rotation
geometry
geometric transformation
reflection
percent

(reworded)

Define two operations on an equilateral triangle in the plane, one of whose sides is parallel to the x-axis and the opposite vertex is above that side. The first operation rotates the triangle 120 degrees clockwise. The second operation reflects the triangle over its vertical axis, in other words it basically switches the bottom two vertices.

In how many ways can we map the triangle to itself, in the same orientation, in 5 operations?

[hide="99 percent chance wrong"]I only found 2 so far, but there's got to be more.[/hide]

[quote="kyyuanmathcount"][quote="13375P34K43V312"]Actually here's a hint

[hide]The answer is not ${5\choose2}$.[/hide][/quote]

Lol.

[hide="Further Hint"]So:

How many rotations return the triangle to it's original state?

How many flips?

What does this tell us about our working transformation sequences?[/hide][/quote]

[hide="to build on that"]we need 3 turns and 2 flips, and we have 5 turns (hey, 3+2=5!). But again, the answer is not ${5\choose2}$.[/hide]

[hide="probably not"]
Now I get these 4:

R=rotation
F=flip

FFRRR
RFFRR
RRFFR
RRRFF
[/hide]

[hide]Is the answer 6[/hide]

I know it is wrong, but I don't know.

[hide="Solution"]Three rotations will return it back to normal, or two flips.

Let F= flip and R=rotation.

Thus, we must have 3 R's and 2 F's.  However, having them scattered among eachother doesn't work, (ex: FRFRR doesn't work).  We must either have the two F's together or the 3 R's together:

FRRRF
FFRRR
RRRFF
RFFRR
RRFFR

That's $\boxed{5}$.  [/hide]

[hide]Oh, right, I added an extra one, that didn't work.

But, I had the same idea too. :mad: [/hide]

Yeah, it actually is quite easy if you're thinking the right way, but if you don't read it right, it could be really hard to get.

[hide]Let the operations be R=rotation, F=reflection.

For the triangle to map itself, there must be either 0 R's, or 3 R's.

If there are 0 R's, then the operations is FFFFF, which doesn't work, because you reflect the bottom vertices with each other each time, but it won't be the same after an odd number of F.

There are 3 R's: So there are 2 F's:
Before the last R, A must be in the bottom left, because after R, we cannot move A to the top.
Thus, before the second R, A must be in either the top or the bottom right - But if it is in the top, then it must have been in the bottom right before the first R, however, we cannot use reflections to get A to the bottom right.
So, A must be in the bottom right before the second R.
We also know that A must be in the top place before the first R, since it is impossible to reflect to a bottom.

Thus, the position of A is fixed right before each R. So the 2 F's can be either in
FFRRR
RFFRR
RRFFR
RRRFF
FRRRF, all the ways such that A is in it's fixed place before each R $\boxed{5}$.[/hide]


The roots of $ x^{4}\minus{}14x^{2}\plus{}24x\minus{}k\equal{}0$ are real and distinct.Find the range of $ k$


Given an integer $ m\geq$ 2, m positive integers $ a_1,a_2,...a_m$. Prove that there exist infinitely many positive integers n, such that $ a_{1}1^{n} \plus{} a_{2}2^{n} \plus{} ... \plus{} a_{m}m^{n}$ is composite.

Take a prime factor $ p$ of $ a_{1} \plus{} 2a_{2} \plus{} ... \plus{} ma_{m}$, then by Fermat's Little Theorem, $ a_{1}1^{(p \minus{} 1)n \plus{} 1} \plus{} a_{2}2^{(p \minus{} 1)n \plus{} 1} \plus{} ... \plus{} a_{m}m^{(p \minus{} 1)n \plus{} 1}$ is divisible by $ p$ for all $ n$ and as they should be all distinct, we can conclude that there exist infinitely many positive integers n, such that $ a_{1}1^{n} \plus{} a_{2}2^{n} \plus{} ... \plus{} a_{m}m^{n}$ is composite.

[quote="KenHungKK"]Given an integer $ m\leq$ 2, m positive integers $ a_1,a_2,...a_m$. Prove that there exist infinitely many positive integers n, such that $ a_{1}1^{n} + a_{2}2^{n} + ... + a_{m}1^{m}$ is composite.[/quote]
I think $ m\ge 2$ and $ x_n=a_1*1^n+a_2*2^n+...+a_m*m^n$ composite for infinetely many $ n$.
Then $ x_{n+m}=\sigma_1x_{n+m-1}-\sigma_2x_{n+m-2}+...+(-1)^{m-1}\sigma_mx_n$. Were $ \sigma_1=1+2+...+m=\frac{m(m+1)}{2}, \sigma_m=m!, \sigma_k=\sum_{i_1<i_2<...<i_k}i_1i_2...i_k$
Let p is prime, then ${ y_n=(x_n\mod p,x_{n+1}\mod p,...,x_{n+m-1}\mod p})$ is periodic. Therefore we can take $ p=x_k$ and have infinetely many $ n$, suth that $ p|x_n$.

let fix n.let a1*1^n+a2*2^n+....+am*m^n=p where p is prime
by little fermat theorem we get p|a1*1^[n+(p-1)k]+a2*2^[n+(p-1)k]+....+am*m^[n+(p-1)k]=b where k integer >0
from this we conclude that b is composite.it's easy to see that problem solved


If $ f:  \mathbb{R}^n\rightarrow \mathbb{R}^m$ is continuous everywhere and $ S\subset \mathbb{R}^n$ is bounded.  Show that $ f(S)$ must also be bounded.

Hint: $ S$ is contained in a compact set.


Hello,
In one of Valentin's .pdf problem sheets/tutorials I stumbled upon the following cute problem:

"There exist arbitrarily long arithmetic progresses of integers, whose terms are all integer powers. No such infinite progresses exist, though"

Any hints or solutions?

Thanks!

--Thanos

You might be interested by this as a hint:

[b]Theorem.[/b] There are arbitrarily large gaps in the series of primes.
[i]Proof.[/i]  Consider the series
(k+1)! + 2, (k+1)! + 3, ..., (k+1)! + k, (k+1)! + k+1.
Every term is not prime, since n divides (k+1)! + n if 2 <= n <= k+1.

It was taken from Ivan Nivens An Introduction to the Theory of Numbers.

It's rather surreal to answer my own message, but the problem I mention is rather easy ( :( ), so don't bother.

The shortest solution to the first (and most interesting, I think) part of the question is the following: Suppose the general term of the arithmetic progress is given by c_n = a + b.n, where n \in Z and put d = gcd(a, b). Then, the sequence is c_n = d(a/d + b/d.n) and the quantity (a/d + b/d.n) assumes infinite prime values by Dirichlet's celebrated theorem. Hence, if every term was a perfect power, then d would be divisible by an infinitude of primes, contradiction!

--Thanos

here is a simpler solution(not using Dirichlet th):

we will show that in any infinite sequence u_k=ak+b,
there is at least one term which is NOT a perfect power.
Let p>(a+b) be a prime number. Then GCD(p^2,a)=1 and then
there exist integers (x,y) so that a*x-p^2*y=1.
Let k=(p-b)*x=(p-b)*(1+p^2y)/a  (which is an integer).
We have
u_k=ak + b
   =(p-b)*(1+p^2y) + b
   =p^2*(p-b)*y + p.
Hence a_k is a multiple of p but not of p^2.
-> a_k is not a perfect power.


If {-2a+1}<13, what is the sum of the distinct possible integer values
of a?

Is there any way to do these types of problems without just plugging in number?
Sry, I'm not good at math  :blush:

yeh, but waht im asking is if there are faster ways to do this kinds of problems than plug in

You can just solve it by using regular absolute value inequality methods.

To solve, break it up into two equations :

$ | \minus{} 2a \plus{} 1| < 13$

and

$ | \minus{} 2a \plus{} 1| < \minus{} 13$

Now simply solve for both cases, yielding $ a \equal{} \minus{} 6$ and $ a \equal{} 6$, or $ a \equal{} \{ \minus{} 6,\,6\}$. Therefore, the sum would be $ \boxed{0}$.

That's wrong

the possible solutions for the problem are   -6<a<7

this means the sum of all possible values of a are -5+(-4)+......4+5+6= [b]6[/b] [u]ans.[/u]

Yeah, that's right

@7h3.D3m0n.117 - you have to reverse the sign on the second inequality

... wow.
err...
If $ |\minus{}2a\plus{}1|<13$, then it means that the expression within the absolute value has to be greater than -13 or less than 13.
$ \minus{}13<\minus{}2a\plus{}1<13$
$ \minus{}14<\minus{}2a<12$
$ 7>a>\minus{}6$

Which means that ilikepotatoes is right...

[quote="alanchou"]@7h3.D3m0n.117 - you have to reverse the sign on the second inequality[/quote]

Yeah for a second there I forgot what the word inequality meant  :ewpu:

Yes there is....

[-2a+1]<13

Use the Addition property of Equality

[-2a+1-1]<13-1

[-2a]<12

You will come to the conclusion that a has to equal less than -6.

that doesn't apply in absollute value signs for example:

|-1|=1     if u add 1 to both sides...
|0|=2      false!

u need 2 split inequalities with absolute values into 2 absolute vales equations.

[hide="erm, i dont know about you guys, but i saw this pretty quickly."]when the abs values are out and you have 3 in an inequality, when you eliminate the 1, it makes the sum $ 2\plus{}(13\minus{}1)\equal{}14$. then when you divide by 2, it is 7.
i will use x in this example to make it clearer
$ \minus{}x\plus{}1<2a<x\plus{}1$
the sum -x+1 and x+1 is 2, same with -x+2 and x, and -x+3 and x-1, and so on, until 0 and 2. this happens 13-1=12 times. that is the sum of the values of 2a. therefore the sum of the values of a is 6
[/hide]


[quote="Incorrect solution"][hide]
First, let's examine the conditions needed to form a right triangle with such points. We can show that if we connect the midpoints of two sides that are not adjacent nor opposite each other, and a third side adjacent to exactly one of them, then we have our triangle.

Now, pick an arbitrary side. The probability that the second side will be neither adjacent nor opposite it is $ \frac {2}{5}$. The chance that the last side is adjacent to exactly one of them is $ \frac {1}{2}$. Thus, the probability is $ \frac {1}{2}\cdot\frac {2}{5} \equal{} \boxed{\frac {1}{5}}$
[/hide][/quote]

EDIT: Whoops, I missed a few cases.  :oops:

[quote="i_like_pie"]Three sides of a regular hexagon are chosen at random, and their midpoints are connected. Find the probability of the resulting triangle being right.[/quote]
[hide="solution"]
It is as if you are choosing vertices of a regular hexagon and connecting them, since the midpoints of all the sizes form another regular hexagon.
There are two arangements for right triangles: (counterclockwise from vertex 1) $ (1,2,4)$ and $ (1,2,5)$
Basically, it's the same right triangle, but one has the smaller side adjacent counterclockwise to the hypoteneuse and the other has the smaller side adjacent clockwise to the hypoteneuse.

There are $ 6$ possible spots for each, so there are $ 12$ possibilities for a right triangle.
$ \frac {12}{\binom{6}{3}} \equal{} \frac {12}{20} \equal{} \boxed{\frac {3}{5}}$
[/hide]

edit: IRT Temperal's solution:
You can have the 2nd side opposite the first, as long as the third side is adjacent to any of them (basically the third side is unconditional).
You can also have the 2nd side adjacent to the first, as long as the third side is opposite to either one of them.
[hide]
so $ \frac {2}{5}\left( \frac {2}{4} \right) \plus{} \frac {1}{5}\left( \frac {4}{4} \right) \plus{} \frac {2}{5}\left( \frac {2}{4}\right) \equal{} \frac {3}{5}$
[/hide]

Basically, anytime the triangle is scalene, it's a right triangle.


AMC
AIME
trigonometry
AMC 12
AMC 10
AMC 10 A

I improved a lot with AOPS Volume 1, so my average AMC 10 Practice score improved from 116 to 140!! Is the AMC 12 Problem Series necessary?

Hi,

I think you'd be better off asking that question here: http://www.artofproblemsolving.com/Forum/index.php?f=140 ;)

Wow, it improved that much? Try the 2007 AMC 10 A, tell me your score... It's my favorite AMC10...

About the classes: AMC 12 class will show you a bunch of problems and stuff, and make your over all score go higher so I suggest the AMC 12 class [b]if, and only if[/b] your goal is scoring high on AMC 12 and making AIME.

[quote="AIME15USAMO"]Umm, what else would be your goal? I'm not trying to insult you, but isn't that what he's asking (AMC10/12/AIME prep)?[/quote]

Actually I'd like to get 8+ on AIME this year.

[quote="topofmath"]Actually I'd like to get 8+ on AIME this year.[/quote]

THen take the AIME class.

I would recommend you to look at previous threads covering some of your questions before you repeat them, such as this one: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=292685[/url].

If you have managed that high score on AMC 10, you will be better off taking AIME series. You will learn more, and if you can handle middle to hard problems on AIME, you should be set for AMC 12 as well.

[quote="Ihatepie"]I'm in a similar situation as him, my average AMC 10 is like 144 and my average amc 12 is about 120. I've nearly (not quite) mastered the C&P and algebra sections (I can get 4/5 on average of these problems in the last 5 right) but i'm only average at NT (i can only get like 1/2 on average of this in the last 5 right), while my knowledge of trig is limited to right angle triangles. Would the AMC 12 class help?[/quote]

The AMC12 class definitely helps with trig, but if you're scoring 120 or more consistently, you're better off just studying your weak parts in Volume 2 or Intermediate Series. Awesome AMC10 score by the way

Sorry to bring up a new question but has anyone done the AMC12 course, if so, is it helpful?


Leaderboard:
theone853-5

1. What is the perimeter of a triange formed by the coordinate axes and the line 12x + 5y = 60?

2. The average of a set of integers is 6.  The sum of the integers is 18.  What is the number of integers in the set?

3. A party of 18 people went to a restaurant.  They each chose a $21 meal but 4 of them forgot to bring their money.  In order to settle the total bill, how much extra did those who brought their money have to pay?

4. The first four triangle numbers are 1,3,6,10.  What is the eighth triangle number?

5. An auto traveling at 50 miles per hour covers the distance between two towns in 6 hours less than a truck which travels at 35 miles per hour.  How many hours does it take the auto to cover the distance?

MathFiend gets #5
MathFiend & Leeyuhang & MysticTerminator get #3.

I'll post the next set in 3 minutes.

numbers 3&5 of the ones topper just posted are in Joel's contest...lucky for me, I worked on his problems earlier today...=)


calculus
integration
analytic geometry
abstract algebra
complex analysis
complex analysis unsolved

Could you please solve the following exercise including all the details?

Suppose that $ f$ is analytic in the open disk $ D(0;1)$. Using polar coordinates, prove that for every $ w\in D(0;1)$ we have

$ f(w)=\frac{1}{\pi}\int\int_{D(0;1)}\frac{f(z)}{(1-\bar{z}w)^{2}}dxdy \ \ (z=x+iy).$

Alexandros

I am so happy to have a complex analysis problem to work on.  Thank you!

Since $ f$ is analytic in the open disk $ D(0;1)$, it has a power series expansion there in the form:  $ f(z)=\sum_{n=0}^{\infty}c_{n}z^{n}$ or in polar coordinates one would write:  $ f(r,\theta)=\sum_{n=0}^{\infty}c_{n}r^{n}e^{i n \theta}$.

Substituting this value of $ f$ into your integral gives:

$ \frac{1}{\pi}\int \int_{D(0;1)}\frac{\sum_{n=0}^{\infty}c_{n}r^{n}e^{i n \theta}}{(1-r e^{-i \theta}w)^{2}}r \hspace{1mm}dr d\theta$

=$ \frac{1}{\pi}\sum_{n=0}^{\infty}c_{n}\int_{0}^{2 \pi}\int_{0}^{1}\frac{r^{n+1}e^{i n \theta}}{(1-r e^{-i \theta}w)^{2}}\hspace{1 mm}dr d\theta$

Now ${ \frac{1}{(1-r e^{-i \theta}w)^{2}}=\sum_{k=0}^{\infty}(k+1) r^{k}e^{-i k \theta}w^{k}}$.

Substituting in that expression gives:

$ \frac{1}{\pi}\sum_{n=0}^{\infty}c_{n}\left( \sum_{k=0}^{\infty}\int_{0}^{2 \pi}\int_{0}^{1}(k+1) r^{k+n+1}e^{i (n-k) \theta}w^{k}\hspace{1 mm}dr d\theta \right)$

$ =\frac{1}{\pi}\sum_{n=0}^{\infty}c_{n}\left( \sum_{k=0}^{\infty}(k+1) w^{k}\frac{1}{k+n+2}\delta_{nk}(2 \pi) \right)$

$ =\frac{1}{\pi}\sum_{n=0}^{\infty}c_{n}\left(\frac{(n+1) w^{n}(2 \pi)}{2n+2}\right)$

$ =\sum_{n=0}^{\infty}c_{n}w^{n}$

$ =f(w)$

QED

An important detail is missing: why can we integrate the series term-by-term? We don't know if the series converges uniformly on the disk. In fact, the "strange double integral" may be divergent: consider $ f(z)=\frac{1}{(1-z)^{2}}$. The statement of the problem is incomplete.

This integral formula is nothing else but the famous Reproducing formula of the Bergman spaces of the unit disk. The missing part is that $ f\in L^{1}_{a}(D)$ namely that $ f$ is an integrable function. The integral operator $ P{f}(z)=\int_{D}\frac{f(w)dA(w)}{(1-z\bar{w})^{2}}$ is the famous Bergman projection which coincides with the identity operator when applied to the set of analytic functions.

Thanks too Didilica.  Your answer to Mlok is much better than what I would have written being that I have never heard of the "famous reproducing formula of the Bergman spaces of the unit disk"  :)

I will have to read more about this...

By the way, I would love to have this problem on my Prelim exam just so I could write that your Bergman kernel observation is an "obvious" alternative solution.


i'm a new mathlinkers. but however, i'm interested in IMO

did there any IMO for middle-school since i am in middle-school

There is no specific IMO for middle-school students. However sufficiently strong middle-school students can participate in the IMO.

Maybe JBMO, some mini IMO's can help you.

davron


induction
inequalities solved
inequalities

If $ a \geq b \geq c>0$ then show that we have:
$a^{n}b(a-b)+b^{n}c(b-c)+c^{n}a(c-a) \geq 0$  $ \forall n \in N$.

made by Lascu M.

by induction on n :
n=2 : it is IMO 1983
assume the ineq is true with n
or a^nb(a-b) +b^nc(b-c) + c^na(c-a)>=0
hence : a^nb^2 + b^(n+1)c(b-c) + c^nab(c-a) >=0
        b^(n+1)c(b-c)>=-a^nb^2 - c^nab(c-a)
Thus : a^(n+1)b(a-b) + b^(n+1)c(b-c)+c^(n+1)a(c-a)>=a^(n+1)b(a-b)-a^nb^2-c^nab(c-a)+c^(n+1)a(c-a)
       = a^nb(a-b)^2 + c^n(c-a)(c-b)>=0
So , the ineq is also true with n+1 ,we have conclusion


I live in New Jersey and it's my second time mailing in solutions. It's been five days since I mailed them in and I haven't seen a check mark yet for Round 2 so I'm getting kind of worried because the last time I mailed in my solutions it only took two days. I know it says explicitly not to panic and that it may take several days before they log receiving my solutions, but is it normal for it to take more than five days until they log it? Thanks for any responses!

Because of the holiday, I don't believe that we have processed very many mailed-in solutions.  We will be doing them in earnest today and they should all be in our system by Wednesday.

Yeah my solutions were logged today. Thanks!


Solve the following equation:
$ x^{5}+10x^{3}+20x-16=0.$


geometry
3D geometry
octahedron
probability
AMC

The labels 1, 2, ... , 8 are randomly placed on the faces of an octahedron (one per face). Find the probability that no two adjacent faces (sharing an edge) have adjacent numbers, where 1 and 8 are also considered adjacent.

Well we know that there are

8 P 8

ways to do it in total... right?

Probably not, because thats not considering rotations.

each face of an octahedron is adjacent to about/exactly three others, I believe.

so we need the probability that, if we just look at the face w. a one on it, that the two and the eight are both on two of the five faces left open... hmm

:: Walks off Pondering ::


algebra
polynomial
inequalities proposed
inequalities

1.

$2a^3 - (a^2+b^2)(2a-b)=b^3-2ab^2+a^2b=b(b^2-2ab+a^2)\geq 0$

Summing cyclically gives the desired result (this was solved by darij grinberg some time back)

Yes,but what about ineq (ii)?

Can some body prove or disprove that ineq:

$\frac{a^4}{2a^3+b^3}+\frac{b^4}{2b^3+c^3}+\frac{c^4}{2c^3+a^3} \ge \frac{a+b+c}{3}$

is it actually symmetric though?  Should that be a cyclic sum instead?

You're right it should be cyclic - I've corrected it - but it doesn't change anything, does it ?

[quote="Megus"]for 2)

write it as:

$\sum_{cyc} \frac{a^3}{2a^2+b^2}-\frac{a+b+c}{9}=\sum_{cyc} \frac{7a^3-2a^2b-2a^2c-b^2a-b^3-b^2c}{18a^2+9b} \geq 0$

now multiplying everything out and getting all elements with minus on the other side, we can use rearrangarement. I know it's a bit bad-looking solution  :?[/quote]

Can you show me a complete and clear solution!

So it is to prove equivalently:

$\sum_{cyc} 7a^3(18b^2+9c)(18c^2+9a) \geq \sum_{cyc} (2a^2b+2a^2c+b^2a+b^3+b^2c)(18b^2+9c)(18c^2+9a)$

which follows from reangarrement applied few times

[b]@Megus:[/b]

Beyound the trivial calculation mistakes ($18b^2+9c$), why should it follow from rearrangement?

[b]@Hungkhtn:[/b]

[quote="hungkhtn"]Can some body prove or disprove that ineq:

$\frac{a^4}{2a^3+b^3}+\frac{b^4}{2b^3+c^3}+\frac{c^4}{2c^3+a^3} \ge \frac{a+b+c}{3}$[/quote]

It already fails for a = 1, b = 2, c = 3...

  darij

Hm, forget about my blabbering, here's much nicer proof but still not quite nice:
let $a+b+c=1$ then:

dividing by 3 and using AM-GM we get:

$\frac{1}{(2a^2+b)(2b^2+c)(2c^2+a)} \geq \frac{1}{9^3}$
so
$9^3 \geq 8a^2b^2c^2+4a^3b^2+4a^2c^4+2a^3c^2+4b^3c^2+2b^3a+2bc^4+bc^2a$
and the terms on the right can be obtained by using enough times $a+b+c=1$ so when we substract everything from the right we should get $something \  \ positive \geq 0$ which is obvious.

Is it okay or am I still saying some crazy things ?

[quote="hungkhtn"]And now

Find k max st:
$\frac{a^3}{ka^2+b^2}+\frac{b^3}{kb^2+c^2}+\frac{c^3}{kc^2+a^2} \ge \frac{a+b+c}{1+k}$

k=2 isn't best constant![/quote]

anybody answers my question?

Ji Chen, where did that answer come from?

I think he used computer with a buffalo calculating with Mixing variables! :D


[quote="Timothy"]In the arithmetic sequence 17, a, b, c, 41, what is the value of b?[/quote]

$b=29$.

[quote="Timothy"]In the arithmetic sequence 17, a, b, c, 41, what is the value of b?[/quote][hide]41-17= 24/4=6  6*2+17=29
b=29[/hide]


geometry
trigonometry
trig identities
Law of Cosines
area of a triangle
Heron\u0027s formula

Alright, we have an equilateral triangle and the three vertexes connect to some point in the middle, not necessarily any of the 4 major points of a triangle (centroid and friends). The segment from each vertex to this point are respectively, 3, 4, and 5. The question is to find the area. I've been having a lot of trouble with this seemingly easy problem (seemed that way for me anyways). Also, another question is to find out what the area would be if it wasnt equilateral? Thanx in advance.

Hmm, well, I haven't actually done it out yet, but it seems that it would be a good idea to apply the Law of Cosines to the three smaller triangles (within the equilateral one) that are created. So your unknowns are the side of the equilateral triangle and the angles around the intersection point in the middle. Solving the simultaneous equations should yield some useful results, if not the answer itself.

Fierytycoon

I actually did that... that was my first course of action, but i got some screwed up answers that didnt work in the end. Any other ideas?

Hmm, actually, now that I think about it, is the answer supposed to be a constant? Because conjuring the image of the diagram in my head, I can see the segments rotating around that one point in the center and forming triangles of varying sizes.

Fierytycoon

yah.... that sounds correct. Can anyone corroborate this? Basically, me and fiery are seeing this point in the middle and as it moves, you still have the constant of the side length no matter what, so therefore the area would be constant. Gonna think about this more a bit later.

yeah, it is constant. i rushed my thinking and posting that last time, completely forgot about the equilateral property of the triangle  :D Im still not sure how to solve the problem though.

Fierytycoon

OK. I think I might have a way of solving this problem, but I don't have the time to actually solve it. (Well, its really cuz I'm lazy).

Anyway, we let each side of the triangle be "x". Then, we apply Heron's formula to the three smaller triangles formed. The sum of these three triangle's areas is equal to the area of the entire equilateral triangle, which is [(x^2*sqrt(3)]/4. Therefore, you have only one variable and one equation. Provided, the equation might get messy, but I think it's doable.

[quote]Provided, the equation might get messy, but I think it's doable.[/quote]
Messy, like with an easy potential to be eighth or sixteenth degree . . . this doesn't sound overly promising.  Also, the trig approach probably leads to very-very difficult equations, I would guess.  I'll edit this when I have something constructive to add, as well.

I solved it using equations of circles. Draw your equilateral triangle with vertices A = (0,0), B = (a/2, a*sqrt3/2), and C = (a,0). And suppose the distances to your interior point P are AP = 3, BP = 4, and CP = 5.

Then think of three circles centered at the vertices and meeting at the point P. The equations of these circles are:

(1) x^2 + y^2 = 9
(2) (x - a/2)^2 + (y - a*sqrt3/2)^2 = 16
(3) (x - a)^2 + y^2 = 25

We want to find an a so that there is a simultaneous solution to all of these equations.

Well taking equations (1) and (3) we get, a^2 - 2ax = 16.
So x = (a^2 - 16)/(2a).

And taking equations (1) and (2) we have, a^2 - ax - aysqrt 3 = 7.
Putting this together with the above equation we get y = (a^2 + 2)/(2asqrt 3).

Putting these two together into x^2 + y^2 = 9 we have:
3(a^2 - 16)^2 + (a^2 + 2) = 9*12a^2
4a^4 - 92a^2 + 772 = 108a^2
4a^4 - 200a^2 + 772 = 0
a^4 - 50a^2 + 193 = 0

So a^2 = 25 + sqrt 432 = 25 + 12*sqrt 3 (I rejected the - sign because I knew it would be too small).

Then the Area of the Equilateral Triangle is:
a^2*sqrt 3/4 = (25*sqrt 3 + 36)/4.


An $ n \times n$ grid has the first column and first row completely filled with the integers from $ 1$ to $ n$, with one integer in each box.  In how many ways can the remainder of the grid be filled with the integers from $ 1$ to $ n$ so that each row and each column contains each integer exactly once?

Equivalent to counting [url=http://en.wikipedia.org/wiki/Latin_square]Latin Squares[/url] after some permutations, which is an unsolved problem.


Our number system has base 10 and uses the digits 0 to 9 inclusive. A number system with base 8 uses the 8 digits 0 to 7 inclusive. If 267 and 135 are numbers in the system using base 8, find their sum.

[hide]in base 8 their sum is 424, in base 10 their sum is 276[/hide]

I'm moving this to the Getting Started forum.


Provide a method for the following transformation:

$ C_6H_5\minus{}COCH_2CH_3 \longrightarrow C_6H_5\minus{}CH_2CH_2CHO$

(compare with OS,P38).

1. Reduce with LAH
2. Dehydrate with alc. KOH
3. Use NBS
4. Use H2/Pt
5. use aq. NaOH and then use PCC

Don't forget that the allyl radical could rearrange to the even more stable allylic-benzylic radical.

This conversion may be long but i think its right

1.Perform Clemmensen reduction to get the alkane.
2.Use Hot alk.KMnO4 to get Benzoic acid
3.Use a mild reducing agent to get benzaldehyde
4.Perform Perkins reaction to get cinnamaldehyde
5.use LAH to reduce both double bond and carbonyl group.
6.Use PCC to get the respective aldehyde

Is this correct :wink: ????????????????

1. Reduce with LAH
2. Dehydrate with alc. KOH
3. PCC(allylic oxidation)
4. protect aldehyde by forming acteal
5. reduce double bond
is this ok?? :maybe:

Both seem to be right.

[quote="Valeriummaximum"]1.Perform Clemmensen reduction to get the alkane.
2.Use Hot alk.KMnO4 to get Benzoic acid[/quote]

You can oxidize directly the ketone to the acid.


Two men starting at a point on a circular 1-mile race track walk in opposite directions with uniform speeds and meet in 6 minutes, but if they walk in the same direction it requires one hour for the faster walker to gain a lap. What is the rate of the slower walker?

MAO 1992

[hide]Suppose the rate of the faster man is x mph and the rate of the slower man is y mph.
If they walk in opposite directions, their combined speed is x+y mph. It takes 1/(x+y) hours to walk 1 mile, or 60/x+y minutes. Since 60/x+y = 6, x+y=10.

If they walk in the same direction, their combined speed is x-y mph (since x>y). Using the same method as before, x-y=1.

Therefore, x+y=10 and x-y=1. Solving for x and y, x=5.5 and y=4.5. The rate of the slower walker is 4.5 mph.[/hide]

A non-algebraic way:

[hide]since they meet up in 6 minutes, we can say that together they cover one mile in 6 minutes.

So in one hour, they'll cover 10 miles combined, and the faster guy will have done 1 mile more than the slower guy. It's easy to see that they will have covered 5.5 miles and 4.5 miles, so the slower guy walked 4.5 miles in one hour.[/hide]

(essentially the same as your soln. though)

Think about how much they travel in 6 minutes combined.  Then think about the rest.

Klebian, your solution sort of has algebra in it.  Just indirectly shown.

[quote="vishalarul"]Suppose the rate of the faster man is x mph and the rate of the slower man is y mph.
If they walk in opposite directions, their combined speed is x+y mph. It takes 1/(x+y) hours to walk 1 mile, or 60/x+y minutes. Since 60/x+y = 6, x+y=10.

If they walk in the same direction, their combined speed is x-y mph (since x>y). Using the same method as before, x-y=1.

Therefore, x+y=10 and x-y=1. Solving for x and y, x=5.5 and y=4.5. The rate of the slower walker is 4.5 mph.[/quote]
hide you solution.
[hide]One mile takes the man an hour when you take the difference so he goes one mph faster. the sum of their rates is ten mph.
$x+y=10$
$x-y=1$
$2x=11$
$x=5.5,y=4.5$
$4.5$[/hide]


[quote="ashegh"]prove that for each positive number we have:

$(a^2)/2+(b^2)/2$ >_  $2ab$[/quote]

Unless I understood this wrong, I think there is a counterexample:
$(a^2)/2+(b^2)/2 \ge 2ab$

$a=1$, $b=1$
$1 \not\ge 2$

Well, I think the RHS should be "ab" ;)

Proof :
$(a-b)^2 \geq 0$ <==> $a^2 + b^2 \geq 2ab$ <==> $\frac{a^2+b^2}{2} \geq ab$

$\mathbb{QED}.$


factorial
floor function
modular arithmetic

[quote="bnick17"]Find the number of zeros in which the decimal expansion of the integer of 2007! ends. Also find its last non-zero digit.[/quote]

Use Legendre's theorem.

[b]Theorem:[/b] The number of times a prime $p$ divides $n!$ is:

$\sum_{k=1}^{\infty}\left[ \frac{n}{p^{k}}\right]$

Now Number of zeros  = number of 10's dividing 2007!

[b]Solution.[/b] The power of $5$ in a factorial is always less than or equal to the power of $2$ in a factorial.  As a result, it suffices to find the factors of $5$ in $2007!$, that is,
\[\sum_{k=1}^{\infty}\left\lfloor\frac{2007}{5^{k}}\right\rfloor=401+80+16+3=500. \]
Therefore, there are $500$ zeros at the end of $2007!$.

Let $L(x)$ represent the last nonzero digit of $x!$.  The last nonzero digit of $2007!$ is equal to
\[\begin{eqnarray*}L(2007)&\equiv& 2^{401+80+16+3}L(3)L(1)L(0)L(1)L(2)\\ &\equiv& 3\cdot2^{502}\pmod{10}. \]
Notice that $2007=31012_{5}$.
Since the residue modulo $4$ of $2^{n}$ for $n\geq 1$ has period $4$, $L(2007)\equiv 3\cdot 2^{502}\equiv 3\cdot 4\equiv 2\pmod{10}$.

And the conclusion follows.$\Box$

[b]Remark.[/b] If I remember correctly, the last nonzero digit of $n!$ can be found on this forum.

[quote="sylow_theory"]Now Number of zeros  = number of 10's dividing 2007![/quote]
Correction: ... = number of 5's ... , of course.

Totally this is
401+80+16+3=500.

[quote="digger"][quote="sylow_theory"]Now Number of zeros  = number of 10's dividing 2007![/quote]
Correction: ... = number of 5's ... , of course.
[/quote]
No, its right. The number of times 10's divide that number is the number of 2's and 5's, which in the same as number of 5's because 5>2.

[quote="sylow_theory"] The number of times 10's divide that number is the number of 2's and 5's, which in the same as number of 5's because 5>2.[/quote]
OK, this is what I've meant.  :lol:

P.S. I've noticed now just by a chance the surprising thing
I haven't seen yesterday the post of [b]boxedexe[/b], otherwice there was no reason for me to send duble.
Compare please the times of posting:
digger Posted: Yesterday, at 11:20 am
boxedexe Posted: Yesterday, at 12:05 am
Do not think, that it is struggle for a priority - the occasion is too small.
Nothing of the kind, it is simply interesting, how it could happen?

boxedexe had originally posted his solution on a different thread (similarly named, from the same poster).  A moderator split it from that thread and re-attached it here, where it belongs.

The formula boxedexe used was derived here:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=577096#577096[/url].


A man was born on February 29, 1928. He died January 29, 2004. How many actual birthdays did he get to celebrate during his life (not counting the day he was born as a birthday)? Hint: February has 29 days in years divisible by four.

(A)18 (B) 19 (C) 36 (D) 75 (E) 76

or...
you can subtract the years... 2004-1928 and get 76
then you divide 76 by four, since leap years are every 4 years...
then, you would get 19, but his birthday has not past in 2004 so
the answer is 18


conics
ellipse
geometry
geometric transformation
dilation

P lies on the ellipse $ \frac{x^{2}}{a^{2}}$+$ \frac{y^{2}}{b^{2}}$ = 1. Find the locus of the midpoint PS where S is a focus

[hide="Geometric description of the locus"]It's a dilation of the elipse by a factor of $ \frac{1}{2}$ centered at $ S$.  In general, if $ S$ is a fixed point and $ C$ is a curve, the set of midpoints of segments joining $ S$ to $ C$ is a dilation of $ C$ by factor $ \frac{1}{2}$ centered at $ S$, and similarly for other points dividing the segments into fixed ratio.[/hide]

hmm ok i see

So if I obtain 2 simultaneuos eqns then I have to square both because the midpoint is a dilation of C(curved path)


What is the largest number $ c$ such that $ 2x^2\plus{}5x\plus{}c\equal{}0$ has at least one real solution? Express your answer as a common fraction.

For this to be real the discriminant, $ b^2\minus{}4ac$, has to be non-negative. So we start by plugging in what we know:

$ 25\minus{}8c$

The maximum value of c that will not make it negative is when it equals 0. Thus:

$ 25\minus{}8c\equal{}0$
$ c\equal{}\frac{25}8$


If it was posted please give a link. The problem

$a,b,c >0$ with $abc=1$

Prove that  $\dfrac{a}{a^2+2}+\dfrac{b}{b^2+2}+\dfrac{c}{c^2+2}\leq 1$


geometry
circumcircle
incenter
vector
trigonometry
geometry unsolved

[color=darkblue][b]Please[/b], can anyone draw for a given constant $ k > 0$ the set $ \mathrm M(k)$ of all points from inside of $ \triangle ABC$ for which

the sum of the distancies from it to the sides of $ \triangle ABC$ is equal to $ k$ ? At least in two cases :

$ \blacktriangleright\ k \equal{} h_a$ , where $ h_a$ is the length of $ A$ - altitude.

$ \blacktriangleright\ k \equal{} R \plus{} r$ , where $ R$ , $ r$ are the lengths of circumradius and  inradius.

[b]Thank you very much ![/b][/color]

[color=darkred][b][u]An easy problem.[/u][/b] Prove that $ R \plus{} r \equal{} h_a\ \Longleftrightarrow\ OA\perp OI$ .[/color]

Let $ I$ be incenter of $ ABC$ and $ A'B'C'$ be contact triangle, we must find the locus $ P$ such that $ PA'' \plus{} PB'' \plus{} PC'' \equal{} k > 0$ when $ A'',B'',C''$ are orthogonal projection from $ P$ to side, it is line
\[ kr \equal{} (PA'' \plus{} PB'' \plus{} PC'')r \equal{} PA''IA' \plus{} PB''IB' \plus{} PC''IC' \equal{} \sum\vec{PA''}\vec{IA'} \equal{} \sum(\vec{PI} \plus{} \vec{IA'} \plus{} \vec{A'A''})\vec{IA'} \equal{} \vec{PI}(\sum\vec{IA'}) \plus{} 3r^2\ (*)
\]
(Note that $ IA'\perp A'A''$ e.t.c)

So let $ G$ be centroid of $ A'B'C'$ from $ (*)$ we have
$ 3\vec{PI}\vec{IG} \plus{} 3r^2 \equal{} kr\Rightarrow \vec{PI}\vec{IG} \equal{} \frac {r}{3}(k \minus{} r) \equal{} const$

So locus $ P$ is a line with normal vector $ \vec{IG}$ choose fix $ P_0$ such that $ \vec{P_0I}\vec{IG} \equal{} \frac {r}{3}(k \minus{} r)$ the line will be connecting $ P_0$ and perpendicular to $ \vec{IG}$.

By this idea we can find the locus $ P$ such that $ xPA''\plus{}yPB''\plus{}zPC''\equal{}t$ when $ x,y,z,t\in\mathbb{R}$ actually they are equation of line in trilinear coordinate.

[color=darkblue] [size=117][b]Very nice proof ![/b][/size]

[b][u]Encyclopedia's proof.[/u][/b] Let $ I$ be the incenter of $ \triangle ABC$ and $ DEF$ be the contact triangle. We must find the locus of $ P$

 such that $ PX \plus{} PY \plus{} PZ \equal{} k > 0$ , where $ X$ , $ Y$ , $ Z$ are the orthogonal projections from $ P$ to sides. Thus,

$ kr \equal{}r\cdot\sum PX \equal{}\sum PX\cdot ID \equal{}$ $ \sum\overrightarrow{PX}\cdot\overrightarrow{ID} \equal{}$ $ \sum\overrightarrow{ID}\cdot\left(\overrightarrow{PI} \plus{} \overrightarrow{ID} \plus{} \overrightarrow{DX}\right)\equal{}$ $ \overrightarrow{PI}\cdot \sum\overrightarrow{ID} \plus{} 3r^2\ (*)$

because $ \overrightarrow {ID}\cdot\overrightarrow {ID}\equal{}r^2$ and $ ID\perp DX$ $ \implies$ $ \overrightarrow {ID}\cdot\overrightarrow{DX}\equal{}0$ . Let $ G'$ be the centroid of $ DEF$. Then $ \sum\overrightarrow {ID}\equal{}3\cdot\overrightarrow{IG'}$ .

The relation $ (*)$ becomes  $ 3\cdot\overrightarrow{PI}\cdot\overrightarrow{IG'} \plus{} 3r^2 \equal{} kr\Longrightarrow \overrightarrow{IP}\cdot\overrightarrow{IG'} \equal{} \frac {r(3r\minus{}k)}{3}$ ([b]constant[/b]). In conclusion,

the required geometrical locus of the point $ P$ for which $ \sum PX\equal{}k$  is a line which is perpendicular on $ IG'$

in the fixed point $ P_0\in IG'$ because $ IP_0\equal{}IP\cdot\cos\widehat {PIG'}\equal{}\frac {r(3r\minus{}k)}{3\cdot IG'}$ (constant) .

In the particular case $ k\equal{}R\plus{}r$ obtain $ \cos\widehat {PIG'}\le 0$ because $ R\ge 2r$ . Since $ \sum\delta_{BC}(O)\equal{}R\plus{}r$ ,

where $ O$ is circumcenter of $ \triangle ABC$  obtain that the angle $ \widehat {OIG'}$ is nonacute.[/color]


number theory
prime numbers
prime factorization

The numbers aren't really that big.  I just remember that $2^{10}=1024$ and can find a lot of the powers of $2$ from there.  Advanced factorization could help if that exponent was large through.  It really would be less work to just divide 1024 by 4, subtract 1 and right off you see the number ends with a $5$.

I just did $16 \times 16$ and then found the factors  :lol:

[quote="7h3.D3m0n.117"]The value $2^{8}-1$ is divisible by 3 prime numbers. What is the sum of the three prime numbers?

I got an answer by computing $2^{8}-1$ then finding the primes is there another way to do this?[/quote]
[hide]
$2^{8}-1=(2^{4}+1)(2^{2}+1)(2+1)(2-1)$
$=(17)(5)(3)(1)$
the 3 prime numbers add up to $25$
it's also helpful to memorize the powers of 2, so you know $2^{8}-1=255$
[/hide]

[quote="7h3.D3m0n.117"]The value $2^{8}-1$ is divisible by 3 prime numbers. What is the sum of the three prime numbers?

I got an answer by computing $2^{8}-1$ then finding the primes is there another way to do this?[/quote]
[hide="Brute Force"]$2^{8}-1=255$

The prime factorization of $255$ is $3\cdot5\cdot17$. The sum is $\boxed{25}$.[/hide]
[hide="Logical"]$2^{8}-1=\left(2^{4}-1\right)\left(2^{4}+1\right)$

$2^{8}-1=\left(2^{2}-1\right)\left(2^{2}+1\right)\left(2^{4}+1\right)$

$2^{8}-1=\left(2-1\right)\left(2+1\right)\left(2^{2}+1\right)\left(2^{4}+1\right)$

$2^{8}-1=(1)(3)(5)(17)$

$3+5+17=\boxed{25}$[/hide]


Find all real numbers a such that the equation $ x^3\minus{}ax^2\minus{}2ax\plus{}a^2\minus{}1 \equal{} 0$ has exactly one real solution in $ x$.

Rewrite the equation like this :
$ a^2\minus{}(x^2\plus{}2x)a\plus{}(x^3\minus{}1)\equal{}0$, then $ a\equal{}\frac{(x^2\plus{}2x) \pm (x^2\plus{}2)}{2}$

As $ a\equal{}\frac{(x^2\plus{}2x)\minus{}(x^2\plus{}2)}{2}\equal{}x\minus{}1$ must give a solution $ x\equal{}a\plus{}1$,
then either $ a\equal{}x^2\plus{}x\plus{}1$ has no solution for $ x$, i.e.$ (1)^2\minus{}4(1\minus{}a)<0 \Leftrightarrow a<\frac{3}{4}$

or $ a\equal{}x^2\plus{}x\plus{}1$ has only one solution for x, and that is $ x\equal{}a\plus{}1$, which is easy to check to be impossible.

So $ a<\frac{3}{4}$ is the solution.

[hide=Shorter solution]$0 = x^2 - ax^2 - 2ax + a^2 - 1 = (x - a - 1)(x^2 + x - a + 1)$. Clearly, this cubic has $x = a + 1$ as a real root, so we must find values of $a$ such that $x^2 + x - a - 1$ has no real roots. For this to happen, the discriminant $1^2 - 4(1)(-a + 1) = 3a - 4 < 0$, so $\boxed{a < \frac{3}{4}}$.


geometry
circumcircle
trigonometry
geometry unsolved

Let $ABC$ a triangle and in the line $CA$ ubicate the point $P$; $D$, the midpoint of $BC$, $AA_{1}$ is an altitud, the intersections of the lines $PD$ and $AB$ is $L$, and $AL=AP$, if $O$ is the circumcenter of the triangle $ALP$, prove that $OA_{1}=OD$

[hide="Hint 1."]Prove easily that the points $A_{1}$ and $D$ have the same power w.r.t. the circumcircle of the triangle $APL\ .$[/hide]
[hide="Hint 2."]Denote the second intersection $A'$ between the bisector of the angle $\widehat{BAC}$ with the circumcircle of the triangle $ABC\ .$
Prove easily that the segment $AA'$ is a diameter in the circumcircle of the triangle $APL\ .$[/hide]

Interesting and nice problem !

[hide="Here is a proof."]
[color=darkred][b]Enunciation.[/b] Let $ABC$ be a triangle. Denote : the middlepoint $M$ of the side $[BC]$ ; the points $N\in AC$ , $P\in AB$ so that $M\in NP$ , $AN=AP\ ;$ the point $D\in BC$ for which $AD\perp BC\ ;$ the circumcenter $O_{1}$ of the triangle $ANP\ .$ Prove that $O_{1}D=O_{1}M\ .$[/color]

[color=darkblue][b]Proof.[/b] Suppose w.l.o.g. that $AC>AB\ ,$ i.e. $c<b\ .$ Denote : the diameters $[AA']$ , $[AR]$ of the circumcircles for the triangles $ABC$ , $ANP$ respectively ; the middlepoint $S$ of the segment $[DM]\ ;$ the point $L\in BC$ for which $\widehat{BAL}\equiv\widehat{CAL}\ ,$ i.e. $L\in AR\cap BC\ .$
$1.\blacktriangleright$ From the well-known relations $AL=\frac{2bc}{b+c}\cdot\cos\frac{A}{2}$ and $AL\cdot AA'=bc$ obtain the relation $\boxed{\ AA'=\frac{b+c}{2\cos\frac{A}{2}}\ }\ \ (1)\ .$
$2.\blacktriangleright$ Apply the [u]Menelaus' theorem[/u] to the transversal $\overline{PMN}$ in the triangle $ABC\ :$ $\frac{PB}{PA}\cdot\frac{NA}{NC}\cdot\frac{MC}{MB}=1$ $\implies$ $BP=NC=\frac{b-c}{2}$ and $AP=AN=\frac{b+c}{2}\ .$ Thus, $AP=AR\cos\frac{A}{2}$ $\implies$ $\boxed{\ AR=\frac{b+c}{2\cos\frac{A}{2}}\ }\ \ (2)\ .$ From the relations $(1)$ and $(2)$ obtain $R\equiv A'\ .$
Therefore, the line $O_{1}S$ is the bisector of the segment $DM\ ,$ i.e. $O_{1}D=O_{1}M$ and $2\cdot O_{1}S=|AD-MR|\ .$

[b]Remark.[/b] Apply the [u]Stewart's theorem[/u] to the cevian $AM$ in the isosceles triangle $APN\ :$ $AM^{2}+MN\cdot MP=AN^{2}$ $\implies$ $4\cdot MN\cdot MP=$ $(b+c)^{2}-4m^{2}_{a}=$ $b^{2}+c^{2}+2bc-2(b^{2}+c^{2})+a^{2}=$ $a^{2}-(b-c)^{2}=$ $4(p-b)(p-c)$ $\implies$ $MN\cdot MP=(p-b)(p-c)\ ,$ i.e. the power of the point $M$ w.r.t. the circumcircle of the triangle $APN$ is $p(M)=(p-b)(p-c)\ .$ If you show that the power of the point $D$ w.r.t. the circumcircle of the triangle $APN$ is $p(D)=(p-b)(p-c)\ ,$ then $p(D)=p(M)\ ,$ i.e. $O_{1}D=O_{1}M\ .$[/color][/hide]

$A'$ is a point where $AR$ and the circumcle of $ABC$ intersect.


Sorry, I have made mistake in posting, I want to post it in [b]Inequalities Proposed & Own Problems box[/b] because it was solved ;) Who can move my topic?

[quote="zaizai-hoang"]$ \bigstar$ Let $ a,b,c \ge 0$ and such that $ a \plus{} b \plus{} c \equal{} 1$. Prove that:
\[ \frac {a}{\sqrt {a \plus{} 2b}} \plus{} \frac {b}{\sqrt {b \plus{} 2c}} \plus{} \frac {c}{\sqrt {c \plus{} 2a}} \le \frac {2\sqrt {3} \minus{} 2}{\sqrt {2\sqrt {3}}} \plus{} \sqrt { \minus{} 1 \plus{} \frac {2}{\sqrt {3}}}
\]
This one is stronger than a well-known result of Hungkhtn  :wink:
$ \bigstar$ Let $ a,b,c \ge 0$ and such that $ a \plus{} b \plus{} c \equal{} 1$. Prove that:
\[ \frac {a}{\sqrt {a \plus{} 2b}} \plus{} \frac {b}{\sqrt {b \plus{} 2c}} \plus{} \frac {c}{\sqrt {c \plus{} 2a}} < \sqrt {\frac {3}{2}}
\]
[/quote]
You can see the solution in my ebook, Hoang. :)

[quote="can_hang2007"]You can see the solution in my ebook, Hoang. :)[/quote]

Where can I see your ebook?
Thank you very much!

Let try solve it before see any proof for this one. I think it is a nice problem  :wink: The equality holds on
\[ \left(\frac{6\minus{}2\sqrt{3}}{3},\frac{2\sqrt{3}\minus{}3}{3},0\right)\]

There are at least 2 solutions for this one  :wink:

[quote="shfdfzhjj"][quote="zaizai-hoang"]Let try solve it before see any proof for this one. I think it is a nice problem  :wink: The equality holds on
\[ \left(\frac {6 \minus{} 2\sqrt {3}}{3},\frac {2\sqrt {3} \minus{} 3}{3},0\right)
\]
There are at least 2 solutions for this one  :wink:[/quote]

I think f(a,b,c)=<f(a-c,b-c,0)
then it is easy[/quote]
Are you sure? It is not easy to do like that. Don't say anything before you solved it  :wink:

[quote="shfdfzhjj"]I mean x=$ {\sqrt {a \plus{} 2b}}$ ,then f(x,y,z)=<f(x-z,y-z,0)

if it is not true,then i can not solve it. :([/quote]

Okie, I have not said that it was not correct but anyway it is just an idea. The idea is the most important when we solve any problems. But it is not all if you can not use this idea to solve the problem. I means you should try solve it  :wink:

I just see Viet Anh's idea for this problem and I think it's very nice.But of course I can't solve it now with my very terible inequalities skill... :blush:


geometry
symmetry
ratio
similar triangles

If two lines through one vertex of an equilateral triangle divide the semicircle drawn outward on the opposite side into three equal arcs, these same lines divide the side itself into three equal segments.

[hide]
Let $\triangle ABC$ be the equilateral triangle with semicircle outside $\overline{BC}$, $M$ be the midpoint of $\overline{BC}$, $P$ be the point one-third of the way from $C$ to $B$ on the semicircle, and $Q$ be the intersection of $\overline{AP}$ and $\overline{MC}$. By symmetry, it will be enough to prove $BC=3QC$.

Since $\angle ACM=\angle CMP=60^\circ$, we get $AC\parallel MP$ and similar triangles $\triangle ACQ,\ \triangle PMQ$. Also, $AC=2MC=2MP$. Applying the side ratio to the similar triangles, $QC=2MQ$. Then $MC=3MQ$ after adding $MQ$ to both sides. Doubling both sides, $BC=3QC$, as desired.
[/hide]


Prove that
$\sum\frac{1+a^{2}b^{2}}{(a-b)^{2}}>\frac{3}{2}$
where $a,b,c$ are distinct real numbers

when a,b,c,a-d,b-d,c-d>0
(keep a-b,b-c,c-a fixed)
you find that f(a,b,c)>f(a-d,b-d,c-d)
 (d>0)
therefore you only the case f(a,b,0)>3/2 needs to be considered.

.....can anyone finish the proof?


geometry
incenter
circumcircle
parallelogram
geometric transformation
reflection
ratio

(A.Myakishev) Let triangle $ A_1B_1C_1$ be symmetric to $ ABC$ wrt the incenter of its medial triangle. Prove that the orthocenter of $ A_1B_1C_1$ coincides with the circumcenter of the triangle formed by the excenters of $ ABC$.

Denote D the midpoint of AB
it's easy to show C',D,I are collinear C'D=DI  AC'BI is a parallelogram
=> C' is the orthocenter of △OcAB  => Oc,C',H are collinear
Analogously Oa,A',H  Ob,B',H are collinear
∠HOcB=∠HOaB=1/2∠ABC     => HOa=HOc
Analogously HOa=HOc=HOb
H is the circumcenter of  △OaObOc

We denote by $S$ the incentre of the medial triangle and by $I,G,H,O$ the incentre,centroid,orthocentre,circumcentre of $\triangle ABC$ respectively.Now we know that $I$ is the orthocentre of the excentric triangle.This can be easily shown by using the fact that the internal bisectors are perpendicular with the external ones and the excentres are collinear with the incentre and the corresponding vertices
Hence $\triangle ABC$ is the orthic triangle of its excentric triangle,which implies that $O$ is the nine point centre of the excentric triangle.Hence the circumcentre of the excentric triangle is the reflection of $I$ over $O$,say $P$.

Now $\triangle A'B'C'$ is homothetic with $\triangle ABC$ with centre $S$ and ratio 1.Hence the orthocentre of $\triangle A'B'C'$ is the reflection of
$H$ over $S$,say $Q$.We have to show that $P$ and $Q$ are the same point.
Now the medial triangle is homothetic with $\triangle ABC$ with centre $G$ and ratio $\frac{-1}{2}$.Hence $I,G,S$ are collinear and $\frac{IG}{GS}=2$.Also we know that $\frac{HG}{OG}=2$ and $G$ lies on $OH$.Hence in triangles  $OGS$ and $HGI$ we have \[\frac{IG}{GS}=\frac{HG}{GO}=2\].Hence these two triangles are similar,which implies $OS\parallel IH$ and$2OS=IH$.Now let $IO$ and $HS$ intersect at $R$.Now from the previous relation we easily get that $IO=OR$ and $HS=SR$,which implies that $P,Q,R$ all are the same points,which was to be shown.

This is just restatement of the well known result:

[i][quote]In any triangle, the Spieker point is the midpoint of the orthocenter and the Bevan point. [/quote][/i]

See [url= http://www.amazon.com/Episodes-Nineteenth-Twentieth-Euclidean-Mathematical/dp/0883856395 ]Episodes in Nineteenth and Twentieth Century Euclidean Geometry , Ross Honsberger[/url] for a proof.


I recently picked this game up again and blew all my high scores away.
Test: 1931
Hard Written math:782
(two of my favorite scores)

Discussion, anyone?

Oh yeah, that game... haven't played in like a year.

I did have 2030 on test, 749 on written math hard, and gold/platinum on all practices.

The ideal test is (for me)
Written math
number memory
pathfinder
missing link
matchmaker

If I get 4/5 of those I can score consistently over 1800.
Do you get anything special for getting 2000+?

Well, I got number memory and written math. For me, that is 1100-1200 right there. As long as I get those two and no missing link, I can usually score 1900+. Eventually, I got lucky with guessing and got 2030.

My best is written math and number memory.  I like the one with the dog bones.

Please don't revive topics that haven't been posted on since last year. It makes a whole entire discussion.
BTW, I only have the DS version.


If $ P$ be any point on the circumcircle of a triangle $ ABC$ whose Lemoine point is $ K$, show that the line $ PK$ will cut the sides $ BC$, $ CA$, $ AB$ of the triangle in points $ X$, $ Y$, $ Z$ so that
\[ \frac{3}{{PK}}\equal{}\frac{1}{{PX}}\plus{}\frac{1}{{PY}}\plus{}\frac{1}{{PZ}}\]
where the segments are directed.

I have get the terms of this problem from the book [i]R.Lachlan-An elementary treatise on modern pure geometry [/i](pag.67, § 119, Ex.9).

Leon

[b]Theorem 1.[/b] Let P be a point on the circumcircle of a triangle ABC, let K be the symmedian point of triangle ABC, and let X, Y, Z be the points where the line PK meets the sidelines BC, CA, AB of triangle ABC. Then, using directed segments, we have

$ \frac{3}{PK}=\frac{1}{PX}+\frac{1}{PY}+\frac{1}{PZ}$.

[i]Proof.[/i] Let $a = BC$, $b = CA$, $c = AB$ denote the sidelengths of triangle ABC. Also, for every point T in the plane of triangle ABC, let $ d_{a}\left(T\right)$ denote the directed distance from the point T to the line BC; here, this directed distance is considered to be positive if the points T and A lie in the same halfplane with respect to the line BC, and negative if the points T and A lie in different halfplanes with respect to the line BC. Similarly, we define the directed distances $ d_{b}\left(T\right)$ and $ d_{c}\left(T\right)$ from the point T to the lines CA and AB, respectively.

Now, as a lemma, we show:

[b]Theorem 2.[/b] For every point P on the circumcircle of triangle ABC, we have

$ \frac{a}{d_{a}\left(P\right)}+\frac{b}{d_{b}\left(P\right)}+\frac{c}{d_{c}\left(P\right)}=0$.

[i]Proof.[/i] We WLOG assume that the point P lies on the arc BC of the circumcircle of triangle ABC. Let D, E, F be the orthogonal projections of the point P on the sidelines BC, CA, AB of triangle ABC. Again we WLOG assume that the projections D and E lie inside the segments BC and CA, respectively, while the projection F lies on the extension of the segment AB beyound B.

In the following, we will work with non-directed segments. Since the points D, E, F are the orthogonal projections of the point P on the sidelines BC, CA, AB of triangle ABC, the lengths of the segments PD, PE, PF are the non-directed distances from the point P to the sidelines BC, CA, AB. Remembering the sign convention for the directed distances, we see that $ d_{a}\left(P\right)=-PD$, $ d_{b}\left(P\right)=PE$, $ d_{c}\left(P\right)=PF$. Thus, the equation we have to prove,

$ \frac{a}{d_{a}\left(P\right)}+\frac{b}{d_{b}\left(P\right)}+\frac{c}{d_{c}\left(P\right)}=0$,

rewrites as

$ \frac{a}{-PD}+\frac{b}{PE}+\frac{c}{PF}=0$.

In other words, we have to show that

$ -\frac{a}{PD}+\frac{b}{PE}+\frac{c}{PF}=0$.

Now, since the point P lies on the circumcircle of triangle ABC, we have < PAC = < PBC, or, in other words, < PAE = < PBD. Also, from the construction of the points D and E, it is clear that < PEA = 90° = < PDB. Thus, the triangles PAE and PBD are similar; consequently, $ \frac{AE}{PE}=\frac{BD}{PD}$. Similarly, $ \frac{BF}{PF}=\frac{CE}{PE}$ and $ \frac{CD}{PD}=\frac{AF}{PF}$. These equations yield

$ -\frac{BD}{PD}-\frac{CD}{PD}+\frac{CE}{PE}+\frac{AE}{PE}+\frac{AF}{PF}-\frac{BF}{PF}=0$;

in other words,

$ -\frac{BD+CD}{PD}+\frac{CE+AE}{PE}+\frac{AF-BF}{PF}=0$,

But since BD + CD = BC = a, CE + AE = CA = b and AF - BF = AB = c, we thus have

$ -\frac{a}{PD}+\frac{b}{PE}+\frac{c}{PF}=0$.

This completes the proof of Theorem 2.

Now we are going to prove Theorem 1, using directed segments again.

We have to establish the equation

$ \frac{3}{PK}=\frac{1}{PX}+\frac{1}{PY}+\frac{1}{PZ}$.

Multiplication by PK transforms this equation into

$ 3=\frac{PK}{PX}+\frac{PK}{PY}+\frac{PK}{PZ}$.

But actually,

$ \frac{PK}{PX}+\frac{PK}{PY}+\frac{PK}{PZ}=\frac{PX-KX}{PX}+\frac{PY-KY}{PY}+\frac{PZ-KZ}{PZ}$
$ =\left(1-\frac{KX}{PX}\right)+\left(1-\frac{KY}{PY}\right)+\left(1-\frac{KZ}{PZ}\right) = 3-\left(\frac{KX}{PX}+\frac{KY}{PY}+\frac{KZ}{PZ}\right)$.

Hence, in order to prove $ 3=\frac{PK}{PX}+\frac{PK}{PY}+\frac{PK}{PZ}$, it is enough to show that

$ \frac{KX}{PX}+\frac{KY}{PY}+\frac{KZ}{PZ}=0$.

But Thales yields $ \frac{KX}{PX}=\frac{d_{a}\left(K\right)}{d_{a}\left(P\right)}$, $ \frac{KY}{PY}=\frac{d_{b}\left(K\right)}{d_{b}\left(P\right)}$, $ \frac{KZ}{PZ}=\frac{d_{c}\left(K\right)}{d_{c}\left(P\right)}$, so that it remains to prove that

$ \frac{d_{a}\left(K\right)}{d_{a}\left(P\right)}+\frac{d_{b}\left(K\right)}{d_{b}\left(P\right)}+\frac{d_{c}\left(K\right)}{d_{c}\left(P\right)}=0$.

Now, it is well-known that the distances from the symmedian point of a triangle to the triangle's sides are proportional to these sides; thus, we have $ d_{a}\left(K\right): d_{b}\left(K\right): d_{c}\left(K\right)=a: b: c$, and consequently, the equation

$ \frac{d_{a}\left(K\right)}{d_{a}\left(P\right)}+\frac{d_{b}\left(K\right)}{d_{b}\left(P\right)}+\frac{d_{c}\left(K\right)}{d_{c}\left(P\right)}=0$

which we must prove turns out to be equivalent to the equation

$ \frac{a}{d_{a}\left(P\right)}+\frac{b}{d_{b}\left(P\right)}+\frac{c}{d_{c}\left(P\right)}=0$.

But this equation follows from Theorem 2.

This completes the proof of Theorem 1.

[b]PS.[/b] Both Theorem 1 and Theorem 2 have converses: For any arbitrary point P in the plane of triangle ABC, each of the equations $ \frac{3}{PK}=\frac{1}{PX}+\frac{1}{PY}+\frac{1}{PZ}$ and $ \frac{a}{d_{a}\left(P\right)}+\frac{b}{d_{b}\left(P\right)}+\frac{c}{d_{c}\left(P\right)}=0$ is a necessary and sufficient condition for the point P to lie on the circumcircle of triangle ABC. Proving the converse of Theorem 2 is harder than proving Theorem 2 itself; but then, after it is proven, the converse of Theorem 1 can be derived from it similarly to the way we derived Theorem 1 from Theorem 2 above.

[b]PPS.[/b] Nice to meet you here again, Leon!

  Darij

Your solution is too long, we can prove it by using 2 lemma:
1) If P is the point lies on circumcircle of ABC x,y,z are the distances from P to BC, CA, AB and a,b,c are the lengths of triangle's side then:
[tex] \frac{a}{x}= \frac{b}{y} +\frac{c}{z} [/tex]

2) If K is Lemoin piont of ABC and p,q,r is the distances from K to BC, CA, AB then we have
[tex] \frac{p}{a} = \frac{q}{b} = \frac{r}{c} [/tex]

Silva, that's exactly what I did in my post, but I wrote it up with more details.

  Darij

[b]Here is my proof[/b] of the [u]second [b]Darij[/b]'s theorem[/u].

[color=darkred][b][u]PP[/u].[/b] Let $\triangle ABC$ with the circumcircle $w$ and $P\in w$ . Denote $\left\{\begin{array}{cc}
X\in BC\ : & PX\perp BC\\\
Y\in CA\ ; & PY\perp CA\\\
Z\in AB\ ; & PZ\perp AB\end{array}\right\|$ .

Suppose w.l.o.g. that $A\in (BZ)$ and $AC$ separates $B$ , $P$ .Prove that $\boxed{\frac a{PX}+\frac c{PZ}=\frac b{PY}}$ .[/color]

[b]Proof ([u]own[/u]).[/b] I"ll use the well-known relation $\boxed{bc=2Rh_a}\ (*)$ in the triangles $\left\{\begin{array}{cc}
\triangle BPC\ : & PB\cdot PC=2R\cdot PX\\\\
\triangle CPA\ : & PC\cdot PA=2R\cdot PY\\\\
\triangle APB\ : & PA\cdot PB=2R\cdot PZ\end{array}\right\|\ (*)$ .

Apply the [u][b]Ptolemy[/b]'s theorem[/u] in $ABCP\ :$ $a\cdot PA+c\cdot PC=b\cdot PB\iff$ $\frac a{PB\cdot PC}+\frac c{PA\cdot PB}=\frac b{PC\cdot PA}\stackrel{(*)}{\iff}$

$\frac a{2R\cdot PX}+\frac c{2R\cdot PZ}=\frac b{2R\cdot PY}\iff$ $\frac a{PX}+\frac c{ PZ}=\frac b{PY}$  .


When a ~ is used alone inside $ \text{\LaTeX}$ here, weird things happen. Putting one over another character with \tilde{stuff} works fine, but using one alone shifts the whole line up and changes how any other $ \text{\LaTeX}$ on that line is rendered.

An example [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=983784#983784]here[/url].

I went back to the post jmerry linked to. I "edited" his post by hitting edit and then submit, not changing anything. It now works.


geometry
trigonometry
trig identities
Law of Cosines

Let x be the measure of angle AMB.  Then 180-x is the measure of angle AMC.  Let $a$ be the length of BM.  Then it's also the length of MC.

By the law of cosines, with triangle AMB:

$8^2=5^2+a^2-2(5)(a)(\cos{x})$
$64=25+a^2-10a\cos{x}$

Also by the law of cosines, with triangle AMC:

$6^2=5^2+a^2-2(5)(a)(\cos{180-x})$
$36=25+a^2+10a\cos{x}$

Adding the two equations,

$100=50+2a^2$
$a=5$
$BC=10$

By the law of cosines, the angle ABC has measure $\cos^{-1}({\frac{4}{5}})$ which is rougly $36.87$ degrees.

[quote="Ceva Theorem"]In triangle ABC, the median through A meets BC at M. Given that AB = 8 cm, AM = 5 cm, and AC = 6 cm, find BC and the measurement of angle ABC

[hide]hint: Use the law of cosine or sine[/hide][/quote]

To make it more challenging, posting direct hints are discouraged.  Putting hints like this is like telling someone the answer.  :D

[hide]ABC is a right triangle 6-8-10. $\sin B=3/5$, so $B\approx 36.87$.[/hide]

[quote="pianoforte"]ABC is a right triangle 6-8-10.[/quote]
What did you use to find this? Law of Cosines?


algebra
polynomial
calculus
integration
Gauss
number theory
prime numbers

Given distinct prime numbers $p$ and $q$ and a natural number $n \geq 3$, find all $a \in \mathbb{Z}$ such that the polynomial $f(x) = x^n + ax^{n-1} + pq$ can be factored into 2 integral polynomials of degree at least 1.

[color=blue][b]Generalization.[/b] Given two integers $ p$ and $ q$ and a natural number $ n \geq 3$ such that $ p$ is prime and $ q$ is squarefree, and such that $ p\nmid q$.
Find all $ a \in \mathbb{Z}$ such that the polynomial $ f(x) \equal{} x^n \plus{} ax^{n \minus{} 1} \plus{} pq$ can be factored into 2 integral polynomials of degree at least 1.[/color]

[i]Solution.[/i] I hope the following solution is correct. It is more or less a straightforward generalization of [url=http://www.kalva.demon.co.uk/imo/isoln/isoln931.html]IMO 1993 problem 1[/url].

The idea behind is an extension of Eisenstein's criterion for irreducible polynomials:

[color=blue][b]Lemma 1.[/b] Let p be a prime number. If a polynomial $ A\left(x\right) \equal{} a_nx^n \plus{} a_{n \minus{} 1}x^{n \minus{} 1} \plus{} ... \plus{} a_1x \plus{} a_0$ with integer coefficients $ a_n$, $ a_{n \minus{} 1}$, ..., $ a_1$, $ a_0$ is reducible in $ \mathbb{Z}\left[x\right]$, and the prime p divides the coefficients $ a_0$, $ a_1$, ..., $ a_{n \minus{} 2}$, but does not divide $ a_n$, and $ p^2$ does not divide $ a_0$, then p does not divide $ a_{n \minus{} 1}$, and the polynomial A(x) must have a rational root.[/color]

[i]Proof of Lemma 1.[/i] Since the polynomial A(x) is reducible in $ \mathbb{Z}\left[x\right]$, we can write it in the form A(x) = B(x) C(x), where $ B\left(x\right) \equal{} b_ux^u \plus{} ... \plus{} b_1x \plus{} b_0$ and $ C\left(x\right) \equal{} c_vx^v \plus{} ... \plus{} c_1x \plus{} c_0$ are non-constant polynomials with integer coefficients $ b_u$, ..., $ b_1$, $ b_0$, $ c_v$, ..., $ c_1$, $ c_0$. Then, for any i, we have $ a_i \equal{} \sum_{k \equal{} 0}^i b_kc_{i \minus{} k}$ (this follows from multiplying out the equation A(x) = B(x) C(x)). Particularly, $ a_0 \equal{} b_0c_0$. But since the integer $ a_0$ is divisible by the prime p, but not by $ p^2$, this yields that one of the integers $ b_0$ and $ c_0$ is divisible by p, and the other one is not. WLOG assume that $ b_0$ is divisible by p, and $ c_0$ is not.

Not all coefficients $ b_u$, ..., $ b_1$, $ b_0$ of the polynomial B(x) can be divisible by p (else, $ a_n \equal{} \sum_{k \equal{} 0}^n b_kc_{n \minus{} k}$ would also be divisible by p, what is excluded). Let $ \lambda$ be the least nonnegative integer such that the coefficient $ b_{\lambda}$ is [i]not[/i] divisible by p. Then, all the integers $ b_k$ with $ k < \lambda$ are divisible by p. Hence, in the sum $ a_{\lambda} \equal{} \sum_{k \equal{} 0}^{\lambda} b_kc_{\lambda \minus{} k}$, all the summands $ b_kc_{\lambda \minus{} k}$ with $ k < \lambda$ are divisible by p, but the summand $ b_{\lambda}c_0$ (this is the summand for $ k \equal{} \lambda$) is not (since $ b_{\lambda}$ is not divisible by p, and neither is $ c_0$). Hence, the whole sum $ a_{\lambda}$ is not divisible by p. But we know that the coefficients $ a_0$, $ a_1$, ..., $ a_{n \minus{} 2}$ are all divisible by p; hence, $ a_{\lambda}$ must be one of the coefficients $ a_{n \minus{} 1}$ and $ a_n$. Thus, either $ \lambda \equal{} n \minus{} 1$ or $ \lambda \equal{} n$.

If $ \lambda \equal{} n$, then it follows, since the integer $ b_{\lambda}$ is defined, that the polynomial B(x) has a coefficient $ b_n$. In other words, the polynomial B(x) has degree n. Since the polynomial A(x) has degree n, too, it follows from A(x) = B(x) C(x) that the polynomial C(x) is a constant. This is a contradiction.

Thus, we must have $ \lambda \equal{} n \minus{} 1$. Hence, the integer $ a_{n \minus{} 1} \equal{} a_{\lambda}$ is not divisible by p. Also, since the integer $ b_{\lambda}$ is defined, it follows that the polynomial B(x) has a coefficient $ b_{n \minus{} 1}$. In other words, the polynomial B(x) has degree $ \geq n \minus{} 1$. Since the polynomial A(x) has degree n and A(x) = B(x) C(x), this yields that the polynomial C(x) has degree $ \leq 1$. The degree cannot be 0, since the polynomial C(x) is not constant; thus, the degree is 1. Hence, the polynomial A(x) has a linear factor, i. e. it has a rational root. Lemma 1 is proven.

Now let us solve the problem: The number $ pq$ is squarefree (since $ p$ is prime and $ q$ is squarefree, and since $ p\nmid q$).

Applying Lemma 1 to the polynomial $ x^n \plus{} ax^{n \minus{} 1} \plus{} pq$, using the prime p, we see that, if this polynomial can be factored into two non-constant integral polynomials, then it must have a rational root. Since it is a monic polynomial with integer coefficients, it thus must have an integer root. If we denote this root by $ r$, then $ r^n \plus{} ar^{n \minus{} 1} \plus{} pq \equal{} 0$, so that $ pq \equal{} \minus{} r^n \minus{} ar^{n \minus{} 1} \equal{} \minus{} \left(r \plus{} a\right) r^{n \minus{} 1}$ is divisible by $ r^2$ (since $ n\geq 3$ yields $ n \minus{} 1\geq 2$, so that $ r^{n \minus{} 1}$ is divisible by $ r^2$), so that $ r \equal{} 1$ or $ r \equal{} \minus{} 1$ (since $ pq$ is squarefree), so that one of the numbers $ 1$ and $ \minus{} 1$ must be a root of the polynomial $ x^n \plus{} ax^{n \minus{} 1} \plus{} pq$.

Hence, we see that, if the polynomial $ x^n \plus{} ax^{n \minus{} 1} \plus{} pq$ can be factored into two non-constant integral polynomials, then one of the numbers $ 1$ and $ \minus{} 1$ must be a root of this polynomial. Conversely, if one of the numbers $ 1$ and $ \minus{} 1$ is a root of this polynomial, then it has an integer root and thus can be factored into two non-constant integral polynomials. Hence, in order to solve the problem, it remains to find all integers a such that one of the numbers $ 1$ and $ \minus{} 1$ is a root of the polynomial $ x^n \plus{} ax^{n \minus{} 1} \plus{} pq$. But in fact, $ 1$ is a root of the polynomial $ x^n \plus{} ax^{n \minus{} 1} \plus{} pq$ if and only if $ 1^n \plus{} a\cdot 1^{n \minus{} 1} \plus{} pq \equal{} 0$, what is equivalent to $ 1 \plus{} a \plus{} pq \equal{} 0$, i. e. to $ a \equal{} \minus{} 1 \minus{} pq$, and $ \minus{} 1$ is a root of the polynomial $ x^n \plus{} ax^{n \minus{} 1} \plus{} pq$ if and only if $ \left( \minus{} 1\right)^n \plus{} a\cdot\left( \minus{} 1\right)^{n \minus{} 1} \plus{} pq \equal{} 0$, what is equivalent to $ a \equal{} 1 \plus{} \left( \minus{} 1\right)^n pq$. So, the two required values of $ a$ are $ a \equal{} \minus{} 1 \minus{} pq$ and $ a \equal{} 1 \plus{} \left( \minus{} 1\right)^n pq$.

The problem is thus solved.

[hide="Old version of the solution, not generalizing the problem"]

[i]Old version of the solution (of the original problem, not of the generalization).[/i]

Applying Lemma 1 to the polynomial $ x^n \plus{} ax^{n \minus{} 1} \plus{} pq$, using the prime p, we see that, if this polynomial can be factored into two non-constant integral polynomials, then it must have a rational root. Since it is a monic polynomial with integer coefficients, it thus must have an integer root, and by a well-known theorem, this integer root then must be a divisor of pq. This means that the root is one of the numbers pq, p, q, 1, -pq, -p, -q, -1. Actually, none of the numbers pq, p, q, -pq, -p, -q can be a root of the polynomial $ x^n \plus{} ax^{n \minus{} 1} \plus{} pq$ (in fact, every of these numbers is divisible by p or by q, and if an integer root r of the polynomial $ x^n \plus{} ax^{n \minus{} 1} \plus{} pq$ would be divisible by p, then $ r^n \plus{} ar^{n \minus{} 1}$ would be divisible by $ p^{n \minus{} 1}$, while $ pq$ wouldn't be because of $ n\geq 3$, so $ r^n \plus{} ar^{n \minus{} 1} \plus{} pq$ couldn't be 0, what yields a contradiction, and similarly we obtain a contradiction if an integer root would be divisible by q). Hence, only 1 and -1 remain as possible candidates for integer roots of the polynomial $ x^n \plus{} ax^{n \minus{} 1} \plus{} pq$.

Hence, we see that, if the polynomial $ x^n \plus{} ax^{n \minus{} 1} \plus{} pq$ can be factored into two non-constant integral polynomials, then one of the numbers 1 and -1 must be a root of this polynomial. Conversely, if one of the numbers 1 and -1 is a root of this polynomial, then it has an integer root and thus can be factored into two non-constant integral polynomials. Hence, in order to solve the problem, it remains to find all integers a such that one of the numbers 1 and -1 is a root of the polynomial $ x^n \plus{} ax^{n \minus{} 1} \plus{} pq$. But in fact, 1 is a root of the polynomial $ x^n \plus{} ax^{n \minus{} 1} \plus{} pq$ if and only if $ 1^n \plus{} a\cdot 1^{n \minus{} 1} \plus{} pq \equal{} 0$, what is equivalent to $ 1 \plus{} a \plus{} pq \equal{} 0$, i. e. to a = - 1 - pq, and -1 is a root of the polynomial $ x^n \plus{} ax^{n \minus{} 1} \plus{} pq$ if and only if $ \left( \minus{} 1\right)^n \plus{} a\cdot\left( \minus{} 1\right)^{n \minus{} 1} \plus{} pq \equal{} 0$, what is equivalent to $ a \equal{} 1 \plus{} \left( \minus{} 1\right)^n pq$. So, the two required values of a are a = - 1 - pq and $ a \equal{} 1 \plus{} \left( \minus{} 1\right)^n pq$.
[/hide]

  Darij

Different way:
if we reduce the polynomial in $\mathbb{Z}/p\mathbb{Z}$, it becomes $x^n+ax^{n-1}=x^{n-1}(x+a)$. Since both x and x+a are irreducible in $\mathbb{F}_p[x]$, suppose f(x)=g(x)h(x). WLOG, $g(x)=x^k+pg_1(X)$ and $h(x)=x^{n-1-k}(x+a)+ph_1(x)$, where $g_1, h_1 \in \mathbb{Z}[x]$. Multiplying these two gives $f(x)=x^{n-1}(x+a)+pg_1(x)x^{n-1-k}(x+a)+ph_1(x)x^k+p^2g_1(x)h_1(x)$. If 0<k<n-1, we get a contradiction when x=0. We know that k cannot equal zero. Thus, the only possible cases are when k is n-1. In this case, f must have an integer root. Thus, our solution is all a such that f has an integer root.

EDIT: This solution is actually identical to my solution of IMO 1993 #1. In fact, I think it can be generalized to any polynomial of the form $x^n+ax^{n-1}+k$, where there exists a prime p such that $v_p(k)=1$.

edit of the edit: I actually think the original problem says irreducible in Q[x]? But then it's just one extra step using gauss' lemma and the fact that it's primitive.


geometry
3D geometry
prism
Pythagorean Theorem

A diagonal of the front face of a rectangular pnsm is 13 inches long, and a diagonal of the top face of the same prism is 15 inches long. The height of the front face of the prism is 5 inches long. How many cubic inches are in the volume of the prism if each of the dimensions is an integer length?

We know that, on the front face, one side is 5 and the diagonal is 13. Using Pythagorean Theorem (the 5-12-13 triple) we get that the other side is 12. On the top face, we have one side that is 12 and the diagonal of 15. Using the 9-12-15 (3-4-5) triple, we find that the last side is 20. We just multiply these to get:
$ 5\times12\times20\equal{}\boxed{1200}$

Thus, 1200 is our answer.

Nice try, PowerOfPi.
$ pythag(5, x, 13)\equal{}12.$
$ pythag(x, 12, 15)\equal{}9.$
$ 5\times9\times12\equal{}540.$
QED


counting
derangement
search
floor function

HI all,

If the digits 1,2,3 4 and 5 are used to put in a row. If 1 is not at 1st, 2 is not at 2nd, 3 is not at 3rd, 4th is not at 4th and 5 is not at 5th position. how many ways of arranging?

There are 4 choices for the 1st place(2,3,4,5) and same number of choices for the other 5 places. Therefore the answer should be 4^5(shouldn't it?).

No, because those choices are not independent: after you've placed 1, you've restricted the placement of most of the other letters.  This is the [[derangement]] problem -- try a search of the forum, the wiki or the internet to learn more.

I give it a try. The total number of arrangements is 5! = 120. I tried to count how many of those would not satisfy the conditions. So I used PIE (Principle of Inclusion and Exclusion).

So first I counted how many arrangements there are that have at least one number appear in its place. that one number can be anyone of the 5 numbers and there are 4! ways to arrange the remaining four numbers so there are 4!*5=120 arrangements.

But we counted twice the arrangements which have 2 numbers appear in their correct place. so we need to subtract them. there are 5!/(2!(5-2)!)=10 ways to chose the 2 numbers which appear in their correct places. there are 3!=6 ways to arrange the remaining 3 numbers so there are 6*10=60 arrangements we need to subtract.

But we will be subtracting twice any arrangements with 3 numbers in their correct placements. so we need to re-add those. There are 5!/((5-3)!3!)=10 ways to chose the 3 numbers that appear in the correct placements. there are 2!=2 ways to arrange the remaining 2 numbers. so we need to add 2*10=20 arrangements.

But we are adding twice any arrangements that feature four numbers in the correct places. so we need to re-subtract those. there are 5 ways to chose the four numbers that sill be in the correct places. there is one way to arrange the remaining number. so there are 5*1=5 arrangements that will be subtracted.

finally we need to re-add the single case where all five numbers are in the correct position.

So there are 120-60+20-5+1=76 arrangements in which one or more numbers are in the correct position. So there are 120-76 = 44 arrangements in which 1 isn't first, 2 isn't second, etc.

[quote="bertram"]There are about $ \frac {5!}{e}$ ;)[/quote]

5!/e=44.1455 which is almost exactly 44. What theorem gives you that answer?

It can be proven using the taylor series for $ e^x$ about $ 0$, evaluated at $ x\equal{}1$. You can prove that you get the correct answer using the lagrange error bound (or probably the alternating series test).

I think he was referring to the Derangement Theorem.
(not sure if this is the proper name)

Loosely stated (VERY loosely) it says that the number of derangements of a set group of objects of cardinality N (where a derangement is defined as no object being in it's initial position) is asymptotically close (as $ n \to \infty$) to $ \left \lfloor \frac{n!}{e} \right \rfloor$

[quote="Altheman"]It can be proven using the taylor series for $ e^x$ about $ 0$, evaluated at $ x \equal{} 1$. You can prove that you get the correct answer using the lagrange error bound (or probably the alternating series test).[/quote]

I realized that later. thanks though.


Let  x_i , y_i>0. Prove  that
[Σ(xi+yi)^2]*[Σ(1/(χi*yi)]>=4n^2
Abel  Contest , Norway  1995 Thanks

Appears to be rather easy, no ?

(xi+yi)^2 >= 4xi*yi + Cauchy-Schwarz : sum ai sum 1/ai >= n


calculus
derivative
videos
calculus computations

A light is mounted on a pole 15 feet high. A six foot tall man is walking in a straight line away from the pole at 5 feet per second. I am asked to find the speed of the leading edge of his shadow as he passes a point 40 feet from the pole.

do i take an equation that defines the position of the man's shadow and then take the derivative?

(Calculus problems shouldn't be posted in this forum).

I remember a problem eerily similar to this being solved with derivatives before. I certainly wouldn't know how to solve it without using calculus in any case, and it looks like the problem was set up for you to use calculus, so, yes, you're certainly along the right lines.

There is a [url=http://www.midnighttutor.com/shadow_tightrope.html]video to that problem I mentioned available here[/url] - because I'm sad enough to remember these things from when I've watched them - that explains a problem almost exactly like it (a guy on a tightrope with a light and his speed), so you might find that useful to watch if you can as a kind of background.

Sorry, i cannot follow the solution on video. Could the moderator post my problem in the calculus & Tutorial forum? Sorry for posting in the wrong forum.


HI
I've a problem, this is took of Stanley's book "a course in enumeration" , I had a solution but I'm not sure...

In how many ways can one list the numbers 1,2,3, .... n such that apart from the leading element the number k can be placed only if either k or k-1 already appears?

For n >= 5 I think the answer is (n-1)2^{n-1} + 1.
Thanks in advance for any suggestion.

[quote="reben"]

In how many ways can one list the numbers 1,2,3, .... n such that... [/quote]

What do you mean by "list"? In most cases, I would assume it means a permutation of the numbers, but here it doesn't seem to be the case...

I mean, put in a list, e.g  3 4 5 2 1  or   4 3 2 1 5 are in the list, but not 2 4 5 3 1.

With silly typo corrected:
[quote="reben"]In how many ways can one list the numbers 1,2,3, ... $ n$ such that apart from the leading element the number $ k$ can be placed only if either $ k \plus{} 1$ or $ k \minus{} 1$ already appears?[/quote]

There's no way that $ (n \minus{} 1)2^{n \minus{} 1} \plus{} 1$ is going to be correct: the number of these permutations starting with $ k$ will be always at most $ 2^{n \minus{} 2}$ (for each of the subsequent numbers, you have at most two choices -- the largest number smaller than all those you've picked, or the smallest number larger than all those you've picked -- and for the last number you only get one choice) and $ n\cdot 2^{n \minus{} 2} < (n \minus{} 1)2^{n \minus{} 1}$ for all positive integers $ n$.

There are basically two natural ways to proceed:

[hide="Recursively"]Let $ a_n(k)$ be the number of such permutations beginning with $ k$, with $ a_n(0): \equal{} 0$.  Then, for example, $ a_n(1) \equal{} a_n(n) \equal{} 1$.  Show that $ a_{n \plus{} 1}(k) \equal{} a_n(k \minus{} 1) \plus{} a_n(k)$.  Use this to express $ \sum_k a_{n \plus{} 1}(k)$ in terms of $ 2\sum_{k} a_n(k)$.[/hide]

[hide="Cleverly"]Try building your permutations starting from the other end -- what can you choose as the last possible element?[/hide]

Edit: Well, I guess there's at least a third nice way as well -- see the next post :)

[quote="reben"][...]
In how many ways can one list the numbers 1,2,3, .... n such that apart from the leading element the number k can be placed only if either k or k-1 already appears[...][/quote]

I think you mean "if either [b]k+1[/b] or k-1 already appears".

[hide="Solution."]
=====
At every step we choose a next number, we can only choose to take the next lower number or the next higher number than the numbers currently in the list (note that one option will not be possible if $ 1$ or $ n$ already appeared). Thus every choice of next number can be represented by either H or L; that is, whether the next number is the next higher number than the highest in the current list or the next lower number than the lowest in the current list. We can therefore express our list as a sequence of $ n \minus{} 1$ Hs and Ls (since the first number does not get one).

Note that, if given a sequence of Hs and Ls, we can determine the first number by taking one more than the number of Ls, since we can only go as low as $ 1$. Conversely, given the first number, the sequence must have one less than this in number of Ls. Therefore every sequence of Hs and Ls corresponds to such a listing, of which there are $ \boxed{2^{n \minus{} 1}}$.
=====
[/hide]


Integers $ a_1,a_2,\ldots a_9$ satisfy the relations $ a_{k\plus{}1}\equal{}a_k^3\plus{}a_k^2\plus{}a_k\plus{}2$ for $ k\equal{}1,2,...,8$. Prove that among these numbers there exist three with a common divisor greater than $ 1$.

mod 5,

5->2->1

3->1

4->1

Easy to show 3 are $ \equiv 0$ mod 5

rofler's representations:
[hide]
$ \text{when}$
 $ a_1\equiv 0\pmod 5 , \implies  a_2\equiv 2\pmod 5 \implies a_3\equiv 1\pmod 5$                                                $ \implies   a_4\equiv 0\pmod 5\implies a_5\equiv 2\pmod 5\implies a_6\equiv 1\pmod 5$
 $ \implies a_7\equiv 0\pmod 5 , \text {i.e. }  a_1,a_4,a_7 \text{has common factor 5}$

$ \text{when}$
$ a_1\equiv 1\pmod 5 , \text {in brief   } 1 \to 0 \to 2 \to 1 \to 0 \to 2 \to 1 \to 0 , \text {i.e. }  a_2,a_5,a_8 \text{has common factor 5}$

$ \text{when}$
$ a_1\equiv 2\pmod 5 ,  \text {in brief   } 2 \to 1 \to 0 \to 2 \to 1 \to 0 \to 2 \to 1\to 0 , \text {i.e. }  a_3,a_6,a_9 \text{has common factor 5}$

$ \text{when}$
$ a_1\equiv 3\pmod 5 , \text {in brief   }  3  \to 1 \to 0 \to 2 \to 1\to 0 \to 2 \to 1 \to 0  ,\text {i.e. }  a_3,a_6,a_9 \text{has common factor 5}$

$ \text{when}$
$ a_1\equiv 4\pmod 5 , \text {in brief   }  4  \to 1 \to 0 \to 2 \to 1\to 0 \to 2 \to 1 \to 0 ,\text {i.e. }  a_3,a_6,a_9 \text{has common factor 5}$

[/hide]


prove the following inequality for all a,b,c real :
$ \frac{a^2(7\minus{}b)(7\minus{}c)}{(a\minus{}b)(a\minus{}c)}\plus{}\frac{b^2(7\minus{}c)(7\minus{}a)}{(b\minus{}a)(b\minus{}c)}\plus{}\frac{c^2(7\minus{}a)(7\minus{}b)}{(c\minus{}b)(c\minus{}a)} \leq 7\plus{}21[\frac{(a\plus{}b)^2}{(a\minus{}b)^2}\plus{}\frac{(b\plus{}c)^2}{(b\minus{}c)^2}\plus{}\frac{(a\plus{}c)^2}{(a\minus{}c)^2}]$

[hide][quote="bertram"][quote="arqady"][quote="HTA"]prove the following inequality for all a,b,c real :
$ \frac {a^2(7 \minus{} b)(7 \minus{} c)}{(a \minus{} b)(a \minus{} c)} \plus{} \frac {b^2(7 \minus{} c)(7 \minus{} a)}{(b \minus{} a)(b \minus{} c)} \plus{} \frac {c^2(7 \minus{} a)(7 \minus{} b)}{(c \minus{} b)(c \minus{} a)} \leq 7 \plus{} 21[\frac {(a \plus{} b)^2}{(a \minus{} b)^2} \plus{} \frac {(b \plus{} c)^2}{(b \minus{} c)^2} \plus{} \frac {(a \plus{} c)^2}{(a \minus{} c)^2}]$[/quote]
Id est, we need to prove that $ \frac {(a \plus{} b)^2}{(a \minus{} b)^2} \plus{} \frac {(b \plus{} c)^2}{(b \minus{} c)^2} \plus{} \frac {(a \plus{} c)^2}{(a \minus{} c)^2}\geq2.$ :D[/quote]

care to explain? ;)[/quote]
basically , this inequality was from 2 known results :
a,b,c,x are real numbers
the first one  :
$ \sum \frac{a(x\minus{}b)(x\minus{}c)}{(a\minus{}b)(a\minus{}c)} \equal{} x^2$
the second one , another equality :
$ \sum \frac{(a\plus{}b)(a\plus{}c)}{(a\minus{}b)(a\minus{}c)} \equal{} \minus{}1$[/hide]


could someone please show me the order of operations for calculating the determinant for a 4X4 and 5X5 matrix, respectively?

I have seen examples of this using elimination techniques and by recursively calculating the determinants of the minors, but what I am looking for is a 'straight arithmetic' solution, eg:

3X3:

abc
def
ghi

a*e*i + b*f*g + c*d*h
- a*f*h - b*d*i - c*e*g

thank you for your time.

would you mind giving me an example for 4X4 and 5X5 (in the form that I posted)? the problem is basically that I have a difficult time grasping mathematical shorthand - so I figured that if I could see the order of operations for these types of matrices, I could probably infer a generic algorithm for an NXN matrix.

In general, you can use elementary row operations to transform your matrix to te echelon form (Gauss' algorithm). Then the determinant of a matrix is a product of the diagonal elements. This is quite fast method to calculate determinants, but if you need to evaluate the determinant of 4x4-matrix or 5x5-matrix, you can first calculate the determinant of general 4x4 or 5x5 matrix and then use this formula.

Let's take the $4\times 4$ matrix
$a_{11}a_{12}a_{13}a_{14}$
$a_{21}a_{22}a_{23}a_{24}$
$a_{31}a_{32}a_{33}a_{34}$
$a_{41}a_{42}a_{43}a_{44}$

For $n=4$ there are $4!=4\times 3\times 2\times 1=24$ different permutations (orderings) of $1234$, with the following swap numbers (number of times you have to swap the position of two numbers to get back to $1234$):
$s(1234)=0$
$s(1243)=1$, $1243\to 1234$
$s(1324)=1$, $1324\to 1234$
$s(1342)=2$, $1324\to 1243\to 1234$
$\dots$

The determinant is the sum of the $4!$ terms like this:
$\pm a_{1x}\times a_{2y}\times a_{3z}\times a_{4t}$,
where $xyzt$ is a permutation of $1234$
and the sign is $+$ if the swap number is even and $-$ if it's odd.

Therefore, the determinant will begin with
$+a_{11}\times a_{22}\times a_{33}\times a_{44}$
$-a_{11}\times a_{22}\times a_{34}\times a_{43}$
$-a_{11}\times a_{23}\times a_{32}\times a_{44}$
$+a_{11}\times a_{23}\times a_{34}\times a_{42}$
$\dots$

The order of calculations for a $n\times n$ matrix is $n!$ products of $n$ numbers, but only if you have previously calculated the permutations and signs.

[quote="lordWings"]The order of calculations for a $n\times n$ matrix is $n!$ products of $n$ numbers, but only if you have previously calculated the permutations and signs.[/quote]

If you use Gaussian elimination to write it as an upper triangular matrix and then take the produce of the diagonal elements, it's  $O(n^{3})$.

You're both right, Franz Joseph and blahblahblah.
But garth explicitly asked for this:[quote="garth"]I have seen examples of this using elimination techniques and by recursively calculating the determinants of the minors, but what I am looking for is a 'straight arithmetic' solution, eg:
a*e*i + b*f*g + c*d*h
- a*f*h - b*d*i - c*e*g[/quote]

Alright Garth,
Here it is.  I was looking for the same solution you were.  I know it is now 2010, but perhaps someone else will also be looking for this.

I adapted the info from this website [url]http://www.euclideanspace.com/maths/algebra/matrix/functions/determinant/fourD/index.htm[/url]

For Matrix
a	b	c	d
e	f	g	h
 i	j	k	l
m	n	o	p
         Basic idea is to take the top row, one element at a time and multiply by the remaining 3x3's determinant.
[u][b]Arithmetic Solution[/b][/u]
[b]Downs[/b]
+a * f * k * p
+a * g * l * n
+a * h * j * o

+b * h * k * m
+b * e * l * o
+b * g * i * p

+c * e * j * p
+c * f * l * m
+c * h * i * n

+d * g * j * m
+d * e * k * n
+d * f * i * o

[b]- Up[/b]s
-a *n * k * h
-a * o * l * f
-a * p * j * g

-b * p * k * e
-b * o * i * h
-b * m * l * g

-c * m * j * h
-c * n * l * e
-c * p * i * f

-d * o * j * e
-d * n * i * g
-d * m * k * f

Here is the complete file [url]http://mrs--nelson.webs.com/4x4ab.doc[/url]


khob man nemidoonam ke ma asan ejaze inkaro darim ya na,vali ghaedatan nabayad iradi dashte bashe,age yeki az aghaye hamidi beporse ke dge che behtar...
be nazare man age rahe hallamoono vase soalaye dastchin bezarim be do dalil mofide:
1)in forum kheili vaghte ke dge bihal shodeo kamkar,vase hamin bayad ye tahavoli toosh dade she,ke ba in kar momkene yekhorde bishtar joon begire...
2)masalehaye tahe dastchin hal nadaran,vase hamin age inja rooshoon bahs konim mitoonim be halleshoon beresimo oonai ke natoonestan yeserishoono hal konan rahesho inja peida konan,shayad in zaminei shod ke dastchinaye badi ba pasokhname bashe...  :wink:

khob man avalisho shoroo mikonam omdivaram baghieam hamrahi konan,age ba esteghbal rooberoo shod ke edame midim ageam na ke hich,pas age movafeghid nazare khodetoono elam konid ta edame bedim...

$ 7)$darbare tabe $ f:\mathbb{N}\cup\{0\}\to\mathbb{N}\cup\{0\}$ midanim:
$ f(|f(n)\minus{}n|)\plus{}n\leq |f(n)\minus{}n|\plus{}1$ ke dar an $ n\in\mathbb{N}\cup\{0\}$
sabet konid moadele $ f(m) \equal{} 0$ binahayat javab darad.

[hide="rahe hal"]
edea mikonim ke be ezae binahayat adade tabiie $ n$ darim $ f(n)\leq n$...
-----------------------
[u]esbate edea[/u]:

farz konid be ezae $ n$i tabii dashte bashim $ f(n) > n$ pas ba tavajoh be moadele soorate masale khahim dasht:

$ f(f(n)\minus{}n)\plus{}n\leq f(n)\minus{}n\plus{}1$

hal gharar dahid $ t \equal{} f(n)\minus{}n$ pas avalan $ t\in\mathbb{N}\cup\{0\}$ hamchenin ba tavajoh be namosavie akhar natije mishe ke:

$ f(t)\plus{}n\leq t\plus{}1$ pas chon $ n\geq 1$ khahim dasht $ f(t)\leq t$...

pas be sadegi moshahede mishavad ke be ezae binahayat $ n\in\mathbb{N}\cup\{0\}$ darim $ f(n)\leq n$...
-------------------------------
amma farz konid baraye $ n_{0}\in\mathbb{N}\cup\{0\}$ dashte bashim:

$ f(n_{0})\leq n_{0}$

pas ba gharar dadane $ n_{0}$ dar moadele khahim dasht:

$ f(n_{0}\minus{}f(n_{0}))\plus{}n_{0}\leq n_{0}\minus{}f(n_{0})\plus{}1$

$ \Rightarrow f(n_{0}\minus{}f(n_{0}))\plus{}f(n_{0})\leq 1$

pas haddeaghal yeki az do adade $ f(n_{0}\minus{}f(n_{0}))$ va $ f(n_{0})$ bayad barabare $ 0$ bashad,chera ke dar gheire in soorat khahim dasht $ f(n_{0}\minus{}f(n_{0}))\plus{}f(n_{0})\geq 2$ ke ba namosavie fogh dar tanaghoze...
pas age be ezae $ n_{0}$i dashte bashim $ f(n_{0})\leq n_{0}$ oon moghe yeki az aadade $ f(n_{0}\minus{}f(n_{0}))$ ya $ f(n_{0})$ barabare sefr khahad shod...

hal kafist sabet konim moadele $ f(n)\minus{}n \equal{} c$ be ezae binahayat meghdare tabie $ n$ nemitavanad javab dashte bashad ($ c$ adadi tabii ya sefr va sabet ast)...

[hide="esbate"]ba borhane kholf farz konid be ezae binahayat $ n$ dashte bashim $ f(n)\minus{}n \equal{} c$ pas ba gharar dadane $ n$ dar rabete masale darim:

$ f(c)\plus{}n\leq c\plus{}1$

pas be ezae binahayayt adade tabiie $ n$ darim:

$ f(c)\plus{}n\leq c\plus{}1$ ke tanaghoz ast...

tanaghoze hasele neshan midahad ke moadele $ f(n)\minus{}n \equal{} c$ binahayat javabe tabiie baraye $ n$ nadarad...[/hide][/hide]

Manam Moafegham . manam javabe soale 8 o mizaram :)
$ 8)a,b,c$ adadhaye haghighie mosbat and va :
$ \frac{1}{a\plus{}1}\plus{}\frac{1}{b\plus{}1}\plus{}\frac{1}{c\plus{}1}\equal{} 2$ .
sabet konid :
$ \frac{1}{4a\plus{}1}\plus{}\frac{1}{4b\plus{}1}\plus{}\frac{1}{4c\plus{}1}\geq 1$

[hide]
khob ye ideaye khoob ine ke hamgenesh konim masalan farz konim $ \frac{1}{a\plus{}1}\equal{}\frac{2x}{x\plus{}y\plus{}z}$ ,  $ \frac{1}{b\plus{}1}\equal{}\frac{2y}{x\plus{}y\plus{}z}$
$ \frac{1}{c\plus{}1}\equal{}\frac{2z}{x\plus{}y\plus{}z}$ , age $ a,b,c$ ro be dast biarim darim :
$ a \equal{}\frac{y\plus{}z\minus{}x}{2x}, b \equal{}\frac{z\plus{}x\minus{}y}{2y}, c \equal{}\frac{x\plus{}y\minus{}z}{2z}$ pas $ \frac{1}{4a\plus{}1}\equal{}\frac{x}{2y\plus{}2z\minus{}x},\frac{1}{4b\plus{}1}\equal{}\frac{y}{2z\plus{}2x\minus{}y}$
, $ \frac{1}{4c\plus{}1}\equal{}\frac{z}{2x\plus{}2y\minus{}z}$ pas bayad sabet konim :
$ \frac{x}{2y\plus{}2z\minus{}x}\plus{}\frac{y}{2z\plus{}2x\minus{}y}\plus{}\frac{z}{2x\plus{}2y\minus{}z}\geq 1$ . amma midanim $ \frac{x}{2y\plus{}2z\minus{}x}\equal{}\frac{x^{2}}{2yx\plus{}2zx\minus{}x^{2}}$. hal agar baraye se kasre dige ham in namosavi haro benevisim va az namosavie Cauchy - Schwartz estefade konim be dast miad :$ \frac{x^{2}}{2xy\plus{}2xz\minus{}x^{2}}\plus{}\frac{y^{2}}{2yz\plus{}2yx\minus{}y^{2}}\plus{}\frac{z^{2}}{2zx\plus{}2zy\minus{}z^{2}}\geq\frac{(x\plus{}y\plus{}z)^{2}}{4(xy\plus{}yz\plus{}zx)\minus{}x^{2}\minus{}y^{2}\minus{}z^{2}}$ . hala esbate inke samte rast az 1 bishtar ya mosavie ba AM-GM badihie :) .

[/hide]

khoshhalam ke mibinam movafeghi faghat too rahe hallet moghe type ye eshkali pish oomade say kon dorostesh koni,ye rahe dge vase hamin soale $ 8$ ine:
[hide="rahe halle dovom"]
farz konid $ x\equal{}\frac{1}{a\plus{}1},y\equal{}\frac{1}{b\plus{}1},z\equal{}\frac{1}{c\plus{}1}$ pas darim $ x\plus{}y\plus{}z\equal{}2$
deghat konid ke $ \frac{1}{a\plus{}4}\equal{}\frac{x}{4\minus{}3x}\equal{}\frac{x}{2y\plus{}2z\minus{}x}$
pas soorate hokmemoon be shkele zir dar miad:
$ \frac{x}{2y\plus{}2z\minus{}x}\plus{}\frac{y}{2z\plus{}2x\minus{}y}\plus{}\frac{z}{2x\plus{}2y\minus{}z}\geq 1$
hala biaid ino makhraje moshtarak begirid!!
bad az sadde kardan,mimoone sabet konim ke:
$ 4(x^{3}\plus{}y^{3}\plus{}z^{3})\plus{}6xyz\geq 3(x^{2}y\plus{}xy^{2}\plus{}y^{2}z\plus{}yz^{2}\plus{}z^{2}x\plus{}zx^{2})$
amma tebghe namosavie shur darim:
$ \boxed{x^{3}\plus{}y^{3}\plus{}z^{3}\plus{}3xyz\geq x^{2}y\plus{}xy^{2}\plus{}y^{2}z\plus{}yz^{2}\plus{}z^{2}x\plus{}zx^{2}(I)}$
hamchenin tebghe miangine hesabi hendesi darim:
$ \boxed{x^{3}\plus{}y^{3}\plus{}z^{3}\geq 3xyz (II)}$
hala age tarafeine namosavie $ (I)$ ro dar $ 3$ zarb konimo ba namosavie $ (II)$ jamesh konim hokmemoon natije mishe...[/hide]

inam az rahe halle oon yeki namosavie:
$ 9)$ $ a,b,c$ adadhaye haghighi va mosbatand va $ abc\geq 1$.sabet konid:
$ \frac{a}{\sqrt{b\plus{}\sqrt{ac}}}\plus{}\frac{b}{\sqrt{c\plus{}\sqrt{ba}}}\plus{}\frac{c}{\sqrt{a\plus{}\sqrt{cb}}}\geq\frac{3}{\sqrt{2}}$
[hide="rahe hal"]
tebghe namosavie cauchy darim:

$ \frac{a}{\sqrt{b\plus{}\sqrt{ac}}}\plus{}\frac{b}{\sqrt{c\plus{}\sqrt{ba}}}\plus{}\frac{c}{\sqrt{a\plus{}\sqrt{cb}}}$

$ \geq\frac{(\sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c})^{2}}{\sqrt{b\plus{}\sqrt{ac}}\plus{}\sqrt{c\plus{}\sqrt{ba}}\plus{}\sqrt{a\plus{}\sqrt{cb}}}$

hala biaiad soorato makhraje in ebarato dar $ 2\sqrt{2}$ zarb konid,tabdil mishe be in:

$ \frac{2\sqrt{2}(\sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c})^{2}}{2\sqrt{2}(\sqrt{b\plus{}\sqrt{ac}}\plus{}\sqrt{c\plus{}\sqrt{ba}}\plus{}\sqrt{a\plus{}\sqrt{cb}})}$

amma midoonim $ x\plus{}y\geq 2\sqrt{xy}$ hala gharar bedid $ x \equal{} 2,y \equal{} b\plus{}\sqrt{ac}$ pas khahim dasht:

$ 2\plus{}b\plus{}\sqrt{ac}\geq 2\sqrt{2}\sqrt{b\plus{}\sqrt{ac}}$

pas samte raste namosavimoon az ebarate zir bozorgtare:

$ \frac{2\sqrt{2}(\sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c})^{2}}{6\plus{}a\plus{}b\plus{}c\plus{}\sqrt{ab}\plus{}\sqrt{bc}\plus{}\sqrt{ca}}$

pas kafie sabet konim in ebarat az $ \frac{3}{\sqrt{2}}$ bozorgtare ke age tarafein vasateyn konim mishe:

$ 4(\sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c})^{2}\geq 18\plus{}3(a\plus{}b\plus{}c)\plus{}3(\sqrt{ab}\plus{}\sqrt{bc}\plus{}\sqrt{ca})$

ke age sadash konim moadele ine:

$ a\plus{}b\plus{}c\plus{}5(\sqrt{ab}\plus{}\sqrt{bc}\plus{}\sqrt{ca})\geq 18$...

pas kafie namosavie balaro ba sharte $ abc\geq 1$ sabet konim...
amma tebghe miangine hesabi hendesi darim:

$ a\plus{}b\plus{}c\plus{}\sqrt{ab}\plus{}\sqrt{ab}\plus{}\sqrt{ab}\plus{}\sqrt{ab}\plus{}\sqrt{ab}\plus{}\sqrt{bc}\plus{}\sqrt{bc}\plus{}\sqrt{bc}\plus{}\sqrt{bc}\plus{}\sqrt{bc}\plus{}\sqrt{ca}\plus{}\sqrt{ca}\plus{}\sqrt{ca}\plus{}\sqrt{ca}\plus{}\sqrt{ca}$
$ \geq 18\sqrt [18]{abc\sqrt{a^{10}b^{10}c^{10}}}\equal{} 18\sqrt [18]{a^{6}b^{6}c^{6}}\equal{} 18\sqrt [3]{abc}\geq 18$...[/hide]

$ 18)$[color=blue] noghteye O markaze dayere mohitiye ABC ast .farz konid D paye ertefae vared az A bashad va E mahale barkhorde AO va BC.farz konid khathaye momas bar dayere mohiti mosalase ABC dar noghate B va C (dar majale eshtebahan A va B gofte shode) yekdigar ra dar T ghat mikonand va AT in dayere ra dar F ghat mikonad.sabet konid dayerehaye mihitie mosalashaye DEF va ABC bar ham momasand.[/color]

[hide="esbat"]
farz konid momas bar dayere dar A , BC ra dar X ghat konad ghotbiye T az X mogozarad pas ghotbiye X niz az T migozarad pas XF bar dayere mohitiye ABC mommas ast .pas $ XF^{2}\equal{} XB.XC$ .va hamchenin darim $ \angle AEB \equal{}\angle DAX$ banabarin darim:$ XA^{2}\equal{} XD.XE \equal{} XF^{2}$[/hide]

cheghad khoob ke in topic ba esteghbal rooberoo shode...  :lol:

$ 16)$adadhaye tabii ra ba dorang rang kardeim.sabet konid dombalei az adadhaye tabii manande $ k_{1}< k_{2}< ... < k_{n}< ...$ vojood darad ke hame adadhaye dombale $ 2k_{1}< k_{1}\plus{}k_{2}< 2k_{2}< k_{2}\plus{}k_{3}< 2k_{3}< ...$ be yek rangand.
[hide="rahe hal"]
do rang ra abio ghermez benamid,ba borhane kholf farz konid chenin nabashad,yani farz konid tedade aadade abie $ 2n_{1},2n_{2},...,2n_{k}$ ke tamame aadade $ n_{i}\plus{}n_{i\plus{}1}$ ham be range abi bashad motanahi bashad...
hamchenin baraye range ghermez...
hal biaid dombalei az aadade abie $ 2x_{1},2x_{2},...,2x_{b}$ ba toole maximal dar nazar begirid be torike aadade $ x_{i}\plus{}x_{i\plus{}1}$ ham abi bashand,pas be ezae har adade abie $ 2t$ ke $ t > x_{b}$,adade $ t\plus{}x_{n}$ ghermez khahad bood,cherake dar gheire in soorat dombalei ba toole bozorgtar be vojood miaiad ke tanaghoz ast...
hamchenin baraye range ghermez estedlali moshabeh ra be kar barid,yeni dombalei az aadade ghermeze $ 2y_{1},2y_{2},...,2y_{r}$ ba toole maximal dar nazar begirid ke aadade $ y_{i}\plus{}y_{i\plus{}1}$ ham ghermez bashand,pas baraye har adade ghermeze $ 2l$ ke $ l > y_{r}$ adade $ l\plus{}y_{r}$ bayad abi bashe...
hala bedoone az das dadne kolliate masale farz konid $ x_{b}> y_{r}$ va gharar bedid $ x_{b}\minus{}y_{r}\equal{} k$ hala farz konid $ e > x_{b}$ oonvagh biaid dombale zir ro dar nazar begirid:
$ 2e,2e\plus{}k,2e\plus{}2k,2e\plus{}3k,2e\plus{}4k,...$
tavajoh dashte bashid ke ba kami deghat moshakhas mishavad ke in dombale be forme khaste shode dar soorate4 masale ast
kafist gharar dahid $ k_{1}\equal{} e,k_{2}\equal{} e\plus{}k,k_{3}\equal{} e\plus{}2k,k_{4}\equal{} e\plus{}3k,...,k_{i}\equal{} e\plus{}(i\minus{}1)k,...$ pas tanha kafist sabet konim aadade in dombale hamrangand...
tavajoh dashte bashid ke tafazole do ozve motavalie in dombale barabare $ k$ hast,hala hokme ghavitario sabet mikonim,sabet mikonim har do adade $ 2p,2q$ ke $ k$ ta ekhtelaf dashte bashan,nahamrang nistan,ya be ebarati hamrangan...
[hide="esbate edeae hamrangi"]
agar farz konim $ p,q$ do adad bashand ke $ 2p$ abio $ 2q$ ghermez betorike $ p > q\geq\max\{x_{b},y_{r}\}$ oon moghe tebghe estedlali ke ghabltar anjam shod avalan natije mishavad ke adade $ 2(2p\minus{}x_{b})$ ghermeze,hamchenin adade $ 2(2q\minus{}y_{r})$ abie...
amma age $ 2p\minus{}2q \equal{} k \equal{} x_{b}\minus{}y_{r}$ bashe oon moghe adade $ S \equal{} 2(2p\minus{}x_{b}) \equal{} 2(2q\minus{}y_{r})$ ham ghermeze ham abi ke tanaghoze...
pas $ 2(p\minus{}q)\not \equal{} k$
[/hide]
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pas har do ozve motavalie dombalamoon,va dar natije har do ozve dombalamoon hamrangano esbat kamel shode...[/hide]

In ham rahe hale masaleye [b]13[/b] ke saale putnam 2006 boode :[url]http://www.mathlinks.ro/viewtopic.php?t=122453[/url]

in hame rahe hale masale[b] 5[/b]: [url]http://www.mathlinks.ro/viewtopic.php?search_id=573364683&t=74746[/url]

donbale $ c_{0},c_{1},c_{2},\dots$ ra be in tartib tarif mikonim:
$ c_{0}\equal{} 1$ , $ c_{n}\equal{} (\frac{a_{n\minus{}1}}{1\plus{}a_{n}})c_{n\minus{}1}$ , $ n\geq 1$ ke mitavan anra besoorate$ c_{n}\equal{} a_{n\minus{}1}c_{n\minus{}1}\minus{}a_{n}c_{n}$ niz nevesht. ba jam zadane aazaye in donbale bedast miaaiad:$ \sum_{i \equal{} 1}^{n}c_{i}\equal{} a_{0}\minus{}a_{n}.c_{n}$ , $ (1)$
kafist sabet konim baraye binahayat $ n$ ,$ \frac{c_{n}}{c_{n\minus{}1}}<\sqrt [n]{\frac{1}{2}}$ farz konid adade $ N$ vojood dashte bashad betorike $ \sqrt [n]{\frac{1}{2}}\leq\frac{c_{n}}{c_{n\minus{}1}}$ , $ N\leq n$. dar in soorat baraye $ N < n$ khahim dasht :
$ c_{n}\geq c_{N}.(\frac{1}{2})^{(\frac{1}{N\plus{}1}\plus{}\frac{1}{N\plus{}2}\plus{}\dots\plus{}\frac{1}{n})}\equal{} C.(\frac{1}{2})^{(1\plus{}\frac{1}{2}\plus{}\dots\plus{}\frac{1}{n})}$ ke $ C$ adadi sabet va mosbat ast. hal agar $ 2^{k\minus{}1}\leq n < 2^{k}$ angah:
$ \sum_{i \equal{} 1}^{n}\frac{1}{i}\leq 1\plus{}(\frac{1}{2}\plus{}\frac{1}{3})\plus{}(\frac{1}{4}\plus{}\dots\plus{}\frac{1}{7})\plus{}\dots\plus{}(\frac{1}{2^{(}k\minus{}1)}\plus{}\dots\plus{}\frac{1}{2^{k}\minus{}1})\leq 1\plus{}1\plus{}\dots\plus{}1 \equal{} k$ banabar in baraye $ 2^{(}k\minus{}1)\leq n < 2^{k}$ , $ C^{(}2\minus{}k)\leq c_{n}$.
hal farz konid $ m$ va $ r$ adaadi bashand ke $ r < m$ , $ 2^{(r\minus{}1)}\leq N < 2^{r}$.dar in soorat:
$ c_{2^{r}}\plus{}c_{2^{r}\plus{}1}\plus{}\dots\plus{}c_{2^{m}\minus{}1}\equal{} (c_{2^{r}}\plus{}\dots\plus{}c_{2^{r\plus{}1}\minus{}1})\plus{}(c_{2^{r\plus{}1}}\plus{}\dots\plus{}c_{2^{r\plus{}2}\minus{}1})\plus{}\dots\plus{}(c_{2^{m\minus{}1}}\plus{}\dots\plus{}c_{2^{m}\minus{}1})\geq C.(\sum_{i \equal{} r}^{m\minus{}1}2^{i}.2^{\minus{}1\minus{}i}) \equal{}\frac{c(m\minus{}r)}{2}$ ke neshan midahad majmoooe $ c_{n}$ha mitavanad be andaze delkhah bozorg sahavad. vali $ (1)$ natije midahad majmoooe $ c_{n}$ha nemitavanad az $ a_{0}$ bozorgtar sahavad.va in tanaghos neshand midahad ke baraye namotenahi $ n$ , $ \sqrt [n]{\frac{1}{2}}>\frac{c_{n}}{c_{n\minus{}1}}$
javae in masale dar short list :[ http://www.mathlinks.ro/resources.php?c=1&cid=17&year=2001&p=119160]

bacheha soorate soale $ 19$ doroste?  :maybe:

are doroste.  :wink:

[color=blue]$ 19)$dayere mohati mosalase $ ABC$ bar zelhaye $ AC$ va $ BC$ be tartib dar noghtehaye $ E,D$ momas ast.dayere mohati khareji(wrt A) bar zel'e BC dar dar noghteye $ A_{1}$ va bar emtedade AB va AC be tartib dar $ C_{1}$ va $ B_{1}$ momas ast.sabet konid noghteye barkhorde khathaye $ DE$ va $ A_{1}B_{1}$ rooye dayere mohitiye mosalase $ A_{1}BC_{1}$ gharar darad. [/color]

[hide="Esbat"]fazr konid $ A_{1}B_{1}\cap DE\equiv F$ ghotbiye C nesbat be do dayere (mohati dakheili va kharaje) az F migozarad pas dayerei be ghotre $ CF$ az varoone $ C$ nesbat be dodayere migozarad pas in dayere be dodayere amood ast , pas markazash yani vasate CF ($ M$) rooye mehvar asliye dodayere gharar darad pas $ M$ ba noghate vasate $ A_{1}D$ va $ EB_{1}$ hamkhat ast ke in khat bar $ II_{a}$ amood ast.
choon vasate $ DA_{1}$ vasate $ BC$ niz hast pas $ BF\perp II_{a}$ dar natije $ \angle ABF \equal{}\angle FA_{1}C_{1}\equal{}\frac{\angle B\plus{}\angle C}{2}$ pas chaharzeliye $ BFA_{1}C_{1}$ mohati ast.

[b]Remark[/b].noghteye F rooye $ II_{a}$ ham gharar darad,va hamchenin ghotbiye $ F$ nesbat be dodayere mohati dakheli va khareji(wrt A) bar ham montabegh hastand.[/hide]