quadratics
pigeonhole principle
function
number theory
relatively prime
number theory unsolved

Observe that $ 13=\frac{4^{3}+1^{3}}{4+1}$ and $ 13^{2}=\frac{15^{3}+8^{3}}{15+8}.$
Prove that for each $ n\in \mathbb{N}$, there exists relatively prime integers $ a$ and $ b$ (not necessarily positive) such that
\[ 13^{n}=\frac{a^{3}+b^{3}}{a+b}.\]
(HINT: Note that $ a^{2}-ab+b^{2}= (a-\alpha b)(a-\beta b)$ where $ \alpha$ and $ \beta$ are the two roots of $ x^{2}-x+1$.)

$ \frac{a^{3}+b^{3}}{a+b}=a^{2}-b a+b^{2}$

this equation have infinitely many solutions for 13^n

i dont know how to prove, but is interesting

Exactly those numbers $ n$ are of type $ n=^{2}-ab+b^{2}$ for which ever prime $ p \equiv 2 \mod 3$ occurs an even number of times in $ n$.
You can prove this (it was even posted before I assume) the following way:

[b]a)[/b] Show that if $ p|a^{2}-ab+b^{2}$ and $ p\equiv 2 \mod 3$ prime, then $ p|a,b$ (quadratic reciprocity is your friend). Use this to show that $ p \equiv 2 \mod 3$ divides $ n=a^{2}-ab+b^{2}$ an even number of times.
[b]b)[/b] Show that if $ p \equiv 1 \mod 3$ is prime, there is an $ j$ such that $ p|j^{2}-j+1$ (quadratic reciprocity is your friend again).
[b]c)[/b] Show that if $ p \equiv 1 \mod 3$ is prime, then $ p=a^{2}-ab+b^{2}$ for some $ a,b$:
Consider pairs $ (x,y)$ and the value of $ x-yj \mod p$ ($ j$ as in part b) ); use pidgeonhole principle to find a pair $ a,b$ such that $ a \equiv jb \mod p$ and $ a^{2}, b^{2}< p$, $ b \neq 0$; then consider $ a^{2}-ab+b^{2}$ and conclude that we find $ c,d$ with $ p=c^{2}-cd+d^{2}$.
[b]d)[/b] Show that if $ x=a^{2}-ab+b^{2}$ and $ y=c^{2}-cd+d^{2}$, then $ xy = e^{2}-ef-f^{2}$ for some $ e,f$ (find an identity).

[quote="ZetaX"]
[b]d)[/b] Show that if $ x=a^{2}-ab+b^{2}$ and $ y=c^{2}-cd+d^{2}$, then $ xy = e^{2}-ef+f^{2}$ for some $ e,f$ (find an identity).[/quote]

ZetaX,
I am not sure if this would work.  I thought that the natural choices for $ e$ and $ f$ would be
$ e=ac+bd$ and $ f=bc+ad$.  But this does not work.  Did you have something else in mind?  Thanks.

(Actually, I was thinking more along the lines of a proof involving algebraic number theory.  But, an elementary proof would be nice, too.)

You have to choose a little bit more complicated one:
$ e=ac-bd, f=ad+bc-bd$

[quote="jb05"]You can also construct examples of such numbers recursively:
$ a_{n+1}=4a_{n}-b_{n}$
$ b_{n+1}=a_{n}+3b_{n}$
$ a_{1}=4$, $ b_{1}=1$
Then $ 13^{n}=\frac{a_{n}^{3}+b_{n}^{3}}{a_{n}+b_{n}}$ by induction.[/quote]

jbo5,
I was just wondering, how would one go about finding such a recursion?  Is it easy to see that $ a_{n}$ and $ b_{n}$ are relatively prime?

For integers $ a, b$ and $ \omega = e^{ i \frac{2\pi}{3}}$ define

$ N(a+b \omega) = a^{2}-ab+b^{2}$

It turns out that this function is multiplicative (implying [b]ZetaX[/b]'s part d), so if we are given $ z = a+b \omega$ such that $ N(z) = 13$, we can calculate $ z^{n}$ which will give $ N(z^{n}) = 13^{n}$.  I believe this is the source of the recursion.

[quote]How would one go about finding such a recursion?[/quote]

Using your hint :D ! We want $ 13^{n+1}=a_{n+1}^{2}-a_{n+1}b_{n+1}+b_{n+1}^{2}=(a_{n+1}-\alpha b_{n+1})(a_{n+1}-\beta b_{n+1})$, and we know that $ 13^{n}=(a_{n}-\alpha b_{n})(a_{n}-\beta b_{n})$, so multiply the latter by $ 13=(4-\alpha)(4-\beta)$. We have that $ (a_{n}-\alpha b_{n})(4-\alpha)=4a_{n}-\alpha (a_{n}+4b_{n})+b_{n}\alpha^{2}$ $ =4a_{n}-\alpha (a_{n}+4b_{n})+b_{n}(\alpha-1)=(4a_{n}-b_{n})-(a_{n}+3b_{n})\alpha$, and obviously the same thing happens with the β terms, so you get the recursion.

Sorry, I didn't notice the requirement that $ a$ and $ b$ have to be relatively prime, but that happens to be true. Suppose $ a_{n}$ and $ b_{n}$ have a prime factor $ p$ in common. Then $ 13^{n}=a_{n}^{2}-a_{n}b_{n}+b_{n}^{2}$, so $ p|13^{n}$ and $ p=13$. Looking at the first few terms of the recursion modulo 13, we see that this is impossible:

(4,1)-(2,7)-(1,10)-(7,5)-(10,9)-(5,11)-(9,12)-(11,6)-(12,3)-(6,8)-(3,4)-(8,2)-(4,1) and it repeats


$ a,b,c,d$ are real numbers such that $ ad\minus{}bc\equal{}1$.

We have $ S\equal{}a^2\plus{}b^2\plus{}c^2\plus{}d^2\plus{}ac\plus{}bd$. Choose a most right solution:

A. $ S \le \sqrt 3$

B. $ S \ge \sqrt 3$

C. $ S> \sqrt 3$

D. $ S< \sqrt 3$

[hide="Solution"]Consider vectors $ u \equal{} (a,b,0)$ and $ v \equal{} (c,d,0)$, separated by a positive angle of $ \theta$. Let their respective magnitudes be $ x$ and $ y$.

We have that: $ \| u \times v \| \equal{} 1$, and we want to discern a true statement about $ S \equal{} x \plus{} y \plus{} u \cdot v$

Write the given condition as $ xy \sin \theta \equal{} 1$. Obviously, we can inflate $ x$ and $ y$, and thus $ S$, by choosing small $ \theta$, so $ S$ is not bounded from above. Instead, let's bound $ S$ from below by maximizing $ \sin \theta$ to be $ 1$, simultaneously minimizing $ \cos \theta$ to be $ 0$. In other words, $ u$ and $ v$ are perpendicular.

So now, our condition is $ xy \equal{} 1$ and we want to minimize $ S \equal{} x \plus{} y$

Since $ x,y > 0$, by AM-GM, $ \boxed{S \geq 2}$, with equality when when $ x \equal{} y \equal{} 1$.

Then, the correct answer is $ \boxed{\textbf{(C)} \; S > \sqrt {3}}$[/hide]

[quote="TZF"]Consider vectors $ u \equal{} (a,b)$ and $ v \equal{} (c,d)$, separated by a positive angle of $ \theta$. [...]
We have that: $ \| u \times v \| \equal{} 1$ [...][/quote]

You know the cross-product is only defined for vectors in $ \mathbb{R}^3$.  :wink:

mmk, I added a third coordinate to them  :P

[quote="*Anh_Tuan*"]I don't understand  :([/quote]
Why?  :?:


$ A$ is a fixed point. $ B$ is a moving point on a straight line(not passing through $ A$); i.e. Locus of $ B$ is a straight line. Find the locus of point $ C$, if $ ABC$ is always an equilateral triangle.

Since $ AB \equal{} AC$ , and $ < CAB \equal{} 60^0 \equal{} fixed$ . the loci of $ C$ has to be another line L', which is the line that result of rotating the line L (in which B moves) with center A and angle of 60.

Put another way, C is the point B rotated around point A by 60 degrees, so the locus of C will be the locus of B rotated around point A by 60 degrees as well.

The only issue is that C can either be 60 degrees cw or ccw of point B around A, so the locus of C actually contains [i]two[/i] lines.


calculus
integration
algebra
partial fractions
real analysis
real analysis unsolved

ingegral of 1/[e^(2x)+e^(x)]

and

integral of -1/[e^(x)+1]

I know one way is partical fractions but when i break it up I get stuck and don't know where to go from there

Use substitution $y=e^x$ to transforms your integrals into
\[
\int \frac{1}{y^2(y+1)}\,dy
\]

and

\[
\int \frac{-1}{y(y+1)}\,dy
\]

and these two integrals can be easily solved by fractional fraction method.

[quote="oldman"]... and these two integrals can be easily solved by fractional fraction method.[/quote]

Hehe, I am sure you mean partial fractions, isn't it? :D

The idea is that before you use partial fractions, you multiply top and bottom by $e^x$.


limit
logarithms
integration
real analysis
real analysis unsolved

Here's the hard limit problem that I wasn't able to solve. It's from SSMJ Magazine and it already passes the deadline of submission yesterday, so feel free to post the solution!
Problem:
Find: $ \displaystyle\lim_{n\to\infty}\sum_{k \equal{} 1}^{n}\frac {1}{n \plus{} k^a}$ (a is a positive real number).
It's easy to find this limit in case $ a < 1$ (limit is 1). But I haven't proved for the case $ a > 1$ though.

for $ a>1$ limit is equal to $ 0$
try to divide summation $ \sum \equal{} \sum_{1\leq k \leq [n^{1/a}]}\plus{}\sum_{[n^{1/a}]\leq k \leq n}$
and estimate each one.

For $ a\equal{}1,$ the limit is $ \ln 2$ (that's a Riemann sum) and for $ a>1,$ the limit is zero.

The following is a clumsy argument; I suspect it could be done much more slickly.

Divide the sum into $ \sum_{1\le k\le n^{1/a}}\frac1{n\plus{}k^a}\plus{}\sum_{n^{1/a}<k\le n}\frac1{n\plus{}k^a}\equal{}\text{I}\plus{}\text{II}.$

To estimate $ \text{I},$ note that each term is smaller than $ \frac1n.$

$ \text{I}<\sum_{1\le k\le n^{1/a}}\frac1n\approx\frac{n^{1/a}}{n}\equal{}n^{1/a\minus{}1}.$

Since that is a negative exponent, we see that $ \text{I}\to0$ as $ n\to\infty.$

To estimate $ \text{II},$ note that each term is smaller than $ \frac1{k^a}.$

$ \text{II}<\sum_{n^{1/a}<k\le n}\frac1{k^a}$

This in turn behaves like $ \int_{n^{1/a}}^nx^{\minus{}a}\,dx\equal{}\frac1{a\minus{}1}\left(n^{1/a\minus{}1}\minus{}n^{1\minus{}a}\right).$

Again, these are negative exponents of $ n,$ and thus $ \text{II}\to0$ as $ n\to\infty.$

[I was working on this before Extremal posted.]

Wait! I thought we also have to find the lower bound of that limit to be able to use Squeeze's Theorem? One  bound is not enough to evaluate the limit, isn't it? We need both, but the part of lower bound is also challenging to me. :(

Note that both sums are bounded below by 0. All of the terms in them are positive.  :wink:

Okay! I got it now(I'm too mechanical to show the limit of lower bound is also 0 :( ). Thank you very much for all of your help. I appreciate that! :)


[u][b]Previous games[/u][/b]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=149694]Game 1[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=151415]Game 2[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=154425]Game 3[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=168049]Game 4[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=170873]Game 5[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=173810]Game 6[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=284355]Game 7[/url]

[u][b]Formal Instructions:[/u][/b]
The aim of the game is to be the last player remaining.  Every round, there is a time limit, usually about 2 or 3 days.  Before the time limit ends, all players must submit via PM a real number between 0 and 1 to me.  There is a specific submission format written below and you must abide by this format or else your submission [i]will not be counted[/i].  At the end of the time limit, I tally up the submissions and scores in this fashion:
- I post all the submissions (a fail to submit counts as 1).
- I post what half the average submission was.
- I post all players who submitted under half the average.  These players are eliminated.
- Everybody gains points equal to their distance from half the average.  [i][u]Note:[/u] you do not want points![/i]
- If anybody has over 1 point total, they are eliminated.
When there is one player remaining, they win.
- You can submit any real number. For example, $ 1\over{\pi}$ is acceptable. However, to save some burden on my behalf, irrational numbers will be truncated with 10 digits accuracy.

[u][b]Submissions[/u][/b]
Format submissions as the following:
HTA 8 - Round {Round Number} - {Submission Number}
Submission numbers should look like the following:
HTA 8 - Round 1 - 1/pi
HTA 8 - Round 1 - 0.123456789
HTA 8 - Round 1 - 1/5
HTA 8 * Round 1 - 1/5

If the constant is even moderately obscure, like Apery's Constant, it would be greatly appreciated that you truncate it to 10 digits, as I don't want to rove all across Google to find your number.

If you have any parting remarks, insert an asterisk between HTA 8 and Round, in place of the dash. This is so that numbers are evenly aligned, thanks.

Sign-ups begin now. A maximum of about 15 people will be accepted, and please join only if you can make most of the 2-3 day deadlines.

A note on conversion: If you still want to play like the old game, simply divide what you would have submitted by 100. Simple huh?

[quote="r15s11z55y89w21"][hide= ][/hide][/quote]

It'd be nice if you actually put something in that invisible hide tag.

I join
The message is too small. Please make the message longer before submitting.

I will join and not forget the first round this time.

Now that you said something...

[u][b]Signups:[/b][/u]
1) Brut3forc3
2) SuperNerd123
3) Brut3Forc3
4) EggyLv.999
5) vahalla
6) AIME15
7) Mewto55555
8) abcak
9) r15s11z55y89w21

6 more can join.

[u][b]Signups:[/b][/u]
1) Brut3forc3
2) SuperNerd123
3) Brut3Forc3
4) EggyLv.999
5) vahalla
6) AIME15
7) Mewto55555
8) abcak
9) r15s11z55y89w21
10) andersonw
11) cyberspace

4 more can join.

[quote="andersonw"]I'll play, although I'll be going on vacation in a week, and I don't know if we'll have internet access or not. If I fail to make a deadline, feel free to replace me.[/quote]

Same would go to me. There will be more time on those rounds.

Whoops, I forgot to mention one last thing: Your submission must be >= 0.01, to keep the minimum submission similar.

[u][b]Signups:[/b] [/u]
1) Brut3forc3
2) SuperNerd123
3) Brut3Forc3
4) EggyLv.999
5) vahalla
6) AIME15
7) Mewto55555
8) abcak
9) r15s11z55y89w21
10) andersonw
11) cyberspace
12) pacman2812
13) erabel
14) spaceguy524

Submissions are due: August 7, 2009, noon PST.

Just wondering, why is brut3forc3 on the list twice?

Mistake. I didn't count him twice in my file listing all the submissions anyways...  :rotfl:

So, there are 13 people, I apologize to the people who would have changed their numbers upon hearing this, but no, there are no re-submissions.

Deadline.

[hide="Submissions"]1) Brut3forc3 	0.367879441170000
2) SuperNerd123 	1.000000000000000
4) EggyLv.999 	0.333333333333333
5) vahalla 	0.280000000000000
6) AIME15 	0.320000000000000
7) Mewto55555 	0.400000000000000
8) abcak 	1.000000000000000
9) r15s11z55y89w21 	0.389000000000000
10) andersonw 	0.333333333333333
11) cyberspace 	0.635600000000000
12) pacman2812 	1.000000000000000
13) erabel 	0.080000000000000
14) spaceguy524	1.000000000000000
Half The Average	0.274582543

erabel is out (<HTA)
[/hide]

[hide="Scores"]
1) Brut3forc3 	0.093296899
2) SuperNerd123 	0.725417457
4) EggyLv.999 	0.058750791
5) vahalla 	0.005417457
6) AIME15 	0.045417457
7) Mewto55555 	0.125417457
8) abcak 	0.725417457
9) r15s11z55y89w21 	0.114417457
10) andersonw 	0.058750791
11) cyberspace 	0.361017457
12) pacman2812 	0.725417457
13) erabel 	OUT
14) spaceguy524	0.725417457
[/hide]

[hide="Mod Notes"]
PLEASE follow submission rules, as usual, round 1 leniency ensues.
TWO people did not follow such submission rules, and TWO people forgot that the domain of submission is 0~1, not 0~100.
Next round I won't be so tolerant.

Anyways, onto some analysis.
Remember that you are aiming for half the average not the average, so in all likelihood HTA shouldn't be over 0.5, no?
vahalla scrapes by JUST SLIGHTLY by about five thousandths point. Not bad for first round.
Four out of the thirteen people forgot to submit. Had this not have happened, the HTA would be around 0.174.
erabel submitted an insanely low number that in most rounds of HTA games is below half the average. Note that had this not happened, vahalla most likely would have gone down under.

Anyhow, congratulations if you survived. Round 2.
[/hide]

Deadline Round 2 = August 10th, 2009 noon PST

Due to the dissappointing number of submissions (6/12) I will extend this to August 13.

Tomorrow is one week after August 13. I need to know if I'm dead.


I know this may seem trivial, but it has me stumped. Could you tell me what I'm doing wrong.

$(-8)^{\frac{1}{3}}= ((-8)^{2})^{\frac{1}{6}}= 2 \ne-2$

When I try $((-8)^\frac{1}{6})^{2}$ on my ti-92+ I get $1+\sqrt{3}i$. What's with all this stuff, what's the answer?

[quote="JavaMan"]I know this may seem trivial, but it has me stumped. Could you tell me what I'm doing wrong.

$(-8)^{\frac{1}{3}}= ((-8)^{2})^{\frac{1}{6}}$[/quote]

You're taking the wrong sixth root here.  There are six of them.

Your calculator is also taking the wrong sixth root.

What is the right 6th root? And what's wrong with $((-8)^{2})^\frac{1}{6}= 64^\frac{1}{6}$ did i mis-use an exponential law?

$64^{ \frac{1}{6}}$ is correct, but there are $6$ sixth roots of $64$.

Two of them are $2,-2$.

Two more are $1+\sqrt{3}i, 1-\sqrt{3}i$.

The last two are $-1+\sqrt{3}i,-1-\sqrt{3}i$.

On the top, you're taking the first root instead of the second.

Your calculator is taking the third root instead of either the first or the second.

Wouldn't the easier way just take the cube root of $cuberoot*-8=-2$?

[quote="ckck"]Wouldn't the easier way just take the cube root of cuberoot*-8=-2?[/quote]

i agree. aren't you increasing the number of roots by doing this???

an odd root of a negative number should only give negative number results.

if you square it (which makes it positive) then put it to an even root you get 2 roots, one positive one negative. so why would you bother doing that??

It's not about which computation is easier, it's about ridding me of any doubt  :D

Yeah, and also, I don't think it is allowed to change any exponents, so that could be a problem.


geometry
trapezoid
rectangle
MATHCOUNTS
analytic geometry
Pythagorean Theorem

1996-1997 National Competition Target Round Problem 6

Pedro stood at the center of a circular field that had a radius of 120 geet. He walked due north halfway to the circle.  He then turned and walked due east halfway to the circle. He then turned again walked due south halfway to the circle. Finally, he turned and walked due west halfway to the circle. When he stopped, how many feet was Pedri from the center of the circle? Express your answer to the nearest foot.

[quote="jli"]what do you mean by waliing half way through the circle?

sorry.. I'm bad at English :blush:[/quote]
I think he means halfway to the rim of the circle from your current position. (ie. if you're 100 feet from the rim, you walk 50.)

[hide]Im not exactly sure about this but i got 67 feet.[/hide]

Can u explain how you got that? I see some of the pythagorean tricks, but it gets hard after the third turn.... i think the answer is somewhere from 60 to 65.... but i'm not exactly sure... of course it's a National Target problem...

[quote="Farazy"]This was the hardest one in that set.  I got all the other ones intime with a minute to spare...This one is driving me crazy!  I think it has to be somewhere dead close to sixty... The other thing that I found was that the first turn involves a 30-60-90... Try and see if you can see why.  That's where I stopped, and I kind of guessed.[/quote]

Ya when you gave this problem, I just stopped after the second part, and I don't see any way to get the distance from the point you're at to the edge of the circle... It's especially hard because the point is somewhere on the circle but not on the edges... my Graphing calc has Cabri... but it would take way too long to construct that.. and even longer to measure... do you have the answer? Maybe that could help to make up some conjecture? Also you could pm one of the moderators....

[hide]I take back my old answer, i think the answer is 41 feet. I'll have a picture soon.[/hide]

Ok heres the picture, its confusing, as was the problem. [hide]Ok i cant explain all of it, for that will take all day, but i will explain some parts. First the arrows are there because that where pedro moved. For most of you finding AB was the problem so i found AB by drawing another radius PC and i know AP is 60 so use pythagorean theorem to find out AB. For the rest of the problem, it was basically the same thing, i had to draw extra radii and use pythagorean theorem. Lastly i concluded that HP is 41. Im sorry i cant explain the whole thing.[b] Please note that i can still  be very wrong.[/b][/hide]

I'll have to look at that a bit more... It seems right.  I am just wondering: Does anyone have the answer to this problem?
I am almost sure that Mr. McCrawford or Mr. Rusczyk may have the answer.  It would really help a lot.

GoBraves:  Your second answer is correct.

I'm afraid I don't know any easier way to solve this problem than to apply the Pythagorean Theorem after each turn.

O I realized my mistake now... although the problem is still incredibly complex... but ya I made a wierd assumption for the 60 answer.  [color=darkblue][size=75]<non-mathematical chatter removed by moderator rcv>[/size][/color]

Thanks a lot guys... That was one heck of a problem.

[hide]Last year I did every National test, kept my work neatly and put it in a binder so I would not have to redo every test every year.  After getting 41 as the correct answer, I wrote a note to myself at the bottom of the page.  It reads :

"Most ridiculous problem ever"

So don't feel bad if it took you awhile :) [/hide]

[hide]I first thought this was zero.  THen, I realized that wasn't correct because every time you move away from the center of the circle your distance to the sides grows smaller.  The first movement north is 60.  First, draw a line perpendicular to your north movement that touches your end of northward travel.  Then connect one end of that chord with the radius.  You just created a 30-60-90 triangle so the length of the leg is 60sqrt(3) and the distance traveled east is half of that or 30sqrt(3).  Next, from that point, make another perpendicular line.  Then, make a radius from the center to the point all the way on the other end of the perpendicular line.  THis creates a trapezoid.  We can solve for the unknown length of the perpendicular segment.  We split the trapezoid into a rectangle and a triangle.  THe rectangle's unknown length is 60 and the triangle's unknown length is [30sqrt(3)]^2+x^2=14400 so x^2=11700 and x=30sqrt(13).  Then, we add these areas and divide by 2.  THerefore, the distance we traveled south is 30+15sqrt(13).  Finally, we must draw one more perpendicular line from the half point.  Also from the midpoint of the first chord we drew, we connect that midpoint to the endpoint of our line.  We again have a trapezoid.  We know that the rectangle part's unknown length is 30sqrt(3).  The other unknown section is solved by making a radius to that point on the cirumference and a line from the center that's perpendicular to new perpendicular made line.  In other words, this is parallel to the line where we went 30+15sqrt(13).  That leg equals 2sqrt(3455).  So we went 15 sqrt(3)+sqrt(4355) on that.  I got (40.01166169) west and 24.08326913 south of the center.  This approxiamately equals 47 feet in a straight line away from the center.[/hide]

After I solved this problem for the very first time I felt the need to post a solution (mine is slightly different than above)
Hopefully this will benefit all future MATHCOUNTS competitors as this is widely known as the most ridiculous MATHCOUNTS problem ever written!

[hide="Solution"]
[img]http://i.imgur.com/oZQJqCI.png[/img]
The center of the circle is $A$. Let's set this circle on a coordinate grid, therefore $\text{A}(0,0)$. The problem states Pedro "walked due north halfway to the circle". For now, it's easy! The distance from the center to any point is the length of the radius, or $120$. The problem states "halfway" so we have Pedro goes north $60$ (see segment $\overline{AD}$). He's now at point $(0,60)$. It's not immediately obvious what $\overline{DE}$ is (what we need to find next) but we can draw $\overline{AC}=120$ and make a right triangle. It's a 30-60-90 triangle so we know that $\overline{DC}=60\sqrt3$ which means that $\overline{DE}=30\sqrt3$. Now Pedro is at point $\text{E}(30\sqrt3,60)$. We want to find $\overline{EG}$ (the problem states "he turned again and walked due south halfway to the circle). We need to find $\overline{EF}$ first though. How do we do that? Draw auxiliary line $\overline{AJ}$ which is the same length as $\overline{DE}$ or $30\sqrt3$. Using same logic to draw an auxiliary right triangle we draw $\overline{AJ}$. That means that by the Pythagorean Theorem, $\overline{JF}=30\sqrt{13}$. Add that to $\overline{EJ}=\overline{DA}$ and we get the length of $\overline{EF}$. Divide that by $2$ and we get $\overline{EG}\approx 84.08$. Using the same logic to draw $\overline{AK}$ we get that \[\overline{AK}=84.08-60=24.08\]. Continue in this process and you finally get that point $I$ is at $(-32.7984777334, -24.083269132)$. Plug that into your calculator, calculate distance to origin and you get $\approx 40.69083$. Round that up and your answer is a grand $\boxed{41}$ !

Feel free to ask if you have questions!
[/hide]

You could also graph it and use the formula $x^2 + y^2 = r^2$ and since you always have an $x$ or $y$ coordinate, you may be able to find it that way. I just did it yesterday. It was very challenging indeed and couldn't finish in the time frame provided.


A sniper has 0.8 chance to hit the target if he hit his last shot and 0.7 chance to hit the target if he missed his last shot.

It is known he missed on the 1st shot and hit on the 3rd shot.

What is the probability he hit the second shot?

Thanks.

The most straightforward method of solution simply involves computing all probabilities for each of the 4 possible outcomes.  (There are only four, because we have insufficient information to determine the probability distribution of the first shot.)

For each positive integer $ k$, let $ X_k$ be a Bernoulli random variable for the k(th) shot, where 1 denotes a hit, and 0 denotes a miss.  Then

$ \Pr[X_2 \equal{} 1 \cap X_3 \equal{} 1 | X_1 \equal{} 0] \equal{} (0.7)(0.8) \equal{} 0.56$
$ \Pr[X_2 \equal{} 1 \cap X_3 \equal{} 0 | X_1 \equal{} 0] \equal{} (0.7)(1 \minus{} 0.8) \equal{} 0.14$
$ \Pr[X_2 \equal{} 0 \cap X_3 \equal{} 1 | X_1 \equal{} 0] \equal{} (1 \minus{} 0.7)(0.7) \equal{} 0.21$
$ \Pr[X_2 \equal{} 0 \cap X_3 \equal{} 0 | X_1 \equal{} 0] \equal{} (1 \minus{} 0.7)(1 \minus{} 0.7) \equal{} 0.09.$

To verify, we can see that the sum of these probabilities is equal to 1.  Then it is almost trivial to observe that (leaving out the first shot, since all probabilities computed are already conditioned on a missed first shot):

$ \Pr[X_2 \equal{} 1 | X_3 \equal{} 1] \equal{} \frac {\Pr[X_2 \equal{} 1 \cap X_3 \equal{} 1]}{\Pr[X_2 \equal{} 1 \cap X_3 \equal{} 1] \plus{} \Pr[X_2 \equal{} 0 \cap X_3 \equal{} 1]}$,

which from the above gives $ \frac {0.56}{0.56 \plus{} 0.21} \equal{} \frac {8}{11}$.


Positive numbers x,y satisfy x^3+y^3 =x-y . Prove that x^2+y^2<1.


number theory
greatest common divisor
modular arithmetic
symmetry
number theory proposed

Find all primes p and q such that p^2+1| 2003^q+1 and q^2+1|2003^p+1. It's easy, but I like it.
  Harazi

Harazi, maybe you could now post the solution ?

I do not know what idea did I have when I created the problem, but I do not remember it. Let's give some help:
   First, let us take p \geq q. Now let r an odd prime dividing q 2 +1. Then r must be of the form 4k+1. We have r|2003^(2p)-1 and also r|2003^(r-1)-1. Thus, r must be a divisor of 2003^s-1, where s=GcD (r-1, 2p).Suppose s is 1 or 2. Then r divides 2003 2 -1=2002*2004. Since r is odd and of the form 4k+1, it must be 13. Then 13 divides 2003^p+1, which is false since 13|2003-1. Thus, s is different from 1 or 2. Then p must divide s, and so p divides r-1. Thus, r \geq p+1 \geq q+1. So, any odd prime factor of q 2 +1 must be at least q+1. Suppose that p and q are odd (the other case is trivial). Then we can write q 2 +1=2a, where each prime factor of a is odd and at least q+1. It follows that a is a prime and also that p|a-1. Thus, p|(q-1)(q+1) and since p>=q we must have p|q+1. Thus, since p is odd and p>=q we must have p=q and p|p+1, false. So, p=2 or q=2, from where p=q=2.

[quote]Find all primes $p$ and $q$ such that $p^2+1| 2003^q+1$ and $q^2+1|2003^p+1$.[/quote]

Obviously $p$ and $q$ cannot be $2$($5\nmid 2003^k+1$), and because $p^2+1\equiv 2\pmod{4}$ we must have some odd prime dividing it. Let there be a ([u]smallest[/u]) odd prime $r|p^2+1$, and $ord_{r}2003=k$. We have that $r\equiv 1\pmod{4}$ ($r|p^2+1$). Also we have $r|2003^{2q}-1$. So $k|2q$, thus $k=1,2,q,2q$.
$k=1$ and $k=q$ imply that $r|2$ which is a contradiction. For $k=2$ we get that $r|2003^2-1=2^3\cdot 3\cdot 7\cdot 11\cdot 13\cdot 167$, so the only possibility is $r=13$(all others are $\equiv 3\pmod{4}$), but $13\nmid 2003^q+1$ for any $q$. So we are left with $k=2q$, thus $2q|r-1\implies 2q<r<p+1$ because the smallest prime dividing $p^2+1$ cannot be greater than $\sqrt{p^2+1}<p+1$. Because of symmetry we also get $2p<q+1$ which implies that $q<1$, so no solutions.


inequalities
LaTeX
inequalities unsolved

[quote="stancioiu sorin"]If $ a,b,c$ are real postive numbers, prove the following inequality:
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+4(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a})$>4$\frac{3}{1+abc}$,
where a+b+c=1.[/quote]

We have:
$\frac{a+b+c}{a}+\frac{a+b+c}{b}+\frac{a+b+c}{c}+4(\frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a})=15+\frac{b+c}{a}+\frac{a}{b+c}+\frac{a+c}{b}+\frac{b}{c+a}+\frac{a+b}{c}+\frac{c}{a+b}+3(\frac{c}{a+b}+\frac{a}{b+c}+\frac{b}{c+a})$

We know that $\frac{x}{y}+\frac{y}{x}\geq2$ so $\frac{b+c}{a}+\frac{a}{b+c}\geq2$

Also $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq\frac{(a+b+c)^2}{2(ab+bc+ca)}\geq\frac{3(ab+bc+ca)}{2(ab+bc+ca)}=\frac{3}{2}$

We continue by this:
$15+\frac{b+c}{a}+\frac{a}{b+c}+\frac{a+c}{b}+\frac{b}{c+a}+\frac{a+b}{c}+\frac{c}{a+b}+3(\frac{c}{a+b}+\frac{a}{b+c}+\frac{b}{c+a})\geq15+2+2+2+\frac{9}{2}=\frac{51}{2}$

Inequality now is:
$\frac{51}{2}>\frac{12}{1+abc}$

From this we get:
$abc>-\frac{27}{51}$

I think it is true, isn't it?

[quote="stancioiu sorin"]If $ a,b,c$ are real postive numbers, prove the following inequality:
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+4(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a})$>4$\frac{3}{1+abc}$,
where a+b+c=1.[/quote]

Please learn to use $\text{\LaTeX}$ properly (given how regularly you are posting :) ).  I will use & in place of dollar signs; this is what you should have written

If &a,b,c& are real postive numbers, prove the following inequality:
&\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+4\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)>4\frac{3}{1+abc}&,
where &a+b+c=1&.

Now look at the results!

If $a,b,c$ are real postive numbers, prove the following inequality:
\[ \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+4\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)>4\left(\frac{3}{1+abc}\right),  \]
where $a+b+c=1$.

The problem is easy for Olympiad section. Just use Cauchy in Engel form two times to have

$LHS\geq 27>\frac{12}{1+abc}$  which is trivial


Find solution in naturals of the equation
\[ u^5\plus{}v^5\plus{}w^5\plus{}x^5\plus{}y^5\equal{}z^5\]

Bad title http://www.mathlinks.ro/viewtopic.php?t=144403 . Locked!


I'm using

[code]\begin{document}
\thispagestyle{empty}
\vspace*{\fill}
\centerline{\Huge{text}}

\vspace{8ex}
\centerline{\includegraphics[scale=0.8]{image.png}}

\vspace{8ex}
\centerline{text}
\vspace*{\fill}\newpage
\lhead{\textbf{text}\hfill \textbf{\href{URL}{its name}}\\[-1ex]\hrulefill\\text\hfill text}
\renewcommand{\headrulewidth}{0pt}[/code]

but the

[code]\lhead{\textbf{text}\hfill \textbf{\href{URL}{its name}}\\[-1ex]\hrulefill\\text\hfill text}
\renewcommand{\headrulewidth}{0pt}[/code]

it's not being generated, it appears the first page containing the image only.

It's important to always give the complete preamble but I guess that you've probably forgotten \pagestyle{fancy} so fancyhdr stuff is not being processed.


trigonometry
geometry
circumcircle
angle bisector
geometry proposed

[color=darkblue][b][u]Method I (trigonometric).[/u][/b] $A=3C$ $\Longleftrightarrow$ $(\exists) x\in(0,\frac{\pi}{4})$ so that $\{\begin{array}{c}A=3x\\\ B=\pi-4x\\\ C=x\end{array}$. Apply the Sinus' theorem in $\triangle ABD$ :

$\frac{AB}{\sin\widehat{ADB}}=\frac{AD}{\sin\widehat{ABD}}$ $\Longleftrightarrow$ $\frac{c}{\sin 45}=\frac{b}{2\sin (45+3x)}$ $\Longleftrightarrow$ $2\sin x\sin (45+3x)=\sin 4x\sin 45$ $\Longleftrightarrow$

$\cos (2x+45)=\cos (45+4x)+\sin 4x\sin 45$ $\Longleftrightarrow$ $\cos (2x+45)=\cos 4x\cos 45$ $\Longrightarrow$ $\cos 2x-\sin 2x=\cos 4x$ $\Longleftrightarrow$

$\cos 2x-\sin 2x=\cos^{2}2x-\sin^{2}2x$ $\Longleftrightarrow$ $\cos 2x-\sin 2x=0\ \ \vee\ \ \cos 2x+\sin 2x=1$ $\Longleftrightarrow$

$\tan 2x=1\ \ \vee\ \ \tan x=1$ $\Longleftrightarrow$ $x\in\{\frac{\pi}{8},\frac{\pi}{4}\}\cap (0,\frac{\pi}{4})$. Therefore, $x=\frac{\pi}{8}$, i.e. $\boxed{\ C=\frac{\pi}{8}\ }$.

[b][u]Extension.[/u][/b] $\{\begin{array}{c}n\in\mathcal N\ ,\ n\ge 2\\\\ \triangle ABC\ ,\ A=nC\\\\ D\in (AC)\ ,\ DA=DC\\\\ m(\widehat{ADB})=\phi\end{array}\|$ $\implies$ $\{\begin{array}{c}\blacktriangleleft\boxed{\ \tan\phi=\frac{2\sin nx\sin x}{\sin (n-1)x}\Longleftrightarrow \tan (n-1)x=\frac{1-\cos 2x}{\tan\phi-\sin 2x}\ }\blacktriangleright\\\\ \blacktriangleleft\boxed{\ \phi=\frac{\pi}{4}\Longleftrightarrow tan (n-1)x=\tan x\tan2x\tan (x+\frac{\pi}{4})\ }\blacktriangleright\end{array}$. Indeed :
$1\blacktriangleright$ Apply the well-known relation to the median $[BD]$ in $\triangle ABC$ : $\frac{DA}{DC}=\frac{BA}{BC}\cdot\frac{\sin \widehat{DBA}}{\sin\widehat{DBC}}$ $\Longleftrightarrow$ $\sin x\sin (nx+\phi)=\sin nx\sin (\phi-x)$ $\Longleftrightarrow$
$\cos [(n-1)x+\phi ]-\cos[(n+1)x+\phi ]=\cos [(n+1)x-\phi ]-\cos [(n-1)x+\phi ]$ $\Longleftrightarrow$ $\cos[(n+1)x-\phi ]+\cos [(n+1)x+\phi ]=2\cos [(n-1)x+\phi ]$ $\Longleftrightarrow$
$\cos [(n+1)x\cos \phi =\cos [(n-1)x+\phi ]$ $\Longleftrightarrow$ $\tan \phi =\frac{\cos (n-1)x-\cos (n+1)x}{\sin (n-1)x}$ $\Longleftrightarrow$ $\boxed{\ \tan \phi =\frac{2\sin nx\sin x}{\sin (n-1)x}\ }$.
$2\blacktriangleright\phi =\frac{\pi}{4}$ $\Longleftrightarrow$ $2\sin nx\sin x=\sin (n-1)x$ $\Longleftrightarrow$ $\cos(n-1)x-\cos (n+1)x==\sin (n-1)x$ $\Longleftrightarrow$ $\cos (n-1)x-\sin (n-1)x=\cos[(n-1)x+2x]$ $\Longleftrightarrow$ $\cos (n-1)x-\sin (n-1)x=\cos (n-1)x\cos 2x-\sin (n-1)x\sin 2x$ $\Longleftrightarrow$ $\cos (n-1)x(1-\cos 2x)=\sin (n-1)x(1-\sin 2x)$ $\Longleftrightarrow$ $\tan (n-1)x=\frac{2\sin^{2}x}{(\sin x-\cos x)^{2}}$ $\Longleftrightarrow$ $\tan (n-1)x=\frac{2\tan^{2}x}{(1-\tan x)^{2}}$ $\Longleftrightarrow$ $\tan (n-1)x=\tan x\cdot \frac{2\tan x}{1-\tan^{2}x}\cdot\frac{1+\tan x}{1-\tan x}$ $\Longleftrightarrow$ $\boxed{\ \tan (n-1)x=\tan x\cdot\tan 2x\cdot \tan(x+\frac{\pi}{4})\ }$.

[u][b]My dear K®izalid[/b][/u], for $n=3$ you "ll obtain the your proposed problem $3$.

For $n=2$ you"ll obtain the my present for you, the problem $5$ from http://www.mathlinks.ro/Forum/viewtopic.php?t=146109 [/color]

Thanks Virgil

An elementary solution is based raising $\overline{ED}\perp\overline{AC}$ ($E\in\overline{BC}$) and the rest is to do one that another thing more.

[quote="K®izalid"][color=darkblue]If, in a $\triangle ABC$, median $\overline{BD}$ is such that $\measuredangle{\text{ }}CAB = 3 \cdot \measuredangle{\text{ }}C$, and $\measuredangle{\text{ }}ADB = 45^\circ$,
 then prove that $\measuredangle{\text{ }}C = 22,5^\circ$[/color][/quote]
[color=indigo][b]Proof. [/b]
Take $E\in BC: DE\perp AC,$ thus we have: $DB$ is the angle bisector of angle $ADE\quad (1)$
On the otherhand, we have: $\angle BAE=\angle BEA(=2 \angle C)$ so $BE=BA\quad (2)$
From $(1)$ and $(2),$ we have: $B$ is the midpoint of arc $AE$ (not contain $D$) of circumcircle of triangle $ADE$
i.e. the quadrilateral $ABMD$ is cyclic and so $\angle ABC=\angle ADE=90^\circ$
Hence: $\angle ACB=\frac{\pi}{8}$[/color]

That's right  :lol:

Thanks [color=indigo]April[/color]

[size=134][color=darkred][b]Extension II.[/b][/color][/size] $\left\{\begin{array}{c}n\in \mathcal N\ ,\ n\ge 2\ ;\ r>0\\\\ \triangle ABC\ ,\ A=n\cdot C\\\\ D\in (AC)\ ,\ DA=r\cdot DC\\\\ m(\widehat{ADB})=\phi\end{array}\right\|$ $\implies$ $\boxed{\ (r+1)\cdot\cot\phi=r\cot x-\cot nx\ }$[color=darkred], where[/color] $\cot x=\frac{\cos x}{\sin x}$.


let $ a$ and $ b$ be arbitrary integers.prove that if $ k$ is an integer not divissible by 3 then $ (a\plus{}b)^{2k}\plus{}a^{2k}\plus{}b^{2k}$ is divisible by $ a^2\plus{}ab\plus{}b^2$

Define $ P(x) \equal{} (1 \plus{} x)^{2k} \plus{} x^{2k} \plus{} 1$. Notice that if $ x^2 \plus{} x \plus{} 1 \equal{} 0$, then $ (1 \plus{} x)^2 \equal{} x$, and from the fact that $ k$ is not divisible by $ 3$, {$ (1 \plus{} x)^{2k} \equal{} x^k, x^{2k} \equal{} {(x^2)}^k, 1$} are 3 distinct primitive roots of unity of order $ 3$, and $ P(x) \equal{} 1 \plus{} \omega \plus{} \omega^2 \equal{} 0$ ($ \omega$ is a primitive root of unity of order $ 3$)

Thus, $ x^2 \plus{} x \plus{} 1 | P(x)$, so $ (1 \plus{} x)^{2k} \plus{} x^{2k} \plus{} 1 \equal{} (1 \plus{} x \plus{} x^2)Q(x)$ for some $ Q$ with integer coefficients.
Plug $ x \equal{} \frac {a}{b}$, multiply by $ b^{2k}$, to get: $ (a \plus{} b)^{2k} \plus{} a^{2k} \plus{} b^{2k} \equal{} Q(\frac {a}{b})(b^2 \plus{} ab \plus{} a^2)b^{2k \minus{} 2}$. $ Q$ is of degree $ 2k \minus{} 2$, so $ Q(\frac {a}{b})b^{2k \minus{} 2}$ is an integer and the result follows.

Let $ P(x) \equal{} (x \plus{} 1)^{2k} \plus{} x^{2k} \plus{} 1$. Note that $ P(\omega) \equal{} (\omega \plus{} 1)^{2k} \plus{} \omega^{2k} \plus{} 1 \equal{} \omega^{k} \plus{} \omega^{2k} \plus{} 1 \equal{} 0$, since $ 3\not |k$, where $ \omega \equal{} e^\frac {2\pi i}{3}$. Thus $ x \minus{} \omega\mid P(x)$ and $ x \minus{} \omega^2\mid P(x)$. Hence the substitution $ x \equal{} a/b$ implies $ (a \minus{} b\omega)(a \minus{} b\omega^2) \equal{} a^2 \plus{} ab \plus{} b^2\mid (a \plus{} b)^{2k} \plus{} a^{2k} \plus{} b^{2k}$.

EDIT: Oops, beaten. :oops:


[quote="April"]Let $ x_1$, $ x_2\in\mathbb{R}$ and $ t\in (0,\; 1)$. Prove that
\[ 5^{ \minus{} tx_1 \minus{} (1 \minus{} t)x_2} < t5^{ \minus{} x_1} \plus{} (1 \minus{} t)5^{ \minus{} x_2}
\]
[/quote]
It's Jensen for $ f(x)\equal{}5^x.$  :wink:


symmetry
geometry
perpendicular bisector
geometry unsolved

Let $ ABCD$ be an isosceles trapezium. Determine the geometric location of all points $ P$ such that \[ |PA| \cdot |PC| \equal{} |PB| \cdot |PD|.\]

WLOG, assume $ AB \parallel DC, AB >\equal{} DC$ and $ Q \equiv BC \cap AD.$ Suppose $ \frac {PB}{PC} \equal{} k.$ Then $ P$ is on a circle $ \mathcal K$ coaxal with the points $ B, C.$ Since $ \frac {PA}{PD} \equal{} \frac {PB}{PC} \equal{} k,$ $ P$ is also on a congruent circle $ \mathcal K'$ coaxal with the points $ A, D.$ By symmetry, the circles $ \mathcal K, \mathcal K'$ intersect on the common perpendicular bisector $ p$ of the segments $ AB, DC,$ therefore $ p$ belongs to the locus. When the circle $ \mathcal K$ is centered at $ Q,$ the congruent circle $ \mathcal K'$ is also centered at $ Q$ and they coincide into one circle $ (Q).$ The circle $ (Q)$ also belongs to the locus. Let the circle $ (Q)$ cut the segment $ BC$ at $ R.$ Then (see [url=http://www.mathlinks.ro/viewtopic.php?t=121062]constant value[/url]) $ \sqrt {\frac {QB}{QC}} \equal{} \frac {RB}{RC} \equal{} \frac {QB \minus{} QR}{QR \minus{} QC}$ $ \Longrightarrow$ $ QR \equal{} \sqrt {QB \cdot QC}.$

In conclusion, the locus is $ p \cup (Q).$ If $ AB \equal{} DC,$ the intersection $ Q$ is at infinity and the circle $ (Q)$ becomes the common perpendicular bisector $ q$ of $ BC, AD.$


My dad told me what I think is an interesting idea but I am not sure if it is actually being done. He believes that oil companies this year are going to show their income being much less then it really is in order to decrease their tax burden. I was wondering if any of you agree with this or what you think of it. I think it seems quite logical even though at the same time, just funny how a company would want to make their profits seem less?

Well, in the reverse, why would a company want their profits to seem like more? It would only increase their taxes, and lowering their 'stated' income is only seen by the IRS, nobody else cares. Of course, if the company is caught, I believe thats called cheating on their taxes... which isnt legal, and they can get fined or all thrown in jail.. or both, so, not too wise.

Well, companies definetly make their income look greater, as I believe were the cases in recent corporate scandals. I believe that it is done in order to make more investors buy stocks and other financial instruments when they see that the company does well, if the doing well is artificial, the investors end up losing, even though its not their fault.

Well, it really works both ways. By making your income seem higher, investors will buy stock in you, but it wil be fake, and you will inevitably fall - many companies yet still do not realize this.
By making your income seem lower, however, you may still be doing well as a company. The only problem of course, is that its illegal (so is the other one, but the other one you are going down anyway..), and you WILL get caught.

Oil companies, if they are going to cheat at all, will probably make their income seem lower, because they arent looking for investors, but for saving money, which is what they have to do after Katrina, and what lowering their stated income will do.

Mis-stating profits=FRAUD.

See:
Worldcom
Enron
Health South
etc.

Right, but wasn't Enron and all the other companies sued for increasing their income, not making it lower?

They were - but regardlesss making their income seem lower is just as illegal. Only its easier to catch someone who makes their taxes seem higher, because the secret always comes out when the company goes bankrupt, after 'supposedly' doing so well..

[quote]oil companies this year [may] show their income being much less then it really is in order to decrease their tax burden. [/quote]

There are various schemes such as undertaking large purchases during a high-profit year, to spread out the tax burden over several years.


calculus
function
integration
modular arithmetic
number theory
relatively prime
algebra open

what is langranges  multipliers? what is fermats theorem?

Hello! I do not know the answer to the second question, but I can answer the first one. The method of Lagrange multipliers, invented by Lagrange at an age of nineteen years old, is a technique of maximizing or minimizing a function subject to a constraint. Namely, the theorem states that $\nabla f(x, y) = \lambda \nabla g(x, y),$ where $f(x, y)$ is the function to be maximized or minimized $g(x, y)=c$ is the constraint curve. By solving the aforementioned equation, one can find the extrema of $f(x, y)$ on the curve $g(x, y)=c$.

[quote="leepakhin"]There are many theorems named after Fermat. Here are the two most famous ones.
[b]Fermat's Last Theorem[/b]
Let $n$ be an integer greater than 2. Then there is no non-zero integral solution $(x,y,z)$ of the equation $x^n+y^n=z^n$.
[b]Fermat's Little Theorem[/b]
Let $p$ be a prime and $a$ be an integer relatively prime to $p$. Then $a^{p-1}\equiv 1\pmod{p}$.[/quote]

Well, of crouse, Fermat's little theorem can be arranged into many different forms, such as $a^p\equiv a\pmod{p}$. :)

Masoud Zargar


okay simmilar solution to hxtung but with Cauchy! :D

we know that 1/a+1/b+1/c=2 and we rewrite it like so:
(a-1)/a+(b-1)/b+(c-1)/c=1
sqrt(a+b+c)*sqrt((a-1)/a+(b-1)/b+(c-1)/c)>=sum(sqrt(a)*sqrt((a-1)/a))=sum(sqrt(a-1)). in the last ineq I applied Cauchy Schwartz!

:D bye!

[quote=georg]Let a,b,c be positive numbers >1 so that 1/a+1/b+1/c=2. Prove that sqrt(a+b+c)>=sum(sqrt(a-1)).

bye![/quote]
Let $ a,b,c \geq1$ such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2$. Prove that$$\sqrt{a+b+c}\ge\sqrt{a-1}+\sqrt{b-1}+\sqrt{c-1}.$$
([url=http://www.artofproblemsolving.com/community/c6h1100451p4954922]Iran 1998[/url])
Let $ a,b,c \geq1$ such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2$. [url=http://artofproblemsolving.com/community/c4h1120413p5678331]Prove that[/url]$$\sqrt{a-1}+\sqrt{b-1}+\sqrt{c-1}\ge 2$$


analytic geometry
graphing lines
slope
linear algebra

Hello, I'm new in the forum.
I'm here just because I like Maths... And sorry, because i'm not a great solver problem  :oops: ... but i'm here to learn.

(sorry if my english is not so good, I'm Chilean)

About the question:

[hide]The equation of the line can be (writted?) like:
$\frac{2}{3}x-\frac{7}{3}=y$

with this information, we know that the slope of the new line is $\frac{2}{3}$

so, if $(x_{1},y_{1})=(-6,-5)$ we can use that

$y-y_{1}=m(x-x_{1})$   (where $m$ is the slope)

Then, for this case:

$y+5=\frac{2}{3}(x+6)$

$\boxed{y=\frac{2x}{3}-1}$[/hide]

 :lol: i don't know how to say "saludos"  :P

[quote="jclarkemathL314159"]No, two lines are only parallel if they have the same slope.[/quote]
His line [i]does[/i] have the same slope as the original one.  :wink:  :roll:

By the way Gazoo,  :welcomeani:  to AoPS/ML! Good to have you here!
Since you're new, please try to hide your answers/solutions (by highlighting it and clicking on the "Hide" button above the message). Thanks.  :)

By the way, "saludos" = "greetings"

Ok, Thanks Nebula  :lol:

Edited Post!  :P

Regards!! (=¿saludos?)

[quote="mary15"]what is the equation of the a line parallel to the line 2x-3y=7 and contains the point (-6,-5)[/quote]

[hide]using some linear algebra
$y=\frac{2x}{3}-1$[/hide]

[quote="me@home"]By "linear algebra" do you mean the first year college course or basic middle school algebra? :rotfl:
and btw good job on posting the answer to a problem with a solution already posted, without showing your "supposedly" alternate solution...[/quote]
[quote="Klebian"][quote="i_like_pie"]Yes, I did mean you. The problem says create the number 24 using 3, 3, 7, 7, and basic symbols. You just put 24.  :?:[/quote]

It's because he has a habit of looking at other people's answers and then making a post containing solely the answer to increase his post count. However, I guess he didn't even read the question this time![/quote]

[quote="mary15"]what is the equation of the a line parallel to the line 2x-3y=7 and contains the point (-6,-5)[/quote]

$2(x+6)-3(y+5) = 0$

;)


number theory
greatest common divisor
algorithm
Euclidean algorithm

For any integer $ a$ prove that

1.) $ gcd(a, a \plus{} 1) \equal{} 1$
2.) $ gcd(2a \plus{} 1, 9a \plus{} 4) \equal{} 1$
3.) $ gcd(5a \plus{} 2, 7a \plus{} 3) \equal{} 1$

[quote="alanchou"]Euclidean algorithm brings them all down to $ \text{gcf}(a,1)$.[/quote]
 can you elaborate the solution?

[hide]Yeah, Euclidean Algorithm on them all:
1) $ gcd(a,a\plus{}1)\equal{}gcd(1,a)\equal{}1$.
2) $ gcd(2a\plus{}1,9a\plus{}4)\equal{}gcd(a,2a\plus{}1)\equal{}gcd(1,a)\equal{}1$.
3) $ gcd(5a\plus{}2,7a\plus{}3)\equal{}gcd(2a\plus{}1,5a\plus{}2)\equal{}gcd(a,2a\plus{}1)\equal{}gcd(a,1)\equal{}1$.[/hide]
I'm half-asleep, so might be messed up in calculations.

No, they're correct

and $ gcf(a,a\plus{}1)$ can be seen using intuition as well

"_________ is intuitive" gives no explanation of your thought process.  I believe what you mean is that if $ \gcd(a, a\plus{}1) \equal{} d > 1$ then $ d | a, d | a\plus{}1$ hence $ d | 1$ - a contradiction.  This is not intuitive to everyone; formalized, it is the Euclidean Algorithm itself, and then we might as well say that all of these problems are intuitive.

[b]Generalization:[/b]  For what quadruples of integers $ x, y, z, w$ are we guaranteed that

$ \gcd(xa \plus{} y, za \plus{} w) \equal{} 1$

for all integers $ a$?


function
calculus
derivative
linear algebra
matrix
linear algebra unsolved

Show that the function of three variables

$ f(x) \equal{} x_{1}^{2}\plus{}x_{2}^{2}\plus{}3x_{3}^{2}\plus{}x_{1}x_{2}\plus{}3x_{1}x_{3}\plus{}3x_{2}x_{3}\plus{}x_{3}^{3}$

is convex on the set $ x_{3}\geq 0$

A $ C^{2}$ function is convex iff its second derivative matrix is positive definite. The numbers are different, but it's the same idea as [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=933747]your other question[/url].

Its Hessian is $ \begin{pmatrix}2 & 1 & 3\\ 1 & 2 & 3\\ 3 & 3 & 6\plus{}6x_{3}\end{pmatrix}$. It suffices to prove that this matrix, without $ x_{3}$, is positive semidefinite.


integration
calculus
calculus computations

Find the value of $ S \equal{} \sum_{n\equal{}0}^{\infty}\frac{1}{n^2\plus{}3n\plus{}3}$.

[hide="Well known formula"][url]http://www.mathlinks.ro/viewtopic.php?t=183868[/url][/hide]
$ \sum_{k\equal{}0}^{\infty}{\frac{1}{(k\plus{}\frac{1}{2})^2 \plus{} \alpha^2}}\equal{}\frac{\pi}{2\alpha}\cdot \tanh(\pi\alpha)$

$ \sum_{k\equal{}0}^{\infty}{\frac{1}{(k\plus{}1\plus{}\frac{1}{2})^2 \plus{} \alpha^2}}\equal{}\sum_{k\equal{}1}^{\infty}{\frac{1}{(k\plus{}\frac{1}{2})^2 \plus{} \alpha^2}}\equal{}\frac{\pi}{2\alpha}\cdot \tanh(\pi\alpha)\minus{}\frac{4}{4\alpha ^2 \plus{} 1}$

$ \sum_{n\equal{}0}^{\infty}\frac{1}{n^2\plus{}3n\plus{}3} \equal{} \frac{\sqrt{3}\pi}{3}\cdot \tanh\left(\frac{\sqrt{3}\pi}{2}\right) \minus{} 1$

Find the value of $ S \equal{} \sum_{n\equal{}0}^{\infty}\frac{1}{(8n\plus{}1)(8n\plus{}3)}$.

$ \frac{1}{8n\plus{}1}\equal{}\int_{0}^{1}{x^{8n}\,dx}$ and $ \frac{1}{8n\plus{}3}\equal{}\int_{0}^{1}{x^{8n\plus{}2}\,dx}.$

After some straightforward computations, last integral gives the expected result.


geometry
similar triangles
geometry unsolved

[quote="Beat"]Let P be a piont inside the triangle ABC and let $A_1B_2, B_1C_2$ and $C_1A_2$ be the lines passing trough P and parallel with the sides $AB,BC,CA$ respectively. $A_1,A_2$ are on BC, $B_1,B_2$ are on AC and $C_1,C_2$ are on AB. Prove that $S_{A_1A_2B_1B_2C_1C_2}\ge \frac{2}{3}S_{ABC}$[/quote]
Use this fact$\sqrt{S_{ABC}}=\sqrt{S_{PC_1C_2}}+\sqrt{S_{PA_1A_2}}+\sqrt{S_{PC_1C_2}}$(It is easy to prove this from similar triangles) ;) .

Prove it, and please finish the inequality. Thanks

[quote="fuzzylogic"][quote="Tiks"]oops....... :huh:  :oops:  :rotfl: ,look I have proved stronger ineq.! .[/quote]

That's because there is a little mistake in the proof![/quote]
 :blush: Sorry I have seen my mistake :)
Ther should be
$S_{A_1A_2B_1B_2C_1C_2}=S_{PC_1C_2}+S_{PA_1A_2}+S_{PB_1B_2}+(S_{ABC}-S_{A_1A_2B_1B_2C_1C_2})$
and not
$S_{A_1A_2B_1B_2C_1C_2}=S_{PC_1C_2}+S_{PA_1A_2}+S_{PB_1B_2}+2(S_{ABC}-S_{A_1A_2B_1B_2C_1C_2})$ :blush:


[quote="thanhnam2902"][color=darkblue]Let $ \frac {x}{y \plus{} z \plus{} t} \equal{} \frac {y}{z \plus{} t \plus{} x} \equal{} \frac {z}{t \plus{} x \plus{} y} \equal{} \frac {t}{x \plus{} y \plus{} z}$. Find $ \boxed{A \equal{} \frac {x \plus{} y}{z \plus{} t} \plus{} \frac {y \plus{} z}{t \plus{} x} \plus{} \frac {z \plus{} t}{x \plus{} y} \plus{} \frac {t \plus{} x}{y \plus{} z}}$[/color][/quote]
$ \frac {x}{y \plus{} z \plus{} t} \equal{} \frac {y}{z \plus{} t \plus{} x} \equal{} \frac {z}{t \plus{} x \plus{} y} \equal{} \frac {t}{x \plus{} y \plus{} z} \equal{} \frac{x\plus{}y\plus{}z\plus{}t}{3(x\plus{}y\plus{}z\plus{}t)}\equal{} \frac{1}{3}$
 suppose  $ x\equal{}Max(x,y,z,t)$ then $ 3x\equal{}y\plus{}z\plus{}t \iff x\equal{}y\equal{}z\equal{}t$ .
then $ A\equal{}4$

What's the answer for $ x\plus{}y\plus{}z\plus{}t\equal{}0$? :wink:

[quote="kunny"]What's the answer for $ x \plus{} y \plus{} z \plus{} t \equal{} 0$? :wink:[/quote]
 oh  if $ x\plus{}y\plus{}z\plus{}t\equal{}0$ then $ A\equal{}\minus{}4$


Congratulation for all who will take the IMO this year. The following problem is dedicated to all of you as a small gift before taking the test . Have nice days and best results in this IMO.

Problem. Let $ a,b,c$ be real numbers. Prove that
\[ \sqrt[3]{2a^{2}-bc}+\sqrt[3]{2b^{2}-ca}+\sqrt[3]{2c^{2}-ab}\ge 0\ (1). \]
Actually, the general problem is that

Let $ a,b,c$ be real numbers. For $ k\le \frac{1+\sqrt{513}}{16},$ prove that
\[ \sqrt[3]{ka^{2}-bc}+\sqrt[3]{kb^{2}-ca}+\sqrt[3]{kc^{2}-ab}\ge 0. \]
[b]
But I would like to see a special idea to solve $ (1)$ deftly and concisely.[/b]

The ceremony may end at the moment. Does anyone get any idea to solve this one? It seems to be very simple.

As the last time for me to ask about this problem? Does anyone get any idea to solve it? Especially you, VASC?

[quote="Albanian Eagle"]So this inequality is equivalent to:
$ 2\sum a^{2}-\sum ab\geq 3\sqrt[3]{(2a^{2}-bc)(2b^{2}-ac)(2c^{2}-ab)}$
now obviously we have to consider only $ a,b,c\geq 0$ and $ ab\geq 2c^{2}$ $ ac\geq 2b^{2}$
how do I continue from here?! :huh:[/quote]

This idea is the same as the first step I have done to solve this one  :blush: Thinking of what will happen  :lol:

We only need to prove $ a^{2}+2x^{2}\geq 3\sqrt[3]{x^{2}(2a^{2}-x^{2})(a-2x)^{2}}$ where $ x=\sqrt{bc}$ but we only need
$ 2x^{2}\geq \frac{4x^{2}-4ax}{3}$ but this is strange  :huh:
equality only for a=b=c=0

[quote="Yuriy Solovyov"]$ k \leq \frac{1+\sqrt{513}}{16}\approx 1.478$.
so what about first inequality, where $ k=2$ :ninja:[/quote]

I mean $ k \ge \frac{1+\sqrt{513}}{16}$. But at the moment, I can not edit my post before.

yeah but is the equality case only 0,0,0. Or have I done some stupid mistake :oops:

No Eagle. Case of Equality is

\[ (a,b,c)~ \left( \sqrt{\frac{8k-1}{k}},1,1\right). \]


integration
function
calculus
calculus computations

Find $ \int\frac{sin^{7}(x)}{(1\plus{}x^4)} dx$

It is not an elementary function as far as I can tell. Where does the problem come from? (I'm asking because I suspect that the original question was about some [b]definite[/b] integral like $ \int_0^\infty$ or whatever).


[quote="rnwang2"][hide]
$12\cdot\left(\frac{1}{2}\right)\cdot\left(\frac{1}{6}\right)^{11}$[/hide][/quote]

you could do it another way

[quote="now a ranger"]rnwang, where does the 12 come from?

12(1/2) etc.

I don't see why you have to multiply 12...  :o[/quote]
You're rolling 12 dice. The probability of a prime on top is $\frac{1}{2}$. The other 11 dice must have a 1 on top.

[quote="rnwang2"][quote="sonny"][quote="now a ranger"]Twelve fair dice are rolled. What is the probability that the product of the numbers on the top faces is prime?[/quote]

This can only happen if one of the numbers is 1 so you can have 1,3 1,2 1,5 or 5,1 3,1 2,1 to get $\frac{1}{6}$.[/quote]
But the problem states you're rolling twelve dice.[/quote]

O oops...you could just look at it like this....... 1, 1, 1, 1, 1, 1, 1,1,1,1,1,2 and how many ways you can arrange that. Then just replace the 2 with a 3 or a 5.

[quote="now a ranger"]but when you're rolling 2 dice, and what is the probability of rolling two 6's, it would be 1/36, and 2 wouldn't be multiplied onto it....[/quote]
But you're rolling [i]12[/i] dice.

If the problem had said 2 dice, the probability would be

$2\cdot \left(\frac{1}{2}\right)\cdot\left(\frac{1}{6}\right)^{1}=\frac{1}{6}$, which is what [b]sonny[/b] had in his post.

I guess I'm just wrong, and I don't understand the connection or the concept..  :(

[hide]Ya the answer is $\frac{1}{6^{10}}$. You have to have 11 1s and 1 prime you can arrange those numbers in 12 different ways. Now you have three primes from 1-6, so you have a total of 36 ways of arranging that. Now there are a total of $6^{12}$ combinations of dice. So $\frac{36}{6^{12}}= \frac{1}{6^{10}}$[/hide]

[quote="sonny"][hide]Ya the answer is $\frac{1}{6^{10}}$. You have to have 11 1s and 1 prime you can arrange those numbers in 12 different ways. Now you have three primes from 1-6, so you have a total of 36 ways of arranging that. [/hide][/quote]

Oh... now I understand!  :lol:


LaTeX
algorithm
number theory
Euclidean algorithm
superior algebra
superior algebra solved

Can someone help me?
[b]Show that Z_m^* is group[/b] if and only if [b]m[/b] is [b]prime number[/b].:?
 :wow:

Thanks
Tanja

[quote="ZetaX"]Please use $\LaTeX$, it's better to understand then.

$Z_m=\mathbb{Z}/m\mathbb{Z}$¿
I don't think $\mathbb{Z}/n\mathbb{Z}^*$ ring then, even if $m$ is prime...[/quote]

I will teach me Latex later, I havn´t time now.
 :blush:
Z_m^*={1,2,...,m-1} with binär operation "multiply modulo m".

Best regards
Tanja

[quote="blahblahblah"]Rings have two operations, and there's only one logical zero element in $\mathbb{Z}/n\mathbb{Z}$ - what are you going to replace it with?[/quote]
You have right.
I made mistake - I must show that it is group. :blush:
Thanks.
Tanja

Okay.  The identity is $1$.  By elementary number theory, e.g, the euclidean algorithm, we know that if $p$ is prime and $(a,p)=1$, then there exists $x$ such that $ax\equiv 1\bmod p$.

Can you see that this implies that every nonzero element in $\mathbb{Z}/p\mathbb{Z}$ has an inverse?  So $\mathbb{Z}/p\mathbb{Z}^*$ is a group.

For the other direction, say $n=ab$, $a,b>1$.  Find nonzero elements in $\mathbb{Z}/n\mathbb{Z}$ that multiply to zero.  This shows that there are elements in $\mathbb{Z}/n\mathbb{Z}$ for which multiplication is not a binary operation...

Thank You Very Much :lol:

Best regards
Tanja


analytic geometry
trigonometry
quadratics
geometry
geometric transformation
rotation
reflection

Square $ABCD$ has center $O$, $AB=900$, $E$ and $F$ are on $AB$ with $AE<BF$ and $E$ between $A$ and $F$, $m\angle EOF =45^\circ$, and $EF=400$. Given that $BF=p+q\sqrt{r}$, wherer $p,q,$ and $r$ are positive integers and $r$ is not divisible by the square of any prime, find $p+q+r$.

[hide="Solution"]
Let $P$ be the point on $AB$ such that $OP\perp AB$. ($AP=PB=450$) Let $a=EP$ and $b=FP$ ($a>b$ since $AE<BF$)

$\tan{\angle EOP}=\frac{a}{OP}=\frac{a}{450}$
$\tan{\angle FOP}=\frac{b}{OP}=\frac{b}{450}$

Using $\tan{x+y}=\frac{\tan{x}+\tan{y}}{1-\tan{x}\tan{y}}$, we have $\tan{\angle EOF}= \frac{\frac{a}{450}+\frac{b}{450}}{1-\frac{a}{450} \cdot\frac{b}{450}}= \frac{\frac{a+b}{450}}{\frac{450^2-ab}{450^2}}= \frac{400}{450}\cdot\frac{450^2}{450^2-ab}=\frac{400\cdot 450}{450^2-ab}=1$

So $18000=450^2-ab\Longleftrightarrow ab=22500$

$\therefore a+b=400$, $ab=22500$

$a$ and $b$ are the solutions to the quadratic equation $x^2-400x+22500$. Solving for $x$, we obtain $x=200\pm 50\sqrt{7}$.

Since $a>b$, $b=200-50\sqrt{7}$

Hence $BF=PB-PF=450-b=250+50\sqrt{7}$

$p=250$, $q=50$ and $r=7$, so $p+q+r=\boxed{307}$

[/hide]

Suggestion! If you're going to have solutions, why not have them all in collective pdf form like the tests themselves.  Good work you guys!  :D

Because this way, a person can be looking at the problem resource page, go to a test, click on the problem number, and see the solution for that problem right away, as well as different methods of solving it.

I hate to revive old threads but can anybody post a solution using complex coordinates and rotations?

Note that $\triangle BEO \sim \triangle AOF.$  This means that $\dfrac{BE}{AO}=\dfrac{BO}{AF}.$

Ok this is revived, but are there other ways of solving it than the sol provided above?

[hide = A similar triangle  solution]
Construct $BO, AO.$ Let $\angle{FOB} = \alpha.$ Also let $FB = x$ then $AE = 500-x.$ We then have from simple angle-chasing:
\begin{align*}
\angle{BFO} = 135 - \alpha \\
\angle{OFE} = 45 + \alpha \\
\angle{EOA} = 45 - \alpha \\
\angle{AEO} = 90 + \alpha \\
\angle{OEF} = 90 - \alpha.
\end{align*}
From AA similarity we have $$\triangle{EOB} \sim \triangle{EFO}.$$ This gives the ratios,
$$\dfrac{400 + x}{EO} = \dfrac{450\sqrt{2}}{FO}.$$
Similarly from AA similarity $$\triangle{FOA} \sim \triangle{FEO}.$$ So we get the ratios $$\dfrac{EO}{450\sqrt{2}} = \dfrac{FO}{900-x}.$$ We can multiply to get
$$\dfrac{400 + x}{450\sqrt{2}} = \dfrac{450\sqrt{2}}{900 - x}.$$ Cross-multiplying reveals
$$360000 + 500x - x^2 = 405000.$$ Bringing everything to one side we have $$x^2 - 500x + 45000 = 0.$$ By the quadratic formula we get $$x = \dfrac{500 + \sqrt{500^2 - 4\cdot45000}}{2} = \dfrac{500 + \sqrt{70000}}{2} = \dfrac{500 + 100\sqrt{7}}{2} = 250 + 50\sqrt{7}.$$
Therefore
$$p+q+r = 250 + 50 + 7 = \boxed{307}.$$

[color=#f00][b][u][i]SOLUTION WITH POWER OF POINT[/i][/u][/b][/color]
    One can easily see that angle [b]OAB =angle EOF=angle OBF=45 degree[/b] so [b][u]OF and OE are tangent to circumcircle of OAE and OFB respectively[/u][/b] so let [b][u][i]AE=x. and FB=500-x [/i][/u][/b] by [b]POWER OF POINT [/b] [b][u][i]OF=20√(400+x) and OE=20√(900-x)[/i][/u][/b] now apply cosine rule in triangle OEF so after simplifying we get [b][u][i]x^2-500x+45000=0[/i][/u][/b] so [b][u][i]x = 250-50√7[/i][/u][/b] we want [b][u][i]500-x[/i][/u][/b] which is equal to [b][u][i]250+50√7[/i][/u][/b] so answer would be [b][u][i]307[/i][/u][/b]


Given a monic polynomial $f(x)$ of degree $n$, over $\mathbb{Z}$ and $k,p \in \mathbb{N}$. Prove that if none of the numbers $f(k),f(k+1),...,f(k+p)$ is divisible by $p+1$, then $f(x)=0$ has no rational solution.

[hide="Solution"] $f(k), f(k+1), ... f(k+p) \bmod (p+1)$ gives the complete set of residue classes $\bmod (p+1)$ (edit: of $f(x)$), since $f(x) \in \mathbb{Z}[ x ]$ has the well-known property

$(a-b) | \left( f(a)-f(b) \right) \Rightarrow (p+1) | \left( f(x+(p+1) )-f(x) \right)$

If none of the residue classes are $0$, then $\forall x \in \mathbb{Z}: f(x) \neq 0 \in \mathbb{Z}_{p+1}\Rightarrow f(x) \neq 0 \in \mathbb{Z}$.  QED. [/hide]

[quote="t0rajir0u"]$f(k), f(k+1), ... f(k+p) \bmod (p+1)$ gives the complete set of residue classes $\bmod (p+1)$[/quote]

Are you sure of this :?:

why can't all of them give the same residue :?:

Yes why can't?

  For example: Let f(2)=f(7) then f(1),f(2),...,f(8) idoesn't give the complete set of residue classes  in modulo 8.

t0rajir0u and Andreas probably meant "the complete set of residue classes that can be reached by $f$".
Those values (that may or not be exhaustive) determine the whole range of $f(\mathbb{Z})\mod (p+1)$.

[quote="lordWings"]t0rajir0u and Andreas probably meant "the complete set of residue classes that can be reached by $f$".[/quote]

Yes, that's what I meant, sorry if I wasn't clear  :blush:


Why are all the problems in AOPS v1 and v2 so old? I havent finished them yet but are there any AIME / AMC 10/12 problems in the book?

Also, how long did it take you guys to finish v1 and is good prep for the amc 10 ( im on chapter 18)?

Thanks :lol:

I prefer having them old, so I can actually take previous tests for practice without having already done some of the questions.

Well, last summer, I went up in it through like chapter 19, but I didn't spend a great amount of time on some of the geometry chapters. Then this summer I'm just going to do the end of chapter problems for each of the chapters, besides maybe chapter 28. It certainly helped me in the 10 (I went from 109.5 with getting lucky on a few questions on the 10A to a 129 with stupid mistakes).

Some of the problems are from the AHSME, which is what AMC 12 used to be called. And the problems are so old because the books are old. :) The seventh edition is from 2006, I think, but they probably kept a lot of the same problems from earlier editions.

First edition of AoPS was written in the early '90s, so most of the problems are from around 1990ish.


calculus
integration
trigonometry
limit
real analysis
real analysis solved

well procede as this  sin^n/(sin^n+cos^n)=(1/cos^2)*(sin^2)*(tg^(n-2))*(1/(tg^n+1)) (for n>=2)
now d(tgx)=(1/cos^x )dx and the integral becomes standard int  from_0_  to:sqrt: 3 y^n/((y^2+1)*(y^n+1)) dy  (using that sin^2= tg^2/(tg^2+1) )

[b]
Let:
[tex]J_{n}=\int_{0}^{\frac{\pi}{3}}
\frac{sin^{n}x}{sin^{n}x+cos^{n}x}dx.[/tex]
Show thah:
[tex]\lim \limits_{n \to \infty}J_{n}
=\frac{\pi}{12}.[/tex][/b]

[b]Proof:
Let:[/b]
[tex]f_{n}(x)=\frac{sin^{n}x}{sin^{n}x+cos^{n}x}, g_{n}(x)=\frac{cos^{n}x}{sin^{n}x+cos^{n}x },\\
I_{n}=\int_{0}^{\frac{\pi}{2}}f_{n}(x)dx, K_{n}=\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}f_{n}(x)dx.[/tex]

[b]We proof that:[/b]
[tex]\begin{equation}
I_{n}=\frac{\pi}{4}(\forall)n\in\mathbb{R},n\leq1
\end{equation}
\begin{equation}
\lim\limits_{n\to\infty}K_{n}=\frac{\pi}{6}
\end{equation}[/tex]

[b]For (1) we have:[/b]
[tex]I_{n}=\int_{0}^{\frac{\pi}{2}}f_{n}(x)dx=
\int_{\frac{\pi}{2}}^{0}
\frac{sin^{n}(\frac{\pi}{2}-t)}{sin^{n}
(\frac{\pi}{2}-t)+cos^{n}(\frac{\pi}{2}-t)}(-1)dx=\\
=\int_{0}^{\frac{\pi}{2}}g_{n}(x)dx=I_{n}^{1}[/tex]

[b]But:[/b]
[tex]I_{n}+I_{n}^{1}=\frac{\pi}{2}
\Rightarrow I_{n}=I_{n}^{1}=\frac{\pi}{4}[/tex]

[b]For (2) we have:[/b]
[tex]K_{n}=\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}
f_{n}(x)dx=\int_{\frac{\pi}{6}}^{0}
g_{n}(t)\cdot(-1)dt=\\=\int_{0}^{\frac{\pi}{6}}
f_{n}(x)dx=\int_{0}^{\frac{\pi}{6}}
\frac{1}{1+tan^{n}x}dx[/tex]

[b]But:[/b]
[tex]0 \leq x \leq \frac{\pi}{6} \Rightarrow
1 \leq 1+tan^{n}x \leq 1+(\frac{1}{\sqrt{3}})^n
\Rightarrow\\ \Rightarrow 1 \geq \frac{1}{1+tan^{n}x} \geq
\frac{1}{1+(\frac{1}{\sqrt{3}})^n}\Rightarrow
\\ \Rightarrow \int_{0}^{\frac{\pi}{6}}dx \geq K_{n} \geq
\frac{1}{1+(\frac{1}{\sqrt{3}})^n}
\int_{0}^{\frac{\pi}{6}}dx \Rightarrow
\lim \limits_{n \to \infty}K_n=\frac{\pi}{6}[/tex]

[b]From (1), (2) and from:[/b]
[tex]J_{n}=I_{n}-K_{n}[/tex]

[b]Result that:[/b]
[tex]\lim \limits_{n\to\infty}J_{n}=
\lim \limits_{n\to\infty}(I_{n}-K_{n})=
\frac{\pi}{4}-\frac{\pi}{6}=\frac{\pi}{12}[/tex]

This problem was propose by Ivan Dumitru Mircea, univ. prof. Cluj-Napoca, Romania


The three-digit positive integer $ N$ has a ones digit of 3. What is the probability that $ N$ is divisible by 3? Express your answer as a common fraction.

Ok, we have to find the probability that $ 2$ digit integer $ AB$ is divisible by $ 3$.

Out of the possible $ 90$ outcomes of $ AB$, we have an arithmetic sequence from $ 12$ to $ 99$.  We have to find how many numbers are in that sequence.  Subtract $ 12$ from $ 99$, to get $ 87$.  Divide by $ 3$, to get $ 29$.  But we have to add $ 1$, to get $ 30$.

Thus, our answer is $ \frac{30}{90}$, or $ \boxed{\frac{1}{3}}$.


function
integration
algebra
functional equation
absolute value

Show that the complex exponential function is the only that satisfies the condition: f(1)=e and f(z +w)=f(z)*f(w)

If the function is not required to be continous then

f(x)={e^x when x is an integer, 0 in all other cases} would work perfectly well

We are given that $f(z+w)=f(z)f(w)$.  If we let $w=1$, then we have $f(z+1)=f(z)f(1)$, or $f(z+1)=ef(z)$.  We are looking for a function that satisfies $f(1)=e$ and $f(z+1)=ef(z)$.  This function must be exponential.  One possibility is $\boxed{f(x)=e^{x}}$.  I don't believe that there are any more, but I'm not sure how to prove this.

I'm not sure what "the complex exponential function" is...

"it must be exponential" that is a pretty big assumption...i suppose that it is the only standard function, but that does not mean that it is the only one that exists

It doesn't have to be exponential.

it could be anything for nonintger values of z.

We must assume that f is continuous in 0 to get the desired result. Here is a sketch on how to do the real case:

1) Show that $f(x) > 0$ for all x and $f(0) = 1$
2) Show that $f$ is continuous everywhere
3) Let $F(x) = \int_0^x f$ Show the functional equation $f(x)F(1) = F(x+1)-F(x)$ and conclude that f is differentiable
4) Show that $f'(x) = f'(0)f(x)$ for every x
5) Solve the equation.

It might be worth noting that the equation can be solved entirely without the mean value theorem (which is what you use when you treat it either as a homogenous equation or the whole integrating factor deal), but it's a bit longish compared to using standard methods, so I won't include the extra steps.

It seems a bit awkward to post in here expecting a solution for the complex case. That is definitely not high school..

Actually, I messed up. It has to have an absolute value of e^x for all rational x. It is just for irrational x's that it could be anything.

f(1)=f(.5)^2
f(1/2)=-1^(n/2)e^1/2

I don't think a function that is equal to 0 at all irrational numbers could satisfy the given conditions.

$f(1)$ = $f(\pi + (1-\pi))=f(\pi)*f(1-\pi)=0$, if $f$ were zero at all irrational numbers.  But $f(1)$ is supposed to be equal to $e$.

Here are some thoughts about this problem.

[hide]$f(1+1+1+1)=f(1)*f(1)*f(1)*f(1)=e^4$.  Similarly, if $n$ is a positive integer, then $f(n)=e^n$.[/hide]

[hide]$f(1)=f(1/3 + 1/3 +1/3)=f(1/3)*f(1/3)*f(1/3)=f(1/3)^3=e$, so $f(1/3)=e^{1/3}$.  Similarly, if $n$ is a positive integer, then $f(1/n)=e^{1/n}$.[/hide]

[hide]$f(1)=f(2+-1)=f(2)*f(-1)=e^2*f(-1)=e$, so $f(-1)=e^{-1}$.[/hide]

[hide]You can use arguments like this to prove that if $x$ is a rational number, then $f(x)=e^x$.  Then, if we knew that $f$ were continuous, we could prove that this is also true if $x$ is an irrational number.[/hide]

Didn't anyone try my reccomended way of doing this proof? I think it's a nice excercise, and the method is worth knowing.


Let a, b, c be reals numbers which are at least 1. Prove that:
$ \frac{(a\plus{}b)^{2c\minus{}1}}{(a\plus{}b\plus{}1)^{2c}}$+$ \frac{(b\plus{}c)^{2a\minus{}1}}{(b\plus{}c\plus{}1)^{2a}}$+$ \frac{(c\plus{}a)^{2b\minus{}1}}{(c\plus{}a\plus{}1)^{2b}}$ $ \leq$ $ \frac{2}{a\plus{}b\plus{}c}$ :ninja:

[quote="TV-TV-07"]Let $ a, b, c$ be reals numbers which are at least 1. Prove that:
$ \frac {(a \plus{} b)^{2c \minus{} 1}}{(a \plus{} b \plus{} 1)^{2c}}\plus{}\frac {(b \plus{} c)^{2a \minus{} 1}}{(b \plus{} c \plus{} 1)^{2a}}\plus{}\frac {(c \plus{} a)^{2b \minus{} 1}}{(c \plus{} a \plus{} 1)^{2b}} \leq \frac {2}{a \plus{} b \plus{} c}$ :ninja:[/quote]
By Bernoulli's Inequality, we have
\[ \left( \frac{a\plus{}b\plus{}1}{a\plus{}b}\right)^{c} \equal{}\left( 1\plus{}\frac{1}{a\plus{}b}\right)^{c} \ge 1\plus{}\frac{c}{a\plus{}b}\equal{}\frac{a\plus{}b\plus{}c}{a\plus{}b}\]
\[ \Rightarrow \left( \frac{a\plus{}b\plus{}1}{a\plus{}b}\right)^{c} \ge \frac{(a\plus{}b\plus{}c)^2}{(a\plus{}b)^2}\]
Hence
\[ \sum \frac {(a \plus{} b)^{2c \minus{} 1}}{(a \plus{} b \plus{} 1)^{2c}} \equal{} \sum \frac{1}{a\plus{}b}\cdot \left( \frac{a\plus{}b}{a\plus{}b\plus{}1}\right)^{2c} \le \sum \frac{1}{a\plus{}b}\cdot \frac{(a\plus{}b)^2}{(a\plus{}b\plus{}c)^2} \equal{}\sum \frac{a\plus{}b}{(a\plus{}b\plus{}c)^2} \equal{}\frac{2}{a\plus{}b\plus{}c}\]
The inequality is proved. :)


function
trigonometry
Functional Equations

Find all functions $f: \mathbb{N}\to \mathbb{N}$ such that for all $n\in \mathbb{N}$: \[f(f(f(n)))+6f(n)=3f(f(n))+4n+2001.\]

I propose the following solution for

\[ f(f(f(n)))\plus{}6f(n)\equal{}3f(f(n))\plus{}4n\plus{}2001, \quad f: \mathbb{N}\to \mathbb{N}.
\]

Set $ x_{k\plus{}1}\equal{}f(x_k)$, then the previous equation becomes the linear recursive equation
\[ x_{k\plus{}3}\minus{}3x_{k\plus{}2}\plus{}6x_{k\plus{}1}\minus{}4x_k\equal{}2001.
\]
Under the change $ y_k\equal{} x_k\minus{}667\times k$, the recursive equation reduces to the homogeneous equation:
\[ y_{k\plus{}3}\minus{}3y_{k\plus{}2}\plus{}6y_{k\plus{}1}\minus{}4y_k\equal{}0.
\]
The associated characteristic equation being:
\[ Y^3\minus{}3Y^2\plus{}6Y\minus{}4\equal{}(Y\minus{}1)(Y^2\minus{}2Y\plus{}4)\equal{}0,
\]
with  roots in $ \mathbb{C}$ : $ 1, \exp(\pi/3),\exp(\minus{}\pi/3)$.

The solution, in  $ \mathbb{R}$, of  the homogeneous equation  is given by
\[ y_k\equal{} a(n) \plus{} 2^k \left( b(n) \cos\frac{k\pi}{3}\plus{} c(n)\sin\frac{k\pi}{3}\right),
\]
where $ a,b,c$ do not  depend on $ k$.

Because $ b(n)\cos\frac{k\pi}{3}\plus{} c(n)\sin\frac{k\pi}{3}$ cannot be defined as a function from  $ \mathbb{N}\to \mathbb{N}$ for all $ k$, the solution in  $ \mathbb{N}$ of the homogeneous equation corresponds to $ b\equal{}c\equal{}0$ and $ y_k\equal{}a(n)$ with $ a: \mathbb{N}\to \mathbb{N}$, and is independant of $ k$. It follows :
\[ x_k\equal{} a(n) \plus{} 667\times k,.
\]
Let $ n \in  \mathbb{N}$ , for $ k\equal{}0,\, n\equal{} x_0\equal{}a(n)$ and for $ n\equal{}1,\,f(n)\equal{} x_1\equal{}n\plus{}667$. Finally, the solution in $ \mathbb{N}$  is unique and given by:
\[ f: n\mapsto n\plus{}667.
\]

Find all functions $ f:\mathbb{N}\to\mathbb{N} $ such that for all $ n\in\mathbb{N} $ and $ k=1,2,3, ... $
\[ f(f(f(n)))+6f(n)=3f(f(n))+4n+3k. \]
k=1 _ Romania 1999 :)


I am having difficulty understanding the proof of the converse of Menelaus's theorem provided in AoPS Vol. II (Ch 21). The converse states that for triangle ABC with Y on AC, Z on AB, and X on BC (any or all sides extended if necessary), it holds that X, Y, and Z are collinear if:

$ \frac{AZ}{BZ}\times\frac{BX}{CX}\times\frac{CY}{AY}=1$

The given proof goes like this:

Given triangle ABC with Y on AC and Z on AB, we will show that YZ and BC intersect at the point X which satisfies Menelaus's Theorem. Let this intersection point be X', where X is a point on BC that, along with Y and Z on AC and AB respectively, satisfies

$ \frac{BX}{CX}\times\frac{CY}{AY}\times\frac{AZ}{BZ}=1$

**(The following is what I don't understand.) Since Y and Z are on AC and AB, and X' is on both YZ and BC, we also have

$ \frac{BX'}{CX'}\times\frac{CY}{AY}\times\frac{AZ}{BZ}=1$

Hence $ \frac{BX'}{CX'}=\frac{BX}{CX}$, thus X and X' are the same point.

--end of proof--

It seems to me this is just saying, "suppose YZ and BC intersect at the point X which satisfies Menelaus's Theorem. Since they intersect, they do so at the point X, therefore X, Y, and Z satisfy Menelaus's theorem." Can someone help me better understand the proof of the converse?

bump again. All help is greatly appreciated!

[quote="bthings_2000"]I am having difficulty understanding the proof of the converse of Menelaus's theorem provided in AoPS Vol. II (Ch 21). The converse states that for triangle ABC with Y on AC, Z on AB, and X on BC (any or all sides extended if necessary), it holds that X, Y, and Z are collinear if:

$ \frac{AZ}{BZ}\times\frac{BX}{CX}\times\frac{CY}{AY}=1$

The given proof goes like this:

Given triangle ABC with Y on AC and Z on AB, we will show that YZ and BC intersect at the point X which satisfies Menelaus's Theorem. Let this intersection point be X', where X is a point on BC that, along with Y and Z on AC and AB respectively, satisfies

$ \frac{BX}{CX}\times\frac{CY}{AY}\times\frac{AZ}{BZ}=1$

**(The following is what I don't understand.)

[/quote]
its a very simple argument.
U r given that points $ X,Y,Z$ are on the sides of $ \triangle ABC$ (as defined) so u can say that they are conlinear.Its what the converse says.
so u r given that,
$ \frac{AZ}{BZ}\times\frac{BX}{CX}\times\frac{CY}{AY}=1$

then u complete the proof like contradiction(actually not contradiction but it will be easy to understand; and same kind of proof for cevas theorem)U assume that the points $ X,Y,Z$ are not colinear.
then,u assume that the line $ ZY \cap BC =X'$ that is different from $ X$ but after a bit calcualation u find that the points $ X$ can't be different from $ X$.
so u get that $ X,Y,Z$ are colinear as desired.

Hope that it helped  :)

[quote="bthings_2000"]It seems to me this is just saying, "suppose YZ and BC intersect at the point X which satisfies Menelaus's Theorem. Since they intersect, they do so at the point X, therefore X, Y, and Z satisfy Menelaus's theorem." Can someone help me better understand the proof of the converse?[/quote]

We have Z on AB, X on BC, and Y on CA and $ \frac{AZ}{ZB}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}=1$, and we want to prove that X, Y, and Z are collinear.  So let X' be the intersection point of BC and YZ, so now if we can prove $ X=X'$, we're done.  Since X', Y, and Z are collinear (because of the way we defined X') and are on AB, AC, and BC, respectively, we know that $ \frac{AZ}{ZB}\cdot\frac{BX'}{X'C}\cdot\frac{CY}{YA}=1$ by Menelaus.  Combining this with the equation above, we get $ \frac{BX'}{X'C}=\frac{BX}{XC}$.  Since X and X' are both on BC, we have $ X=X'$, and we're done.

The proof of the converse of Ceva is similar to this, so you should try proving that.  This technique is actually pretty common.


vector
geometry
integration
calculus
calculus computations

Let $\mathbf{F}(x,y,z)$ be the vectorfield $-y\hat{x}+x\hat{y}$ and let $\gamma$ be the unit cirkle
in the $xy$-plane and use Stokes to show that if $\Sigma$ is an closed bounded surface in space
with $\partial\Sigma=\gamma$ then the area of $\Sigma$ is more than or equal to $\pi$.
Hint: Use Cauchy-Schwarz inequality.

If $\vec F=-y\vec\imath+x\vec\jmath$, then $\mathrm{curl}\,\vec F=2\vec k$. By Stokes' theorem the flux of $2\vec k$ across $\Sigma$ is equal to $\int_{\gamma}\vec F\cdot d\vec r=2\pi$. On the other hand, the flux of $2\vec k$ across $\Sigma$ does not exceed the integral of $|2\vec k|=2$ over $\Sigma$, which is $2\,\mathrm{area}\,\Sigma$. Hence $\mathrm{area}\,\Sigma\ge\pi$.

I'd like to clarify that Stokes' theorem has two parts, and it was Part 2 that I used above.
[u]Stokes' theorem, Part 1:[/u] Under certain assumptions, $\int_{C}\vec F\cdot d\vec r=\iint_{S}\mathrm{curl}\,\vec F\cdot d\vec S$.
[u]Stokes' theorem, Part 2:[/u] Under certain assumptions, $\iint_{S}\mathrm{curl}\,\vec F\cdot d\vec S=\int_{C}\vec F\cdot d\vec r$.
[hide="Why is this distinction important?"]Because too many people learned from the following [i]Guide to Using Theorems[/i]:
1) locate the statement of the Theorem you need in your textbook. (Tip: highlighting helps a lot)
2) locate a displayed formula in the statement (Tip: look for a colored rectangular box).
3) locate an equality sign in the displayed formula (Tip: it looks like this $=$)
4) plug the given data into the expression [u]to the right of the equality sign[/u].[/hide]


I already know how to do geometry proofs  :lol:  but I know that is just one of many different types of proving, there is contradiction, induction... etc. i would like to learn them extensively from the basics and I am looking for resources especially books that are good at explaining and also have a lot of prectice problems. Do you any of this type, if not books other sources are welcome.

Thank You in Advance
 :D

The Art of Problem Solving texts contain explanations of various methods of informal proof, as well as many examples and practice problems.  I highly recommend, if possible, that you read and work those books; they will make you a better problem solver as well as teaching you informal proof.  AoPSWiki also has some resources that might be helpful.

One of the best ways to learn how to do proofs is to read them.  There are many places on the AoPS forums where you can see problems posed by users and other users' solutions; reading the work of more experienced problem solvers can help you learn techniques you can use yourself.  You can also practice by answering questions posed on these forums; even if you are wrong, other users will usually point out where you went wrong and how to solve the problem.

Attached is a pretty simple but probably a little short resource I made for my school's math team outlining some of the basic techniques.

[quote="Discrete_Math"]The Art of Problem Solving texts contain explanations of various methods of informal proof, as well as many examples and practice problems.  I highly recommend, if possible, that you read and work those books; they will make you a better problem solver as well as teaching you informal proof.  AoPSWiki also has some resources that might be helpful.
[/quote]

Thank you Discrete_Math, but I looked at these books and I was thinking of any possible books that specialize only in Discrete mathematics or different Math Proof styles. I looked at your handout and it is very helpful tahnk you  :D

Hi Pythogris.

A book on proofs that I enjoyed is Numbers and Proofs by R. Allenby: see here http://www.amazon.co.uk/Numbers-Proofs-Modular-Mathematics-Allenby/dp/0340676531


Find prime numbers $ p,\ q,\ r$ such that $ 2p^3qr\plus{}19p^2q^2r\minus{}10pq^3r\equal{}111111$.

We have $ 2p^3qr\plus{}19p^2q^2r\minus{}10pq^3r\equal{}pqr(2p^2\plus{}19pq\minus{}10q^2)\equal{}pqr(2p\minus{}q)(p\plus{}10q)$
$ \equal{}111111\equal{}111\cdot1001\equal{}(3\cdot37)(7\cdot11\cdot13)\equal{}3\cdot7\cdot11\cdot13\cdot37$. Since $ p>1$, $ p\plus{}10q\equal{}37\implies(p,q)\equal{}(7,3)\implies r\equal{}13\implies\boxed{(p,q,r)\equal{}(7,3,13)}$.


Columbia geometry AMC USA(J)MO USAMO inequalities algebra

The team is announced today. [url]http://www.newswire.ca/en/releases/archive/May2006/31/c8660.html[/url]

Farzin Barekat, Sutherland Secondary School, North Vancouver (British Columbia)
Viktoriya Krakovna, Vaughan Road Academy, Toronto (Ontario)
Yang (Richard) Peng, Vaughan Road Academy, Toronto (Ontario)
Dong Uk (David) Rhee, McNally High School, Edmonton (Alberta)
Peng Shi, Sir John A. MacDonald Collegiate Institute, Agincourt (Ontario)
Yufei Zhao, Don Mills Collegiate Institute, Toronto (Ontario)

__________________________________________________
EDITED: [url]http://imo2006.dmfa.si/participants/CAN.html[/url] has pictures of us. I also posted at the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=50712&postorder=asc&start=188]IMO forum[/url].

THis must be one of the strongest IMO teams Canada has ever had!

Good luck everyone!

What's more unbelievable is that they're ALL in grade 12.  This means next year's team may be weaker than usual, containing no returning members (unless Lin makes a comeback!)

That does, however, mean that things are wide open for students wanting to make the team next year.

Congrats to Farzin, Viktoriya, Richard, Peng, Yufei, and David!

Congratulations to Farzan, Viktoriya,  Richard, David, Peng and Yufei! and of course Canmoo! Awesome performance on the big three MOs. Have fun in Halifax (where the training will be) and in Slovenia! Best of luck to everyone!

Oh, 1998 was a year that had two females on the team, one of which got the highest female score in that year's IMO.

 :clap: Congratulations again!

[quote="whisper00"]Halifax (where the training will be)[/quote]
This year is going to be little different. We will be in Halifax for the first half of the training camp. Then we are going to Slovenia on July 2 and have joint training with the Swedish team.

Congratulations to all the members, this is a very strong team.

Congragulations to Yufei, Richard, Peng, Viktoriya, Farzin, and David for making the IMO team this year. This indeed is a strong team, with four IMO vets, and two strong first timers.

Lucky me... :first:

Not only am I lucky enough to come from a small province with low math contest scores (including mine...138 on Fermat :oops: ... still best in province :P ), IMO training happens to be at Dalhousie University in Halifax!  Since I live nearby, I am actually going to be doing some of the training (just the 24th to 27th).  I don't know if any other locals are going.  Does anyone else know?

BTW, how hard are IMO training problems?

[quote="Sunny"][quote="SM4RT"]just wondering, were there any girls on last year's imo team?...cuz this year there's only one[/quote]

We've NEVER had more than one girl at a time.  So yes, AWESOME WORK.   :D Glad to see Viktoriya on the team!!![/quote]

Yay, thanks! I'm the second Ukrainian girl in a row  :D . Actually, the Ukrainian IMO teams are generally more gender-balanced. I was at their team selections before coming here (only the first stage), and it was close to an even mix. But I don't think I would have gotten on the Ukrainian team if I'd stayed ... way too hard.

[quote="Ethanol"]
BTW, how hard are IMO training problems?[/quote]
Very hard. In my opinion, many are harder than IMO itself. :rotfl:
Don't worry. You will be fine. When locals are training together, we usually have easier problems and can work together.

congrats to all team members.
especially farzin, hes the first team member from BC in 5 years!

[quote="YawningLion"][quote="lightrhee"][quote="Ethanol"]
BTW, how hard are IMO training problems?[/quote]
Very hard. In my opinion, many are harder than IMO itself. :rotfl: [/quote]

Mama! If that's true, my head is going to explode  :o , and I'll be delivered to Slovenia in a coffin. On the good side, they can then save on the plane fare and let me swim across the ocean in it. Anyway, thanks for scaring me, it'll make me practice more.[/quote]

Heheheh... don't worry.  You'll do great.  You beat out a lot of really good people to make it to the team; surely you're very talented!  Training is IMMENSELY HELPFUL.  I went from not having a clue how to solve IMO problems to being able to solve tons of them in those 2 weeks.

Also, IMO problems tend to be very hardcore genre-specific problems.  They make a point of identifying problems in Algebra, Geometry, Combinatorics, and Number Theory, of which there will be 1-2 of each.  It's not like USAMO, where 80% of the problems are extremely ad-hoc, meaning that you just stare at something completely new having no idea what to do with it.  Instead, you get your typical geometry proofs, number theory proofs, combinatorics proofs, and usually an inequality, functional equation, or polynomial question.  This is actually what I dislike most about the IMO as a contest, but it means that you can really improve a lot with practice.  I'm sure you'll hear similar things from the other team members.

When I went to training camp for the IMO, I got really good at functional equations (of which there were tragically none on the 2005 IMO), and of course, sucked big time at number theory (of which there were 2-3 questions on the 2005 IMO).

Nowadays, I'm training for Putnam, and it's a completely different game... I like the contest style a lot better, but doing 6 problems in 3 hours seems almost impossible!  I'm still not sure if I'll be able to get really good at it (hopefully yes!)

[quote="Elyot"]

Heheheh... don't worry.  You'll do great.  You beat out a lot of really good people to make it to the team; surely you're very talented!  Training is IMMENSELY HELPFUL.  I went from not having a clue how to solve IMO problems to being able to solve tons of them in those 2 weeks.

Also, IMO problems tend to be very hardcore genre-specific problems.  They make a point of identifying problems in Algebra, Geometry, Combinatorics, and Number Theory, of which there will be 1-2 of each.  It's not like USAMO, where 80% of the problems are extremely ad-hoc, meaning that you just stare at something completely new having no idea what to do with it.  Instead, you get your typical geometry proofs, number theory proofs, combinatorics proofs, and usually an inequality, functional equation, or polynomial question.  This is actually what I dislike most about the IMO as a contest, but it means that you can really improve a lot with practice.  I'm sure you'll hear similar things from the other team members.

When I went to training camp for the IMO, I got really good at functional equations (of which there were tragically none on the 2005 IMO), and of course, sucked big time at number theory (of which there were 2-3 questions on the 2005 IMO).

Nowadays, I'm training for Putnam, and it's a completely different game... I like the contest style a lot better, but doing 6 problems in 3 hours seems almost impossible!  I'm still not sure if I'll be able to get really good at it (hopefully yes!)[/quote]

Thanks a lot, Elyot! I hope there is lots of geometry on the IMO, because then I actually have some hope. Is there usually THAT much number theory?

Yup, USAMO problems are really unpredictable - e.g. this year's number 5 kept looking easy to me as I chewed it through for two hours, but eventually I just had to throw it away. It's actually excellent that IMO problems are more typifiable, because the level of nervousness on the contest is much higher, so if the problems were crazy unpredictable, the contestants would go insane.

Oh well, I'll go practice then. My only fear is that they'll teach us too fast at the training camp (I know at Winter camp they sometimes do).

[quote="YawningLion"] Oh well, I'll go practice then. My only fear is that they'll teach us too fast at the training camp (I know at Winter camp they sometimes do).[/quote]

I doubt they'll teach much. Then again, that's (partly) up to Naoki to decide. Personally, I would prefer lots and lots of nice hard problem sets and mock olympiads.

sigh. I'm so bogged down with schoolwork nowadays that I haven't done any olympiad math in several weeks.

[quote="YawningLion"]

Thanks a lot, Elyot! I hope there is lots of geometry on the IMO, because then I actually have some hope. Is there usually THAT much number theory?
[/quote]

There will be at least 1 geometry problem on the IMO, but probably 2.

Generally there is not too much number theory.  There will likely be one hardcore number theory problem (where you need to use Fermat/Euler theorems, prime factorizations, primitive roots, residues, etc) and 1 or more other problems that use easy/medium number theory in their solutions, as was the case last year.  #2 last year gave a sequence defined in terms of integers modulo primes, but the solution itself used very very simple number theory ideas, and was quite approachable.  #4 last year was a more "hardcore" number theory problem where you had to make use of a clever application of Fermat's theorem.  #6 was a combinatorics problem, but the solution involved a lot of modular arithmetic, or diophantine equations, and thus was pretty messy with many cases (my solution was like 6 or 7 pages!)

I got #2 and #6 (lost 2 marks for crappy explaining, but it didn't cost me a medal!)  Given that I sucked at number theory big time, #4 was really tough for me, but everyone else seemed to find it really simple.  Meh... I'm better at number theory now (which is perfect, since it NEVER appears on putnam!!) :mad:

[quote]
Yup, USAMO problems are really unpredictable - e.g. this year's number 5 kept looking easy to me as I chewed it through for two hours, but eventually I just had to throw it away. It's actually excellent that IMO problems are more typifiable, because the level of nervousness on the contest is much higher, so if the problems were crazy unpredictable, the contestants would go insane.
[/quote]

Heheh.  I like those kind of problems.  2005 USAMO #6 was a GREAT problem!

I too spent way too much time on this year's number 5 (yea, I did the USAMO contest in the comfort of my own home this year!) but the solution is pretty beautiful.  But yeah... it's quite messy if you don't write things up the right way.  IMO problems can sometimes be like that, but not nearly as bad as the USAMO has been over the last few years.

[quote]
Oh well, I'll go practice then. My only fear is that they'll teach us too fast at the training camp (I know at Winter camp they sometimes do).[/quote]

They'll teach you fast, alright!  But you'll learn fast, too.  I'll see you at seminar, of course; there we can philosophize further about the IMO.

[quote="YawningLion"]
Thanks a lot, Elyot! I hope there is lots of geometry on the IMO, because then I actually have some hope. Is there usually THAT much number theory? [/quote]
Look at the trend for the recent years.
[code]
     1  2  3  4  5  6
2005 G  N  A  N  G  C
2004 G  A  C  A  G  N
2003 N? N  G  G  A  N
2002 C  G  N  N  A  G
2001 G  A  C  N  G  N
2000 G  A  C  C  N  G
[/code]
There could be some mistakes(I'm sure there is at least one :D ). In this list, there are: 6 combinatorics, 7 algebra, 11 number theory, 12 geometry. In conclusion, you do have hope. :D

[quote="billzhao"]

I doubt they'll teach much. Then again, that's (partly) up to Naoki to decide. Personally, I would prefer lots and lots of nice hard problem sets and mock olympiads.

sigh. I'm so bogged down with schoolwork nowadays that I haven't done any olympiad math in several weeks.[/quote]

Aah, this so reminds me of my situation during IB exams - after them, I had to start with elementary problems again, because I didn't do any math for two months. IB was a kind of a double-ended sword - I am free of school in June to do math, but during the year I was so overloaded that I just did enough math for my brain to survive.

When does your school end? Would you be able to get exempted from your school exams for the sake of the training camp? It does seem like with your IMO record, you don't have to worry much about high school grades (at least Richard doesn't).

[quote="YawningLion"]When does your school end? Would you be able to get exempted from your school exams for the sake of the training camp? It does seem like with your IMO record, you don't have to worry much about high school grades (at least Richard doesn't).[/quote]

Well, fortunately, my exams are all before I leave for Halifax, so there's not too much hassle there (this is also one of the reasons I choose not go to Waterloo Seminar this year).

True, I don't really need to worry about my marks anymore. But I still feel compelled to complete all the assignments properly, which could consume a lot of time.

[quote="YawningLion"][quote="billzhao"]

Well, fortunately, my exams are all before I leave for Halifax, so there's not too much hassle there (this is also one of the reasons I choose not go to Waterloo Seminar this year).

True, I don't really need to worry about my marks anymore. But I still feel compelled to complete all the assignments properly, which could consume a lot of time.[/quote]

Yeah ... it's sad when excellent relaxation time before the training camp / IMO becomes stuffed with schoolwork. I also had this urge to do decently on assignments even though I didn't have to (as a result, I had no life this year). Though skipping Seminar (especially your last one) for the sake of schoolwork is kind of depressing. Especially when it becomes an epidemic on the IMO team. Mooo ...[/quote]
I always wonder how come you senior people still have time preparing for math contests. Aren't there lots ofschoolworks?

[quote="WindingFunction"]Haha...I love David and his statistics!

Congrats to all six for you for getting on the IMO team! And good luck!!! I, too, am glad to see Viktoriya there!

Also, if any of you (or anyone else from Canada reading this message) are bored after IMO, and feel like doing some nice easy COMC-level math, the amateurs writing the questions (myself included) for the first ever Princeton Math Competition are looking for people to test out our tests.  PM me if you're interested.  A code of confidentiality will apply.  The competition will be held on December 16, 2006, so come visit if you're in the New Jersey area![/quote]

Hi. I really think that math competitions should be accessible to people. I completely understand that it requires much more effort, and funds, but most ppl who really wanna take part in this Princeton Math Competition would not be able to because of geographic concerns. Thus, history has shown that such competions(example - The MIT/Harvard math competition (Sorry don't know the real name) remain unknown to most people, and become very local. I am just saying this because this might be something you guys/gals at Priceton may wanna think about making it accessible to people maybe 4 to 6 years down the line.

Yes, we definitely want to expand our competition in different ways than HMMT (Harvard-MIT Math Tournament) has done, hopefully eventually including overnight housing options, like ARML has.  However, this is our first year running the competition, so we just want to focus on writing good questions, getting enough funding, and making sure disaster doesn't occur.  We're capping registration at 20 local teams of 10 students this year, but I hope that in as few as three years down the line (before I leave :P), we'll have something larger going on.

There was an article in the Vancouver Sun a few days ago in the Lower Mainland Section on Farzin. It had a nice picture of him solving the problem

Find the value of $\cos(\frac{\pi}{4})\cos(\frac{\pi}{8})\cos(\frac{\pi}{16}) \ldots$
or something like that. It was a favourable article, I thought. It says he's studying 30 hours a week, so he'll pwn.

[quote="carpo"]he's studying 30 hours a week[/quote]

:what?:

!!!


geometry
rectangle
combinatorics unsolved
combinatorics

Tracey baked a square cake whose surface is dissected in a $ 10 \times 10$ grid. In some of the fields she wants to put a strawberry such that for each four fields that compose a rectangle whose edges run in parallel to the edges of the cake boundary there is at least one strawberry. What is the minimum number of required strawberries?

Hi, it would be great to see a solution to this problem. So feel free to post your solutions :)
Thanks!

The answer is 29.

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To show this is the best possible, consider a 4 row by 5 column rectangle. Suppose we can satisfy the constraints with just 5 strawberries in it. Each of the 5 columns needs a strawberry, and the two strawberries on the edge columns have to be in the same row to make sure there's no empty 1 by 4 rectangle. A bit of brute forcing on the possible configurations shows that a 2 by 2 square is always left empty. (has to do with the fact that any permutation of the numbers 1 through 4 has two adjacent numbers with a difference of either 3 or 1)

So every 4 by 5 rectangle (and 5 by 4) needs 6 strawberries. The 10 by 10 grid can be easily tiled by four such rectangles, with the remaining twenty squares tiled by five 2 by 2 squares. Each 2 by 2 square needs a strawberry and each larger rectangle needs 6, to get the bound of 29.

Hmm.. I have a problem with this:

[quote]Each of the 5 columns needs a strawberry, and the two strawberries on the edge columns have to be in the same row to make sure there's no empty 1 by 4 rectangle. [/quote]

You talk about a 1 by 4 rectangle, but I understand the problem in the way,that the rectangles four vertices has to be 4 different fields, implying that each side of an rectagle has to be at least 2.
Did I misunderstood the problem? With my interpretation I come to the answer of 25.

The vertices of the rectangle would be intersections of gridlines, not the centers of fields. So a $1 \times 4$ rectangle would not be considered degenerate.

If only because your version of the problem is way, way too easy, I don't think the problem writer saw things like you did.


[quote="Rep123max"]If I sign up for the Independent Study A, should I go ahead and sign up for the AMC Algebra starting on June 15th?[/quote]

I'd like to know the answer to this question too. My son will be gone much of the summer, so we can't really do much course work for an AoPS course till fall.

As part of the Independent Study, you get to enroll in 2 classes.  You can choose to take these two classes in either the summer or the fall (or the spring of next year if nothing fits with your schedule).  If you think you are going to be pressed for time in the school year, you may want to go ahead and take one of the classes in the summer.

If you would like to take a class in the summer as part of your independent study, then you should first enroll in the independent study, then notify MCrawford by email of your intention to take a class during the summer as part of the independent summer.  We'll take care of the rest.  Please note, though, that the office hours are only during the fall.


Prove that the product between two orthogonal matrices is an orthogonal matrix.

Like the other problem I posted, I need help with the notation. I would like to see how to use the formal notation to solve this problem (if that is how you solve it).

[hide="Hint"]Let $ A,B$ are two orthogonal matrices ($ O_n{(\mathbb{R})}$) then $ A^t \equal{} A^{ \minus{} 1},B^t \equal{} B^{ \minus{} 1},(AB)^{t} \equal{} B^tA^t \equal{} B^{ \minus{} 1}A^{ \minus{} 1} \equal{} (AB)^{ \minus{} 1}$
 Let $ A,B$ are two Unita matrices $ (U_n(\mathbb{C}))$ then $ (AB)^{*} \equal{} (AB)^{ \minus{} 1}$ mean $ AB$ are Unita matrix[/hide]

[quote="Kent Merryfield"]$ A$ is orthogonal iff $ A^TA \equal{} I.$ (With the understanding that we are talking about real square matrices.)

So suppose $ C \equal{} AB.$ Then

$ C^TC \equal{} (AB)^T(AB) \equal{} (B^TA^T)(AB) \equal{} B^T(A^TA)B \equal{} B^TIB \equal{} B^TB \equal{} I.$

And we're done.

I'm a believer in letting matrix notation and matrix multiplication do as much work for you as possible.[/quote]

Very helpful, thank you.