An ant is crawling at a rate of $48$ cm per minute along a strip of rubber which can be infinitely and uniformly streched. The strip is initially one meter long and one cm wide and is stretched an additional one meter at the end of each minute. Assume that when the strip is stretched, the ratio of the distances from each end of the strip remains the same before and after the stretch. If the ant starts at one end of the strip, find the number of minutes until it reaches the other end.

This problem is fairly easy if you just go minute by minute, but i was wondering if anyone could find a more elegant solution.

[hide]The strip is 100cm long.  After 1 minute, the ant walks 48cm along the strip, or 48% of the way.  When stretched after the first minute, the ant is 96cm from where he begun on a 200cm strip, still 48%.  Next minute, the ant walks another 48cm, or 24% of the way, making 72%.  It stretches to 3 meters, but the percentage doesn't change.  After a 3rd minute, the ant walks another 48cm, or 16% of the way, making 88%.  The rubber stretches to 4 meters.  After the fourth minute, the ant walks another 48cm, now 12% of the way, finally ending up at 100%.  The answer is $\boxed4$.[/hide]
I think that's the best way to go.

[hide]
If the strip was moving constantly instead of after each minute, the percentage traveled would be proportional to the logarithm of the time elapsed.  SInce it moves every minute, I can't think of any except method, altohugh you could approximate it for sufficiently long strips.[/hide]

[b]EDIT: The below is based on my initial incorrect reading of the problem
(I used a constant expansion)[/b]

----------------------------------------------

[hide]Let $x$ be the ant's position at time $t$. The length of the rubber at time $t$ is $100(t+1)$. The expansion of the rubber is $100$ cm/min. Since the component of the ant's velocity caused by the rubber's expansion is it's fraction of the way across times the total rate of expansion, this component is $100\frac{x}{100(t+1)}=\frac{x}{t+1}$ cm/min. The ant is also moving an additional $48$ cm/min. So, the overall velocity
$\frac{dx}{dt}=48+\frac{x}{t+1}$ with initial condition $(t,x) = (0,0)$
Solving, $x=48(t+1) ln(t+1)$.

We want the time at which the ant's position is equal to the length of the rubber.
$48(t+1) ln(t+1)=100(t+1) \to ln(t+1)=\frac{25}{12} \to t+1=e^{\frac{25}{12}}$

$\boxed{t=e^{\frac{25}{12}}-1}$ min, which is slightly over 7.

[hide="Alternate, but similar"]The effect of the growth of the rubber on the ant's velocity can be emulated by dividing the length and the ant's velocity as explained in the previous explanation both by $t+1$, since $d=vt$ and $t$ remains unchanged.

Thus, the ant's velocity is $\frac{48}{t+1}$ and the rubber's length is a constant $100$. Integrate the velocity with the initial condition $(t, x)=(0,0)$. To get that the ant's position is always $x=48ln(t+1)$. We want it to reach position $x=100$, so solve $48ln(t+1)=100$.[/hide]

[/hide]

[quote="Chocoboom"]...and is stretched an additional one meter [u]at the end of each minute[/u].[/quote]

It is stretched another meter at the end of each minute, so you don't even need calculus to do so.

But if it was stretched at a constant rate, well... I don't know calculus, so I'll just take your word for it.

Oh ok, I read it wrong then. I guess in that case you have no choice but to take it minute by minute.

i guess my question is what if the numbers weren't so nice that he got to the end in exactly 4 minutes. What if i change it to .. lets say, he moves at a speed of 47 cm per min. and the strip extends 99 cm at the end of each minute. How would you do that?

I think you should still go minute by minute.  Use the fact that the percentage remains the same when it stretches.  The only difference now is that we deal with some nasty rational numbers.

Here is my answer:
[hide]First minute:
The ant moved 47 centimeters, and is now 47 cm from where he started.  The rubber stretches to 199cm, so the ant is now 47/100*199 cm from where he started, or 93.53 cm.
Second minute:
The ant moved 47 centimeters, and is now 140.53 cm from where he started.  The rubber stretches to 298 cm, so the ant is now 140.53/199*298 cm from where he started, or $\frac{2093897}{9950}$ cm (about 210.442).  Now we're into some nasty numbers.
Third minute:
The ant moved 47 centimeters, and is now approximately 257.442 cm from where he started.  The rubber stretches to 397 cm, so the ant is now 257.442/298*397 cm from where he started, or $\frac{1016934159}{2965100}$ cm (about 342.968).
Fourth minute:
The ant moves 47 centimeters, and is now approximately 389.968 cm from where he started.  The rubber stretches to 496 cm, so the ant is now 389.968/397*496 cm from where he started, or $\frac{573521754064}{1177144700}$ (about 487.21432).
Fifth minute:
This was a pain to do, but we're at the home stretch.  The ant moves $\frac{2585504284}{294286175}$ cm, or about 8.785 cm.  We divide that by 47 to get the number of minutes it took to travel that distance, and we get $\frac{2585504284}{13831450225}$.  We add 4 to that to get our final answer: $\boxed{\frac{57911305184}{13831450225}}$, or a little less than 4.187 minutes.[/hide]

Of course, if I were wrong, who would know? :D


geometry
inradius
incenter
analytic geometry
similar triangles

here's a start...

$\triangle{PQR}=\frac{90\cdot\120}{2}=5400$
$r_{1}=\frac{5400}{s}=\frac{5400}{180}=30$

Since $\overline{ST}\perp\overline{PR}$ therefore $\overline{PS}=2r=60$ and $\overline{SR}=120-60=60$
$\to\triangle{TSR}$ is $45: 60: 75\Rightarrow{r_{2}}=15$

Similarly $\overline{PU}=2r=60$ and $\overline{QU}=90-60=30$
$\to\triangle{QUV}$ is $30: 40: 50\Rightarrow{r_{3}}=10$

EDIT $r_{3}=10$..

[quote="4everwise"]Let $\triangle{PQR}$ be a right triangle with $PQ=90$, $PR=120$, and $QR=150$. Let $C_{1}$ be the inscribed circle. Construct $\overline{ST}$ with $S$ on $\overline{PR}$ and $T$ on $\overline{QR}$, such that $\overline{ST}$ is perpendicular to $\overline{PR}$ and tangent to $C_{1}$. Construct $\overline{UV}$ with $U$ on $\overline{PQ}$ and $V$ on $\overline{QR}$ such that $\overline{UV}$ is perpendicular to $\overline{PQ}$ and tangent to $C_{1}$. Let $C_{2}$ be the inscribed circle of $\triangle{RST}$ and $C_{3}$ the inscribed circle of $\triangle{QUV}$. The distance between the centers of $C_{2}$ and $C_{3}$ can be written as $\sqrt{10n}$. What is $n$?[/quote]

[hide]Let P(0,0), Q(0,90) and R(120,0). The inradius of $\triangle PQR=30$, so that S(60,0). Then using similar triangles, $\triangle STR$ has sides of length 45,60,75, so the incenter of $\triangle STR$ has coordinates (75,15). Also, U(0,60) so that $\triangle UQT$ has lengths 30,40,50, so the incenter has coordinates (10,70). The distance between (10,70) and (75,15) is $\sqrt{7250}$ so $n=\boxed{725}$.[/hide]

Do you (or anyone) know how to do it non-analytically...

for some reason i get..

[hide]
$\sqrt{7400}\Rightarrow\boxed{740}$
[/hide]

ya i miscalculated $r_{3}$..so i get that answer now..

[hide="Solution"]We calculate the inradius:
$A=rs\implies A=180r$

And by Heron's, $A=\sqrt{(180)(30)(60)(90)}$, so $A=5400$.

Thus, $5400=180r\implies r=30$

We know the inradius is $30$, and since we constructed $ST$ as perpendicular to $PR$, we know $PS=60$.

This means $SR=120-60$, so $\triangle TSR\sim \triangle QPR$ in a $1:2$ scale factor.  Thus, the inradius of $SR$ is $30/2=15$, so we find the coordinates of $C_2$ to be $(75,15)$.

Next, we apply a similar process to $\triangle QUV$, and find $\triangle QUV\sim \triangle QPR$ in a $1:3$ scale factor.  Thus, the inradius of $\triangle QUV$ is $10$, so the coordinates of $C_3$ are $(10,70)$.

Now, we apply the distance formula:

\[C_2C_3 =\sqrt{(75-10)^2+(15-70)^2}=\sqrt{4225+3025}=\sqrt{7250}\]

We write in the form $\sqrt{10n}$, giving $\sqrt{725(10)}$, so our answer is $\boxed{725}$.
[/hide]

[quote=Subh108]Use the $s-a$ formula to find the inradii for both the circles,create a coordinatee system originating at the right angle and coordinate bash.[/quote]

I also used the $s-a$ formula but I used similar triangles instead. This question wasn't that good imo but its nice to see some geo appear

[b]Solution:[/b] Even though this question is easy, I love it. It feels like a good AIME problem. Notice that the two smaller triangles are similar to the overall one. Then using the way the incircle splits up the $3-4-5$ similar triangles, you can extract the answer $\sqrt{65^2 + 55^2} = \sqrt{7250} \Rightarrow \boxed{725}$.


Like the topic before, the one about Germany '06, I have 2 questions:

a) Which team do you support?

b) Which team, in your opinion, is stronger, and has more chance to win the cup?

Here we can discuss anything about this WC, not only my questions (just to make more interesting). Thanks in advance for replying.

My opinion:

a) Go Argentina!!!

b) USA seems to be the strongest team nowadays (or maybe the country with best players, even though the best ones are not taking part), but, like in Indianapolis 2002 and Athens 2004, with the strongest team, they didn't get the 1º place. However, for me they're the best. Another teams with chances will be Argentina and Spain. I don't know how Serbia and Montenegro (are they taking part? [do they play in only one team?]) is playing in this moment, but they had a good team in the past "strong" sompetitions.

[quote="rem"]There is no doubt that US will win. I want this team to win, too. US has the best best best players in the world like Iverson, James, and Wade. :)[/quote]

I have the same opinion that you. But, remember, that they said the same in 2002 and 2004, and, on both tournamentes, they even reached the final (We were the 1º team to beat them in an official competition!)

They do NOT have Allen Iverson, as he was not invited to play.

I believe the US will still dominate at the FIBA World Championship (as shown in there scrimmage game against Puerto Rico)

I support USA

I think they will win. They have the best players. Argentina has a shot, though.

I am looking forward to seeing the defensive skills of Bruce Bowen.

[quote="shobber"][quote="José"][quote="kheg_k"]he is really good at defense. at the top of the NBA.[/quote]
Yes, he's one of the best.
However, the best defender in the last 3 years is Ben Wallace-he's far away the best![/quote]
One cannot see anything from the Statistics of Bowen, unlike Wallace ( You can find the blocks and rebounds, ie.). Also, Wallace do not have to face Briant, McGrady, Iverson, Dirk, Wade and so on...[/quote]

True Shobber, but he was chosen by the NBA the best defense man in the last 3 years. I know it's not the same position, but he was chosen.

That's very true that US has a big chance, but that does not mean they will win. This can apply to basically all sports: in the World Cup, Brazil was out in quarter-finals(and they were supposed to easily win); Real Madrid, which has the best soccer superstars also does not end up winning championships; the New York Yankees, which is a totally stacked team, do not win the World Series every year; etc. etc. etc.
But I agree that US team has the BEST point guards, forwards, and defensive players. They also have one of the best coaches that have ever lived -- coach K(Kryzewski). It's a very strong team.

[quote="surge"]They do NOT have Allen Iverson, as he was not invited to play.
[/quote]

YES, WHAT IS THIS?! :mad:  :mad:

But oh well, the US team will still do very well. They completely slaughtered Puerto Rico yesterday and the best part about their roster is that the team is young and still growing and becomming better and better (one of the reasons they did not choose iverson although his stats are increasing with age :wink:). I think USA will win this year because the talent on their team is so "spread out" unlike two years ago (in which it was pretty much just Duncan and A.I.). Carmelo, Lebron, Wade, (maybe Amare if he plays), are obviously the main go-to guys but the other players on the roster: Jamison, Johnson, etc are vital supporting players.

Also, about the defense thing. I agree with Shobber, Ben Wallace may have the blocks and rebounds, but defending a 25 ppg threat every single night is much harder work than defending a 10 ppg guy. Ben Wallace is a VERY good defender, but Bruce Bowen is the BEST because his defense is usually a deciding factor in the Spurs games when facing people such as Kobe, Vince Carter, McGrady, and all the other go-to Shooting Guards/Small Forwards in the league.

Yes Iverson, there's sth I have to agree: Lots of games won by Spurs in last years' playoffs were because of good defenses in important situations: Almost all of they were due to Bruce Bowen, but some were by Ginobili. However, I agree that Bowen is a very strong defense, and he has to denfende 30 ppg people, which is more difficult than Wallace's job.

Bruce Bowen definitely is better def than wallace. Wallace isn't even a true center, which is a great drawback. Although we are talking about def, his offense is horrendous. Bowen has to defend the 20-25 ppg kobe or ray allen or whoever.

Btw Bowen plays dirty.

[quote]
YES, WHAT IS THIS?!
[/quote]

They already announced this at the last Olympics. The two main problems with Iverson are:

1) He's aging. This team is going to last from 2006-2008. Although he is only 31, he would probably be the oldest guy on the team. This isn't much of a problem... but it leads us to our next point

2) He is a [i]scoring[/i] point guard. With this many all-stars on the field they do not need a scoring point guard, but rather one with more passing attentiveness (note that I didn't say "...with more passing ability").

This year, I believe that Bruce Bowen performed at a superior level (defensively) when compared to Ben Wallace (and should have recieved defensive player of the year).

EXHIBIT A: [url=http://www.nba.com/news/alldefensive_060511.html]All Defensive NBA Team Selections[/url]. Bruce Bowen recieved one more vote than Ben Wallace.

EXHIBIT B: [url=http://youtube.com/watch?v=d3jNf5nlvLk]Rasheed Wallace speaks on Ben Wallace's 4th "Defensive Player of the Year" selection[/url]
Although was probably kidding, he makes a good point.

BUT Ben Wallace is much cooler than Bruce Bowen so he deserves the award.

[quote="surge"]

2) He is a [i]scoring[/i] point guard. With this many all-stars on [/quote]

*cough* gilbert arenas

*cough* 7 assists per game

http://www.nba.com/sixers/features/iverson_060306.html

That's a pro-Iverson site.

Who do you guys think should start?

WORLD CHAMPIONSHIP TEAM FINALISTS
Name 	Pos. 	Height 	Team/Univ
[b]Carmelo Anthony [/b]	F 	6-8 	Denver Nuggets
[b]Gilbert Arenas[/b] 	G 	6-4 	Washington Wizards
[b]Shane Battier[/b] 	F 	6-8 	Houston Rockets
[b]Chris Bosh 	[/b]F 	6-10 	Toronto Raptors
[b]Bruce Bowen [/b]	F 	6-7 	San Antonio Spurs
[b]Elton Brand [/b]	F 	6-8 	Los Angeles Clippers
[b]Kirk Hinrich [/b]G 	6-3 	Chicago Bulls
[b]Dwight Howard[/b] 	F-C 	6-11 	Orlando Magic
[b]LeBron James 	[/b]F 	6-8 	Cleveland Cavaliers
[b]Antawn Jamison[/b] 	F 	6-9 	Washington Wizards
[b]Joe Johnson [/b]	G-F 	6-7 	Atlanta Hawks
[b]Brad Miller 	[/b]C 	7-0 	Sacramento Kings
[b]Chris Paul[/b] 	G 	6-0 	NO/Okla. City Hornets
[b]Amare Stoudemire[/b] 	C 	6-10 	Phoenix Suns
[b]Dwyane Wade [/b]	G 	6-4 	Miami Heat

I think it should be

[b]PG- Chris Paul/Kirk Hinrich
SG- Dwyane Wade
SF- Lebron James
PF- Elton Brand/Dwight Howard
C- Brad Miller[/b]

I know you guys are going to be like "Where is Carmelo/Paul/Arenas, etc.?" But you don't want too many scorers on the field at one time, that's why James/Wade combination should take care of that. Howard/Brand rebound play pretty well defense and rebound.

I agree with your starters except I put Kirk over Chris Paul (they need a 3 point shooter on the court and because Kirk is more experienced) and Elton Brand over Howard (again with the more experienced).

P.S. Amare was cut from the team or I woulda put him as the center :(

Also, what happened with Kobe? I know he got surgery or something but is he still on the team?  :maybe:. If he is, shift Wade up to PG and let Kobe be SG.

Yeah, I was unsure about the Brand/Howard and Hinrich/Paul. Kobe Bryant will not be playing this summer (at FIBA, due to surgery).

I actually think Paul would be a better starter, but Hinrich has a little more experience and range.

Same with the Brand/Howard situation.

Brand should definitely be in starting lineup. One of my fav players.

I think Iverson should have been on the team. He is an amazing player who has much more talent than any other point gurad on the US team. The reason why many people think he does not have passing ability is because the team he is on lacks really good players( the team has some, like Iguodala, but is not full of good players) who would be there to receive passes, so the guy has to do everything on his own. Also, the fact that he is a really good shooting guard should have made him be on the team, because every team needs shooters: just consider the case when the opponent's team has somebody like Ben Wallace who will do anything to prevent a forward(like Wade or James) scoring. However it's barely possible to guard Iverson's shots, and there is a high chance he will get them in.

I don't know much about other teams, but can anyone comment on teams other than US?(like their players, etc)

I'll have to disagree on a few points.

1) Wade is a guard, also.

2) Iverson (.447) has a lower field-goal percentage than Wade (.495) and James (.480).

3) How does being shorter than James/Wade make him harder to block? He doesn't have any special "type" of shot. Like he's not shooting the skyhook (mastered by Kareem and sometimes Magic, nearly impossible to block due to the fact that is released practically behind the shooter at the tip of there fingers).

Actually, Iverson probably gets his shots blocked 3 or 4 times a game BECAUSE of his height.

Coach K said that the team might not have real starters. Instead he's going to rotate the players based on the opponent they're playing. I'd really like to see Bron play the 5 against a smaller, shorter team.  :)

I don't think Brad Miller should be a starter over Amare, even though he is much more experienced. The point is to have a young team that will grow to better in Shanghai (Beijing?) '08 and the World Championships in '10.

I believe that the reason why Iverson wasn't chosen is because he would just think that he's 10 times better than everyone else and not play like a point guard.

[quote="José"][quote="sstutter"]The sole reason I don't want Bowen to be cut is for the US-Argentina game. I want him to guard his usual teammate Ginobili and piss him off.  :roll:[/quote]

This is what I call the usual north-americans arrogance[/quote]

what

[color=blue][size=200]Bowen has been cut!!![/size][/color]

Bruce Bowen of the United States was cut

games started today, I know that US and Argentina won.

[quote="José"]what do you think?[/quote]

I think you didn't understand the post.

Typical South American buffoonery.

[quote="José"][quote="blahblahblah"]Typical South American buffoonery.[/quote]

Well, it seems that you're always against me, but well, that's your problem[/quote]

No, I'm obviously against your [b]entire continent[/b], don't you get it?

As usual you miss the point.

[quote="ricardotb"]iverson should have been the best player of nba if he hadn't hurt his shoulder[/quote]

uh...I wouldn't stretch it, though I wish that was true.

This year was Kobe's and Lebron's year. They were the two best players statistically. Iverson and Wade were probably right after them.

I personally think that Melo will be one of the best in the NBA.

Iverson would have been really good if he was taller...

[quote]Also, the fact that he is a really good shooting guard should have made him be on the team, because every team needs shooters[/quote]
We had an abundance of shooters in the very beginning w/ J Johnson, Kobe, Arenas, Billups, and Redd although most of em couldn't play. And ppl like Melo and LBJ have just as nice of a jumper as AI. So we needed a pure pass-first PG (C Paul) and we all know AI isn't a pure PG. Plus, he already had his chance in 2004 as the leader and look what happened.

[quote]It seems that the americans are afraid of Argentina.... why? You have better team, and, if we play against, you'll beat us. However, I fail to understand your fear[/quote]
What makes you think the US is afraid of Argentina? Nobody else is seeing any signs of fear.

[quote="sstutter"]Plus, he already had his chance in 2004 as the leader and look what happened.

[/quote]

hm yea so i guess duncan sucked too :wink:

[quote]hm yea so i guess duncan sucked too[/quote]

Sorry i'm a lil dumb. :roll: Please explain?

[quote="José"][quote="sstutter"]o ok nvm. Yesssss US sonned Senegal, even though it's nothing to be really proud of.  :)[/quote]

What I said before, americans' arrogance[/quote]

ur damn right son we're arrogant BUT we can live up to it


let 0< x,y <1
prove that
$x^{y}+y^{x}\geq 1$

[hide="solution"]http://www.mathlinks.ro/Forum/viewtopic.php?t=484  :D [/hide]


search
combinatorics proposed
combinatorics

Let $ S$ be a set of points in a plane,with distance betweem any two of them is at least $ 1$.Prove that one can choose subset of $ S$,with not less than $ \frac{|S|}{7}$ elements,such that distance between any two elements from new subset is at least $ \sqrt{3}$.

Probably something like that, but I'm too lazy to verify that we may adapt the proof :
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=133562414&t=5579

Pierre.


trigonometry
induction
algebra
functional equation
algebra unsolved

Find all $ f: \mathbb{R}\to \mathbb{R}$ such that $ f(x \plus{} \cos(2007y)) \equal{} f(x) \plus{} 2007\cos(f(y))$.

if $ y=\frac {(2\alpha +1)\pi}{2}$ so $ \phi$ exists such that $ f(\frac {(2\alpha +1)\pi}{2})=\frac {(2\phi +1)\pi}{2}$, with$ \alpha, \phi \in Z$

so, $ f(\frac {(2\alpha +1)\pi}{2}+\cos{(2007y)})=\frac {(2\phi +1)\pi}{2}+2007\cos(f(y)), \forall y\in R$; so, if $ y=\frac {(2\beta +1)\pi}{2}, \beta \in Z$ so $ f(\frac {(2\beta +1)\pi}{2})=0 \forall \beta \in Z$.

is it absurd?

[quote="bboypa"]
 if $ y \equal{} \frac {(2\beta \plus{} 1)\pi}{2}, \beta \in Z$ so $ f(\frac {(2\beta \plus{} 1)\pi}{2}) \equal{} 0 \forall \beta \in Z$.

is it absurd?[/quote]

a false idea  :(

$ f(x)\equal{}2007x\plus{}2k\pi, x\in\mathbb{R}$
$ f(x)\equal{}\minus{}2007x\plus{}(2k\plus{}1)\pi, x\in\mathbb{R}$
or
$ f(x)\equiv {2k\plus{}1\over 2}\pi$
for some $ k\in\mathbb{Z}$.

[color=green]Hi olorin, long time no see your post :D.

http://www.mathlinks.ro/viewtopic.php?t=185662

Welcome  :)[/color]

[quote="olorin"]$ f(x) \equal{} 2007x \plus{} 2k\pi, x\in\mathbb{R}$
$ f(x) \equal{} \minus{} 2007x \plus{} (2k \plus{} 1)\pi, x\in\mathbb{R}$
or
$ f(x)\equiv {2k \plus{} 1\over 2}\pi$
for some $ k\in\mathbb{Z}$.[/quote]
Well, are you planning to write a solution? I'm curious to see it.

Well, since nobody else seems interested in doing it, here is my solution.

Set $ d: \equal{}f(0)$ and $ g(x): \equal{}f(x)\minus{}d, x\in\mathbb{R}$ and $ c: \equal{}g(1)\equal{}f(1)\minus{}f(0)$.
Then $ g(0)\equal{}0$ and
$ g(x\plus{}\cos 2007y)\equal{}g(x)\plus{}2007\cos (g(y)\plus{}d)$ for all $ x,y\in\mathbb{R}$.

$ x\equal{}0$ gives
$ (1):$    $ g(\cos 2007y)\equal{}2007\cos (g(y)\plus{}d)\le 2007$ for all $ y\in\mathbb{R}$
and then
$ g(x\plus{}\cos 2007y)\equal{}g(x)\plus{}g(\cos 2007y)$ for all $ x,y\in\mathbb{R}$.

This gives
$ (2):$    $ g(y)\le 2007$ for all $ y\in [\minus{}1,1]$
and
$ (3):$    $ g(x\plus{}y)\equal{}g(x)\plus{}g(y)$ for all $ x\in\mathbb{R},y\in [\minus{}1,1]$.

Then by induction $ g(x\plus{}ny)\equal{}g(x)\plus{}ng(y)$ and
$ g(ny)\equal{}g(0)\plus{}ng(y)\equal{}ng(y)$ for all $ x\in\mathbb{R},n\in\mathbb{N},y\in [\minus{}1,1]$.
So $ g(x\plus{}ny)\equal{}g(x)\plus{}g(ny)$ for all $ x\in\mathbb{R},n\in\mathbb{N},y\in [\minus{}1,1]$, which gives
$ (4):$    $ g(x\plus{}y)\equal{}g(x)\plus{}g(y)$ for all $ x,y\in\mathbb{R}$.

So $ g$ satisfies Cauchy's functional equation $ (4)$ and by $ (2)$ has an upper bound on some interval,
which gives $ g(x)\equal{}cx$ and $ f(x)\equal{}cx\plus{}d$ for all $ x\in\mathbb{R}$.

Then $ (1)$ gives
$ (5):$    $ c\cos 2007y\equal{}2007\cos (cy\plus{}d)$ for all $ y\in\mathbb{R}$
and for $ c\not \equal{}0$ we get
$ \left |c\right |\equal{}\sup_{y\in\mathbb{R}}\, c\cos 2007y\equal{}\sup_{y\in\mathbb{R}}\, 2007\cos (cy\plus{}d)\equal{}2007$.
So $ c\in\{0,\pm 2007\}$ and $ y\equal{}0$ in $ (5)$ gives $ \cos d\equal{}{c\over 2007}$.

So either $ c\equal{}0$ and $ d\equal{}{2k\plus{}1\over 2}\pi$ for some $ k\in\mathbb{Z}$
or $ c\equal{}2007$ and $ d\equal{}2k\pi$ for some $ k\in\mathbb{Z}$
or $ c\equal{}\minus{}2007$ and $ d\equal{}(2k\plus{}1)\pi$ for some $ k\in\mathbb{Z}$.
Done


IS there any way to find the intersection of two lines without having to calculate it? And also is there any way to do square roots for lengths?

intersectionpoint(P,Q) gives the intersection point of paths P and Q.
abs(A-B) gives the distance between the pairs $ A$ and $ B$.

[quote="mewto55555"]IS there any way to find the intersection of two lines without having to calculate it?[/quote]
[asy]size(6cm);
import geometry_dev;

pair A=(1,1), B=(3,3);
line l1=line(A,B);

pair C=(2,-1), D=(1,-3);
line l2=line(C,D);

draw(l1^^l2);
dot(A^^B^^C^^D);

dot(intersectionpoint(l1,l2),red);[/asy]
[url=http://svnweb.tuxfamily.org/listing.php?repname=piprim/geometrydoc&path=%2Fbranches%2Fen%2F&rev=0&sc=0]Documentation[/url] of geometry_dev.
[url=http://piprim.tuxfamily.org/asymptote/travaux/index.html]Others examples and download[/url]

[quote="dragon96"]Can we do [b]intersectionpoint(A,B,C);[/b] to find the intersection of the three objects A, B, and C?[/quote]
If you want...
[code]point P=intersectionpoint(A,B);
if(P@C) P\in C else P\not\in C[/code]
You may take a look to [url=http://piprim.tuxfamily.org/asymptote/travaux/index.html#fig0026]fig0026[/url].
According to [url=http://piprim.tuxfamily.org/asymptote/travaux/modules/geometry_dev.asy.index.sign.html]this page[/url] the following operators are available
[code]bool @(point,arc)
bool @(point,conic)
bool @(point,explicit circle)
bool @(point,line)[/code]

The main question is why you would need anything like that: two curves are already enough to determine the point :? (in the worst case you'll have to use [b]intersectionpoints(P,Q)[2][/b] or something like that if the point is not unique).

What if you wrote the intersetion point of a line and a circle? Couldn't there be two i.p.s?

Yeap, that is what the intersectionpoits() command is for. It returns the array containing all intersection points and you choose the one you need by using the appropriate index. intersectionpoint() is basically just an abbreviation for intersectionpoints()[0].


inequalities
function
advanced fields
advanced fields unsolved

I read an article in which there is a line like this

"\[ \|u\|_{H^{2}(\Omega)}\leq \|\Delta u\|_{L^{2}(\Omega)}, |\Omega|<+\infty\quad (1)\]
We note that (1) only holds when $ \Omega\subset \mathbb{R}.$ If $ \Omega\subset \mathbb{R}^{n},n\geq 2,$ (1) doesn't hold."

In my opinion, we can duduce (1) from Poicare's inequality. I wonder what the authors meant?

Would you like to help me?

Thanks.

[quote="mlok"]There had to be more assumptions on $ u$ in that article, even in one dimension. A function can have zero Laplacian without being zero itself.[/quote]

I think that $ u$ must satisfy \[ u\in H^{2}_{0}\]. And that inequality may hold for $ n$ is arbitrary.

Any comments? Thanks

[quote="NguyenThanh_545"]I read an article in which there is a line like this

"
\[ \|u\|_{H^{2}(\Omega)}\leq \|\Delta u\|_{L^{2}(\Omega)}, |\Omega|<+\infty\quad (1) \]
We note that (1) only holds when $ \Omega\subset \mathbb{R}.$ If $ \Omega\subset \mathbb{R}^{n},n\geq 2,$ (1) doesn't hold."

In my opinion, we can duduce (1) from Poicare's inequality. I wonder what the authors meant?

Would you like to help me?

Thanks.[/quote]
I think
\[ \|u\|_{H^{2}(\Omega)}\leq C\|\Delta u\|_{L^{2}(\Omega)}, |\Omega|<+\infty\quad (1) \]
Why $ C=1$?


number theory
relatively prime
number theory unsolved

Show that if k is an integer then $ 21ab \plus{} 14bc \plus{} 34ca \equal{} k$ has a solution where  $ a$, $ b$ and $ c$ are integers.

Since $ 21$ and $ 34$ are relatively prime, from Bezout's relation we can find (infinitely many pairs of) integers $ b, c$, such that $ 21b \plus{} 34c \equal{} 1$. Take then $ a \equal{} k \minus{} 14bc$.


function
Functional Analysis
real analysis
real analysis unsolved

It is easy to do a construction to show that $\chi_{\mathbb{Q}}$ is in the second Baire class $B_{2}$ of the real line. With the first Baire category theorem I can conclude that a function in $B_{1}$ must have a point of continuity, so surely $\chi_{\mathbb{Q}}$ is not in $B_{1}$. However, for such a simple function as $\chi_{\mathbb{Q}}$ it should be possible to give an argument not using involved functional analysis, right? Can anyone think of something?

No one interested in this problem? Maybe I should rephrase it in case you don't know what the baire classes are. I want to show that the function on the real line which is 1 on $\mathbb{Q}$ and 0 elsewhere is not the pointwise limit of a sequence of continuous functions on the real line.

Well, this is exactly a problem Baire category is the most natural tool for. Of course, you can reduce everything to the nested interval lemma in the following way. Assume that such a sequence  $f_{n}$ of continuous functions eqists. Let $q_{1},q_{2},\dots$ be some enumeration of rational points. Let $I_{1}$ be any non-degenerate closed interval not containing $q_{1}$. Let $p_{1}$ be any rational number in $I_{1}$. Let $n_{1}$ be an index such that $f_{n_{1}}(p_{1})>\frac{1}{2}$. Let $I_{2}$ be a non-degenerate closed subinterval of $I_{1}$ near $p_{1}$ such that $q_{2}\ne I_{2}$ and $f_{n_{1}}\ge \frac{1}{2}$ on $I_{2}$. Let $p_{2}$ be a rational point in $I_{2}$. Let $n_{2}>n_{1}$ be an index such that $f_{n_{2}}(p_{2})>\frac{1}{2}$. Let $I_{3}$ be a non-degenerate closed subinterval of $I_{2}$ such that $q_{3}\ne I_{3}$ and $f_{n_{2}}\ge \frac{1}{2}$ on $I_{3}$. And so on. Then any point $c$ in the intersection of $I_{j}$ cannot be rational while, on the other hand, $f_{n_{j}}(c)\ge \frac{1}{2}$ for all $j$, so $f_{n}(c)$ cannot converge to $0$. But does this argument really avoid Baire category? Isn't Baire category just a simple buildup over the nested ball lemma going in exactly the same direction as this proof? Haven't we used here the fact that $\mathbb Q$ is a countable union of nowhere dense sets? On the other hand, I doubt that an approach based on some completely different idea is possible (or, at least, I'm sure it won't be easy to find).

Your post was really helpful for me fedja, although I forgot to tell you. Thanks.

Just a remark: the graph of a Baire-one function is connected.

Are you sure? This topic http://www.mathlinks.ro/Forum/viewtopic.php?t=137798 deals with exactly this question for the special case of a derivative.

$f_{n}(x) = \frac{1}{1+nx^{2}}$ is a counterexample. The limit function is Baire one and its graph has an obvious disconnection


Nomic is a self-modifying game. This means that during the game, players have the opportunity to propose new rules and amend or repeal existing ones. Rules are changed by majority votes or unanimous votes, depending on the rule. No rule is exempt from change, even ones that deal with the rule-making and rule-chaning process.

Here is a summary of the (initial) rules:

There are two kinds of rules: mutable and immutable. Every new rule is initially mutable. A mutable rule may be amended or repealed, but not an immutable rule. A rule may also be transmuted from mutable to immutable or vice versa. Transmuting from immutable to mutable requires a unanimous vote, while all other rule changes require only a majority vote. Each player proposes one new rule, amendment, or repealing of a rule, per turn. The player then gets points based on how many people voted for their proposal. Players with unsuccessful proposals lose points. The first player to reach 200 points wins. However, all these rules may be changed during the course of the game. Some of them, such as how to gain points, are even mutable, so that they may be easily changed.

If you would like to play Nomic, sign up here.

Currently signed up:
GoBraves
WindSlicer
math92
baseballfreak2
paladin8
chess64
Hexaditidom

[quote="GoBraves"]I dont really get it, but i'll sign up.[/quote]

Exactly what I was going to say.

Can't people just not vote for anything and everyone would lose points?

That's why you should try to make a proposal that other people will like enough to vote aye on.

Sure, I'll play. Sounds intense. And very, very difficult on the brain :P

If you're confused, this might clear it up: [url=http://www.earlham.edu/~peters/writing/nomic.htm]Nomic[/url]

There are currently enough people to play a game, but there is still plenty of room if anyone else would like to join.

I'll play...

I'll probably lose though  :blush:

I would like to play. After finding out today what nomic is, I've been searching everywhere for a live game. This should be fun.

I'll play even though im not quite sure how this actually works...

May I ask as to what date this game shall begin?

[quote="math92"]i think one game already started[/quote]
I saw your game and noticed that it has remained stagnant since Thursday. Is it possible to start a new game for those interested and currently not playing?

Since we are still on the first proposal in the game, you may still join the game.

nonie, Doubtable, and k2c901_1 have been added.

i'm in.
whatever happened to the other game? have people still not voted on that first proposal (cuz i have)

i'm in too

booyah again...take that L_Li   :P


let C denote the cantor set ( the middle thirds cantor as some ppl might refer to). show that C+C=[0,2] , where it is understood that $C+C=\{x+y: x,y\in C\}$

Think of the ternary (base 3) representation of the numbers in $[0, 1]$. Which ones get removed by the Cantor set construction? From this, what can you say about $C+C$?

ok so C consists of those numbers in $[0,1]$ whose ternary representations dont consist of 1's. so when you do $C+C$ you will get numbers which have 0's,1's and 2's. i'm a little weak in these sort of things. so i cant see where this argument leads to. can someone help me?
the way i know it and the way my prof told it is like this take $C_{n}$ to be the union of the intervals at the nth stage of the construction. now if $s\in [0,2]$ first we show that there exists $x_{1},y_{1}$ belonging to $C_{1}$ such that $x_{1}+y_{1}=s$. in general i show this for all n. then using some sequential argument i get the reqd result. these 2 things must amount to the same thing but i cant see how. please help


what is the number of real numbers ,A, such that (a,1),(1,a) and (a-1,a-1) are three distinct collinear points ?

(p.s - the correct answer is infinitely many)
 how to solve it ?

what is the number of real numbers ,[b]a[/b], such that (a,1),(1,a) and (a-1,a-1) are three distinct collinear points ?
 note the correct question!

[hide=" click to reveal content"]

I make this $ a \equal{} 3$.

$ (1, a)$ and $ (a, 1)$ lie on the line $ y \equal{} \minus{} x \plus{} (a \plus{} 1)$.

If $ (a \minus{} 1, a \minus{} 1)$ is also on this line then $ a \minus{} 1 \equal{} \minus{} a \plus{} 1 \plus{} a \plus{} 1 \implies a \equal{} 3$.

The other possibility is $ (1, a) \equal{} (a, 1) \equal{} (1, 1)$ with $ (a \minus{} 1, a \minus{} 1) \equal{} (0, 0)$ but then the three points on not distinct.

So the number of real numbers $ a$ such that we have three distinct collinear points is one.

[/hide]


Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ ab \plus{} ac \plus{} bc\neq0.$ Prove that
\[ \frac {a^2}{b \plus{} c} \plus{} \frac {b^2}{a \plus{} c} \plus{} \frac {c^2}{a \plus{} b}\geq\frac {3}{2}\sqrt {\frac {a^3 \plus{} b^3 \plus{} c^3}{a \plus{} b \plus{} c}}
\]
[hide="remark"]The following inequality ( with a same conditions ) is wrong already.
$ \frac {a^2}{b \plus{} c} \plus{} \frac {b^2}{a \plus{} c} \plus{} \frac {c^2}{a \plus{} b}\geq\frac {3}{2}\sqrt {\frac {a^4 \plus{} b^4\plus{} c^4}{a^2 \plus{} b^2 \plus{} c^2}}.$
[/hide]

[quote="arqady"]Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ ab \plus{} ac \plus{} bc\neq0.$ Prove that
\[ \frac {a^2}{b \plus{} c} \plus{} \frac {b^2}{a \plus{} c} \plus{} \frac {c^2}{a \plus{} b}\geq\frac {3}{2}\sqrt {\frac {a^3 \plus{} b^3 \plus{} c^3}{a \plus{} b \plus{} c}}
\]
[hide="remark"]The following inequality ( with a same conditions ) is wrong already.
$ \frac {a^2}{b \plus{} c} \plus{} \frac {b^2}{a \plus{} c} \plus{} \frac {c^2}{a \plus{} b}\geq\frac {3}{2}\sqrt {\frac {a^4 \plus{} b^4 \plus{} c^4}{a^2 \plus{} b^2 \plus{} c^2}}.$
[/hide][/quote]

Bu Cauchy Schwarz we have
\[ 2\left( {\sum a } \right)\left( {\sum {\frac{{a^2 }}{{b \plus{} c}}} } \right) \equal{} 2\sum {a^2 }  \plus{} 2\sum {\frac{{a^3 }}{{b \plus{} c}}}  \ge 2\sum {a^2 }  \plus{} \frac{{2\left( {\sum {a^3 } } \right)^2 }}{{\sum {ab\left( {a^2  \plus{} b^2 } \right)} }}
\]
By AM-GM inequality we have
\[ 2\sum {a^2 }  \plus{} \frac{{2\left( {\sum {a^3 } } \right)^2 }}{{\sum {ab\left( {a^2  \plus{} b^2 } \right)} }} \ge 3\sqrt[3]{{\frac{{2\left( {\sum {a^3 } } \right)^2 \left( {\sum {a^2 } } \right)^2 }}{{\sum {ab\left( {a^2  \plus{} b^2 } \right)} }}}}
\]
Hence it suffices us show that
\[ \left[ {\frac{{2\left( {\sum {a^3 } } \right)^2 \left( {\sum {a^2 } } \right)^2 }}{{\sum {ab\left( {a^2  \plus{} b^2 } \right)} }}} \right]^2  \ge \left[ {\left( {\sum {a^3 } } \right)\left( {\sum a } \right)} \right]^3
\]
\[ \Leftrightarrow 4\left( {\sum {a^3 } } \right)\left( {\sum {a^2 } } \right)^4  \ge \left( {\sum a } \right)^3 \left[ {\sum {ab\left( {a^2  \plus{} b^2 } \right)} } \right]^2
\]
which is true since
\[ 9\left( {\sum {a^3 } } \right) \ge \left( {\sum a } \right)^3
\]
\[ 2\left( {\sum {a^2 } } \right)^2  \ge 3\left[ {\sum {ab\left( {a^2  \plus{} b^2 } \right)} } \right]
\]
We have done. :)

[quote="nguoivn"]The stronger has been solved by SOS:
$ \sum\ \frac {a^2}{b \plus{} c} \geq\ \frac {3(a^3 \plus{} b^3 \plus{} c^3)}{2(a^2 \plus{} b^2 \plus{} c^2)}$[/quote]

My solution, I hope it's right. :)

By Cauchy Schwarz we have
\[ \left( {\sum {ab} } \right)\left( {\sum {\frac{{a^2 }}{{b \plus{} c}}} } \right) \equal{} \sum {a^3 }  \plus{} abc\left( {\sum {\frac{a}{{b \plus{} c}}} } \right) \ge \sum {a^3 }  \plus{} \frac{{abc\left( {a \plus{} b \plus{} c} \right)^2 }}{{2\left( {ab \plus{} bc \plus{} ca} \right)}}
\]
Therefor it suffices us to show that
\[ \left( {\sum {a^2 } } \right)\left( {2\sum {a^3 }  \plus{} \frac{{abc\left( {a \plus{} b \plus{} c} \right)^2 }}{{\left( {ab \plus{} bc \plus{} ca} \right)}}} \right) \ge 3\left( {\sum {ab} } \right)\left( {\sum {a^3 } } \right)
\]
\[ \Leftrightarrow 2\left( {\sum {a^2 } } \right)\left( {\sum {a^3 } } \right) \plus{} \frac{{abc\left( {\sum a } \right)^2 \left( {\sum {a^2 } } \right)}}{{ab \plus{} bc \plus{} ca}} \ge 3\left( {\sum {ab} } \right)\left( {\sum {a^3 } } \right)
\]
Now we assume $ a\plus{}b\plus{}c\equal{}1$, $ q \equal{} ab \plus{} bc \plus{} ca$ and $ r\equal{}abc$ then the inequality becomes
\[ 2\left( {1 \minus{} 2q} \right)\left( {1 \minus{} 3q \plus{} 3r} \right) \plus{} \frac{{r\left( {1 \minus{} 2q} \right)}}{q} \ge 3q\left( {1 \minus{} 3q \plus{} 3r} \right)
\]
\[ \Leftrightarrow \left( {1 \minus{} 3q} \right)\left[ {\left( {7q \plus{} 1} \right)r \plus{} 2q \minus{} 7q^2 } \right] \ge 0
\]
Because we have $ q \le \frac{1}{3}$ so it suffices to check $ {\left( {7q \plus{} 1} \right)r \plus{} 2q \minus{} 7q^2  \ge 0}$

Now if $ q \le \frac{2}{7} \Rightarrow 2q \minus{} 7q^2  \ge 0 \Rightarrow \left( {7q \plus{} 1} \right)r \plus{} 2q \minus{} 7q^2  \ge 0$

And if $ q \ge \frac{2}{7} > \frac{1}{4} \Rightarrow r \ge \frac{{4q \minus{} 1}}{9}$
\[ \Rightarrow \left( {7q \plus{} 1} \right)r \plus{} 2q \minus{} 7q^2  \ge \frac{{\left( {7q \plus{} 1} \right)\left( {4q \minus{} 1} \right)}}{9} \plus{} 2q \minus{} 7q^2  \equal{} \frac{{ \minus{} 35q^2  \plus{} 15q \minus{} 1}}{9} \ge 0
\]
We have done. :)

Equality holds if and only if $ a\equal{}b\equal{}c$.

[quote="nguoivn"]I think we can expand and usings pqr easily, don't have to use Cauchy Schwart.
Because equality occurs only when $ a \equal{} b \equal{} c$, I think it's not hard  :)[/quote]

Post your solution for proving what you say my friend.  :wink:

In my opinion, the most important is not using Cauchy Schwarz in the proofs, it is how to use it.

And $ p,q,r$ is also.  :)

I don't like the solution that use expanding and $ pqr$ too.  :)

Finally, did you use Cauchy Schwarz to prove your inequality.

[quote="nguoivn"][quote="nguoivn"]I think we can expand and usings pqr easily, don't have to use Cauchy Schwart.
Because equality occurs only when $ a \equal{} b \equal{} c$, I think it's not hard  :)[/quote]
Sorry because indeed, I haven't try for this silly hint  :(
But I said that:
[quote="nguoivn"]The stronger has been solved easily by SOS:
$ \sum\ \frac {a^2}{b \plus{} c} \geq\ \frac {3(a^3 \plus{} b^3 \plus{} c^3)}{2(a^2 \plus{} b^2 \plus{} c^2)}$[/quote]
And here is my solution, my friend  :)[/quote]

Anyway, thanks for sharing your solution.  :)

The following inequality is still true.
Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ a^2\plus{}b^2\plus{}c^2\neq0.$ Prove that
\[ \frac{a^2}{a\plus{}9b\plus{}9c}\plus{}\frac{b^2}{b\plus{}9a\plus{}9c}\plus{}\frac{c^2}{c\plus{}9a\plus{}9b}\geq\frac{3(a^3\plus{}b^3\plus{}c^3)}{19(a^2\plus{}b^2\plus{}c^2)}\]
But the following ( with the same conditions ) is wrong already.
$ \frac{a^2}{a\plus{}8b\plus{}8c}\plus{}\frac{b^2}{b\plus{}8a\plus{}8c}\plus{}\frac{c^2}{c\plus{}8a\plus{}8b}\geq\frac{3(a^3\plus{}b^3\plus{}c^3)}{17(a^2\plus{}b^2\plus{}c^2)}.$

[quote="arqady"]The following inequality is still true.
Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ a^2 \plus{} b^2 \plus{} c^2\neq0.$ Prove that

$ \frac {a^2}{a \plus{} 9b \plus{} 9c} \plus{} \frac {b^2}{b \plus{} 9a \plus{} 9c} \plus{} \frac {c^2}{c \plus{} 9a \plus{} 9b}\geq\frac {3(a^3 \plus{} b^3 \plus{} c^3)}{19(a^2 \plus{} b^2 \plus{} c^2)}$
[/quote]
Am-Gm solved it easily  :)

nguoivn, I have proved it by SOS and AM-GM.


Find the least integer $N$ such that $N/2$ is a perfect square and $N/3$ is a perfect cube.

 :)

[hide]648[/hide]

not really sure, i dont have a calculator at my disposal

first post btw, hope im not wrong

[hide]
$2a^2=3b^3$
So b divisible by 2 and a divisible by 3
so $2a^2$ divisible by 18 and $3b^3$ divisible by 24.

so N divisible by 18 and 24, so divisible by 72. [edit: from here $b^3$ divisible by 3 and hence $3b^3$ by $3\cdot2^3\cdot3^3=648$]

Trying the first few multiples of 72 gives 648, as Seiken said[/hide]

[hide]
A perfect square requires multiples of two as powers in the prime factorization.
A perfect cube requires multiples of three as powers in the prime factorization.

$N$ has to be divisible by 2 and 3, and you want the smallest number possible.

$2^{a-1=2}\cdot 3^{b-1=3}=N$

$2^3\cdot 3^4=N$

$N=\boxed{648}$[/hide]

[hide="well..."]$0$ also works but if you want positive integers, $648$ would indeed be the least.[/hide]

[quote="Seiken"][hide]I also pondered the idea of the answer being zero but threw it out cuz of the terms "cube" a"nd "square," which are 2-D and 3-D and thus require quantity to their dimension[/hide][/quote]

I disagree. When we're talking about numbers being perfect cubes, we include 0. However, when we talk about actual cubes and square, what you say is true.

Find the least positive integer $N$ such that $2N$ is a perfect square and $3N$ is a perfect cube :D

[quote="JavaMan"][hide]
$2^3 * 3^2 = 72$
[/hide][/quote]

72/3 is not a cube

[quote="Seiken"][hide]0 if you agree with Hamster, 72 if you do not. This of course is responding to User's "new" problem[/hide][/quote]

The question asks for a positive integer,so i can't be 0(talking 'bout 10000thuser's question).

edit-@ jli--new question

heh not reading the question thoroughly was always the reason I missed questions on a math test if i missed any.

[quote="shadysaysurspammed"]Find the[b] least integer[/b] $N$ such that $N/2$ is a perfect square and $N/3$ is a perfect cube.

 :)[/quote][quote="shadysaysurspammed"]

The question asks for [b]a positive integer[/b],so i can't be 0...
 [/quote]

Am I missing something here, or the original question needs an edit? :)
(By the way, for fun, one can modify it as by adding  "... and $N/5$ is perfect fifth power"  :) , and make it a little more fun.)

I think shadysaysurspammed is talking here about 10000th User's question, where a positive integer [b]is[/b] asked for, not the original question


if today was Friday the 13th in January, and it was NOT a leap year, how many more Friday the 13th's will there be in the rest of the year and in which months will they occur?

I would do the number of days in each month, divide it by 28, and add the remainder to the days. (Friday + 3 = Monday, etc) I think that October would be another Friday the 13th.

correct, october will be the only other month in which a Friday the 13th will occur for the rest of the year.


geometry rectangle rotation trigonometry AMC AIME search

1) In rectangle ABCD, AB = 6 and diagonal BD = 10. Circle O (with center O) is inscribed in triangle CBD, and circle P (with center P) is inscribedin triangle BAD. Find OP.

By noting that [BAD] = (1/2)(AB)(AD) = rs, where s is the semiperimeter, and solving for r, we can deduce that r = 2. I then managed to solve this problem by placing the figure onto the cartesian plane, placing AB along the y-axis and BC along the x-axis. Is there a way of doing this problem without plotting everything on a plane?

2) A triangle has sides 13, 14, 15. It is rotated through 180 degrees about its centroid to form an overlapping triangle. Find the area of the union of the two triangles.

well, there is a way to do it without plotting the points. I mean, you can create a right triangle with one leg 2, the hypotenuse OP, and the other leg a segment on the diagonal, and then use trig to find out that length on the diagonal and then finally use Pythagorean Theorem.

"I promise it will work. I promise it will not be nice."

So you're saying that my original method would be more doable (without a calculator)? Ok. Anyone know how to do the second problem?

Well, I don't think my method is that bad, becasue the angles involved are rather nice, 45/2, for example.

As for the second one, I'll post my solution in a while, but is it 2003 AIME B by any chance?

Oh, that's what I thought you meant by "I promise it will work. I promise it will not be nice." By the way, how could a right triangle with OP as its hypotenuse possibly have a leg BD? They would intersect at the center of the rectangle, which is not a right angle unless ABCD is a square.

And yes, the second one is from the 2003 AIME B.

Ok, this is what i meant:

Drop a perpendicular from P to BD, call it X, and drop another perpendicular from O to BD, call it Y. Now, we know that Triangle BPX is a right triangle with leg 2 and one angle 45/2. So the length of BX is 2/tan(45/2). The length of XY is therefore 10-4/tan(45/2), since DY=BX. So by Pythagorean Theorem, OP^2=2^2+XY^2, and since we already know XY, we can solve for OP. Provided, this could get a little messy, but it shouldn't be bad, not bad at all. Just turn tangent into sin/cos, and you get the half angles right there, and since your going to end up squaring anyway, I think it should work out...

Ok, I understand all of it except your assertion that angle PBX = 45/2. How did you arrive at this?

eww, oops

it's not 45/2. But you can find out what it is rather easily, since its half of angle DBC, which you can find out since it is part of a 3-4-5 Triangle

actaully, you don't need to find out wat angle PBX is exactly.

LEt angle DBC be B. Then, CosB is 4/5. Then, you know angle PBX is half of angle B, since point P lies on the angle bisector of angle B. THus, apply the half angle formula to Sin(B/2) and you can find out what its sine is.

Yeah. Except that it wouldn't be so nice of an angle. :( I will just assume that the way the test-makers expected students to solve the problem on an actual non-calculator competition would be by placing everything on a plane.  :D Thanks for your help!

[hide]

For #2 : is-it 4/3 of the area of the triangle alone ?

Strange it does not depend on the dimensions ...

[/hide]

Yep, that's the answer. How did you get it though?

It's a cute problem, perhaps we should let people search a little bit more

[hide]

I used these facts in my proof :

- the medians cut the triangle into 6 parts of equal area

- the centroid cut the median in the ratio 2:1[/hide]

Ah, I get it now.  :D I wonder why I didn't get it before though--I was looking at those same two properties. Maybe I just didn't draw my diagram well. Oh well. Thanks.

Great !

Btw, the triangle 13,14,15 has a nice (and unique I think) property. Do you know which ?

[hide]

The "altitude from side 14" (= 12) and the three sides are consecutive integers !

Nice isn't it ?[/hide]

Cool. Maybe that coupled with the fact that its area is an integer is why triangles with sides 13, 14, and 15 often appear on competitions.

[quote="i/3"]
Btw, the triangle 13,14,15 has a nice (and unique I think) property. Do you know which ?
[/quote]

Well, you can break it up into two right triangles: a 5-12-13 and a 9-12-15.  Putting the two together with a coincident 12-side leaves us with a side of 13, a side of 15, and a side of 5+9=14.


floor function
real analysis
real analysis unsolved

Given $ \alpha$  irrational, prove that there are infinitely many fractions $ {\displaystyle \frac{m}{n}}$
such that:

\[ {\displaystyle \left| \alpha \minus{}\frac{m}{n}  \right| \leq \frac{1}{n^2}}\]

I know it is a simple application of Kronecker aproximation theorem, but I failed in trying to solve it. :oops:

Let $ \theta \in [0,1]$ and suppose that an $ \epsilon > 0$ is given.

Then $ \exists n \in \mathbb{N}$ such that $ |n \alpha \minus{} \lfloor n \alpha\rfloor \minus{} \theta | < \epsilon$ by the theorem in question.

Letting $ \epsilon \equal{} \frac{1}{n}$ and $ m \equal{} \theta \plus{} \lfloor{n \alpha} \rfloor$ gives the result.

[quote="akech"]Let $ \theta \in [0,1]$ and suppose that an $ \epsilon > 0$ is given.

Then $ \exists n \in \mathbb{N}$ such that $ |n \alpha \minus{} \lfloor n \alpha\rfloor \minus{} \theta | < \epsilon$ by the theorem in question.

Letting $ \epsilon \equal{} \frac {1}{n}$ and $ m \equal{} \theta \plus{} \lfloor{n \alpha} \rfloor$ gives the result.[/quote]

Is it right???
Because $ m$ should be an integer!
And, since $ \epsilon$ is given and $ n$ is such that $ |n \alpha \minus{} \lfloor n \alpha\rfloor \minus{} \theta | < \epsilon$ we cannot say that $ \epsilon \leq \frac {1}{n}$. Maybe it just happens when $ n > \frac {1}{\epsilon}$  !!


Given a positive integer $n$, consider the sets $A \subset \{1,2,...,n\}$ in which the congruence $x+y \equiv u+v (mod n)$ has no other solutions but the trivial $x=u, y=v$, and $x=v, y=u$. Let $f(n)$ denote the maximum number of elements in such sets. a) Prove that $f(n)< \sqrt{n}+1$. b) Give an example for infinitely many n, such that $f(n)> \sqrt{n}-1$

You are wrong, let $A=\{1,2,4,8\}$. I think, that x and y distinct.
If A has m terms we have $C_m^2=m(m-1)/2$ distinct sums (if may be x=y, then m(m+1)/2).
Therefore $m\le \frac{1+\sqrt{8n+1}}{2}$ (or $m\le \frac{-1+\sqrt{8n+1}}{2}$).
$f(n)>\sqrt n -1$ can be proved for n=pq (p,q)=1, because we have $f(pq)\ge f(p)f(q)$.


induction
modular arithmetic
linear algebra
matrix
vector
algebra
system of equations

Let $ X: \equal{} \{x_1,x_2,\ldots,x_{29}\}$ be a set of $ 29$ boys: they play with each other in a tournament of Pro Evolution Soccer 2009, in respect of the following rules:
[list]i) every boy play one and only one time against each other boy (so we can assume that every match has the form $ (x_i \text{ Vs } x_j)$ for some $ i \neq j$);
ii) if the match $ (x_i \text{ Vs } x_j)$, with $ i \neq j$, ends with the win of the boy $ x_i$, then $ x_i$ gains $ 1$ point, and $ x_j$ doesn’t gain any point;
iii) if the match $ (x_i \text{ Vs } x_j)$, with $ i \neq j$, ends with the parity of the two boys, then $ \frac {1}{2}$ point is assigned to both boys.
[/list]
(We assume for simplicity that in the imaginary match $ (x_i \text{ Vs } x_i)$ the boy $ x_i$ doesn’t gain any point).
Show that for some positive integer $ k \le 29$ there exist a set of boys  $ \{x_{t_1},x_{t_2},\ldots,x_{t_k}\} \subseteq X$ such that, for all choice of the positive integer $ i \le 29$, the boy $ x_i$ gains always a integer number of points in the total of the matches $ \{(x_i \text{ Vs } x_{t_1}),(x_i \text{ Vs } x_{t_2}),\ldots, (x_i \text{ Vs } x_{t_k})\}$.
[i](Paolo Leonetti)[/i]

I managed to get two different interpretations of this problem... sadly I did not get much progress on either one, had to go sleep, and failed to solve it. :/

[hide="Intrepetation #1"]View it as a graph of 29 nodes, with an edge going between two nodes if and only if they tie. We want to show we can pick a subset of the nodes such that the for all nodes $ n$, the parity of the number of vertices in our set is even.

I think we may be able to use induction, but I'm not sure. [/hide]

[hide="Interpretation #2"]Enumerate the 29 boys by $ a_1, a_2, \ldots, a_{29}$. Let $ e_1, e_2, \ldots, e_{29}$ be residue classes mod 2. Let $ v_{i,j}$ be 1 if there is an edge between $ a_i$ and $ a_j$, and 0 otherwise (or if $ i \equal{} j$). We want to show that the system of equations $ \sum_{k\equal{}1}^{29} e_k v_{i,k} \equiv 0 \pmod 2$ for all $ i$ from 1 to 29 has a nonzero solution mod for $ e_1, \ldots, e_{29}$ for all possible sets of $ v_{i,j}$. To each solution, put $ e_i \in S$ if and only if $ e_i \equal{} 1$.

I tried using Chevalley on this interpetation, but the number of variables I had was always the same as the number of equations. I tried to eliminate one, but failed to do so. I'm inexperienced with Chevalley (and didn't really try hard on this approach) so if anyone solves it using this approach, please tell me.

I also tried interpreting this as a linear algebra problem. I [i]think[/i] that we want to show that the coefficient matrix is not invertible in $ F_2$. Sadly I know no linear algebra and this is all speculation, and I could not progress on this problem at all with linear algebra, considering how I didn't know any.[/hide]

So I only found out about this today and didn't have enough time to try the questions, but perhaps this works?
[hide]
Let $ A \in M_{29 \times 29}(\mathbb{F}_2)$ with column vectors $ A^1, A^2, \ldots, A^{29}$ denote the adjacency matrix of the graph in Zhero's first interpretation; it is symmetric and has a main diagonal with elements zero. We want to show that there exists $ O \neq X \in (\mathbb{F}_2)^{29}$ such that $ AX \equal{} O$ (if $ X_i \equal{} 1$ then player $ x_i$ is in our set), or that is there is a statement of linear dependency between $ A^1, A^2, \ldots, A^{29}$, or equivalentally that $ \det(A) \equal{} 0$.

Then for all permutations $ \sigma: J_n \to J_n$, we use the determinant formula $ \det (A) \equal{} \sum_{\sigma} \epsilon(\sigma) \prod_{i \equal{} 1}^{29} a_{i,\sigma(i)} \equiv \sum_{\sigma} \prod_{i \equal{} 1}^{29} a_{i,\sigma(i)} \pmod{2}$. We can ignore all permutations $ \sigma$ with an index $ 1 \le i \le n$ such that $ i \equal{} \sigma(i)$, because $ a_{i,\sigma(i)} \equal{} 0$. For the remaining permutations $ \sigma$, each can be paired with a distinct inverse $ \sigma^{ \minus{} 1} \neq \sigma$ (otherwise $ \sigma$ is an involution and pairs up an even number of elements; this generalizes to all odd numbers, but not evens) such that $ \prod_{i \equal{} 1}^{29} a_{i, \sigma(i)} \plus{} \prod_{i \equal{} 1}^{29} a_{i, \sigma^{ \minus{} 1}(i)} \equal{} \prod_{i \equal{} 1}^{29} a_{i, \sigma(i)} \plus{} \prod_{i \equal{} 1}^{29} a_{\sigma^{ \minus{} 1}(i) , i} \equiv 0$ because $ A$ is symmetric. Hence $ \det(A) \equal{} 0$ as desired.

[edit: amusingly enough, I didn't even read your second interpretation, haha    :roll: ]
[/hide]

oh god I was randomly asking people if they knew how to take determinants of symmetric matrices with odd dimensions whose entires were all 1's and 0's with a main diagonal of 0.

I should really learn linear algebra...


Solve in natural numbers:

x 2 +xy+y 2 =7631

rewrite as

$(x-y)^2 + 3xy = 7631$ which implies that $(x-y)^2 \equiv 2 \pmod 3$

but 2 isn't a square mod 3, so there are no solutions...


我的中學有兩位老師是夫妻: 妻子教語文, 丈夫教數學. 今年二月十四將近, 正傷神如何祝福這對"良師眷侶". 今天靈機一動, 想到一副上聯, 但下聯卻毫無頭緒. 現希望能集思廣益, 覓得妙聯. 我的上聯是:

[size=117][b]婚姻無三角[/b][/size]

("婚姻"可以"戀愛"代替; 三角又即trigonometry)

nice. :P

婚姻無三角
恋爱富几何

what should be the summarization line of the two then?


Team $ A$ and team $ B$ play a series. The first team to win three games wins the series. Each team is equally likely to win each game, there are no ties, and the outcomes of the individual games are independent. If team $ B$ wins the second game and team $ A$ wins the series, what is the probability that team $ B$ wins the first game?

$ \textbf{(A)}\ \frac{1}{5}\qquad
\textbf{(B)}\ \frac{1}{4}\qquad
\textbf{(C)}\ \frac{1}{3}\qquad
\textbf{(D)}\ \frac{1}{2}\qquad
\textbf{(E)}\ \frac{2}{3}$

grn_trtle. Listing them out, I can only find four possibilities in total for A to win. They are:

$ BBAAA, ABABA, ABBAA, ABAA$

Since there is only 4 possibilities, and only work works, the answer is $ \frac {1}{4}$

So the answer would be

$ \boxed{\textbf{(B)}\ \frac {1}{4}}$

You would be right if every outcome was equally likely, but it's not; the ABAA case has a higher chance of occurring.

ABAA has a $ \left ( \frac{1}{2} \right )^4$ chance, while the rest have $ \left ( \frac{1}{2} \right )^5$ chance.  This is because ABAA is equivalent to ABAAA+ABAAB, even though the series would already be over before the fifth game.

[quote=grn_trtle][hide="Solution"]If B wins the first game, we have BBAAA.
If A wins the first game and we suppose 5 games were played, we have AB, then any permutation of {B, A, A}, of which there are $ \frac{3!}{2!}\equal{}3$.
If A wins the first game and we suppose 4 games were played, we have ABAA.

Out of 5 outcomes, 1 works, so we have $ \boxed{\textbf{(A)}\ \frac{1}{5}}$.[/hide][/quote]

Sorry I'm confused: how is this valid, because  it assumed the series will continue even if A wins?
Could someone reexplain?

Thanks

Edit never mind I finally understood it:

It is just an expected value prob, so you have to multiply the 4 games success and total, 5 games by 1/32, and then proceed like a normal problem :)

[quote=BakedPotato66]Technically, wouldn't the answer be 1/5?

Two conditions: 1) B wins the second game 2) A wins the series.
-----
BBAAA: (1/2)^5=1/32

ABBAA: (1/2)^5=1/32
ABABA: (1/2)^5=1/32
ABAA: (1/2)^4=1/16
-----
Total probability is 5/32, and probability of B winning the first game is 1/32, hence, [b]1/5[/b].

Why is this wrong though?[/quote]

It's not.

[quote=klip2008][quote=BakedPotato66]Technically, wouldn't the answer be 1/5?

Two conditions: 1) B wins the second game 2) A wins the series.
-----
BBAAA: (1/2)^5=1/32

ABBAA: (1/2)^5=1/32
ABABA: (1/2)^5=1/32
ABAA: (1/2)^4=1/16
-----
Total probability is 5/32, and probability of B winning the first game is 1/32, hence, [b]1/5[/b].

Why is this wrong though?[/quote]

Since it is guranteed A to be won the series, you dont multiply the probability by 1/2 again[/quote]

Oooooooh, that makes sense.

[quote=BakedPotato66]Multiply which probability?[/quote]

The last A. Since A has to win the last game.

How is it not $\frac{1}{2}$ ik you use states to solve but still it said each team is equally likely to win

From the condition [i]"[b]The first team to win three games wins the series"[/b][/i], we know this is a series with 5-games.

Given the condition that [i][b]"team $ B$ wins the second game and team $ A$ wins the series"[/b][/i], then team $ A$ must win $3$ out of $4$ remaining games other than the second. There are total $\frac {4!}{3!1!} = 4$ possible outcomes.

There is only one successive outcome for [i][b]"team $ B$ wins the first and second games, and team $ A$ wins the series"[/b][/i]. That is, team $ A$ wins all $3^{rd}, 4^{th} \text { and } 5^{th}$ games.

Thus, the probability is $$\mathbb P = \frac {\text{Successive Outcomes}}{\text{Total Possible Outcomes}}=\boxed {\frac{1}{4}}$$


In any set of $181$ square integers, prove that one can always find a subset of $19$ numbers, sum of whose elements is divisible by $19$.

It is a well-known result, which appeared many times on ML (See Erdös-Ziv-Ginzburg), that  for each $n$, in any set of $2n-1$ integers we may find $n$ whose sum is divisible by $n$.

Your problem follows trivially from it.

Pierre.

The Erdos--- theorem puts a GREAT and Difficult restriction on the no of integers ( $37$ as opposed to $181$ 'square' integers as my question asks!)....

how do you think this problem ( mine) can be solved by using just Pigeon hole Principle??

Direct checking shows that  any square is $0,1,4,9,16,6,17,11,7,5 \mod [19].$
Thus there are exactly $10$ distinct possibilities. From the pigeon-hole principle, since you used $181$ numbers at least one of the class contains at least $19$ of your squares. Adding these $19$ squares from the same class leads to the result.

Pierre.

I have come across a similar problem which is the following -
Given a set of 4n+1 square integers(n odd), there always exists a subset of n integers such that the sum of those integers is divisible by n.

Again the point is not to use EGZ. I could not solve this one and I do not know whether the problem is correct.

Just observe that

$x^{2}\equiv 0,1,4,5,6,7,9,11,16,17\mod 19$

Then after applying the pigeonhole Principle the result follows

We know that,
$a^2\equiv 0,1,4,9,16,6,17,11,7,5 (mod 19)$
We have 181 square numbers and 10 possible remainders .So, we have atleast 19  numbers which have same remainders.So, we can easily find a set of 19 numbers whose sum is divisible by 19.


modular arithmetic
algebra
polynomial
number theory

[hide]
By Fermat's Little Theorem, $ a^p \equiv a \pmod{p}$.
So $ (a\plus{}1)^p \equiv (a\plus{}1) \pmod{p}$ and $ a^p \equiv a \pmod{p}$.
Therefore, $ (a\plus{}1)^p\minus{}a^p \equiv 1 \pmod{p}$
We know that $ q \equiv 0 \pmod{(a\plus{}1)^p\minus{}a^p}$, so $ q \equiv 1 \pmod{p}$.
Therefore, $ (q\minus{}1) \equiv 0 \pmod{p}$.
[/hide]

Nice problem!

[quote="Discrete_Math"][hide]...
We know that $ q \equiv 0 \pmod{(a \plus{} 1)^p \minus{} a^p}$, so $ q \equiv 1 \pmod{p}$....
[/hide]

[/quote]

Could someone explain this?

That argument appears to show that if $ q$ divides any number $ n$ that is $ 1 \pmod{p}$, then $ p \mid q \minus{} 1$. This is not true; take $ p \equal{} 5, n \equal{} 21, q \equal{} 3$.

As for how to actually solve this, use the same ideas to prove that any prime divisor of $ \Phi_n(x)$ ($ n$th cyclotomic polynomial) either divides $ n$ or is $ 1 \pmod{n}$. [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?p=873886#873886]See here[/url] for how to do this.

[quote="carlebee"]İf $ p$ is prime, $ n$ is a natural number and $ q$ divides $ (n \plus{} 1)^p \minus{} n^p$, prove that $ p$ divides $ q \minus{} 1$[/quote]
[hide="Solution"]
Notice that $ (n \plus{} 1)^p \equiv n^p \bmod q$.  Thus, $ (\frac {n \plus{} 1}{n})^p \equiv 1\bmod q$.  Let the order of $ \frac {n \plus{} 1}{n}\bmod q$ be $ k$.  Then, we have that $ k|p$, so $ k \equal{} 1$ or $ k \equal{} p$.  If $ k \equal{} 1$, then $ \frac {n \plus{} 1}{n}\equiv 1\bmod q$, so $ n \plus{} 1\equiv n\bmod q$, so $ 1\equiv 0\bmod q\implies q \equal{} 1$, so $ p|0 \equal{} 1 \minus{} 1$ and we are fine.  If $ k \equal{} p$, then since $ k|(q \minus{} 1)$, we have that $ p|(q \minus{} 1)$ and we are done. [/hide]

EDIT:

[b]MellowMelon[/b]: Sorry, I assumed that $ q$ was prime.  I think that the argument could be changed a bit to work:

[hide="Hopefully Correct Solution"]
Let $ r$ be an arbitrary prime factor of $ q$.  Notice that $ (n \plus{} 1)^p \equiv n^p \bmod r$.  Thus, $ (\frac {n \plus{} 1}{n})^p \equiv 1\bmod r$.  Let the order of $ \frac {n \plus{} 1}{n}\bmod r$ be $ k$.  Then, we have that $ k|p$, so $ k \equal{} 1$ or $ k \equal{} p$.  If $ k \equal{} 1$, then $ \frac {n \plus{} 1}{n}\equiv 1\bmod r$, so $ n \plus{} 1\equiv n\bmod r$, so $ 1\equiv 0\bmod r$, which is impossible.  If $ k \equal{} p$, then since $ k|(r \minus{} 1)$, we have that $ p|(r \minus{} 1)$.  Thus, every prime factor of $ q$ is congruent to $ 1\bmod p$.  This implies that $ q\equiv 1\bmod p$, so we are done.  [/hide]

[b]Kyoshiro[/b]: I considered $ \frac{n\plus{}1}{n}\bmod q$ to be $ (n\plus{}1)n^{\minus{}1}\bmod q$, where $ n^{\minus{}1}$ is the inverse of $ n\bmod q$.

[hide]Your last statement should be $ p \mid \phi(q)$, which is not necessarily $ q\minus{}1$.[/hide]

[quote="The QuattoMaster 6000"][quote="carlebee"]İf $ p$ is prime, $ n$ is a natural number and $ q$ divides $ (n \plus{} 1)^p \minus{} n^p$, prove that $ p$ divides $ q \minus{} 1$[/quote]
[hide="Solution"]
Notice that $ (n \plus{} 1)^p \equiv n^p \bmod q$.  Thus, $ (\frac {n \plus{} 1}{n})^p \equiv 1\bmod q$.  Let the order of $ \frac {n \plus{} 1}{n}\bmod q$ be $ k$.  Then, we have that $ k|p$, so $ k \equal{} 1$ or $ k \equal{} p$.  If $ k \equal{} 1$, then $ \frac {n \plus{} 1}{n}\equiv 1\bmod q$, so $ n \plus{} 1\equiv n\bmod q$, so $ 1\equiv 0\bmod q\implies q \equal{} 1$, so $ p|0 \equal{} 1 \minus{} 1$ and we are fine.  If $ k \equal{} p$, then since $ k|(q \minus{} 1)$, we have that $ p|(q \minus{} 1)$ and we are done. [/hide][/quote]
$ (\frac {n \plus{} 1}{n})^p\notin Z$, how can $ (\frac{n \plus{} 1}{n})^p\equiv 1\bmod q$??? :huh:


analytic geometry
superior algebra
superior algebra unsolved

Describe the colimits of the $ Set$ category.

which category?

EDIT: Rendering problem -- didn't see "Set" when I posted.

If we have a diagram, we take the disjoint union of all the sets in the diagram, and then we declare two elements to be equivalent if there is a morphism that takes one to the other. We then take the smallest equivalence relation with all of these relations, and the quotient set is our colimit.

Yeps.

That being said, the limit consists of all elements of the *product* of all the sets in the diagram, such that when there is a morphism from $ S \to T$ in your diagram, the element in coordinate $ S$ is taken to that in coordinate $ T$.

your identifications are not fine enough. the elements ai , aj are identified if there are i -> k , j -> k such that ai and aj are mapped to the same element. in general, (co)limits can be contructed from (co)equalizers and (co)products.

The identification should be fine enough, since he took the smallest equivalence relation generated by the mapping equivalences.


[color=blue]it is a very well-known problem and i wanted to know where i can find a proof without using inversion or $ (x,y)$ coordinates.
Let $ AB$ be a segment and $ C$ be a point on it.
we draw the circles with diameter $ AB,AC,BC$
let the perpendicular from $ C$ to $ AB$ intersects the circle $ (BA)$ at $ X$ and $ Y$
consider the circle which is tangent internally to $ (AB)$ and externally to $ (AC)$ and tangent to the line $ XY$ and denote its center by $ O_1$
consider the circle constructed like $ (O_1)$ but this time,tangent externally to $ (BC)$ and denote its center by $ O_2$
prove that the circles $ (O_1)$ and $ (O_2)$ have equal radius
thx in advance for your help. :) [/color]

Dear shoki.

I think this is known (by Archimedes) as shoemaker's knife.

[hide]
$ O_{1}H_{1}\bot{AB}(H_{1}\in{AB})$,   $ O_{2}H_{2}\bot{AB}(H_{2}\in{AB})$

$ D$= center of circle $ (AB)$,  $ E$= center of circle $ (BC)$

$ DO_{1}^{2} \minus{} DH_{1}^{2} \equal{} OO_{1}^{2} \minus{} OH_{1}^{2}$ (Pythagoras)

therefore radius of $ O_{1} \equal{} \frac {AD\cdot{BE}}{AD \plus{} BE}$

symmetrically, radius of $ O_{2} \equal{} \frac {BE\cdot{AD}}{BE \plus{} AD}$

thus two radii are equal.[/hide]

I hope it is correct.


logarithms
integration
calculus
calculus computations

Find the sum:

1). $ \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{1^{2}+2^{2}+\cdots+n^{2}}$.

Let $ k\geq 0$ be a natural number. Find the sum of the following series

2). $ \sum_{n\geq 1}^{\infty}\frac{(-1)^{n-1}}{1^{2}+2^{2}+\cdots+(n+k)^{2}}$.

Let $ k$ be a natural number.

Calculate:

3) $ \sum_{n=1}^{\infty}\frac{1}{2n(2n+1)(2n+2)\cdots(2n+k)}$.


Let $ a,b,c$ be nonnegative real numbers. Prove that

$ \sum\sqrt{\frac{a^3}{a^2\plus{}8b^2}}\ge \sqrt{\frac{a^2\plus{}b^2\plus{}c^2}{a\plus{}b\plus{}c}}$

[quote="dduclam"]Let $ a,b,c$ be nonnegative real numbers. Prove that

$ \sum\sqrt {\frac {a^3}{a^2 \plus{} 8b^2}}\ge \sqrt {\frac {a^2 \plus{} b^2 \plus{} c^2}{a \plus{} b \plus{} c}}$[/quote]
I proved that
\[ \sum\sqrt{\frac{a^3}{a^2\plus{}8b^2}} \ge \sqrt{\frac{a^3\plus{}b^3\plus{}c^3}{a^2\plus{}b^2\plus{}c^2}}\]
But my solution is very ugly :(


There is a cube and all of its faces are divided into 3x3 squares.
each side of the square is 1cm, so each side of the cube is 3cm.
A fly starts at one corner of the cube and must travel to the other corner of the cube (diagonally opposite corner i.e. 9cm away)
If every move the fly is moving towards the other corner (i.e. total length travelled is 9cm) then how many routes are there?
We consider only 3face of the cube and assume that the fly can only move on those 3 faces.
The 3faces are touching one another.

[quote="Truffles"]So the corner that the fly is trying to reach and the corner the fly starts on are on the same face?[/quote]

I took it to mean the fly can only move along the edges of the small squares, and the destination is the antipode to the starting point, so [i]not[/i] on the same face.

But "consider only 3 faces" ... I don't know what that means.

well basically what transseries said
and 3faces part was me messing up my attempt at clearer definition.
it's cause u can see only 3faces when u look at the cube and u only need to consider them as the other 3 faces would give repeats....
basically going from A to B in the picture

[quote="Truffles"]Oh ok, thanks.  So since it's a fly can it fly over without going on the "lines" on the cube?[/quote]
No. He means for it to only crawl along the given lines.

[hide="answer?"]Then is the answer $ \binom{9}{3} \cdot \binom{6}{3}$?[/hide]

Edit: Sorry it's hidden now.

Hide your answer.

[hide]No, I don't think that is correct because, if you're counting the way I think you are, some of those paths would pass through the interior of the cube.[/hide]

[hide="Possible Hint, if I understood the problem correctly"][geogebra]bc92522c9531be87661b43c0ac403781efe32750[/geogebra]

There may be an easier way to solve this, but: start at $ B$ and count how many different routes the fly has from each vertex, as you move toward $ A$; for example, adjacent vertices only have one route, and those vertices collinear with the adjacent ones and $ B$ also have one route. Keep doing this until you reach $ A$ [/hide]

[hide="Answer"]By the method mentioned above, the answer is $ \boxed{148}$[/hide]

Metfor's answer is correct, though his explanation may be a little hard to follow. [hide]To read up on how to solve these kinds of problems - look in [u]The Art of Problem Solving, Vol. 2[/u] under Block Walking.[/hide]
A followup problem would be to solve this for walks along all six sides of the cube.
Another followup would be to find a general formula for walks along the faces of $ n \times n \times n$ cubes and $ m \times n \times p$ rectangular solids.

Ok, I got a different answer, I think I messed up somewhere.

If you start from A and go to the corner right above there (3 choose 3) ways. And then it takes (6 choose 3) ways to get to B. So doing that for each point along the first faces edge. I get

(3 choose 3) * (6 choose 3) + (4 choose 1) * (5 choose 2) + (4 choose 1) * (5 choose 2) + (4 choose 1) * (5 choose 2) + (4 choose 1) * (5 choose 2) + (3 choose 3) * (6 choose 3).

And thats 220.[/code]

You're overcounting some paths. For example, all of the paths that go up 3 units, then right one unit, then whatever from there - you've counted in both your (3 choose 3)*(6 choose 3) term and also in your (4 choose 1)*(5 choose 2) term.

:?:  How does metafor's method take into account the routes where the fly doesn't reach any vertices of the cube until B?

Metafor's explanation is pretty brief, but he's trying to explain the execution of a technique that's discussed in the AoPS Vol. 2 book called Block Walking. It's basically a bash-method of enumerating the number of paths leading up to each point - starting with the ones closest to A and ending at B.

The points along the edges emanating from A all have one path leading to them, so mark them all with 1. Then for any unmarked points, add the numbers of the points that immediately predeed it along a path to get the total number of paths leading to that point. Continue this through the whole diagram until you get to point B. It may seem like it would take a long time, but it doesn't.

Ohh I see.  I guess AOPS volume 2 is really useful huh?


Let $ ABC$ be a triangle and $ \omega$ be its circumcircle. Construct a point $ P$ lies on $ \omega$ such that $ PA\equal{}PB\plus{}PC$.

[hide="Comment."] I think $ P$ is the intersection of $ \omega$ with the Euler circle of triangle $ I_aBC$, where $ I_a$ is the ex-center on the side $ BC$ of triangle $ ABC$.[/hide]

P is the tangency point of (O) with the incircle (I') of the supermedial triangle $ \triangle A'B'C',$ having medial $ \triangle ABC$ and 9-point circle (O). See
[url]http://www.mathlinks.ro/viewtopic.php?t=100242[/url]
[url]http://www.mathlinks.ro/viewtopic.php?t=85415[/url]
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=24959[/url]


logarithms
induction
inequalities unsolved
inequalities


$ n\in N*$ .What's the maximum value of $ \sum^{n}_{k\equal{}1} \frac{a_{k}}{k}$
where $ a_{1},...,a_{n}$ are real numbers such that
 $ \sum^{n}_{k\equal{}1} a_{k} \equal{}0$ and $ \sum^{n}_{k\equal{}1}\left|a_{k}\right| \equal{}1$

Have anybody got a hint? :)
Urgently waited. :maybe:
Thanks!

Hint: [hide]in the extremal case, have a look at how many of the $ a_k$ can be strictly $ > 0$ and how many $ < 0$ :)[/hide]


Q1 show that 3^2008+4^2009 can be written as product of two positive integers each of which is larger than 2009^182.
i try to use binomial but looks typical.

try using the Sophie Germain identity:

\[ a^4\plus{}4b^4\equal{}(a^2\plus{}2b^2\minus{}2ab)(a^2\plus{}2b^2\plus{}2ab)\]

[quote="pleurestique"]try using the Sophie Germain identity:
\[ a^4 \plus{} 4b^4 \equal{} (a^2 \plus{} 2b^2 \minus{} 2ab)(a^2 \plus{} 2b^2 \plus{} 2ab)\]
[/quote]thanks my friend


Given a,b,c>0.Prove that:
(a) $ \frac{b\plus{}c}{a}\plus{} \frac{c\plus{}a}{b}\plus{} \frac{a\plus{}b}{c} \ge 6\plus{} \frac{(a^2\plus{}b^2\plus{}c^2\minus{}ab\minus{}bc\minus{}ca)^2}{a^3b\plus{}b^3c\plus{}c^3a}$
(b)$ (a\plus{}b)(b\plus{}c)(c\plus{}a) \ge 8abc\plus{} \frac{(a\minus{}b)^2(b\minus{}c)^2(c\minus{}a)^2}{a^3\plus{}b^3\plus{}c^3}$

[quote="secrets"]Given a,b,c>0.Prove that:
(b)$ (a \plus{} b)(b \plus{} c)(c \plus{} a) \ge 8abc \plus{} \frac {(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2}{a^3 \plus{} b^3 \plus{} c^3}$[/quote]
See its here:http://www.mathlinks.ro/viewtopic.php?t=263960 :wink:


[quote="imathsvn"]:welcomeani:
Show that for any positive integer n one can find integer a,b,c so that a and b are odd and c is non-negative and
2n=a^19 +b^99 +c2^1999
(i dont use latex,HELP ME :rotfl: )[/quote]

$2n=a^{19} +b^{99} +c*2^{1999}$

This is what you mean, right?


calculus
integration
algebra
function
domain
polynomial
Ring Theory

Well... page 32, theorem 5.2

Let $k$ be a field and $A=k[a_{1},...,a_{n}]$ an integral domain with $a_{i}$ elements of an overring. Write $r=tr.deg_{k}A$. If $r>0$, then $A$ is not a field.

This how I would 'prove' this...

The statement is equivalent to showing that $A$ is a field implies that $r=0$. Assume that $A$ is a field and $r>0$. Then there exist $r$ algebraically independent elements, say $b_{1},...,b_{r}$. Since there are no polynomial relations among them, they are indeterminates, and $k[b_{1},...,b_{r}]$ can serve as a polynomial ring (which is a subring of $A$). But a polynomial ring certainly has non-units, so $A$ can't be a field. Contradiction.

So... where did I go wrong... Matsumura is "pretty :D:D" concise and he proved it on whole page, so I really doubt my proof is correct :)

I do not see your argument. You construct a subring of a field, and this subring is not a field. So what? owk

It didn't make sense indeed....  :blush:


[color=darkblue] Find all positive integers $ a$ and $ b$, for which $ a^4 \plus{} 4b^4$ is prime.
[/color]

Sophie Germain?
[hide="Solution"]
$ a^4\plus{}4b^4\equal{}a^4\plus{}4a^2b^2\plus{}4b^4\minus{}4a^2b^2\equal{}(a^2\plus{}2b^2\minus{}2ab)(a^2\plus{}2b^2\plus{}2ab)$, so $ a^2\minus{}2ab\plus{}2b^2\equal{}1$, or $ a\equal{}b\equal{}1,p\equal{}5$ [/hide]


Find all continous functions $ f,g,h: \mathbb R\rightarrow \mathbb R$ satisfying $ f(x\plus{}y)\equal{}g(x)\plus{}h(y)$ for every $ x,y\in\mathbb R$

for $ x \equal{} y \equal{} 0$ we have $ h(0) \equal{} f(0) \minus{} g(0)$

for$ y \equal{} 0$ we have for every $ x$ , $ g(x) \equal{} f(x) \minus{} h(0) \equal{} f(x) \plus{} g(0) \minus{} f(0)$and  for $ x\equal{} 0$ we have $ h(y) \equal{} f(y) \minus{} g(0)$.

now let $ u(x) \equal{} f(x) \minus{} f(0)$ we have$ u(x \plus{} y) \equal{} u(x) \plus{} u(y)$ and $ u$ is continous function so $ u(x) \equal{} u(1)x$ for every $ x$.


Congruent equilateral triangles are used to tessellate the plane. How many triangles share a common vertex?

[hide]Six triangles make a hexagon, so the answer is $6$.[/hide]


geometry
analytic geometry
graphing lines
slope
ratio
circumcircle
perpendicular bisector

How do you prove that the medians and angle bisectors are concurrent.

Draw the incircle of the triangle. This should help you prove that the angle bisectors are concurrent. As for the medians, place $\triangle ABC$ in the first quadrant of the coordinate plane with $AC$ lying on the positive x-axis. Let $A$ be at $(0,0)$, $B$ at $(b,0)$, and $C$ at $(a,c)$. Now express the midpoints of the sides in terms of $a, b, c$ and then find the line equation for each median. Lastly, find the intersection of the medians two at a time and you will see that each two intersect at the same point.

You can use Ceva to prove that the medians are concurrent.  Of course, I haven't done any geometry in a year and a half, so take that with a grain of salt.

for the medians you have to draw perpendiculars to the sides from the intersection of $2$ medians and then prove the other line passing through the intersection bisects the base.

I know what you guys mean, I have tried some methods as well.
However, can somebody show me a working method that's easy and clean?

Let $C$ be at $(0,0)$, $B$ at $(b,0)$, and $A$ at $(a,c)$. The intersection of the medians with sides $AC$, $BC$, and $AB$ will be $\left(\frac{a}{2}, \frac{c}{2} \right)$, $\left(\frac{b}{2},0 \right)$, and $\left(\frac{a+b}{2}, \frac{c}{2} \right)$ respectively. Call the median from $A$ $M_1$, the median from $B$ $M_2$, and the median from $C$ $M_3$. The slopes of the three medians are

\[M_1=\frac{2c}{2a-b}\]
\[M_2=\frac{c}{a-2b}\]
\[M_3=\frac{c}{a+b}\]

The equations of the medians are

\[M_1 \Rightarrow y=\frac{2c}{2a-b}\left(x-\frac{b}{2} \right)\]
\[M_2 \Rightarrow y=\frac{c}{a-2b}(x-b)\]
\[M_3 \Rightarrow y=\frac{c}{a+b} \cdot x\]

We find the points of intersection of these two lines two at a time

\[M_1, M_2 \Rightarrow \frac{2c}{2a-b}\left(x-\frac{b}{2} \right)=\frac{c}{a-2b}(x-b)\]
\[\Leftrightarrow x=\frac{a+b}{3}, y=\frac{c}{3}\]
\[{M_2, M_3 \Rightarrow \frac{c}{a-2b}(x-b)}=\frac{c}{a+b} \cdot x\]
\[\Leftrightarrow x=\frac{a+b}{3}, y=\frac{c}{3}\]
\[M_1, M_3 \Rightarrow \frac{2c}{2a-b}\left(x-\frac{b}{2} \right)=\frac{c}{a+b} \cdot x\]
\[\Leftrightarrow x=\frac{a+b}{3}, y=\frac{c}{3}\]

Hence all three medians intersect at $\left(\frac{a+b}{3}, \frac{c}{3} \right)$ which by definition is the centroid.

For many triangle lines (not necessarily cevians) or even circles, the concurrency follows simply from the transitivity of equivalence.

[color=blue][b](1)[/b]  [u]Perpendicular bisectors of the sides. [/u][/color]
The perpendicular bisector $p_a$ of the side a = BC is the locus of points P with equal distances from the vertices B, C and similarly for the perpendicular bisectors $p_b, p_c$ of the sides b = CA, c = AB. Let the perpendicular bisectors $p_a, p_b$ of the sides BC, CA meet at a point O. Since $O \in p_a$, OB = OC. Since $O \in p_b$, OC = OA. From the transitivity of equivalence, OB = OA, which means that $O \in p_c$.

[color=blue][b](2)[/b]  [u]Internal angle bisectors. [/u][/color]
The internal bisector $l_a$ of the angle $\angle A$ is the locus of points P with equal perpendicular distances d(P, b), d(P, c) from the lines b = CA, c = AB and similarly for the internal bisectors $l_b, l_c$ of the angles $\angle B, \angle C$. Let the angle bisectors $l_a, l_b$ of the angles $\angle A, \angle B$ meet at a point I. Since $I \in l_a$, d(I, b) = d(I, c). Since $I \in l_b$, d(I, c) = d(I, a). From the transitivity of equivalence, d(I, b) = d(I, a), which means that $I \in l_c$.

[color=blue][b](3)[/b] [u]Medians.[/u][/color]
Let D, E, F be the midpoints of the sides a = BC, b = CA, c = AB. The median $m_a = AD$ from the vertex A is the locus of points P, such that the areas $\triangle_{ABP}, \triangle_{CAP}$ of the triangles $\triangle ABP, \triangle CAP$ are equal and similarly for the medians $m_b = BE, m_c = CF$ from the vertices B, C. This is because these areas are the differences of the triangle areas $\triangle_{ABP} = \triangle_{ABD} - \triangle_{PBD},\ \triangle_{CAP} = \triangle_{ACD} - \triangle_{PCD}$ and the areas $\triangle_{ABD} = \triangle_{ACD},\ \triangle_{PBD} = \triangle_{PCD}$ are equal (equal bases BD = CD and equal altitudes d(A, a) or d(P, a) to these bases). Let the medians $m_a, m_b$ from the vertices A, B meet at a point G. Since $G \in m_a$, $\triangle_{ABG} = \triangle_{CAG}$. Since $G \in m_b$, $\triangle_{BCG} = \triangle_{ABG}$. From the transitivity of equivalence, $\triangle_{CAG} = \triangle_{BCG}$, which means that $G \in m_c$.

[color=blue][b](4)[/b] [u]Altitudes.[/u][/color]
Let D, E, F be the midpoints of the sides a = BC, b = CA, c = AB and (D), (E), (F) circles on the diameters BC, CA, AB. Let the circle (D) cut the lines CA, AB at points Y, Z. Since BC is the diameter of the circle (D), the angles $\angle BYC = \angle BZC = 90^\circ$ are right, hence, the points Y, Z are the feet of the altitudes $h_b = BY, h_c = CZ$ from the vertices B, C. Similarly, the circle (E) passes through the feet Z, X of the altitudes $h_c = CZ, h_a = AX$ from the vertices C, A and the circle (F) through the feet X, Y of the altitudes $h_a = AX, h_b = BY$ from the vertices A, B. Consequently, the altitudes $h_a = AX,\ h_b = BY,\ h_c = CZ$ are the radical axes of the circle pairs ((E), (F)); ((F), (D)); ((D), (E)). Let the altitudes $h_a, h_b$ from the vertices A, B meet at a point H. Since $H \in h_a$, the powers of the point H to the circles (E), (F) are equal, p(H, (E)) = p(H, (F)). Since $H \in h_b$, the powers of the point H to the circles (F), (D) are equal, p(H, (F)) = p(H, (D)). From the transitivity of equivalence, p(H, (E)) = p(H, (D)), the powers of the point H to the circles (E), (D) are equal, which means that $H \in h_c$.

[color=blue][b](5)[/b] [u]Apollonius circles.[/u][/color]
The A-Appolonius circle of a triangle $\triangle ABC$ is the locus of points P with a constant ratio of the distances $\frac{PB}{PC} = \frac c b$ from the vertices B, C and similarly for the B- and C-Apollonius circles. Let K, L, M be the intersections of the internal bisectors $l_a, l_b, l_c$ of the angles $\angle A, \angle B, \angle C$ with the opposite sides a = BC, b = CA, c = AB and let K', L', M' be the intersections of the external bisectors $l_a', l_b', l_c'$ of the angles $\angle A, \angle B, \angle C$ with the opposite sides a = BC, b = CA, c = AB, respectively. Obviously, the A-, B-, C-Apollonius circles $c_a, c_b, c_c$ are the circumcircles of the right angle triangles $\triangle AKK', \triangle BLL', \triangle CMM'$ with the diameters KK', LL', MM'. Let the A- and B-Apollonius circles $c_a, c_b$ meet at a point P. Since $P \in c_a$, we have $\frac{PB}{PC} = \frac c b$. Since $P \in c_b$, we have $\frac{PC}{PA} = \frac a c$. From the transitivity of equivalence,

$\frac{PA}{PB} = \frac{PA}{PC} \cdot \frac{PC}{PB} = \frac c a \cdot \frac b c = \frac b a$

which means that $P \in c_c$. Since the circles $c_a, c_b$ meet at the point P, they also meet at one other point Q. For exactly the same reasons as for the point P, $Q \in c_c$, which means that the circles $c_a, c_b, c_c$ all pass through the points P, Q, they belong to the same pencil of circles and their centers are collinear. The 2 points P, Q are called the 1st and 2nd isodynamic points of the triangle $\triangle ABC$.

wooow! what are those "Apponolious circles"????? Anyway cant you just proof the third line which passes through the centroid bisects the base???? That is much easier.

Of course it is easier in the case of medians. I was just trying to demonstrate a general method. While general methods can be applied in many different cases, special cases often admit simpler solutions.

The Apollonius circles are the circles defined exactly as in my previous post. The A-Apollonius circle of a triangle $\triangle ABC$ is the locus of points P with the constant ratio of distances $\frac{PB}{PC} = \frac{AB}{AC} = \frac c b$ from the vertices B, C. The locus of points with a constant ratio of distances from 2 given points is always a circle (I am not trying to prove this claim here). Obviously, $\frac{AB}{AC} = \frac c b$, so the vertex A belongs to the locus circle. The internal angle bisector of the angle $\angle A$ cuts the opposite triangle side BC[b] internally[/b] in the ratio of the adjacent sides, i.e., the cutting point K is inside of the segment BC. Hence, $\frac{KB}{KC} = \frac{AB}{AC} = \frac c b$ and the point K belongs to the locus circle.  The external angle bisector of the angle $\angle A$ cuts the opposite triangle side BC [b]externally[/b] in the ratio of the adjacent sides, i.e., the cutting point K' is outside of the segment BC. Hence, $\frac{K'B}{K'C} = \frac{AB}{AC} = \frac c b$ and the point K' belongs to the locus circle. If it is known that the locus of some points is a circle and if we find 3 points of this locus, the points A, K, K' in the case of the A-Apollonius circle, the locus circle is the circumcircle of the triangle formed by these 3 points, the circumcircle of the triangle $\triangle AKK'$ in the case of the A-Apollonius circle. The internal and external bisectors AK, AK' of the angle $\angle A$ are perpendicular to each other, $AK \perp AK'$. Hence, the triangle $\triangle AKK'$ is right and its circumcenter is the midpoint of its hypotenuse KK'. Most likely, the Apollonius' circles are named after Apollonius of Perga, one of the best old Greek geometers after Euclid.


Find all triplets $ (x,y,z)$, that satisfy:
$  \{\begin{array}{c}\ \ x^2 - 2x - 4z = 3\
y^2 - 2y - 2x = - 14 \
z^2 - 4y - 4z = - 18  \end{array}$

[quote="Ahiles"]Find all triplets $ (x,y,z)$, that satisfy:
$  \{\begin{array}{c}\ \ x^2 - 2x - 4z = 3 \
y^2 - 2y - 2x = - 14 \
z^2 - 4y - 4z = - 18 \end{array}$[/quote]
by suming all the inequalities and adding to form squares wa have:
(x-2)²+(y-3)²+(z-4)²=0
so x=2,y=3 and z=4. :wink: .

[quote=Ahiles]Half of students made this mistake. Try to put your answer in one of equations...
So, system doesn't have real roots.[/quote]

Oh. So, are there any hints?

i believe the question is asking for real roots, so the answer is $\varnothing$. if it asks for complex roots i dont think its possible to find them


abstract algebra
linear algebra
matrix
vector

Let $ \mathsf{A}$ be a real $ 3 \times 3$ matrix, and let $ \mathbf{d}$ be a 3-component column vector. Explain briefly how the general solution of the matrix equation $ \mathsf{A}\mathbf{x} \equal{} \mathbf{d}$, where $ \mathbf{x}$ is a 3-component column vector, depends on the kernel and image or the linear map $ \mathbf{x} \mapsto \mathsf{A}\mathbf{x}$.
Consider the case
\[ \mathsf{A} \equal{} \left(\begin{array}{ccc} 1 & 1 & 1\\ 1 & a & b \\ 1 & a^2 & b^2 \end{array} \right), \qquad \mathbf{x} \equal{} \left( \begin{array}{c} x \\ y\\ z \end{array} \right), \qquad \mathbf{d} \equal{} \left( \begin{array}{c} 1\\ \minus{}1\\ 1 \end{array} \right).\]Find the kernel and image of the corresponding map, noting the different possibilities according to different values of $ a$ and $ b$.

For which values of $ a$ and $ b$ do the equations $ \mathsf{A}\mathbf{x} \equal{} \mathbf{d}$ have (i) a unique solution, (ii) more than one solution, (iii) no solution? For each pair $ (a,b)$ satisfying (ii), give the solutions as the sum of a fixed solution and the general solution of the corresponding homogeneous equations.

Let $ f(\mathbf{x}) \equal{} \mathsf{A} \mathbf{x}$.

Im $ f \equal{} \{w : \exists v$ with $ f(v) \equal{} w\}$.

So, the first thing we can say is $ \mathsf{A} \mathbf{x} \equal{} \mathbf{d}$ has at least one solution $ \iff \mathbf{d} \in$ Im $ f$.

Suppose this is the case. Take $ w, w'$ (not necessarily distinct) such that $ f(w) \equal{} f(w') \equal{} \mathbf{d}$. Since $ \mathsf{A} w \equal{} \mathsf{A} w' \equal{} \mathbf{d}$, $ \mathsf{A} (w\minus{}w') \equal{} 0 \rightarrow w\minus{}w' \in \ker f$. So, if $ w$ is a particular solution of $ f(w) \equal{} \mathbf{d}$, any other solution is of the form $ w \plus{} v$ with $ v \in \ker f$.

Now we have to find the kernel and image of $ g(\mathbf{x}) \equal{} \mathsf{A} \mathbf{x}$, where $ \[ \mathsf{A}\equal{}\left(\begin{array}{ccc}1 & 1 & 1\\ 1 & a & b\\ 1 & a^{2}& b^{2}\end{array}\right)$.

Suppose $ g(\mathbf{x})\equal{}0$ for some $ \mathbf{x} \equal{} (x,y,z)$. Then, $ \mathsf{A} \mathbf{x} \equal{} \left(\begin{array}{c}x\plus{}y\plus{}z\\ x \plus{} ay \plus{} bz\\x \plus{} a^2y \plus{} b^2z\end{array}\right) \equal{} \left(\begin{array}{c}0\\ 0\\0\end{array}\right)$.

From the first equation we know $ x \equal{} \minus{}y\minus{}z$. We make the corresponding substitution in the second and third equations and we get
$ (a\minus{}1)y \plus{} (b\minus{}1)z \equal{} 0$
$ (a^2\minus{}1)y \plus{} (b^2\minus{}1)z \equal{} 0$.

Suppose $ a\equal{}1$. The first equation would become $ (b\minus{}1)z \equal{} 0$, and the second one $ (b^2\minus{}1)z \equal{} 0$.
*If $ b \equal{} 1$, then every real number $ z$ satisfies both equations. So in this case, we would have $ \ker \mathsf{A} \equal{} \{\mathbf{x} \in \mathbb{R}^3 : x \equal{} \minus{}y \minus{} z\} \equal{} <(\minus{}1,1,0); (\minus{}1,0,1)>$.
*If $ b \neq 1$, then the first equation yields $ z\equal{}0$. So, in this case, $ \ker \mathsf{A} \equal{} \{\mathbf{x} \in \mathbb{R}^3 : x \equal{} \minus{}y$ and  $ z\equal{}0\} \equal{} <(\minus{}1,1,0)>$.

Now suppose $ a \neq 1$. From the first equation we get $ y \equal{} \frac{1\minus{}b}{a\minus{}1}z$. Replacing in the second equation gives

$ \left((a\plus{}1)(1\minus{}b) \plus{} (b^2\minus{}1)\right) z \equal{} (1\minus{}b)\left((a\plus{}1)\plus{}(\minus{}1\minus{}b)\right) z \equal{} (1\minus{}b)(a\minus{}b) z \equal{} 0$.
*If $ a \equal{} b$, then every real number $ z$ satisfies the equation. Also, replacing we get $ y\equal{}\minus{}z$, so $ x\equal{}0$. In this case, $ \ker \mathsf{A} \equal{} \{\mathbf{x} \in \mathbb{R}^3 : y \equal{} \minus{}z$ and  $ x\equal{}0\} \equal{} <(0,\minus{}1,1)>$.
*If $ b \equal{} 1$, then every real number $ z$ satisfies the equation. Also, replacing we get $ y\equal{}0$, so $ x\equal{}\minus{}z$. In this case, $ \ker \mathsf{A} \equal{} \{\mathbf{x} \in \mathbb{R}^3 : x \equal{} \minus{}z$ and  $ y\equal{}0\} \equal{} <(\minus{}1,0,1)>$.
*If none of the above happens, then $ z\equal{}0$, which implies $ y\equal{}0$ and $ x\equal{}0$. $ \ker \mathsf{A}$ in this case is only $ \{(0,0,0)\}$.

To be continued...


In the figure, $ \triangle ABC$ is a right triangle with right angle at $ B$.  $ BD$ is altitude of $ \triangle ABC$,  and $ BE$ is a median of $ \triangle ABC$.  Find the ratio of the area of the $ \triangle BEA$ to the area of $ \triangle BDA$.  Express your answer as a common fraction.

[asy]pair A=(0,0), B=(6.4, 4.8), C=(10,0), D=(6.4,0), E=(5,0);

draw(A--B--C--cycle);
draw(B--E);
draw(B--D);
label("$A$", A , W);
label("$B$", B, N);
label("$C$", C,E);
label("$D$", D, S);
label("$E$", E, S);
label("$8$", A--B, NW);
label("$6$", B--C, NE);[/asy]

Is $ \frac {25}{32}$ right?

It's not unique. The ratio is $ \frac {\frac {AC}2}{AD} \equal{} \frac {AC}{2AD}$...

[b]Edit:[/b] In your particular diagram, that is correct.

[quote="math154"]It's not unique. The ratio is $ \frac {\frac {AC}2}{AD} \equal{} \frac {AC}{2AD}$...[/quote]

I thought the ratio would be $ \frac {AE}{AD}$...

Bleh, I spent a long time trying to figure out the sides while I could have just as easily done this.

[b]Edit:[/b] Yay!!!

It is $ \frac{AE}{AD}$, since $ AE\equal{}\frac{AC}2$...

Could you also figure that out this way:
\[ \frac {\frac {AE \times BD}{2}}{\frac {BD \times AD}{2}} \equal{} \frac {AE}{AD}
\]

Do most MATHCOUNTS problems requiring ratios have an easy way like this one where you can simplify the whole thing and just find a few things like $ AE$ and $ AD$ rather than the whole area?

Yeah, that's pretty much why my method works?

[quote="mathemagician1729"]@007math

You forgot to include the sides of the right triangle, in this problem they were 6 & 8. So it was
25/32.[/quote]

Fixed. :)


I am pretty sure that this cannot be solved analytically, but I found it very easy to 'see' that x=1 was a solution by inspection (and maple confirms that there aren't any other). All terms are 2 when x = 1.


linear algebra
matrix
linear algebra unsolved

If $ A\in M_{n}(\mathbb{R})$ such that $ A^{3}\equal{}A^{2}$.
Prove:$ Rank(A)\ge Trace(A)$

[quote="JC_math"]$ A$ can be written as an upper triangular matrix with $ 1$s and $ 0$s on its main diagonal, hence $ \mbox{rank}(A)\geqslant\mbox{trace}(A)$.[/quote]
Oh,that a obvious way,we can also prove  $ \mbox{rank}(A) \equal{}\mbox{trace}(A)$ when $ A^{2}\equal{} A$,Sorry for my trivial problem :blush: