Hi. I was just wondering at what time / with which flights do IMO teams arrive on 14th? Will someone arrive by bus or train (Portugal maybe?). Of course, one team will probably arrive by public transportation (Spain, of course  :P ).

Croatian team arrives on 2:40pm; flight LX2026 from Zurich. Is there some other IMO team on that flight?

EDIT: Check further below (somewhere close to the end of the thread) to view an updated list of arrivals in Madrid. Join in with your information and make the list longer!

Well, even Portugal will arrive by plane  :D , even though it's this close. Our flight arrives on 2:55pm. So I guess only Spain won't go by plane, possibly... :P

As you probably guessed, i won't be going in your flight...
See you in Spain!

My flight is like 20 hours long :( Maybe I'll have some of the US team members with me on the plane but I guess that will be it.

Alright, thanks for responding. I just thought it would be nice if some team members posted when and how they arrive in Madrid so people know if they are sharing a flight with somebody.

[quote="Jorge Miranda"]Our flight arrives on 2:55pm. [/quote]

Well, maybe we'll share a ride to our housing, because
[quote="http://www.imo-2008.es/travelinfo.html"]Bus transfer to housing: Bus service has been arranged to transfer participants to their housing site, according to estimated arrival times. There will be various services gathering participants arriving within reasonable range of timings, therefore allowing in some cases a reasonable waiting time.[/quote]

Speaking of housing, no information as to what it is yet, huh?

Well, hana, if you come from Paris at around 2pm, it can't be too hard to track down the flight you're on... just go here and see:
[url=http://www.aeroportsdeparis.fr]Aeroports de Paris[/url]
I already checked and it seems that, if you're coming from Charles De Gaulle and not Orly (I think no flights from Belgrade to Orly exist), you will probably be on one of these three flights:
11:35 - 13:35 Air Comet A7 235 (which I doubt considering Air Comet is not a large airline)
12:10 - 14:10 Iberia IB 3435
12:35 - 14:35 Air France	AF 1600 (Air Europa "flight" at the same time is just a codeshare - not a real flight)

Hope this helps!  :)

[quote="hana"]Serbia team should also arrive  around 2 pm, from Paris ...We know just this..
Does somebody come from Paris ?  :)[/quote]
The Hong Kong team will go to Paris first  :)

[hide="EDIT: See updated information below"]
[size=150][b]Known team arrivals:[/b][/size]

[i][b]Team[/b] - City - Flight Number - Arrival Time[/i]
[b]Serbia[/b] - Paris - ? - around 2pm (see [url=http://www.mathlinks.ro/viewtopic.php?p=1178919#1178919]post 6[/url])
[b]Croatia[/b] - Zurich - LX 2026 - 2:40pm
[b]Portugal[/b] - Lisbon - IB 3109 - 2:55pm
[b]Hong Kong[/b] - Paris - ? - ?

[size=75]Disclaimer: I tried compiling these information both from Mathlinks member posts and from airport and airline websites. Therefore, I'm not 100% sure all of these times, cities and flights are correct. Please, correct me if I'm wrong somewhere.[/size][/hide]
Thanks. lchserious, do you know when you arrive in Madrid?

Everybody, please, if you want, post some information about the flight which you will arrive in Madrid with. This will give some teams a chance to meet before even landing. This statement is (almost - I believe there are actually a lot more than 100 flights to Madrid daily, but some flights from smaller airports or with weird times can probably be discarded) provable by Dirichlet.  :)  Please, fill in the gaps on the current table, if you can, as well.

i think i can help lchserious to answer~~

Hong Kong team: Paris AF2100
Time: 1915 - 2115

pretty late...

OK, brianchung11, thanks for answering. Here's a new list:

[hide="EDIT: See updated information below"][size=150][b]Known team arrivals:[/b][/size]

[i][b]Team[/b] - City - Flight Number - Arrival Time[/i]
[b]Serbia[/b] - Paris - ? - around 2pm (see [url=http://www.mathlinks.ro/viewtopic.php?p=1178919#1178919]post 6[/url])
[b]Croatia[/b] - Zurich - LX 2026 - 2:40pm
[b]Portugal[/b] - Lisbon - IB 3109 - 2:55pm
[b]Hong Kong[/b] - Paris - AF 2100 - 9:15pm

[size=75]Disclaimer: I tried compiling these information both from Mathlinks member posts and from airport and airline websites. Therefore, I'm not 100% sure all of these times, cities and flights are correct. Please, correct me if I'm wrong somewhere.[/size][/hide]
Slightly OT: It seems stupid for me that after every single one (or at least every couple) of these responses I'd have to post a table in a new post. I know I can edit posts, but is there a time limit on how long I can edit the opening post? That way I could just update a table in the first post - this way I kinda feel like I'm doing this for my post count. I'm worried that if I put the table up there now, I won't be able to edit it after a while and it will contain old information.

Still OT: Well, as far as I know, the edit limit is 48 hours. After that, only the mods can edit posts (right?).

Israeli team:
Frankfurt - Madrid, departure 11:55, arrival 14:25.

Thanks for answering, Fuaran.

Right, here's the new list:

[hide="EDIT: See updated information below"][size=150][b]Known team arrivals:[/b][/size]

[i][b]Team[/b] - City - Flight Number - Arrival Time[/i]
[b]Serbia[/b] - Paris - ? - around 2pm (see [url=http://www.mathlinks.ro/viewtopic.php?p=1178919#1178919]post 6[/url])
[b]Israel[/b] - Frankfurt - LH 4412 - 2:25pm
[b]Croatia[/b] - Zurich - LX 2026 - 2:40pm
[b]Portugal[/b] - Lisbon - IB 3109 - 2:55pm
[b]Hong Kong[/b] - Paris - AF 2100 - 9:15pm

[size=75]Disclaimer: I tried compiling these information both from Mathlinks member posts and from airport and airline websites. Therefore, I'm not 100% sure all of these times, cities and flights are correct. Please, correct me if I'm wrong somewhere.[/size][/hide]
Keep 'em coming.

[quote="mornik"]hsiljak, do you know when team Bosnia and Herzegovina arrives?[/quote]

I checked it: they should depart to Munich at 11:45, and arrive in Madrid at about 17:30.

Wow, this has become quite a project - people really started posting after this making the front page. OK, time for expanding the list once again. cyshine, it's JJ, but never mind... I looked the flight up in the timetable anyway. Thanks for posting. the.sceth, you are already in Madrid? I wish I was there already. What's the housing like (I mean, where the contestants will be staying)? That is, I presume you are staying there at the moment, not in some other hotel or whatever. Thanks a lot. Also, Erken, never mind, thanks for the information you gave us.  aev5peru, your flight gave me a lot of trouble - there are two flights leaving Lima only 10 minutes apart. I hope I got the correct one - you said 7.45pm!

[hide="EDIT: See updated information below"][size=150]Known team arrivals:[/size]

[i]Team - City - Flight Number - Arrival Time[/i]
[b]Trinidad and Tobago[/b] - New York - DL 126 - Thurday, July 10, 9:35 [u](already arrived)[/u]
[b]Kazakhstan[/b] - ? - ? - some time on Monday (see post [url=http://www.mathlinks.ro/viewtopic.php?p=1185123#1185123]36[/url])
[b]Brazil[/b] - Sao Paulo - JJ 8064 - 11:25
[b]Georgia[/b] - Munich - LH 4420 - 11:25
[b]Mexico [size=75](4 contestants)[/size][/b] - Mexico City - AM 642 - 12:10
[b]Hungary[/b] - Budapest - MA 580 - 13:15
[b]Germany[/b] - Frankfurt - LH 4412 - 14:25
[b]Israel[/b] - Frankfurt - LH 4412 - 14:25
[b]Peru[/b] - Lima - LP 706 - 14:25
[b]Serbia[/b] - Paris - AF 1600 - 14:35
[b]Croatia[/b] - Zurich - LX 2026 - 14:40
[b]Portugal[/b] - Lisbon - IB 3109 - 14:55
[b]Macedonia[/b] - Vienna - IB 3577 or JK 160 - 15:25 or 16:00 (see post [url=http://www.mathlinks.ro/viewtopic.php?p=1181646#1181646]23[/url])
[b]Bosnia and Herzegovina[/b] - Munich - LH 4426 - 17:20
[b]Mexico [size=75](2 contestants)[/size][/b] - Paris - AF 1900 - 18:40
[b]Belgium[/b] - Brussels - IB 3219 - 19:00
[b]Poland[/b] - Amsterdam - KL 1705 - 20:30
[b]Hong Kong[/b] - Paris - AF 2100 - 21:15
[b]Moldova[/b] - Budapest - MA 582 - 22:20

[size=75]Disclaimer: I tried compiling these information both from Mathlinks member posts and from airport and airline websites. Therefore, I'm not 100% sure all of these times, cities and flights are correct. Please, correct me if I'm wrong somewhere. If not specifically emphasized, all arrival times are on Monday, July 14.[/size][/hide]

As i can remember, Turkey team will have been there about 14:00, Monday. We are flying directly from İstanbul. See you there.

Well it seems tentative, but we´re in Colegio Mayor Santa Mónica and the rooms are singles with personal bath and shower and verandah. If there are four people in the strict interior, then at least one pair will be separated by a distance not exceeding three feet (over-exaggeration). So far food has been moderate (a plate of boiled cauliflower?) to splendid (paella :) ). Nights can get hot, and while this might seem really weird, the verandah is quite cool. The elevator cannot really hold more than two persons and one person´s baggage. Lunch is at 2pm  :| , breakfast from 7 to 9 am, dinner from 9pm but there´s a cafeteria.

Nice, finally we've got some information :D

the.sceth, thank you very much for all the information! The housing seems great... Although, could you please, if you want, of course, elaborate a bit on what you ment by "tentative"? Oh yeah, I won't mind the temperature - I'm bringing a fan to keep it cool in my room (if I'll have enough space in the bag)!  :ninja: Umut, the only thing I found is a Turkish Airlines flight which arrives at 13:25, so I'm guessing this is what you ment by "around 14:00".

[hide="EDIT: See updated information below"][size=150]Known team arrivals:[/size]

[i]Team - City - Flight Number - Arrival Time[/i]
[b]Trinidad and Tobago[/b] - New York - DL 126 - Thurday, July 10, 9:35 [u](already arrived)[/u]
[b]Kazakhstan[/b] - ? - ? - some time on Monday (see post [url=http://www.mathlinks.ro/viewtopic.php?p=1185123#1185123]36[/url])
[b]Brazil[/b] - Sao Paulo - JJ 8064 - 11:25
[b]Georgia[/b] - Munich - LH 4420 - 11:25
[b]Mexico [size=75](4 contestants)[/size][/b] - Mexico City - AM 642 - 12:10
[b]Hungary[/b] - Budapest - MA 580 - 13:15
[b]Turkey[/b] - Istanbul - TK 1857 - 13:25
[b]Germany[/b] - Frankfurt - LH 4412 - 14:25
[b]Israel[/b] - Frankfurt - LH 4412 - 14:25
[b]Peru[/b] - Lima - LP 706 - 14:25
[b]Serbia[/b] - Paris - AF 1600 - 14:35
[b]Croatia[/b] - Zurich - LX 2026 - 14:40
[b]Portugal[/b] - Lisbon - IB 3109 - 14:55
[b]Macedonia[/b] - Vienna - IB 3577 or JK 160 - 15:25 or 16:00 (see post [url=http://www.mathlinks.ro/viewtopic.php?p=1181646#1181646]23[/url])
[b]Bosnia and Herzegovina[/b] - Munich - LH 4426 - 17:20
[b]Mexico [size=75](2 contestants)[/size][/b] - Paris - AF 1900 - 18:40
[b]Belgium[/b] - Brussels - IB 3219 - 19:00
[b]Poland[/b] - Amsterdam - KL 1705 - 20:30
[b]Hong Kong[/b] - Paris - AF 2100 - 21:15
[b]Moldova[/b] - Budapest - MA 582 - 22:20

[size=75]Disclaimer: I tried compiling these information both from Mathlinks member posts and from airport and airline websites. Therefore, I'm not 100% sure all of these times, cities and flights are correct. Please, correct me if I'm wrong somewhere. If not specifically emphasized, all arrival times are on Monday, July 14.[/size][/hide]

Hi, I am BGD 3.
We are going to Madrid via Doha by Qatar Airways, flight 67 (as it is written on the ticket)
Our time of arrival is 19:55 (Madrid local time), oh..and date is 14th July,Monday and we shall arrive at Terminal 1.

Hope to see you all there :)

p.s. I hope that, our guide will be present at the airport, last year our guide was absent :maybe:

Yes, it must be that. Thank you for the correction :)

[quote="mornik"]Oh yeah, I won't mind the temperature - I'm bringing a fan to keep it cool in my room (if I'll have enough space in the bag)!  :ninja:[/quote]
ha ha...if it is more than $ 35^\circ$ then we'll be almost toasted without Air Conditioner. :wink:
Anyway, I am not that afraid of hot weather (as our weather is almost always very hot), rather I am afraid of cold weather. If it goes down to $ 15^\circ$ or less (at night), it won't be very good for me.  :(

[hide="@mornik"] you are Croatia 5,right? :) [/hide]

Another point is that everyone can't have enough space to carry a fan.(i.e. me :D) Also it can be very disturbing in the exam, unless the organizators pay attention to cooling system.

Apsolutely, Umut, I agree with you. If I bring my fan, I will just be able to use it (of course) in my room, when I sleep. During the exam, it will be a totally different story - the heat really doesn't help one's concentration so I believe some kind of air-conditioning will have to be provided during the exam.
Moonmathpi496, yup, that's me. That was hard to guess, right? I don't have anything against cold nights - I usually find I sleep more easily when it's a bit cold than when it's really hot. Of course, I wouldn't like any extremely hot or cold temperatures during the day, I believe none of us would.  :)

[size=150]Known team arrivals:[/size]

[i]Team - City - Flight Number - Arrival Time[/i]
[b]Trinidad and Tobago[/b] - New York - DL 126 - Thurday, July 10, 9:35 [u](already arrived)[/u]
[b]Kazakhstan[/b] - ? - ? - some time on Monday (see post [url=http://www.mathlinks.ro/viewtopic.php?p=1185123#1185123]36[/url])
[b]Brazil[/b] - Sao Paulo - JJ 8064 - 11:25
[b]Georgia[/b] - Munich - LH 4420 - 11:25
[b]Mexico [size=75](4 contestants)[/size][/b] - Mexico City - AM 642 - 12:10
[b]Hungary[/b] - Budapest - MA 580 - 13:15
[b]Turkey[/b] - Istanbul - TK 1857 - 13:25
[b]Germany[/b] - Frankfurt - LH 4412 - 14:25
[b]Israel[/b] - Frankfurt - LH 4412 - 14:25
[b]Peru[/b] - Lima - LP 706 - 14:25
[b]Serbia[/b] - Paris - AF 1600 - 14:35
[b]Croatia[/b] - Zurich - LX 2026 - 14:40
[b]Portugal[/b] - Lisbon - IB 3109 - 14:55
[b]Macedonia[/b] - Vienna - IB 3577 or JK 160 - 15:25 or 16:00 (see post [url=http://www.mathlinks.ro/viewtopic.php?p=1181646#1181646]23[/url])
[b]Bosnia and Herzegovina[/b] - Munich - LH 4426 - 17:20
[b]Mexico [size=75](2 contestants)[/size][/b] - Paris - AF 1900 - 18:40
[b]Belgium[/b] - Brussels - IB 3219 - 19:00
[b]Bangladesh[/b] - Doha - QR 67 - 19:55
[b]Poland[/b] - Amsterdam - KL 1705 - 20:30
[b]Hong Kong[/b] - Paris - AF 2100 - 21:15
[b]Moldova[/b] - Budapest - MA 582 - 22:20

[size=75]Disclaimer: I tried compiling these information both from Mathlinks member posts and from airport and airline websites. Therefore, I'm not 100% sure all of these times, cities and flights are correct. Please, correct me if I'm wrong somewhere. If not specifically emphasized, all arrival times are on Monday, July 14.[/size]

[quote="mornik"]Moonmathpi496, yup, that's me. That was hard to guess, right?[/quote]Did you mean, "It was not hard to guess"?  :wink: Anyway, It was not hard. Valentin wrote that, you are from Croatia and only one in that team has name similar to your nick.

[quote="mornik"]Of course, I wouldn't like any extremely hot or cold temperatures during the day, I believe none of us would.
[/quote] Yes, the weather which is going to be OK for everyone, should be the best. Anyway, now the websites are showing that, the temperature is just $ 14^\circ$  :ninja:

It seems from the [url=http://www.imo-2008.es/images.html]images[/url] that Canada is already there, too.

Moonmathpi496, I was merely being ironic. I know it wasn't hard.  :wink:

Alright, I wanted to wish everyone a good trip, see you all in Spain.

GIL Publishing House will be in Madrid during 17.07.2008 (Thursday) - 21.07.2008 (Monday). Our publications in English will be available for sale and you can save the shipping charges. Any of you interested in acquiring books please contact us using this username or using imo@gil.ro e-mail address.

Thank you!

[quote="mornik"]the.sceth, thank you very much for all the information! The housing seems great... Although, could you please, if you want, of course, elaborate a bit on what you ment by "tentative"? Oh yeah, I won't mind the temperature - I'm bringing a fan to keep it cool in my room (if I'll have enough space in the bag)!  :ninja: Umut, the only thing I found is a Turkish Airlines flight which arrives at 13:25, so I'm guessing this is what you ment by "around 14:00".
[/quote]

1.
The housing has indeed changed from the pre-IMO spots, and it´s not the same place for all the teams.

2.
It was only "warm" for me because I kept the windows closed at first. Otherwise it´s quite cool.

I shouldn't be on Mathlinks now...


algorithm
real analysis
logarithms
function
Support

Given a four digit number $ n$, how to find the smallest positive integer $ m$ such that the digits of $ mn$ forms either increasing or decreasing sequence.

This algorithm is very easy to program in any programming language.

[quote="miller4math"]can't you just generate all increasing numbers / decreasing numbers using basic recursion (there are quite few of them i think < 2000?) and check for divisibility?[/quote]

This problem can be easy or hard depending on whether or not it is *explicitly* decreasing.

If a sequence like 987332 is valid, then so can 987333333332, 987333333333333333333333333333333333333333333333333333332, and so on. An easy constructive algorithm does exist for this but it will be even slower than incrementing 0 by n repeatedly and checking for satisfaction of conditions.

Okay. I try to make some huge precision program or try to learn using GMP. I think the fastest method is to consider first every four digit numbers with digits monotonous order, then five digit numbers and so on. $ 50$ digits should be enough.

The following may work a bit faster. ;) Let's discuss the (non-strictly) increasing digit sequences. Let $ k_j$ be the remainder of the division of $ 10^j$ by $ n$. Let $ K_j \equal{} \sum_{i \equal{} 0}^j k_i$. You need to find the sum of a few $ K_j$ (from one to nine) that is divisible by $ n$ and among all possible choices, you need to take the one that minimizes $ \sigma \equal{} \sum_j K_j(10^{j \plus{} 1} \minus{} 1)$. Now just keep track of all possible sums you can get and the corresponding ways to minimize the corresponding $ \sigma$s considering first sums of one term, then of two terms, etc. up to nine terms. That'll require an array of length $ n$ and at most $ 9n^2$ steps, which is about $ 10^{11}$ operations at worst. Your PC should be fine with that but my old one would not survive it, so I would need to think how to make the convolutions faster (FFT is a good way, but how will I know which particular two numbers added to something? Remember, I need not only the possibility to get a particular sum, but also the minimal cost of getting it :?).

The decreasing case should be similar but some technical details still evade me :(. Can anybody modify this algorithm to cover it properly?

I can barely read the previous post.. lol... i got a heart attack when i saw FFT.

This problem is pretty easy with dynamic programming.
Where the state space is (last digit,current modulus N).
If you do traditional dp, it will never finish, because the last digit will stay the same.
SO.. you should keep updating the array until one of the moduli turns to 0 or it stops changing.
If you want to actually find the whole number, you can keep pointers to the previous state.
The speed is O(10*#ofdigits * N).

PS: you realize that my previous algorithm is O(#ofdigits^3) right? It's fast enough to finish in less than 1 sec. (all of you critics)

Yes, it is dynamic programming and the running time is $ O(\text{(number of digits)}\cdot n\)$ for both your and my approaches but what makes you think that the minimal number of digits is much smaller than $ n$? :P The running time of $ 9n^2$ is just too much for me, I wanted guaranteed $ O(n\log n)$. Hence the FFT (though I still do not know whether it'll help me or not :( ).

I completely agree that your realization is somewhat better than mine if $ O(n^2)$ is a satisfactory running time but if you take some time to understand what I wrote, you'll realize that our approaches are not so different. :)

Anyway, let's see if I got your algorithm right. You start with the 9 "finite" states $ (x,x;x)$ and plenty of "infinite" states $ (x,y,\infty)$ where the first position is the currently last digit, the second is the remainder modulo $ n$ and the last is the "cost" of this state, i.e., the smallest number compatible with the first 2 positions. You can go from $ (x,y,z)$ to $ (k,(10y+k)\mod n; 10z+k)$ for $ k\le x$ (if we are talking about the decreasing digit sequences) and update the cost at $ (k,(10y+k)\mod n)$ to $ 10z+k$ if it is less than the current cost at that position.

Note that just having a pointer to the previous state as you suggested is not enough. You'll have to compare the numbers quickly and traveling along the pointers to figure out the current cost is too expensive. Of course, $ z$ should be encoded as an array of length 9 telling you where you switch from one digit to the next one or something like that.

Now you go over the array making such updates. The traditional DP tells you to stop when no cost updates have been made after a full pass (i.e., after trying all pairs $ (x,y)$ and all admissible $ k$'s for each of them). Actually, it makes sense to stop earlier one pass after the one in which a state with $ 0$ in the second position is updated to a finite one, say, if it happens earlier (this may be a slight overkill, but it'll not increase the running time by any noticeable amount).

My (very likely completely stupid) question about FFT was the following:

I have two arrays $ a[i]$ and $ b[i]$ of length $ n$ with positive entries (those terrible cost numbers and infinities). I want to build the array $ c[i]=\min_{j+k\equiv i\mod n}a[j]+b[k]$ in time $ n\log n$ instead of trivial $ n^2$. Is there any way to do it? It seems like it is hopeless when cost numbers are arbitrary, but the ones we have are of very special type. If the allowed costs were just $ 1$ and $ +\infty$, it would be possible using FFT because in that case we would be talking about the convolution of characteristic functions of 2 sets.

I think you people are overthinking the problem.
i dont think the number of digits ever exceeds 30 (i'm almost 100% sure #ofdigits < n), so the constant overhead induced by FFT (plus all of those other computations) will likely make it slower.
[code]#include <stdio.h>
#include <stdlib.h>
typedef struct STATESPACE{
    char mark;
    int preva;
    int prevb;//these are shifted one 0:NULL 1:0 2:1 etc.
} stsp;
void inita(stsp ***a,int num)
{
    int i;
    (*a)=(stsp**)calloc(10,sizeof(stsp*));
    for(i=0;i<10;i++)
	(*a)[i]=(stsp*)calloc(num,sizeof(stsp));
    for(i=1;i<10;i++)
	(*a)[i][i%num].mark=1;
}
void freea(stsp ***a)
{
    int i;
    for(i=0;i<10;i++)
	free((*a)[i]);
    free((*a));
}
void increasing(int num)
{
    int i,j,k;
    stsp **olda,**newa,**tpt;
    int backa,backb,ta;
    inita(&olda,num);
    inita(&newa,num);
    while(olda[0][0].mark==0 && olda[1][0].mark==0 && olda[2][0].mark==0 && olda[3][0].mark==0 &&
	  olda[4][0].mark==0 && olda[5][0].mark==0 && olda[6][0].mark==0 &&
	  olda[7][0].mark==0 && olda[8][0].mark==0 && olda[9][0].mark==0)
    {
	for(i=0;i<10;i++)
	    for(j=0;j<num;j++)
		if(olda[i][j].mark)
		    for(k=i;k<10;k++)
			if(newa[k][(10*j+k)%num].mark==0)
			{
			    newa[k][(10*j+k)%num].mark=1;
			    newa[k][(10*j+k)%num].preva=i+1;
			    newa[k][(10*j+k)%num].prevb=j+1;
//             printf("%d %d: %d %d\n",i,j,k,(10*j+k)%num);fflush(stdout);//debug
			}
	for(i=0;i<10;i++)
	{
	    for(j=0;j<num;j++)
		if(newa[i][j].mark!=olda[i][j].mark)
		    break;
	    if(j!=num)
		break;
	}
	if(i==10)
	{
	    printf("increasing:NO SOLUTION!\n");
	    return;
	}
	tpt=olda;//swap olda & newa
	olda=newa;
	newa=tpt;
    }
    for(i=0;olda[i][0].mark==0;i++);
    backa=i+1;
    backb=1;
    printf("increasing (flip it around):");
    while(backa)
    {
	printf("%d",backa-1);
	ta=olda[backa-1][backb-1].preva;
	backb=olda[backa-1][backb-1].prevb;
	backa=ta;
	if(backa==0)
	    break;
	printf("%d",backa-1);
	ta=newa[backa-1][backb-1].preva;//go to newa (not exactly clean)
	backb=newa[backa-1][backb-1].prevb;
	backa=ta;
    }
    printf("\n");
    freea(&olda);
    freea(&newa);
}
void decreasing(int num)
{
    int i,j,k;
    stsp **olda,**newa,**tpt;
    int backa,backb,ta;
    inita(&olda,num);
    inita(&newa,num);
    while(olda[0][0].mark==0 && olda[1][0].mark==0 && olda[2][0].mark==0 && olda[3][0].mark==0 &&
	  olda[4][0].mark==0 && olda[5][0].mark==0 && olda[6][0].mark==0 &&
	  olda[7][0].mark==0 && olda[8][0].mark==0 && olda[9][0].mark==0)
    {
	for(i=0;i<10;i++)
	    for(j=0;j<num;j++)
		if(olda[i][j].mark)
		    for(k=0;k<=i;k++)
			if(newa[k][(10*j+k)%num].mark==0)
			{
			    newa[k][(10*j+k)%num].mark=1;
			    newa[k][(10*j+k)%num].preva=i+1;
			    newa[k][(10*j+k)%num].prevb=j+1;
//             printf("%d %d: %d %d\n",i,j,k,(10*j+k)%num);fflush(stdout);//debug
			}
	for(i=0;i<10;i++)
	{
	    for(j=0;j<num;j++)
		if(newa[i][j].mark!=olda[i][j].mark)
		    break;
	    if(j!=num)
		break;
	}
	if(i==10)
	{
	    printf("decreasing:NO SOLUTION!\n");
	    return;
	}
	tpt=olda;//swap olda & newa
	olda=newa;
	newa=tpt;
    }
    for(i=0;olda[i][0].mark==0;i++);
    backa=i+1;
    backb=1;
    printf("decreasing (flip it around):");
    while(backa)
    {
	printf("%d",backa-1);
	ta=olda[backa-1][backb-1].preva;
	backb=olda[backa-1][backb-1].prevb;
	backa=ta;
	if(backa==0)
	    break;
	printf("%d",backa-1);
	ta=newa[backa-1][backb-1].preva;//go to newa (not exactly clean)
	backb=newa[backa-1][backb-1].prevb;
	backa=ta;
    }
    printf("\n");
    freea(&olda);
    freea(&newa);
}
int main()
{
    int num;
    printf("this program prints out the minimum decreasing / increasing number that is divisble by your number\n");
    scanf("%d",&num);
    decreasing(num);
    increasing(num);
    return 0;
}
[/code]
running times are miniscule.
EDIT: it now supports no solution[/code]

That code does not give always the smallest solution. Try 5353.

Oh yeah.. well it gets the minimum # of digits for sure.
What is the answer for 5353?
Edit: well you can use the #ofdigits^3 algorithm after the O(n*#ofdigits) algorithm (since you know the min # of digits). and since #ofdigits<30, it should be fine

Unfortunately I don't know. There was a small competition on the net where I found this problem. Competition is over now but one can check their answers. An optimal solution gives 10 points, a number which digits are on monotonous order and which is divisible by a given number but it's not optimal gives 5 points and else you get 0 points. I got 5 points.

Yes, I remembered to flip the number around.

My program gives the same answer for 5353 (111155558888); that's definitely optimal. (And yes, you never get that many digits in a solution.)

[code]#include <iostream>
using namespace std;
bool ans[10000][10][40][2];
int N;
void print(int first, int numdigs, int decr) {
    printf("%d",first);
    int a = first, b = first, c = numdigs-1;
    while (c) {
        int low = decr?0:b;
        int high = decr?b:9;
        for (int i=low; i<=high; i++) {
            if (ans[(10*a+i)%N][i][c-1][decr]) {
                printf("%d",i);
                a = (10*a+i)%N;
                b = i;
                c--;
                break;
            }
        }
    }
    printf("\n");
}
void solve() {
    for (int j=0; j<10; j++) {
        ans[0][j][0][0]=ans[0][j][0][1]=true;
    }
    for (int k=1; k<40; k++) {
        for (int j=0; j<10; j++) {
            for (int i=0; i<N; i++) {
                if (i>0 && i<10) {
                    if (ans[i][i][k-1][1]) {
                        print(i,k,1);
                        return;
                    } else if (ans[i][i][k-1][0]) {
                        print(i,k,0);
                        return;
                    }
                }
                ans[i][j][k][0]=ans[i][j][k][1]=false;
                for (int m=0; m<10; m++) {
                    if (m>=j && ans[(i*10+m)%N][m][k-1][0]) ans[i][j][k][0]=true;
                    if (m<=j && ans[(i*10+m)%N][m][k-1][1]) ans[i][j][k][1]=true;
                }
            }

        }
    }
}
int main() {
    scanf("%d",&N);
    solve();
}
[/code]

In my computer and compiler, Miller4math's program outputs
this program prints out the minimum decreasing / increasing number that is divisble by your number
5353
decreasing (flip it around):111122228888
increasing (flip it around):333333332222

I think it should output 888855551111, which is an optimal solution.

Oops, you're right, his program was wrong after all. So 111155558888 is definitely the answer for 5353.

wait.. are you sure that your program can guarantee the smallest output?
I don't see any checking for that condition.

Yeah, it does. My code generates them from front to back (unlike back to front which I think yours does); and since it checks in order of number of digits followed by the order of the first digit, as soon as it finds a solution, that one is optimal (since it checks the decreasing number starting with that digit first).

DOH! lol i missed that trivial observation :(
a couple of lines could fix my code


it would be nice if we can be  updated with possible and official results of members of mathlinks in the 2005 IMO. Post here latest news!!

any news on the possible scores of some participants for the 1st day?

I guess the contestants are kept offline so they can't tell you  :)  Or am I wrong?  :?

last year therewas a internet cafe next to the hotel...

Hi!

actually they wanted to keep us away from the internet, but they cant really control that we are not doing it, so right now, darij, zetax, noname, werwulff and me are right here in some internet cafe. so, our estimated results for the first day:

me: 21 points (if i didnt do some stupid miscalculation like last year) ( *GGG* :D :D :D)
darij: 15 points
georg: 10 points
christopher(werwulff): 10 points
daniel(zetax): 9 points
christian(noname): 6 points (but he tends to underestimate his results)

the third problem was quite hard; the only solutions during the exam i heard of were solution using muirhead and schur, like mine :). but for example, iura, dust and sasha did not solve problem 3(actually noone i talked to, but sasha told me that he found some guys who solved it)... and darij as well did not solve it, although he is a specialist in inequalities(besides geometry).

So, thats all for now, tomorrow we can give more precise results :) (and i hope i can get 42 points :D ).

Peter

Good luck to all participants! :)

kunny

[quote="Peter Scholze"]Hi!

actually they wanted to keep us away from the internet, but they cant really control that we are not doing it, so right now, darij, zetax, noname, werwulff and me are right here in some internet cafe. so, our estimated results for the first day:

me: 21 points (if i didnt do some stupid miscalculation like last year) ( *GGG* :D :D :D)
darij: 15 points
georg: 10 points
christopher(werwulff): 10 points
daniel(zetax): 9 points
christian(noname): 6 points (but he tends to underestimate his results)

the third problem was quite hard; the only solutions [b]during the exam i heard of [/b]were solution using muirhead and schur, like mine :). but for example, iura, dust and sasha did not solve problem 3(actually noone i talked to, but sasha told me that he found some guys who solved it)... and darij as well did not solve it, although he is a specialist in inequalities(besides geometry).

So, thats all for now, tomorrow we can give more precise results :) (and i hope i can get 42 points :D ).

Peter[/quote]

Indeed, Peter has busdriven me and Georg ;-)
And it's interesting to hear that Peter is cheating...

[quote="ZetaX"][quote="Peter Scholze"]
And it's interesting to hear that Peter is cheating...[/quote][/quote]
Good observation  :lol:  :lol:  :lol:  :rotfl:

[quote="ZetaX"][quote="Peter Scholze"]
And it's interesting to hear that Peter is cheating...[/quote]
Good observation  :lol:  :lol:  :lol:  :rotfl:[/quote]
I think we can call him a predictor!  :D  Lol, I am just joking....

[quote="Peter Scholze"]

me: 21 points (if i didnt do some stupid miscalculation like last year) ( *GGG* :D :D :D)
darij: 15 points
georg: 10 points
christopher(werwulff): 10 points
daniel(zetax): 9 points
christian(noname): 6 points (but he tends to underestimate his results)

the third problem was quite hard; the only solutions during the exam i heard of were solution using muirhead and schur, like mine :). but for example, iura, dust and sasha did not solve problem 3(actually noone i talked to, but sasha told me that he found some guys who solved it)... and darij as well did not solve it, although he is a specialist in inequalities(besides geometry).

So, thats all for now, tomorrow we can give more precise results :) (and i hope i can get 42 points :D ).

Peter[/quote]

Congrats to the two perfect guys. But also an excellent job by the others.  :)

[quote="Valentin Vornicu"]I've met many MathLinkers yesterday at the big hurricane party :heli: I will upload the pictures I took soon  :lol:[/quote]
And their results!  :D


Let $ x_i\in[0,1000], \ i\equal{}1\dots 11$. Prove that there exist 2 points such that $ |x_m\minus{}x_n|\leq 1 \plus{}3\sqrt[3]{x_m x_n}$.

Sorry i have some misstake , i will edit it after i go to school  :D

[quote="quangpbc"][quote="Ronx"]Let $ x_i\in[0,1000], \ i \equal{} 1\dots 11$. Prove that there exist 2 points such that $ |x_m \minus{} x_n|\leq 1 \plus{} 3\sqrt [3]{x_m x_n}$.[/quote]

$ x_{m \plus{} 1}\ge x_{m} \plus{} 2 \plus{} 3\sqrt [3]{x_mx_{m \plus{} 1}}$

[/quote]
Why this holds? Actually, it is very hard to understand what you write..you give no comments.

[quote="Ronx"][b]quangpbc[/b], i used your ideas to fix the solution: suppose $ x_{k \plus{} 1} \minus{} x_k > 1 \plus{} 3\sqrt [3]{x_kx_{k \plus{} 1}}$. Then we have $ \sqrt [3]{x_{k \plus{} 1}} \minus{} \sqrt [3]{x_k} \equal{} \frac {x_{k \plus{} 1} \minus{} x_k}{\sqrt [3]{x_{k \plus{} 1}^2} \plus{} \sqrt [3]{x_k^2} \plus{} \sqrt [3]{x_k x_{k \plus{} 1}}} > \frac {1 \plus{} 3\sqrt [3]{x_{k \plus{} 1}x_k}}{(\sqrt [3]{x_{k \plus{} 1}} \minus{} \sqrt [3]{x_k})^2 \plus{} 3\sqrt [3]{x_k x_{k \plus{} 1}}}$
where we added and subtracted $ 3\sqrt [3]{x_k x_{k \plus{} 1}}$ in denominator to get this inequality. From the above it is obvious that $ \sqrt [3]{x_{k \plus{} 1}} \minus{} \sqrt [3]{x_k} > 1$, otherwise we have
$ \sqrt [3]{x_{k \plus{} 1}} \minus{} \sqrt [3]{x_k} > \frac {1 \plus{} 3\sqrt [3]{x_k x_{k \plus{} 1}}}{(\sqrt [3]{x_{k \plus{} 1}} \minus{} \sqrt [3]{x_k})^2 \plus{} 3\sqrt [3]{x_k x_{k \plus{} 1}}} > \frac {1 \plus{} 3\sqrt [3]{x_k x_{k \plus{} 1}}}{1 \plus{} 3\sqrt [3]{x_k x_{k \plus{} 1}}} \equal{} 1$, so $ \sqrt [3]{x_{11}} > 10$, which gives the contradiction.[/quote]

 I think you're wrong! While $ \sqrt [3]{x_{k \plus{} 1}} \minus{} \sqrt [3]{x_k} > 1$, otherwise we have
$ \sqrt [3]{x_{k \plus{} 1}} \minus{} \sqrt [3]{x_k} > \frac {1 \plus{} 3\sqrt [3]{x_k x_{k \plus{} 1}}}{(\sqrt [3]{x_{k \plus{} 1}} \minus{} \sqrt [3]{x_k})^2 \plus{} 3\sqrt [3]{x_k x_{k \plus{} 1}}} > \frac {1 \plus{} 3\sqrt [3]{x_k x_{k \plus{} 1}}}{1 \plus{} 3\sqrt [3]{x_k x_{k \plus{} 1}}} \equal{} 1$,
ps:Using Diricle's theorem AND IT'LL will work OUT!

[b]ghjk[/b], you can post your solution also. Btw, i'm not wrong: if you suppose that $ \sqrt[3]{x_{k\plus{}1}}\minus{}\sqrt[3]{x_k}<1$, from the inequalities i wrote you get $ \sqrt[3]{x_{k\plus{}1}}\minus{}\sqrt[3]{x_k}>1$, which is contradiction, so we must have $ \sqrt[3]{x_{k\plus{}1}}\minus{}\sqrt[3]{x_k}>1$.

It's not what I meant! I meant you're wrong with this inequality:
\frac {1 + 3\sqrt [3]{x_k x_{k + 1}}}{(\sqrt [3]{x_{k + 1}} - \sqrt [3]{x_k})^2 + 3\sqrt [3]{x_k x_{k + 1}}} > \frac {1 + 3\sqrt [3]{x_k x_{k + 1}}}{1 + 3\sqrt [3]{x_k x_{k + 1}}} = 1
PS:Can you check carefully before telling me wrong? i JUST WANT TO DISCUSS but you seem to be so serious that it isn't neccessary

[quote="ghjk"]It's not what I meant! I meant you're wrong with this inequality:
$ \frac {1 \plus{} 3\sqrt [3]{x_k x_{k \plus{} 1}}}{(\sqrt [3]{x_{k \plus{} 1}} \minus{} \sqrt [3]{x_k})^2 \plus{} 3\sqrt [3]{x_k x_{k \plus{} 1}}} > \frac {1 \plus{} 3\sqrt [3]{x_k x_{k \plus{} 1}}}{1 \plus{} 3\sqrt [3]{x_k x_{k \plus{} 1}}} \equal{} 1$
PS:Can you check carefully before telling me wrong? i JUST WANT TO DISCUSS but you seem to be so serious that it isn't neccessary[/quote]
Yes, of course it is wrong in general, but it becomes correct of we suppose $ \sqrt [3]{x_{k \plus{} 1}} \minus{} \sqrt [3]{x_k} < 1$.


Prove or disprove:

For positive integers m and n, (mn)!=m!n! and (m+n)!=m!+n!
Thanks. :)

Do you maybe mean that we need to find integers $ m$ and $ n$ so the equations hold  :)?
Because I dont see the point in proving or disproving this since we can easily check case for m=2, n=2 or something like that and see that it does not hold.

In order to disprove a statement, it is only necessary that you provide a single counter-example.  You have done this, so you disproved the statement (although you might want to state that $ (2\plus{}2)!\equal{}4!\equal{}24\neq 2\plus{}2\equal{}2!\plus{}2!$ or something to justify your claim).

This problem does not ask you to find all solutions to the system; it is simply giving a hypothetical formula and asking you to show whether or not that formula holds in general.

That's exactly what I had done.But I was not sure.Thank you for clearing my doubt.


ΠΡΟΤΑΣΗ  5. Σημείο Ρ κινείται επί ημιπεριφέρειας και έστω Σ το εσωτερικό σημείο τομής της διαμέτρου της ΑΒ, από το τόξο κύκλου με κέντρο το Α και ακτίνα ΑΡ. Προσδιορίστε τη θέση του σημείου Ρ, ώστε το ευθύγραμμο τμήμα ΡΣ να είναι μέγιστο και αποδείξτε ότι τότε, η μεσοκάθετος του ΑΒ, το χωρίζει σε δύο τμήματα, το ένα τριπλάσιο του άλλου.

Ενδιαφέρουσα παρατήρηση, αλλά δεν ξέρω την απάντηση. Μετά τη λύση του προβλήματος ( γεωμετρικά αλλά με τους αναγκαστικούς αλγεβρικούς μόνο υπολογισμούς ), δεν σκέφτηκα ούτε από περιέργεια έστω, να δω πόσο είναι αυτή η γωνία.

     Με μία γρήγορη ματιά πάντως από το σχέδιο ( με σχεδιαστικό πρόγραμμα υπολογιστή ), φαίνεται ελαφρώς μεγαλύτερη από 48º. Θα το κοιτάξω πιο προσεκτικά, πριν η απάντησή μου να είναι οριστική.

     Έχω βρει ότι μέγιστο ΡΣ = (4R√2 )/(3√3 ), όπου R, είναι η ακτίνα της δοσμένης ημιπεριφέρειας και ελπίζω να μην έχω κάνει λάθος στις πράξεις. Η τιμή αυτή του μέγιστου ΡΣ, προκύπτει από τη θέση του Ρ όταν ΑΡ = 4R/3.

     ( √2 = ρίζα του 2 και συγνώμη για τα ακατάλληλα σύμβολα )

     Με εκτίμηση.
     Κώστας Βήττας.

[quote="Ramanujan"][quote="silouan"]Αν οι υπολογισμοί μου είναι σωστοί το ΡΣ είναι μέγιστο όταν η γωνία ΡΑΣ είναι 48 μοίρες ...Έχω άδικο κύριε Κώστα ??[/quote]

Silouan νομίζω ότι είναι   $\frac{180 \cdot \arctan \left( \frac{\sqrt{5}}{2}\right) }{ \pi }\approx 48.1897$

 άρα σωστός πρέπει να είσαι... :)[/quote]

Ακριβώς...... :lol:
Απλά έπρεπε να γράψω περίπου 48

[quote="vittasko"]Άρα έχουμε δύο διαφορετικές προσεγγίσεις και θα είχε ενδιαφέρον να γίνουν ευρύτερα γνωστές κάποια στιγμή ( νομίζω ότι δεν έχει νόημα να κρατάμε τις λύσεις μας μυστικές, γιατί κάτι ενδιαφέρον μπορεί να προκύψει από ένα άλλο μάτι ).

     Μπορείτε σας παρακαλώ να επαληθεύσετε αν η τιμή αυτή της γωνίας <ΡΑΣ ≈ 48.1897,
αντιστοιχεί στην τιμή ΑΡ = 4R/3, για να σιγουρευτώ και εγώ ότι δεν έχω λάθος?

     Σας ευχαριστώ και τους δύο.
     Κώστας Βήττας[/quote]

Ναι τόσο βγαίνει . :)
 Η λύση μου είναι αρκετά άσχημη και δεν την γράφω ακόμα.
 Θα δοκιμάσω πρώτα να την λύσω γεωμετρικά.  :D

[quote="vittasko"]Για τη λύση ενός γεωμετρικού προβλήματος, μία απλή απόδειξη κατανοητή στο μέσο νου, έχει ίσως μεγαλύτερη αξία ( αξιολογείται με καλύτερο βαθμό ), πρέπει να είναι στόχος μας και μπράβο σε κάποιον όταν το καταφέρνει.

     Πιστεύω όμως πως μας είναι χρήσιμη κάθε απόδειξη έστω και μακροσκελής, γιατί μπορεί να συνεισφέρει το ίδιο στην έμπνευση κάποιας ιδέας που θα βοηθήσει στην λύση άλλων προβλημάτων.

     Τελικά νομίζω ότι δεν υπάρχουν άσχημες αποδείξεις.

     Κώστας Βήττας.[/quote]

Σε γενικές γραμμές έχετε δίκιο.

Πιστεύω ότι είναι άσχημη η απόδειξη γιατί χρησιμοποιώ μια μέθοδο
λύσης του προβλήματος που ξέρω ότι θα μου δώσει απάντηση
χωρίς να έχω καταλάβει την ουσία του προβλήματος. :D

Όμως αφού θέλετε  θα σας στείλω την απόδειξη με pm να της ρίξετε μια ματιά .  :)

[quote="silouan"]Δημιούρησες μια συνάρτηση και μετά βρήκες το μέγιστο μέσω ανάλυσης έτσι δεν είναι Ramanajuan ??  :)[/quote]

 :blush:  :D

Μάλλον δουλέψαμε το ίδιο  :blush:  :lol:

Χαχα, αυτό μας κάνει τρεις! :lol: Σνομπάρουμε τη γεωμετρία μου φαίνεται! :P

Πιθανον όμως χάνουμε το νόημα της Ευκλείδιας γεωμετρίας  :maybe:

δοκιμασε κανεις αναλυτικη γεωμετρια? γιατι με ευκλειδια γεωμετρια δεν την καταφερα., με αναλυση βγαινει ευκολα ?

Νομίζω ότι είναι καιρός, να θέσω υπό την κρίση σας τη λύση που έχω υπόψη μου, για την άσκηση αυτή.

[b][size=117]ΠΡΟΤΑΣΗ 5. [/size][/b] - Σημείο $P$ κινείται επί ημιπεριφέρειας και έστω $\Sigma$ το εσωτερικό σημείο τομής της διαμέτρου του $AB,$ από το τόξο κύκλου με κέντρο το $A$ και ακτίνα $AP.$ Προσδιορίστε τη θέση του σημείου $P,$ ώστε το ευθύγραμμο τμήμα $P\Sigma$ να είναι μέγιστο και αποδείξτε ότι τότε, η μεσοκάθετος του $AB,$ το χωρίζει σε δύο τμήματα, το ένα τριπλάσιο του άλλου.

[b][size=117]ΛΥΣΗ. [/size][/b]– Έστω $P$ το ζητούμενο σημείο και αρκεί να εκφραστεί το μέτρο του $AP,$ σε συνάρτηση με το μέτρο του σταθερού δοσμένου τμήματος $AB.$

Έστω $A',$ το σημείο τομής $($εξωτερικό των $A, B$$)$ της ευθείας $AB,$ από το ημικύκλιο $(\alpha)$ με κέντρο το $A$ και ακτίνα $AP.$

Στο ημικύκλιο $(\kappa),$ λόγω διαμέτρου $AB,$ έχουμε ότι η $AB$ είναι κάθετη στην $BP$ και άρα η ευθεία $BP,$ είναι εφαπτομένη του ημικυκλίου $(\alpha)$ στο σημείο $P.$ Επομένως προκύπτει η ισότητα των γωνιών $\angle\Sigma A'P = \angle\Sigma PB, (1)$ και επειδή $\angle\Sigma A'P = \angle AA'P = \angle APA',$ $(2)$ $($από ισοσκελές τρίγωνο $\bigtriangleup AA'P),$ συμπεραίνουμε ότι $\angle APA' = \angle\Sigma PB = \angle\omega,$ $(3).$

Στο τρίγωνο $\bigtriangleup PA'B$ λόγω της $(3),$ $\Longrightarrow$  $\frac{(PA')^{2}}{(PB)^{2}}= \frac{AA'}{AB}\cdot\frac{\Sigma A'}{\Sigma B}$ $,$ $(4).$

Από ορθογώνια τρίγωνα $\bigtriangleup PA'\Sigma$ $(\angle A'P\Sigma = 90^{o})$ και $\bigtriangleup PAB$ $(\angle APB = 90^{o}),$ έχουμε:

$(PA)^{2}= (\Sigma A')^{2}-(P\Sigma)^{2},$ $(5)$ και $(PB)^{2}= (AB)^{2}-(AP)^{2},$ $(6).$

Έστω $AP = AA' = A\Sigma = R_{1}$ και $AB = 2R,$ οπότε $\Sigma B = 2R-R_{1}$ και $\Sigma A' = 2R_{1}$ και άρα η $(4)$ λόγω

και των $(5),$ $(6),$ γίνεται $\frac{4{R_{1}}^{2}-(P\Sigma)^{2}}{4{R}^{2}-{R_{1}}^{2}}$ $=$ $\frac{R_{1}}{2{R}}\cdot\frac{2{R_{1}}}{2{R}-R_{1}}$ $,$ $(7).$ $\Longrightarrow$ $P\Sigma = \frac{R_{1}\sqrt{2{R}-R_{1}}}{\sqrt{R}}$ $,$ $(8).$

Θέτουμε $R_{1}= \kappa R$, οπότε η $(8)$ γίνεται $P\Sigma = R\cdot(\kappa \sqrt{2-\kappa})$ , $(9).$

Από την ισότητα $(9),$ συμπεραίνουμε $($ αφού $R =$ σταθερό $)$ ότι το $P\Sigma$ γίνεται μέγιστο όταν γίνεται μέγιστη η τιμή της παράστασης $\kappa\sqrt{2-\kappa}$ . Επειδή τώρα $\kappa > 0$  και  $2-\kappa > 0$ , $($ γιατί αλλιώς θα είχαμε $P\Sigma = 0$ $)$ και ισχύει $\kappa+(2-\kappa) = 0 =$  σταθερό,

$\Longrightarrow$ το γινόμενο $\kappa\sqrt{2-\kappa}$ γίνεται μέγιστο όταν $\frac{\kappa}{1}= \frac{2-\kappa}{\frac{1}{2}}$ από την οποία προκύπτει ότι $\kappa = \frac{4}{3}$. Επομένως το ΡΣ γίνεται μέγιστο

όταν $R_{1}= \frac{4{R}}{3}$ και το πρόβλημα έχει λυθεί. Για $R_{1}= \frac{4{R}}{3},$ από την ισότητα $(8)$ προκύπτει η τιμή $max P\Sigma = \frac{4{R}\sqrt{2}}{3\sqrt{3}}$

$\bullet$  Για την απόδειξη ότι η μεσοκάθετη του $AB$ χωρίζει το $P\Sigma,$ όταν αυτό γίνεται μέγιστο, σε δύο τμήματα το ένα τριπλάσιο του άλλου, εργαζόμαστε ως εξής:

Έστω $\Lambda,$ η προβολή του $P$ επί την $AB$ και $M$ το σημείο του $P\Sigma,$ που έχει για προβολή του το κεντρο $K$ του δοσμένου ημικυκλίου $(\kappa).$ Θα αποδείξουμε ότι $\Sigma M = 3(MP)$ και αρκεί για αυτό να αποδειχθεί ότι ισχύει $\Sigma K = 3(K\Lambda).$

Αποδεικνύεται εύκολα ότι η $P\Sigma$ είναι διχοτόμος της γωνίας $\angle \Lambda PB$ και άρα σύμφωνα με το θεώρημα της διχοτόμου, στο ορθογώνιο τρίγωνο $\bigtriangleup \Lambda PB$

$(\angle P\Lambda B = 90^{o})$ $\Longrightarrow$ $\frac{\Sigma \Lambda}{\Sigma B}= \frac{P\Lambda}{PB},$ $\Longrightarrow$ $\Sigma \Lambda = \frac{(\Sigma B)\cdot(P\Lambda)}{PB}$ , $(1).$

Από όμοια ορθογώνια τρίγωνα $\bigtriangleup P\Lambda \Sigma$ και $\bigtriangleup A'P\Sigma,$ $\Longrightarrow$ $\frac{P\Lambda}{PA'}= \frac{P\Sigma}{\Sigma A'}$ $\Longrightarrow$ $P\Lambda = \frac{(PA')\cdot(P\Sigma)}{\Sigma A'}$ , $(2).$

Από τις $(1),$ $(2),$ αντικαθιστώντας με τα ίσα τους τα διάφορα ευθύγραμμα τμήματα και λαμβάνοντας υπόψη ότι $P\Sigma = \frac{4R}{3}$ προκύπτει ότι $\Sigma \Lambda = \frac{4R}{9}$ και επειδή $K\Sigma = \frac{3R}{9}$ $(= R-\Sigma B = R-\frac{2R}{3})$ συμπεραίνουμε ότι $\Sigma K = 3 (K\Lambda)$ και άρα $\Sigma M = 3 (MP)$ και το $(\beta)$ ζητούμενο της πρότασης έχει αποδειχθεί.

Κώστας Βήττας.


linear algebra
matrix
linear algebra unsolved

Prove that if $A$ is an $n\times n$ matrix, then $A$ and $A^{T}$ are similiar.

Not suprisingly, the problem has been discussed [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=similar&t=72724]before[/url].

To prove it for Jordan blocks $J_{k}= \left( \begin{matrix}\lambda & 1 & 0 & \ldots & 0 \\ 0 & \lambda & 1 & \ldots & 0 \\ \ldots & \ldots & \ldots & \ldots \\ 0 & 0 & 0 & \ldots & \lambda \end{matrix}\right)$, let $S_{k}= \left( \begin{matrix}0 & \ldots & 0 & 1 \\ 0 & \ldots & 1 & 0 \\ \ldots & \ldots & \ldots & \ldots \\ 1 & \ldots & 0 & 0 \end{matrix}\right)$, so that $J_{k}^{T}= S_{k}^{-1}J_{k}S_{k}.$

Now it is easy to prove that the conclusion holds for $A$ in Jordan form, from where we get that it holds for all $A$.

I first found this exercise in a book, where a hint was given (to consider that $S_{k}$).


algebra
polynomial
trigonometry
induction
algebra proposed

Show that there exists a polynomial $p(x)$ (not the zero polynomial) such that $(x - \tan{1^{\circ}})\cdot p(x) \in \mathbb{Z}[x]$.

All you have to do is to show tan(1 degree) is algebraic. But you know tan(30 degrees) is algebraic, and you can use:
tan(a+1) = (tan(a)+tan(1))/(1-tan(a)tan(1)). By induction, you will then show that tan(1 degree) is algebraic, which ends the proof.

Identifying the real and imaginary parts in the equality
\[(\cos\alpha +i\sin\alpha)^n=\cos{n\alpha}+i\sin{n\alpha}
\]
we obtain
\[\cos{n\alpha}=\cos^n\alpha-\binom{n}{2}\cos^{n-2}\alpha \sin^2 \alpha+\ldots
\]
\[
\sin{n\alpha}=\binom{n}{1}\cos^{n-1}\alpha \sin\alpha-\binom{n}{3}\cos^{n-3}\alpha \sin^3 \alpha+\ldots
\]

and hence
\[\tan{n\alpha}=\frac{\binom{n}{1}\tan\alpha -\binom{n}{3}\tan^{3}\alpha +\ldots}{1-\binom{n}{2}\tan^{2}\alpha +\ldots}
\]
Setting $\alpha=1^{\circ}$ and $n=45$, we obtain, after clearing denominators, that $\tan {1^{\circ}}$ is the root of an integer polynomial, as desired.


[quote="indianamath"]tell your friends and other people you know from IN on here. we are a strong math state and we have a LOT of AoPsers on here. We should be one of the most active communities! :jump:  :trampoline:  :wow:[/quote]

I'm all for it!

[quote="indianamath"]tell your friends and other people you know from IN on here. we are a strong math state and we have a LOT of AoPsers on here. We should be one of the most active communities! :jump:  :trampoline:  :wow:[/quote]

ive already invited one person, but im not sure he's coming.


number theory
prime numbers
number theory unsolved

do exist an infinit set S from natural numbers that for two member of that we have a^2-ab+b^2|(ab)^2

Let $ x,y$ any integer. Consider $ a\equal{}xc(x^2\minus{}xy\plus{}y^2),b\equal{}yc(x^2\minus{}xy\plus{}y^2)$.
Obviosly $ a^2\minus{}ab\plus{}b^2\equal{}(x^2\minus{}xy\plus{}y^2)^3c^2|(ab)^2\equal{}(x^2\minus{}xy\plus{}y^2)^4c^4x^2y^2$.

[quote="Rust"]Let $ x,y$ any integer. Consider $ a \equal{} xc(x^2 \minus{} xy \plus{} y^2),b \equal{} yc(x^2 \minus{} xy \plus{} y^2)$.
Obviosly $ a^2 \minus{} ab \plus{} b^2 \equal{} (x^2 \minus{} xy \plus{} y^2)^3c^2|(ab)^2 \equal{} (x^2 \minus{} xy \plus{} y^2)^4c^4x^2y^2$.[/quote]
From what I see, the problem should mean for any two members $ a$ and $ b$, $ \frac {a^2b^2}{a^2 \minus{} ab \plus{} b^2}$ should be an integer.  Your construction only lets it work for when the $ c$s are equal in $ a$ and $ b$, I believe.

I hope that the following solution is valid:
[hide="Solution"]
Assume that an infinite set exists.  Notice that $ \frac {a^2b^2}{a^2 \plus{} b^2 \minus{} ab}$ is an integer, so we can add $ ab$ to this and it will still be an integer.  Then, $ \frac {ab(a^2 \plus{} b^2)}{a^2 \plus{} b^2 \minus{} ab}$ is an integer.  Letting $ \gcd (a, b) \equal{} g$,  we can let $ a \equal{} a_1g$ and $ b \equal{} b_1g$.  This gives that $ \frac {g^2a_1b_1(a_1^2 \plus{} b_1^2)}{a_1^2 \plus{} b_1^2 \minus{} a_1b_1}$ is an integer.  Yet, $ \gcd (a_1, a_1^2 \plus{} b_1^2 \minus{} a_1b_1) \equal{} \gcd (a_1, b_1^2) \equal{} 1$ since $ a_1$ and $ b_1$ are coprime.  This implies that $ \frac {g^2b_1(a_1^2 \plus{} b_1^2)}{a_1^2 \plus{} b_1^2 \minus{} a_1b_1}$ is an integer.  We can do the same thing and cancel off $ b_1$, so $ \frac {g^2(a_1^2 \plus{} b_1^2)}{a_1^2 \plus{} b_1^2 \minus{} a_1b_1}$ is an integer.  Then, we can subtract $ g^2$ from this and get an integer, so $ \frac {g^2a_1b_1}{a_1^2 \plus{} b_1^2 \minus{} a_1b_1}$ is an integer.  We can cancel off the $ a_1$ and the $ b_1$ as we did before, so $ \frac {g^2}{a_1^2 \plus{} b_1^2 \minus{} a_1b_1}$ is an integer.  We choose the minimal number from the set, say $ a$, and a number $ b$, which is arbitrarily large.  This means that $ \gcd (a,b)\le a$.  Also, $ a_1^2 \plus{} b_1^2 \minus{} a_1b_1$ can become arbitrarily large, so it can surpass $ a^2$, which is fixed.  Hence, $ a_1^2 \plus{} b_1^2 \minus{} a_1b_1 > a^2\ge g^2$ for large enough $ b$, which means that $ \frac {g^2}{a_1^2 \plus{} b_1^2 \minus{} a_1b_1}$ cannot be an integer for minimal $ a$ and large enough $ b$.  This is a contradiction, so no infinite set exists.  [/hide]

it appears valid

another question arises here. can set S have arbitrary many elements (if we start with the smallest number big enough)
 or if not what is the maximum.

to answer my own question. Yes, set S can have arbitrary many elements. an example for $ n$ elements can be generated in a following way.

as The QuattoMaster 6000 has shown $ \frac {a^2b^2}{a^2 \minus{} ab \plus{} b^2}$ is an integer only if $ \frac {g^2}{a_1^2 \minus{} a_1b_1 \plus{} b_1^2}$ is an integer. ($ g \equal{} gcd(a,b)$ and $ a \equal{} ga_1,b \equal{} gb_1$)

now let $ p_1,p_2...p_n$ be first $ n$ prime numbers ($ p_1<p_2....<p_n$).
define $ a_i \equal{} (p_{n}^2 \minus{} p_{n}p_{n \minus{} 1} \plus{} p_{n \minus{} 1}^2)!\cdot p_i$

obviously set $ (a_1,a_2....,a_n)$ has the desired properties


function
limit
topology
real analysis
real analysis unsolved

Let $C_0$ be the space of functions $f : \mathbb R \rightarrow \mathbb R$ such that

$\lim_{x \to \infty} f(x) = \lim_{x \to -\infty} = 0$

The space $C_0$ becomes a metric space with the sup-norm:

$\Vert f \Vert = \sup_{x \in \mathbb R} |f(x)|$

Let $A$ be a family of functions in $C_0$ that is uniformly bounded and equicontinuous. Prove that every sequence of functions from $A$ has a convergent subsequence in the sup-norm.

I can't get anywhere on this problem. Does someone know how this argument goes?

[quote="grobber"]These functions can be regarded as continuous functions on the circle: the one-point compactification of $\mathbb R$. After that, just apply the Arzela-Ascoli Theorem: a subset of $C(X,\mathbb R)$, where $X$ is a compact Hausdorff topological space, has compact closure iff it's uniformly bounded and equicontinuous.

As for the theorem itself, its proof is not that short, and I'd have to spend some time on it to reconstruct it, so I won't post it, but I'm sure you can find it in any self-respecting book on the subject :).[/quote]
The theorem is incorrect in this form. Take $f_n(x) = \frac{1}{\cosh(x-n)}$, then the $f_n$ are equicontinuous and uniformly bounded, but have no convergent subsequence in the sup-norm.

What's needed is that the functions are equicontinuous also at $\pm \infty$, i.e. $\forall \epsilon \, \exists M \, \forall f_n \, |x| > M \implies |f_n(x)| < \epsilon$, by Arzela-Ascoli.


We roll a dice until we get $ 6$ two times. Let $ Y$ - number of all rolls and $ X$ - number of "$ 1$". Find $ E(Y \minus{} X|X \equal{} 4)$.

i think.. i might have picked the probability wrong...  $ \frac{j\plus{}k\minus{}1!}{j!k\minus{}2!}$ might be the right coeffecient..?

hmmm something is wrong in this probability. Correct answer should be 12 :/

Maybe i'm wrong, but...

You wait for two "6" and just don't count "1". So you are counting only "2",... , "6".
To get one "6" (when throwing "2"..."6" only) you (on average) wait 5 turns.

So the answer is 10.


Let $ABCD$, $A'B'C'D'$ be quadrilaterals incsribed in the same circle $O$. The four diagonals all pass through a point $P$. The angle between the diagonals of the first quadrilateral is equal to that of the second quadrilateral. Moreover, $AB=A'B'$ and $CD=C'D'$. Show that the two quadriaterals must be congruent.


calculus integration function probability geometry derivative topology

I know that advanced math is a beautiful thing. And that most of us are passionnate about math. However, some people find that combinatorics are the most beautiful math one can study, while others are more partial to complex analysis.

As for me, definite integrals! More so when one does double/triple integrals (even if one of the variables was bound by a function)

Can you explain what you find so beautiful about definite integrals?

Oh, and I have a thing for linear algebra.

I find the main foundations of complex analysis (or really anything having to do with complex numbers) to be very beautiful. But I also really appreciate probability and combinatorics and calculus/differential geometry. To me, though, the divisions between the areas of mathematics are just in our minds (and our books). I believe that all of the areas of mathematics are connected. We just have to find those connections.

And those said connections are what makes mathematics as a whole beautiful.

The clichéd phrase "Beauty is in the mind of the beholder" applies well here.

Here follows nonsense.

All mathematics arises from the same root of logic, and it is all beautiful. Even if a field seems particularly ugly to one, it must be beautiful, being mathematics. Beauty is not in the eye of the beholder here. Mathematics transcends human reactions, preferences and emotions - it is beautiful, period.

My own lowly feelings, which have no meaning, significance, or relevance whatsoever, prefer combinatorics.

I find definite integrals the prettiest because its applications transcend feelings and fields.

According to me, the most interesting and beautiful area of mathematics has to be Geometry. You can always arrive at the answer by performing calculations no doubt, but it is only in the branch of Geometry that in most cases you can actually [b][i]see[/i][/b] the answer. That’s what is beautiful about Geometry. To get to the solution of good geometry problem one has to depend on one’s ability to see through the intricate designs made up of lines and curves and in some cases solid shapes (3D geometry) and discover the patterns that lie deep within. However, once you discover this hidden pattern, there is an instant realization of the beauty of the problem solved and makes you wonder what other beautiful patterns exist in this universe, which is, mostly made up of shapes and designs. You realize the ubiquitous nature of the geometry in the universe.

The cliché [i][b]“A picture is worth a thousand words”[/b][/i] aptly applies to Geometry. Just a glimpse at the diagram is all that’s needed to understand the whole purport of the theorem or problem. That is what fascinates me about geometry.

I always seem to draw a clear analogy between music and mathematics and especially in case of Geometry and Mozart.
To me Geometry is separate class of mathematics as beautiful and pure as Mozart’s compositions are compared to other classical music.
There are many geometrical patterns my fascination over which never ceases.
Take for instance the Nine-point circle theorem. This is one of the many gems that exist in nature by design. There’s absolutely nothing superfluous about it. So very “self-sufficient”, a quality that applies very much to Mozart’s compositions. I could continue to talk a lot about other associations of music and mathematics in general.

However, I always believe that mathematics is made up of different elements to make one whole unit. You cannot separate elements from one another because the beauty of mathematics is that all the separate elements are in fact interrelated in a very intricate manner almost to the extent that one element is a derivative of the other. So you cannot exclusively deal with one element without overlapping or burgeoning into the other. Not even with geometry.

Zetax Wrote:
[quote]
Geometry is old, boring and unimportant. There is noting to research, nothing unexpected, etc.

Things change if you talk about modern geometries, but your post sounds more like that ancient boring greek thing.[/quote]

One can never really achieve anything of value, unless he has appreciation and realization of the value of the things already discovered.  :wink:

Hmm... it seems I disagree with nearly every post in this topic...

Mathematics is beautiful not because of any human experience, such as awe/wonder, thrill/satisfaction of problem solving, etc., but rather because of its own structure. It would be beautiful even with any being to know of it. (OK, that probably sounds like Pythagorean nonsense...)

I don't understand where you're coming from. Beauty is necessarily subjective. Subjective means that someone had to form an opinion

Elementary math - Combinatorics...

Advanced - Topology...

[quote="JBL"][quote="Erken"]Elementary math - Combinatorics...[/quote]  Quite a lot of combinatorics (both of the graph theory/probabilistic method variety and the enumerative/algebraic variety) is "non-elementary."  Also, simply listing two branches of mathematics doesn't seem like a very interesting addition to the discussion without some argument about why those branches are the ones being listed.[/quote]

okay, I meant those branches of combinatorics which are related to elementary math. Why?

hmm...I personally consider combinatorics to be the most beautiful branch of [b]elementary[/b] math simply because in comparison with other branches, combinatorics "insistently" demands an exceptional talent - a man has to be quite richly endowed. Whereas, for example - profound knowledge is enough to solve elementary algebra problems.

I think there's some value in distinguishing between fields which have 'beautiful results' and 'beautiful solutions'.  For example, I study probability theory and feel that it is a field which is filled with beautiful results, but some of the solutions are too technical to be called beautiful.  Combinatorics jumps to mind as a field which has some beautiful results and even some very beautiful solutions (see: proofs utilizing the probabilistic method).

There's a very big personal component in any sort of aesthetic considerations; for example, I don't find much of algebra to be that interesting or beautiful, although I'm sure many would disagree vehemently (and with good reason!).

The ideas are beautiful. The beauty of a proof comes from the unexpectedness of an idea that leads to the argument. It's the "how did he think of that?" thing that matters. So we find beauty in math, in physics, in poetry, in music, and so on. Of course math is totally made of ideas, so we have plenty of material to admire.

All mathematical concepts have some intuitive meaning in their roots. We start if intuitive concepts, define some things in a precise way and generalize, generalize, generalize...but the intuition is still there. The ideas are still there.

Of course the personal component is very strong. Because of it some people like algebra, other prefer geometry, etc. I don't know nothing of advanced math, for now I'm just playing with elementary math. For me, the most beautiful math is geometry, probably because it is the most intuitive of the subjects. Maybe I will change my ideas when I develop more abstraction capacity. I have really hated number theory, but after seeing some nice results and proofs I changed my view and now look at NT in a more friendly way. Following, I think combinactorics is nice. Algebra is the worst subject for me, but this may change in the future of course.

Topology.

Topology is a whole other view, a different perspective of mathematics. It requires imagination and intuition, yet still requires the skills of integral calculus. The most amazing intuitive idea that i've seen would be ricci flow and surgery. To use purely physical ideas (such as entropy and thermodynamics) and intuitively use them in a purely mathematical way is elegant.


Show that for $p>1$ we have \[\lim_{n\rightarrow+\infty}\frac{1^p+2^p+...+(n-1)^p+n^p+(n-1)^p+...+2^p+1^p}{n^2} = +\infty\]
Find the limit if $p=1$.

[quote="Peter VDD"]Show that for $ p>1 $ we have \[ \lim_{n\rightarrow+\infty}\frac{1^p+2^p+...+(n-1)^p+n^p+(n-1)^p+...+2^p+1^p}{n^2} = +\infty \]
Find the limit if $ p=1 $.[/quote]

1)Using power mean inequality $ \left(k^p+(n-k)^p\right)\ge 2\left(\frac{n}{2}\right)^p $ so $ 1^p+2^p+...+(n-1)^p+n^p+(n-1)^p+...+2^p+1^p\ge 2n\cdot\left(\frac{n}{2}\right)^p\longrightarrow \infty $ as $ n\longrightarrow \infty $
2)For the second case it is easy to calculate $ \frac{1^p+2^p+...+(n-1)^p+n^p+(n-1)^p+...+2^p+1^p}{n^2}=1 $

[quote=seamusoboyle]We could also use that whatyeamaycallit theorem, LaGrange, and diff the top and bottom twice...[/quote]

L'hospital you mean?


algorithm
number theory
Euclidean algorithm

Prove that there two and only two fractions with denominators less than 19 which lie between $\frac{10}{13}$ and $\frac{4}{5}$

Let the desired fraction be $r$. Then

${10\over 13}<r<{4\over 5}$

${5\over 4}<{1\over r}<{13\over 10}$

${1\over 4}<\frac{1-r}{r}<{3\over 10}$

${10\over 3}<\frac{r}{1-r}<4$

${1\over 3}<\frac{4r-3}{1-r}<1$

$1<\frac{1-r}{4r-3}<3$

$0<\frac{4-5r}{4r-3}<2$

${1\over 2}<\frac{4r-3}{4-5r}$

The smallest possible denominators of $r$ will be obtained if the last inequation's RHS is integer. Therefore $\frac{4r-3}{4-5r}=n\iff r=\frac{4n+3}{5n+4}$ where $n$ is an integer not less than $1$. Obviously, the two desired fractions are ${7\over 9}$ and ${11\over 14}$, for $n=1$ and $n=2$. Using Euclidean algorithm, it's easily proven that $(4n+3,5n+4)=1$, hence we can have no other solutions in a non-reduced form.

Alternatively, we can put $\frac{4r-3}{4-5r}={p\over q}\iff r=\frac{4p+3q}{5p+4q}$ where ${p\over q}>{1\over 2}\iff q<2p$. For $p=1$ we can have only $q=1$, giving $r={7\over 9}$. For $p=2$ we can have $q=1,2,3$, giving $r={11\over 14},{14\over 18},{17\over 22}$. The second fraction is equal to ${7\over 9}$ and the third doesn't satisfy the initial condition. For $p\geqslant 3$ the denominator is necessarily greater than $18$. Again, Euclidean algorithm shows $(4p+3q,5p+4q)=(p,q)$, hence no non-reduced solutions are obtained for coprime pairs of $p$ and $q$.

NOTE: The claim and the proof are valid if we take for granted that the fractions must be reduced. Otherwise, there's another one, ${14\over 18}$


If $n$ is a positive integer greater than $1$, let $h$ be the hundreds digit and $u$ be the units digit of $\frac{1 + 5^{2n+1}}{6}$. Find all possible ordered pairs $(h,u)$. (no calculator)

[hide="Solution"]
$f(n) = \frac{5^{2n+1} + 1}{5 + 1} = 5^{2n} - 5^{2n-1} + 5^{2n-2} +-... -5 + 1$

Plainly, it only has possible values of $u = 1$ (there are always an odd number of terms, and the $5$s all cancel out.)  As for the values of $h$, we will have to evaluate this expression $\bmod 125$ and $\bmod 8$.

Plainly, all exponents above $5^3 \equiv 0 \bmod 125$ so the only possible variance $\bmod 125$ is in the first few terms.  We calculate $f(1) = 1, f(2) = 21$.  We conclude, for all $n \ge 2, f(n) \equiv 21 \bmod 125$.

Now, $f(n) \equiv (-3)^{2n} - (-3)^{2n-1} +-... -(-3) + 1 \equiv 3^{2n} + ... \bmod 8$.  For $n > 2$, every increment of $n$ adds an exponent of the form $3^{2k} + 3^{2k-1} = 3^{2k-1} (3 + 1) \equiv 4 \bmod 8$, so we know $f(n) \equiv 1, 5 \bmod 8$.

Case 1:  $f(n) \equiv 1 \bmod 8$.  This gives us $f(n) \equiv 521 \bmod 1000$ (for $n > 2$), so we know $h = 5$.

Case 2:  $f(n) \equiv 5 \bmod 8$.  This gives us $f(n) \equiv 21 \bmod 1000$, which gives us $h = 0$.

Finally:  $(h, u) = \boxed{(5, 1), (0, 1)}$.
[/hide]

[quote="gotztahbeazn"]hmm why do u evaluate it $\bmod 8$ again?[/quote]
Because 8*125=1000 ...  :?

[quote="Acadeeme"]Hi all

can one give more explain of this?   please

Case 1:  $f(n) \equiv 1 \bmod 8$.  This gives us $f(n) \equiv 521 \bmod 1000$ (for $n > 2$), so we know $h = 5$.

Case 2:  $f(n) \equiv 5 \bmod 8$.  This gives us $f(n) \equiv 21 \bmod 1000$, which gives us $h = 0$.[/quote]

This is a straightforward application of Chinese Remainder Theorem, or CRT.  In the first case, $f(n) \equiv 1 \bmod 8$ and $f(n) \equiv 21 \bmod 125$ (as has been established earlier); then if we evaluate $\bmod 1000$ then from the first congruence we know $f(n) \equiv 1, 9, 17, ... \bmod 1000$ and from the second case we know $f(n) \equiv 21, 146, 276, ... \bmod 1000$, and the Chinese Remainder Theorem states that there is a unique intersection of these two congruences, which is at $f(n) \equiv 521 \bmod 1000$.

Similar logic applies to the second; for more information, consult

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

ooh
i have never thought of CRT  :oops:

thank u  t0rajir0u

[quote="matt276eagles"]Is there a simple way to find the intersection of the two congruences?  It seems like guess-and-check to me...  :([/quote]

It's sorta complicated. Say you have

$x \equiv a \pmod{m}$
$x \equiv b \pmod{n}$

and you want to find $x \pmod{mn}$ (assume $(m,n) = 1$). So $x = km+a \equiv b \pmod{n}$ and $km \equiv (b-a) \pmod{n}$. Then there is only one congruence to solve, but I guess it's not simple and still sort of guess-and-check.


N2 + 2H2 => N2H4

If N2H4 produced as a gas rather than a liquid, how would the sign and magnitude of delta(S) for the reaction be affected? Why?

After the process of dissolving silver nitrate in pure water, What's the sign of delta(G) and delta(S), why?

1. The sign of $ \bigtriangleup S$ is not affected, since all substances are in gaseous form. Yet there will be a change in magnitude, becoz what was a mixture of free gases has now been converted into one single molecule (gas). If a liquid is formed, $ \bigtriangleup S < 0$.(decrease in randomness)

2. $ \bigtriangleup G < 0$, since the rxn is spontaneous. (Dissociation into ions). I think $ \bigtriangleup S > 0$ , becoz randomness has increased (ions are mobile in water.

[quote="Carcul"]At first sight, the sign of $ \Delta S$ would not be affected, since disorder decreases in both cases, but its magnitude will decrease.[/quote]

How does magnitude decrease??
When gas is formed,the difference between entropys of reactants and products is lesser than that when liquid is formed??
If product has magnitude of entropy as a,
gas product=b,
liquid prod=c,
then magnitude wise,
a>b>c.

Is it right??

actually,

1. fewer moles of gas are produced than reacted. The net effect then is a decrease in entropy. If N2H4 becomes a gas, then the magnitude decreases because you have more disorder coming out with another gaseous product.

2. delta G is indeed less than 0 because AgNO3 is soluble. delta S also increases in this case since you have formation of ions, which have more "freedom," but formation of ions does not always increase entropy. Small highly charged particles like Al3+ form a cage-like structure in solution, which ends up decreasing entropy.


Solve for what? Second, should there be sines and stuff in the middle school forum?


Tracy has been baking a rectangular cake whose surface is dissected by grid lines in square fields. The number of rows is $ 2^n$ and the number of columns is $ 2^{n \plus{} 1}$ where $ n \geq 1, n \in \mathbb{N}.$ Now she covers the fields with strawberries such that each row has at least $ 2n \plus{} 2$ of them. Show that there four pairwise distinct strawberries $ A,B,C$ and $ D$ which satisfy those three conditions:

(a) Strawberries $ A$ and $ B$ lie in the same row and $ A$ further left than $ B.$ Similarly $ D$ lies in the same row as $ C$ but further left.

(b) Strawberries $ B$ and $ C$ lie in the same column.

(c) Strawberries $ A$ lies further up and further left than $ D.$

This is http://www.mathlinks.ro/viewtopic.php?t=287863 with one of the strengthened bounds.

  darij


let incircle and A-excircle of triangle $ ABC$ be tangent to BC at D and D' . and $ M$ and $ H$ be midpoint of BC and feet of altitude from $ A$ to BC.
the in-tangent of $ (I)$ and $ I_a$ ($ \not \equal{}\ BC$)meet the cirlce that passes through $ D$ and $ D'$ at $ E$ and $ F$ prove that points $ E,F,M,H$ lie on a cirlce.

$ EF$ is symetric to $ DD'$ with respect to $ II_a$ (bisector of $ A$). So $ EF$, $ DD'$ and $ II_a$ concurr in a point $ K$. So $ EK \cdot KF \equal{} DK\cdot KD'$.

Now:
$ E,F,M,H$ lie on a cirlce $ \Leftrightarrow EK \cdot KF \equal{} HK\cdot KM \Leftrightarrow DK \cdot KD' \equal{} HK\cdot KM \Leftrightarrow \frac {KD'}{KM} \equal{} \frac {HK }{DK}$

[b]1)[/b] Looking at triangles $ \triangle AHK$ and $ \triangle IDK$:
$ \frac {HK}{DK} \equal{} \frac {AH}{IR} \equal{} \frac {h_a}{r}$
Where $ h_a$ is the altitud at $ A$ and $ r$ the inradius. If $ (ABC)$ is the are of the triangle one has:
$ \frac {h_a}{r} \equal{} \left(\frac {\frac {2(ABC)}{a}}{\frac {2(ABC)}{a \plus{} b \plus{} c}}\right) \equal{} \frac {a \plus{} b \plus{} c}{a}$
So:
$ \boxed{\frac {HK}{DK} \equal{} \frac {a \plus{} b \plus{} c}{a}}$

[b]2)[/b] Some easy to show facts (Leting WLOG $ B$ closer to $ H$ than $ C$):
$ BK \equal{} \frac {ac}{b \plus{} c}$
$ BD \equal{} s \minus{} b$
$ BM \equal{} \frac {a}{2}$
$ MD' \equal{} DM$
$ KD' \equal{} KM \plus{} MD'$ (Because $ B$ closer to $ H$ than $ C$)
Then:
$ \frac {KD'}{KM} \equal{} \frac {KM \plus{} MD'}{KM} \equal{} \frac {KM \plus{} MD}{KM} \equal{} \frac {(BM \minus{} BK) \plus{} (BM \minus{} BD)}{(BM \minus{} BK)}$
So
$ \frac {KD'}{KM} \equal{} \frac {(\frac {a}{2} \minus{} \frac {ac}{b \plus{} c}) \plus{} (\frac {a}{2} \minus{} s \plus{} b)}{(\frac {a}{2} \minus{} \frac {ac}{b \plus{} c})} \equal{} \frac {(\frac {a(b \minus{} c)}{2(b \plus{} c)}) \plus{} (\frac {b \minus{} c}{2}}{(\frac {a(b \minus{} c)}){2(b \plus{} c)})} \equal{} \frac {a \plus{} b \plus{} c}{a}$
So:
$ \boxed{\frac {KD'}{KM} \equal{} \frac {a \plus{} b \plus{} c}{a}}$

$ \frac {KD'}{KM} \equal{} \frac {HK}{DK}$ and so $ E,F,M,H$ lie on a cirlce.

Any nicer solution?

[size=117][color=darkred][b]Nice and short proof of this nice and easy problem, [u]Skuge[/u] ![/b][/color] [/size]

[color=darkblue]Here is a similar metrical proofs.

[b][u]Method I.[/u][/b] Define the point $ A'\in II_a$ for which $ MA'\perp BC\ .$

$ \frac {KD}{KH} = \frac {r}{h_a} = \frac {ar}{ah_a} = \frac {ar}{2S} = \frac {ar}{2pr}$ $ \implies$ $ \boxed {\frac {KD}{KH} = \frac {a}{2p}}\ \ (1)\ .$

$ S = pr = (p - a)r_a$ $ \implies$  $ p(r_a - r) = ar_a\ .$

$ \{\begin{array}{c} MA' = \frac {r_a - r}{2} = \frac {ar_a}{2p} \\
 \\
\frac {KM}{KD'} = \frac {MA'}{r_a}\end{array}$ $ \implies$ $ \boxed {\frac {KM}{KD'} = \frac {a}{2p}}\ \ (2)\ .$

$ (1)\ \wedge\ (2)$ $ \implies$ $ \frac {KM}{KD'} = \frac {KD}{KH}$ $ \implies$ $ KD\cdot KD' = KM\cdot KH\ .$

[b][u]Method II.[/u][/b] Using "[b]brute-force[/b]" obtain :

$ \{\begin{array}{ccc} KD = \frac {p - a}{b + c}\cdot |b - c| & , & KD' = \frac {p}{b + c}\cdot |b - c| \\
 \\
KM = \frac {a}{2(b + c)}\cdot |b - c| & , & KH = \frac {2p(p - a)}{a(b + c)}\cdot |b - c|\end{array}$ $ \implies$

 $ KD\cdot KD' = KM\cdot KH = \frac {p(p - a)(b - c)^2}{(b + c)^2}\ .$[/color]

let In-Tan lines of incricle and A_excircle intersect T.easy to see that $ T$ lies on $ A$-angle bisector.
we know that $ (AT,II_a)\equal{}\minus{}1$ hence perpendiculars from these points to $ BC$ are harmonic hence $ (DD',TH)\equal{}\minus{}1$ from a well-know realation(at least for me) for harmonic points we have:$ TM.TH\equal{}TD.TD'$ also we have $ TD.TD'\equal{}TE.TF$ hence four points $ H,E,M,F$ are on a circle.


[hide="Hint"]A simple induction shows $a_{n}= 2\cos{\left (\frac{\pi}{2^{n+1}}\right )}.$ The rest follows from there.
[/hide]


Factorize the polynomial $ x^{256}\plus{}y^{256}$ completely over the real polynomials.

I think that expressions like $ a^2 \plus{} b^2$ can't be factored any more, and $ x^{256} \plus{} y^{256}$ is of that form(I mean that it can't be factored in rationals)

But it asked to be factored over the reals :)

I think it is well-known that the number of linear factors of a polynomial is the number of real zeros it contains. So for example, the expression $ x^2 \plus{} a^2$ has no linear factors because the function $ y \equal{} x^2 \plus{} a^2$ is always positive for all (real) values of $ a$, which implies that the function containes no zeroes, and thus contains no linear factors. This comes from the Fundamental Theorem of Algebra.

Likewise, the function $ y \equal{} x^{256} \plus{} a^{256}$ is always positive for all (real) values of $ a$, so it cannot be factored into linear factors. However, all polynomials can be factored over the complex numbers. For this one, you'd go like
\[ x^{256} \plus{} a^{256} \equal{} (x^{128} \plus{} a^{128}i)(x^{128} \minus{} a^{128}i)
\]
and so on.

I think you meant the problem to be to factor $ a^{256} \minus{} b^{256}$?

I edited this so I don't look like a buffoon :D (But be aware that I didn't solve the problem - I just inserted "linear" where appropriate)

[quote="Yongyi781"]But it asked to be factored over the reals :)

I think it is well-known that the number of "proper" real factors of a polynomial is the number of real zeros it contains. So for example, the expression $ x^2 \plus{} a^2$ has no "proper" factors because the function $ y \equal{} x^2 \plus{} a^2$ is always positive for all (real) values of $ a$, which implies that the function containes no zeroes, and is thus not factorable over the reals. This (I think) comes from the Fundamental Theorem of Algebra.

Likewise, the function $ y \equal{} x^{256} \plus{} a^{256}$ is always positive for all (real) values of $ a$, so it cannot be factored further over the reals. However, all polynomials can be factored over the complex numbers. For this one, you'd go like
\[ x^{256} \plus{} a^{256} \equal{} (x^{128} \plus{} a^{128}i)(x^{128} \minus{} a^{128}i)
\]
and so on.

I think you meant the problem to be to factor $ a^{256} \minus{} b^{256}$?[/quote]

If a polynomial has no real roots, then it has no [i]linear[/i] factors, but factors of higher degrees may exist. For example, $ x^4\plus{}3x^2\plus{}2\equal{}(x^2\plus{}1)(x^2\plus{}2)>0$ for all real $ x$. What you wrote is true for second- and third-degree polynomials, since if they can be factored as a product of two lower degree polynomials, then at least one of the factors must be linear, and the polynomial would have a real root.

(It can be factor into quadratics. I factored it into a lot of terms like
$ (x^2\minus{}y^2 \plus{} 2xy$ cos$ (\frac{a\pi}{256}))$
where a ranges in 1, 3, 5, 7...511.

[quote="Cassi"]Factorize the polynomial $ x^{256} \plus{} y^{256}$ completely over the real polynomials.[/quote]

note that $ x^4 \plus{} y^4 \equal{} (x^2\plus{}xy\sqrt2 \plus{}y^2)(x^2 \minus{} xy\sqrt2 \plus{} y^2)$

[quote="Yongyi781"]the function $ y \equal{} x^{256} \plus{} a^{256}$ is always positive for all (real) values of $ a$, so it cannot be factored further over the reals. [/quote]

This is quite false.  For example, $ x^4 \plus{} 4y^4 \equal{} (x^2 \plus{} 2xy \plus{} 2y^2)(x^2 \minus{} 2xy \plus{} 2y^2)$ (from which it is possible to derive Lazarus' example).  A nontrivial real factoring must in fact exist.  One way to state the fundamental theorem of algebra is that [i]any[/i] polynomial can be written as the product of linear and quadratic (real) factors.  (This has some interesting [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=152452]geometric[/url] consequences.)

pythag011 is correct.  The point here is that the roots are easy to describe, and then one simply pairs each root with its conjugate.


AMC
USA(J)MO
USAMO
analytic geometry
geometry

Can you still get full credit for a solution if it is extremely unelegant?

Yes, you can - inelegance, such as coordinate bashing a geometry problem, simply makes the grader more likely to give you 1 or 0 if you make a minor error.


Hi,
Can (n-1)+(n-2)+(n-3)+...+1 be written in a shorter form?

[color=darkblue]URL removed by moderator. [/color]

[hide="Different Approach"]\[ S \equal{} (n \minus{} 1) \plus{} (n \minus{} 2) \plus{} \ldots \plus{} (n \minus{} (n \minus{} 2)) \plus{} (n \minus{} (n \minus{} 1))
\]The sum can be rewritten like this:\[ S \equal{} \underbrace{(n \plus{} \ldots \plus{} n)}_{n\ \text{times}} \minus{} (1 \plus{} 2 \plus{} 3 \plus{} \ldots \plus{} n)
\]The first part is simply $ n^2$, and the second part is $ \frac {n(n \plus{} 1)}{2}$. Thus, the sum is\[ S \equal{} n^2 \minus{} \frac {n(n \plus{} 1)}{2} \equal{} \frac {2n^2 \minus{} (n^2 \plus{} n)}{2} \equal{} \boxed{\frac {n(n \minus{} 1)}{2}}
\][/hide]

Since $ 1\plus{}2\plus{}3\plus{}\hdots\plus{}n\equal{}\frac{n(n\plus{}1)}{2}$, We plug in

$ n\minus{}1$ to get $ \boxed{\frac{n(n\minus{}1)}{2}}$.

btw, this is why $ 1\plus{}2\plus{}3\plus{}\hdots\plus{}n\equal{}\frac{n(n\plus{}1)}{2}$.
------

Let the sum $ 1\plus{}2\plus{}3\plus{}\hdots\plus{}n$ be equal to $ S$. We can then

write $ S$ two different ways.

$ S\equal{}1\plus{}2\plus{}3\plus{}\hdots\plus{}n$
$ S\equal{}n\plus{}(n\minus{}1)\plus{}(n\minus{}2)\plus{}\hdots\plus{}1$.

Adding, we see that there are $ n$ terms of $ n\plus{}1$. Therefore,

$ 2S\equal{}n(n\plus{}1) \implies S\equal{}\frac{n(n\plus{}1)}{2}$.

QED.

Suppose we have n people in a line and we want to pick two of them. We will count the number of ways of doing this in two different ways:

[b]Counting Method #1:[/b] There are n ways to pick one person, and n-1 ways to pick the second person. However, this counts each pairing twice, since there are two possible orders of picking each pair. So, the number of ways is n(n-1)/2

[b]Counting Method #2:[/b] We will first pick the leftmost of the two people, and then pick one other person to the right of this person.

If the leftmost person is the 1st person, there are n-1 people to the right to pick for the other person.
If the leftmost person is the 2nd person, there are n-2 people to the right to pick for the other person.
...
If the leftmost person is the ath person, there are n-a people to the right to pick for the other person
...
If the leftmost person is the second to last person, there is only one person to the right to pick for the other person.

Adding up each of our cases, there are (n-1)+(n-2)+...+2+1 ways of picking 2 people

[b]Conclusion:[/b] Since we counted the same thing in two valid ways and got two expressions, the two expressions must be equal. In other words:

(n-1)+(n-2)+...+2+1 = n(n-1)/2


calculus
integration
geometry
trapezoid
algebra
polynomial
calculus computations

The question asks:
Use the midpoint rule to evaluate the integral (ranging from 10 to 17) (-3x-8x^2) dx
n = 3

So here is what I did (though my answer nevertheless is incorrect):
The midpoints of these intervals are, respectively, 11 1/6, 13½, and 15 1/6.

Let f(x) = -3x - 8x²

(7/3) [f(11 1/6) + f(13½) + f(15 1/6)]

f(11 1/6) ≈ -1031.05556
f(13½) = -1498.5
f(15 1/6) ≈ -1885.72222
So the estimate of the integral ≈ -10302.315

Any help is appreciated.

[quote="fedja"][quote="daneeyah"]
The midpoints of these intervals are, respectively, 11 1/6, 13½, and 15 1/6.
[/quote]
Are you sure it is 15 1/6, not 15 5/6 ?[/quote]

I tried that as a point but I'm still getting the wrong answer.

Well, I do not see any other mistake in your solution.

With the correction to 15 5/6, you should be getting an approximate integral of -10692.759. Is that what you have?

I would add that the trapezoid rule with the same intervals gives -10768.981. Doing the weighted average of those (2/3 of the midpoint rule plus 1/3 of the trapezoid rule) gives -10718.167, which is (except for roundoff) exactly correct - as expected for a polynomial of degree $ \le$ than 3.


suppose $ \{ X_{k} \}_{k\geq 1}$  be sequence of independent random variables such that :
\[ \mathbb{P}\{ X_{k} \equal{} 1\} \equal{} \mathbb{P}\{ X_{k} \equal{} \minus{} 1\} \equal{} \frac {1}{2}\left(1 \minus{} \frac {1}{2^{k}}\right)
\]
and
\[ \mathbb{P}\{ X_{k} \equal{} 2^{k}\} \equal{} \mathbb{P}\{ X_{k} \equal{} \minus{} 2^{k}\} \equal{} \frac {1}{2^{k\plus{}1}}
\]
Satisfy or not such sequence to the strong law of large numbers? :huh:

yes, the law of strong numbers applies here since it works for a sequence of Bernoulli rv $ P(Y_n\equal{}\minus{}1)\equal{}P(Y_n\equal{}\plus{}1)$ and the events $ \{|X_n| > 1\}$ will only occur a finite number of time (Borel Cantelli), so will disappear after taking the mean.

For example you can define $ Y_n\equal{}1$ if $ X_n>0$ and $ Y_n\equal{}\minus{}1$ otherwise. You know that $ X_n\equal{}Y_n$ for all $ n > N(\omega)$ (Borel Cantelli), and one can apply the law of strong numbers to $ Y_n$, hence the conclusion.

yes  :) , all of this is true.
Indeed, solution of this problem  is based on the Lemma about equivalence by Khinchin.
We say that sequences $ \{X_{n}\}$ and $ \{Y_{n}\}$  of i.i.d r. v is equivalence by Khinchin if :
$ \sum \mathbb{P}\{X_{k}\neq Y_{k}\}<\infty$
and lemma as you already know is next :
 if $ \{X_{n}\}$ and $ \{Y_{n}\}$  of i.i.d r. v is equivalence by Khinchin then if $ X_{n}$ satisfy the strong law of large numbers then $ Y_{n}$ satisfay  to the strong law of large numbers (it's simple consequence of Borel Cantelli )

oh, I did not know this lemma, but this is a very common technique (this is a very simple case of a coupling argument: see for example a proof of the convergence of Markov chain theorem that is often proved thanks to such a coupling argument)


Let $ABC$ be a triangle. Maximize $\sum{\frac{\sqrt{cosA}}{cos{\frac{B}{2}}cos{\frac{C}{2}}}}$

According to Cauchy-Schwarz, your sum will be maximum when
cos B/2 cos C/2 root(cos A) = cyclic = cyclic, hence when ABC is equilateral
The maximum value will be 2.root(2)

Dear jpe,
     I don't think your solution is right because my solution has two possible cases in which the equality holds (hope my computation is right) :)

I have just checked my computation again and find a mistake there. There is only one triangle that can reach the maximum value : the equilateral triangle.
  I will post my solution tomorrow.

Here is the solution.

 [hide="Solution"]
    Let 's first notice that $\frac{cosA}{sinBsinC} = 2$ and $tan{\frac{A}{2}}tan{\frac{B}{2}} = 1$.
    By Cauchy-Schwarzt, we have
    $LHS = 2(\sqrt{\frac{cosA}{sinBsinC}}). (\sqrt{tan{\frac{B}{2}}tan{\frac{C}{2}}}) \leq 2\sqrt{2}$
    The equality holds when $cosA sin^2{\frac{A}{2}} = cosB sin^2{\frac{B}{2}} = cosC sin^2{\frac{C}{2}}$. It can be proved that this condition is equivalent to triangle $ABC$ is equilateral by letting $x = sin{\frac{A}{2}}, y = sin{\frac{B}{2}}, z = sin{\frac{C}{2}}$ and noticing that $x^2+y^2+z^2 + 2xyz = 1$.
                              [/hide]


I think is is famous problem and maybe solved here many times but I cannot find link so :
On a circle are written 100 positive integers $a_{1},...,a_{100}$ in this sequence and $GCD(a_{1},...,a_{100})=1.$ We are allowed to do the following operation in one step: choose some $a_{j}$ and we replace it with $a_{j}+GCD(a_{j-1},a_{j+1}).$ Prove that after finite number of steps we can make $GCD(a_{i},a_{j})=1$ for all $i\not=j.$
We have that $a_{101}=a_{1}$


is composed of:

Claudiu Raicu           111p
Ana Caraiani            110p
Andrei Stefanescu  100.5p
Andrei Negut            100p
Gabriel Kreindler      95.5p
Ioan Manolescu        90.5p

Congrats! and Good Luck at the IMO!

I have seen  the "Romanian Olympiad 2003 IMO Team Selection Tests" problems, will there ever be here posted the "official" solutions?

Good luck to the IMO everyone.. bring us 6 gold medals!!!! :D

i don't think so ... at least right now i'm not allowed to post them on the web site but you can discuss them on the forum

so far the results for the romanian team are

1      2     3     4     5     6
------------------------------
7      7     3     7     7     7    38p ROM1 Caraiani Ana
5      1     0     7     7     -    20p ROM2 Kreindler Gabriel
7      7     0     0     1     -    15p ROM3 Manolescu Ioan
7      0     0     7     7     -    21p ROM4 Negut Andrei
7      7     0     7     1     -    22p ROM5 Raicu Claudiu
7      5     0     7     5     -    25p ROM6 Stefanescu Andrei

Probably China will win the 1st place, Russia the 2nd and (maybe) Bulgaria 3rd.

I'll be back with more details ASAP.

wait a second , i think i remember one of the members in your team

ana cariani, she's participating for the third time isn't she?

wasn't she the only female contestant with gold in glasgow?

i missed the chance of talking to her, which is a shame

now she seems to have become even better, 38 is not far from perfect

ana caraiani has a fanclub

every math nerds adore her :D

she's got a fanclub ?  didn't know that .... could you spare the adress of the web site if this "fan-club" has one ?

she's coming to Princeton, so watch yourselves :D

go Ana go!

hey!

i'm just saying that i recognise the name and that i'm glad she did well, and as probably for many others contestants, a shame i didn't meet her

that does not mean i am a fan, nor that :(  that i am a math nerd

:) nobody said you're a math nerd .. in fact i personally think that this category doesn't exist: let me be more specifically: a math nerd is, by definition, a person that doesn't have a clue about anyhing else than maths. Since obviously at the IMO after the contest (almost) no one was talking about maths, but rather singing and drinking, and doing all sorts of other stuff, I have drawn the conclusion above. :)

anyways, I belive that this term really applies (for some reasons) to those countries which have a separation in school, such like: the football guys, the cheerleaders (which date the football guys), the hippies, etc. and finally the nerd, people which because of their occupation don't have enough time (or charm) to get a girl. In some countries the reverse phenomenon happens, i.e. girls appreciate more the fact that you are smart than the fact that you can score an essay from 50 meters.

... long talk for nothing, gotta run now cuz I gotta learn for a drivers' license :D

is the manolescu in this year's team somehow related to triple perfect scorer Ciprian Manolescu?

:D not quite - in fact it seems that there isn't any long relative relationship between the two of them - besides that they live in different cities (used to live, anyway)

[quote="math8dragons"]ahir, this has not been posted for 9 years. Stop....[/quote]

OK, OK.

You go back, revive the oldest thread in Fun Factory that is not locked and think you get a +1 in post count.

[quote="ZZmath9"]BTW, your FF post count is always zero. So you don't get post count.

Ahir, your posts are all fine except for the bumps. Otherwise, you post nothing wrong.

So, rules of thumb on bumping:

[u]1. Usually try to only bump games you play in.
2. A thread can be bumped after approximately 12 hours have passed since the last post.
3. Don't bump threads that you don't post in.[/u]

Once you get the hang of it, you'll post excellent![/quote]
OK! Thanks!

And once the double post count goes down.

you

bumped

a

17

year

old

thread

wow that's actually quite an accomplishment

wait actually this is the oldest unlocked thread

"RT-2PM2 Topol-M Silo-based intercontinental ballistic missile strikes on streak"


$n \in \mathbb{N}$ a fixed integer. For what complex number $\lambda$ does there exist a doubly stochastic matrix $M$ with eigenvalue $\lambda$ ?

Alekk,

You showed (http://www.artofproblemsolving.com/Forum/viewtopic.php?t=105426) that if $\lambda$ is an eigenvalue, and $|\lambda|=1$ then $\lambda \in \bigcup_{m=2}^{n}\{ e^{2i\pi \frac{k}{m}}, k=0,\dots,m-1\}$.

It looks to me that set you are looking for is $\bigcup_{m=2}^{n}\mbox{Conv}\left( \{ e^{2i\pi \frac{k}{m}}, k=0,\dots,m-1\}\right)$. Where $\mbox{Conv}()$ is the convex hull.
But I can't prove it rigorously with Krein-Milman.

Actually, what I wrote above can be proven for $n=2,3$, but it is wrong for $n=5$.
To see it, try the following \[\begin{pmatrix}0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 1/2 & 0 & 1/2 \\ 0 & 1/2 & 1/2 & 0 & 0 \\ 0 & 1/2 & 0 & 0 & 1/2 \\ 1 & 0 & 0 & 0 & 0 \end{pmatrix}\] Does anyone have a more complete answer? It would be nice :!:


[b]Let x, y, z be real numbers greater than - 1. Prove that the following inequality takes place:[/b]

[tex]\frac{{1 + x^2 }}{{1 + y + z^2 }} + \frac{{1 + y^2 }}{{1 + z + x^2 }} + \frac{{1 + z^2 }}{{1 + x + y^2 }} \ge 2[/tex]

[b](7-th Junior Balkan Math. Olympiad, Panaitopol)[/b]

You can find a solution in the topic:" BMO 2002 /2003 inequalities !!"  in Solved Problems Section.


A circle is tangent to the sides AB and AC and to the circumcircle at points P, Q and R, respectively. If the lines PQ and AR intersect at point S, prove that angles SBA and SCA are congruent.

[hide="Hide"]$ \triangle SBP \sim \triangle SCQ$ because $ SP: SQ\equal{}BR: CR\equal{}BP: CQ$[/hide]

Sorry, but why $ \dfrac{BP}{CQ}\equal{}\dfrac{BR}{CR}$ ? :?:  Thanks


superior algebra
superior algebra unsolved

$G$ is an infinite group that for each non-obvious sungroup $H$, $[G: H]<\infty$. Prove that $G$ is cyclic.


Ok, great. Sorry, on my browser it looked a little fuzzy and I wanted to clarify.

Working on a proof. Fun problem!
 :lol:


I'm using the common application.  In some colleges' supplement essay, they don't specify their supplement essay's length.

I was wondering if it's best to keep these essays at one page or is it fine if I wrote little over one page?

thank you

Try the print preview; sometimes the essay length isn't specified on the supplement form but it's specified on the .pdf.

If that isn't the case, generic advice:  the essay should be as long as you need to say what you want to say.

I see, thank you.

and I was also wondering: some of the supplement essays, allows you to either upload file or type in the box provided.  the Box has a characters limit but the file upload allows more characters.

Does this mean that the file such as word i upload should also be in the characters length?

I realize that "about 300 words" means "a few insightful paragraphs telling us about yourself".  But is 420 words too long?

I believe the character limit is approximate (if you're uploading).  As for the other question, if you're using 420 words to answer a 300-word prompt, consider being less wordy instead of saying less.


If you want to know how :
clearly, 1 is a root, so, you can factor by (x-1), then, you have at least 2 methods to get the other term ;)

[quote="wolf"]Factorise: 13x^2 + 14x - 27 = 0[/quote]
[hide]
$13x^2+14x-27$
Obvioulsy 1 is a root.
so we have to divide $13x^2+14x-27$ by $x-1$.
We use sythetic division, to get 13x+27. Thus we have:
(x-1)(13x+27) as our solution.
[/hide]

[quote="Silverfalcon"][quote="Fermat -Euler"][hide="ansewr"] $(13x+27)(x-1)$ [/hide][/quote]

Here is a few comments that go to you, Fermat -Euler.

1. Your level is beyond Getting Started. Don't answer this type of basic problems.
2. I saw this in couple of your posts regardless of level. Why do you simply post the answer? If you are going to post something, post it with the solutions. It's much more helpful and it's pretty annoying if you keep doing it.

Thanks.

Moderator Silverfalcon[/quote]
OK  Silverfalcon