geometry AMC AIME USA(J)MO USAMO Stanford college

Ok, I got this from my school's math club, and I'll post the important information here.

This is NOT sorted by date, sadly.
Let me know if this is helpful for any of you. I tried to collect as much information as I could.
-------------------------------------------------------------------

First Contest List, 2007/2008:

[b]BAMO(Bay Area Mathematics Olympiad)[/b]
Individual Exam
February 27, 2008
_______________________________________________________
[b]CAML(California Math League)[/b]
Six rounds, individual exam.  Top five scores yield Menlo’s team score.  “Official” dates given.  Exam can be scheduled for up to one week before the official dates.
October 23, 2007
November 20, 2007
December 18, 2007
January 15, 2008
February 12, 2008
March 18, 2008

____________________________________________________
[b]The Mandelbrot Competition (New Format)[/b]
Exam “windows” given below.
We decide on date and time within each window.

Individual Competition (5 rounds):
October 15-19
December 3-7
January 7-11
February 4-8
March 3-7

Team Competition(3 rounds, two official teams)
January 7-18
February 4-15
March 3-14

_____________________________________________________
[THE preeminent national competition, open to all interested students.]
[b]AMC 10/12A Tuesday, February 12, 2008[/b]
[b]AMC 10/12B Wednesday, February 27, 2008[/b]

[An invitational exam, for high scorers on the AMC 10/12.]
[b]AIME Tuesday, March 18, 2008[/b]
[For students with an official school excuse for missing this exam, a makeup version will be given Wednesday, April 2, 2008]

[Also invitational, for students with exceptionally high combined scores on the AMC 10/12 and the AIME.]
[b]USAMO April 29 & 30, 2008[/b]

____________________________________________________________
Competitions at local universities that we have attended:
Santa Clara High School Mathematics Contest
(TBA, at SCU, Saturday in early November)
Stanford Mathematics Tournament
(TBA, at Stanford, last year held on Sunday, March 4)
BAMM(Bay Area Math Meet)
(TBA???  This much-beloved event was last held at USF on Saturday, April 23, 2005.  Run by the legendary Paul Zeitz, whose The Art and Craft of Problem Solving is universally admired.)
Julia Robinson Mathematics Festival
(Sponsored by MSRI and held on the Google campus, this was a HUGELY popular event.  This year we have registered to participate.  The new format is supposed to involve a series of problem sets sent to each school and a big mathematics challenge day held at a central site, perhaps Google.]

____________________________________________________________
Programming/Computer Science Contests

[b]ACSL(American Computer Science League)[/b]
(Short answer and programming problems. Individual, with top five providing a team score. Four rounds.)
Friday, December 14, 2007
Friday, February 8, 2008
Friday, March 7, 2008
Friday, April 11, 2008

A more challenging high school contest which is independent of school affiliation:
[b]USACO (USA Computing Olympiad)[/b]
See: http://oldweb.uwp.edu/academic/mathematics/usaco/
for a discussion of format, training exercises, etc.

___________________________________________________________
The list of other mathematics contests is nearly endless.

[b]Santa Clara Valley Mathematics Association Field Day[/b]
(A large competition with many levels and schools usually held in early March at San Jose State University.)

[b]ARML(American Regions Mathematics League)[/b]
Website: http://www.arml.com
May 30 & 31, 2008
(From their website:  This competition involves teams of 15 students, often from several different schools.  Teams travel to a competition site and some fees for travel, etc. are required.)

[b]The Stanford Math Circle[/b]
(This is the closest of several bay area math circles, which include Berkeley, San Jose, and San Francisco.)
Meeting time: 2 pm – 4 pm, Sundays in 380-C in the Stanford Mathematics Building.  Interesting math lectures aimed at high school students and lots of problems to solve.  (You may have an opportunity to meet the faculty sponsor, Professor Ravi Vakil, one of the world’s most gifted problemists.) See: http://www.stanfordmathcircle.org for more information and directions.
-------------------------------------------------------------

Just wondering if that is of use to anyone.

I guess I'll take that at face value, junggi :-)

Yep, eric, that's a nifty table too.

I might have missed some of the more obscure contests, like the Tokutomi one (well, I thought it was obscure...)

My name is Eric? :rotfl:

I think you've confused me with [url=http://www.artofproblemsolving.com/Forum/profile.php?mode=viewprofile&u=17840]daermon[/url].

Oh DAMN! I wasn't paying attention xD

Yea, sorry man.

I'll read people's posts more carefully next time (note to self: stare at avatar for at least ten minutes before posting ANYTHING)

Hmm for the comp sci competitions I saw somewhere that they don't allow java. Is that true?

[quote="Ubemaya"]:omighty:
Wow... that's all I can say.

Are any of you actually doing the computer science stuff?

EDIT: I'm confused now... I had three smilies, and I couldn't submit my post cause it said I had six...[/quote]

Well obviously your avatar counts as 3 smileys.
EDIT: Yes I shouldn't be talking, my avatar is an animated smiley.

@discreet:
Where do we sign up for BAMO? I wanna join I wanna join I wanna join  :D

[quote="CircleSquared"]
EDIT: Yes I shouldn't be talking, my avatar is an animated smiley.

Where do we sign up for BAMO? I wanna join I wanna join I wanna join  :D[/quote]

ROFLL!!!!!!

For BAMO, you have to register with some site that's taking it. I'm surprised that Harker isn't registered for it... I thought they were. If not, the San Jose Math Circle offers it. So does Redwood/Saratoga High, but you'll want to stay away from there ;) .

Well, I guess you can come to my school and take it if you want.
I'll see if I can get you an invitation.

PM me for more details on that one.

I believe SMT is being scheduled for February 23 this year.  Don't take that as gospel, but we're pretty sure that'll be the date.

Also, there's a new series of contests we're hoping to get running this year.  They're going to be an extension of the Great Plains Math League ([url]http://www.mathleague.org[/url]), on the same dates, same formats, and everything.  The championship meet in April will be at Stanford.

[quote="DiscreetFourierTransform"]Oh -! I wasn't paying attention xD

Yea, sorry man.

I'll read people's posts more carefully next time (note to self: stare at avatar for at least ten minutes before posting ANYTHING)[/quote]
hehe silly matt trix are for kids!!
gosh did you confuse me with nakita?  don't insult me like that! (hehe sry nakita...couldn't resist)
MY AVATAR WILL STARE RIGHT BACK AT YOU WITH ITS OOGLY BOOGLY EYES
anyways, i never really post on AOPS anymore outside classes...i actually have a life now :D

hmm i should say something pertaining to contests now so this isnt spam...ooh stanford math circle is back?  i thought that died when Sam left
i'll try to come a few times when school+social life+sleep will allow, aka never

[quote="junggi"][quote="daermon"]
i actually have a life now :D
[/quote]
dude
i read that as "i quit math"[/quote]

Lol. "Quitting math is not acceptable. You must try your hardest, and learn more to get good grades." ~ what my parents probably would say if I tried to quit.

[quote="junggi"][quote="daermon"]
i actually have a life now :D
[/quote]
dude
i read that as "i quit math"[/quote]

"he died" according to Erdos

[quote="humanbeing"]I am new to the Bay Area I have just come here from Vietnam.
I wonder how can I participate in those competitions.

Thanks for any help.[/quote]

Chances are that there is some kind of club that takes you to the contests. If not just try to get in contact with the administrators of the tests.

But dunno I am a freshman and it may be different in each high school.

[quote="daermon"]anyways, i never really post on AOPS anymore outside classes...i actually have a life now
i'll try to come a few times when school+social life+sleep will allow, aka never[/quote]
: P  Congratulations.  Good job setting your priorities in order.

[quote="davidyko"]Chances are that at least the AMCs will be offered at your school; even at such an un-mathy school as mine, it happens every year (and I was the only one to qualify for AIME last year, but that's another story.)

For ARML, just check out the SFBA-ARML group.[/quote]
Actually, more schools than you'd think have never even heard of the AMC's.

@ Eric - Some [Matt] confuse me with you because of our avatars, and some [you] can't spell it. :D

@ humanbeing - Try to find out where the AMC contests are being held near you, and then you will be informed about the AIME and beyond if you "pass" the AMC(s). :)

Those damn penguins with their damn googly eyes confuse me xD

They're brainwashing you of all math knowledge slowly.  :P
Anyway, great list, I'm impressed.

[quote="i_like_pie"]
@ humanbeing - Try to find out where the AMC contests are being held near you, and then you will be informed about the AIME and beyond if you "pass" the AMC(s). :)[/quote]

So I can participate in the AMC eventhough my school don't offer it?
I've had a look at the form and there's something like: High School CEEB, Contest Manager, ... and I really don't understand.

Any updates for this thread? All the contests on this list have passed.

[quote="DiscreetFourierTransform"]I'll release one for next year. I'm pretty big into organization, so I can probably get the list up before school starts next year.[/quote]
Awesome! Thanks for all your hard work.  I'll be sure to email our math department with this list.

I'll make sure you get word when SMT is scheduled so you can put that on the list.  I'm hoping we'll have our house in order earlier this year and get a date set by the end of October or so.

Yea, I'd like to show up to that one this year :P
I've never been, but it looks cool.

So, reading this, I realized that our school has a similar list. It's really, really long, and not all that organized. I could sent that to you if you like, matt...


function
search
calculus
calculus computations

I need the value of this infinite sum ( and solution , too , of course :) ) :
${ \sum\limits_{n=1}^{\infty}\frac{1}{n^{2}}}$

I know that it is well-known  , but ...  :roll:

If anybody know where I can find similar relations, please tell me ... I am very interested in it  :)

Enjoy reading:

Mathworld: http://mathworld.wolfram.com/RiemannZetaFunctionZeta2.html
Excellent paper by R. Chapman: http://www.secamlocal.ex.ac.uk/people/staff/rjchapma/etc/zeta2.pdf

More on Zeta function:

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

Or you can simply search the ML forum-or google it.

A few threads on ML which may be of some interest to you:

http://www.mathlinks.ro/viewtopic.php?t=159424
http://www.mathlinks.ro/viewtopic.php?t=180928

See here as well for more on Riemann-Zeta: http://en.wikipedia.org/wiki/Riemann_zeta_function


Is anyone here doing it? I have a team at my school that wants to do it. What should we expect. I read the problem and solutions from last year and I've realized it's a lot of modeling and writing. Did anyone do it last year? Did you have fun with it?

Wow, I didn't see this thread until now. Of course, you've already figured out what it's about by now! I hope you enjoyed it. Best of luck!

If you make it to NYC and have any questions about that, feel free to PM me. You can also check out videos of last year's presentations online.


Find $ x$ such that $ 243x\plus{}17\equiv 101 (\mod725)$.

Then $ 243x\equiv 84 \mod 725$. Now, $ 725y\plus{}84\equiv 0 \mod 243$ so $ \minus{}4y\equiv \minus{}84 \mod 243$. 4 and 243 are coprime, so $ y\equal{}21\mod 243$, so $ y\equal{}22\plus{}243t$, and $ 243x\equal{}725y\plus{}84$, so $ x\equiv 63\mod 725$


A policeman is trying to catch a robber on a board of $2001\times2001$ squares. They play alternately, and the player whose trun it is moves to a space in one of the following directions: $\downarrow,\rightarrow,\nwarrow$.

If the policeman is on the square in the bottom-right corner, he can go directly to the square in the upper-left corner (the robber can not do this). Initially the policeman is in the central square, and the robber is in the upper-left adjacent square. Show that:

$a)$ The robber may move at least $10000$ times before the being captured.
$b)$ The policeman has an strategy such that he will eventually catch the robber.

Note: The policeman can catch the robber if he reaches the square where the robber is, but not if the robber enters the square occupied by the policeman.

Could you please give me some guidance ?
Thank you.

Solution with [b]MinosFx[/b]

Let's color the $4001$ diagonals that have positive slope with [color=#00f]blue[/color], [color=#f00]red[/color], and [b]black[/b] starting from the bottom-right diagonal (which is the bottom-right corner) and ending with the upper-left diagonal (the upper-left corner). Here is the coloring for a $3\times 3$ board:

[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(3cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -9.55652892561984, xmax = 1.5509090909090903, ymin = -2.6157024793388435, ymax = 7.36776859504132;  /* image dimensions */

draw((-4,2)--(-4,3)--(-3,3)--(-3,2)--cycle, linewidth(2) + blue);
draw((-5,2)--(-5,3)--(-4,3)--(-4,2)--cycle, linewidth(2) + red);
draw((-5,4)--(-5,5)--(-4,5)--(-4,4)--cycle, linewidth(2) + blue);
draw((-4,3)--(-4,4)--(-3,4)--(-3,3)--cycle, linewidth(2) + red);
draw((-5,3)--(-6,3)--(-6,4)--(-5,4)--cycle, linewidth(2) + blue);
draw((-6,4)--(-6,5)--(-5,5)--(-5,4)--cycle, linewidth(2) + red);
/* draw figures */
draw((-4,2)--(-4,3), linewidth(0.4) + blue);
draw((-4,3)--(-3,3), linewidth(0.4) + blue);
draw((-3,3)--(-3,2), linewidth(2) + blue);
draw((-3,2)--(-4,2), linewidth(2) + blue);
draw((-5,2)--(-5,3), linewidth(2) + red);
draw((-5,3)--(-4,3), linewidth(2) + red);
draw((-4,3)--(-4,2), linewidth(2) + red);
draw((-4,2)--(-5,2), linewidth(2) + red);
draw((-5,4)--(-5,5), linewidth(0.4) + blue);
draw((-5,5)--(-4,5), linewidth(2) + blue);
draw((-4,5)--(-4,4), linewidth(2) + blue);
draw((-4,4)--(-5,4), linewidth(2) + blue);
draw((-4,3)--(-4,4), linewidth(2) + red);
draw((-4,4)--(-3,4), linewidth(2) + red);
draw((-3,4)--(-3,3), linewidth(2) + red);
draw((-3,3)--(-4,3), linewidth(2) + red);
draw((-5,3)--(-6,3), linewidth(2) + blue);
draw((-6,3)--(-6,4), linewidth(2) + blue);
draw((-6,4)--(-5,4), linewidth(0.4) + blue);
draw((-5,4)--(-5,3), linewidth(2) + blue);
draw((-3,4)--(-3,5), linewidth(2));
draw((-5,2)--(-6,2), linewidth(2));
draw((-6,4)--(-6,5), linewidth(2) + red);
draw((-6,5)--(-5,5), linewidth(2) + red);
draw((-5,5)--(-5,4), linewidth(2) + red);
draw((-5,4)--(-6,4), linewidth(2) + red);
draw((-6,2)--(-6,3), linewidth(2));
draw((-4,5)--(-3,5), linewidth(2));
/* dots and labels */
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]

Now, note that, only performing $\nwarrow$, $\downarrow$ or $\rightarrow$, the colors of the cells in which the policeman and the robber are on repeat periodically:

[color=#00f]blue[/color] $\mapsto$ [b]black[/b] $\mapsto$ [color=#f00]red[/color] $\mapsto$ [color=#00f]blue[/color]

Thus, as the policeman starts at [b]black[/b] and the robber at [color=#f00]red[/color], the policeman will never catch the robber without using his "special move" (moving from the bottom-left corner directly to the upper-left corner). Morerover, he must use it twice. When he uses it for the first time, he goes from a [color=#00f]blue[/color] cell to a [color=#f00]red[/color] cell, while the robber would be in a [b]black[/b] cell, according to the cycle. Hence, the policeman must perform the trick again: the policeman would go from [color=#00f]blue[/color] to [color=#f00]red[/color] again, but, remembering he has skipped [b]black[/b] twice, now the robber would be in a [color=#f00]red[/color] cell, too.

Now, we explain how the robber can move $10000$ times. The strategy of the robber would be reaching the bottom-left corner and doing this repeatedly:

[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(10cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -6.745758138056621, xmax = 8.36871694688661, ymin = -1.7830630832992078, ymax = 10.688443680779603;  /* image dimensions */

draw((-4,3)--(-5,3)--(-5,4)--(-4,4)--cycle, linewidth(2) + blue);
draw((-4,5)--(-4,4)--(-5,4)--(-5,5)--cycle, linewidth(2) + red);
draw((-5,4)--(-5,3)--(-6,3)--(-6,4)--cycle, linewidth(2) + red);
draw((0,3)--(-1,3)--(-1,4)--(0,4)--cycle, linewidth(2) + blue);
draw((-1,3)--(-2,3)--(-2,4)--(-1,4)--cycle, linewidth(2) + red);
draw((-1,4)--(-1,5)--(0,5)--(0,4)--cycle, linewidth(2) + red);
draw((4,3)--(3,3)--(3,4)--(4,4)--cycle, linewidth(2) + blue);
draw((3,3)--(2,3)--(2,4)--(3,4)--cycle, linewidth(2) + red);
draw((3,4)--(3,5)--(4,5)--(4,4)--cycle, linewidth(2) + red);
draw((8,3)--(7,3)--(7,4)--(8,4)--cycle, linewidth(2) + blue);
draw((7,3)--(6,3)--(6,4)--(7,4)--cycle, linewidth(2) + red);
draw((7,4)--(7,5)--(8,5)--(8,4)--cycle, linewidth(2) + red);
/* draw figures */
draw((-3.608128237093616,3.98299760787552)--(-2.409463905013026,3.98299760787552), linewidth(2),EndArrow(6));
draw((0.3825469172722511,3.98299760787552)--(1.5958291070611412,3.9976154655838196), linewidth(2),EndArrow(6));
draw((4.387839929346418,3.9976154655838205)--(5.557268546010409,3.9976154655838205), linewidth(2),EndArrow(6));
draw((-4,3)--(-5,3), linewidth(2) + blue);
draw((-5,3)--(-5,4), linewidth(0.4) + blue);
draw((-5,4)--(-4,4), linewidth(0.4) + blue);
draw((-4,4)--(-4,3), linewidth(2) + blue);
draw((-4,5)--(-4,4), linewidth(2) + red);
draw((-4,4)--(-5,4), linewidth(2) + red);
draw((-5,4)--(-5,5), linewidth(2) + red);
draw((-5,5)--(-4,5), linewidth(2) + red);
draw((-5,4)--(-5,3), linewidth(2) + red);
draw((-5,3)--(-6,3), linewidth(2) + red);
draw((-6,3)--(-6,4), linewidth(2) + red);
draw((-6,4)--(-5,4), linewidth(2) + red);
draw((0,3)--(-1,3), linewidth(2) + blue);
draw((-1,3)--(-1,4), linewidth(0.4) + blue);
draw((-1,4)--(0,4), linewidth(0.4) + blue);
draw((0,4)--(0,3), linewidth(2) + blue);
draw((-1,3)--(-2,3), linewidth(2) + red);
draw((-2,3)--(-2,4), linewidth(2) + red);
draw((-2,4)--(-1,4), linewidth(2) + red);
draw((-1,4)--(-1,3), linewidth(2) + red);
draw((-1,4)--(-1,5), linewidth(2) + red);
draw((-1,5)--(0,5), linewidth(2) + red);
draw((0,5)--(0,4), linewidth(2) + red);
draw((0,4)--(-1,4), linewidth(2) + red);
draw((4,3)--(3,3), linewidth(2) + blue);
draw((3,3)--(3,4), linewidth(0.4) + blue);
draw((3,4)--(4,4), linewidth(0.4) + blue);
draw((4,4)--(4,3), linewidth(2) + blue);
draw((3,3)--(2,3), linewidth(2) + red);
draw((2,3)--(2,4), linewidth(2) + red);
draw((2,4)--(3,4), linewidth(2) + red);
draw((3,4)--(3,3), linewidth(2) + red);
draw((3,4)--(3,5), linewidth(2) + red);
draw((3,5)--(4,5), linewidth(2) + red);
draw((4,5)--(4,4), linewidth(2) + red);
draw((4,4)--(3,4), linewidth(2) + red);
draw((8,3)--(7,3), linewidth(2) + blue);
draw((7,3)--(7,4), linewidth(0.4) + blue);
draw((7,4)--(8,4), linewidth(0.4) + blue);
draw((8,4)--(8,3), linewidth(2) + blue);
draw((7,3)--(6,3), linewidth(2) + red);
draw((6,3)--(6,4), linewidth(2) + red);
draw((6,4)--(7,4), linewidth(2) + red);
draw((7,4)--(7,3), linewidth(2) + red);
draw((7,4)--(7,5), linewidth(2) + red);
draw((7,5)--(8,5), linewidth(2) + red);
draw((8,5)--(8,4), linewidth(2) + red);
draw((8,4)--(7,4), linewidth(2) + red);
draw((6,4)--(6,5), linewidth(2));
draw((7,5)--(6,5), linewidth(2));
draw((2,4)--(2,5), linewidth(2));
draw((2,5)--(3,5), linewidth(2));
draw((-2,4)--(-2,5), linewidth(2));
draw((-2,5)--(-1,5), linewidth(2));
draw((-6,4)--(-6,5), linewidth(2));
draw((-6,5)--(-5,5), linewidth(2));
label("$R$",(2.2981490849012145,3.8125964085308546),SE*labelscalefactor);
label("$R$",(-1.7282479664156294,4.8037095288549985),SE*labelscalefactor);
label("$R$",(-4.701587327388068,3.8332445985376076),SE*labelscalefactor);
label("$R$",(7.2950110665354515,3.8125964085308546),SE*labelscalefactor);
/* dots and labels */
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]

As the policeman has to reach the bottom-right corner from being in the center cell, use the trick, go from the upper-left corner to the bottom-left corner, do the trick again, and reach from the upper-left corner to the bottom-left $2\times 2$ square in order to possibly catch the robber, the robber can move at least $$(1000+1000)+1+(2000+2000)+1+(1999+1999)=10000$$times. This proves $a)$.

Now, let's see that the policeman, after having done the trick two times and being in the upper-left corner, indeed has an strategy for trapping the robber. The strategy is as follows:

$\bullet$ The policeman uses $\rightarrow$ or $\downarrow$ until he is in the same diagonal as the robber. (He performs the move that gets him nearer to the robber's diagonal in the given turn).

$\bullet$ After this happens, if the robber does $\rightarrow$ or $\downarrow$, the policeman does exactly the same thing.

$\bullet$ If it is the policeman's turn and the robber is on the same diagonal as him, he uses $\downarrow$. Note that, while both of them are in the same diagonal, there is at least one free square between the policeman and the robber in the diagonal (or the policeman would have already caught the robber), as the policeman began above the robber and the robber must be in a black square during the policeman's turn. Thus, the robber never gets to be above or in the same cell as the policeman. The next time the robber performs $\nwarrow$, the policeman does $\rightarrow$, positioning himself in the same diagonal as the robber, again.

[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(10cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -6.928254722806678, xmax = 4.704143906950972, ymin = -7.0005807623688625, ymax = 3.612665117349596;  /* image dimensions */

draw((-2,-1)--(-3,-1)--(-3,0)--(-2,0)--cycle, linewidth(2) + blue);
draw((-3,-1)--(-4,-1)--(-4,0)--(-3,0)--cycle, linewidth(2) + red);
draw((-2,0)--(-3,0)--(-3,1)--(-2,1)--cycle, linewidth(2) + red);
draw((-5,-1)--(-6,-1)--(-6,0)--(-5,0)--cycle, linewidth(2) + blue);
draw((-4,0)--(-5,0)--(-5,1)--(-4,1)--cycle, linewidth(2) + blue);
draw((-3,1)--(-4,1)--(-4,2)--(-3,2)--cycle, linewidth(2) + blue);
draw((-2,2)--(-3,2)--(-3,3)--(-2,3)--cycle, linewidth(2) + blue);
draw((-5,0)--(-6,0)--(-6,1)--(-5,1)--cycle, linewidth(2) + red);
draw((-4,1)--(-5,1)--(-5,2)--(-4,2)--cycle, linewidth(2) + red);
draw((-3,2)--(-4,2)--(-4,3)--(-3,3)--cycle, linewidth(2) + red);
draw((-5,2)--(-6,2)--(-6,3)--(-5,3)--cycle, linewidth(2) + blue);
draw((3,-1)--(2,-1)--(2,0)--(3,0)--cycle, linewidth(2) + blue);
draw((2,-1)--(1,-1)--(1,0)--(2,0)--cycle, linewidth(2) + red);
draw((3,0)--(2,0)--(2,1)--(3,1)--cycle, linewidth(2) + red);
draw((0,-1)--(-1,-1)--(-1,0)--(0,0)--cycle, linewidth(2) + blue);
draw((0,0)--(-1,0)--(-1,1)--(0,1)--cycle, linewidth(2) + red);
draw((0,2)--(-1,2)--(-1,3)--(0,3)--cycle, linewidth(2) + blue);
draw((0,1)--(0,2)--(1,2)--(1,1)--cycle, linewidth(2) + red);
draw((1,2)--(1,3)--(2,3)--(2,2)--cycle, linewidth(2) + red);
draw((1,1)--(1,0)--(0,0)--(0,1)--cycle, linewidth(2) + blue);
draw((1,1)--(1,2)--(2,2)--(2,1)--cycle, linewidth(2) + blue);
draw((2,2)--(2,3)--(3,3)--(3,2)--cycle, linewidth(2) + blue);
draw((-2,-6)--(-3,-6)--(-3,-5)--(-2,-5)--cycle, linewidth(2) + blue);
draw((-3,-6)--(-4,-6)--(-4,-5)--(-3,-5)--cycle, linewidth(2) + red);
draw((-3,-5)--(-3,-4)--(-2,-4)--(-2,-5)--cycle, linewidth(2) + red);
draw((-2,-2)--(-2,-3)--(-3,-3)--(-3,-2)--cycle, linewidth(2) + blue);
draw((3,-6)--(2,-6)--(2,-5)--(3,-5)--cycle, linewidth(2) + blue);
draw((3,-5)--(2,-5)--(2,-4)--(3,-4)--cycle, linewidth(2) + red);
draw((3,-3)--(2,-3)--(2,-2)--(3,-2)--cycle, linewidth(2) + blue);
draw((2,-6)--(1,-6)--(1,-5)--(2,-5)--cycle, linewidth(2) + red);
draw((0,-6)--(-1,-6)--(-1,-5)--(0,-5)--cycle, linewidth(2) + blue);
draw((0,-3)--(-1,-3)--(-1,-2)--(0,-2)--cycle, linewidth(2) + blue);
draw((2,-3)--(1,-3)--(1,-2)--(2,-2)--cycle, linewidth(2) + red);
draw((2,-4)--(1,-4)--(1,-3)--(2,-3)--cycle, linewidth(2) + blue);
draw((1,-4)--(0,-4)--(0,-3)--(1,-3)--cycle, linewidth(2) + red);
draw((1,-5)--(0,-5)--(0,-4)--(1,-4)--cycle, linewidth(2) + blue);
draw((0,-5)--(-1,-5)--(-1,-4)--(0,-4)--cycle, linewidth(2) + red);
draw((-5,-3)--(-6,-3)--(-6,-2)--(-5,-2)--cycle, linewidth(2) + blue);
draw((-3,-3)--(-4,-3)--(-4,-2)--(-3,-2)--cycle, linewidth(2) + red);
draw((-3,-4)--(-4,-4)--(-4,-3)--(-3,-3)--cycle, linewidth(2) + blue);
draw((-4,-5)--(-5,-5)--(-5,-4)--(-4,-4)--cycle, linewidth(2) + blue);
draw((-5,-4)--(-5,-3)--(-4,-3)--(-4,-4)--cycle, linewidth(2) + red);
draw((-5,-5)--(-6,-5)--(-6,-4)--(-5,-4)--cycle, linewidth(2) + red);
draw((-5,-6)--(-6,-6)--(-6,-5)--(-5,-5)--cycle, linewidth(2) + blue);
/* draw figures */
draw((-2,-1)--(-3,-1), linewidth(2) + blue);
draw((-3,-1)--(-3,0), linewidth(2) + blue);
draw((-3,0)--(-2,0), linewidth(2) + blue);
draw((-2,0)--(-2,-1), linewidth(2) + blue);
draw((-3,-1)--(-4,-1), linewidth(2) + red);
draw((-4,-1)--(-4,0), linewidth(2) + red);
draw((-4,0)--(-3,0), linewidth(2) + red);
draw((-3,0)--(-3,-1), linewidth(2) + red);
draw((-2,0)--(-3,0), linewidth(2) + red);
draw((-3,0)--(-3,1), linewidth(2) + red);
draw((-3,1)--(-2,1), linewidth(2) + red);
draw((-2,1)--(-2,0), linewidth(2) + red);
draw((-5,-1)--(-6,-1), linewidth(2) + blue);
draw((-6,-1)--(-6,0), linewidth(2) + blue);
draw((-6,0)--(-5,0), linewidth(2) + blue);
draw((-5,0)--(-5,-1), linewidth(2) + blue);
draw((-4,0)--(-5,0), linewidth(2) + blue);
draw((-5,0)--(-5,1), linewidth(2) + blue);
draw((-5,1)--(-4,1), linewidth(2) + blue);
draw((-4,1)--(-4,0), linewidth(2) + blue);
draw((-3,1)--(-4,1), linewidth(2) + blue);
draw((-4,1)--(-4,2), linewidth(2) + blue);
draw((-4,2)--(-3,2), linewidth(2) + blue);
draw((-3,2)--(-3,1), linewidth(2) + blue);
draw((-2,2)--(-3,2), linewidth(2) + blue);
draw((-3,2)--(-3,3), linewidth(2) + blue);
draw((-3,3)--(-2,3), linewidth(2) + blue);
draw((-2,3)--(-2,2), linewidth(2) + blue);
draw((-5,0)--(-6,0), linewidth(2) + red);
draw((-6,0)--(-6,1), linewidth(2) + red);
draw((-6,1)--(-5,1), linewidth(2) + red);
draw((-5,1)--(-5,0), linewidth(2) + red);
draw((-4,1)--(-5,1), linewidth(2) + red);
draw((-5,1)--(-5,2), linewidth(2) + red);
draw((-5,2)--(-4,2), linewidth(2) + red);
draw((-4,2)--(-4,1), linewidth(2) + red);
draw((-3,2)--(-4,2), linewidth(2) + red);
draw((-4,2)--(-4,3), linewidth(2) + red);
draw((-4,3)--(-3,3), linewidth(2) + red);
draw((-3,3)--(-3,2), linewidth(2) + red);
draw((-5,2)--(-6,2), linewidth(2) + blue);
draw((-6,2)--(-6,3), linewidth(2) + blue);
draw((-6,3)--(-5,3), linewidth(2) + blue);
draw((-5,3)--(-5,2), linewidth(2) + blue);
draw((-5,-1)--(-4,-1), linewidth(2));
draw((-2,1)--(-2,2), linewidth(2));
draw((-4,3)--(-5,3), linewidth(2));
draw((-6,1)--(-6,2), linewidth(2));
draw((3,-1)--(2,-1), linewidth(2) + blue);
draw((2,-1)--(2,0), linewidth(2) + blue);
draw((2,0)--(3,0), linewidth(2) + blue);
draw((3,0)--(3,-1), linewidth(2) + blue);
draw((2,-1)--(1,-1), linewidth(2) + red);
draw((1,-1)--(1,0), linewidth(2) + red);
draw((1,0)--(2,0), linewidth(2) + red);
draw((2,0)--(2,-1), linewidth(2) + red);
draw((3,0)--(2,0), linewidth(2) + red);
draw((2,0)--(2,1), linewidth(2) + red);
draw((2,1)--(3,1), linewidth(2) + red);
draw((3,1)--(3,0), linewidth(2) + red);
draw((0,-1)--(-1,-1), linewidth(2) + blue);
draw((-1,-1)--(-1,0), linewidth(2) + blue);
draw((-1,0)--(0,0), linewidth(2) + blue);
draw((0,0)--(0,-1), linewidth(2) + blue);
draw((0,0)--(-1,0), linewidth(2) + red);
draw((-1,0)--(-1,1), linewidth(2) + red);
draw((-1,1)--(0,1), linewidth(2) + red);
draw((0,1)--(0,0), linewidth(2) + red);
draw((0,2)--(-1,2), linewidth(2) + blue);
draw((-1,2)--(-1,3), linewidth(2) + blue);
draw((-1,3)--(0,3), linewidth(2) + blue);
draw((0,3)--(0,2), linewidth(2) + blue);
draw((0,2)--(1,2), linewidth(2) + red);
draw((1,2)--(1,1), linewidth(2) + red);
draw((1,1)--(0,1), linewidth(2) + red);
draw((1,2)--(1,3), linewidth(2) + red);
draw((1,3)--(2,3), linewidth(2) + red);
draw((2,3)--(2,2), linewidth(2) + red);
draw((2,2)--(1,2), linewidth(2) + red);
draw((1,1)--(1,0), linewidth(2) + blue);
draw((1,0)--(0,0), linewidth(2) + blue);
draw((0,0)--(0,1), linewidth(2) + blue);
draw((0,1)--(1,1), linewidth(2) + blue);
draw((1,1)--(1,2), linewidth(2) + blue);
draw((1,2)--(2,2), linewidth(2) + blue);
draw((2,2)--(2,1), linewidth(2) + blue);
draw((2,1)--(1,1), linewidth(2) + blue);
draw((2,2)--(2,3), linewidth(2) + blue);
draw((2,3)--(3,3), linewidth(2) + blue);
draw((3,3)--(3,2), linewidth(2) + blue);
draw((3,2)--(2,2), linewidth(2) + blue);
draw((-1.7816706774971747,0.9775970110052327)--(-1.226249843103488,0.9775970110052327), linewidth(2),EndArrow(6));
draw((-1.144968745387338,-1.1898989280920769)--(-1.8358580759746068,-1.8266008602019115), linewidth(2),EndArrow(6));
draw((-2,-6)--(-3,-6), linewidth(2) + blue);
draw((-3,-6)--(-3,-5), linewidth(2) + blue);
draw((-3,-5)--(-2,-5), linewidth(2) + blue);
draw((-2,-5)--(-2,-6), linewidth(2) + blue);
draw((-3,-6)--(-4,-6), linewidth(2) + red);
draw((-4,-6)--(-4,-5), linewidth(2) + red);
draw((-4,-5)--(-3,-5), linewidth(2) + red);
draw((-3,-5)--(-3,-6), linewidth(2) + red);
draw((-3,-5)--(-3,-4), linewidth(2) + red);
draw((-3,-4)--(-2,-4), linewidth(2) + red);
draw((-2,-4)--(-2,-5), linewidth(2) + red);
draw((-2,-5)--(-3,-5), linewidth(2) + red);
draw((-2,-2)--(-2,-3), linewidth(2) + blue);
draw((-2,-3)--(-3,-3), linewidth(2) + blue);
draw((-3,-3)--(-3,-2), linewidth(2) + blue);
draw((-3,-2)--(-2,-2), linewidth(2) + blue);
draw((3,-6)--(2,-6), linewidth(2) + blue);
draw((2,-6)--(2,-5), linewidth(2) + blue);
draw((2,-5)--(3,-5), linewidth(2) + blue);
draw((3,-5)--(3,-6), linewidth(2) + blue);
draw((3,-5)--(2,-5), linewidth(2) + red);
draw((2,-5)--(2,-4), linewidth(2) + red);
draw((2,-4)--(3,-4), linewidth(2) + red);
draw((3,-4)--(3,-5), linewidth(2) + red);
draw((3,-3)--(2,-3), linewidth(2) + blue);
draw((2,-3)--(2,-2), linewidth(2) + blue);
draw((2,-2)--(3,-2), linewidth(2) + blue);
draw((3,-2)--(3,-3), linewidth(2) + blue);
draw((2,-6)--(1,-6), linewidth(2) + red);
draw((1,-6)--(1,-5), linewidth(2) + red);
draw((1,-5)--(2,-5), linewidth(2) + red);
draw((2,-5)--(2,-6), linewidth(2) + red);
draw((0,-6)--(-1,-6), linewidth(2) + blue);
draw((-1,-6)--(-1,-5), linewidth(2) + blue);
draw((-1,-5)--(0,-5), linewidth(2) + blue);
draw((0,-5)--(0,-6), linewidth(2) + blue);
draw((0,-3)--(-1,-3), linewidth(2) + blue);
draw((-1,-3)--(-1,-2), linewidth(2) + blue);
draw((-1,-2)--(0,-2), linewidth(2) + blue);
draw((0,-2)--(0,-3), linewidth(2) + blue);
draw((2,-3)--(1,-3), linewidth(2) + red);
draw((1,-3)--(1,-2), linewidth(2) + red);
draw((1,-2)--(2,-2), linewidth(2) + red);
draw((2,-2)--(2,-3), linewidth(2) + red);
draw((2,-4)--(1,-4), linewidth(2) + blue);
draw((1,-4)--(1,-3), linewidth(2) + blue);
draw((1,-3)--(2,-3), linewidth(2) + blue);
draw((2,-3)--(2,-4), linewidth(2) + blue);
draw((1,-4)--(0,-4), linewidth(2) + red);
draw((0,-4)--(0,-3), linewidth(2) + red);
draw((0,-3)--(1,-3), linewidth(2) + red);
draw((1,-3)--(1,-4), linewidth(2) + red);
draw((1,-5)--(0,-5), linewidth(2) + blue);
draw((0,-5)--(0,-4), linewidth(2) + blue);
draw((0,-4)--(1,-4), linewidth(2) + blue);
draw((1,-4)--(1,-5), linewidth(2) + blue);
draw((0,-5)--(-1,-5), linewidth(2) + red);
draw((-1,-5)--(-1,-4), linewidth(2) + red);
draw((-1,-4)--(0,-4), linewidth(2) + red);
draw((0,-4)--(0,-5), linewidth(2) + red);
draw((-5,-3)--(-6,-3), linewidth(2) + blue);
draw((-6,-3)--(-6,-2), linewidth(2) + blue);
draw((-6,-2)--(-5,-2), linewidth(2) + blue);
draw((-5,-2)--(-5,-3), linewidth(2) + blue);
draw((-3,-3)--(-4,-3), linewidth(2) + red);
draw((-4,-3)--(-4,-2), linewidth(2) + red);
draw((-4,-2)--(-3,-2), linewidth(2) + red);
draw((-3,-2)--(-3,-3), linewidth(2) + red);
draw((-3,-4)--(-4,-4), linewidth(2) + blue);
draw((-4,-4)--(-4,-3), linewidth(2) + blue);
draw((-4,-3)--(-3,-3), linewidth(2) + blue);
draw((-3,-3)--(-3,-4), linewidth(2) + blue);
draw((-4,-5)--(-5,-5), linewidth(2) + blue);
draw((-5,-5)--(-5,-4), linewidth(2) + blue);
draw((-5,-4)--(-4,-4), linewidth(2) + blue);
draw((-4,-4)--(-4,-5), linewidth(2) + blue);
draw((-5,-4)--(-5,-3), linewidth(2) + red);
draw((-5,-3)--(-4,-3), linewidth(2) + red);
draw((-4,-3)--(-4,-4), linewidth(2) + red);
draw((-4,-4)--(-5,-4), linewidth(2) + red);
draw((-5,-5)--(-6,-5), linewidth(2) + red);
draw((-6,-5)--(-6,-4), linewidth(2) + red);
draw((-6,-4)--(-5,-4), linewidth(2) + red);
draw((-5,-4)--(-5,-5), linewidth(2) + red);
draw((-5,-6)--(-6,-6), linewidth(2) + blue);
draw((-6,-6)--(-6,-5), linewidth(2) + blue);
draw((-6,-5)--(-5,-5), linewidth(2) + blue);
draw((-5,-5)--(-5,-6), linewidth(2) + blue);
draw((-1.7991049342473708,-4.535285403222338)--(-1.0839408493060767,-4.535285403222338), linewidth(2),EndArrow(6));
draw((-6,-4)--(-6,-3), linewidth(2));
draw((-5,-2)--(-4,-2), linewidth(2));
draw((-5,-6)--(-4,-6), linewidth(2));
draw((-2,-4)--(-2,-3), linewidth(2));
draw((-1,1)--(-1,2), linewidth(2));
draw((0,3)--(1,3), linewidth(2));
draw((3,1)--(3,2), linewidth(2));
draw((1,-1)--(0,-1), linewidth(2));
draw((0,-2)--(1,-2), linewidth(2));
draw((3,-4)--(3,-3), linewidth(2));
draw((1,-6)--(0,-6), linewidth(2));
draw((-1,-4)--(-1,-3), linewidth(2));
draw((0,1)--(0,2), linewidth(2));
draw((1,-6)--(1,-5), linewidth(2));
label("$P$",(-5.715814382242815,2.769228358696473),SE*labelscalefactor);
label("$P$",(-0.6903370286012918,1.6973608112414633),SE*labelscalefactor);
label("$P$",(-5.645527985688388,-3.2754017449842387),SE*labelscalefactor);
label("$R$",(0.36395891971511163,-3.257830145845632),SE*labelscalefactor);
label("$P$",(0.10038493263601078,-3.257830145845632),SE*labelscalefactor);
label("$R$",(1.3655400706156948,0.7484944577567003),SE*labelscalefactor);
label("$R$",(-3.6950804813030413,0.7133512594794869),SE*labelscalefactor);
label("$R$",(-4.749376429619445,-3.257830145845632),SE*labelscalefactor);
/* dots and labels */
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]

It is clear that the policeman can always play in that way. As he only uses $\downarrow$ and $\rightarrow$, the amount of moves must end, and the only way this can happen is if the policeman is already on the same cell as the robber, proving $b)$. We are done $\square$.

[quote=Modesti]Solution with [b]MinosFx[/b]

Let's color the $4001$ diagonals that have positive slope with [color=#00f]blue[/color], [color=#f00]red[/color], and [b]black[/b] starting from the bottom-right diagonal (which is the bottom-right corner) and ending with the upper-left diagonal (the upper-left corner). Here is the coloring for a $3\times 3$ board:

[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(3cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -9.55652892561984, xmax = 1.5509090909090903, ymin = -2.6157024793388435, ymax = 7.36776859504132;  /* image dimensions */

draw((-4,2)--(-4,3)--(-3,3)--(-3,2)--cycle, linewidth(2) + blue);
draw((-5,2)--(-5,3)--(-4,3)--(-4,2)--cycle, linewidth(2) + red);
draw((-5,4)--(-5,5)--(-4,5)--(-4,4)--cycle, linewidth(2) + blue);
draw((-4,3)--(-4,4)--(-3,4)--(-3,3)--cycle, linewidth(2) + red);
draw((-5,3)--(-6,3)--(-6,4)--(-5,4)--cycle, linewidth(2) + blue);
draw((-6,4)--(-6,5)--(-5,5)--(-5,4)--cycle, linewidth(2) + red);
/* draw figures */
draw((-4,2)--(-4,3), linewidth(0.4) + blue);
draw((-4,3)--(-3,3), linewidth(0.4) + blue);
draw((-3,3)--(-3,2), linewidth(2) + blue);
draw((-3,2)--(-4,2), linewidth(2) + blue);
draw((-5,2)--(-5,3), linewidth(2) + red);
draw((-5,3)--(-4,3), linewidth(2) + red);
draw((-4,3)--(-4,2), linewidth(2) + red);
draw((-4,2)--(-5,2), linewidth(2) + red);
draw((-5,4)--(-5,5), linewidth(0.4) + blue);
draw((-5,5)--(-4,5), linewidth(2) + blue);
draw((-4,5)--(-4,4), linewidth(2) + blue);
draw((-4,4)--(-5,4), linewidth(2) + blue);
draw((-4,3)--(-4,4), linewidth(2) + red);
draw((-4,4)--(-3,4), linewidth(2) + red);
draw((-3,4)--(-3,3), linewidth(2) + red);
draw((-3,3)--(-4,3), linewidth(2) + red);
draw((-5,3)--(-6,3), linewidth(2) + blue);
draw((-6,3)--(-6,4), linewidth(2) + blue);
draw((-6,4)--(-5,4), linewidth(0.4) + blue);
draw((-5,4)--(-5,3), linewidth(2) + blue);
draw((-3,4)--(-3,5), linewidth(2));
draw((-5,2)--(-6,2), linewidth(2));
draw((-6,4)--(-6,5), linewidth(2) + red);
draw((-6,5)--(-5,5), linewidth(2) + red);
draw((-5,5)--(-5,4), linewidth(2) + red);
draw((-5,4)--(-6,4), linewidth(2) + red);
draw((-6,2)--(-6,3), linewidth(2));
draw((-4,5)--(-3,5), linewidth(2));
/* dots and labels */
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]

Now, note that, only performing $\nwarrow$, $\downarrow$ or $\rightarrow$, the colors of the cells in which the policeman and the robber are on repeat periodically:

[color=#00f]blue[/color] $\mapsto$ [b]black[/b] $\mapsto$ [color=#f00]red[/color] $\mapsto$ [color=#00f]blue[/color]

Thus, as the policeman starts at [b]black[/b] and the robber at [color=#f00]red[/color], the policeman will never catch the robber without using his "special move" (moving from the bottom-left corner directly to the upper-left corner). Morerover, he must use it twice. When he uses it for the first time, he goes from a [color=#00f]blue[/color] cell to a [color=#f00]red[/color] cell, while the robber would be in a [b]black[/b] cell, according to the cycle. Hence, the policeman must perform the trick again: the policeman would go from [color=#00f]blue[/color] to [color=#f00]red[/color] again, but, remembering he has skipped [b]black[/b] twice, now the robber would be in a [color=#f00]red[/color] cell, too.

Now, we explain how the robber can move $10000$ times. The strategy of the robber would be reaching the bottom-left corner and doing this repeatedly:

[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(10cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -6.745758138056621, xmax = 8.36871694688661, ymin = -1.7830630832992078, ymax = 10.688443680779603;  /* image dimensions */

draw((-4,3)--(-5,3)--(-5,4)--(-4,4)--cycle, linewidth(2) + blue);
draw((-4,5)--(-4,4)--(-5,4)--(-5,5)--cycle, linewidth(2) + red);
draw((-5,4)--(-5,3)--(-6,3)--(-6,4)--cycle, linewidth(2) + red);
draw((0,3)--(-1,3)--(-1,4)--(0,4)--cycle, linewidth(2) + blue);
draw((-1,3)--(-2,3)--(-2,4)--(-1,4)--cycle, linewidth(2) + red);
draw((-1,4)--(-1,5)--(0,5)--(0,4)--cycle, linewidth(2) + red);
draw((4,3)--(3,3)--(3,4)--(4,4)--cycle, linewidth(2) + blue);
draw((3,3)--(2,3)--(2,4)--(3,4)--cycle, linewidth(2) + red);
draw((3,4)--(3,5)--(4,5)--(4,4)--cycle, linewidth(2) + red);
draw((8,3)--(7,3)--(7,4)--(8,4)--cycle, linewidth(2) + blue);
draw((7,3)--(6,3)--(6,4)--(7,4)--cycle, linewidth(2) + red);
draw((7,4)--(7,5)--(8,5)--(8,4)--cycle, linewidth(2) + red);
/* draw figures */
draw((-3.608128237093616,3.98299760787552)--(-2.409463905013026,3.98299760787552), linewidth(2),EndArrow(6));
draw((0.3825469172722511,3.98299760787552)--(1.5958291070611412,3.9976154655838196), linewidth(2),EndArrow(6));
draw((4.387839929346418,3.9976154655838205)--(5.557268546010409,3.9976154655838205), linewidth(2),EndArrow(6));
draw((-4,3)--(-5,3), linewidth(2) + blue);
draw((-5,3)--(-5,4), linewidth(0.4) + blue);
draw((-5,4)--(-4,4), linewidth(0.4) + blue);
draw((-4,4)--(-4,3), linewidth(2) + blue);
draw((-4,5)--(-4,4), linewidth(2) + red);
draw((-4,4)--(-5,4), linewidth(2) + red);
draw((-5,4)--(-5,5), linewidth(2) + red);
draw((-5,5)--(-4,5), linewidth(2) + red);
draw((-5,4)--(-5,3), linewidth(2) + red);
draw((-5,3)--(-6,3), linewidth(2) + red);
draw((-6,3)--(-6,4), linewidth(2) + red);
draw((-6,4)--(-5,4), linewidth(2) + red);
draw((0,3)--(-1,3), linewidth(2) + blue);
draw((-1,3)--(-1,4), linewidth(0.4) + blue);
draw((-1,4)--(0,4), linewidth(0.4) + blue);
draw((0,4)--(0,3), linewidth(2) + blue);
draw((-1,3)--(-2,3), linewidth(2) + red);
draw((-2,3)--(-2,4), linewidth(2) + red);
draw((-2,4)--(-1,4), linewidth(2) + red);
draw((-1,4)--(-1,3), linewidth(2) + red);
draw((-1,4)--(-1,5), linewidth(2) + red);
draw((-1,5)--(0,5), linewidth(2) + red);
draw((0,5)--(0,4), linewidth(2) + red);
draw((0,4)--(-1,4), linewidth(2) + red);
draw((4,3)--(3,3), linewidth(2) + blue);
draw((3,3)--(3,4), linewidth(0.4) + blue);
draw((3,4)--(4,4), linewidth(0.4) + blue);
draw((4,4)--(4,3), linewidth(2) + blue);
draw((3,3)--(2,3), linewidth(2) + red);
draw((2,3)--(2,4), linewidth(2) + red);
draw((2,4)--(3,4), linewidth(2) + red);
draw((3,4)--(3,3), linewidth(2) + red);
draw((3,4)--(3,5), linewidth(2) + red);
draw((3,5)--(4,5), linewidth(2) + red);
draw((4,5)--(4,4), linewidth(2) + red);
draw((4,4)--(3,4), linewidth(2) + red);
draw((8,3)--(7,3), linewidth(2) + blue);
draw((7,3)--(7,4), linewidth(0.4) + blue);
draw((7,4)--(8,4), linewidth(0.4) + blue);
draw((8,4)--(8,3), linewidth(2) + blue);
draw((7,3)--(6,3), linewidth(2) + red);
draw((6,3)--(6,4), linewidth(2) + red);
draw((6,4)--(7,4), linewidth(2) + red);
draw((7,4)--(7,3), linewidth(2) + red);
draw((7,4)--(7,5), linewidth(2) + red);
draw((7,5)--(8,5), linewidth(2) + red);
draw((8,5)--(8,4), linewidth(2) + red);
draw((8,4)--(7,4), linewidth(2) + red);
draw((6,4)--(6,5), linewidth(2));
draw((7,5)--(6,5), linewidth(2));
draw((2,4)--(2,5), linewidth(2));
draw((2,5)--(3,5), linewidth(2));
draw((-2,4)--(-2,5), linewidth(2));
draw((-2,5)--(-1,5), linewidth(2));
draw((-6,4)--(-6,5), linewidth(2));
draw((-6,5)--(-5,5), linewidth(2));
label("$R$",(2.2981490849012145,3.8125964085308546),SE*labelscalefactor);
label("$R$",(-1.7282479664156294,4.8037095288549985),SE*labelscalefactor);
label("$R$",(-4.701587327388068,3.8332445985376076),SE*labelscalefactor);
label("$R$",(7.2950110665354515,3.8125964085308546),SE*labelscalefactor);
/* dots and labels */
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]

As the policeman has to reach the bottom-right corner from being in the center cell, use the trick, go from the upper-left corner to the bottom-left corner, do the trick again, and reach from the upper-left corner to the bottom-left $2\times 2$ square in order to possibly catch the robber, the robber can move at least $$(1000+1000)+1+(2000+2000)+1+(1999+1999)=10000$$times. This proves $a)$.

Now, let's see that the policeman, after having done the trick two times and being in the upper-left corner, indeed has an strategy for trapping the robber. The strategy is as follows:

$\bullet$ The policeman uses $\rightarrow$ until he is in the same diagonal as the robber.

$\bullet$ After this happens, if the robber does $\rightarrow$ or $\downarrow$, the policeman does exactly the same thing.

$\bullet$ If it is the policeman's turn and the robber is on the same diagonal as him, he uses $\downarrow$. Note that, while both of them are in the same diagonal, there is at least one free square between the policeman and the robber in the diagonal (or the policeman would have already caught the robber), as the policeman began above the robber and the robber must be in a black square during the policeman's turn. Thus, the robber never gets to be above or in the same cell as the policeman. The next time the robber performs $\nwarrow$, the policeman does $\rightarrow$, positioning himself in the same diagonal as the robber, again.

[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(10cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -6.928254722806678, xmax = 4.704143906950972, ymin = -7.0005807623688625, ymax = 3.612665117349596;  /* image dimensions */

draw((-2,-1)--(-3,-1)--(-3,0)--(-2,0)--cycle, linewidth(2) + blue);
draw((-3,-1)--(-4,-1)--(-4,0)--(-3,0)--cycle, linewidth(2) + red);
draw((-2,0)--(-3,0)--(-3,1)--(-2,1)--cycle, linewidth(2) + red);
draw((-5,-1)--(-6,-1)--(-6,0)--(-5,0)--cycle, linewidth(2) + blue);
draw((-4,0)--(-5,0)--(-5,1)--(-4,1)--cycle, linewidth(2) + blue);
draw((-3,1)--(-4,1)--(-4,2)--(-3,2)--cycle, linewidth(2) + blue);
draw((-2,2)--(-3,2)--(-3,3)--(-2,3)--cycle, linewidth(2) + blue);
draw((-5,0)--(-6,0)--(-6,1)--(-5,1)--cycle, linewidth(2) + red);
draw((-4,1)--(-5,1)--(-5,2)--(-4,2)--cycle, linewidth(2) + red);
draw((-3,2)--(-4,2)--(-4,3)--(-3,3)--cycle, linewidth(2) + red);
draw((-5,2)--(-6,2)--(-6,3)--(-5,3)--cycle, linewidth(2) + blue);
draw((3,-1)--(2,-1)--(2,0)--(3,0)--cycle, linewidth(2) + blue);
draw((2,-1)--(1,-1)--(1,0)--(2,0)--cycle, linewidth(2) + red);
draw((3,0)--(2,0)--(2,1)--(3,1)--cycle, linewidth(2) + red);
draw((0,-1)--(-1,-1)--(-1,0)--(0,0)--cycle, linewidth(2) + blue);
draw((0,0)--(-1,0)--(-1,1)--(0,1)--cycle, linewidth(2) + red);
draw((0,2)--(-1,2)--(-1,3)--(0,3)--cycle, linewidth(2) + blue);
draw((0,1)--(0,2)--(1,2)--(1,1)--cycle, linewidth(2) + red);
draw((1,2)--(1,3)--(2,3)--(2,2)--cycle, linewidth(2) + red);
draw((1,1)--(1,0)--(0,0)--(0,1)--cycle, linewidth(2) + blue);
draw((1,1)--(1,2)--(2,2)--(2,1)--cycle, linewidth(2) + blue);
draw((2,2)--(2,3)--(3,3)--(3,2)--cycle, linewidth(2) + blue);
draw((-2,-6)--(-3,-6)--(-3,-5)--(-2,-5)--cycle, linewidth(2) + blue);
draw((-3,-6)--(-4,-6)--(-4,-5)--(-3,-5)--cycle, linewidth(2) + red);
draw((-3,-5)--(-3,-4)--(-2,-4)--(-2,-5)--cycle, linewidth(2) + red);
draw((-2,-2)--(-2,-3)--(-3,-3)--(-3,-2)--cycle, linewidth(2) + blue);
draw((3,-6)--(2,-6)--(2,-5)--(3,-5)--cycle, linewidth(2) + blue);
draw((3,-5)--(2,-5)--(2,-4)--(3,-4)--cycle, linewidth(2) + red);
draw((3,-3)--(2,-3)--(2,-2)--(3,-2)--cycle, linewidth(2) + blue);
draw((2,-6)--(1,-6)--(1,-5)--(2,-5)--cycle, linewidth(2) + red);
draw((0,-6)--(-1,-6)--(-1,-5)--(0,-5)--cycle, linewidth(2) + blue);
draw((0,-3)--(-1,-3)--(-1,-2)--(0,-2)--cycle, linewidth(2) + blue);
draw((2,-3)--(1,-3)--(1,-2)--(2,-2)--cycle, linewidth(2) + red);
draw((2,-4)--(1,-4)--(1,-3)--(2,-3)--cycle, linewidth(2) + blue);
draw((1,-4)--(0,-4)--(0,-3)--(1,-3)--cycle, linewidth(2) + red);
draw((1,-5)--(0,-5)--(0,-4)--(1,-4)--cycle, linewidth(2) + blue);
draw((0,-5)--(-1,-5)--(-1,-4)--(0,-4)--cycle, linewidth(2) + red);
draw((-5,-3)--(-6,-3)--(-6,-2)--(-5,-2)--cycle, linewidth(2) + blue);
draw((-3,-3)--(-4,-3)--(-4,-2)--(-3,-2)--cycle, linewidth(2) + red);
draw((-3,-4)--(-4,-4)--(-4,-3)--(-3,-3)--cycle, linewidth(2) + blue);
draw((-4,-5)--(-5,-5)--(-5,-4)--(-4,-4)--cycle, linewidth(2) + blue);
draw((-5,-4)--(-5,-3)--(-4,-3)--(-4,-4)--cycle, linewidth(2) + red);
draw((-5,-5)--(-6,-5)--(-6,-4)--(-5,-4)--cycle, linewidth(2) + red);
draw((-5,-6)--(-6,-6)--(-6,-5)--(-5,-5)--cycle, linewidth(2) + blue);
/* draw figures */
draw((-2,-1)--(-3,-1), linewidth(2) + blue);
draw((-3,-1)--(-3,0), linewidth(2) + blue);
draw((-3,0)--(-2,0), linewidth(2) + blue);
draw((-2,0)--(-2,-1), linewidth(2) + blue);
draw((-3,-1)--(-4,-1), linewidth(2) + red);
draw((-4,-1)--(-4,0), linewidth(2) + red);
draw((-4,0)--(-3,0), linewidth(2) + red);
draw((-3,0)--(-3,-1), linewidth(2) + red);
draw((-2,0)--(-3,0), linewidth(2) + red);
draw((-3,0)--(-3,1), linewidth(2) + red);
draw((-3,1)--(-2,1), linewidth(2) + red);
draw((-2,1)--(-2,0), linewidth(2) + red);
draw((-5,-1)--(-6,-1), linewidth(2) + blue);
draw((-6,-1)--(-6,0), linewidth(2) + blue);
draw((-6,0)--(-5,0), linewidth(2) + blue);
draw((-5,0)--(-5,-1), linewidth(2) + blue);
draw((-4,0)--(-5,0), linewidth(2) + blue);
draw((-5,0)--(-5,1), linewidth(2) + blue);
draw((-5,1)--(-4,1), linewidth(2) + blue);
draw((-4,1)--(-4,0), linewidth(2) + blue);
draw((-3,1)--(-4,1), linewidth(2) + blue);
draw((-4,1)--(-4,2), linewidth(2) + blue);
draw((-4,2)--(-3,2), linewidth(2) + blue);
draw((-3,2)--(-3,1), linewidth(2) + blue);
draw((-2,2)--(-3,2), linewidth(2) + blue);
draw((-3,2)--(-3,3), linewidth(2) + blue);
draw((-3,3)--(-2,3), linewidth(2) + blue);
draw((-2,3)--(-2,2), linewidth(2) + blue);
draw((-5,0)--(-6,0), linewidth(2) + red);
draw((-6,0)--(-6,1), linewidth(2) + red);
draw((-6,1)--(-5,1), linewidth(2) + red);
draw((-5,1)--(-5,0), linewidth(2) + red);
draw((-4,1)--(-5,1), linewidth(2) + red);
draw((-5,1)--(-5,2), linewidth(2) + red);
draw((-5,2)--(-4,2), linewidth(2) + red);
draw((-4,2)--(-4,1), linewidth(2) + red);
draw((-3,2)--(-4,2), linewidth(2) + red);
draw((-4,2)--(-4,3), linewidth(2) + red);
draw((-4,3)--(-3,3), linewidth(2) + red);
draw((-3,3)--(-3,2), linewidth(2) + red);
draw((-5,2)--(-6,2), linewidth(2) + blue);
draw((-6,2)--(-6,3), linewidth(2) + blue);
draw((-6,3)--(-5,3), linewidth(2) + blue);
draw((-5,3)--(-5,2), linewidth(2) + blue);
draw((-5,-1)--(-4,-1), linewidth(2));
draw((-2,1)--(-2,2), linewidth(2));
draw((-4,3)--(-5,3), linewidth(2));
draw((-6,1)--(-6,2), linewidth(2));
draw((3,-1)--(2,-1), linewidth(2) + blue);
draw((2,-1)--(2,0), linewidth(2) + blue);
draw((2,0)--(3,0), linewidth(2) + blue);
draw((3,0)--(3,-1), linewidth(2) + blue);
draw((2,-1)--(1,-1), linewidth(2) + red);
draw((1,-1)--(1,0), linewidth(2) + red);
draw((1,0)--(2,0), linewidth(2) + red);
draw((2,0)--(2,-1), linewidth(2) + red);
draw((3,0)--(2,0), linewidth(2) + red);
draw((2,0)--(2,1), linewidth(2) + red);
draw((2,1)--(3,1), linewidth(2) + red);
draw((3,1)--(3,0), linewidth(2) + red);
draw((0,-1)--(-1,-1), linewidth(2) + blue);
draw((-1,-1)--(-1,0), linewidth(2) + blue);
draw((-1,0)--(0,0), linewidth(2) + blue);
draw((0,0)--(0,-1), linewidth(2) + blue);
draw((0,0)--(-1,0), linewidth(2) + red);
draw((-1,0)--(-1,1), linewidth(2) + red);
draw((-1,1)--(0,1), linewidth(2) + red);
draw((0,1)--(0,0), linewidth(2) + red);
draw((0,2)--(-1,2), linewidth(2) + blue);
draw((-1,2)--(-1,3), linewidth(2) + blue);
draw((-1,3)--(0,3), linewidth(2) + blue);
draw((0,3)--(0,2), linewidth(2) + blue);
draw((0,2)--(1,2), linewidth(2) + red);
draw((1,2)--(1,1), linewidth(2) + red);
draw((1,1)--(0,1), linewidth(2) + red);
draw((1,2)--(1,3), linewidth(2) + red);
draw((1,3)--(2,3), linewidth(2) + red);
draw((2,3)--(2,2), linewidth(2) + red);
draw((2,2)--(1,2), linewidth(2) + red);
draw((1,1)--(1,0), linewidth(2) + blue);
draw((1,0)--(0,0), linewidth(2) + blue);
draw((0,0)--(0,1), linewidth(2) + blue);
draw((0,1)--(1,1), linewidth(2) + blue);
draw((1,1)--(1,2), linewidth(2) + blue);
draw((1,2)--(2,2), linewidth(2) + blue);
draw((2,2)--(2,1), linewidth(2) + blue);
draw((2,1)--(1,1), linewidth(2) + blue);
draw((2,2)--(2,3), linewidth(2) + blue);
draw((2,3)--(3,3), linewidth(2) + blue);
draw((3,3)--(3,2), linewidth(2) + blue);
draw((3,2)--(2,2), linewidth(2) + blue);
draw((-1.7816706774971747,0.9775970110052327)--(-1.226249843103488,0.9775970110052327), linewidth(2),EndArrow(6));
draw((-1.144968745387338,-1.1898989280920769)--(-1.8358580759746068,-1.8266008602019115), linewidth(2),EndArrow(6));
draw((-2,-6)--(-3,-6), linewidth(2) + blue);
draw((-3,-6)--(-3,-5), linewidth(2) + blue);
draw((-3,-5)--(-2,-5), linewidth(2) + blue);
draw((-2,-5)--(-2,-6), linewidth(2) + blue);
draw((-3,-6)--(-4,-6), linewidth(2) + red);
draw((-4,-6)--(-4,-5), linewidth(2) + red);
draw((-4,-5)--(-3,-5), linewidth(2) + red);
draw((-3,-5)--(-3,-6), linewidth(2) + red);
draw((-3,-5)--(-3,-4), linewidth(2) + red);
draw((-3,-4)--(-2,-4), linewidth(2) + red);
draw((-2,-4)--(-2,-5), linewidth(2) + red);
draw((-2,-5)--(-3,-5), linewidth(2) + red);
draw((-2,-2)--(-2,-3), linewidth(2) + blue);
draw((-2,-3)--(-3,-3), linewidth(2) + blue);
draw((-3,-3)--(-3,-2), linewidth(2) + blue);
draw((-3,-2)--(-2,-2), linewidth(2) + blue);
draw((3,-6)--(2,-6), linewidth(2) + blue);
draw((2,-6)--(2,-5), linewidth(2) + blue);
draw((2,-5)--(3,-5), linewidth(2) + blue);
draw((3,-5)--(3,-6), linewidth(2) + blue);
draw((3,-5)--(2,-5), linewidth(2) + red);
draw((2,-5)--(2,-4), linewidth(2) + red);
draw((2,-4)--(3,-4), linewidth(2) + red);
draw((3,-4)--(3,-5), linewidth(2) + red);
draw((3,-3)--(2,-3), linewidth(2) + blue);
draw((2,-3)--(2,-2), linewidth(2) + blue);
draw((2,-2)--(3,-2), linewidth(2) + blue);
draw((3,-2)--(3,-3), linewidth(2) + blue);
draw((2,-6)--(1,-6), linewidth(2) + red);
draw((1,-6)--(1,-5), linewidth(2) + red);
draw((1,-5)--(2,-5), linewidth(2) + red);
draw((2,-5)--(2,-6), linewidth(2) + red);
draw((0,-6)--(-1,-6), linewidth(2) + blue);
draw((-1,-6)--(-1,-5), linewidth(2) + blue);
draw((-1,-5)--(0,-5), linewidth(2) + blue);
draw((0,-5)--(0,-6), linewidth(2) + blue);
draw((0,-3)--(-1,-3), linewidth(2) + blue);
draw((-1,-3)--(-1,-2), linewidth(2) + blue);
draw((-1,-2)--(0,-2), linewidth(2) + blue);
draw((0,-2)--(0,-3), linewidth(2) + blue);
draw((2,-3)--(1,-3), linewidth(2) + red);
draw((1,-3)--(1,-2), linewidth(2) + red);
draw((1,-2)--(2,-2), linewidth(2) + red);
draw((2,-2)--(2,-3), linewidth(2) + red);
draw((2,-4)--(1,-4), linewidth(2) + blue);
draw((1,-4)--(1,-3), linewidth(2) + blue);
draw((1,-3)--(2,-3), linewidth(2) + blue);
draw((2,-3)--(2,-4), linewidth(2) + blue);
draw((1,-4)--(0,-4), linewidth(2) + red);
draw((0,-4)--(0,-3), linewidth(2) + red);
draw((0,-3)--(1,-3), linewidth(2) + red);
draw((1,-3)--(1,-4), linewidth(2) + red);
draw((1,-5)--(0,-5), linewidth(2) + blue);
draw((0,-5)--(0,-4), linewidth(2) + blue);
draw((0,-4)--(1,-4), linewidth(2) + blue);
draw((1,-4)--(1,-5), linewidth(2) + blue);
draw((0,-5)--(-1,-5), linewidth(2) + red);
draw((-1,-5)--(-1,-4), linewidth(2) + red);
draw((-1,-4)--(0,-4), linewidth(2) + red);
draw((0,-4)--(0,-5), linewidth(2) + red);
draw((-5,-3)--(-6,-3), linewidth(2) + blue);
draw((-6,-3)--(-6,-2), linewidth(2) + blue);
draw((-6,-2)--(-5,-2), linewidth(2) + blue);
draw((-5,-2)--(-5,-3), linewidth(2) + blue);
draw((-3,-3)--(-4,-3), linewidth(2) + red);
draw((-4,-3)--(-4,-2), linewidth(2) + red);
draw((-4,-2)--(-3,-2), linewidth(2) + red);
draw((-3,-2)--(-3,-3), linewidth(2) + red);
draw((-3,-4)--(-4,-4), linewidth(2) + blue);
draw((-4,-4)--(-4,-3), linewidth(2) + blue);
draw((-4,-3)--(-3,-3), linewidth(2) + blue);
draw((-3,-3)--(-3,-4), linewidth(2) + blue);
draw((-4,-5)--(-5,-5), linewidth(2) + blue);
draw((-5,-5)--(-5,-4), linewidth(2) + blue);
draw((-5,-4)--(-4,-4), linewidth(2) + blue);
draw((-4,-4)--(-4,-5), linewidth(2) + blue);
draw((-5,-4)--(-5,-3), linewidth(2) + red);
draw((-5,-3)--(-4,-3), linewidth(2) + red);
draw((-4,-3)--(-4,-4), linewidth(2) + red);
draw((-4,-4)--(-5,-4), linewidth(2) + red);
draw((-5,-5)--(-6,-5), linewidth(2) + red);
draw((-6,-5)--(-6,-4), linewidth(2) + red);
draw((-6,-4)--(-5,-4), linewidth(2) + red);
draw((-5,-4)--(-5,-5), linewidth(2) + red);
draw((-5,-6)--(-6,-6), linewidth(2) + blue);
draw((-6,-6)--(-6,-5), linewidth(2) + blue);
draw((-6,-5)--(-5,-5), linewidth(2) + blue);
draw((-5,-5)--(-5,-6), linewidth(2) + blue);
draw((-1.7991049342473708,-4.535285403222338)--(-1.0839408493060767,-4.535285403222338), linewidth(2),EndArrow(6));
draw((-6,-4)--(-6,-3), linewidth(2));
draw((-5,-2)--(-4,-2), linewidth(2));
draw((-5,-6)--(-4,-6), linewidth(2));
draw((-2,-4)--(-2,-3), linewidth(2));
draw((-1,1)--(-1,2), linewidth(2));
draw((0,3)--(1,3), linewidth(2));
draw((3,1)--(3,2), linewidth(2));
draw((1,-1)--(0,-1), linewidth(2));
draw((0,-2)--(1,-2), linewidth(2));
draw((3,-4)--(3,-3), linewidth(2));
draw((1,-6)--(0,-6), linewidth(2));
draw((-1,-4)--(-1,-3), linewidth(2));
draw((0,1)--(0,2), linewidth(2));
draw((1,-6)--(1,-5), linewidth(2));
label("$P$",(-5.715814382242815,2.769228358696473),SE*labelscalefactor);
label("$P$",(-0.6903370286012918,1.6973608112414633),SE*labelscalefactor);
label("$P$",(-5.645527985688388,-3.2754017449842387),SE*labelscalefactor);
label("$R$",(0.36395891971511163,-3.257830145845632),SE*labelscalefactor);
label("$P$",(0.10038493263601078,-3.257830145845632),SE*labelscalefactor);
label("$R$",(1.3655400706156948,0.7484944577567003),SE*labelscalefactor);
label("$R$",(-3.6950804813030413,0.7133512594794869),SE*labelscalefactor);
label("$R$",(-4.749376429619445,-3.257830145845632),SE*labelscalefactor);
/* dots and labels */
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]

It is clear that the policeman can always play in that way. As he only uses $\downarrow$ and $\rightarrow$, the amount of moves must end, and the only way this can happen is if the policeman is already on the same cell as the robber, proving $b)$. We are done $\square$.[/quote]

I have been wondering for a while, but is it not the case that the robber can move just 9999, and the one moving 10 000 is the police man?

Oh Thanks, I couldn't read from the post who starts, and I didn't realize the robber started on the upper left diagonal. I was following the statement from the iberoamerican site ([url]https://www.oei.es/historico/oim/xviioimprob.htm[/url]), in which actually the policeman starts and the initial position of the robber is upper right from the policeman instead.

[quote=Modesti][quote=Ozc]

I have been wondering for a while, but is it not the case that the robber can move just 9999, and the one moving 10 000 is the police man?[/quote]

Hi! No, that can't happen because the robber is the one who moves first. He will always have moved as many times or one more time as the policeman.[/quote]


calculus
integration
trigonometry
symmetry
calculus computations

1. int[(cos(pi/x)) / (x^2)]dx

2. int[(sin x)(cos(cos x))dx]

3. int_-1^1[(sin x) / (1 + x^4)]dx

4. int_0^2[y^2 * sqrt(1 + y^3)]dy

Pardon my formatting... any help would be appreciated

1. $\int \frac{\cos \pi x}{x^2}dx=\int \left(-\frac{1}{x}\right)' \cos \pi xdx.$ You can use integral by parts.

2. $\int \sin x\cos (\cos x) dx=-\int \cos (\cos x)(\cos x)' dx.$

3. Remark the interval of integration.

4. $\int_ 0^2 y^2\sqrt{1+y^3} dy=\frac{1}{3}\int \sqrt{1+y^3} (1+y^3)' dy.$

It's readable- at least you use brackets and parentheses where appropriate.

#1,2,4 are all the same idea; substitute for the argument of the nasty function.
For #3, you won't find an antiderivative; try symmetry instead.

Great... thanks for the help with those... my prof doesn't exactly have a knack for explaining things, so I'm trying to learn everything out of the textbook....

What do you mean exactly about number three when you said to use symmetry?

Let $f(x)$ such that $f(x)=-f(x),$  Calculate $\int_{-a}^a f(x)dx.$

Hmm ok... also, the answer you gave for number one, you interpreted the question as cos(pi*x), but it is cos(pi / x)... will this change the process that you outlined, aside from what follows cos?

I have to hand the assignment with these questions in tomorrow... if it would be possible for somebody to post solutions for me to check my answers to, it would be greatly appreciated!


inequalities
search
LaTeX
inequalities unsolved

And how is the topic's title "a,b,c,d>0" related to it?

Anyway, we are to show that $ \sum _{i=1}^{4}x_i^3 \ge \sum_{i=1}^{4}x_i $ and  $\sum _{i=1}^{4}x_i^3 \ge \sum _{i=1}^{4} \frac{1} {x_i}$. The general form of the first one was discussed two days ago (see http://www.mathlinks.ro/Forum/viewtopic.php?t=32719 )and the second one is just $\sum _{i=1}^{4}x_i^3 \ge \sum _{i=1}^{4} x_ix_jx_j$, i.e. pure Murheid.

By the way, this inequality is from Iran 1998.

Yes, if I am to trust the Swedish IMO team-leader. He writes in his text on inequalities that this is from Iran '98.

[b]EDIT[/b]: This link confirms that it is from Iran:  [url=http://www.unl.edu/amc/a-activities/a4-for-students/problemtext/mc97-98-01feb.pdf]www.unl.edu/amc/a-activities/a4-for-students/problemtext/mc97-98-01feb.pdf[/url]. Have a look at page 190.

[quote="payman_pm"]yes,it's for Iran second exam![/quote]

Please correct all your previous postings and complete the source field !

a.) Search the forum before posting, e.g. search for Iran *. Your problem might be already on the forum.

b.) Specify  your sources.  :)

[quote="orl"][quote="payman_pm"]yes,it's for Iran second exam![/quote]

Please correct all your previous postings and complete the source field ! You know your posts can be gone pretty quickly if you

a.) don't search the forum before posting, e.g. search for Iran *

b.) don't specify  your sources.[/quote]
Ok i try to correct my posts but why shoudlnt i specify my sources? and what do u mean by
don't search the forum before posting, e.g. search for Iran .

[quote="payman_pm"]$x_1,x_2,x_3,x_4 $ are positive  such that $x_1 x_2 x_3 x_4=1$ prove that
$\sum _{i=1}^{4}x_i^3 \ge max\{\sum_{i=1}^{4}x_i ,\sum _{i=1}^{4} \frac{1} {x_i}\}$[/quote]

Hmm... if we replace the $x_i$ with their reciprocals we can get the following stronger inequality?

\[\text{min}\{\sum _{i=1}^{4}x_i^3, \sum _{i=1}^{4} \frac{1} {x_i^3} \} \geq \text{max}\{\sum_{i=1}^{4}x_i ,\sum _{i=1}^{4} \frac{1} {x_i}\}\]

[quote="payman_pm"] Ok i try to correct my posts but why shoudlnt i specify my sources? and what do u mean by
don't search the forum before posting, e.g. search for Iran .[/quote]

I changed my posting above, Read my instructions again.  :)

[quote]Hmm... if we replace the $x_i$ with their reciprocals we can get the following stronger inequality?

\[\text{min}\{\sum _{i=1}^{4}x_i^3, \sum _{i=1}^{4} \frac{1} {x_i^3} \} \geq \text{max}\{\sum_{i=1}^{4}x_i ,\sum _{i=1}^{4} \frac{1} {x_i}\}\][/quote]

It wouldn't be any stronger - replace x_i by 1/x_i.

I have a solution to the first part:

[hide]
I have no LaTeX skills, so I'm calling x_1, x_2, x_3, x_4  a, b, c, d.

By AM-GM, (a+b+c+d) / 4   >=   1.
By the Power-Mean inequality, cubed_root ( (a^3 + b^3 + c^3 + d^3) / 4 )   >=   (a+b+c+d)/4   >=  1.

Since the left hand side is more than or equal to one, cubing it can only make it bigger, so

( a^3 + b^3 + c^3 + d^3 ) / 4   >=   ( a+b+c+d ) / 4, or
a^3 + b^3 + c^3 + d^3   >=   a+b+c+d.
[/hide]

Edit: Oh, wait, you already had the exact solution.  :blush:


Determine all positive integers n for which the equation\[x^n+(2+x)^n+(2-x)^n=0\]has an integer as a solution.

I believe the proposer of this problem wouldn't expect it became extremely trivial 1 year(or 2) after APMO 1993 :D

Many thanks to Andrew Wiles :P

all you have to do is check $n=1,2,3$ so wat is the problem ?

as you said after Andrew Wiles proves FLT its easy

Of course, we can easily solve it without using of FLT. ;)

Actually, is there any old problems before 1994 which are very hard but can be solved by FLT easily?


algebra
polynomial
superior algebra
superior algebra unsolved

Suppose R is a commutative ring (with unity) and $ a\in R$. If $ a$ is $ \underline{not}$ a zero divisor in R, does it then follow that it is not a zero divisor in $ R[x]$?

Yes. Take an arbitrary $ p(x)\equal{}\sum_n c_nx^n$, and look at the coefficients of $ x^n$ in $ ap(x)$. $ ac_n\equal{}0$ if and only if $ c_n\equal{}0$, so $ ap\equal{}0$ if and only if $ p\equal{}0$.

More generally a theorem of McCoy [1] implies that if a polynomial f in R[x] is a nonzero zero-divisor then rf = 0 for some nonzero r in R, i.e. there is a nonzero constant r that simultaneously kills [u]all [/u]the coef's of f.

[1] sci.math, 4 May 2004, problem with rings of polynomials
http://google.com/group/alt.algebra.help/msg/d3179d6f17afa48c
http://google.com/groups?selm=y8z3c6fwpme.fsf%40nestle.csail.mit.edu


algebra
polynomial
USAMTS
function
functional equation

This one is not very difficult.

Let p(x) be a polynomial of degree 2002 such that

p(k) = (k - 1)/(2k)

for k = 1, 2, 3, ..., 2003. Determine all possible values of p(2004).

The last one before returning to real life  :cry:

2X P(X) - X + 1 has degree 2003 and has roots 1, 2, .. 2003.

So 2X P(X) - X + 1 = a * prod (X - i), i = 1 .. 2003

Identifying element of degree 0 : 1 = a * prod (-i)

a = -1/2003!

After it's easy to compute P(2004)  :wink:

Arne wrote:Indeed. Actually I like this one ...

Generalisation. Let p(x) be a polynomial of degree N such that

p(k) = (k - 1)/(2k) for k = 1, 2, ..., N + 1.

Find p(N + 2).

didn't we do this on a USAMTS mathjam or someat? anyways, [hide]we form a new function q(k)=2kp(k)-k+1. The roots of q(k) are 1, 2, ..., N+1, so it can be factored as a(x-1)(x-2)...(x-N-1). We plug in 0 into the functional equation q(k)=2kp(k)-k+1 to get 1=q(0)=a*(N+1)!*(-1)^(N+1). Therefore a=(-1)^(N+1)/(N+1)!. We now plug in N+2 into the functional equation q(k)=2kP(k)-k+1 to find p(N+2)=(q(N+2)+N+1)/2(N+2)=((-1)^(N+1)+N+1)/2(N+2). Thus if N is odd, the answer is N/(2N+4) and if N is even, the answer is 1/2. Correct?[/hide]

We COULD use the Lagrange formula for polynomials of degree <= n s.t. f(ai)=bi for i from 0 to n, but it's WAY too complicated. However, this works for similar but harder prbs. One of them was posted on this forum and I gave a soln based on that. I can't exactly remember where it was posted.

yeah...it was something with 2^n for each n. Could you say exactly what Lagrange polynomials are?

I think I said what Lagrange polynomials are in that post. Could you check it out over there? It would take a lot of time and effort to write it all again because it involves a firce-looking sum.  :D


integration
trigonometry
calculus
derivative
calculus computations

Note: $ \sin (t^2 \minus{} t \plus{} x) \equal{} \sin (t^2 \minus{} t) \cos x \plus{} \sin x \cos (t^2 \minus{} t)$

Then we get:

$ f(x) \equal{} \int_{0}^{x}\sin (t^{2}\minus{}t\plus{}x) dt$

$ \equal{} \int_{0}^{x}\sin (t^2 \minus{} t) \cos x dt \plus{} \int_{0}^{x}\sin x \cos (t^2 \minus{} t)dt$

$ \equal{} \cos x \int_{0}^{x}\sin (t^2 \minus{} t) dt \plus{} \sin x\int_{0}^{x} \cos (t^2 \minus{} t)dt$

Now apply the product rule of differentiation together with the Fundamental Theorem of Calculus and you are done.


Σήμερα κατασκεύασα αυτή την άσκηση με το θείο μου βάσει μιας  άλλης .
Έστω τρίγωνο ΑΒΓ και σημείο D στην πλευρά ΒΓ τέτοιο ώστε 3ΒD=ΒΓ .Επίσης  $\hat{B}=45^o$ και $\widehat{BAD}=15^o$
Βρείτε τη γωνία $\widehat{A\Gamma B}$

Έστω σημείο $F\in AD$ τέτοιο ώστε $\widehat{DBF}=30^{o}$.

Το τρίγωνο $DBF$ είναι ισοσκελές $(DB=DF)$

Επίσης, $\widehat{FBA}=45-30=15^{o},$ άρα και το $\triangle FAB$ είναι ισοσκελές $(FA=FB)$

ΘΔΟ το $F$ είναι περίκεντρο του τριγώνου $ABC$.
Σαν αποτέλεσμα, η γωνία $\widehat{BCA}$ θα είναι ίση με το μισό της αντίστοιχης επίκεντρης γωνίας $\widehat{BFA}$. Άρα θα είναι $\widehat{BCA}=75^{o}$

Έστω $E$ το μέσον του $DC$. Θα είναι $BD=DE=EC = \frac{1}{3}BC$

Προφανώς οι μεσοκάθετοι των $DE,BC$ ταυτίζονται.

Το τρίγωνο $FED$ είναι ισόπλευρο, αφού $DF=DE$ και $\widehat{EDF}=60^{o}$. Έτσι, το $F$ ανήκει στη μεσοκάθετο του $DE$, άρα και στη μεσοκάθετο του $BC$. Έτσι λοιπόν θα είναι $FB=FC=FA$ ο.ε.δ.


Well... in calculus, we're doing Delta-Epsilon proofs, and I don't get the whole point.

We get to the point where we proved our objective statement, but we do everything backwards to get back to our original statement...

I don't get how Delta-Epsilon proofs work...
anyone care to explain?

The only way to answer your question is to go over a few proofs with you and see what exactly is confusing. It is better to do sitting face to face rather than online, but let's try nevertheless. Let's start with something really simple. Here is a typical "$ \varepsilon-\delta$ limit proof":

Problem: show that $ \lim_{x\to 2}x^2=4$.

Solution: We need to show that for every $ \varepsilon>0$, there exists $ \delta>0$ such that $ |x^2-4|<\varepsilon$ whenever $ 0<|x-2|<\delta$.
[color=magenta]Note that we do not need to find an optimal $ \delta$ here, just some that works![/color]
Since we can control $ |x-2|$, it would be nice to see it in $ |x^2-4|$. But we can just factor: $ |x^2-4\=|x-2|\cdot|x+2|$. Now let us start choosing $ \delta$. We can make it as small as we wish, so let us start with assuming that it is not greater than my favorite positive number, which happens to be $ 239$. Then the second factor is at most $ 241$ and it is enough to choose $ \delta\le 239$ such that $ 241\delta<\varepsilon$. Take $ \delta=\min(239,\varepsilon/241)$. We are done.

Now is there anything confusing in this proof? If yes, which part of it gives you trouble? If no, try to mimic it to show that $ \lim_{x\to 1}x^3=1$.

After you answer fedja's questions, I think the next place to go after that is the proof of a general theorem. First, $ \lim_{x\to a}cf(x)\equal{}c\cdot\lim_{x\to a}f(x),$ where $ c$ is a nonzero constant. Second, "the limit of a sum is the sum of the limits, provided they both exist." Then maybe we'll look at products.


If you were standing, would you expect your blood pressure to be higher in your arm or in your leg?

I would guess arm since its closer to the heart.. but not 100% sure

Just standing, I would think it would be higher in your legs, since if one started lying down with arms and legs at the same pressure and rapidly stood up, one would feel faint (I know, I've done this, as have many gentle readers).  The reason this doesn't happen is because of the contraction in the legs, which would increase their pressure.

I think this is something that a couple of minutes with a plood pressure cuff and several repetitions of the experiement.  If I do it tonight, I'll let you know.

The blood pressure would be higher in your leg.

According to Pascal's Law, $%Error. "bigdelta" is a bad command.
P = \rho gh$, where h indicates depth.  Since the depth of the leg is greater (i.e. its' lower than the arm), pressure will be higher there.

Yes..and just think about the caliber of the vessels.The femural artery is wider then the arms arteris.And also considering the hydrostatic factor you'll get that the blood has a higher presure in the arm.

I think the blood preasure is higher in legs because of them when we stand the blood is comes to our legs.


Bob chooses 5 cards from a deck of cards.  What is the probability that he picks one club and two red cards?

I thought it would be [hide]$\frac{\displaystyle\binom{13}{1}\binom{26}{2}\binom{49}{2}}{\displaystyle\binom{52}{5}}$ since you can pick one club in $\displaystyle\binom{13}{1}$ ways, two red cards in $\displaystyle\binom{26}{2}$ ways, and the last two cards in $\displaystyle\binom{49}{2}$ ways, since 3 cards have already been drawen, divided by $\displaystyle\binom{52}{5}$, the total number of ways Bob can choose the 5 cards.[/hide]but when I try my method, the probability is over one.  Can someone tell me how to do this problem and what is wrong with my solution?

[hide]The sequence must be $CRRDD \cdot 5$ because it can be permuted five different ways. C represents club, R a red card, and D doesn't matter. So $\frac 14 \cdot \frac {26}{51} \cdot \frac 12 \cdot 5 = \frac {65}{204}$. I think that's the answer. [/hide]

Yes it can be permutated five ways... but wouldn't the probabilities be different for drawing cards at different times?  ie. probability of drawing club first is $\frac 1 4$ but probability of drawing club fifth would be different, depending on if clubs were drawn before.  Also, the question isn't very clear on if it's ONLY one club and two reds, or AT LEAST one club and two reds.  That would obviously change things, however I believe it's asking the former.
I still don't know where to start on solving it.  I can't solve the most basic probability/combinatorics problems...

Yeah that doens't matter. Even if you worked out all of the possibilities, it -should- come out the same. I'm pretty sure it's that times 5.

[quote="deej21"]  Also, the question isn't very clear on if it's ONLY one club and two reds, or AT LEAST one club and two reds. [/quote]The way I interpreted it, it means at least one club and two red cards.  To tarquin, doing the question your method, wouldn't it be permutated $5!$ ways, not $5$.  Probability/counting is one of my weakest areas.  :(

eryaman,
$\frac {5!}{2!2!} = 30$ and I still made a mistake.
For some reason now it's coming out over 1... ?

[quote="amirhtlusa"]since you're problem has not stated whether you want excatly those things or minimum or maximum of those numbers. i assum that it means that you exactly want one club, and one red. if this is true then you're solution is wrong. because if you have only one club you have to have 3 cards that are not clubs. so you need to include a C(39,3) as well. then it's a very  simple problem. :D[/quote]I think you read the question wrong...it's actually two red cards that we must draw, not one.  Anyway, this is my solution if you must pick up exactly one club and exactly two red cards:
[hide]$\frac{\displaystyle\binom{13}{1}\binom{26}{2}\binom{13}{2}}{\displaystyle\binom{52}{5}}\approx .1268$
We can choose the one club in 13 ways, the two red cards in 26 ways (since clubs are black cards, not red cards), and the remaining 2 cards out of all the cards that are not clubs or red, the spades, of which there are 13.  So I used 13C2.  Multiplying all those together to see how many total ways match those conditions over the total number of ways the cards can be drawn.[/hide]
Can someone tell me if my solution for this is correct?

Well to test your solution I wrote a quick program that goes thru all the possibilities (quite a lot...) and checked.  I might have done something wrong, that's always a possibility, but I got $\frac{323297}{2399040} \approx .134761$  So you were quite close.  Or maybe I was.

I got 13/34 .... The work was kind of complicated so I did it on paper. What's the answer?

I don't have the answer...that's why I posted it on this forum  :D

[quote]We can choose the one club in 13 ways, the two red cards in 26 ways (since clubs are black cards, not red cards), and the remaining 2 cards out of all the cards that are not clubs or red, the spades, of which there are 13. So I used 13C2. Multiplying all those together to see how many total ways match those conditions over the total number of ways the cards can be drawn. [/quote]

Can someone tell me if my solution for this is correct?
yeah i made a mistake in my explanation(actually two). but  i think your solution is right. atleast it makes sense. :lol:

I found the same answer, so unless I did a mistake in the original number of the cards, your solution is correct!! ;)


let the sequence  $ (a_n)$ such that :
$ a_1\equal{}a_2\equal{}a_3\equal{}1$ and $ a_{n\plus{}1}\equal{}\frac{1\plus{}a_n a_{n\minus{}1}}{a_{n\minus{}2}}$ $ n\equal{} 3,4, ...$
prove that all terms of  $ (a_n)$   are $ \in Z$

That one is quite nice I think.
We define $ u_{n}$ as follows:
$ u_{n}\equal{}a_{n\plus{}1}\minus{}a_{n}$
First to understand how to come to a conclusion observe first few examples:
$ u_{1}\equal{}u_{2}\equal{}0 \\ u_{3}\equal{}u_{4}\equal{}1 \\ u_{5}\equal{}u_{6}\equal{}4\equal{}4*1\minus{}0 \\ u\_{7}\equal{}u_{8}\equal{}15\equal{}4*4\minus{}1 \\ u_{9}\equal{}u_{10}\equal{}56\equal{}4*15\minus{}4$ etc.

Now just prove inductively that:
$ \forall_{k\in \mathbb{N}} \ u_{2k\plus{}4}\equal{}u_{2k\plus{}3}\equal{}4*u_{2k\plus{}2}\minus{}u_{2k\plus{}1}$ and with that the thesis of our problem is obvious :wink:


Let $ a,b,c$ be positive real number satisfying $ a^3 \plus{}b^3 \plus{}c^3 \equal{}4$ .Prove that

$ \frac{1}{a^4 \plus{}3} \plus{}\frac{1}{b^4 \plus{}3} \plus{}\frac{1}{c^4 \plus{}3} \geq \frac{2}{3}$

[hide="solution"]
we have to prove ,
$ \sum_{cyclic} \frac {3}{a^4 \plus{} 3} \ge 2$
 $ \sum_{cyclic} \frac {3}{a^4 \plus{} 3} \equal{} 3 \minus{} \sum_{cyclic} \frac {a^4}{a^4 \plus{} 1 \plus{} 1 \plus{} 1} \ge 3 \minus{} \sum_{cyclic} \frac {a^4}{4a} \equal{} 2$ hence proved[/hide]


Suppose that $0<a_1<a_2<...<a_n\leq  2n$ are the natural number such that $1\leq i<j \leq n ,LCM(a_i,a_j)>2n $ .Prove that $a_1>[\frac{2n}{3}]$.(where [$x$ ] is the greatest or equal that $x$)

we know that from $n+1$ number between $1$ to $2n$ there exist two of them such that one of them is divisible by the other so if $3a_1<2n+1$ then $2a_1,3a_1,a_2,\dots,a_n$ apply above argument and yields contradiction.


trigonometry
inequalities solved
inequalities

Three angles of triangle ABC satisfy condition
     cosA + cosB = 2cosC
Show that
a) a^2 + b^2 <= c^2 + ab <= ac + bc
b) c >= (8/9)max{a, b}

=========
Namdung

here my solution for b)
we have cosA+cosB=2cosC
<=>(a+b)(c-a+b)(c+a-b)=2c(a 2 +b 2 -c 2 ) (*)
let
a=x+y
b=y+z
c=z+x
(*)<=>4xz(x+y)=4(x+z)(y 2 +xy+yz-xz)
let x+y+z=1 then
y 2 -y+2xz=0
<=>y 2 -(x+y+z)y+2xz=0
<=>8y=16xz/(x+z)
we need to prove that
x+z \geq 8/9(x+y) and x+z \geq 8/9(y+z)
<=>x 2 +9z 2  \geq 6xz and 9x 2 +z 2  \geq 6xz.  :D


Hi guys,i need some help regarding the following problems:
Q. A string of negligible thickness is wrapped several times around a cylinder kept on a horizontal surface.A man standing at a distance l from the cylinder holds one end of the string and pulls the cylinder towards him.There is no slipping anywhere.the length of the string that has passed through the hand of the man while the cylinder reaches his hands is?
Q.A metal cube is placed in an empty vessel.When water is filled in the vessel,so that the cube is completely immersed in the water,the force on the bottom of the vessel in contact with the cube:-
a)will increase b)will decrease c)remains the same d)will become zero
Please reply soon with ur detailed solutions :D

ok Emmanual Bonfils,
i think i have figured out the first one....just see it is correct or not.
Well intially,the man is at a distance of l from the cylinder.The cylinder rolls without slipping and reaches the man.So here,the length of string passing thru his hands is l(because the cylinder does not slip).But as it is,he was at a distance l from the cylinder,so the length of string passing thru his hands is:l+l=2l.
For the second one,i am not able to understand......there is a buoyant force acting on the cube in the second place and so...if u draw the FBD,the Normal force it exerts on the vessel's floor should decrease.....it is not so though,why?

Remember the action and reaction law, and let us know about your progress, please.


Let $ ABC$ be a triangle with $ C\equal{}90^{\circ}$ and $ CA\equal{}CB$ . Consider  the point $ D\in (AC)$ for which

$ DA\equal{}2\cdot DC$ and the point $ M\in (BD)$ for which $ \widehat {MCB}\equiv\widehat {MBA}$ . Prove that $ MA\perp MC$ .

Hello!

Let $ E$ be the midpoint of $ AB$ and $ \left\{F\right\} \equal{} BD\cap CE$.
$ CA \equal{} CB \equal{} x\Rightarrow AB \equal{} x\sqrt {2}\Rightarrow CE \equal{} AE \equal{} BE \equal{} \frac {x\sqrt {2}}{2}$
$ DA \equal{} 2\cdot DC\Rightarrow DC \equal{} \frac {x}{3}$ and $ DA \equal{} \frac {2x}{3}$
$ \frac {FC}{FE} \equal{} \frac {DC}{DA}\cdot\frac {BA}{BE} \equal{} \frac {1}{2}\cdot 2 \equal{} 1\Rightarrow FC \equal{} FE \equal{} \frac {CE}{2} \equal{} \frac {x\sqrt {2}}{4}$
$ FB^{2} \equal{} FE^{2} \plus{} BE^{2}\Rightarrow FB \equal{} \frac {x\sqrt {10}}{4}$
$ \frac {FD}{FB} \equal{} \frac {CD}{CA}\cdot\frac {EA}{EB} \equal{} \frac {1}{3}\cdot 1 \equal{} \frac {1}{3}\Rightarrow FD \equal{} \frac {1}{3}\cdot FB\Rightarrow FD \equal{} \frac {x\sqrt {10}}{12}$
$ \angle CMD \equal{} \angle MCB \plus{} \angle MBC \equal{} \angle MBA \plus{} \angle MBC \equal{} \angle ABC \equal{} 45^{\circ}\Rightarrow$
$ \Rightarrow\angle CMD \equal{} 45^{\circ}$     (1)
$ \Delta MFC\sim \Delta CFB\Rightarrow FC^{2} \equal{} FM\cdot FB\Rightarrow FM \equal{} \frac {FC^{2}}{FB}\Rightarrow FM \equal{} \frac {x\sqrt {10}}{20}$
$ DM \equal{} FD \plus{} FM \equal{} \frac {x\sqrt {10}}{12} \plus{} \frac {x\sqrt {10}}{20} \equal{} \frac {2x\sqrt {10}}{15}\Rightarrow DM \equal{} \frac {2x\sqrt {10}}{15}$
$ \frac {DM}{DA} \equal{} \frac {\frac {2x\sqrt {10}}{15}}{\frac {2x}{3}} \equal{} \frac {\sqrt {10}}{5}$     (2)
$ BD \equal{} FB \plus{} FD \equal{} \frac {x\sqrt {10}}{4} \plus{} \frac {x\sqrt {10}}{12} \equal{} \frac {x\sqrt {10}}{3}\Rightarrow BD \equal{} \frac {x\sqrt {10}}{3}$
$ \frac {DA}{DB} \equal{} \frac {\frac {2x}{3}}{\frac {x\sqrt {10}}{3}} \equal{} \frac {\sqrt {10}}{5}$     (3)
(2),(3)$ \Rightarrow\frac {DM}{DA} \equal{} \frac {DA}{DB}$
$ \Delta MDA\sim \Delta ADB\Rightarrow\angle AMD \equal{} \angle BAD \equal{} 45^{\circ}$     (4)
(1),(4)$ \Rightarrow\angle AMC \equal{} 90^{\circ}\Rightarrow MA\bot MC$

Petrisor Neagoe :)

The circumcircle $ (K),$ of the right isosceles triangle $ \bigtriangleup OAC,$ where $ O$ is the circumcenter of $ \bigtriangleup ABC,$ intersects the line segment $ BD$ at point so be it $ M$ and because of $ MA\perp MC,$ it is enough to prove that $ \angle MAB \equal{} \angle MBC.$

Let be the point $ N\equiv OC\cap BD$ and applying the [b]Menelaos theorem[/b], in triangle $ \bigtriangleup OAC,$ taken $ BD$ as its transversal, we have $ \frac {NO}{NC}\cdot \frac {DC}{DA}\cdot \frac {BA}{BO} \equal{} 1$ $ \Longrightarrow$ $ NO \equal{} NC$ $ ,(1)$

Because $ NK\parallel AB,$ where $ K$ is the midpoint of $ AC$ and $ NK \equal{} \frac {OA}{2} \equal{} \frac {OB}{2},$ we conclude that $ EN \equal{} NB$ $ ,(2)$ and $ EK \equal{} KO$ $ ,(3)$ where $ E\equiv BN\cap OK.$

So, from $ (3)$ we have that the point $ E$ lies on the circle $ (K)$ and then, from $ (1),$ $ (2),$ it is easy to show that $ OE\parallel BC.$

Hence, we conclude that $ \angle MAB \equal{} \angle OEB \equal{} \angle MBC$ and the proof is completed.

Kostas Vittas.

It sufficient to prove that $ M$ is the [b]Brocard point[/b] for $ \triangle ABC.$ So it's enough to prove that $ cot\alpha\equal{}cotA\plus{}cotB\plus{}cotC\equal{}2$. But $ cot\alpha\equal{}\frac{AB^2\plus{}BD^2\minus{}AD^2}{4S_{ABD}}\equal{}$ $ \frac{a^2\plus{}\frac{a^2}{9}\plus{}2a^2\minus{}\frac{4a^2}{9}}{\frac{4}{3}S_{ABC}}\equal{}2$ ($ CA\equal{}CB\equal{}a$) and we are done :)

Call E the common point of AB and CM, see that the triangles ADB and BEC are similar, hence BE/AD = BC/AB and, with AD = 2 CD, we get DE and BC parallel and angle DEA = angle ABC = 45 degs ( 1 ). Also from their similarity we get DAEM cyclic, with angle DMC = angle BAC = 45 degs ( 2 ). From (1) and (2) AM and CM are perpendicular.

Best regards
sunken rock


geometry
inradius
inequalities
geometric transformation
reflection
trigonometry
geometry proposed

Let $ ABC$ be an acute triangle Prove that
$ tanA\plus{}tanB\plus{}tanC\geq \frac{s}{r}$
where s and r are semiperimeter and inradius of triangle ABC respectively

My solution to this problem is turn the inequality into this form
$ (a\plus{}b\minus{}c)^2(a\minus{}b\plus{}c)^2(\minus{}a\plus{}b\plus{}c)^2\geq(a^2\plus{}b^2\minus{}c^2)(a^2\minus{}b^2\plus{}c^2)(\minus{}a^2\plus{}b^2\plus{}c^2)$
which is brute force and ugly
can i have your beautiful solution please  :blush:

I think it isn't a difficult problem..here is a solution:
[hide]In every $ \triangle ABC$ we have $ \frac{a^2}{tgA}\plus{}\frac{b^2}{tgB}\plus{}\frac{c^2}{tgC}\equal{}4S(\star)$, where $ S$ is the area of $ \triangle ABC$.
With CBS(in Engel form), we have that $ tgA\plus{}tgB\plus{}tgC\geq\frac{(a\plus{}b\plus{}c)^2}{4S}\equal{}\frac{s^2}{S}\equal{}\frac{s}{r}$ and we have done.:)

ps:the relation $ (\star)$ is obtained with this: $ tgA\equal{}\frac{4S}{b^2\plus{}c^2\minus{}a^2}$ and $ 16S^2\equal{}2((ab)^2\plus{}(bc)^2\plus{}(ca)^2)\minus{}(a^4\plus{}b^4\plus{}c^4)$[/hide]

thank a lot i just know that

$ \sum_{cyc}\frac{a^2}{tanA}\equal{}4S$
 :lol:

:o  this problem is from the running edition of math reflections, you should not have posted this... any moderator please lock the topic...

$ \left\|\begin{array}{c}
\tan A\equal{}\frac {2\sin^2A}{\sin 2A}\equal{}\frac {1}{2R^2}\cdot\frac {a^2}{\sin 2A}\\\\
\sum\sin 2A\equal{}4\cdot\prod\sin A\equal{}\frac {2pr}{R^2}\end{array}\right\|$ $ \implies$ $ \sum\tan A\equal{}\frac {1}{2R^2}\cdot\sum\frac {a^2}{\sin 2A}$ $ \stackrel{(C.B.S.)}{\ \implies\ }$ $ \frac {1}{2R^2}\cdot\frac {\left(\sum a\right)^2}{\sum \sin 2A}\equal{}$ $ \frac {1}{2R^2}\cdot\frac {4p^2}{\frac {2pr}{R^2}}\equal{}\frac pr$ .

Campos, this inequality is really too old to transform this topic into a threat for Math. Reflections.

From this ineq derives this : $ a^2\plus{}b^2\plus{}c^2\leq 8R^2\plus{}4r^2$..:D


Substitute $a^{2}+b^{2}+c^{2}=3$ in Right Side to get:
$(a^{3}+b^{3}+c^{3})(ab+bc+ca) \ge 9abc$
$\Leftrightarrow (a^{3}+b^{3}+c^{3})(ab+bc+ca) \ge 3(a^{2}+b^{2}+c^{2})abc$
$\Leftrightarrow \sum_{sym}a^{4}b+\frac{1}{2}\sum_{sym}a^{3}bc\ge \frac{3}{2}\sum_{sym}a^{3}bc$
$\Leftrightarrow \sum_{sym}a^{4}b\ge \sum_{sym}a^{3}bc$
which is true by Muirhead.


logarithms
calculus
calculus computations

This is a relatively simple problem, but my friend thinks that the answer is something different than what I figured it to be. What do you guys think?

Population $y$ grows according to the equation $\frac{dy}{dt} = ky$ where $k$ is a constant and $t$ is measured in years. If the population doubles every $10$ years, then what is the value of $k$?

The solution to your equation is $Ce^{kx}$ and we know that $Ce^{k(x+10)}/Ce^{kx}=2$, so $e^{10k}=2$. Therefore, $k=\frac{1}{10}\ln 2$.

thanks myth. yeah thats what i was thinking, jsut got a good old math brain fart. it happens. thanks again man.


:( I need a sequence formula for the following numbers i.e. I need the formula for the following map of numbers:

2 - 5
3 - 10
4 - 15
5 - 20
6 - 25
7 - 30
8 - 35
9 - 40
10 - 45
11 - 50
etc etc

The first number needs to map to the second number. I need a formula that'll map all the first numbers to the second numbers(the multiples of 5)

Thanks


Does anyone here write well? I know that this is a math forum, but I am sure that there are many talented writers here also. If you are one of them, do you have any advice on how to write an essay or other paper? Are there analogs of the AoPs series for writing somewhere? I honestly learn nothing from my English class about how to write. My English teachers only assign you an essay to write, but they never discuss the mechanics of what makes a good essay. Sometimes when I read an article, editorial, op-ed from the New York Times I am amazed at how well they write, and I cannot possibly conceive of even coming near that. Any suggestions?

Almost anything that the late Rudolf Flesch has written about writing is pretty good. The key idea is to write naturally ("like you talk"), but he gives many specific examples of good and bad writing and discusses how to practice the principles of the good examples.

Ehh I'm okay.

The main thing is to be interesting.  If the paper isn't interesting to the reader its nothing.  I have learned that you need to express emotion in your writing.  That helps.

Pm aalindsey100 her papers go national contest level.

Well, I got a 12/12 on the SAT II writing essay and an 11/12 on the state assesment one...  Anyway it depends what sort of writing you are talking about, standard generic essays or creative essays.

For the standard generic one, it basically follows a formula.  First you write an intro paragraph, and at the end of this paragraph you must have a thesis statement.  Then, you write 3 paragraphs giving different examples/arguments to support your thesis.  Make sure to use transitions and intro and conclusion sentences in your paragraphs.  Then you write a conclusion, where you restate your thesis and then summarize the three points that you made.  This essay always works.  The most important part of writing a good essay like this is that it is coherent.  YOU MUST MAKE SENSE, AND HAVE A LOGICAL AND UNDERSTANDABLE THOUGHT PROCESS.  (sorry, that's important)

Then if you want to get more creative, that will only come w/practice.  Try using a wider vocabulary, using literary techniques, adding conversation.  Some of these will be derivatives of the regular 5-paragraph one, but others will not.

Neal just sent me this:

http://www.mang.canterbury.ac.nz/courseinfo/AcademicWriting/Flesch.htm

One way to improve your writing is to combine short sentences but keep a couple short sentences in. Use descriptive words.
examples:
The  lightning struck the tree. The tree fell. A girl screamed.

could become

Crack! Lightning flashes thruogh the sky and hits a nearby tree. The electricuted tree falls to the wet ground and a young girl lets out a shriek.

I'm a decent writer (when I try). I find that the best method is simply follow a thought through its entirety. When you feel that your thought is completed, and properly represented on paper, it is time to move on to another idea.

I personally think I'm pretty good writer since I won several awards at writing when I was younger and well, you get the picture.

The main thing about the writing is, you have to have the feel of it.  Whether it's expository, narrative, or persuasive, you should feel something about your writing if YOU read it after you wrote it.  Without that feeling, your writing will be a dead body without a soul.  Interesting beginning is important too but the main thing is to keep the constant point you said at the beginning to end.

I don't know about the high school yet but I think you should never use words as I put:

Like, cuz, w/e, and,and,et... You get the picture..

It is true that it's good to write as you say but if you do, try to keep up right grammar.

Hope this helps!

Re-read -- if it doesn't make sense to you, it won't to other people :)

[quote="yif man12"]never ever ever ever trust peer editing. if someone says that your paper is really good, they're just not looking hard enough. always look for yourself.[/quote]

Best advice yet.

My lit teacher believed in practice. Lots of essays make better essays. She was right, at least...as far as my writing goes.

Writing, I have sucked at writing most of my life. Luckily though, my 12th grade English Lit AP teacher was awesome and she taught me how to write well. Tonight I have to work on my graduation speech! AHH! Writing an essay is one thing but writing a speech that one has to read in front of 5,000 people. *Tear*

I'm scared.

[quote="Rep123max"]Any advice on how not to write too converstionally?[/quote]

Really, you want your writing to sound conversational. The best expository writing about mathematics is like a mathematician having a conversation with another mathematician. But the quality of your conversation will improve over time if you READ, READ, and READ, because you will build up new ways of expressing yourself.

[quote="nr1337"][quote="yif man12"]never ever ever ever trust peer editing. if someone says that your paper is really good, they're just not looking hard enough. always look for yourself.[/quote]

My teacher sometimes forces us to let our classmates read and edit our work, but I always re-check it myself.  Most of my classmates hate math and english, so I don't trust them with those things.[/quote]
tell me about it. my friend has gotten 6 As in a row with writing (and our teacher grades very tough). he's the best writer in our class. of our 7 prompts, 3 times he said mine had no problems. my grades for those prompts? A, C+,C. Ouch.

Always make your first couple of sentences hook the reader. I usually do this by asking a question that relates to the paper or by starting off 'Imagine if...'  Just be sure not to go off topic by carrying those sentences too far. The intorduction is the most important part of a paper--it gets the reader interested if they can relate to it so that they will want to keep reading.


limit
function
algebra
polynomial
calculus
calculus computations

Exist or no:$ \lim_{x\to \plus{}\infty}(x\minus{}e^x)$

Well, as you know that exponential function grows significantly faster than any polynomial function. Hence, the limit is $ \minus{}\infty$.

Here's my solution
Set$ x\equal{}\minus{}t\Rightarrow L\equal{}\minus{}\lim_{t\to \minus{}\infty}(t\plus{}\frac{1}{e^t})\equal{}\minus{}\lim_{t\to \minus{}\infty}\frac{te^t\plus{}1}{e^t}$
And$ \lim_{t\to \minus{}\infty} te^t\equal{}\minus{}\lim_{u\to \plus{}\infty}\frac{u}{e^u}\equal{}0$
$ \Rightarrow L\equal{}\minus{}\infty$


geometry
cyclic quadrilateral
geometry unsolved

[quote="communist"]i am new to this site and found good problems by many users. so i have posed my own first problem which looks fairly easy.
ABCD is a cyclic quadrilateral such that the diagonals intersect at O. An arbitary line PQ  with P on AD and Q on BC passes through O. such that PD, AP, QC and BQ are in continued proportion. prove that AO.BQ=AP.BO. thanks!![/quote]
Because $ \Delta ADO$~$ \Delta BCO$ then $ \frac{AD}{BC}\equal{}\frac{DO}{CO}$
Since $ \frac{PD}{AP}\equal{}\frac{QC}{QB}$ we obtain $ \frac{AP}{AD}\equal{}\frac{BQ}{BC}$
So $ \frac{AP}{BQ}\equal{}\frac{AD}{BC}\equal{}\frac{DO}{CO}$
Hence $ \frac{AP}{OD}\equal{}\frac{BQ}{CO}$ or $ \frac{AP.BO}{OB.OD}\equal{}\frac{BQ.AO}{OA.OC}$
We have $ OA.OC\equal{}OB.OD$ therefore $ AP.BO\equal{}AO.BQ$


The 20 pupils in a class each send 10 cards to 10 (different) class members. [size=92][i][note: you cannot send a card to yourself.][/i][/size]
(a) Show at least 2 pupils sent each other a card.
(b) Now suppose we had $n$ pupils sending $m$ cards each. For which $(m,n)$ is the above true? (That is, find minimal $m(n)$ or maximal $n(m)$)

Solve the part b only.

We claim iff $m > (n-1)/2$ then it occurs.

Necessity:  if $m \le (n-1)/2$ then let $A_j$ send to $A_{1+j}, \dots, A_{m+j}$ where indices are mod n.  Then $A_i$ and $A_j$ for any i<j don't send to each other obviously, because j-i and n-(j-i) cant both be $\ge m$.

Sufficency:  $m > (n-1)/2$.  Suppose A sends to "B" people and recieves from "C" people; then the total number of recieves and sends is greater than the number of targets, which means that someone necessarily got a card from A and sent one to A as well.

obviously, if it is a class, there is bound to be ppl who know eachother better than others. These will post cards to eachother. It is psycologically and mathematically impossible to det another answer than the one we are trying to prove

Singular is correct. I don't understand what you mean though, dead fish? Is it a joke? :)

My best guess for solution is pigeon whole principle , Hope anybody can provide me with solution.

The total number of possible links between two people would be $ \binom{20}{2} \equal{} \frac{20(19)}{2} \equal{} 190$. (Think of each person as a vertex, and the directed segment connecting two vertices as the letter one person sends to the other.) The total number of letters being sent is $ 20(10) \equal{} 200$. Since we have more letters being sent than links, then by the Pigeonhole Principle, at least one link will have two letters--thus one link will have directed segments in both directions. This implies that at least two people send letters to each other.


How would you synthesize PVA (poly vinyl alcohol)?

i don't know if this would work out but $ \text{Vinyl Alcohol}\plus{}\text{BrN}_3\plus{} \Delta/h\nu$

That would be impossible. Can you discover why?

poor reactivity of vinylic group  :rotfl:

No, that's not the reason: a double bond has a substantial reactivity.

Vinyl Alcohol is an enol and hence once synthesized tautomerization occurs and the equilibrium lies predominantly to the side of ethanal. Hence you cannot produce Poly Vinyl Alcohol this way.

However fortunately you can use vinyl acetate which is not subjected to the tautomerization of vinyl alcohol and polymerization yields P.V.A (Poly Vinyl Acetate).

P.V.A the undergoes hydrolysis with $ NaOH$ to obtain Poly Vinyl Alcohol.


abstract algebra
category theory
group theory
superior algebra
superior algebra unsolved

Provide examples for each of the following statements ($ R$ is a ring with identity):

[b](a)[/b] $ M$ is a right $ R$-module and $ N_i$ are left $ R$-module:
\[ M\otimes _R ( \prod _{i\in I} N_i )\ncong_{\mathbb{Z}} \prod_{i\in I}  (  M \otimes _R N_i)\]
[b]In (b), (c), (d) and (e) $ M,N,M_i,N_i$ are left $ R$-module:[/b]

[b](b)[/b] $ Hom_R ( \prod _{i\in I}M_i,N  )\ncong_{\mathbb{Z}} \sum_{i\in I}Hom_R ( M_i,N  )$

[b](c)[/b] $ Hom_R ( \prod _{i\in I}M_i,N  )\ncong_{\mathbb{Z}} \prod_{i\in I}Hom_R ( M_i,N  )$

[b](d)[/b] $ Hom_R ( M,\sum_{i\in I}N_i )\ncong_{\mathbb{Z}} \sum_{i\in I}Hom_R ( M,N_i  )$

[b](e)[/b] $ Hom_R ( M,\sum_{i\in I}N_i )\ncong_{\mathbb{Z}} \prod_{i\in I}Hom_R ( M,N_i  )$

Thanks

Most of these are pretty easy for R=Z.  You can take M=Q and Ni=Z/iZ in (a).  For (b) and (c) take Mi=Z/iZ and N=Q.  For (e) take M to be free of finite rank.

My examples for (d) are too difficult to verify.  Perhaps someone can find simpler ones.

I think we need to say that we should only consider the natural (group) homomorphisms which are deemed not to be an isomorphism in each case.

For (d) take $ R$ a field, $ I$ infinite, $ N_i\equal{}R$ for each $ i\in I$ and $ M\equal{}\oplus_iR$. Then the natural injection read from right to left does not have the identity on $ M$ in its range, since it cannot be represented by finitely many linear functionals.

Apparently these sorts of questions are not always relegated to homework exercises (as in Rotman's Intro to Hom Alg. 1ed).  Here are some references to more serious studies of these isomorphisms as well as a reference for an easy solution to (d):

For (d) you can take M=Ni to be any infinite torsion abelian group, by proposition 3.1 of:

Arnold, D. M.; Murley, C. E. "Abelian groups, A, such that Hom(A,-) preserves direct sums of copies of A."
Pacific J. Math. 56 (1975), no. 1, 7-20.
http://www.ams.org/mathscinet-getitem?mr=376901
http://projecteuclid.org/getRecord?id=euclid.pjm/1102906578

In general, if M is a module such that Hom(M,-) preserves direct sums, then M is called "small", though there is some possible discrepancy on whether Hom is a functor from R-modules to Z(R)-modules or from R-modules to abelian groups.  This does not affect the example I give for (d) as R=Z(R)=Z.

Groups M,N_i for which (b) never holds are studied in:

Göbel, Rüdiger. "On stout and slender groups."
J. Algebra 35 (1975), 39-55.
http://www.ams.org/mathscinet-getitem?mr=376879
http://dx.doi.org/10.1016/0021-8693(75)90034-4

Groups N for which (c) never holds are called cosmall, and are shown in some sense not to exist in:

Goldsmith, B.; Kolman, O. "On cosmall abelian groups."
J. Algebra 317 (2007), no. 2, 510-518.
http://www.ams.org/mathscinet-getitem?mr=2362928
http://dx.doi.org/10.1016/j.jalgebra.2007.05.009

In this same paper, groups N in which (b) never holds are called strongly slender and a few remarks are given.  In the same paper, it is mentioned that any torsion-free group of infinite rank also provides an example for (d).  This is credited to a fairly early paper:

Rentschler, Rudolf. "Sur les modules M tels que Hom(M,-) commute avec les sommes directes."
C. R. Acad. Sci. Paris Sér. A-B 268 1969 A930-A933.
http://www.ams.org/mathscinet-getitem?mr=241466

This paper and some followups such as:

Žemlička, Jan; Trlifaj, Jan. "Steady ideals and rings."
Rend. Sem. Mat. Univ. Padova 98 (1997), 161-172.
http://www.ams.org/mathscinet-getitem?mr=1492975
http://www.numdam.org/item?id=RSMUP_1997__98__161_0

focus on rings where the small modules (or "dually slender") are exactly the finitely generated modules.  Such rings are called "steady" and include noetherian and perfect rings.