Isosceles right triangles $ AA'B, BB'C, CC'A$ are constructed outwardly on the sides of a triangle
$ ABC$,with the right angles at $ A',B',C'$.Prove or disprove that
\[ \sin A'+\sin B'+\sin C'\ge\cos\frac{A}{2}+\cos\frac{B}{2}+\cos\frac{C}{2}\]

If $ A',B',C'$ are right angles then $ \sin A'=\sin B'=\sin C'=1$ and the inequality is obvious.

Sorry I mean
Isosceles right triangles $ AA'B, BB'C, CC'A$ are constructed outwardly on the sides of a triangle $ ABC$,with the right angles at $ A',B',C'$.We will be able to contruct a triangle $ A'B'C'$ and let $ A',B',C'$ be angles of it. Prove or disprove that
\[ \sin A'+\sin B'+\sin C'\ge\cos\frac{A}{2}+\cos\frac{B}{2}+\cos\frac{C}{2}\]


geometry
Gauss
vector
cyclic quadrilateral
geometry solved

Let $ABCD$ be a cyclic quadrilateral, and let $E$ and $F$ be the feet of the perpendiculars from the point of intersection of its diagonals $AC$ and $BD$ to the lines $AB$ and $CD$, respectively. Prove that the line $EF$ is perpendicular to the line through the midpoints of the segments $AD$ and $BC$.

This must have been posted plenty of times before, but here's a solution I'm sure I've never seen before :).

Let $P$ be the intersection of the diagonals $AC,BD$ of $ABCD$, and let $H_E,H_F$ be the orthocenters of $PAB,PCD$ respectively. $PAB,PDC$ are similar, so $\dfrac{PH_E}{PE}=\dfrac{PH_F}{PF}$, meaning that $H_EH_F\|EF$. Now, the line connecting the midpoints of $AD,BC$ is the Newton-Gauss line of the quadrilateral $ABDC$ (notice the order of the vertices; the sides of this quadrilateral are $AB,BD,DC,CA$, and its diagonals are $AD,BC$), while $H_EH_F$ is its Aubert line. It's well-known (and it has also been discussed several times on the forum) that the Newton-Gauss line and the Aubert line of a quadrilateral are perpendicular to each other, so we're done.

See http://www.mathlinks.ro/Forum/viewtopic.php?t=16761 post #3 and http://www.mathlinks.ro/Forum/viewtopic.php?t=30926 , and, if you read Romanian ;) , http://www.mathlinks.ro/Forum/viewtopic.php?t=41648 .

See also http://www.mathlinks.ro/Forum/viewtopic.php?t=39107 .

  darij

posted many times. the idea is see that $EG = FG$ and $EH = FH$.
(where $G$ and $H$ are the midpoints of $AD$ and $BC$).

:P ,from bulgarian spring 83 competition! :idea: take pt. of intersection of diagonals as origin,then standard vector computations do the job,there is one more nbice eucleadaean sol due to s.savchev


I'm not sure whether this has come up before, but it hasn't been fixed. In the resources section, when you click "print pdf version", all of the images are displayed as code. is there a way to make the images come out as images?

It won't be fixed until the next big update, which is still a while away.


calculus
geometry
circumcircle
trigonometry
perpendicular bisector
trig identities
Law of Sines

Let $A_1, A_2, \ldots, A_n$ be an $n$ -sided regular polygon. If $\frac{1}{A_1 A_2} = \frac{1}{A_1 A_3} + \frac{1}{A_1A_4}$, find $n$.

$n$ should be $7$.  We can get $A_1A_4=A_1A_5$ from the equation.

So what is $A_1$ , $A_2$ anyway  ? :roll:

[quote="shobber"][quote="varun"]How do you get it.[/quote]
Ptolemy.  Use the fact that $A_1A_2=A_2A_3$ and $A_1A_3=A_2A_4$ and $A_1A_4=A_2A_5$ and so on...[/quote]

I want to know how you prove $A_1A_4=A_1A_5$

[quote="FMako"][quote="shobber"][quote="varun"]How do you get it.[/quote]
Ptolemy.  Use the fact that $A_1A_2=A_2A_3$ and $A_1A_3=A_2A_4$ and $A_1A_4=A_2A_5$ and so on...[/quote]

You are assuming that the $n$-gon is regular.

$A_xA_y = z$, where $z$ is the distance between the points $A_x$ and $A_y$.

Correct me, if I am wrong.[/quote]

You are correct, but still no satisfactory answer has been provided. :(

Are $A_1 , A_2 ,....$ the vertices of a regular polygon. If so the the problem becomes very easy.
Rushil, can we use Calculus in solving the problems at RMO.

yes they are vertices.
But, I don't think use of calculus is allowed.
anyway you asked Rushil. :D

I dont really know whether Calculus is allowed at RMO - the local RMO coordinator said it was OK ( to a 'limited' extenet) but one of my friends says its use is not fully credited even though it is allowed!

So the best answer is : Calculus is allowed but you wont get full marks on solving the whole question using Calculus!

I think the problem is true only when the polygon is regular.
As we all know, a regular heptagon satisfy the conditions. Let the vertices be $A_1$ to $A_7$. If we add as many points as we like on the side of $A_1A_4$ not containing $A_2$ and $A_3$, the equation $\frac{1}{A_1A_2}=\frac{1}{A_1A_3}+\frac{1}{A_1A_4}$ is still valid. Therefore the polyogn must be regular.

I know the answer is $7$.
This is a problem of Indian National Mathematics Olympiad.
You have to prove why it is satisfied for $n=7$

I did it with one method. Wanted to know other solutions and so I've asked it. :blush:

Please prove it.

Until now, nobody has ever posted their full solutions so I am posting mine. :)
(The polygon should be regular).
Obviously the polygon has at least 5 sides. Since it is regular, it is cyclic.
Let $A_1A_2=A_2A_3=A_3A_4=A_4A_5=x$, $A_1A_3=y$, $A_1A_4=z$.
$\frac{1}{x}=\frac{1}{y}+\frac{1}{z}$
$\Rightarrow yz=zx+xy$
By Ptolemy's Theorem (applying to quadrilateral $A_1A_3A_4A_5$),
$A_1A_5\cdot x+xy=A_3A_5\cdot z=yz=zx+xy$
$\Rightarrow A_1A_5=z=A_1A_4$
Therefore, $A_1$ lies on the perpendicular bisector of $A_4A_5$ which passes the centre of the circle.
$\therefore$ The polygon has $7$ sides.

Of course there's the 'long, dirty and algebraic solution' using the expression for sides and the lengths involved in terms of angles they subtend at centre of polygon! Of course, that's for the regular polygon! I'll check what the problem says  - regular or any polygon!?!?!?!?

The problem can't possibly be true for ANY polygon. I'm SURE they meant a regular polygon.

Ya! I feel the same way! I think I'll change the statement now! If someone else finds the result true for any polygon,  or the correct verified statement , please tell!!

I just checked one of my book. The $\Longleftarrow$ part was a problem of 21st All USSR Math Olympiad.

Try this one:

Find all regular polygons $A_1A_2..A_n$ such that $\frac{1}{A_1A_2}=\sum_{k=2}^p \frac{1}{A_1A_k}$ for some p.

WLOG let us assume that the circumradius of the regular n gon be 1.
Let the angle subtended by each side onto the center be $\theta$
Then applying the [b]law of sines[/b] to triangles[b] $OA_1A_2,OA_1A_3,OA_1A_4$[/b] we get
  \[A_1A_2=2sin(\theta/2)\]
  \[A_1A_3=2sin(\theta)\]
  \[A_1A_4=2sin(3\theta/2)\]

Plugging these back in the equation gives
\[cosec(\theta/2)=cosec(\theta)+cosec(3\theta/2)\]
Simple manipulations of this yeild $\theta=2\pi/7$ whence $n=7$.

It can be solved using complex numbers.
Let Ao be the centre of the regular polygon which gives us Ar=Ao+|A1-Ao|e^2i(r-1)(pi)/n
Now |Ar-A1|=|A1-Ao|2sin2(r-1)(pi)/n
Putting this in given identity and simplifying gives sin(3pi/n)=sin(4pi/n) which gives n=7

[quote=nik9796]It can be solved using complex numbers.
Let $A_o$ be the centre of the regular polygon which gives us $A_r=A_o+|A_1-A_o|e^{\frac{2i(r-1)(pi)}{n}}$
Now $|A_r-A_1|=|A_1-A_o|2\sin \frac{2(r-1)(pi)}{n}$
Putting this in given identity and simplifying gives $\sin\frac{ 3pi}{n}=\sin \frac{4pi}{n} $which gives $n=7$[/quote]

Number of diagonals in $n-gon$ also tells the number of quadrilaterals,right?
So starting with minimal $n=4$, this is not a possibility because then $A_1A_2=A_1A_4$ which we lead $\dfrac{1}{A_1A_3}=0$ which is pretty inappropriate.


In the triangle $ABC$ the following equality holds:
\[\sin^{23}{\frac{A}{2}}\cos^{48}{\frac{B}{2}}=\sin^{23}{\frac{B}{2}}\cos^{48}{\frac{A}{2}}\]
Determine the value of $\frac{AC}{BC}$.

We will prove that $ AC = BC$.

Suppose not, for example, assume that $ \angle A > \angle B$. Then $ \sin \frac{\angle A}{2}> \sin \frac{\angle B}{2}$, so $ \sin^{23}\frac{\angle A}{2}> \sin^{23}\frac{\angle B}{2}$, and $ \cos \frac{\angle B}{2}> \cos \frac{\angle A}{2}$, so $ \cos^{48}\frac{\angle B}{2}> \cos^{48}\frac{\angle A}{2}$. Contradiction, since the given equality cannot hold.


A right triangle with integer side lengths is such that its area is 2007 times its perimeter. How many such right triangles exist?(the lengths of the three sides uniquely determine a triangle)

[hide]
$\text{area}=2007\cdot \text{perimeter}\iff \text{inradius}=4014\iff a+b-c=8028$ where $a,b$ are the legs, $c$ the hypotenuse. Isolate $c$ to get $a+b-8028=c$. Squaring,

\[a^{2}+b^{2}+8028^{2}+2ab-2\cdot 8028(a+b)=a^{2}+b^{2}\]
\[ab-8028(a+b)+2\cdot 4014^{2}=0\]
\[(a-8028)(b-8028)=2\cdot 4014^{2}\]
Since $2007=3^{2}\cdot 223$, the right hand side is equal to $2^{3}\cdot 3^{4}\cdot 223^{2}$. But since $b<c$ and $a+b-c=8028$, we must have $a>8028$. Similarly, $b>8028$. Therefore both factors on the left hand side are positive. The number of positive divisors on the right hand side is $(3+1)(4+1)(2+1)=60$, and accounting for the fact that $(a,b)$ and $(b,a)$ produce the same triangle, we get $\boxed{30}$ solutions.
[/hide]

Think about another way to represent the sides of the triangle ($s^{2}+t^{2}, 2st, s^{2}-t^{2}$)

[quote="scorpius119"][hide]
$\text{area}=2007\cdot \text{perimeter}\iff \text{inradius}=4014\iff a+b-c=8028$ where $a,b$ are the legs, $c$ the hypotenuse.
[/hide][/quote]

Probably a dumb question (I'm not good with geometry), but how do you get from the inradius being 4014 to $a+b-c=8028$?

The radius of a circle inscirbed in a right triangle with sides $a,b,c$ (the [b]inradius[/b]) is $\frac{a+b-c}{2}$.

So $r=4014 \Rightarrow \frac{a+b-c}{2}=4014 \Rightarrow a+b-c=8028$.

[quote="1234567890"]The radius of a circle inscirbed in a right triangle with sides $a,b,c$ (the [b]inradius[/b]) is $\frac{a+b-c}{2}$. [/quote]

A short demonstration:

We can drop perpendiculars from the incenter to each side, which divide the sides into lengths

$a = x+y$
$b = y+z$
$c = x+z$

(By the creation of a few similar triangles.)  $y$ is the length shared by the two legs.  Because $\angle C$ is right, we have a square with side lengths $r, r, y, y$; hence $r = y$ and

$r = y = \frac{a+b-c}{2}$


Show that for every integer $ m>1$ there are positive integers $ x$ and $ y$ such that $ x^{2}-y^{2}=m^{3}$

[hide="Generalization"] $ \exists x, y : x^{2}-y^{2}= n \Leftrightarrow n \not \equiv 2 \bmod 4$ [/hide]

Answer

$ (x-y)(x+y) = m^{3}$

let $ x-y = m$ and $ x+y = m^{2}$

$ \Rightarrow 2x = m+m^{2}$
$ \Rightarrow 2x = m(m+1)$

$ m(m+1)$ is the product of 2 consecutive integers $ \Rightarrow 2|m(m+1) \Rightarrow x \in Z^{+}$

$ y = x-m \Rightarrow y = \frac{1}{2}m(m+1)-m = \frac{1}{2}m(m-1) > 0 \Rightarrow y \in Z^{+}$


Find $ y$ if $ \dfrac{y^2 \minus{} 9y \plus{} 8}{y\minus{}1} \plus{} \dfrac{3y^2 \plus{}16y\minus{}12 }{3y \minus{}2} \equal{} \minus{}3$.

hello, note that the identity $ \frac{y^2\minus{}9y\plus{}8}{y\minus{}1}\equal{}y\minus{}8$ is true for all $ y$ with $ y\ne1$ and $ \frac{3y^2\plus{}16y\minus{}12}{3y\minus{}2}\equal{}6\plus{}y$ for all $ y$ with $ y\ne \frac{2}{3}$.
Sonnhard.


Let f(x), xε[0,1] with f'(x)>=M>0
Show that there is a space I, in the [0,1] , with length 1/4 : |f(x)|>=M/4 for all x whose belongs to I

If $ f\left(\frac{1}{2}\right) \ge 0$, then by the Mean Value Theorem there exists a $ c \in \left(\frac{1}{2}, \frac{3}{4}\right)$ such that \[ \frac{f(\frac{3}{4}) \minus{} f(\frac{1}{2})}{\frac{3}{4} \minus{} \frac{1}{2}} \equal{} f'(c).\] Since $ f'(c) \ge M$, we have \[ f\left(\frac{3}{4}\right) \ge f\left(\frac{1}{2}\right) \plus{} \frac{1}{4}M \ge \frac{1}{4}M.\] $ f(x)$ is increasing because $ f'(x) > 0$, so $ f(x) \ge f\left(\frac{3}{4}\right) \ge \frac{1}{4}M$ for all $ x \in \left[\frac{3}{4}, 1\right]$.

If $ f\left(\frac{1}{2}\right) < 0$, then we can use a similar argument to show that $ f(x) \le f\left(\frac{1}{4}\right) \le \minus{}\frac{1}{4}M$ for all $ x \in \left[0, \frac{1}{4}\right]$.

thank you very mutch Carpo for the solution,thanks a lot! :)


inequalities
induction
inequalities proposed

If $x_1,x_2,...,x_n$ are non-negative numbers, then
$\sum x_1^3+3\sum x_1x_2x_3 \geq \sum x_1x_2(x_1+x_2)$.

Moreover, I think that for $k \geq 3$ we have
$(k-2)\sum x_1^k+k\sum x_1x_2...x_k \geq \sum x_1x_2(x_1^{k-2}+x_2^{k-2})$,
or, equivalently,
$(k-1)\sum x_1^k+k\sum x_1x_2...x_k \geq (\sum x_1)(\sum x_1^{k-1})$.

[quote="Vasc"]If $x_1,x_2,...,x_n$ are non-negative numbers, then
$\sum x_1^3+3\sum x_1x_2x_3 \geq \sum x_1x_2(x_1+x_2)$.
[/quote]

We will prove it by Induction .
For $n=3$ it is Schur .

Assume the inequality is true for $k$ variable .

To prove $k+1$ variable , we can WLOG let $x_1\geq x_2\geq \cdots x_k\geq x_{k+1}$ , then by hypothesis

$(x_1^3+x_2^3+\cdots x_{k-1}^2+(x_k+x_{k+1})^3)+\left(3\sum_{<>k,k+1} x_1x_2x_3\right)+\left(3(x_k+x_{k+1})\sum^{k-1} x_1x_2\right)\geq \left(\sum_{<>k,k+1} x_1x_2(x_1+x_2)\right)+\left((x_k+x_{k+1})\sum^{k-1} x_1^2\right)+\left((x_k+x_{k+1})^2\sum^{k-1}x_1\right)$

(we just replace the $x_k$ with $x_k+x_{k+1}$)

now we add $(x_k+x_{k+1})\sum_{i<>j=3}^{k-1}x_ix_j$ to both side and notice that at LHS the $(x_k+x_{k+1})^3\leq x_k^3+x_{k+1}^3+3x_kx_{k+1}(x_1+x_2)$

So now we have LHS $\leq \sum^{k+1} x_1^3+3\sum^{k+1} x_1x_2x_3$

As for RHS , after some simplifying job , we notice that $2x_kx_{k+1}\sum^{k-1} x_1+(x_k+x_{k+1})\sum_{i<>j=3}^{k-1}x_ix_j\geq x_kx_{k+1}(x_k+x_{k+1})$

RHS $\geq \sum^{k+1} x_1x_2(x_1+x_2)$

and our induction complete .

The inequality holds when 2 variable or 3 variable are equal while other is $0$


function
number theory theorems
number theory

It is well known that $ \varphi(mn)\equal{}\varphi(m) \varphi(n)\quad if \quad (m,n)\equal{}1$. I was trying to argue about it the following way, I know there is a error in it. Please point out the error.

Let $ \varphi_e(k)$ be the set of all the positive integers not greater than $ k$ which co prime to $ k$. By definition,$ |\varphi_e(k)|\equal{}\varphi(k)$. Since $ (m,n)\equal{}1$ we have $ \varphi_e(m) \cap \varphi_e(n)\equal{}1$. It is true that $ \forall \quad a\in \varphi_e(m),b \in \varphi_e(n) \quad ab\in \varphi_e (mn)$. But when I try to prove the inclusion the other way, I think, there could exist a prime $ p\in (b,ab)$ (assuming wlog, $ a<b$). It is clear that $ p\in \varphi_e(mn)$ but it does not belong to set $ S\equal{}\{ab|a\in \varphi_e(m),b \in \varphi_e(n)\}$.

why is $ \varphi_e(m) \cap \varphi_e(n)\equal{}1$ ?


I need help with this...

I followed the link that AoPS Wiki has posted, but it's messed up and I can't download gsview.exe.
What's going on?
Where can I download it? :huh:  :?:

Oh yeah, I forgot to mention that I have downloaded Asymptote and Ghostscript, and that I use Windows.
PLZ HELP ME ASAP!!!!!!!!!!!!