Why does current flow through a diode in the n to p direction if the potential difference in that direction gets very high?

Which current are you talking about? The direction that electrons flow or the "standard" which is opposite that? Also, you can forward and reverse bias a diode and that will make a difference. Darn this is hard to talk about without pictures...

the answer is: with a very high potential, electrons can be driven across the base into the n-type semiconductor, against the opposing field set up by the excess electrons

This is a Zener Diode and the two main ways this can happen is through "avalanche multiplication" and "Zener breakdown."

what are "avalanche multiplication" and "Zener breakdown" ?

There may be other names for these, but my book says:

Avalanche multiplication: when an electron near the junction is sufficiently accelerated by the electric field to ionize atoms it collides with, thereby creating fresh electron-hole pairs. The new electrons in their turn continue the process to produce a flood of charge carriers in the diode.

Zener breakdown: this is less common than avalanche multiplication (this occurs in heavily doped diodes). Valence electrons on the p side tunnel to the conduction band on the n side even though they do not have enough energy to first enter the conduction band on the p side.

It is from my modern physics book. I took this as my third class. It comes right after Physics I and II are done.


function
geometry
rectangle
real analysis
real analysis unsolved

[img]http://farm4.static.flickr.com/3293/3079781387_7b6bc5d8bd_o.png[/img]

my try on problem2 was take the partial devaritive eaqual to zero, get the C.P.  then take the boundary point (3,2) , to find the max and min.  it that right?

also, how to do problem#4?[/img]

You need to find the Hessian to classify the critical points correctly.

[quote="JoeBlow"]In problem 2, the boundary is not a discrete set of points as in the one dimensional case, but rather four line segments, so you would have to check the value along each side of the rectangle.  Since you are looking for the global maximum, there is no need to look at the Hessian nor to classify the critical points, as you really only need the value of the function not its local behavior.  In this case, you will find that the maximum is at $ (3,0)$ and the minimum is at $ (0,0)$.

In problem 4, the second equality applied to the first implies that all points of intersection satisfy $ z^2 \equal{} \minus{} xy$, so you are trying to minimize $ \sqrt {x^2 \plus{} y^2 \plus{} z^2} \equal{} \sqrt {1 \minus{} xy}$ subject to $ x^2 \plus{} y^2 \equal{} 1$ and $ xy \leq 0$.  Clearly, therefore, the distance is minimal if $ xy \equal{} 0$, so the four points closest to the origin are $ \left \{(0,\pm 1,0),(\pm 1,0,0)\right \}$.[/quote]

so it is just the 4 pts
(0,0) (3,0) (3,2)  (0,2)

I used the Lagrange Multipliers,  just got crazy.. cant solve the system

[quote="yzc717"]so it is just the 4 pts
(0,0) (3,0) (3,2)  (0,2)[/quote]No, the boundary of a rectangle is four line segments, not four points.  You need to check every point on those lines, not just the corners.

You do not need Lagrange multipliers for problem 4.  To see that $ \sqrt {1 \minus{} xy}$ constrained by $ xy\leq 0$ has a minimum when $ xy\equal{}0$ requires only arithmetic, not calculus.

[quote="JoeBlow"]To see that $ \sqrt{1\minus{}xy}$ constrained by $ xy\leq 0$ has a minimum when $ xy\equal{}0$ requires only arithmetic, not calculus.[/quote]
I would say more common sense that arithmetic.  Since $ xy\leq 0$, $ 1\minus{}xy\geq 1\minus{}0\equal{}1$, so $ \sqrt{1\minus{}xy}\geq\sqrt{1}\equal{}1$.


This is sad. Last year, Cupertino MS had a team score on the AMC-8 of 73. Yay us!

This year, apparently we didn't even get 66 to get on the honor roll. :o  :(
EDIT: It's probably because vishalarul, i_like_pie, and I all left. Sad.

Anyone else want to talk AMC8 scores?

I notice that JLS made the honor roll this year though...

EDIT: Perfect scores are out.

Last year JLS was 3 points away, this year the cutoff was lower...since JLS made it and the top 2 scores were 25 and 19.
WOW this is really good, that Jesse Kim guy at terman scored perfect as well? He must've improved a lot just like Steven. Yay more math people for gunn next year!

[quote="CheeseIsGood"]Wait, I never said 19 was the second score. All I said was the I knew 19 was A score that was received.[/quote]
Oh, lol. Well AMC8 is easy, but I've been quite disappointed at the performance of "top students" recently including myself (although of course I suck at math FOR REAL).

[quote="Ubemaya"]You're still no match for SHS :D .

j/k... (maybe)[/quote]
Hmmm...:) If I work super hard I think I have a chance of making red MOP but its pretty small I think...I got 130.5 on the 2005 AMC12A when I did it for practice today but I dunno if crazy scores like that will pop up when I take the real thing. Gotta try my hardest to make the Gunn math team I guess you could call it more godly.

Hey this is good, today the forum is quite lively! Keep up the posting. (eh I wish it was more lively still, making 75 posts in one day was so fun...)

[quote="CheeseIsGood"]I hear JLS has a total score of 73. 25, 24, 24.[/quote]
GASP pretty good hehe. Who are the other 2 people?
Last year Jordan apparently had a total score of 75 with boyle pinsker and kevin all getting perfect. Or maybe Kevin only got 24 since I didn't see him on the perfect score list but whatever.

[quote="Ubemaya"][quote="xscapezaer"][quote="Ubemaya"]You're still no match for SHS :D .

j/k... (maybe)[/quote]

Whoa, holy #%#@#!

Where did that come from? :o

Don't you mean that SHS probably now sucks because you go there? :rotfl:[/quote]

yeah but there's another freshmen there

*cough, cough*
don't deny it Gay[/quote]

Oh, so now you're saying that someone else is there who's l337 enough to "cancel out" your "failing?"

Yeah, I wish I "fail" as much as you do...

Seems like someone might have said that already

[quote="bookaholic"]Whoa--we're that good?! I honestly didn't expect that, but YAY CONGRATULATIONS Steven and whoever got the 24s, tell them congrats from me too![/quote]
I'm definitely coming next thursday (finals but who gives a crap), and the other 2 24s BETTER be there...

One of them probably won't. Not sure about the other. And just remember, all of this information I am giving is from another student in the Alg2/Geo class, so if I am wrong, don't kill me.  :P Although it is most likely correct.

[quote="bookaholic"]Wait. Someone who got a 24 on the AMC-8 "probably won't" be showing up to the school target round?! What madness is this?

Also, I checked with Mr. Peterson, and apparently the 07-08 school tests just hadn't arrived by the scheduled time of the sprint. The target round will be from this year.

The team won't have any time to train, though. We didn't exactly expect the chapter contest to be suddenly moved up by three weeks.[/quote]
No time to train? 3 weeks is plenty. Well not thoroughly train but cover most stuff.
I'll come every week then, but I'll sorta be focusing on Steven since he's the only person who actually cares about math. I wanna see how this turns out hehehe.
edit: oops I thought you said it was three weeks from now

[quote="CheeseIsGood"]Wait, when is the chapter competition?[/quote]
Don't worry about that, just get your geo down within 2 weeks and you'll be fine.

[quote="CheeseIsGood"]Lol, geo in 2 weeks... I don't think I will retain much if I rush.

Also, how is the JLS team going to be determined? All the Alg2/Geo people got the things inviting them to take the target, but some of them didn't even take the sprint. Theirs were different too. Instead of top scorer, it said Alg2/Geo.[/quote]
Yea its unfair, a lot of geo/alg2 people suck :( What did you get on sprint?

[quote="CheeseIsGood"]I don't know... it wasn't given like last year.

Oh no, the paper says "The top 8 performers in the next round will be invited to be part of the JLS MathCounts team."

If they don't consider the sprint for this, the Alg2/Geo people have a big advantage. Big. They automatically pass the sprint.

Evil. My chances of making the team died right there. If they considered the sprint score also, I might have had a small chance... even though I have no idea what my score was.[/quote]
:P you have no idea how bad those geo/alg2 people are...really. What have you been scoring on your practices? If you're getting over like 24s every time then you'll definitely beat them. Even though mathcounts is easy math its surprising how much a lot of people suck at it.


x,y,z>0  ABC  triangle. ==> [x/(y+z)]b 2 c 2  + [y/(z+x)]c 2 a 2  + [z/(x+y)]a 2 b 2  \geq 8[ABC] 2
THANKS!

x\leq y \leq z
c \leq b \leq a
From Cebasev we have:
 \sum x/y+z * \sum a 2 b 2  *1/3 \geq 8S 2
 \sum a 2 b 2  \geq 16S 2
 \sum a^4 \geq  \sum a 2 b 2

I don't think you can order all 6 variables.


How's Guantanamo Bay...

ignore this

uber boredom

That  mightve been because Missouri is a boring state with no activity.

We should seriously have an AOPS brawl tournament.

[quote="mewto55555"][quote="xyzzy"]Except for the whole "Brawl is an incredibly boring game after the first two minutes" thing.[/quote]

Only if you are a n00b. Or have a short attention sp-Ooh look a butterfly![/quote]

Hey make your 1000th post here! You're at 999!!

I'm sorry! I wasnt keeping track (also i made like 995 here)

Well, I still have 100+ to go. Who cares.

Idk, post-counting people? Did I ever get around to holding that brawl tournament described above?

[quote="math154"]Well, Zev started it!!! :rotfl:  :|  :ninja:[/quote]

Zev has started a lot of things. Anyway, we will not fail Trajectory again. I=beast.

Well, explain the physics of your trebuchet. :wink:

Bah trebuchet=Noob! I'll make a rough pic:

\
..\
....\0
.....\/ <---rubber band

Ive made the chute and created a slit to attach the rubber band. I'll show yuou some time.

Uh huh. Time to study for the vocabulary quiz tomorrow.

The aloof boy bolted out of the school door with contempt and indignation, as his nonchalant, deleterious, credulous teacher who, he wistfully dreamt, would rue the incident, had, for some unfathomable reason, dissected his trebuchet and wasted his time.

Gah that was too easy. Matching? Nice job on SRI btw :rotfl:

Excuse: Like 1 month ago, my account was screwed up and it didn't log in. So I waited 20 minutes, and they finally let me go to Mr. Lackey. Then, I wasted 20 more minutes, and went back. I had about 2 minutes to take the test. I ended up having to guess, thinking they wouldn't let me take the test. I ended up failing, AND not finishing.  :wink:

=Person set your first question at 1499.

No I have SS 5th period. Stupid independent study. Although better than being with people who pronounce the word on the map uhsuh yet say el oh el.

I tested out with a 73%. Now I have to take the revolutionary war pretest.

1. Was the revolutionary war an example of

a. A revolution
b. A war on terror
c. A war to destroy/weaken ones enemies

While the kids in my class miss the obvious reference to the war in Iraq, these same kids then ask what revolution means. And then get a 20 on a test filled with these quesitons.


calculus
integration
trigonometry
calculus computations

Evaluate $ \int_\frac{\pi}{4}^{\frac {\pi}{3}} (\sin x \plus{} \cos x)\sqrt {\frac {e^x}{\sin x}}\ dx$.

Integration by parts, non?

I feel lazy to compute the value but I wrote it out and I think the idea behind it is more important.  :lol:

[quote="kunny"][quote="§outh§tar"]Integration by parts, non?:[/quote]

Oui, c'est vrai. :lol:[/quote]

Bon appetit mon ami!  :|

Assuming no mistakes: $ 2e^{{\textstyle{x \over 2}}} \sqrt {\sin x} \left| {_{{\textstyle{\pi  \over 4}}}^{{\textstyle{\pi  \over 3}}} } \right.$

hello, i have found $ \minus{}\sqrt{2}\cdot3^{\frac{1}{4}}\cdot e^{\frac{\pi}{6}}\plus{}e^{\frac{\pi}{8}}\cdot2^{\frac{3}{4}}$.
Sonnhard.

Sonnhard probably just reversed the bounds (it should have been $ \sqrt{2}\cdot3^{\frac{1}{4}}\cdot e^{\frac{\pi}{6}}\minus{}e^{\frac{\pi}{8}}\cdot2^{\frac{3}{4}}$): the solution is equivalent to one §outh§tar gave (and it's correct).

By setting $ u \equal{} e^{x} \sin x$, I get the following:
$ \int_\frac {\pi}{4}^{\frac {\pi}{3}} (\sin x \plus{} \cos x)\sqrt {\frac {e^x}{\sin x}}\ dx \equal{} 2\left(\sqrt {\frac {\sqrt {3}}{2}e^{\frac {\pi}{3}}} \minus{} \sqrt {\frac {\sqrt {2}}{2}e^{\frac {\pi}{4}}} \right )$.

hello kunny, we are all waiting for your next integral!
Sonnhard.


algebra
polynomial
linear algebra
matrix

Let $ f_{A}(\lambda)$ be the characteristic polynomial of the matrix $ A$, $ f_{A}(\lambda) = \det (\lambda I-A)$.  Prove $ f_{AB}= f_{BA}$ for all same size square matrices.

This is a pretty common question.  See here:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=64324[/url]


superior algebra
superior algebra unsolved

Is there any group statisfying the condition :
The group has exactly $ 5$ elements of order $ 5$.
Thanks in advance.

If $ p|n$ and $ p\not \in \{2,3\}$ then there is no group containing exactly $ n$ elements of order $ n$

The number of elements or order $ p$ (prime) is always $ \equiv \minus{}1 \mod p$ (or just $ 0$). More generally, $ x^n\equal{}e$ has a number of solutions that is divisble by $ n$ if $ n$ divides the group's order.


algebra
polynomial
real analysis
real analysis unsolved

Evaluate the sum
$ \displaystyle\sum_{n\equal{}1}^{\infty}\frac{n^2\plus{}1}{2^{n}n!}x^n$.

Make the substitution $ x \mapsto 2x$ to kill off the $ 2^n$.  Then [hide]we have \begin{eqnarray*}\sum_{n \geq 1} \frac{n^2 + 1}{n!}x^n & = & \sum_{n \geq 1} \frac{n(n-1)}{n!}x^n + \sum_{n \geq 1} \frac{n}{n!}x^n + \sum_{n \geq 1} \frac{1}{n!}x^n \\
& = & x^2 \sum_{n \geq 0} \frac{x^n}{n!} + x\sum_{n \geq 0} \frac{x^n}{n!} + \left(-1 + \sum_{n \geq 0} \frac{x^n}{n!}\right) \\
& = & (x^2 + x + 1)e^x - 1.
\end{eqnarray*}

Any similar problem can be solved identically: it simply amounts to changing bases in the space of polynomials in $ n$, from the standard basis $ \{1, n, n^2, n^3, \ldots\}$ to the basis of falling powers, $ \{1, n, n(n-1), n(n-1)(n-2), \ldots \}.$[/hide]


[quote="arqady"][quote="CSS-MU"]Let $ a,b,c$ be nonnegative numbers and $ ab \plus{} bc \plus{} ca \equal{} 1$. Prove:
\[ \frac {1}{(1 \plus{} a^2)^2} \plus{} \frac {1}{(1 \plus{} b^2)^2} \plus{} \frac {1}{(1 \plus{} c^2)^2} \le 2\]
[/quote]
$ \sum_{cyc}\frac {1}{(1 \plus{} a^2)^2} \le 2\Leftrightarrow\sum_{cyc}\frac {1}{(a \plus{} b)^2(a \plus{} c)^2}\leq\frac {2}{(ab \plus{} ac \plus{} bc)^2}\Leftrightarrow$
$ \Leftrightarrow(ab \plus{} ac \plus{} bc)^2\sum_{cyc}(a^2 \plus{} ab)\leq\prod_{cyc}(a \plus{} b)^2.$
Let $ a \plus{} b \plus{} c \equal{} 3u,$ $ ab \plus{} ac \plus{} bc \equal{} 3v^2$ and $ abc \equal{} w^3.$
Hence, $ (ab \plus{} ac \plus{} bc)^2\sum_{cyc}(a^2 \plus{} ab)\leq\prod_{cyc}(a \plus{} b)^2\Leftrightarrow w^3\leq9uv^2 \minus{} \sqrt {27(3u^2 \minus{} v^2)v^4}.$
But $ w^3$ gets a maximal value when two numbers from $ \{a,b,c\}$ are equal.
Id est, it remains to prove that
$ (ab \plus{} ac \plus{} bc)^2\sum_{cyc}(a^2 \plus{} ab)\leq\prod_{cyc}(a \plus{} b)^2$ for $ b \equal{} c \equal{} 1,$
which gives $ 4a^3 \plus{} 3a^2 \plus{} 2a \plus{} 1\geq0.$[/quote]
Thanks for your solution, Michael Rozenberg! From this problem we have another:
$ (a\plus{}b)^2(b\plus{}c)^2(c\plus{}a)^2 \ge (a^2\plus{}b^2\plus{}c^2\plus{}ab\plus{}bc\plus{}ca)(ab\plus{}bc\plus{}ca)^2$ with $ a,b,c \ge 0$
 :lol:

to $ (a+b)^{2}(b+c)^{2}(c+a)^{2}\ge (a^{2}+b^{2}+c^{2}+ab+bc+ca)(ab+bc+ca)^{2} $

it is
$(a+b+c-\frac{abc}{ab+bc+ca})^2\ge a^2+b^2+c^2+ab+bc+ca$

but
$(a+b+c-\frac{abc}{ab+bc+ca})^2-(a^2+b^2+c^2+ab+bc+ca)$
$\ge ab+bc+ca-2\frac{abc(a+b+c)}{ab+bc+ca}$
$=\frac{a^2b^2+b^2c^2+c^2a^2}{ab+bc+ca}$

hence we are done with the case of equality a=b=0,or b=c=0,c=a=0

[quote="CSS-MU"]Let $ a,b,c$ be nonnegative numbers and $ ab\plus{}bc\plus{}ca\equal{}1$. Prove:
\[ \frac{1}{(1\plus{}a^2)^2}\plus{}\frac{1}{(1\plus{}b^2)^2}\plus{}\frac{1}{(1\plus{}c^2)^2} \le 2\][/quote]
The inequality equivalent to
\[\sum_{cyc}(a^{2}+ab)\leq\prod_{cyc}(a+b)^{2}.\]
Putting $p=a+b+c, \ q=ab+bc+ca, \ r=abc.$
It become $(p-r)^2\geq p^2-1\Leftrightarrow r^2-2pr+1\geq 0.$
It is true because $1=(ab+bc+ca)^2\geq 2(a+b+c)abc.$
Don't have equality. :(

Let $ a=tan\frac{A}{2} , $and so on,$ A,B,C $ are angles of  triangle$ ABC $

the inequ become $ (cosA+1)^{2}+(cosB+1)^{2}+(cosC+1)^{2}\leq 8 $

or $ cos^{2}A+cos^{2}B+cos^{2}C\leq 5-2(cosA+cosB+cosC) $


hm...
[hide="hm..."]
the equation is

x(1) + y(5) + z(10) = 23

x + y + z = 6

hm...

to solve this equation, you need one piece more piece of information
hm....

wait, if you try without dimes, it will be impossible to get the answer
and if you have two dimes, it's still impossible

so we have 1 dime,

eliminating from the equation, we get a new set

(1)x + (5)y = 13
x + y = 5
now there you go
so we need
2 nickels
[/hide]

[quote="jimli"][quote="chess64"][quote="Silverfalcon"]If Suzie has 6 coins worth 23 cents, how many nickels does she have?[/quote]
The greatest amount she can have is 4 nickels. That's impossible.
If she has 3 nickels, then 5 non-nickel coins must add up to 8 cents. Impossible.
If she has 2 nickels, then 5 non-nickel coins must add up to 13 cents. Impossible.
If she has 1 nickel, then 5 non-nickel coins must add up to 18 cents. Impossible.
If she has 0 nickels, then 5 non-nickel coins must add up to 23 cents. Possible:2 dimes and 3 pennies.

She has 0 nickels.[/quote]

huh? i found that 2 nickels a dime and 3 pennies add to 23 cents.  2 nickels.[/quote]
Thats what i got too. It is correct.

[quote="chess64"][quote="Silverfalcon"]If Suzie has 6 coins worth 23 cents, how many nickels does she have?[/quote]
The greatest amount she can have is 4 nickels. That's impossible.
If she has 3 nickels, then 5 non-nickel coins must add up to 8 cents. Impossible.
If she has 2 nickels, then 5 non-nickel coins must add up to 13 cents. Impossible.
If she has 1 nickel, then 5 non-nickel coins must add up to 18 cents. Impossible.
If she has 0 nickels, then 5 non-nickel coins must add up to 23 cents. Possible:2 dimes and 3 pennies.

She has 0 nickels.[/quote]

Its two.. chess got that cuz he kept saying '5 non-nickel coins' - its not. when its 4 nickels its [b]2[/b] non nickel, and when its 3 nickels, its [b]3[/b] non nickel, and when its 2 nickels, its [b]4[/b] non - nickels, etc... answer is 2

[quote="chess64"]Oops that was stupid!  :blush:[/quote]

lol
stupid
lol

[quote="ToBe"][quote="Silverfalcon"][quote="ToBe"]silverfalcon got that cuz he kept saying '5 non-nickel coins' - its not. when its 4 nickels its 2 non nickel, and when its 3 nickels, its 3 non nickel, and when its 2 nickels, its 4 non - nickels, etc... answer is 2[/quote]

What the heck you talking about?

I didn't say ANYTHING about this problem's answer yet.

[hide="View"]

1*10+2*5+3*1 = 10+10+3 = 23.

Therefore, 1 dime, 2 nickels, and 3 pennies. Thus, 2 nickels is our answer.[/hide][/quote]

sorry.. when i was quoting it i accidentally looked at the wrong part of the quote since he quoted u - its chess, i know. i changed it 20 min ago[/quote]

No prob. I just wondered what you meant. :)

[quote="4everwise"]2 nickels.

1 dime, 2 nickels, and 3 pennies adds up to 23 cents.[/quote]

Thats the easiest way.  Theres obviously 3 pennies. now we have 3 that add up to 20.  Obviously, 1 dime, 2 nickels


modular arithmetic
number theory unsolved
number theory

Show that the equation $ y^{37}\equiv x^3\plus{}11 \pmod p$ is solvable for every prime $ p$, where $ p\leq100$.

It is at least (but not at most) solvable for all primes $ p \not\equiv 1 \mod 37$ and $ p \not\equiv 1 \mod 3$, thus by all primes not $ \equiv 1 \mod 111$.
By the way, it is solvable for all but finitely many primes.


I read somewhere that it is more difficult for seniors to qualify for the USAMO than students in other grades.  Is this true?  If so, how?

for seniors and juniors you need the index score that is your amc12 score + 10 times your aime score.

It's considerably more difficult to qualify for MOP. You must be in Black MOP.


I'm doing an biology internship that's taking up six hours a day.  It's interesting but sometimes rather tedious and I wish I could fit in more time for math...is it worth the effort?

thanks!

of course you're using your time wisely. The internship is only for the summer, but you can do math all year long. Just enjoy the experience and remember that lab experience is invaluable


A climber of weight 650 N is rappelling down a cliff. The rope makes an angle of 18 degrees with the vertical and the cliff exerts a force Fn on the feet of the climber; this force is directed 18 degrees above the horizontal. Determine the tension in the rope.

Im not exactly sure how to find Fn and tension.

It seems to me that we have a system of equations.
T = tension in rope and N = normal force exerted by the cliff and W = weight force.

T * sin 72 + N * sin 18 - W= 0
T * cos 72 - N * cos 18 = 0

We know this because there is no velocity in either the x or y direction when the climber's feet is on the cliff.

We know W and a s a result we can just solve for T and N.

Another way to do this would be to rotate your axis coutner clockwise by 18 degrees.
Then  we solve it by the following:
T - W cos 18 = 0
N - W sin 18 = 0

Ideally both methods will give you the same answer.


Google ARML and try to get to the page from there. Interesting...

Also, I tried to go there on Firefox and Firefox blocked it, listing it as an attack site.

And then, the next time I visited it, Firefox crashed.

The above incidence may have just been a random thing (as when I rebooted Firefox and went there again, it just listed it as being an attack site and didn't do anything.)

EDIT: Then, I chose to ignore the warning and skip to the site just to see what would happen (Google said it'd been safe in the last 90 or so days.) And Firefox instantly closed itself.

EDIT2: Darn I misread it. It resulted in malicious content being installed on 10/24/09.  Scanning computer now...

Comments?

I get blocked...

EDIT: Could it have been hacked?

Yeah, I noticed this too.

It probably was hacked. I wonder why they targeted ARML and not something else...

Does anyone know of a new website that ARML has migrated to, in the meantime?

I got the following response to an email message:

[quote="Mike Curry"]Hi. We are aware of it and are working to correct it.  Thanks for the
feedback.  The problem should be corrected soon.   Thanks again.  Mike.[/quote]

I doubt it. We should wait until ARML organizers tell us it is safe to use.

See [url=http://arml.com/]here[/url]; apparently "the malicious code has been removed".

People that are malicious people use codes that are malicious in nature.

While we are on the subject of the website: Are (were) there any past problems other than last year's?

There are, but I think ARML has a copyright on them. They're probably going to get released in a book soon (if not already)...

There WERE, but ARML took them down after they made the new book.


$ABCD$ is a rectangle of area 2.  $P$ is a point on side $CD$ and $Q$ is the point where the incircle of $\triangle ABC$ touches the side $AB$.  The product $PA \cdot PB$ varies as $ABCD$ and $P$ vary.  When $PA \cdot PB$ attains its minimum value,
a) Prove that $AB \geq 2BC$,
b) Find the value of $AQ \cdot BQ$.

a)
Denote $x=AB, y=BC, u=PC, v=PD$, Then
$(PA\times PB)^2=\sqrt{(u^2+y^2)(y^2+v^2)}\geq uy+yv=xy=2$
If and only if $uv=y^2$, then equality holds.
Because $(u+v)^2\geq 4uv$, i.e, $x\geq 2y$


geometry
Harry Potter
IMO Shortlist
Spiral Similarity
IMO 2005
IMO
Waldemar Pompe

Let $ABCD$ be a fixed convex quadrilateral with $BC=DA$ and $BC$ not parallel with $DA$. Let two variable points $E$ and $F$ lie of the sides $BC$ and $DA$, respectively and satisfy $BE=DF$. The lines $AC$ and $BD$ meet at $P$, the lines $BD$ and $EF$ meet at $Q$, the lines $EF$ and $AC$ meet at $R$.

Prove that the circumcircles of the triangles $PQR$, as $E$ and $F$ vary, have a common point other than $P$.

The point is the intersection of the perpendicular bisectors of $AC,BD$, I think.

Let $T$ be the point described above. It's clear that $TA=TC,TB=TD$, and $\angle ATC=\angle BTD$, so the triangles $TAC,TDB$ are similar (and $TAC$ is the image of $TDB$ through a spiral similarity centered at $T$). It's also easy to see that $TFE$ is similar to these two triangles.

Sorry if I skip some steps, but they all fall in the wide category of angle chasing :).

$\angle TFR=\angle TAR$, so $TAFR$ is cyclic, so $\angle TRA=\angle TFA$. In the exact same way we show that $\angle TQB=\angle TEB=\pi-\angle TEC=\pi-\angle TFA$ (the figures $TBCE,TDAF$ are congruent), and from these two relations we find that $\angle TRP+\angle TQP=\pi$, i.e. $T$ lies on all the circles $PQR$.

Oh, such easy geometry problem, still not commented by Darij.

It may be solved by very standard methods. For proving that the circles which have one point have another we must prov that their centres lie on one line (which does not contain point $P$, what I am not going to check). Coordiantes of a centre in appropriate (not-Descartues, but skew) coordinate system is $(PR/2,PQ/2)$ (segments $PR,PQ$ are oriented of course). These are just projections of a vector $PO$ to diagonals of our initial quadrilateral. So, we are searching for a linear dependence between $PR$ and $PQ$, or, what is the same, a dependence between $CR$ and $DQ$.

Let $BC$ intersect $DA$ in point $K$, wlog $B$ lies betwee $K$ and $C$, $A$ between $K$ and $D$. By Menelaus for $KAC$ we have $AR\cdot CE\cdot KF=RC\cdot EK\cdot AF$. Since $CE=AF$, we get $AR/RC=EK/KF$, $(AC-RC)/RC=EK/KF$, $AC/RC=1+EK/KF$, $RC=AC\cdot \frac{KF}{KF+EK}$. $KF+EK$ is constant and equals $KA+KB+BC$, so $RC$ is proportional to $KF$. Analogously, $DQ$ is proportioal to $KE$. But since $KE$ and $KF$ have fixed sum, it implies a linear relation for $CR$ and $DQ$.

First, BE/EC = DF/FA is sufficient. Let T be the center of the spiral similarity S that sends D to B and A to C.
By a well-known lemma, T is the second point of intersection of the circumcircles of triangles APD and BPC. Clearly, S also sends F to E.
Therefore, the circumcircles of triangles FQD and BEQ also pass through T. Now, because T lies on the circumcirles of triangles FQD and APD,
by Miquel's theorem, it also lies on the circumcircles of triangles PQR and ARF. Therefore, the circumcircle of triangle PQR always passes through
the second point of intersection of the circumcircles of triangles APD and BPC.

:D   Similar solution...
I'd like to say about something interesting...Here!A picture of this problem!

[img]http://www.sjiedu.com/blog/uploadfile/2005715183143277.gif[/img]

Several circles go through the point(S in the picture) we want.
I hope this could make someone think of a special solution...

P.S.  Has the friend upstair read the book Morden Geometry By R.A.Johnson?Such similar thought we had...exactly the same...

Nice problem ! :)
  My solution use vector and Simson's theorem :
 Let $ (APD) $ intersect $ (BPC) $ at $ K $ . We'll prove $ K \in (PQR) $
 It's easy to show : $ KAC , KBD , KEF $ are isosceles triangles .
 Let $ T , U , V $ be the feet of the perpentdiculars from $ K $ to $ BD, EF, AC $ , respectively .
  So $ T, U , V $ are the midpoints of $ BD , EF , AC $ , respectively . Using vectors , we can easily show that : $ T, U ,V $ are collinear .
  By Simson's theorem , we have $ K \in (PQR) $ . Q.E.D  :)  :lol:

Dear Valentin , which country proposed this problem ?  :)

As far as I know, this problem is from Poland, also problem 4.

I solved this problem this way:

By three good diagrams, I guessed that the second intersection point $S$ is the intersection of the circumcircles of $APC$ and $BPD$. Well, check Skyward_Sea's diagram above.

I just did this one (on Paint, so it's not really accurate)

[img]http://www.geocities.com/cyshine/problema5.jpg[/img]

Let's prove that $S$ is indeed the desired point, that is, that $PQSR$ is a cyclic quadrilateral. Since $S$ does not depend on $E$ and $F$, $S$ is a fixed point, so if we manage to prove that $S$ belongs to the circumcircle of $PQR$ regardless of the positions of $E$ and $F$ we are done.

Let $\angle CPS = \alpha$ and $\angle DPS = \beta$. Doing some angle chasing (look at the diagram!), we see that $ADS$ and $CBS$ are congruent triangles. Moreover, $ADS$ can be obtained from a rotation of $CBS$ with rotation center $S$. Notice that the image of $E$ in this rotation is $F$, so $\angle ESF = \angle BSD$ (both are the rotation angle). Thus, since $ESF$ and $BSD$ are isosceles, they are similar and, in particular, $\angle QFS = \angle EFS = \angle BDS = \angle QDS$. Thus $QFDS$ is a cyclic quadrilateral and hence $\angle RQS = \angle FDS = \alpha = \angle RPS$. So $PQRS$ is cyclic, as we wanted to prove.

Oh, I just noticed that my solution is similar to grobber's... but I'll post it anyway.

A friend of mine sent me the following super fast and beatyfull solution, in just 4 lines.
(In fact his solution generalizes this problem!! Can you see how ? ) Let M,N be the midpoints of the diagonals.
We'll prove that the readical axis of the circles C1=(P,Q,R) and C2=(P,M,N) is constant line and so we are done. Particularly, we'll prove that this line cuts DC at L, which is a constant point.
From Cayley's Theorem for these circles and the points D,C we have that :
1. $2 \cdot O_1 O_2 \cdot h_1 = DM \cdot DP - DQ\cdot DP $.
2. $2 \cdot O_1 O_2 \cdot h_2 = CR \cdot CP - CN\cdot CP $. where $h_1$ and $h_2$ are the projections of D and C to the radical axis. Then we get that : $\frac{h_1}{h_2}=\frac{LD}{LC}=\frac{DP}{CP} \cdot \frac{DM-DQ}{CR-CN}$. So we only need to prove that the quantity $\frac{DM-DQ}{CR-CN}$ is constant, which obvious since from Menelaous theorem with respect to the triangles APD and PBC and the lines F,Q,N and Q,R,E respectivly, we get $\frac{QD}{DB}=\frac{AR}{AC}=s$. So $QD=s\cdot DB$ and $RC=(1-s)\cdot AC$.
So we are done.

Isn't it true that the possible lines EF are all tangent lines to a parabola?
Does anyone have such a solution?
(Then the rest is easy, i.e. showing that the centres of the circles lie on a line..)

Ok, guys, here is my solution to this problem (I think it's very nice and simple).Suppose we have two such triangles $PQR$ and $PLT$ ($Q$ and $L$ are on $AC$). Suppose that the intersection point of the circumcircles of this triangles is$ X$. Now due to the similarity of triangles $PRT$ and $PQL$ we have that $\frac{RT}{QL}=\frac{\sin{TPX}}{\sin{QPX}}$.But on the other hand $\frac{RT}{QL}=\frac{BD}{AC}$. So $\frac{\sin{TPX}}{\sin{QPX}}$ is constant so $\angle{TPX }$is fixed, so all this circles have one common point besides $P$.
Nice and short, isn't it? :)

[quote="martin99"]Isn't it true that the possible lines EF are all tangent lines to a parabola?[/quote]

Yes, this is true. Actually, it is a very nice observation, leading to a different solution of this problem:

We will work in the projective plane. Let t be a varying parameter from $\mathbb{R}\cup\left\{\infty\right\}$. Consider the points $E_t$ and $F_t$ on the lines BC and DA dividing the segments BC and DA in the equal ratios $\frac{BE_t}{E_tC}=\frac{DF_t}{F_tA}=t$ (where the segments are directed, of course). Thus, if $\mathcal{T}$ is the affine transformation mapping the points B and C to the points D and A, respectively, then this transformation $\mathcal{T}$ will also map the point $E_t$ to the point $F_t$. Now, affine transformations are a particular case of projective transformations, and if we have a projective transformation which maps a line g to another line h, then the lines joining the points of g with their respective images (which lie on h) envelope a conic (this is a well-known fact from projective geometry). Thus, for varying t, the lines joining the points $E_t$ on the line BC with their images $F_t$ under the transformation $\mathcal{T}$ envelope a fixed conic. In other words, for varying t, the lines $E_tF_t$ envelope a fixed conic.

We can easily see that this conic is a parabola: In fact, this conic must touch the line $E_{-1}F_{-1}$; but $E_{-1}$ and $F_{-1}$ are points at infinity (since they divide the segments BC and DA in the ratio $\frac{BE_{-1}}{E_{-1}C}=\frac{DF_{-1}}{F_{-1}A}=-1$), so the line $E_{-1}F_{-1}$ is the line at infinity, and a conic touching the line at infinity must be a parabola.

Thus, for varying t, the lines $E_tF_t$ envelope a fixed parabola p.

The point B divides the segment BC in the ratio $\frac{BB}{BC}=0$; thus, $B=E_0$. Similarly, $D=F_0$. The point C divides the segment BC in the ratio $\frac{BC}{CC}=\infty$; thus, $C=E_{\infty}$. Similarly, $A=F_{\infty}$. Finally, since BC = DA and BE = DF, we have EC = BC - BE = DA - DF = FA, and thus $\frac{BE}{EC}=\frac{DF}{FA}$. If we denote $\frac{BE}{EC}=\frac{DF}{FA}=k$, then we thus have $E=E_k$ and $F=F_k$.

Hence, the lines $BD=E_0F_0$, $CA=E_{\infty}F_{\infty}$ and $EF=E_kF_k$ are three of the lines $E_tF_t$ which envelope the parabola p. Hence, the triangle PQR, being formed by these lines BD, CA and EF, is a triangle formed by three tangents to the parabola p.

But according to Ross Honsberger, [i]Episodes in Nineteenth and Twentieth Century Euclidean Geometry[/i], MAA 1995, Chapter 5, §5, the circumcircle of a triangle formed by three tangents to a parabola always passes through the focus of this parabola. Thus, the circumcircle of triangle PQR, formed by three tangents to the parabola p, must pass through the focus of the parabola p. This focus is clearly a fixed point, and it is different from P (else, the focus of the parabola would be the point of intersection of two of its tangents, what is impossible on the real plane).

So the problem is solved. Okay, this is not the shortest solution, but quite a good example of the use of projective geometry on the IMO. Still, I don't know anyone who solved the problem in this way on the exam (before you start wondering: Grobber didn't participate :D ). My own solution was practically the same as Lego's one. Maybe it will later be published in German on the BWM website ( http://www.bundeswettbewerb-mathematik.de/imo/aufgaben/aufgaben.htm ).

  Darij

[quote="darij grinberg"][quote="martin99"]Isn't it true that the possible lines EF are all tangent lines to a parabola?[/quote]

Yes, this is true. Actually, it is a very nice observation, leading to a different solution of this problem:

...
Okay, this is not the shortest solution, but quite a good example of the use of projective geometry on the IMO. Still, I don't know anyone who solved the problem in this way on the exam (before you start wondering: Grobber didn't participate :D ). My own solution was practically the same as Lego's one. Maybe it will later be published in German on the BWM website ( http://www.bundeswettbewerb-mathematik.de/imo/aufgaben/aufgaben.htm ).

  Darij[/quote]

Hello,

thank you for writing down a complete solution using that observation, Darij.
Since I didn't have to write down a solution (my days at the imo are passed a long time...)
I just thought about the problem and didn't formulate a solution.

[img]http://mathworld.wolfram.com/images/eps-gif/ParabolaEnvelope.gif[/img]
So my solution "consists" of this wonderful parabola that comes out by drawing all lines EF
(and I found a picture for all of you on the web...).
I also saw that the rest of the solution is not so short and not so wonderful. (Actually
I used coordinates to do the rest, which is very easy; maybe here your alternative is nicer
(but I don't know the proof from that book).)

Martin

I think the most difficult part of the problem is actually finding the common point, as opposed to showing that the point works.

Quite trivial

Let $X = (APD)\cap (BPC)$. By spiral sim, $(XREC)$ and $(XQFD)$ are cyclic. Now, we have
\[\angle RXQ = \angle RXP + \angle PXQ = (\angle PXC - \angle RXC) + (\angle PXD - \angle QXD)\]
\[= (180 - \angle CBP - \angle BER) + (180 - \angle PAD - \angle RFA) = \angle ARF + \angle BQE = 180 - \angle RPQ\]
Thus $X\in (RPQ)$. Since $X$ is fixed, we conclude the problem.

Let $M=(ADP)\cap(BCP)$ and note that it is the Miquel point of $ABDC.$ Since $M$ is the center of the spiral similarity mapping $\overline{BC}$ to $\overline{AD},$ it is also the center of the spiral similarity mapping $\overline{BE}$ to $\overline{AF}$ and another that maps $\overline{EC}$ to $\overline{FD}.$ Hence, $M$ is also the Miquel point of $BFED$ and $AFEC,$ implying that $AFRM$ and $MFDQ$ are cyclic. Therefore, $$\measuredangle MRP=\measuredangle MRA=\measuredangle MFA=\measuredangle MFD=\measuredangle QMD,$$ and $MRPQ$ is cyclic. $\square$

Okay
Let $M$ be the miquel point of quadrilateral $ABCD$. By [url=https://artofproblemsolving.com/community/u584086h84559p23726057]USAMOOO 2006 #6[/url] we know that $M \in (RAF)$
$$\measuredangle MPQ = \measuredangle MPD = \measuredangle MAD = \measuredangle MAF = \measuredangle MRF \ \text{(Tyres)} = \measuredangle MRQ \implies M \in (PRQ) \ \blacksquare$$

Whattt! Why it's P5?
Let $(BCP)\cap (ADP)=T$. So $T$ is center of spiral similarity $\mathcal{S}$ that sends $BC$ to $DA$. Since $\frac{BE}{EC}=\frac{DF}{FA}$ we get $\mathcal{S}$ sends $E$ to $F$. Also $\triangle TBE \cong \triangle TDF$.
Now from things that I found above we get $QBET, ARTF$ are cyclic $\implies \angle PRT=\angle TFD=\angle TEB=180-\angle PQT \implies PQRT$ is cyclic. So $(PQR)$ passes through $T$ and since $T$ is fixed point we get the desired result.

Define point $X$ as intersection of $\odot(BPC)$ and $\odot(APC)$. We will prove that point $X$ lies on $\odot(PQR)$, which will obviously solve the problem.

[b]Claim: [/b] $\triangle XBC \sim \triangle XDA$
[b]Proof: [/b] This is easy angle chasing. Indeed:
$$ \angle CBX = \angle CPX = \angle XDA \quad \text{and} \quad \angle XAD = \angle XPD = \angle BCX $$
This proves the claim since both of these $2$ triangles has $2$ equal angles.

From the claim, it is to deduce that $X$ is the centre of spiral similarity sending $DA$ to $BC$. Note that:
$$ \frac{DF}{FA} = \frac{BE}{EC} $$
Therefore point $X$ is also the centre of spiral similarity sending $FA$ to $EC$. This implies that point $R=AC \cap EF$ lies on both $\odot(AFX)$ and $\odot(XEC)$. This is enough to conclude since:
$$ \angle PRX = \angle ARX = \angle XFD = \angle XQD $$
implying that quadrilateral $PQRX$ is cyclic as desired.

Consider $X$, the intersection of the perpendicular bisectors of $AC$ and $BD$. Note that $AXC$ and $BXD$ are isosceles. Note that $\triangle AXD\cong \triangle CXB$, so $\angle BXC=\angle AXD$ which implies $AXC=\angle BXD$ so $\angle DBX=\angle ACX$ which implies $PBCX$ cyclic.
\\$~$\\
Similarly, $\triangle AXF\cong \triangle CXE$ which implies $\angle AXF=\angle CXE$ so $\angle EXF=\angle AXC$, which implies $\angle FEX=\angle ACX$, implying $ECXR$ cyclic. Thus, $X$ is miquel point of $BERP$ which implies $PQRX$ cyclic as desired.

Let $M$ be the Miquel Point of self-intersecting quadrilateral $ADBC$. We will show that $M$ is the desired fixed point.

[b][color=#f00]Claim.[/b][/color] $M$ is the Miquel Point of $PRBE$.

Notice that $M$ takes the spiral congruence $\overline{AD} \to \overline{BC}$, so it sends $\overline{DF} \to \overline{BE}$ as well. Thus $MQBE$ is cyclic. By definition, $MPBC$ is cyclic as well, so the claim is proven. $\blacksquare$

Thus $PQRM$ is cyclic by definition.

Let $T$ be the intersection of the perpendicular bisectors of $AC$ and $BD$

[b]Claim: $T$ is the center of a spiral sending $A \rightarrow C$, $D \rightarrow B$, $F \rightarrow E$ [/b]
Proof:

Note that by definition, $TB=TD$, $TA=TC$, $AD=BC$

So, $\triangle DTA \cong \triangle BTC$

Which implies that $T$ is the center of the spiral

Since $EC=AF$, it means that $T$ sends $ F \rightarrow E$

[b]Claim: $AFTR, ECTR, FDTQ,BQTE$ are all cyclic quadrilaterals[/b]

Proof:

Sicne $T$ is the center sending $F \rightarrow E$, $A \rightarrow C$,
$\triangle FTE  \sim \triangle ATC$

Which implies that  $\angle RFT = \angle EFT = \angle TAC = \angle TAR$

As such, $RAFT$ is cyclic. By symmetry, the other three quadrilaterals are also all cyclic

[b]Claim: $PQTR$ cyclic [/b]

Proof:

From the earlier claim, $$\angle ART= \angle TFD $$
$$=\angle TEB$$
$$= 180 - \angle BQT$$
$$=180- \angle PQT $$

So, $PQTR$ concyclic

Significantly misplaced problem, it took me 10 minutes altogether and spiral sim trivializes everything lol
By spiral sims, we get free AF->CE, DF->EB, AD->BC all with center S, the second intersection of circles ADP and CBP. Hence we get cyclic quads ADPS, BCPS, and because spiral similarities are unique, if S' is the intersection of (AFR) and (BCR), S' takes AF->CE and so S=S', etc. with the other cyclic quads to get AFRS, BCRS, CDFS, BEQS. Then $$SQP=SCD=180-SFD=SFA=SRA=180-SRP,$$ so PQSR is cyclic, meaning that X is the fixed point that is independent other than P (since it's only dependent on ADP and BCP), as desired. $\blacksquare$

We claim that the intersection of the circumcircles of $APD$ and $BPC$ is the common point of the circumcircles of $PQR$.
Define $(APD)\cap(BPC)=Y$. We will first observe that it suffices to prove that $PQYR$ is a cyclic quadrilateral, as this would mean that for every position of $E$ and $F$, $(PQR)$ passes through $Y$.
In addition, if $Y$ is the Miquel point of $APQF$, $PQYR$ is a cyclic quadrilateral, and vice versa. Thus, it is sufficient to show that $Y$ is the Miquel point of $APQF$.

There exists a spiral similarity that sends $DA$ to $BC$, and since $\frac{DF}{FA}=\frac{BE}{CE}$, $\triangle EYC \sim \triangle FYA$. An obvious result of this is that $\frac{EY}{YC} = \frac{FY}{YA}$.
We then do some simple angle chasing:
\begin{align*}
\angle AYC &= \angle AYE + \angle EYC \\
&= \angle AYE + \angle FYA \\
\angle AYC &= \angle FYE
\end{align*}
Then, by $SAS$ similarity, $\triangle AYC \sim \triangle FYE$, which means that $Y$ is the spiral center of a spiral similarity sending $FE$ to $AC$.
Because of this, $ARYF$ is a cyclic quadrilateral, and $(ARF)$ passes through $Y$. From here, it is easy to see that $Y$ is the Miquel point of $APQF$, and we are done. $\blacksquare$

Let the intersection of (APD) and (BPC) be point X. It is easy to see X is the spiral similarity center of quadrilateral ABCD, thus AXD and CXB are similar, and since AD=CB this implies they are congruent. Thus it is deduced that AXF and CXE are also congruent. Now, let the intersection of (AXF) and (CXE) be R'. Then, note <ER'X=<ECX=<FAX=<FR'X implying E, R', and F are collinear, and similarly, C, R', and A are collinear, implying R'=R. This means that AFXR is a cyclic quadrilateral, but APXD is also a cyclic quad, thus <XPQ=<XPD=<XAD=<XAF=<XRF=<XRQ, thus PQXR is cyclic, and since X clearly stays fixed as E and F vary, we are done.

Let $M = (APD) \cap (BPC)$ so that $M$ is the center of the spiral similarity which maps $\overline{AD} \mapsto \overline{CB}$. The length conditions imply this spiral similarity also maps $\overline{AF} \mapsto \overline{CE}$, which implies that $FARM$ is cyclic. Since $APMD$ is also cyclic, this means that $M$ is Miquel point of $APQE$, so $M$ also lies on $(PQR)$.

We claim the desired point is the Miquel point of degenerate quadrilateral $ACBD$, or the center of spiral similarity mapping $AC \rightarrow FE \rightarrow DB$, which exists through the length condition.

Consider Miquel on quadrilateral $APQF$. Three of the circles already pass through $M$ from the spiral similarity mentioned above; hence the final circle, $(PQR)$, also will always pass through this fixed point. $\blacksquare$

Very easy Problem
 [hide=Sketch] Let O be center of spiral similarity that sends AD to BC.
Therefore O also sends DF to BE.
 The intersection of EF and BD is one of the intersection of these circles.
Therefore DFQO and BPOE cyclic. Now just angle chase and we are done [/hide]


In the net drawn below, in how many ways can one reach the point $ 3n\plus{}1$ starting from the point $ 1$ so that the labels of the points on the way increase?

call each pair of numbers, (3n, 3n-1) a tier. to get to a number 3n+1 you must have just passed through the tier (3n, 3n-1) so this problem is equivalent to how many ways can you go from 1 to the nth tier on a strictly increasing path. It is obvious that any path to the Nth tier must pass through all previous tiers. we can also see that there exactly 3 paths from tier n to tier n+1, If we start at 3n the paths are (3n)-(3n+3), (3n)-(3n+1)-(3n+3), (3n)-(3n+1)-(3n+2), if we start at 3n-1 the paths are (3n-1)-(3n+2), (3n-1)-(3n+1)-(3n+2), (3n-1)-(3n+1)-(3n+3). also there are obviously 2 ways to go from 1 to the first tier.

so the number of ways to get to the nth tier and consequently he number of ways to get to 3n+1 is 2*3^(n-1)

The solution given by latios315 using multiplication rule is right. We may also follow a different approach by partitioning the number of ways of reaching $ 3n\plus{}1$ by the number of intermediate points of form $ 3n\plus{}1$ reached while a certain journey. Between these intermediate points, we either move on the upper line or on the lower line. Hence we have

\[ f(n)\equal{}{n\minus{}1\choose 0}2^1 \plus{} {n\minus{}1\choose 1}2^2 \plus{} \dots \plus{} {n\minus{}1\choose n\minus{}1}2^{n\minus{}1}\equal{}2.3^{n\minus{}1}\]

PS:
Its interesting to note that,
$ f(1)\equal{}2$,
$ f(2)\equal{}2\plus{}f(1)f(1)$,
$ f(3)\equal{}2\plus{}f(1)f(2)\plus{}f(2)f(1)\plus{}f(1)f(1)f(1)$,

and so on where every composition appears in the recursion.


I don't like to use LaTeX, partially because I don't know all the controls and I'm not very fast with it. I know it makes it much more clear, but it's a pain to test out the latex.

It's optional.
You can write your solutions and fax them.

The solutions can be mailed or sent in a printable electronic format; the graders will see a hard copy, and that's what matters.

There are typically a few student solutions they reproduce; only solutions in $ \text{\LaTeX}$ are considered for that, since it's a lot easier for them to format that way.
In every other respect, the format doesn't matter as long as it's readable.

Using LaTeX is [b]not[/b] required.  You can handwrite your solutions.  We request that if you are handwriting your solutions, that you print out and use [url=http://www.usamts.org/Downloads/USAMTS_solutions.pdf]our Solutions form[/url], and that you write in [b]black ink[/b].

We can also print out solutions onto the solutions form, instead of writing it, and mail, right?


algorithm
modular arithmetic
number theory
number theory unsolved

Find all positive integer pairs $(a,n)$ such that $\frac{(a+1)^n-a^n}{n}$ is an integer.

$(a+1)^n=a^n (mod \ n) \Longrightarrow (a,n)=1,b^n=1(mod \ n)$, were a(b-1)=1(mod n). It give algoritm:
If (c(n),n)=d>1 we have not trivial solutions b^d=1(mod n). If (b-1,n)=1, it give solutions a=1/(b-1) (mod n).

ehm...i don't understand;here is my solution:

at first, let's suppose $(a;n)=1$

take the smaller prime factor of $n$ and let's call it $p$, with $n=p^ck$

so, it must be

$(a+1)^{p^ck}\equiv a^{p^ck}$ $(\mod p)$

by Fermat's theorem

$(a+1)^{k}\equiv a^{k}$ $(\mod p)$

$[(a+1)\cdot a^{-1}]^k\equiv 1$ $(\mod p)$

so the smaller number $x>0$ such that $[(a+1)\cdot a^{-1}]^x\equiv 1$ $(\mod p)$ is a divisor of $k$.
we now also that that this number x is a divisor of $\phi(p)=p-1$, but all divisors of $k$ are greater than $p$ and so $x=1$.

$(a+1)\cdot a^{-1}\equiv 1$ $(\mod p)$

$a+1\equiv a$ $(\mod p)$

$1\equiv 0$ $(\mod p)$

so this smaller prime does'n't exist,and there can be only $n=1$.
in fact, with $n=1$ we have solution for every $a$.

it remain only the case of $(a;n)=d>0$

but in this case the numerator becomes $\equiv 1$ $(\mod d)$ and so it cannot be a multiple of $n$.

the only solution are so $(a;1)$ for each $a\in\mathbb{N}$

[quote="Omid Hatami"]Suppose $p$ is the smallest prime divisor of $n$ the $p|(a+1)^n-a^n\Rightarrow p|(a+1)-a=1$. So contradiction proves that $n=1$[/quote]

So simple and elegant, excellent!!!!!

Another thing... These three problems are the only 3 or there is another round in China TST?? And another question: These 3 problems are really difficult to solve, aren´t they?????? (as any China TST) or any high-school olympiad student can solve them?????

dear omid :
 I dont follow your solution could you put more details in?

  p.s. your solution is very  interesting.

$p|(a+1)^n-a^n\Rightarrow p|(a+1)-a=1$ is definitely correct, but I think you have jumped too many steps [i]Omid Hatami[/i].  :P

To see why this is correct, we first let $b$ be the unique inverse such that $ab\equiv 1 \pmod{p}$. We have $(a+1)^n \equiv a^n \pmod{p}$, and multiplying both sides by $b^n$, we get $((a+1)b)^n \equiv 1 \pmod{p}$. Letting $d$ be the order of $((a+1)b) \mod p$, we have $d\mid n$. Yet, by Fermat's Little Theorem, we have $((a+1)b)^{p-1}\equiv 1 \pmod{p}$, so $d\mid p-1 \Rightarrow d < p \Rightarrow \gcd (d,n)=1$, due to the definition of $p$ as the [u]smallest[/u] prime divisor of $n$. Thus $d=1$, implying that $(a+1) \equiv a \pmod{p}$, which is the contradiction we want.

obviously $n$ isn't even. Let $p$ be the smallest prime divisor of $n$.and let $g$ be a primitive root modulo $p$ . Let $a+1\equiv g^k,a\equiv g^l \pmod{p}$ so ve have :
$g^{kn} \equiv g^{ln}\pmod{p}$ so $p-1|n(k-l)$ so $p-1|k-l$.so$p|g^k-g^l$ so $p|(a+1)-a$ so $p=1$.

In fact there are 8 round in the China TST and in the first six round every problem is worth 7 points and in the last 2 days every one 21.So when we say China TST,we often means the last two days.And this 3 problem is the first round of 8!
First two problems is very easy but I think no one can complete the third one in the exam.
This 3 problems is from Zuming Feng,the IMO team leader of USA in 2005.

I see jin... and what's the difficulty of this problem????? (really hard?)Is it like what number of an IMO problem?????? or it is easier like an IMO problem?

This one?the second one is really easy, I think almost every one in the exam can solute it.
But I made a very silly mistake(forgot n=1) :(

Hmmm ... I am really new to number theory and I can not understand most of what you guys did and when I can I am not always able to figure out how you got there so I tried to go for it myself. Anyways, this is what I got and please tell me where any errors occur.

((a+1)^n - a^n) / n == 0 (mod m)                              (== means congruent to)

multiply both sides of congrunce by n to get

(a+1)^n - a^n == 0 (mod m)

now we can add a^n to both sides to produce

(a+1)^n == a^n (mod m)

I think that this means n must equal 1, therefore we say

(a+1) == a (mod m)

substracting a from both sides we see that

1 == 0 (mod m) which is the same as saying (1-0)/m is an integer

this is only true if m =1 (we are dealing with only positive integers in this problem)

so going back to our original equation we plug in n=1 and m=1 for the integer our equation must yield

((a+1)^1 - a^1)/1 = 1      =>       a+1 - a = 1     =>     1=1
since this is true for any 'a' our solution is

(a,1)

Thank you for any help you can provide me with!

Here are some of my comments on your solution.

From the first line of your solution, you have defined $m$ as some divisor of $\frac{(a+1)^n-a^n}{n}$. Even if you eventually show that $m=1$, you have only shown that $\frac{(a+1)^n-a^n}{n}$ is indeed an integer, which was what you have assumed right from the start. For this problem, you do not want to investigate the divisors of $\frac{(a+1)^n-a^n}{n}$; rather, you want to investigate the divisors of $n$. So you should just let your $m$ be any divisor of $n$, and what you want to show in the end is that this arbitrary divisor of $n$ has to be $1$. Your first line of the solution should just be ${(a+1)^n-a^n \equiv 0 \pmod{m}}$, where $m\mid n$.

Next, we arrive at your step of $(a+1)^n \equiv a^n \pmod{m}$ implying $(a+1) \equiv a \pmod{m}$. Though this is true in reality, you did not provide any justification to this step.

Remember, you want to show that $m=1$ in the end, so we want to assume $m>1$ and derive a contradiction. Now, if we know $m>1$, definitely, $m$ must have at least one prime divisor. The trick we use here is to let $p$ be the smallest such prime divisor of $m$ to derive the contradiction we want. Now we have $(a+1)^n \equiv a^n \pmod{p}$.

We notice that $p$ does not divide both $(a+1)$ and $a$ (Why? Look back at our original problem..), so there must exist a unique inverse of $a$ (Why?). letting $b$ be that unique inverse, we have $ab \equiv 1 \pmod{p}$ by definition. Thus, back to our congruence $(a+1)^n \equiv a^n \pmod{p}$, by multiplying both sides by $b^n$, we get $((a+1)b)^n \equiv 1 \pmod{p}$.

For the rest of the solution, refer to my previous post, or [i]Omid Hatami[/i]'s solution ([i]behzad[/i]'s solution uses a slightly different approach, but it is still similar in idea). Recall that if $d$ is the order of $x \mod y$, then $d$ is the smallest positive integer such that $x^d \equiv 1 \pmod{y}$. If we have $x^M \equiv 1 \pmod{y}$, then we must necessarily have $d\mid M$. (Why? Try convincing yourself on this first.)

I hope you can understand better now.  :)

I really understand the solutions now. Using be was pure genius and I understand why m=1 by the contradiction. Thanks for all the help dblues!

Clearly $\boxed{n=1}$ works. We now show this is the only solution.

If $n>1$, let $p$ be the smallest prime divisor of $n$.

We need $p\mid (a+1)^n -a^n$, so $(a+1)^n\equiv a^n\pmod p$.

Clearly this doesn't hold true if $p\mid a$.

Otherwise, let $b$ be the inverse of $a$ mod $p$.

We have $1\equiv (ab)^n \equiv ((a+1)b)^n \pmod{p}$.

Clearly the order of $(a+1)b$ mod $p$ divides $\gcd(n,p-1)$, so it must be $1$.

Thus, $(a+1)b=ab+b\equiv 1\pmod p$. So $b\equiv 0\pmod p$, a contradiction.

[hide = Proof of China TST 2006 Day 1/2]
[b][u]Claim[/u]: [/b] All the solutions are of the form $(a, n) \in \{(a, 1) : a \in \mathbb{Z^+} \}$.

$\bullet$ Let $p$ be the least prime divisor of $n$.
$\bullet$ There then exists a unique inverse $b \in \mathbb{Z^+}$ such that $ab \equiv 1 \pmod p. \implies$ $(a + 1)^n \equiv a^n \pmod p, \implies ((a + 1)b)^n \equiv (ab)^n \pmod p$.
$\bullet$ Let $m = \operatorname{ord}_p((a + 1)b)$. Then $m \mid n$ and $m \mid p - 1$. Now $\operatorname{gcd} (m, n) = 1$. (This is because as $m \mid p -1$, then $p = mk + 1$ for some $k$, and so $1 = \operatorname{gcd}(p, m) = \operatorname{gcd}(n, m)$).
$\bullet$ Now $m$ must be $1$ as $m \mid n$ and $\operatorname{gcd}(m, n) = 1$.

Now as $m = 1$, we have that \[(a + 1)b \equiv ab \pmod p \implies a + 1 \equiv a \pmod p \implies 1 \equiv 0 \pmod p\], which is a contradiction. Hence $n = 1$, and we are done. [/hide]

Basically the same as taking the smallest prime, but with infinite descent.

We claim that the only possible family of solutions is $(a,n)=(k,1)$ for some positive integers $k$.

Suppose for the sake of contradiction that there are some solutions such that $n>1$.
Let $(a,n)$ be a solution with $n>1$ such that $n$ is minimal among such "bad " solutions.
Then, we define $r \equiv \frac{a+1}{a} \pmod n$ ($a$ and $a+1$ are clearly coprime to $n$)
Also let $m$ be the order of $r$ mod $n$.

Note that $1<m<n$ since if $m=1$, $\frac{a+1}{a}\equiv 1 \pmod n\implies $ contradiction and if $m=n$, $n\mid \varphi(n)$ which is impossible since $\varphi(n)<n$ for $n>1$.

We have $n\mid r^n-1\implies m \mid n$
For all primes $p$,
So, $m\mid n\mid  r^m-1$
Recall that $1<m \mid n$ and $$ a+1 \equiv ar \pmod n \implies a+1 \equiv ar \pmod m \implies (a+1)^m\equiv (ar)^m \equiv a^m \pmod m$$
In summary, $m\mid (a+1)^m-a^m$ and $1<m<n$ which is a contradiction to minimality of $n$.

Suppose $n>1$, and pick the smallest prime divisor $p$ of $n$. We have $(a+1)^n\equiv a^n\pmod{p}$, therefore $(\tfrac{a+1}{a})^n\equiv 1\pmod{n}$. Let $x=\mathrm{ord}_p(\tfrac{a+1}{a})$. Then $x\mid n$, and we also have $x\mid p-1$, so $x\mid \gcd(n,p-1)$. The divisors of $\gcd(n,p-1)$ are all divisors of $n$ and less than $p$, so there are no prime divisors of $\gcd(n,p-1)$ due to minimality of $p$. This happens if and only if $\gcd(n,p-1)=1$. Then $x\mid 1$, so $x=1$, i.e. $\tfrac{a+1}{a}\equiv 1\pmod{p}$, so $1\equiv 0\pmod{p}$, contradiction. Therefore $n=1$ and $a$ can be any positive integer, resulting in the solutions $(a,1)$ for all $a\in \mathbb{N}$, which we can easily check.

Clearly $n=1$ works. Let $p$ be the smallest prime diving $n$ so that $p | n | (a+1)^n-a^n$, i.e.
	\[(a+1)^n\equiv a^n\mod p.\]
	We will show that thus implies $a+1\equiv a\mod p$, which would show $p=1$, a contradiction. Let $b$ be the unique inverse of $a$ mod $p$. Then,
	\[1\equiv (ab)^n\equiv ((a+1)b)^n\mod p\]
	Now $ord_p((a+1)b) | n$ and $ord_p((a+1)b) | p-1$, hence $ord_p((a+1)b) | \gcd(n,p-1)=1$, i.e $ord_p((a+1)b)=1$. This implies
	\[(a+1)b\equiv 1\mod p\Longleftrightarrow a+1\equiv a\mod p\]
	so we are done.

Clearly $n=1$ works. Let $p$ be the smallest prime diving $n$ so that $p | n | (a+1)^n-a^n$, i.e.
	\[(a+1)^n\equiv a^n\mod p.\]
	We will show that thus implies $a+1\equiv a\mod p$, which would show $p=1$, a contradiction. Let $g$ be a primitive root modulo $p$, so that $a=g^r$ and $a+1=g^s$. Then
	\[g^{rn}\equiv g^{sn}\mod p\Longleftrightarrow g^{n(r-s)}\equiv 0\mod p\]
	and thus $p-1 | n(r-s)\implies p-1 | r-s$, i.e.
	\[g^{r-s}\equiv 0\mod p\Longleftrightarrow g^r\equiv g^s\mod p\]
	so $a+1\equiv a\mod p$, and we are done.

ayy noice bruvv, thanks oreo [hide=sar]omegaorsmaxlavilprowgawdsaar[/hide] for teaching the smallest prime divisor idea :o :omighty:

[hide=Page 1][img]https://i.imgur.com/z4WUhfq.jpg[/img][/hide] [hide=Page 2][img]https://i.imgur.com/s6jGeP8.jpg[/img][/hide]

handwritten solution images attached

The answer is $(1, a)$ only. The idea is to pick a minimal prime $p \mid n$, such that $a+1 \equiv a \pmod p$ for order reasons. This implies $p=1$, thus only $n=1$ can work.

The condition is essentially just $$(a+1)^n\equiv a^n\pmod{n}$$

Case 1: $n$ does not divide $a$. Then, let $p$ be the smallest positive integer that divides $n$ but not $a$. Then, we have $$(\frac{a+1}{a})^n\equiv 1\pmod{p}.$$ Therefore, the order of $(a+1)/a$ mod $p$ divides $n$. However, this is smaller than $p$, and by definition, all divisors of $n$ smaller than $p$ are also divisors of $a$. Thus, unless the order of $(a+1)/a$ is 1 (i.e. $a+1\equiv a\pmod{n}$ so $n=1$, contradiction since $n$ does not divide $a$), $a$ and $n$ are not relatively prime, since this order divides both of them. Suppose $q\mid a$ and $q\mid n$. Then, taking $(a+1)^n\equiv a^n\pmod n$ mod $q$ reveals that $$1\equiv 0\pmod{q},$$ contradiction.\\

Case 2: $n$ divides $a$. Then, this is just $$(a+1)^n\equiv a^n\pmod n$$ $$1\equiv 0\pmod{n},$$ so $n=1$. Thus, the answer is $n=1$ and we are done.

Some months later solve:

Clearly, $n=1$ and $a\in\mathbb{Z}^+$ work. Therefore, assume $n>1$ from now on. Let $p$ be the smallest prime divisor of $n$, so that
	\[(a+1)^n\equiv a^n\mod p\]
	Let $g$ be a primitive root mod $p$, and let $0\le\alpha,\beta\le p-2$ such that $g^{\alpha n}\equiv(a+1)^n\equiv a^n\equiv g^{\beta n}\mod p$. We get
	\[g^{n(\alpha-\beta)}\equiv 1\mod p,\]
	meaning $p-1\mid n(\alpha-\beta)$. However, as $p$ is the smallest prime dividing $n$, $\gcd(p-1,n)=1$, so $p-1\mid \alpha-\beta$. As $0\le\alpha,\beta\le p-2$ we must have $\alpha=\beta$. We get get $a+1\equiv g^{\alpha}\equiv g^{\beta}\equiv a\mod p$, implying $0\equiv 1\mod p$, which is impossible, so we have no more solutions.

Here's an attempt at a different solution

Let $a=\prod_{i=1}^{k}p_i^{\alpha_i}$. Now using this observe that $$(a+1)^n=\left(\prod_{i=1}^{k}p_i^{\alpha_i}+1\right)^n\equiv 1\pmod {p_1}\text{ thus }\operatorname{ord}{p_1}(a+1)=\operatorname{ord}{p_1}\left(\prod_{i=1}^{k}p_i^{\alpha_i}+1\right)=n\mid p_1-1\Longleftrightarrow p_1\equiv1\pmod n$$
$$\text{ Furthermore } (a+1)^n=\left(\prod_{i=1}^{k}p_i^{\alpha_i}+1\right)^n\equiv1\pmod {p_2}\text{ thus }\operatorname{ord}{p_2}(a+1)=\operatorname{ord}{p_2}\left(\prod_{i=1}^{k}p_i^{\alpha_i}+1\right)=n\mid p_2-1\Longleftrightarrow p_2\equiv1\pmod n$$
$$\vdots$$
$$\text{ Now } (a+1)^n=\left(\prod_{i=1}^{k}p_i^{\alpha_i}+1\right)^n\equiv1\pmod {p_k}\text{ thus }\operatorname{ord}{p_k}(a+1)=\operatorname{ord}{p_k}\left(\prod_{i=1}^{k}p_i^{\alpha_i}+1\right)=n\mid p_k-1\Longleftrightarrow p_k\equiv1\pmod n$$
$$\text{ Using all these together we obtain that }\prod_{i=1}^{k}p_i^{\alpha_i}\equiv1\pmod n$$
Moreover notice that our original condition is equivalent to $\frac{(a+1)^n-a^n}{n}\Longleftrightarrow(a+1)^n-a^n\equiv0\pmod n$
However $(a+1)^n-a^n=\left(\prod_{i=1}^{k}p_i^{\alpha_i}+1\right)^n-\left(\prod_{i=1}^{k}p_i^{\alpha_i}\right)^n\equiv (1+1)^n-1^n\equiv 2^n-1\pmod n$
Therefore the condition transforms into $2^n-1\equiv0\pmod n\Longleftrightarrow 2^n\equiv 1\pmod n$

[b][color=#f00]Claim:[/color][/b] $2^n\equiv 1\pmod n$ if and only if $n=1$
[i][b]Proof:[/b][/i]
Notice that if $n\equiv 0\pmod 2$ we have that $2^n-1\equiv 1\pmod 2\text{ while }n\equiv0\pmod 2$ which is clearly a contradiction, thus $n\equiv 1\pmod 2\Longleftrightarrow\gcd(2,n)=1$
Now since $\gcd(2,n)=1$ we have that $2^n\equiv1\pmod n\Longrightarrow\operatorname{ord}_{n}(2)=n\mid\phi(n)\Longrightarrow n\mid\phi(n)\Longleftrightarrow\frac{\phi(n)}{n}\in\mathbb{Z}$
Furthermore let $n=\prod_{i=1}^{t}q_i^{\beta_i}$ and recall that $\frac{\phi(n)}{n}=\frac{\phi(\operatorname{rad} (n))}{\operatorname{rad}(n)}$ thus $\frac{\phi(n)}{n}=\frac{\phi\left(\prod_{i=1}^{t}q_i^{\beta_i}\right)}{\prod_{i=1}^{t}q_i^{\beta_i}}=\frac{\phi\left(\operatorname{rad} \left(\prod_{i=1}^{t}q_i^{\beta_i}\right)\right)}{\operatorname{rad}\left(\prod_{i=1}^{t}q_i^{\beta_i}\right)}=\frac{\phi\left(\prod_{i=1}^{t}q_i\right)}{\prod_{i=1}^{t}q_i}$
Furthermore since $\gcd(q_1,q_2,\dots,q_t)=1$ we have that $\phi\left(\prod_{i=1}^{t}q_i\right)=\prod_{i=1}^{t}\phi(q_i)=\prod_{i=1}^{t}q_i-1$
Therefore we need to have $\frac{\prod_{i=1}^{t}q_i-1}{\prod_{i=1}^{t}q_i}\in\mathbb{Z}$ which is clearly only the case when $n=1$ $\square$.

So plugging in $n=1$ into our original equation yields $\frac{a+1-a}{1}=1$ which is clearly an integer. Thus $(a,n)=(a,1)$ for any integer $a$ is the only solution $\blacksquare$.

I claim the only solutions are $(a, n) = (x, 1).$ Henceforth assume $n > 1.$

[color=#f00][b]Claim:[/b][/color] $\gcd(a, n) = \gcd(a + 1, n) = 1$
[i]Proof.[/i] Let $\gcd(a, n) = d > 1,$ then we have $d \mid a^n,$ but we must have $d \nmid a + 1,$ since if $d \mid a + 1,$ $d \mid \gcd(a, a + 1) = 1,$ contradiction. So, $d \nmid (a + 1)^n - a^n,$ but $d \mid n \mid (a + 1)^n - a^n,$ contradiction. Similarly, $\gcd(a + 1, n) = 1.$

Thus, there exist inverse of $a$ modulo $n$, so $(a + 1)^n \equiv a^n \pmod{n} \implies \left(\frac{a + 1}{a}\right)^n \equiv 1 \pmod{n},$ let $\frac{a + 1}{a} \equiv a^{*} \pmod{n},$ so let $p$ be the smallest prime divisor of $n$, this gives that $\mathrm{ord}_p(a^{*}) \mid \gcd(n, p - 1) = 1,$ so $a^{*} \equiv 1 \pmod{p},$ so $\frac{a + 1}{a} \equiv 1 \pmod{p},$ a contradiction, since $p > 1.$

Had solved this way back. Felt like posting now
Its clear that $\gcd(a, n) = 1$.Its also clear that $n$ is odd. Let $p$ be the minimal prime dividng $n$.
$(a+1)^n \equiv a^n (mod p) $. But we also have that, $(a+1)^{p-1} \equiv a^{p-1} (mod p) $. Its easy to see that due to this $p=2$. A contradiction.
This implies $n=1$. Thus $(a,1)$ is the only solution :D

We claim the solutions are $(a, 1)$ for all positive integers $a$.
Note that the condition reduces to $(a+1)^n \equiv a^n$ mod $n$.
Assume that $n > 1$. In this case, let $p$ be the smallest prime divisor of $n$.
Clearly $(a+1)^n \equiv a^n$ mod $p$.
Also observer that $gcd(a, n) = gcd(a+1, n) = 1$, so from Fermat's Little Theorem, $(a+1)^{p-1} \equiv a^{p-1}$ mod $p$.
Combining these two and applying the Euclidean Algorithm, we get
$a^{gcd(p-1, n)} \equiv (a+1)^{gcd(p-1, n)} \text{ mod } p \implies a+1 \equiv a \text{ mod } p \implies p = 1$
But $p$ is prime so this is not possible. Therefore $n = 1$.
But it is clearly visible that any $a$ works for $n = 1$, so we have our solutions. $\square$

I claim that all solutions of the form $\boxed{(a, 1)}$ where $a \in \mathbb{Z}^+$.

Assume then $n \neq 1$, as $n = 1$ is trivial. Let $p$ be the smallest prime dividing $n$. Also (as the inverse exists) let $b$ denote the inverse of $a + 1$. Then we have $(a + 1)b \equiv 1 \pmod p \implies (a + 1)^nb^n \equiv 1 \pmod p$, so that \[\implies o((a + 1)b)_p \mid p - 1 \text{ and } o((a + 1)b)_p \mid n \implies o((a + 1)b)_p \mid \gcd(n, p - 1).\] However as $p$ is the smallest prime that divides $n$, we must have $\gcd(n, p - 1) = 1$, so that $o((a + 1)b)_p = 1$. Therefore if $b$ is to be $a^{-1}$, then $a + 1 \equiv a \pmod p \implies 1 \equiv 0 \pmod p$, which is a contradiction. Hence $\boxed{n = 1}$ is the only such solution. $\blacksquare$


$ x_1 \equal{}2 \quad x_{n\plus{}1} \equal{}\left[ \frac{3}{2} x_n \right] \quad \text{for } \; n\equal{}1,2,3,\ldots$

Prove that
sequence $ \left( (\minus{}1)^{x_n} \right)$ is not periodic .

It has a flavor of the Collatz problem ...


The product of the digits of 3214 is 24. How many distinct
four-digit positive integers with unit digit not equal to 1 are such that the product of their
digits equals 18?

18=1*1*3*6
=1*1*2*9
=1*2*3*3
so the answer is 12+12+12=36.
but you have to subtract (6+6+3) so the final answer is [b]21[/b].

1. First I though, 18=1x2x3x3. So, I just need to determine where 1 and 2 are. So, I had 4x3=12. I entered 12.

2. Then, I look at the problem again, 1 can not be the last digit. So, I entered 12-3 =9.

3. Still wrong! the solution pop out. I missed big: there are other possibility: 18=1X1X2X9. etc.

Conclusion, need to slow down, read carefully and think twice.

So, I tried stars-and-bars. C(6,4)=20 but the answer should technically be less because all the 1s shouldn't be in the units digit. Why?


calculus
integration
trigonometry
analytic geometry
real analysis
real analysis unsolved

It's liable it should be $ \sqrt{x}$ instead of $ \sqrt[4]{x}$ . Substitute $ x\equal{}r^4$ and $ y\equal{}\sin\varphi$ then

$ I\equal{}\int_0^{\frac{\pi}{2}} \int_0^1 \frac{1}{r^2\, \sqrt{1\minus{}\sin^2 \varphi} \, \sqrt{1\minus{}r^2 \, \sin^2 \varphi}} \, 4r^3\, dr \; \cos\varphi \, d\varphi\equal{}4\,\int_0^{\frac{\pi}{2}} \int_0^1 \frac{1}{\sqrt{1\minus{}r^2\, \sin^2\varphi}}\, r\, dr\, d\varphi$ .

After changing from polar to Cartesian coordinates $ I\equal{}4\, \int_0^1 \int_0^{\sqrt{1\minus{}y^2}} \frac{1}{\sqrt{1\minus{}y^2}}\, dx\, dy$ .


Something is completely wrong with the board. A lot of the icons don't show up.  :(  This probably has something to do with the CSS problem in the blogging system.

Can you please post a print screen?
What icons don't show up? (I see nothing wrong with the board right now).

Also what browser are you using?

Strange... I couldn't see the icons next to the "Quote", "WWW" and other things in a post a few minutes ago, but now I can...
I'm using FF2

Anyone else had this happening?  :maybe:


ratio
geometry
function
geometry unsolved

Let $ABCD$ be a quadrilateral.
Choose $P \in (AB)$ and $Q \in (CD)$ such that $AP/AB=DQ/DC=x$.
Compute $area(APQD)/area(ABCD)$ in  function of $x$.

Perhaps I'm wrong, but AB and CD oppose each other.  If CD is pivoted about the point P for example, you can have several different situations; several different quadrilaterals still satisfying these conditions.  That is,as stated, it is indeterminate.

Otherwise, let AB = m and CD = n.  Let AP = mx, then PB = m(1-x).  Similarly let CQ = nx, then QD = n(1-x).  But that will do no further good if the figure is not stable [precisely determined.]


I was in the Intermediate Algebra class this past year, and I forgot to download the Final Challenge solutions before the page closed (assuming that there were solutions posted, as there have been for the other classes I have taken).  Is it possible for me to get the solutions now?

why don't you ask your "classmates" on the class board? One of them is bound to have a copy and happy to send it to you...


inequalities
High School Olympiads
algebra
JBMO

this one is very hard without the following method :D (e.g. expanding it out ) (BTW this isnt my proof)
x <= 1+x^2 / 2, etc. so it suffices to prove, after the substitution a=1+x^2, b=1+y^2, c=1+z^2 that a/c+(b/2) + b/a+(c/2) + c/b+(a/2) >= 2, which is easy :)

You said "... which is easy..."Why?After applying the AM/GM twice we have 3 {abc/[(2c+b)(2a+c)(2b+c)]}^1/3 which we should prove is [i][b]>eq 2[/b][/i].How does one prove that?

I'll try to clarify. As $x \leq \frac{1 + x^2}{2}$ we have
\[\sum \frac{1 + x^2}{1 + y + z^2} \geq \sum \frac{2(1 + x^2)}{(1 + y^2) + 2(1 + z^2)}.\]
Denoting $1 + x^2 = a$ and so on we have to prove that
\[\sum \frac{a}{b + 2c} \geq 1\]
but Cauchy tells us
\[\sum \frac{a}{b + 2c} \sum a(2b + c) \geq \left(\sum a\right)^2\]
and as
\[\left(\sum a\right)^2 \geq 3(ab + bc + ca) = \sum a(2b + c)\] we have the result.

it is not new. It was posted in Junior Balkan Mathematical Olympiad 2003 and it was proposed by Laurentiu Panaitopol.

[quote="mecrazywong"]Posted before.
The essential idea is to make use of the inequality $\frac{x^2+1}{2}\ge x$ and then Cauchy-Schwarz.[/quote]

Can you show?

OK,thanks! Are there any other solutions?

Man, i have already told you. Another solution is to make the substitutions $2a+b=m, 2b+c=n, 2c+a=p$ and after that the inequality reduces to a direct application of AM-GM. ;)

[quote="Cezar Lupu"]Well, I'm tired of saying it again and again ,but the initial inequality reduces to this well-known one:

  $\frac{a}{2b+c}+\frac{b}{2c+a}+\frac{c}{2a+b}\geq 1$ which can be easily solved by using Cauchy or by AM-GM. ;)[/quote]

Can any one show me how to solve it using AM-GM? :|

See also : http://www.mathlinks.ro/Forum/viewtopic.php?t=64654 . (for $\frac a{2b+c} + \frac b{2c+a} + \frac c{2a+b} \ge 1$, and a bit more ;) )

[quote="perfect_radio"]After those substitutions, I can't get AM-GM to work.   :(  :([/quote]

a,b,c>0...so is m,n,p>0

so when you substitute those values you will have 9a= 4m-2n+p....etc.

putting them together in the inequality you'll have something like this....

4(r+1/r)+4(s+1/s)+4(t+1/t)>=.... ; where r=m/n, s=n/p, t=p/m...etc.

AM-GM from there

Is the condition $ x,y,z>\minus{}1$ necessary? I can't seem to see where it is used.

Yes, it is: taking $ z \equal{} 0$ and $ y$ just slightly smaller than $ \minus{} 1$ allows the left-hand side to take arbitrarily big negative values.  It appears that it's used in the proof in the form of the (implicit) assumption that the fractions in the question are positive.

The logic of the very first step is as follows:
"$ 0 \neq a \leq b \neq 0$ and therefore $ \frac{c}{a} \geq \frac{c}{b}$."
This is not true for arbitrary reals -- in particular, it can fail if $ a$ is negative.

Another way of saying the same thing: just go through the whole proof with $ z \equal{} 0$ and $ y \equal{} \minus{}1 \minus{} \varepsilon$ for some very small $ \varepsilon > 0$ and see which statements fail.

[quote="Shanku"][quote="Arne"]I'll try to clarify. As $ x \leq \frac {1 \plus{} x^2}{2}$ we have
\[ \sum \frac {1 \plus{} x^2}{1 \plus{} y \plus{} z^2} \geq \sum \frac {2(1 \plus{} x^2)}{(1 \plus{} y^2) \plus{} 2(1 \plus{} z^2)}.\]
Denoting $ 1 \plus{} x^2 \equal{} a$ and so on we have to prove that
\[ \sum \frac {a}{b \plus{} 2c} \geq 1\]
but Cauchy tells us
\[ \sum \frac {a}{b \plus{} 2c} \sum a(2b \plus{} c) \geq \left(\sum a\right)^2\]
and as
\[ \left(\sum a\right)^2 \geq 3(ab \plus{} bc \plus{} ca) \equal{} \sum a(2b \plus{} c)\]
we have the result.[/quote]

Where is the (implicit) assumption that the fractions are positive ? The only fact used in the above solution is $ x \leq \frac {1 \plus{} x^2}{2}$ , which is true for all real $ x$ . ($ a,b,c$ are clearly positive for any real $ x,y,z$ ). I didn't get where the condition is used.[/quote]

[quote="JBL"]The logic of the very first step is as follows:
"$ 0 \neq a \leq b \neq 0$ and therefore $ \frac {c}{a} \geq \frac {c}{b}$."
This is not true for arbitrary reals -- in particular, it can fail if $ a$ is negative.

Another way of saying the same thing: just go through the whole proof with $ z \equal{} 0$ and $ y \equal{} \minus{} 1 \minus{} \varepsilon$ for some very small $ \varepsilon > 0$ and see which statements fail.[/quote]

Just now my eyes opened :idea: ! I posted stupid question. Sorry.
The condition is to -
[1] avoid zero in the denominators.
[2] keep the fractions in the positive zone only.

Cheers :lol:

can also be proved above like this
$\sum_{\text{cyc}}\frac{a^2}{(2c+b)a} $
$\ge \frac{(a+b+c)^2}{3(ab+bc+ac)}\ge 1$

Here is my solution i used the main idea and simply solved so yeah i did it and i know above are also correct solution i am sending here to make this threat a life(Active) a bit  :rotfl: \\
 Main Idea is here to see that with the help of AM-GM that \[(y^2+1)\geq 2y\]\[(z^2+1)\geq 2y\]\[(1+x^2)\geq 2x\] we can use this in main problem \[\sum_{cyc}\frac{1+x^2}{1+y+z^2}\geq \sum_{cyc}\frac{1+x^2}{1+\frac{1+y^2}{2}+z^2}=\sum_{cyc}\frac{2(1+x^2)}{(1+y^2)+2(1+z^2)}\]So we can see that by again using that \[\implies \sum_{cyc}\frac{2(1+x^2)}{(1+y^2)+2(1+z^2)}\geq \frac{4x}{2y+4z}=\frac{2x}{y+2z}\]
So its suffice to show \[\implies\sum_{cyc}\frac{x}{y+2x}\geq 1\]
Now we can use cauchy inequlity it makes thing simple that i suppose so \[\implies\left(\sum_{cyc}\frac{x}{y+2z}\right)\left(\sum_{cyc}x(y+2z)\right)\geq (x+y+z)^2\]So again it suffice to show that \[\implies(x+y+z)^2\geq \sum_{cyc}x(y+2z)=3(xy+yz+zx)\]\[\implies x^2+y^2+z^2\geq xy+yz+zx\] So its simple AM-GM and we are done Demon Killed mission accomplished. :-D

\[\implies \sum_{cyc}\frac{2(1+x^2)}{(1+y^2)+2(1+z^2)}\geq \frac{4x}{2y+4z}=\frac{2x}{y+2z}\]
This step is wrong @above

Other way to prove the below inequality
$$\sum_{cyc} \frac{a}{b+2c} \ge 1$$
Is to notice that the expression is equivalent to
$$ S = \sum_{cyc} \frac{a^2}{ab+2ac}$$
By C.S Engel or Titu's lemma, we have
$$ S \ge \frac{(a+b+c)^2}{3(ab+bc+ca)} \ge 1$$
Which is easy to prove.


calculus
integration
logarithms
complex analysis

[quote="Vinxenz"]Prove that
\[ I\equiv\int_{ \minus{} \infty}^{\infty}\frac {e^{\frac x3}}{1 \plus{} e^x}\ \mathrm {dx} \equal{} \frac {2\pi}{\sqrt {3}} \ .
\]
[/quote]
$ \left\|\begin{array}{c} x: \equal{} 3\ln x \\
 \\
\mathrm {dx}: \equal{} \frac 3x\cdot \mathrm {dx}\end{array}\right\|$ $ \implies$ $ I \equal{} 3\cdot J$, where $ J \equal{} \int_0^{\infty}\frac {1}{1 \plus{} x^3}\ \mathrm {dx}\ .$

$ \left\|\begin{array}{c} x: \equal{} \frac 1x \\
 \\
\mathrm {dx}: \equal{} \minus{} \frac {1}{x^2}\cdot \mathrm {dx}\end{array}\right\|$ $ \implies$ $ J \equal{} \int_0^{\infty}\frac {x}{1 \plus{} x^3}\ \mathrm {dx}$ $ \implies$ $ 2\cdot J \equal{} \int_0^{\infty}\frac {1 \plus{} x}{1 \plus{} x^3}\ \mathrm {dx}\ .$

$ J \equal{} \frac 12\cdot\int_0^{\infty}\frac {1}{x^2 \minus{} x \plus{} 1}\ \mathrm {dx}$ $ \implies$ $ J \equal{} \frac {2\pi}{3\sqrt 3}$ $ \implies$ $ I \equal{} \frac {2\pi}{\sqrt 3}\ .$

Hey Vinxenz, if you saw my earlier post, I made a mistake in stating a sum converged.  It didn't by the most basic test for convergence.  Sorry if I misled you.Thought it would just be cleaner to delete it. :oops:

Integrate on polygonal $ R,R \plus{} 2\pi i, \minus{} R \plus{} 2\pi i, \minus{} R,R$ $ R > 0$

[quote="Vinxenz"]Prove:
\[ \int_{ \minus{} \infty}^{ \plus{} \infty}\frac {e^{x/3}}{1 \plus{} e^x}dx \equal{} \frac {2\pi}{\sqrt {3}}
\]
[/quote]
put $ t\equal{}e^{\frac{x}{3}}$
$ I \equal{} \int_0^{ \plus{} \infty}\frac {1}{1 \plus{} t^3}dt$


in thinking about it, I can't really come up with something that specifically relates time to another area of mathematics. Of course all you do in 2nd or 3rd grade math is learn to tell time, but beyond that, math and time just don't really interact too much that I can see. Many a good problem comes from rate-time-distance formulas or functions, as you mentioned, but in my experience time is usually used as a source of inspiration for a problem, not an actual concept in one.

[quote="maskman"]hello everyone , in 2 months i will have to do a report about the topic of time (scientific report) which should be as original as possible , i found a lot of related  things (relativity , time units and their definitions...) however i have no ideas with a mathematical viewpoint (except curves defined by x(t) , y(t)...) so what can we say about time in mathematics?[/quote]

Well, since the framework of theoretical physics is mathematics you could say almost anything within a mathematical viewpoint. However, time particularly have been subject of much debate. I wonder how advanced you want to make your report. The final word about how physicists understand time is given by general relativity and differential geometry, at least, until a quantum gravity theory is accepted. Time and space in GR are represented by coordinates on a four dimensional manifold. The actual mathematical definition of time is pretty precise and it separates a lot from what your intuition tells you. I don't know what's your knowledge of mathematics, but I suggest you to find a book on general relativity that suits it. There are a lot made for a wide public and where things are explained in simple words. If you don't find anything like that, place this post on top again and surely I or somebody else will be able to help you more.
But you really should try by yourself

Ursula

Cover special relativity from the point of view of Lorentz transformations.

[quote="maskman"]Thanks to all your ideas are really useful
To tell you more , i am in the second year of university (maths and physics section) so the documentations i use can be as advanced as far i can perfectly understand,by the way i've just decided to title my report : "how can we understand time?" ; any other suggestions :)?[/quote]

well, you could start with spacetime (minkowski has done it very well)

here some links:
[url=http://en.wikipedia.org/wiki/Minkowski_space]Minkowski space[/url]
[url=http://en.wikipedia.org/wiki/Mathematics_of_general_relativity]math in GR[/url]


calculus
integration
vector
function
analytic geometry
calculus computations

On Wikipedia, I found the formulas:

1) (work) $\displaystyle W = \int \mathbf{F} \cdot \mathrm{d}\mathbf{s}$

2) (kinetic energy) $\displaystyle E_k = \int \mathbf{v} \cdot \mathrm{d}\mathbf{p}$

How does one actually calculate these integrals? I never encountered vectors as variables to integrate with, nor dot products in integrals.

It also said there that $E_p = -\int f(x) \, dx$. ($f$ is the force) Does that mean that $E_p=mgh$ or $E_p=-mgh$?

It is not so easy to compute such integrals.
Our prof in the university showed us a way to
find this particular integral...

In general, however, we need calculus with more than one variables knowledge....

 :)

Yeah, this is multivariable calculus.

Integrals like (1) or (2) have little meaning in physics at least, unless you specify the curve of integration. Typically, to calculate such an integral (here work) you have given $\vec{F}$ as a function of position, that is $\vec{F}=\hat{x}F_x(x,y,z)+\hat{y}F_y(x,y,z)+\hat{z}F_z(x,y,z)$. Then you need to parametrize the curve $C$ in one way or another: $\vec{s}(t)=\hat{x}s_x(t)+\hat{y}s_y(t)+\hat{z}s_z(t)$ for $t_a<t<t_b$. Then you have:

$\vec{F} \cdot d\vec{s} = \vec{F} \cdot \frac{d\vec{s}}{dt} dt$

So

$\int_C \vec{F} \cdot d\vec{s} = \int_{t_a}^{t_b} \vec{F} \cdot \frac{d\vec{s}}{dt} dt$

Which again is

$\int_{t_a}^{t_b} ( F_x(s_x(t),s_y(t),s_z(t))s_x'(t) + F_y(s_x(t),s_y(t),s_z(t))s_y'(t) + F_z(s_x(t),s_y(t),s_z(t))s_z'(t)) dt$

Hopefully (depending on $F_x,F_y,F_z,s_x,s_y,s_z$) the last integral comes out not too ugly. :)

$E_p$ is usually defined as the work required to put a system into a position from a position where $E_p=0$ by definition. In the case of raising a particle from altitude zero to altitude $h$ we have:

$F=-mg\hat{z}$, $C$ the straight line from $z=0$ to $z=h$ giving $s_x=s_y=0$ and $s_z=ht$ for $0<t<1$, so the integral becomes (just insert and see):

$E_p=-\int_0^1 (-mgh) dt = mgh$.

The minus sign is because we want work done on the particle, not [i]by[/i] the particle.

Hope I made it a little clearer... lol.

Edit: If you haven't had any multivariable calculus there is a free text availible [url=http://www.math.gatech.edu/~cain/notes/calculus.html]here[/url].

What about the potential energy of a spring? What direction of the force do we consider?

If we have a spring with spring constant $k$ with a particle of mass $m$ attached to it, we can calculate the potential energy as follows. Suppose $x=0$ is the equilibrium position of the spring (where no force acts on the particle). We define $E_p=0$ for this position. We want to find $E_p$ as a function of $x$, that is, $E_p(x)$ is the potential energy of the system when the spring is pulled a length $x$ in a direction. (Which direction $x$ is doesn't matter, it may be positive or negative).

To push or pull the string a length $dx$ we need to apply work $dW=Fdx$. Here, $F$ is the force we are acting on the particle with, which is the same force acting on the particle from the spring, but oppositely directed. So, $F=kx$. This is Hooke's law. Note that $F$ acts in the same direction as $x$ (the force from the spring acts in the other diretion, pulling $x$ back to $0$). So $dW=kxdx$.

The rest is straightforward:

$E_p = W = \int dW = \int_0^x kx' dx' = \frac{1}{2}kx^2$

Things to note:

1. $E_p$ is independent of $m$.
2. $E_p(x)$ is an even function. Pushing or pulling the spring is equivalent energy-wise.

I still don't understand :(

If we plug in $F=kx$ in $E_p = -\int F(x) \, dx$ we get the opposite of what you said. (or do we?)

Why does this happen? (according to your first post, $F$ is considered to be the force we act on the object, if I understood correctly)

In my first post, $\vec{F}$ is the force acting on the body from the system. When we "manipulate" the system we apply a new force $\vec{F_n}$. Now, since these two are equal in strength but opposite in direction at all times ($\vec{F} = -\vec{F_n}$), they both perform equal amount of work on the system. So we can use either to calculate $E_p$:

$E_p = -\int \vec{F} \cdot d\vec{s} = \int \vec{F_n} \cdot d\vec{s}$

In my second post I just used the outside force instead. That's why the minus vanished...

It's getting really hard for me to understand.

In the spring case: $F_n = kx$ - this is the force we act on the spring, it has the same direction as the elongation; $F=-kx$ is the elastic force, which makes the spring want to go back.

In the case of raising a particle: To raise it we apply a force opposite to $G$ $\left( -\overrightarrow G \right)$ , so this must be $F_n$; the opposite force is actually $G$, so this is $F$; hence a minus sign   :(  :(

I hope that you now understand what I don't understand.

When talking about some issues, we need to specify the POSITIVE direction.

I'll try again. :)

Energy is what we call a [i]state function[/i]. This means that the only energy that matters is [i]change[/i] in energy. Let's say we have a system in two states, $A$ and $B$. We cannot really determine the energy of the system $E(A)$ and $E(B)$ in those states, but we [i]can[/i] determine the [i]difference[/i] $E(A)-E(B)=\Delta E$.

The practical upshot is that when we are making calculations on the energy of a system we should define one particular state of the system where $E=0$.

Once you have done that you can calculate the potential energy of any other state. The potential energy of [i]any[/i] state is defined as the work you have to apply [i]to the system[/i] to get from the zero-state to that state.

This is where you have to start doing math, and this is where it can get really complicated. I can do some simple examples to show what I mean:

[b]1: The potential energy of a spring[/b]

First we need to define what the zero-state is. The obvious choice is the "equilibrium" state. This now has $E=0$. Now we want to calculate the energy in any other state. Here "any other state" means that the spring has been contracted or elongated a distance $x$. The energy we will find must thus be a function of $x$.

Now we must make it clear what we mean by $x$. I.e. we have to make a coordinate system (with one axis). The $x$-axis will have it's zero in the equilibrium position, and will have positive direction away from the wall. That is, when $x=0$ we have equilibrium and $E=0$. When $x>0$ the spring is elongated and when $x<0$ the spring is contracted.

Our mission is now to calculate $E(x)$. The definition of $E$ tells us that this is the work we have to apply to move the spring a distance $x$. Now, according to mechanical physics, work is equal to force times distance. Distance is clearly $x$. What is the force?

To make matters simple we will move the spring at a constant velocity. According to Newton's laws, then, the net force on the spring edge must be zero. However, there are [i]two[/i] forces in work here. One from the spring ($F_s$) and one from us ($F_u$). As mentioned, they must add up to zero:

$F_s+F_u=0\Rightarrow F_u=-F_s$

Hooke's law states that $F_s=-kx$. Or in other words, [i]the force from the spring is proportional with the distance displacement from the equilibrium, and acts in the direction opposite of the displacement[/i]. $k$ is called the [i]spring constant[/i]. We then get:

$F_u=kx$

Since $E(x)$ is the work we need to apply [i]to the system[/i] it is equal to the work done by the force we apply to the system. What we should ask ourselves is, how much work does $F_u$ do over the distance $x$?

Since $F_u$ is variable with $x$, we must apply calculus to answer this question. Over an infitesimally small distance $dx$ the infitesimally small work done $dW$ is equal to (force times distance) $F_u \cdot dx$, or $dW = kx \cdot dx$.

We can now calculate:

$E(x)=E(x)-E(0)=\Delta W=\int_{x'=0}^{x'=x} dW = \int_{x'=0}^{x'=x} kx' dx' = \frac{1}{2}kx^2$

And that's it.

If you wanted the minus sign thing explained:

$E(x)=E(x)-E(0)=\Delta W=\int_{x'=0}^{x'=x} dW = \int_{x'=0}^{x'=x} F_u dx' = - \int_{x'=0}^{x'=x} F_s dx'$

The two last integrals are equal. One uses the [i]outside[/i] force with [i]no[/i] minus sign. The other uses the "[i]inside[/i]" force (in this case, the force from the spring) [i]with[/i] a minus sign.

[b]2: Potential energy in a gravitational field[/b]

I'll skim over this one a bit quicker. We have an object of mass $m$ in the gravitational field of another object (say, a planet) of mass $M$ at a distance $r$. Here the $r$-axis has the zero point at the center of the large mass and positive direction outwards. What is the potential energy of the system?

Usually the zero-state in this case is set when the objects are infinitely far away. The reason for this is that the two have "no interaction" and it "fits" well with the idea of zero energy.

Now we will be moving the small mass from infinitely far away towards the large mass at constant velocity. Again, there are two forces acting on the small mass. The gravitational attraction from the large object, $F_g$ and the outside force from us $F_u$. And again, requiring constant velocity, we get:

$F_u=-F_g$

According to Newton's gravity law, the gravity force is $\gamma\frac{mM}{r^2}$, where $\gamma$ is the gravity constant. This points in the negative $r$-direction (towards the center of the large mass). We thus have $F_g=-\gamma\frac{mM}{r^2}\Rightarrow F_u=\gamma\frac{mM}{r^2}$.

Again, the force is varying with $r$, so we must apply calculus:

On $dr$, the work $dW$ is $F_u\cdot dr=\gamma mM \cdot \frac{dr}{r^2}$. So,

$E(r)=E(r)-E(\infty)=\Delta W=\int_{r'=\infty}^{r'=r} dW=\gamma mM \int_{r'=\infty}^{r'=r} \frac{dr'}{(r')^2} = -\gamma mM \left[ \frac{1}{r'} \right]_{r'=\infty}^{r'=r} = -\gamma\frac{mM}{r}$

As you can see, $E(r)<0$, as is required in this case ($E$ increases as we get further away, so if $E(\infty)=0$, $E(r)<0$).

Hope it got a little clearer now...


Aurillus was born in 23 B.C. and died 87 years later. In what year did he die?

-23+87=64, but we have to subtract one for the year ZERO, so 63.

No...0 AD doesn't exist, so it is $ \minus{} 23 \plus{} 1 \plus{} 87 \equal{} \boxed{65}$. It's really stupid, but whatever...

seriously? oh wait, my answer was wrong anyways...
but didn't know that. weird...math requires general knowledge now.

Oh well, it's 1992. I don't think this is required nowadays. (For example, look at the increase in the average description of each problem from back then to now.) :wink:


Mandvil has one standard quarter and one special quater with a Head on both sides. He selects one of these two coins at random, and without looking at it first, he flips the coin three times. If you flips a Head three straight times, what is the probability that he selected the special quarter? Express your answer as a common fraction.

Ithink its 8/9 from conditoiinal probalbitlity

[quote="i_like_pie"][hide="Solution"]The probability of getting 3 consecutive heads with a normal quarter is $ \tfrac{1}{8}$. The probability of getting them with the special quarter is $ \tfrac{1}{1}$. Therefore, if you get 3 consecutive heads, the probability that you used the special quarter is $ \frac{1}{\tfrac{1}{1}+\tfrac{1}{8}}=\boxed{\tfrac{8}{9}}$.[/hide][/quote]

well technically you have to note that picking each coin is equal problability, 1/2..but it donest really matter becuase it will cancel out


In the coordinate plane, a circle which passes through $ (-2,-3)$ and $ (2,5)$ cannot also pass through $ (1978,y)$. What is the value of $ y$?

y=3957 ( If y=3957 these 3 points belongs in a straight line ).


What is the smallest composite number whose prime factorization contains a number other than 2, 3, 5, or 7?

it should be next smallest prime number^2.

so it's 11^2, which is 121.

answer ; 121

Actually, it says [b]a[/b] number besides 2, 3, 5, or 7, meaning that only one factor has to not be included in these. Therefore, the answer is 2*11=22.


calculus
trigonometry
AMC
AIME
geometry
algebra
system of equations

[url=http://mathjunk.googlepages.com/MockAMCFProblems.pdf]Problems (PDF)[/url]
[url=http://mathjunk.googlepages.com/AnswerKey.pdf]Answers[/url]
[color=blue][u]Solutions[/u][/color] - Please post all solutions here.  I already have a partial solutions document, but I need all the solutions before I can post it.

[hide="1st Place"]
Elemenop - 144
[/hide]
[hide="2nd Place"]
knexpert - 139.5
[/hide]
[hide="3rd Place"]
SplashD - 138
[/hide]
[hide="Honorable Mention"]
4th - Iversonfan2007 - 132
5th - lotrgreengrapes7926 - 130.5
6th - white_horse_king88 - 129
[/hide]
[hide="Highest Score Under 8th Grade"][/hide]
[hide="Highest Scoring 8th Grader"]SplashD - 138[/hide]
[hide="Highest Scoring Freshman"]knexpert - 139.5[/hide]
[hide="Highest Scoring Sophomore"]DarkArchon - 126
Xantos C. Guin - 126[/hide]
[hide="Highest Scoring Junior"]beta - 127.5[/hide]
[hide="Highest Scoring Senior"]Elemenop - 144[/hide]
[hide="30 Participants"]
In no particular order:
Bobtheball
1=2
cosinator
bpms
h_s_potter2002
1337h4x
adalton
mustafa
E^i*pi=-1
Calculus
Fraxia
pianoforte
rnwang2
Leao
calc rulz
junggi
EFuzzy
1234567890
arjunlandes
ccy
10000th User
Xantos C. Guin
DarkArchon
beta
white_horse_king88
lotrgreengrapes7926
Iversonfan2007
SplashD
knexpert
Elemenop
[/hide]
[hide="List of Scores"]
52.5, 73.5, 90, 91.5, 96, 96, 102, 102, 103.5, 103.5, 103.5,106.5, 109.5, 144, 114, 115.5, 115.5, 117, 120, 124.5, 124.5, 126, 126, 127.5, 129, 130.5, 132, 138, 139.5, 144

24 out of 30, or 80%, would have qualified for AIME.
[/hide]
[hide="Average Score"]
112.25
[/hide]
[hide="Trickiest Problems"]
The number of people who DID NOT omit the problem, and got it wrong:
1 - 6
2 - 8
3 - 1
4 - 0
5 - 1
6 - 3
7 - 0
8 - 5
9 - 8
10 - 1
11 - 3
12 - 7
13 - 8
14 - 5
15 - 5
16 - 0
17 - 9
18 - 9
19 - 7
20 - 2
21 - 1
22 - 3
23 - 7
24 - 0
25 - 2
[/hide]
[hide="Most Omitted Problem"]
17 people omitted problem #24, making it the most omitted problem.
[/hide]

Solution for #24 -

[hide] All the answer choices are larger than the diameter, so choose E.

Other than that it's just system of equations with power of a point.

[/hide]

24 was the only big problem on the test.  About 3 people noticed the obvious reason why it was E, and most people just omitted it.  Other than that I think it went really smoothly, as I caught the only real error in the first hour and corrected it.

Can I get a solution for 18? As far as I can see, 0, 1, 2, 3 all work for n meaning more than 3. I didn't bother checking beyond that.

Number 24:
Unluckily, I spent five minutes finding the answer before realizing how obvious the solution was from the choices.

[hide]Draw the perpendicular from $E$ to the point $E'$ on $AB$.
Also the perpendicular from $O$ to the point $F'$ on $XY$.

The similarity $ABC \approx AEE'$ gives us $EE' = \frac{3}{5}$ and $AE' = \frac{4}{5}$.

So $E'F = \frac{21}{5}$, $OF = \frac{5}{2}$, and by Pythagorean theorem, $EF = 3\sqrt{2}$.

Notice similar triangles $EE'F \approx OF'F$.
So $OF' = \frac{1}{2\sqrt{2}}$.

So half of $XY$, $XF' = \sqrt{(OX)^{2}-(OF')^{2}}$, which can be calculated, but I'm too lazy right now to repeat the calculation.

This is when I realized how 'off' my solution was.[/hide]

Yeah, I thought number 24 was really funny, because I wasted a considerable amount of time on it before realizing the obvious answer.

[hide="10"]Assume $\sqrt{7+\sqrt{40}}=\sqrt{a}+\sqrt{b}$ where $a$ and $b$ are rational. Then $7+\sqrt{40}= a+b+2\sqrt{ab}$. Equate rational parts of both sides and irrational parts of both sides to find that $a+b=7$ and $ab=10$. By inspection, $a=5$ and $b=2$. From here it is a question of whether you can count, which for me can sometimes be just as difficult as solving the meat of the problem :D. [/hide]

[quote="Fraxia"]Could someone post a solution to 25 or tell me what's wrong with mine? I got 380, which was close enough to 400 that I put down E  :P

[hide]
Put one of the vertices of the 20-gon at $(1,0)$. Then the other vertices are at the $20th$ roots of unity:
\[(\cos 18, \sin 18), (\cos 36, \sin 36), \cdots (\cos342, \sin 342) \]
The square of the distance from $(1, 0)$ to $(\cos 18, \sin 18)$ is
\[(1-\cos 18)^{2}+(\sin 18)^{2}= 2-2\cos 18 \]
Similarly, the distance from $(1, 0)$ and the other points are
\[2-2\cos 36, 2-2\cos 57, \cdots, 2-2\cos341 \]
So taking the sum of all these distances, multiplying by $20$ since there are $20$ points, and dividing by $2$ to fix the double counting will be equal to the desired sum
\begin{eqnarray*}S &=& \frac{20}{2}(2-2\cos 18+2-\cos36+\cdots+2-2\cos 342)\\ &=& 10 \Biggl( 2 \times 19-2(\cos 18+\cos 36+\cdots+\cos 342 ) \Biggr) \end{eqnarray*}
Pairing up the $\cos$ terms, $\cos 18+\cos 162 = 0$, through $\cos 72+\cos 108$, and $\cos 90 = 0$, and similarly for $\cos 198, \cos 216, \cdots \cos 342$, so the sum of all the $\cos$ terms becomes zero, and
\[S = \frac{20}{2}\times 38 = 380? \]
I can't find my mistake  :maybe:
[/hide]

Could someone also explain how to solve 10?[/quote]

you forgot the BIG diagonals

:wallbash:

I should have noticed 180 was missing. Thanks for pointing that out, and the solution to 10  :)

Solution to number 25:

Use the Pythagorean Theorem.  :rotfl:
Try and it and you'll see what's funny.

[quote="mysmartmouth"][hide=Highest Score Under 8th Grade][/hide][/quote]

uuhh.... I'm under 8th grade.  Isn't there anyone else under 8th grade??? :huh:

I've seen problems like number 10 a few times on here, but I always have a hard time with them. What is a good site/book that will help improve these radical simplification problems?

Anyone want to post a solution for 18? =/

Hey can someone post solutions to
15
18
20
21
22
23

Thanks.

My answer doesn't agree with the answer key. What did I do wrong for number 23?

[hide]$\sum_{a>b>0}\frac{a}{2^{a}3^{b}}$
(I'm assuming $a$ and $b$ are integers... otherwise I'm not sure how that would work...)

$\sum_{a=1}^{\infty}\frac{a}{2^{a}}\sum_{b=1}^{a-1}\frac{1}{3^{b}}\\ = \sum_{a=1}^{\infty}\frac{a}{2^{a+1}}\left(1-\frac{1}{3^{a}}\right) \\ = \frac{1}{2}\sum_{a=1}^{\infty}\frac{a}{2^{a}}-\frac{1}{2}\sum_{a=1}^{\infty}\frac{a}{6^{a}}\\ = 1-\frac{1}{5}= \frac{4}{5}$[/hide]

Apparently, it's what I got squared. I'm wondering where I messed up.

Number 22:

Notice that $x^{2}+x+1 = 0$ implies $x^{3}= 1$, and also that $x+1+\frac{1}{x}= 0$.

So from that we see that $x+\frac{1}{x}=-1$, and $x^{2}+\frac{1}{x^{2}}= \left(x+\frac{1}{x}\right)^{2}-2 =-1$.
And also that $x^{3}+\frac{1}{x^{3}}= 2$.
And since $x^{3}= 1$, we know that $x^{k}+\frac{1}{x^{k}}= x^{k-3}+\frac{1}{x^{k-3}}$, so that entire sum is $9\left((-1)^{2}+(-1)^{2}+2^{2}\right) = 54$. (C)

Hmm... I just looked at the test for fun... #25 is surprisingly easier than many of the questions. (or it's just me  :P )

It's actually almost identical to an old AIME problem from 1990. Except that this time the pythagorean theorem makes it almost trivial.

[quote="white_horse_king88"]My answer doesn't agree with the answer key. What did I do wrong for number 23?

[hide]$\sum_{a>b>0}\frac{a}{2^{a}3^{b}}$
(I'm assuming $a$ and $b$ are integers... otherwise I'm not sure how that would work...)

$\sum_{a=1}^{\infty}\frac{a}{2^{a}}\sum_{b=1}^{a-1}\frac{1}{3^{b}}\\ = \sum_{a=1}^{\infty}\frac{a}{2^{a+1}}\left(1-\frac{1}{3^{a}}\right) \\ = \frac{1}{2}\sum_{a=1}^{\infty}\frac{a}{2^{a}}-\frac{1}{2}\sum_{a=1}^{\infty}\frac{a}{6^{a}}\\ = 1-\frac{1}{5}= \frac{4}{5}$[/hide]

[/quote]

a cannot be 1

you evaluated the geometric series wrong for 1/3^b

=1/3*3/2(1-1/3^{a-1})=1/2(1-3^{1-a})

Ah thanks, I completely forgot about that  :oops:

Also can someone tell me what I'm doing wrong on 22?
[hide]$R1 \cap R2$ is $\frac{1}{4}$ of a circle with radius $\frac{2}{\sqrt{3}}$ so the area is $\frac{4\pi}{3 * 4}$ which is $\frac{\pi}{3}$ which is $\frac{0\sqrt{3}+3\pi}{9}$
The answer would be $0+3$, but the answer is actually $4$ instead of $3$.[/hide]

[quote="ark_xm"][quote="white_horse_king88"]

Number 22:

Notice that $x^{2}+x+1 = 0$ implies $x^{3}= 1$, and also that $x+1+\frac{1}{x}= 0$.

So from that we see that $x+\frac{1}{x}=-1$, and $x^{2}+\frac{1}{x^{2}}= \left(x+\frac{1}{x}\right)^{2}-2 =-1$.
And also that $x^{3}+\frac{1}{x^{3}}= 2$.
And since $x^{3}= 1$, we know that $x^{k}+\frac{1}{x^{k}}= x^{k-3}+\frac{1}{x^{k-3}}$, so that entire sum is $9\left((-1)^{2}+(-1)^{2}+2^{2}\right) = 54$. (C)[/quote]

Oh nice, but how did you know that $x^{2}+x+1 = 0$ implies $x^{3}= 1$?[/quote]

$(x^{2}+x+1)=0 \implies (x-1)(x^{2}+x+1)=0\implies x^{3}-1=0$

Comment: steps not reversable  :!:

so $x^{2}+x+1=0\implies x^{3}=1$, but we cannot say $x^{3}=1\implies x^{2}+x+1=0$

As to: "How would I think of this?" The roots of $x^{2}+x+1=0$ are 3rd roots of unity...but in general, you would have probably have to have seen something similar before.

[quote="Altheman"]

a cannot be 1

you evaluated the geometric series wrong for 1/3^b

=1/3*3/2(1-1/3^{a-1})=1/2(1-3^{1-a})[/quote]

Oh, ouch. Yeah. Thanks.


What percent of a 48-minute basketball game is over if the time
remaining is one-third the time played?

The time remaining is $ \frac {1}{4} \equal{} 25\%$, so $ 75\%$ has been played.


n is a non-square natural number, then, are there exist infinitely many primes who have n as a primitive root?
If not, for which n that statement above holds?

I thougt about this, but in vain. Help.