trigonometry
function
advanced fields
advanced fields unsolved

This problem would seem too basic for this particular section, but I don't have a really good clue where it would fit. Anyways, I was hoping to solve this problem(Just as a precursor to more general cases of adding trig functions with exponential components), and using the sum-to-product function, I eventually obtained two answers for two different cases. However, some sample answers didn't quite check out with the original formula and I was wondering what exactly did I do wrong. The trig functions are in radians.

sin(x)+sin(x^2)= 2
=>(sum-to-product conversion)
2sin([x+x^2]/2)cos([x-x^2]/2)=2

sin([x+x^2]/2)cos([x-x^2]/2)=1

So either

sin([x+x^2]/2)=cos([x-x^2]/2)=1 or sin([x+x^2]/2)=cos([x-x^2]/2)=-1

In the first case, where k and j is some integer

(x+x^2)/2 = pi/2+2*pi*k and (x-x^2)/2 = 2*pi*k;

Adding both equations,

2x=pi+8*pi*k so x=pi/2+4*pi*k.

Via similar arithmetic, I found x was equal to (5/2)*pi+4*pi*j in the second case.

The major problem is is that these solutions apparently don't work(When I tried out 1 for k, for example, it obviously didn't work when squaring x). Why? My arithmetic seems fine. I'm utterly perplexed.

There are no real solutions. The problem comes from adding the equations: you went from two constraints to one, destroying information. While any solution would have to satisfy your new equation, it would also have to satisfy another- and the two are inconsistent. We need [b]both[/b] $ x$ and $ x^2$ to of the form $ \frac{\pi}{2}\plus{}2k\pi$, and this says that $ \left(\frac{\pi}{2}\plus{}2k\pi\right)^2\equal{}\frac{\pi}{2}\plus{}2j\pi$, or $ \pi(1\plus{}4k)^2\equal{}2\plus{}8j$ and $ \pi$ is rational.

since $ \sin \theta$ has maximum value of 1, $ \sin x \plus{} \sin x^2 \equal{} 2$ gives $ \sin x \equal{} 1$ and $ \sin x^2 \equal{} 1$

This means both $ x$ and $ x^2$ are of the form $ (4n \plus{} 1) \frac {\pi}{2}$ where $ n \in \mathbb N$

This is impossible as it would mean that $ \pi$ is a rational number

PS: sorry jmerry sir, i guess i was typing in the answer at about the same time

[quote="jmerry"]There are no real solutions. The problem comes from adding the equations: you went from two constraints to one, destroying information. While any solution would have to satisfy your new equation, it would also have to satisfy another- and the two are inconsistent. We need [b]both[/b] $ x$ and $ x^2$ to of the form $ \frac {\pi}{2} \plus{} 2k\pi$, and this says that $ \left(\frac {\pi}{2} \plus{} 2k\pi\right)^2 \equal{} \frac {\pi}{2} \plus{} 2j\pi$, or $ \pi(1 \plus{} 4k)^2 \equal{} 2 \plus{} 8j$ and $ \pi$ is rational.[/quote]

Hah. I came up with a vaguely similar equation two days ago, and thought I proved that an analytic solution couldn't exist. But it seems it turns out that it essentially proved no real solution exists, which becomes evident with the way you formulated it. Do you just happen to know if this proof readily extends to a particular class of other exponents(like sin(x)+sin(x^n) or, even more, sin(x^a)+sin(x^n))? With the seeming celerity that these solutions came forth, I'm almost led to believe this type of problem(and corresponding solutions) is more common than I originally thought. Also, if these types of problem fit into a more fitting category on mathlinks.ro, please don't hesitate to suggest in case of further future inquiry. :)

It's mostly just that this one isn't very hard. It's also strongly dependent on the exact value $ 2$; we get arbitrarily close to it infinitely often.
The variants you mentioned would generate similar expressions for $ \pi$ as the root of an integer polynomial.


Δίνω ένα πρόβλημα που συνάντησα σήμερα καθώς έψαχνα να βρω τις λύσεις στη γεωμετρία του Kedlaya.

 Θα τη στείλλω σε εικόνα. \ Με την ευκαιρία : Μήπως έχεις κανείς σας το σύνδεσμο για τις λύσεις ; Τον έχω χάσει , έχω όμως το αρχείο. Απλά ψάχνω να τον βρω μήπως έχουν κάνει βελτιώσεις. ΑΝ τον βρελιτε , παρακαλώ να τον στείλλετε εδώ.

[quote="savvas_michael"]Έστω $ M$ το μέσο του τόξου $ AB$,$ L \equal{} EM\cap AB$ και ακόμα $ \angle ACB \equal{} 2\varphi$. Το τρίγωνο $ MEC$ είναι ισοσκελές, άρα $ \angle EMC \equal{}\angle ECM \equal{}\varphi$, οπότε $ \angle FEM \equal{}\angle EMC \equal{}\varphi$, $ \angle MLD \equal{}\angle BLE \equal{}\angle BEL \equal{} 2\varphi$ και $ \angle BFE \equal{} 3\varphi$, $ \angle MDL \equal{} 180\minus{}3\varphi$. Συνεπώς τα τρίγωνα $ MDL, BFE$ έχουν μια γωνία ίση ($ \angle BEF \equal{}\angle LMD \equal{}\varphi$) και μία παραπληρωματική και άρα $ \frac{BE}{ML}\equal{}\frac{BF}{DL}$. Όμως $ \angle LMB \equal{}\angle LBM \equal{}\varphi$, οπότε $ ML \equal{} BL \equal{} BE$ και επομένως $ BF \equal{} DL$, δηλαδή $ FD \equal{} DL\plus{}LF \equal{} LF\plus{}BF \equal{} BL \equal{} BE$.

[color=red]Moderator[/color]

 I wrote on your post  $ \angle ACB \equal{} 2\varphi$. There was a typo in the first text.[/quote]

[b]Μιχάλη, είσαι σε φόρμα :)  !!![/b]

[color=red][b] ΣΧΟΛΙΟ[/b][/color]

  Από την άλλη εβδομάδα που για μένα αρχίζουν τα μαθήματα και οι ποικίλες υποχρεώσεις και αναγκαστικά  δε θα είμαι τόσο συχνά στη σελίδα αυτή , θέλω να συνεχίσετε με τα άλλα παιδιά την ωραία αυτή δουλειά. Γύρισε και ο Νίκος Ράπανος από την Αμερική , οπότε αναλάβετε εσείς  τα ηνία και ανεβάστε τον πήχυ μέχρι εκεί που θέλετε ! Αν μπουν όλοι στον αγώνα , είναι εύκολο.Έχετε όρεξη και μπορείτε. Δεν χρειάζεται να είναι εξωντοτικός ο ρυθμός . Να προτείνετε σχετικά  λίγα προβλήματα , ώστε να προλαβαίνετε να τα επεξεργάζεστε . Αυτός που προτείνει κάποιο θέμα , να βάζει και τη λύση ύστερα από 5 μέρες το πολύ, άν δεν έχει ήδη λυθεί. Με αυτόν τον τρόπο δεν θα μένουν άλυτα θέματα στη σελίδα. ΑΝ το προτεινόμενο θέμα είναι άλυτο , να το αναφέρετε , ώστε να το προσπαθούν επίμονα όλοι μαζί οι  εμπλεκόμενοι .

Εγώ [b]θα περιοριστώ στο βιβλίο [/b]για να δω μήπως κάποια στιγμή αρχίσουμε την πληκτρολόγηση. Η υπόθεση Ολυμπιάδα και η ασκησεολογία  ανήκει σε σας !

[b]Μπάμπης[/b]

Ωραία λύση Μιχάλη είχα βρει και εγώ κάτι ανάλογο (λίγο πιο σύνθετα) . Να σημειώσω ότι το πρόβλημα λύνεται εύκολα με αναλυτική γεωμετρία (παίρνουμε χχ' την ΑC και thn mesokatyheto tis AC yy' )(oi parkseis einai liges logo ton pollon katheton) .


geometry
3D geometry
induction
modular arithmetic
number theory proposed
number theory

Prove that every integer can be written as sum of $5$ third powers of integers.

i just can show that this hold for third powers.
$x=2k+1\rightarrow (2k+1)^3=(2k-1)^3+(k+4)^3+(4-k)^3+(-5)^3+(-1)^3$
if$x=2k$ use induction and write $k$ as desired $k=k_1^3+\dots+k_5^3$
now $2k_i$are suitable for showing $x$.

I'm quite sure that this has been discussed before, but I cannot find the post.
Anyway, here's the solution:
We note that $(x+1)^3 + (x-1)^3 + (-x)^3 + (-x)^3 = 6x$, thus every multiple of $6$ can be expressed as a sum of four cubes. To add another cube so that every integer can be expressed as a sum of five cubes, just note that $(\pm 1)^3 \equiv \pm 1 \pmod{6}$, $(\pm 2)^3 \equiv \pm 2 \pmod{6}$, $3^3 \equiv 3 \pmod{6}$ and add the suitable cube accordingly.

SOrry to bump, but this post was just redirected to by ComplexZeta...

What is the motivation during a contest for deriving such a neat factorization? There is certainly a fundamental reason it exists, could someone elaborate?


We consider that the series of real pozitive numbers (a_n)n>=1 defined like so:(a_n+a_(n+1))^2=n(a_n-a_(n+1)), for any natural n\{0}, a1>0 is given.
i) compute lim n-->oo a_n.
ii) compute lim n-->oo  a_n/(a_(n+1)+a_n).

thank you!

Here's a solution for i):

We can see that an is strictly decreasing, and since the terms of the sequence are positive an must tend to l<infinity, l>=0. If we take the sum of the equalities (a_k+a_(k+1))^2=k(a_k-a_(k+1)) from k =1 to n and we divide their sum by n we get (by replacing the terms with l) 4*l^2=0, so an tends to 0 as n tends to infinity (you can do the calculations yourself and see if it works like that)

About ii):

Divide (a_n+a_(n+1))^2=n(a_n-a_(n+1)) by n*(a_n+a_(n+1)). Considering that an->0 we get (a_n-a_n+1)/(a_n+a_(n+1))->0 as n->infinity. From this we can easily find the limit we're looking for. I didn't actually do all the calculations, but it looks like the limit is 1/2 (we eventually get a_n/a_(n+1)->1)


prove that in triangle we have:
$ab+ac+bc=p^2+r^2+4Rr$

$(p=\frac{a+b+c}{2})$

It is easy but very important when we want
to transform $\displaystyle ab+ac+bc$....

Just use $r=\frac{S}{p}$ and $R=\frac{abc}{4S}$
where $S$ denotes the area,
and do calculations.

A good problem concerning the identity above is in

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=40857&start=180[/url] (post #191)

 :)  :)

Easy, indeed. It follows from $2Rr=\frac{abc}{a+b+c}$ and $r=\frac{S}{p}$ and calculations :blush: .I'm lazy to do all calculations. :P

See also http://www.mathlinks.ro/Forum/viewtopic.php?t=36699 .

  darij


I can't solve this and I have no clue what level it belongs in.
1+1
  -----
    2+(1)
      ------
        (3)+...
Remember this is an infinite fraction so it's 2+1/3 not (2+1)/3.  I just don't know LaTeX which would allow more accurate portrayal of the problem.

Can you rewrite the expression with the first few terms using parentheses?

Like this:

1 + (1/(2 + 1/(2 + 1/(2 + ...

for

$1 +\cfrac{1}{2+\cfrac{1}{2 +\cfrac{1}{2 + \cdots}}}$

Also, you can look at my latex code:

1 +\cfrac{1}{2+\cfrac{1}{2 +\cfrac{1}{2 + \cdots}}}

to see how I made that continued fraction.

There is another way I have seen used in older textbooks

$1+\frac{1}{2+}\frac{1}{2+}\frac{1}{2+}\cdots$

which is equivalent to Mr. Rusczyk's continued fraction


The post [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=98451[/url]  by Bert titled "Is this bigger?" appears in the College Playground Unsolved and Proposed Problems list but, when clicking on it, displays the "No posts match your search criteria" message. I'm not  sure it is related to the new server in any way, but it is clearly a bug :)

I second that question... it happened to me in the Chemistry section

It will be fixed. Please keep a record of these topics in this topic fedja.

How about http://www.artofproblemsolving.com/Forum/viewtopic.php?t=98427 ...

Yes I know about that, that is why I moved the topic inside this forum. I am fixing some other thing now, so I will deal with this later.

Sometimes that happens when you delete a topic.


How many pairs of positive integers have greatest common divisor 5! and least common multiple 50!

Okay first of all, I know how to do this one but I dont know how to set it up algebraically. I have taken to learning the roots of problems and I was just wondering how you might set this one up.

remember that the product of those two numbers = 50!5!

Not quite -- they still have to satisfy the given conditions, which, say, (1, 50!*5!) does not.

yeah, we can let one number be 5!m and the other 5!n, and then we get mn=50*49*48*47*...*7*6

This problem was basically solved so I'll just finish it off (I hope)

Continuing from rep123max and s0mp:

[hide]

50/5! = 2^44 * 3^21 * 5^11 * 7^8 * 11^4 * 13^3 * 17^2 * 19^2 * 23^2 * 29 * 31 * 37 * 41 * 43 * 47. We want to find the number of m,n which are coprime and multiply to the above.

Note that since they are coprime, if one is divisible by 3, its divisible by 3^21, etc.

Now, since we consider the pair (a,b) to be the same as (b,a), assume we write the even number first. There are 14 other prime factors of 50!/5!, and each one we can choose to put it as a or as b. Thus the final answer is 2^14.

[/hide]


geometry trigonometry perimeter function rectangle calculus limit

[b]A regular octagon is inscribed within a circle, and has a side measure of 6. [i]Find the area of this octagon.[/i][/b]

Sadly this problem was addressed easily... First, because it was multiple choice ,and second, because the answer did not require trig to be identified.
[hide="On the test I..."]
made the assumption that the perimeters differed by a negligible margin, giving me;
$48 = 2\pi r$
$r = \frac{24}{\pi}$
$ area = \pi r^{2}= \pi \left(\frac{24}{\pi}\right)^{2}$
$\approx 183$... :whistling:
The answer was [i]B, 184[/i]. Mind you, the real answer is not a pretty number.[/hide]I got it, but I wanna see how you guys would solve this with trig.  Hella kudos if you express the answer in its exact value. I'll post my solution later today. For those of you new to the subject, here's a [hide="hint."]For a regular octagon, $Area = 3r^{2}$[/hide]

[hide="hint"]
It doesn't matter that it's inscribed in a circle.
If it's regular, then that condition is implied.

We know that we can break it up into 8 mini triangles.
Now connect each vertex of the octagon to the center.
We have 8 isoceles triangles with a base of 6, and two congruent angles of 67.5 degrees.
The central angle on these triangles is then 45 degrees.
Let the missing sidelength (segment from centroid to vertex of polygon) be denoted by $x$.

By the law of sines,
$\frac{6}{\sin 45}= \frac{x}{\sin 67.5}$
We know $\sin 45 = \frac{\sqrt{2}}{2}$, so $x = 6\sqrt{2}\sin 67.5$.

Now we can use the fact that the area of a triangle can be expressed as $0.5ab\sin \theta$.
We pick the sides as $x$ and $x$, and the angle between them is $\sin 45$
We have that the area of one individual triangle is $(0.5)(6\sqrt{2}\sin 67.5)^{2}(\sin 45) = 18\sqrt{2}\sin^{2}67.5$

See if you can solve it from here.
[/hide]

[quote="nayel"]a regular $n$ gon has area of $\frac{n}{4}a^{2}\tan\left(\frac{(n-2)\cdot 90^{\circ}}{n}\right)$ (where $a$ is the side length of the polygon), a result i discovered. i can give a proof if you want. but i think you don't want this cause it uses trig.

also here's another result of mine: a regular octagon of side length $a$ has an area of $4(1+\sqrt 2)a^{2}$. and guess what! i didn't use trig this time! :)[/quote]

I can do it. I wanted new stuff like THAT! not extraordinary, but nice to have.

pm me proof sir?

EDIT 1: Typoooooos
EDIT 2: I didn't look past that solution on the test, I figured i would do it later and save time on the test lol.

[quote="igiul"][b]A regular octagon is inscribed within a circle, and has a side measure of 6. [i]Find the area of this octagon.[/i][/b][/quote]
[hide="Solution"]Draw a triangle whose vertices are:
1. the center
2. a vertex of the polygon
3. an adjacent vertex to (2)

It shouldn't be too hard to figure out that this is an isosceles triangle with vertex angle $45^{\circ}$, and the area of the octagon is $A_{O}= 8A_{T}$, where $A_{T}$ is the area of the small isosceles triangle

Draw the altitude from the center to the shorter side. This is $h$. Note that we have formed two right triangles, with one angle $22.5^{\circ}$. Using trig,

$h = 3\cot(22.5^{\circ}) = 3(1+\sqrt{2})$

So:

$A_{T}= \frac{bh}{2}= 9(1+\sqrt{2})$

Thus, $A_{O}= 72(1+\sqrt{2})$[/hide]

[quote="igiul"][hide="hint."]For a regular octagon, $Area = 3r^{2}$[/hide][/quote]

Not useful (not an explicit formula). We want the area in terms of $R$, not $r$.

Expanding from Visharul's idea to a regular n-gon, we can prove the area for the n-gon with geometry and some elementary trig.

[hide="Proof"]
Circumscribe the regular n-gon with a circle. The triangle formed by each "sector" of the n-gon has the angles of $\frac{360}{n}, \ 90-\frac{180}{n}, \ 90-\frac{180}{n}$. We'll call this $\triangle OAB$, O being the center and A, B being the adjacent vertices of the n-gon.
Draw a perpendicular bisector  from the center of the n-gon to the opposite side of the triangle. Call the interection C. This forms a right triangle with the sides $\overline{OC}= \sin{\angle ABO}\cdot \overline{BO}, \ \overline{CB}= \sqrt{\overline{OB}^{2}-\overline{OC}^{2}}, \ \overline{BO}= R$ where R is the radius of the circle. The area of this triangle is $\frac{\overline{CB}\cdot \overline{OC}}{2}$ Multiply this by 2 to get the total area of the triangle sector, so the area of the triangle sector is $\overline{CB}\cdot \overline{OC}$ To get the total area of the n-gon, simply multiply by n to get:
$\overline{CB}\cdot \overline{OC}\cdot n$
$= \sqrt{R^{2}(1-\sin(90-\frac{180}{n}))}\cdot \sin(90-\frac{180}{n}) \cdot n$

This can probably reduced quite a bit, since it is rather unwieldy at the moment. Perhaps using tangent in place of sine would've worked better. Also, it works with the radius instead of the side length.
[/hide]

you can also extend every-other side of the octagon to get a square, and since an angle of a regular octagon is 135, you get 4 isosceles right triangles.  so, u know what to do from there.

Question: [hide="I derived.."]$A = \frac{nb^{2}}{4}tan\left(\frac{90^{\circ}(n-2)}{n}\right)$

EDIT:  :blush: I don't want to talk about it.

A as a function of $n$ and $b$ ($n$ = number of sides and $b$ is the side length). As $n$ approaches $\infty$, and $b$ becomes infinitely small, but is greater than zero, does this yield $\pi r^{2}$ somehow?[/hide]

You could also derive the formula for the area of a regular octagon with side $s$ by drawing some diagonals:

[hide][asy]pair A,B,C,D,E,F,G,H; A=dir(22.5); B=dir(67.5); C=dir(112.5); D=dir(157.5); E=dir(202.5); F=dir(247.5); G=dir(292.5); H=dir(337.5); draw(A--B--C--D--E--F--G--H--A); draw(A--D); draw(E--H); draw(B--G); draw(C--F);[/asy]
We have four isosceles right triangles, four rectangles, and one square.  The length of the legs of the triangles is $\frac{s\sqrt{2}}{2}$, so they each have area $\frac{s^{2}}{4}$.  The square has side length $s$, so its area is $s^{2}$.  The rectangles have side lengths $\frac{s\sqrt{2}}{2}$ and $s$, so they have area $\frac{s^{2}\sqrt{2}}{2}$.  Hence the area of the octagon is $1\left(s^{2}\right)+4\left(\frac{s^{2}}{4}\right)+4\left(\frac{s^{2}\sqrt{2}}{2}\right)$ or $\boxed{2s^{2}\left(1+\sqrt{2}\right)}$.[/hide]

Thats a nice one too. I like the $n,n,n\sqrt{2}$ triangles.

[quote="igiul"]$A = \frac{nb^{2}}{4}tan\left(\frac{90^{\circ}(n-2)}{n}\right)$

...

does this yield $\pi r^{2}$ somehow?[/quote]

Yes.  The first thing to note is that $90 \frac{n-2}{n}= 90-\frac{180}{n}$, and therefore that

$A = \frac{nb^{2}}{4}\cot \frac{180}{n}$

[hide="From here..."] We convert to radians.  That is,

$A = \frac{nb^{2}}{4}\cot \frac{\pi}{n}$

Now, $nb$ is the perimeter $p$, so we can rewrite this as

$A = \frac{p^{2}}{4n}\cot \frac{\pi}{n}$

It's a well-known result in calculus that ([b]in radians[/b]) $\lim_{x \to 0}\frac{\sin x}{x}= 1$, from which it follows that

$\lim_{n \to \infty}\frac{p^{2}}{4n}\cot \frac{\pi}{n}= \frac{p^{2}}{4 \pi}$

If we accept that $\lim_{b \to 0}p = 2 \pi r$, the circumference of the circle, the conclusion follows. [/hide]
Of course, this would be a little easier if you had written the formula in terms of the radius $r$ of the circle instead.  We would have

$A = \frac{nr^{2}}{2}\sin \frac{360}{n}$

Where $b = 2r \sin \frac{180}{n}$.

[hide="And then we would proceed similarly..."] Specifically,

$A = \frac{nr^{2}}{2}\sin \frac{2\pi}{n}\implies$
$\lim_{n \to \infty}A = \frac{r^{2}}{2}\pi$ [/hide]

it is easier to just extend the sides so you form a square, then you subtract the triangle corners to get the octagon

[quote]$A = \frac{nb^{2}}{4}tan\left(\frac{90^{\circ}(n-2)}{n}\right)$[/quote]

with this we can also derive $\pi=n\sin\theta\cos\theta$, if $n$ is sufficiently large, where $\theta=90^{\circ}\frac{n-2}{n}$.

Cool beans. Anything else worth milking out of this subject?


trigonometry
function
geometry
geometric transformation

Find an equation for the cosine function that has one maximum at the point (4,6) and its next minimum at the point (5.5,1).

Basically from those you can find its amplitude and its period pretty easily, but note that there are infinitely many such functions.

i don't need infinity solutions. i would just like to see one.

then it's not too hard if you just find its period and amplitude, then from there determine translations from y=cosx

y(x) = A*cos(k*x + (phi))

y'(x) = -A*k*sin(k*x + (phi))

y''(x) = -A*k^2*cos(k*x + (phi)) = -k^2*y(x)

So we know that (4, 6) is a maximum point. That is, we know that
y'(4) = -A*k*sin(4k + (phi)) = 0
y''(4) = -k^2*y(4) = -A*k^2*cos(4k + (phi)) < 0 (So y(4) > 0)

This means:
sin(4k + (phi)) = 0

4k + (phi) = n*(pi) <-- n is some integer.

We know that (11/2, 1) is a minimum. That is, we know that
y'(11/2) = -A*k*sin(11k/2 + (phi)) = 0
y''(11/2) = -k^2*y(11/2) = -A*k^2*cos(11k/2 + (phi)) > 0 (So y(11/2) < 0)

Thus
sin(11k/2 + (phi)) = 0

11k/2 + (phi) = m*(pi) <-- m is some integer.

So we have two equations in k and (phi):
4k + (phi) = n*(pi)
11k/2 + (phi) = m*(pi)

The solution is:
k = 2(m - n)(pi)/3
(phi) = (11n - 8m)(pi)/3

So
y(x) = A*cos([2(m - n)/3](pi)*x + [(11n - 8m)/3](pi))
Where
y(4) > 0 and y(11/2) < 0

since there are many possibilities for this. Simply by playing around with m and n I find that m = 2 and n = 1, with A = -1 gives a good function. That is, one solution is
y(x) = -cos((2(pi)/3)*x - (5(pi)/3))

am i correct??

[quote="1=2"]I graphed it, and you are incorrect. I don't think that trigonometric functions have a maxima and a minima at only 1 point.....[/quote]

Well, they do in terms of the y axis.

$\sin{\theta}$ and $\cos{\theta}$ will always have their min and maximum at $-1$ and $1$ respectively, unless they are vertically stretched or compressed. ie $2\cos{\theta}$ has its max at $2$ and its min at $-2$.

The min and max however occur periodically (which is why these are called periodic functions) and they occur $180^{\circ}$ or $\pi$ radians from each other, unless there is a horizontal stretch or compression. The max point occurs $360^{\circ}$ or $2\pi$ radians from the previous max point, and the same follows for the minimum point.

But the maximum and the minimum can be reached at multiple points, which we DON'T want.

[hide]
The amplitude is $(6-1)/2 =5/2$.
The period is $2\pi/3$.
The vertical shift is $5/2+1=7/2$.
So one solution is:
$y=(5/2)cos((2\pi/3)(x-4))+7/2$
[/hide]

Sorry, but you made a mistake with your function.

The K coefficient of that function can not be 2pi/3 becuase that is the same as the period.  To solve for K you must do 2pi over 3 is equal to 2pi.  K is equal to 3.  Therefore the coefficient of the angle or x is 3.

The correct function I believe is

y= (5/2)cos(3x-3)+7/2

but I am not sure.

Have you tried substituting (4,6) and (5.5,1) in your function?
K is actually equal to the frequency of the function, where frequency multiplied by the period equals $2\pi$.
My function only works in radians, but you could easily change it to degrees (and then the frequency and period would multiply to 360).
[hide]
$y=\frac{5}{2}cos(120(x-4))+\frac{7}{2}$ (if $x$ is in degrees)
[/hide]


Yeah, I checked it too, it's 4  :)

For the second one I have just looked at it and my intuition told me that there's too few conditions (my intuition is very bad these times :D ). So I decided (understand, It's absolutely bad approach :D) to f(1)=1 and f(2)=3 and f(3)=2. If this hold then conditions will be fullfiled. So I compute a,b,c,d and for example here's one result: $f(x)=\frac{\frac{5}{3}x-\frac{7}{3}}{x-\frac{5}{3}}$. Maybe it's just happiness that it holds :)

There is enough information for number 2. Maybe not to find f, but to find f(1).

I'll give a hint (theres actually no messy calculations required at all).

[hide="Hint 1"]What does the function f(f(f(x))) look like?[/hide]
[hide="Hint 2"]How often can f(f(f(x))) = x?[/hide]


I have a very evil hmwrk question:

A glass rod of diameter 12mm and refractive index 1.5 is bent into a circular shape of outer radius r and center Q. A ray OA along the middle of the rod is incident at angle i at the boundry with air. calculate r if the ray is just totaly reflected at A.

[img]http://hometown.aol.co.uk/HL%20Giltrap/fig+17+2.jpg[/img]

OK. First, you need to calculate what can be i. Or let's just call the angle in your picture $\alpha$. As you probably know, the ray is totally internally reflected if:

$n\sin\alpha\geq1$
$\sin\alpha\geq\dfrac{1}{n}$

Now what does alpha have to do with r? Draw a line perpendicular to OA giong through Q. This line intersects OA in some point B and the outer circle in point C. QC and QA are both r, BC is 6mm, therefore QB is r-6mm. In right triangle ABQ,

$\sin\alpha=\dfrac{QB}{QA}=\dfrac{r-\text{6 mm}}{r}$

so you get:

$\dfrac{r-\text{6 mm}}{r}=\sin\alpha\geq\dfrac{1}{n}$
$r-\text{6 mm}\geq\dfrac{r}{n}$
$r(1-\dfrac{1}{n})\geq\text{6 mm}$
$r\geq\dfrac{n\times\text{6 mm}}{n-1}$

r has to be at least 18 mm.

I find this result quite amazing!!!! Using TIR to find Radius!!!!! Great!!


calculus
integration
limit
function
logarithms
trigonometry
calculus computations

$ \int_0^\infty \frac{1}{1\plus{}x^4} \, dx$

and i got another factorization problem, see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=245106]link[/url]

[hide]Let's first show that this integral converges. We have

$ \int_0^{ \plus{} \infty}\frac {1}{1 \plus{} x^4}\,dx \equal{} \int_0^1 \frac {1}{1 \plus{} x^4}\,dx \plus{} \int_1^{ \plus{} \infty}\frac {1}{1 \plus{} x^4}\,dx$.

The first integral on the right is definite. Regarding the second one, observe that, for $ x \geq 1$, $ 1 \plus{} x^4 \geq 1 \plus{} x^2 \equal{} > \frac {1}{1 \plus{} x^4} \leq \frac {1}{1 \plus{} x^2}$, which means that

$ \int_1^{ \plus{} \infty}\frac {1}{1 \plus{} x^4}\,dx \leq \int_1^{ \plus{} \infty}\frac {1}{1 \plus{} x^2}\,dx$.

But

$ \int_1^{ \plus{} \infty} \frac {1}{1 \plus{} x^2}\,dx \equal{} \lim_{a \to \plus{} \infty} \int_1^a \frac {1}{1 \plus{} x^2}\,dx \equal{} \lim_{a \to \plus{} \infty} [\arctan x]^a _1 \equal{} \frac {\pi}{2} \minus{} \frac {\pi}{4} \equal{} \frac {\pi}{4}$,

and so $ \int_1^{ \plus{} \infty}\frac {1}{1 \plus{} x^4}\,dx$ also converges. The final conclusion is that $ \int_0^{ \plus{} \infty}\frac {1}{1 \plus{} x^4}\,dx$ converges. Let's now compute the value of this integral. Calculating the primitive of the function $ \frac {1}{1 \plus{} x^4}$, we get

$ \int\frac {1}{1 \plus{} x^4}\,dx \equal{} \int \frac {1}{(x^2 \plus{} 1)^2 \minus{} 2x^2}\,dx \equal{} \int \frac {1}{(x^2 \minus{} x \sqrt {2} \plus{} 1)(x^2 \plus{} x \sqrt {2} \plus{} 1)}\,dx$

$ \equal{} \int \left(\frac { \minus{} \frac {1}{2 \sqrt {2}} x \plus{} \frac12}{x^2 \minus{} x \sqrt {2} \plus{} 1} \plus{} \frac {\frac {1}{2 \sqrt {2}} x \plus{} \frac12}{x^2 \plus{} x \sqrt {2} \plus{} 1}\right)\,dx$

$ \equal{} \frac {1}{4 \sqrt {2}} \ln \left(\frac {x^2 \plus{} x \sqrt {2} \plus{} 1}{x^2 \minus{} x \sqrt {2} \plus{} 1}\right) \plus{} \frac {1}{2 \sqrt {2}} [\arctan(x \sqrt {2} \plus{} 1) \plus{} \arctan(x \sqrt {2} \minus{} 1)] \plus{} C$,

after some straightforward computations. So now we can write

$ \int_0^{ \plus{} \infty} \frac {1}{1 \plus{} x^4}\,dx \equal{} \lim_{a \to \plus{} \infty} \left[\frac {1}{4 \sqrt {2}} \ln \left(\frac {x^2 \plus{} x \sqrt {2} \plus{} 1}{x^2 \minus{} x \sqrt {2} \plus{} 1}\right) \plus{} \frac {1}{2 \sqrt {2}} [\arctan(x \sqrt {2} \plus{} 1) \plus{} \arctan(x \sqrt {2} \minus{} 1)] \right]^a_0$

$ \equal{}> \int_0^{ \plus{} \infty} \frac {1}{1 \plus{} x^4}\,dx \equal{} \frac {\pi}{2 \sqrt {2}}$.[/hide]

this reminds me of the famous integral $ \int\sqrt{\tan x}\text{d}x$  :)

[hide]As an alternative, we can compute the value of this integral through Complex Analysis. For that purpose, just observe that

$ \int_0^{\plus{} \infty} \frac{1}{1\plus{}x^4}\,dx \equal{} \frac12 \int_{\minus{} \infty}^{\plus{} \infty} \frac{1}{1\plus{}x^4}\,dx \equal{} \int_C \frac{1}{1\plus{}z^4}\,dz$

$ \equal{} 2 \pi i \sum \{\text{residues of}\, \frac{1}{1\plus{}z^4}\,\,\text{in the upper half plane} \}$

where $ C$ is the contour made up of an half circle of radius R in the upper half plane (and we consider $ R \to \plus{} \infty$).[/hide]


Javier is biking at 15 miles per hour. There are 5280 feet in a mile. How many feet does he travel in one minute?

15 miles per hour = 1/4 miles per minute = 1320 feet per minute


Ok. Could somone please answer this, or give me a link to a web site that can help me...

if a + b = 4j - 1 and a - b = i + 3j find a and b algebraically

Thanks

The Olympiad section isn't the place for problems like this.  You want [url=http://www.artofproblemsolving.com/Forum/index.php?f=300]Classroom Math[/url].

Anyway, note that

$a = \frac{(a+b)+(a-b)}{2}$
$b = \frac{(a+b)-(a-b)}{2}$


linear algebra
matrix
linear algebra unsolved

Find the nxn complex matrices which are product of nilpotent matrices. An answer of the form: "the non-invertible matrices" does not suffice.

What about finite product of triangular superior matrices with diagonals only $0$

Triangular matrices with diagonal 0? Sorry if I hurried and wrote non-senses.

Last week my teacher told me that he's seen this problem on AMM: (I myself haven't seen it)
Prove that any non-invertible matrix can be written as a product of four nilpotent matrices.

Indeed it is answered in AMM Oct 1993 page 807.


geometry
3D geometry
sphere
quadratics
calculus
calculus computations

A cylinder is inscribed in a given sphere of unit radius. Find the dimensions of the cylinder so that the surface area (including the top and bottom) of the cylinder is maximal.

Let's try Lagrange multipliers. We are to maximize $ A=2\pi rh+2\pi r^{2}$ for $ h,r\ge0$ subject to the constraint $ \frac{h^{2}}4+r^{2}=1.$

Lagrange multipliers gives us

$ 2\pi h+4\pi r=2\lambda r$

$ 2\pi r=\frac{\lambda}2\,h$

Eliminating $ \lambda$ between these two equations gives us

$ \frac{h^{2}}4+\frac{rh}2=r^{2}$

Now, we use that $ \frac{h}2=\sqrt{1-r^{2}}$ to get

$ 1-r^{2}+r\sqrt{1-r^{2}}=r^{2}$

Isolating the square root on one side and squaring, we get

$ 5r^{4}-5r^{2}+1=0$

This quadratic for $ r^{2}$ has two roots, $ r^{2}=\frac{5\pm\sqrt{5}}{10}.$ However, checking this in the previous equation reveals that one of those is a spurious root produced by squaring the equation. The only root is then $ r^{2}=\frac{5+\sqrt{5}}{10}.$

Now we step back to review the overall problem. We have two extreme cases and one intermediate critical point. One extreme case is $ r=0,h=1.$ For that, we get an area of zero. The other extreme case is $ r=1,h=0,$ with an area of $ A=2\pi.$

The intermediate critical point has $ r=\sqrt{\frac{5+\sqrt{5}}{10}}$ and $ h=2\sqrt{\frac{5-\sqrt{5}}{10}}.$

For that, we note that $ rh=\frac{4\sqrt{5}}{10}$ and the area is

$ A=2\pi\cdot\frac{4\sqrt{5}}{10}+2\pi\cdot{\frac{5+\sqrt{5}}{10}= \pi\left(\sqrt{5}+1\right)}$

Since this is greater than $ 2\pi,$ it gives the maximum area.


Methinks you are missing an exponent sign and some parentheses.

Edit:  all better now.

lol, oops, when i copied it from the site i guess it didnt go through with the superscript and stuff, sorry about that. its fixed now...

so the 3/2's are exponents? if so then:

[hide] let x :^2:=52+6 :sqrt: 43, and y :^2:=52-6 :sqrt: 43. then the expression becomes x^3-y^3. factor, becomes (x-y)(x^2+xy+y^2). the latter factor is 104+34=138, and then let x-y=z. then z^2=104-2(34)=36, z=6, times 138=828?  [/hide]

While you're editing, you might change the word "factorize," which only exists in British English.  Here in America, we use "factor."

well he lives in new england if that means anything

I noticed that

52 + 6 :sqrt: 43 = ( :sqrt: 43 + 3) :^2:

and

52 - :sqrt: 43 = (:sqrt: 43 - 3) :^2:

and i went from there to get the same answer.

thats just off topic, so is this, but pointing it out


1.) Find the value of, $30C0*30C10-30C1*30C11+30C2*30C12-...+30C20*30C30$

2.)A fair dice (6 faced) is rolled repetedly till 1 comes for the first time. Find the  probability that the number of times the dice is rolled is an even number.

Could you explain your solution paladin? I don't get how you got those numbers.

I guess I'll try to explain it then because I did it the exact same way.

Notice that the probability of rolling a 1 on the second roll is $\displaystyle(\frac{5}{6})\displaystyle(\frac{1}{6})$.

The probability of rolling a 1 on the fourth roll is $\displaystyle(\frac{5}{6})^3\displaystyle(\frac{1}{6})$. Since the first three rolls cannot be a 1.

This can be expressed as:

$\displaystyle(\frac{5}{6})\displaystyle(\frac{1}{6}) + \displaystyle(\frac{5}{6})^3\displaystyle(\frac{1}{6}) + \displaystyle(\frac{5}{6})^5\displaystyle(\frac{1}{6}) + \ldots$

$ = \displaystyle(\frac{5}{6})\displaystyle(\frac{1}{6})\displaystyle(1 + \displaystyle(\frac{5}{6})^2 + \displaystyle(\frac{5}{6})^4 + \ldots )$

${ = \frac{5}{36}\displaystyle(\frac{36}{11}}) = \frac{5}{11}$

And we have that as our answer. I hope this is enough explanation.

Sorry. Yeah, white_horse_king88 has most of it. I'll give a quick explanation too:

Probability that the first roll is not $1$ and the second roll is: $\left(\frac{5}{6}\right)\cdot\frac{1}{6}$.
Probability that the first three rolls are not $1$ and the fourth is: $\left(\frac{5}{6}\right)^3\cdot\frac{1}{6}$
Probability that the first five rolls are not $1$ and the sixth is: $\left(\frac{5}{6}\right)^5\cdot\frac{1}{6}$
.
.
.
Continuing this on forever, we have the sum of the infinite geometric series with first term $\left(\frac{5}{6}\right)\cdot\frac{1}{6} = \frac{5}{36}$ and common ratio $\left(\frac{5}{6}\right)^2 = \frac{25}{36}$, with which we can apply the well know formula $S = \frac{a}{1-r} = \frac{\frac{5}{36}}{1-\frac{25}{36}} = \frac{5}{11}$.

Thanks,
Anyone know first one? (xCy is just x choose y)
Just by looking at it it must have something to do with binomial expansion but im not sure wat the binomial would be.


Can you find some sentences like this :
[b]WAS IT A CAT I SAW[/b]
i think these sentences are very beautiful ( do you know why?)
 :D [hide]this sentences are incredible because if you read the letters from left to right or reat the letters from right to left you will read same sentence.[/hide]

Yep they are, except it's going to the Puzzles and Brainteasers section.

Palindromic sentences you want? Palindromic sentences you get!  :D
http://www.mockok.com/

This has already been discussed on AoPs.  Heres a link:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=40010

thank you everyone.thank you so much.  :D


Given an obtuse triangle,

1) Show that its circumcenter lies outside the triangle.

2) Show that its orthocenter lies outside the triangle.

Coxeter and Greitzer don't give very rigorous proofs for these two results. Anyone have any ideas?

[quote="Scrambled"]im probably seriously misunderstANDING[/quote]

What do you mean?

[hide]

1) One of the altitudes lies completely outside the triangle (with the exception of intersecting a vertex) the orthocenter must lie on this line.

2) Look at arcs. The obtuse angle must subtend a segment closer to the vertex than the diameter parallel to it. However, the center lies on this diameter, so the circumcenter is outside the triangle.

[/hide]


trigonometry
Trigonometric Identities
algebra
equation
IMO Shortlist
IMO Longlist

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]


geometry
circumcircle
incenter
geometric transformation
reflection
Euler
geometry proposed

Given is a $ \triangle ABC$ with circumcenter $ O$ and incenter $ I$. Let $ X$, $ Y$ and $ Z$ be midpoints of sides $ BC$, $ CA$, $ AB$, respectively. Denote by $ N$ and $ K$ feet of perpendiculars from $ A$ and $ C$ to $ OI$[color=red],respectively,[/color] and let $ P$ be symmetric point to $ N$ across $ XY$.    Prove that $ ZP \equal{} ZK$.

It is correct,moreover it is very nice problem

[quote]Please check your problem again, I think your problem is wrong![/quote]

Problem is surely correct The thing that I missed is to write "respectively" (red one in edited post). I thought it is obvious from order of text but now I have added that part.

Please try to solve it now.

I think it is still wrong. Isn't $ P$ is reflection of $ N$ in the $ YZ$ and problem is showing $ YP \equal{} YK$?

Most probably you are wrong[b] Leonhard Euler[/b] The statement of problem is true now.

In this configuration it is simple work to prove that $ YP\equal{}YK\equal{}YN$ but I ask to prove $ ZP\equal{}ZK$ which is also true.

[quote="canada"]Most probably you are wrong[b] Leonhard Euler[/b] The statement of problem is true now.

In this configuration it is simple work to prove that $ YP \equal{} YK \equal{} YN$ but I ask to prove $ ZP \equal{} ZK$ which is also true.[/quote]
But in Zhang Fangyu's picture, $ ZP\equal{}ZK$ doen't hold

The problem is wrong. To fix it, $ K$ and $ N$ would have to be the feet of perpendiculars from $ A$ and $ C$ to $ OI,$ respectively. Then $ ZP \equal{} ZK$ is trivial for any line through $ O$, not just $ OI.$

In Zhang Fangyu's picture perpendiculars are $ CK$ and $ AN$ which is not case.

[b]yetti[/b] Firstfull read question correctly I can agree that $ YP\equal{}YK$ holds for arbitrarily line trough $ O$ but I can not see why should $ ZP\equal{}ZK$ be true for any line trough $ O$.

In your problem (the first post), the perpendiculars to $ OI$ are also $ CK$ and $ AN.$ So, what is the case ?

Sorry I have edited my post some hours ago

This picture was generated by current(GMT=8:28am) problem.

[color=red]Moderator EDIT: Attachment replaced by an identical image visible without downloading, with the same size and resolution, and with file size reduced from 245 to 7 kB.[/color]

Thanks for drawing[b] Akashnil[/b] It is just an example that question statement is true

Please can anyone post proof

To canada:
However do understand your problem rightly?  in Akashnil's picture, $ ZP \equal{} ZK$? what do differences have between Akashnil's picture and mine?

That was my point. Your problem is wrong. Both mine and Zhang Fangyu's constructions are correct. I only magnified the difference between $ ZP$ and $ ZK$


A company sells detergent in three different sized boxes: small
(S), medium (M) and large (L). The medium size costs 50\% more
than the small size and contains 20\% less detergent than the
large size. The large size contains twice as much detergent as the
small size and costs 30\% more than the medium size. Rank the
three sizes from best to worst buy. \\ \\
$ \text{(A)}\ SML \qquad \text{(B)}\ LMS \qquad \text{(C)}\ MSL \qquad \text{(D)}\ LSM \qquad \text{(E)}\ MLS$ \\ \\
Select the most appropriate letter.

____________________ S______ M ________________ L
cost:________________ 1 _ 1*(100+50%) _ (100+30%)(1(100+50%))
amount of detergent: __ 1 _ 2*(100-20%) __________ 2
the cost of one unit of detergent in the small size = 1
the cost of one unit of detergent in the medium size = 150%/160% = 15/16
the cost of one unit of detergent in the large size = (1.3*1.5)/2 = 1.95/2 = 39/40
Therefore: the cost is SMALL>LARGE>MEDIUM
The best buy = Medium, then Large, then Small
Ans: MLS = (E)


Find n.

n, 170,27,50,28,164,164,199 and 170
range = 181               median = 164

For range
I reordered.
27,28,50,164,164,170,170,199
199-27= 172
181-172= 9
199+9=208; 208-27=181 or
27-9= 18; 199-18=181
so couldn't n be 208 or 18?

For median
I reordered as above.
164+164= 328/2= 164
so couldn't n be 208 or 18? here also

I know 18 is the answer but I don't understand why?

Thanks for any help to clear this up for me.

Yea when you do it both 18 and 208 work?  Try giving us the exact wording of the problem, there may be something you missed.

Were there any resrtictions for n? (n is less than 200, etc.)


function
calculus
integration
real analysis
real analysis unsolved

[quote="harazi"]Let $f,g$ be nonnegative functions defined on the set of nonnegative real numbers, both integrable on this set and $a,b>0$ such that $g'(x)\leq a+b g^2(x)+f(x)$ for all $x$. Then $g$ goes to 0 at infinity.[/quote]

Manifesly false ! Take $f(x)=g(x)=e^x$
 :cool:

I'd like to see why the exponential function is integrable over $R_{+}$.  :roll:  I think you misunderstood the statement: it means that the integral of $f,g$ over the set of positive numbers is finite.


hey does anyone by any chance have the answers for the sprint test of 92-93 nationals? i wrote down the answers but not i cant read my handwriting...thanks it would be greatly appreciated!


geometry
geometric transformation
rotation
inequalities
logarithms
inequalities proposed

[quote="epsilonist"]And what about the geometrical Proof?  :?:[/quote]
The hint:
 If $ CD$ is altitude of $ \Delta ABC$ and $ \measuredangle  ACB\equal{}90^{\circ}$ then $ CD^2\equal{}AD\cdot BD.$  :wink:  :D  :)

[quote="arqady"][quote="epsilonist"]And what about the geometrical Proof?  :?:[/quote]
The hint:
 If $ CD$ is altitude of $ \Delta ABC$ and $ \measuredangle  ACB\equal{}90^{\circ}$ then $ CD^2\equal{}AD\cdot BD.$  [/quote]

Dear arqady,

I think five years is not too late for a solution to a problem! (You may have forgotten this problem already though...)
Let's see if your "geometric" solution matches mine.

[u]My solution, following your hint[/u]
Let's consider the graph of $f(x)=e^x-1$ and name it $\mathbb{T}$.
Let $O=(0,0), P=(a,0)$ for any $a>0$.
Let the rotation of $\mathbb{T}$ about $O$ by $-\frac{\pi}{2}$ be $\mathbb{T}'$.
Let the intersection between the line $x=a$ and $\mathbb{T},\mathbb{T}'$ be $Q,R$, respectively.
Now rotate $\Delta OPQ$ about $O$ by $-\frac{\pi}{2}$ to become a new triangle $\Delta OP'Q'$.
Look carefully at points $Q', R$ which are both one $\mathbb{T}'$. Since $e^a-1\ge a$ (a known inequality), $Q'$ is further to the right than $R$.
Notice that the equation of $\mathbb{T}'$ is $y=-\ln{(x+1)}$ which is convex.
Therefore, the intersection $I$ between $OQ'$ and $PR$ is above $R$.
Now we can wrap everything up.
\[a^2 = OP^2 = QP \cdot PI < QP \cdot PR = (e^a -1)\ln{(1+a)}\]
:D

PS. I like your idea of using graph to establish an inequality here.

[quote="arqady"]For all $ x>0$ prove that
\[ (e^x\minus{}1)\ln(1\plus{}x)>x^2\]
There is a nice geometrical proof. :wink:[/quote]
See here[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=504441[/url]


Find all positive integers $ (x,y,z)$ such that:
$ 3^x\plus{}4^y\equal{}7^z$.

if y=1 see http://www.mathlinks.ro/viewtopic.php?t=150672
and if y>1 then 8 divides 4^y so x and z are even(check it).  it gives
4^y=((7^(z/2))-(3^(x/2)))*((7^(z/2))+(3^(x/2)))
we must have
2^m=(7^(z/2)) - (3^(x/2)) , 2^n=(7^(z/2)) +( 3^(x/2))   (m+n=2y)
adding up the two equalities we understand that m=1;
it means that there exist k and l such that 7^k - 3^l=2 ( z/2=k and x/2=l)
 7^k - 3^l mod8 4,-2,0 . so if y>1 we don't  have any other solution.


Find the sum of the terms in the sequence 1,4,7,...193

There must be an easier way than adding them all up!

Notice the first and last added up equals 194.  The second added up with the second to last equals 194, and so on.  This should help...

I know the basic formula i figured is 194 times half of the number of numbers, but I just cant figure out the number of numbers  :(

[quote="isabella2296"]There are $ 193 \minus{} 1 \div 3 \plus{} 1 \equal{} 65$ numbers.[/quote]

You mean $ (193 \minus{} 1)\div3 \plus{} 1$...Your's really equals a mixed number.

[quote="tcjy"]AW MANNN!!!
There's an odd number of numbers so I cannot define the solution to this problem! :noo:[/quote]

What do you mean by 'cannot define the solution'?

And guys, do we need 5 posts for this problem?

There are 65 numbers, and since the sum of $ n$th number and $ 66 \minus{} n$th number is 197, the answer is $ \frac {65*197}{2} \equal{} 6305$.

Answer: 6305


$ f$ is a function in real numbers
$ f(0)\equal{}1/2$
For any $ x,y$ there exist $ a$ satisfying this equation
$ f(x\plus{}y)\equal{}f(x)f(a\minus{}y)\plus{}f(y)f(a\minus{}x)$
Prove that $ f$ is constant

[hide="Solution"]
Set $ x \equal{} 0, \ y \equal{} 0$. Then, by the given, we have

$ f(0) \equal{} f(0) f(a) \plus{} f(0) f(a)$

Using $ f(0) \equal{} \frac{1}{2}$,
$ \Rightarrow f(a) \equal{} \frac{1}{2}$
[/hide]

caralebee,I think that you forgot to say that a real number fixes.  then resolution of stuck is wrong.

[hide]

=>(*)

then:  =>(**)

therefor:  =>(***)

(*)=>

(**)=>

then

(***) =>

therefor the only solution is  [/hide]

Denote $ f(x\plus{}y) \equal{} f(x)f(a\minus{}y) \plus{} f(y)f(a\minus{}x)$ by (*).

stuck's interpretation: For every $ x,y,a$, we have (*).
peine's interpretation: Given a fixed real $ a$, for every $ x,y$ we have (*).
My interpretation: For every $ x,y$, there is some $ a$ for which we have (*).

I find this problem most likely false if my interpretation is the correct one (if not then it is exceedingly difficult). Can the OP clarify?

I think peine's interpretation is the correct one, because:
- It makes the problem solvable
- It has been posted before [url=http://www.mathlinks.ro/viewtopic.php?p=822867#822867]here[/url]

Also, peine, I think your solution is incomplete, because $ f(x)^2\equal{}\frac{1}{4}$ only allows you to conclude that $ f(x)\equal{}\frac{1}{2}$ or $ f(x)\equal{}\minus{}\frac{1}{2}$, for each $ x$

why?? he gives $ f(0)\equal{}\frac{1}{2}>0$ then the negative function is not verified. therefore the only solution is f(0)=\frac{1}{2}

It could happen that for some subset $ A$ of $ \mathbb{R}$ we have $ f(x) \equal{} \frac {1}{2},\forall x\in A$, and $ f(x) \equal{} \minus{} \frac {1}{2}\forall x\not\in A$. What you are saying is that $ 0\in A$, but that doesn't imply that $ A \equal{} \mathbb{R}$.


[url]http://www.vf11.com/math92/[/url] its another cool forum, not as cool as Aops of course  :P but we need more memebers. it was created by math92 so, u should at least look at it.

I sense somehow that valedictorian and math92 are not different persons!

Let me add another cool site :[url=http://www.zedboy.vf11.com]the generation zed[/url]


Divisibility
algebra
IMO
IMO 2003
number theory
IMO Shortlist

Determine all pairs of positive integers $(a,b)$ such that  \[ \dfrac{a^2}{2ab^2-b^3+1}  \]  is a positive integer.

if b = 1, a is an even number.
so case(i) may be (2k,1)

why the denominator has to be positive? couldnt it be negative?

Oh, yes...when b=1, then a must be even so case (i) becomes (n,1) with n even. Sorry about that!

The denominator must be positive since the numerator is positive and the ratio must be a positive integer.

sorry I didnt know that the ratio has to be positie integer. all i know was it has to be integer...

:D Be careful next time.
a 2 >=0 so DENOminator >=0

Who can help me with Dmitri Shiryayevs elegant solution :
[color=red][russian team member imo 2003][/color]

Let $n=2ab^2-b^3+1$ than we have $2ab^2 \equiv b^3-1 (modn)$ , also we have
$2a^2b^2 \equiv ab^3-a (modn)$ but $a^2 \equiv 0 (modn)$. Thus we have $ab^3-a \equiv 0 (modn)$,
$2ab^3-2a \equiv 0 (modn)$ also $2ab^3 -b^4+b \equiv 0 (modn)$ since n divides n.
So we have $2a \equiv b^4 -b (modn)$.
That gives: $m=b^4-b-2a \equiv 0 (modn)$.
And, $n=(2a-b)b^2+1>0$ since $\frac{a^2}{2ab^2-b^3+1}$ is positive and $a^2$ postitive, too. So that means denominator is also positive. By the way we have $2a-b \geq 0$
1. $2a=b$ gives $(a,b)=(k,2k)$
2. $2a-b>0$ ... $2a-b \geq 1$. It follows that $a^2$ divisible by $n$.
$a^2 \geq n = (2a-b)b^2+1$ ... $a^2 \geq b^2+1$ than we have $a>b^2$ $(*)$
And  $n=(2a-b)b^2+1>(2b^2-b)b^2+1=2b^4-b^3+1>b^4>b^4-b-2a=m$
From here we have $0 \geq m$.
a-) $b^4-b-2a=0$ gives series $(a,b)=(8k^4-k,2k)$
b-)If $b^4-b-2a<0$ so we have $-(b^4-b-2a) \geq n=2ab^2-b^3+1$ than
$2a(1-b^2) \geq b^4-b-b+1=(b-1)(b-1)^3$
So $b=1$ which gives $(2k,1)$.

So we are done.

Please explain me : $(*)$

davron

[quote="Achilleas Sinefakopoulos"]First note that we must have 1=< 2ab^2-b^3+1=b^2(2a-b)+1
and so, 1<=b<=2a.
Now, we'll prove that the pairs that solve this problem are
(i) $(a,b)=(n,1),$
(ii) $(a,b)=(n,2n)$, and
(iii) $(a,b)=(8n^4-n,2n)$
where $n$ is a positive

Indeed, if $b=1$, we get the pairs in (i), while if b=2a we get the
pairs in (ii). Now assume that $1<b<2a$ and set $r=a^2/( 2ab^2-b^3+1)$
Then the quadratic equation

(*)    $x^2-2xrb^2+r(b^3-1)=0$

has two solution that are positive integers, since one of them is a and their product is $r(b^3-1)>0$. Evaluating the discriminant D we have
$D=4(r^2b^4-rb^3+r)$. Thus $r^2b^4-rb^3+r=m^2$ for some natural
number and the above equation has solution $rb^2+m$ and $rb^2-m$. Now
 note that

$m^2 = r^2b^4-rb^3+r > r^2b^4-2rb^3+b^2=(rb^2-b)^2,$

with $rb^2-b>0$. Hence, $b>rb^2-m$.
However, since $a^2> b^2(2a-b\geq b^2$, we have $a>b$. Therefore,
$a=rb^2+m > b$ and the other (also positve) root, say c, is $c=rb^2-m < b$.
Furthermore, if 2c>b, then we have

$r=c^2/(b^2(2c-b)+1 \leq  c^2/b^2 <1$,

a contradiction. Thus $b=2c$ and $r=c^2$. Since $ac=r(b^3-1)$, we get
$a=c(b^3-1)=8c^4-c$. It is easy to verify that if $c\geq 1$, then the integers
b=2c and a=c(b^3-1) solve the problem, since

$a^2/(2ab^2-b^3+1) = c^2(b^3-1)^2 / [(b^3-1)(2cb^2-1)]=c^2$.

The proof is complete.[/quote]

what do you think about that,
but it might be true since i read that from KVANT magazine
[ i hope that someone will help]

davron

[quote="Davron"][quote="bilarev"][quote="Davron"]
$a^2 \geq b^2+1$ than we have $a>b^2$
[/quote]
I cant understand this... :|[/quote]

ok now i got this point :-) bilarev look it is really nice :-)

we know that $a^2\geq 2ab^2-b^3+1=b^2(2a-b)+1>b^2a$ thus we have $a^2>b^2a$ implies to $a>b^2$
nice right ?

davron[/quote]

I don't understand how from \[ a^2\geq b^2(2a-b)+1 \] you get \[ a^2>b^2a \]

ok i will explain this more clearly :

we know that $2ab^2-b^3+1$ is positive since $a^2\geq 0$
Also we know that $a^2 \geq b^2(2b-b)+1>b^2+1$ from here we can easily get that $a>b$.
ok ?
Then we again rewwrite the inequality $a^2 \geq b^2(2a-b)+1=b^2[a+(a-b)]+1>b^2a+1$ since $a>b$
Thus we can easily see that $a^2>b^2a$ which leasds to the desired $a>b^2$

[please if you understand this let me know :-)]

davron

This problem is very similar with 2007kmo first round,,,
So I v got it  :)

I was looking Davron's solution and I didn't understand the part the says that "From here $ m\le0$" .
Why?

When they ask for positive prove it to be positive!

The problem is more interesting if the problem is to find all positive integer pairs $ (a,b) $ such that \[ \frac{a^2}{2ab^2-b^3+1} \] is an integer

[quote=jgnr][hide]If $ b\equal{}1$, then $ (a,b)\equal{}(2l,1)$. Now, assume $ b>1$. Let $ \frac{a^2}{2ab^2\minus{}b^3\plus{}1}\equal{}k$. We have $ a^2\minus{}2b^2ka\plus{}(b^3\minus{}1)k\equal{}0$. The discriminant of this quadratic equation is a perfect square. So $ D\equal{}(2b^2k)^2\minus{}4(b^3\minus{}1)k$ is a perfect square. We can easily verify that $ (2b^2k\minus{}b\minus{}1)^2<D<(2b^2k\minus{}b\plus{}1)^2$. Therefore $ D\equal{}(2b^2k\minus{}b)^2$. Thus we have $ b^2\equal{}4k$. $ b$ must be an even number, let $ b\equal{}2l$. So $ a\equal{}l$ or $ a\equal{}8l^4\minus{}l$. Therefore, the solutions are $ (2l,1),(l,2l),(8l^4\minus{}l,l)$, where $ l$ is any positive integer.[/hide][/quote]

What a great soloution!
Wonderful:)

[quote=Anonymous]First note that we must have 1=< 2ab^2-b^3+1=b^2(2a-b)+1
and so, 1<=b<=2a.
Now, we'll prove that the pairs that solve this problem are
(i) (a,b)=(n,1),
(ii) (a,b)=(n,2n), and
(iii) (a,b)=(8n^4-n,2n)
where n is a positive

Indeed, if b=1, we get the pairs in (i), while if b=2a we get the
pairs in (ii). Now assume that 1<b<2a and set r=a^2/( 2ab^2-b^3+1)
Then the quadratic equation

(*)    x^2-2xrb^2+r(b^3-1)=0

has two solution that are positive integers, since one of them is a and their product is r(b^3-1)>0. Evaluating the discriminant D we have
D=4(r^2b^4-rb^3+r). Thus r^2b^4-rb^3+r=m^2 for some natural
number and the above equation has solution rb^2+m and rb^2-m. Now
 note that

m^2 = r^2b^4-rb^3+r > r^2b^4-2rb^3+b^2=(rb^2-b)^2,

with rb^2-b>0. Hence, b>rb^2-m.
However, since a^2> b^2(2a-b>=b^2, we have a>b. Therefore,
a=rb^2+m > b and the other (also positve) root, say c, is c=rb^2-m < b.
Furthermore, if 2c>b, then we have

r=c^2/(b^2(2c-b)+1 <= c^2/b^2 <1,

a contradiction. Thus b=2c and r=c^2. Since ac=r(b^3-1), we get
a=c(b^3-1)=8c^4-c. It is easy to verify that if c>=1, then the integers
b=2c and a=c(b^3-1) solve the problem, since

a^2/(2ab^2-b^3+1) = c^2(b^3-1)^2 / [(b^3-1)(2cb^2-1)]=c^2.

The proof is complete.[/quote]

Can you or someone else latex this?

Let $\frac{a^2}{2ab^2-b^3+1}=k$ Then rewriting this as a quadratic in $a$,
$$a^2 -2b^2ka +k(b^3-1)=0 \dots (*)$$
$\Delta = 4 (b^4k^2 - k(b^3-1))$ is a perfect square. Hence $b^4k^2-k(b^3-1)$ is a perfect square.

$$\left( b^2k-\frac{b}{2}- \frac{1}{2} \right)^2<b^4k^2-kb^3+k <\left( b^2k-\frac{b}{2}+1 \right) ^2$$
Since $\frac{b^2}{4}-b+1+2b^2k \geq 2b^2k>k$
And $b^2 \left( k-\frac{1}{4} \right) - \frac{b}{2}+k- \frac{1}{4} \geq  b^2 \left( 1-\frac{1}{4} \right)-\frac{b}{2}+1-\frac{1}{4} >0$ implying the LHS inequality.

If $b$ is even
Hence $b^4 k^2 -kb^3 +k = \left( b^2k -\frac{b}{2} \right)^2 \Longrightarrow k=\left( \frac{b}{2} \right)^2$.
Plugging this into (*) gives the pairs $\boxed{(a,b)=(r,2r), \ (8r^4-r,r)}$

If $b$ is odd
$b^4k^2-kb^3+k = \bigg( b^2k -\frac{b-1}{2}\bigg) ^2  \Longrightarrow k = \bigg( \frac{b-1}{2} \bigg)^2$
We get $\boxed {(a,b) = (2r,1) }$
Checking our 3 solutions, we are done.

[hide=Cool]

Notice that any pair $(a,b)$ that satisfies this equation must satisfy either $2a=b$ or $a>b$. To see this, note that $(b^2(2a-b)+1)|a^2$ so therefore if $b>a$ we reach a contradiction unless $2a-b=0$.

Suppose the fraction equals $k$; we expand as a quadratic and obtain $$a^2-2akb^2+b^3k-k=0$$ Now suppose that we have a pair of solutions $(a_1,b)$ and $(a_2,b)$ for some $k$ such that $a_1\ge a_2>b$. We have $$\frac{a_1^2}{2a_1b^2-b^3+1}=\frac{a_2^2}{2a_2b^2-b^3+1}$$ Expanding and rearranging we obtain $$2a_1a_2b^2+a_1+a_2=b^3(a_1+a_2)$$ Write $a_1=b+k$ and $a_2=b+j$. Then, we notice $$2a_1a_2b^2=2b^2+2bk+2bj+2kj>2b^2+bk+bj=b(a_1+a_2)$$ contradiction, so therefore at least one of $a_1,a_2$ (in this case WLOG $a_2$) we have $2a_2=b$.

Plug in $a_2$ for $a$; we obtain that $k=\frac{b^2}{4}$. Plugging this back into the quadratic equation and solving for $a_1$, we obtain that $a_1=\frac{b^4-b}{4}$. Therefore, the solution set is $$(a,b)=(\frac{b}{2},b), (\frac{b^4-b}{4},b)$$ for any even $b$. $\blacksquare$

[/hide]

We consider the case $b = 1$ to see that any pair $(a,1)$ will have $\frac{a}{2}$ be an integer, so we have all pairs $(a,1)$ where $a$ is even.

Consider any satisfactory pair $(a,b)$ with minimal $a$, fixed $b \ge 2$, and fixed $k$. We have that $\frac{a^2}{2ab^2-b^3 + 1} = k$. Then, we have that $a^2 - 2kab^2 + kb^3-k = 0$, which is a quadratic in $a$. Thus, $2kb^2-a$ is also a solution to this quadratic. Since $a$ is minimal, we have that $kb^2 \ge a$. This means we can write $a = kb^2 - m$ for some nonnegative $m$. Plugging this in, we see that
\[0 = (kb^2-m)^2 - 2kb^2(kb^2 -m) + kb^3 - k = \]
\[k^2b^4 - 2mkb^2 + m^2 - 2k^2b^4 + 2mkb^2 + kb^3 - k = \]
\[m^2-k^2b^4 + kb^3-k.\]
Then, we have that
\[m^2 = k^2b^4-kb^3+k = \left(kb^2-\frac{b}{2}\right)^2 - \frac{b^2}{4} + k.\]
Now, we observe that changing $kb^2-\frac{b}{2}$ by $\frac{1}{2}$ will change $\left(kb^2-\frac{b}{2}\right)^2$ by $kb^2-\frac{b}{2} \pm \frac{1}{4}$, which must have absolute value greater than $k-\frac{b^2}{4}$ because $kb^2-\frac{b}{2} \pm \frac{1}{4}$ is clearly positive and
\[kb^2-\frac{b}{2} \pm \frac{1}{4} = k(b^2-1)-\frac{b}{2} \pm \frac{1}{4} + k \ge \frac{3kb^2}{4}-\frac{b}{2} \pm \frac{1}{4} + k \ge \]
\[\frac{3b^2}{4}-\frac{b}{2} \pm \frac{1}{4} + k \ge \frac{3b^2}{4}-\frac{b^2}{8} \pm \frac{1}{4} + k = \frac{5b^2}{8}\pm \frac{1}{4} + k \ge \]
\[\frac{5b^2}{8}- \frac{1}{4} + k \ge \frac{9b^2}{16} + k > \frac{b^2}{4} + k \ge \left|-\frac{b^2}{4}+k\right|.\]
Thus, we have that $m = kb^2 - \frac{b}{2}$ and $k = \frac{b^2}{4}$. This gives $a = \frac{b}{2}$, so $k = a^2$. Then, we note that $8a^4 = 2a^2(2a)^2$, so we have the other solution of $(8a^4-a,2a)$. To conclude, we see that all the pairs can be expressed as one of $(2a, 1)$, $(a, 2a)$, and $(8a^4-a,2a)$ for some integer $a$.

Note that if $b=1$ then the expression reduces to $\frac{a}{2}$ and all even $a$ will work. So suppose $b>1$, and let $\frac{a^2}{2ab^2-b^3+1}=n$, where $n$ is a positive integer. This equation becomes $a^2-(2b^2n)a+(b^3n-n)=0$ which is a quadratic in $a$, so by quadratic formula $a=\frac{2b^2n\pm \sqrt{4b^4n^2-4b^3n+4n}}{2}$, meaning that $4b^4n^2-4b^3n+4n$ (which we will henceforth call $d$) is a square.

Computing, we find $d-(2b^2n-b-1)^2=4n+4b^2n-(b+1)^2>4b^2n-(b+1)^2=(2b)^2n-(b+1)^2>0$, so $d>(2b^2n-b-1)^2$. Also, we can find that $(2b^2n-b+1)^2-d=4b^2n-4n+(b-1)^2=4n(b^2-1)+(b-1)^2>0$, so $d<(2b^2n-b+1)^2$. Therefore $d=(2b^2n-b)^2$, which reduces to $b^2=4n$. Thus we can let $n=k^2$ and $b=2k$, and plugging this into the quadratic formula we find that $a=k, 8k^4-k$. Plugging this back into the original expression we find that all of these solutions work, so our final solutions for $(a, b)$ are $(2k, 1)$, $(k, 2k)$, and $(8k^4-k, 2k)$.

$\textbf{\underline{Answer}} $ I claim that $(a,b)$ are in form of $(l,2l),(2l,1),(8l^4-l,2l)$ for $l \in Z^+$. Obviously these pairs works .

$\underline{\emph{Proof}} .$

  At first put $b=1$ to eleminate $b$ in denominator of $\frac{a^2}{2ab^2-b^3+1}$. which is $ \frac{a}{2}$. Indeed it's an integer if $a=2I \in Z^+$.\\
[list]

[*]Suppose $k=\frac{a^2}{2ab^2-b^3+1}$ .
Since $k \in Z^+$ so $2ab^2-b^3+1>0$ .
 which means $ b^2(2a-b) \ge 0$.

which gives $2a\ge b$.
 since $k\ge 1$ we must have $a^2 \ge b^2 (2a-b)\ge 0$ .
$2a-b \ge 0$ so either $2a-b=0$ or $2a-b\ge 1$.

 which means $a^2>b^2$ or $a>b$.

[*]we can set $a,b$ such that  $a+b$ is the minimal  that the sum can take and  the quadratic equation $x^2-2kb^2x +k(b^3-1)=0....(1)$ has a root as $a$ .
 so another root would be $2kb^2-a=\frac{k(b^3-1}{a}$.
It's not hard to show that $2kb^2-a=\frac{k(b^3-1)}{a} \in Z^+$.
so, we get $a\le b$  using the minimal condition .

[/list]

From the above statement we can confirm that $b=2a$ staisfy our equation.

$\underline{\text{Another pair }}$ $(a,b)$ is $ (l,2l)$.

Equation $(1)$ has a root $a$ suppose another root is $\alpha $ so, we must have $\alpha =\frac{k(b^3-1)}{a} \le \frac{k(b^3-1)}{kb^2} <b$.

if we set $\alpha =l$ where $b=2l \in Z^+$.
$a=8l^4-l$ .

$\underline {\text{Last Pair is }} (8l^4-l,2l)$.

The solutions are $(n,1)$, $(n,2n)$, and $(8n^4-n,2n)$ for all positive integers $n$.

Let $\frac{a^2}{2ab^2 - b^3 +1} = k$. Therefore, we can write this as a quadratic in $a$: $a^2 -2b^2k \cdot a + (b^3-1)k=0$. By the quadratic formula,
\[ a = \frac{2b^2k \pm \sqrt{4b^4k^2 - 4(b^3-1)k} }{2} .\]
Note that the discriminant is a perfect square, and we claim that it must be equal to $(2b^2k-b)^2$ for $b>1$. Note that
\[ 4b^4k^2 - 4(b^3-1)k  - (2b^2k-b-1)^2 = 4b^2k + 4k + (b+1)^2 > 0  .\]
Also,
\[ 4b^4k^2 - 4(b^3-1)k  - (2b^2k-b+1)^2 = -4b^2k + 4k + (b-1)^2 < 0 .\]
when $b > 1$. When $b=1$, we have the solution $(n,1)$ for all positive integers $n$. For $b>1$, since it is bounded between these two squares, $4b^4k^2 - 4(b^3-1)k = (2b^2k - b)^2$. This yields $b^2 = 4k$, so we can let $b = 2n$, which gives $a = n, 8n^4 - n$ from the quadratic formula. Thus, the other two solutions are $(n,2n)$ and $(n, 8n^4 - n)$.

yikes my approach is very different (I knew there had to be a solution with the discriminant :/ I guess I'm bad at bounding)

I claim the only solutions are $(2n,1), (n,2n)$, and $(8n^4-n,2n)$ where $n$ is a positive integer.

If $b=1$, it's easy to see that $a$ must be even to get a positive integer, giving $(2n,1)$. Now suppose $b>1$.

Let the expression equal $k$. Note that $n|a^2$. If $a^2=n$, then \[2ab^2-b^3+1=1 \iff b^2(2a-b)=0 \iff b=2a\] since $b>0$, and this gives the solution $(n,2n)$.

Now, suppose $a^2>k$.We can cross multiply and do a bunch of bounding \[a^2=2ab^2k-kb^3+k\iff a^2-k= b^2(2ak-kb) \implies b^2|a^2-k\implies a>b\] Also, from completing the square, \[a^2=2ab^2k-kb^3+k \iff (a-kb^2)^2=k^2b^4-kb^3+k\] One can easily verify that the RHS is bounded above by $(kb)^2$ and bounded below by $(kb^2-b)^2$.

If $a<kb^2$, then we get that $0<a<b$, a contradiction. This means $a>kb^2$, which gives $2kb^2-b<a<2kb^2$. Let $a=2kb^2-t$ for $0<t<b$. Using the fact that $b^2|a^2-k$, \[b^2|a^2-k=(2kb^2-t)^2-k \implies b^2|t^2-k\] If $t^2>k$, then $b^2\leq t^2-k$, which is a contradiction since $b>t$. Thus, $k\geq t^2$.

If we plug in this new expression for $a$ into the original expression, we have that \[a^2=(2kb^2-t)^2=kb^4-kb^3+6 \iff t^2=k(2tb^2-b^3+1)\implies k|t^2\] This means $t^2\geq k$, or $t^2=k$. Plugging this in and solving gives $b=2t$ and $a=8t^4-t$, the last valid solution set. Having exhausted all cases, we are done.

Since the top is positive, so is the bottom. Hence, the denominator is upper bounded by the numerator so\[a^2 \geq (2a-b)b^2 + 1\]so in fact, either $2a = b$ or $a^2 \geq b^2 + 1 \implies a > b$. Indeed, we check that all pairs $\boxed{(n, 2n)}$ work, so we need only consider the remaining $a > b$ case.

Consider $\tfrac{a^2}{2ab^2 - b^3 + 1} = k$ for a fixed $k$; then, we have\[a^2 = kb^2(2a - b) + k \implies a^2 - 2kb^2a + k(b^3 - 1)\]so for any given solution $(a, b)$, so is $(2kb^2 - a, b) = \left(\tfrac{k(b^3 - 1)}{a}, b\right)$.

Therefore, for all solution pairs $(n, 2n)$ with $k = n^2$, it follows that all $\boxed{(8n^4 - n, 2n)}$ are also solutions.

Lastly, note that for all fixed $(b, k)$, there are at most two solutions for $a$, say $a_1$ and $a_2$. If neither of them are $\tfrac b2$, then since we have $a_1 + a_2 = 2kb^2$ and $a_1a_2 = k(b^3 - 1)$, but since $\min(a_1, a_2) > b$, it follows that\[a_1a_2 > b(2kb^2 - b) = 2kb^3 - b^2 > kb^3 - k\]for all $b \geq 2$, so for all $b \geq 2$ it follows that all solutions are $(n, 2n)$ and $(8n^4 - n, 2n)$. A quick final check on the edge case $b = 1$ shows that all solutions of the form $\boxed{(2n, 1)}$ work as well.

These are all of our solutions, and they clearly work, so we are done. $\blacksquare$

Nice !

Suppose the value of the expression is some $k\in \mathbb{N}$. Then
\begin{align*}
    a^2-2b^2ka + (kb^3-k)=0 \implies a=b^2k \pm \sqrt{b^4k^2 - (kb^3-k)}.
\end{align*}
Let $D=b^4k^2-b^3k+k$. We must have $D$ be a square. Notice
\begin{align*}
    (2b^2k-(b+1))^2<4D=4(b^4k^2 - b^3k+k)
    \iff 4k + 4kb^2 > (b+1)^2,
\end{align*}
which is true since $4kb^2 \ge 4b^2 > (b+1)^2$. Also,
\begin{align*}
    (2b^2k-(b-1))^2 > 4D=4(b^4k^2 - b^3k+k)  \iff 4k < 4b^2k+(b-1)^2,
\end{align*}
which is true since $4b^2k>4k$. Since $4D$ is a square, by the above we must have $4D = (2b^2k-b)^2$, which becomes $b^2=4k$ after simplification. Therefore, $b=2x$ for some $x \in \mathbb{N}$, so $k=x^2$, so $a=x, 8x^4-x$. Therefore, $(a,b)=(x,2x),(8x^4-x,2x)$ for any $x\in \mathbb{N}$.

[hide=Remarks] I thought this was really tricky.
Vieta jumping from the initial equation involving $k$ does not work since the equation is not symmetric in $a$ and $b$; in order to Vieta jump we need to be able to jump in both elements of the pair to keep decreasing. The intuition then is to see what $b$ can be. It is easy to see from size that $b$ must be rather small compared to $a$, so for example if $b=2$ we want to find when $\frac{a^2}{8a-7}$ is a positive integer. From here, the obvious approach is to use the quadratic formula and discriminant.
[/hide]

Let $k\in\mathbb N$ be equal to the given expression. Consider a solution $(a,b)$. \\
If $b=1$ then $k=\dfrac a2$ so $a$ is even. We can verify that all solutions of the form $(a,b)=(2n,1)$ for any $n\in\mathbb N$ work. \\
Assume that $b\ge2$. \\
We have $a^2-2kb^2a+kb^3-k=0$, a quadratic in $a$. \\
Now $\exists a_0$ which is another solution to this equation. By Vieta’s: \\
$a_0+a=2kb^2$ \\
$a_0\cdot a=kb^3-k$ \\
Therefore $\underbrace{a_0}_{\mathbb N}=\underbrace{2kb^2-a}_{\mathbb Z}=\underbrace{\dfrac{kb^3-k}a}_{>0}$. \\
Since $k\in\mathbb N$, it must be true that $a^2\ge2ab^2-b^3+1>0\implies a^2-1\ge b^2(2a-b)>-1$. If $2a-b=0$ then we have that $(n,2n)$ is a solution $\forall n\in\mathbb N$ (which works). If $2a-b<0$ then $b^2(2a-b)<0$, contradiction. Now we must have $2a-b\ge1$, so $a^2\ge b^2+1>b^2$ and $a>b$.
Since for fixed $b$ and $k$, $a$ has two solutions (namely $a$ and $a_0$), another solution is $a_0=n$ and $b=2n\forall n\in\mathbb N$. Then: \\
$a_0=n=8n^2\dfrac{a^2}{8an^2-8n^3+1}-a=\dfrac{8an^3-a}{8an^2-8n^3+1}$ \\
$\implies a=8n^4-n$ \\
And we see that $(a,b)=(8n^4-n,2n)\forall n\in\mathbb N$ is a solution. \\
I claim that the only solutions are $\boxed{(n,2n)}$, $\boxed{(8n^4-n,2n)}$, and $\boxed{(2n,1)}$ for arbitrary $n\in\mathbb N$. \\
Assume that $b\ge2$. Here, a contradiction proves our claim. \\
WLOG $a+b$ is minimal. \\
$a_0<a$ \\
$\iff kb^2<a$ \\
$\iff k<\dfrac a{b^2}$ \\
Now, we have that: \\
$k=\dfrac{a^2}{2ab^2-b^3+1}=\dfrac{a^2}{ab^2+\left(b^2(a-b)+1\right)}$ \\
Since $b^2(a-b)+1>0$, we see that: \\
$k<\dfrac{a^2}{ab^2}=\dfrac a{b^2}\bot$, which is a contradiction, thus proving our claim. $\square$

[quote=jgnr][hide]If $ b\equal{}1$, then $ (a,b)\equal{}(2l,1)$. Now, assume $ b>1$. Let $ \frac{a^2}{2ab^2\minus{}b^3\plus{}1}\equal{}k$. We have $ a^2\minus{}2b^2ka\plus{}(b^3\minus{}1)k\equal{}0$. The discriminant of this quadratic equation is a perfect square. So $ D\equal{}(2b^2k)^2\minus{}4(b^3\minus{}1)k$ is a perfect square. We can easily verify that $ (2b^2k\minus{}b\minus{}1)^2<D<(2b^2k\minus{}b\plus{}1)^2$. Therefore $ D\equal{}(2b^2k\minus{}b)^2$. Thus we have $ b^2\equal{}4k$. $ b$ must be an even number, let $ b\equal{}2l$. So $ a\equal{}l$ or $ a\equal{}8l^4\minus{}l$. Therefore, the solutions are $ (2l,1),(l,2l),(8l^4\minus{}l,l)$, where $ l$ is any positive integer.[/hide][/quote]

that's wonderful

The solutions are $(x, 2x), (8x^4 -x, 2x), (2x, 1)$ for all positive integers $x$, which can be easily verified to work.

Let $a^2 = k(2ab^2-b^3+1)$ for some positive integer $k$. We can solve to get: $$a^2 = k(2ab^2-b^3+1) \implies a^2-a(2kb^2)+k(b^3-1) = 0 \implies a = \frac{2kb^2\pm\sqrt{(2kb^2)^2-4k(b^3-1)}}{2},$$ so the discriminant is $4k^2b^4-4kb^3+4k.$ For $a$ to be an integer this must be a square, so set this equal to $n^2$ for some positive integer $n$. We have that $$4k^2b^4-4kb^3+4k = (2kb^2-b)^2-(b^2-4k) = n^2.$$ We have three cases: $b^2-4k > 0$, $b^2-4k < 0$, or $b^2-4k = 0$.

[color=#3D85C6][b]Case 1:[/b][/color] $b^2-4k > 0.$

We have $$b^2-4k = (2kb^2-b)^2-n^2 \ge (2kb^2-b)^2-(2kb^2-b-1)^2 = 4kb^2-2b-1 > b^2-4k,$$ a contradiction.

[color=#3D85C6][b]Case 2:[/b][/color] $b^2-4k < 0$.

We have $$4k-b^2 = n^2-(2kb^2-b)^2 \ge (2kb^2-b+1)^2-(2kb^2-b)^2 = 4kb^2-2b+1.$$ Notice that if $b \ge 2$, then $4kb^2-2b+1 > 4k-b^2$, a contradiction. Only if $b = 1$, then $4kb^2-2b+1 \le 4k-b^2$, and we actually achieve equality when $n = 2kb^2-b+1$. Plugging in $b = 1$ gives $a = 2x$ for all positive integers $x$, and we have our first set of solutions: $(2x, 1)$.

[color=#3D85C6][b]Case 3:[/b][/color] $b^2-4k = 0$.

This means $n = 2kb^2-b$ and $4k = b^2$, so solving for $a$, we get $$a = \frac{2kb^2 \pm (2kb^2-b)}{2},$$ so $a = \frac{b}{2}$, so we get our second set of solutions $(x, 2x)$, or $a = \frac{4kb^2-b}{2} = \frac{b^4-b}{2}$ for even $b$. Substituting $b = 2c$ gives $a = 8c^4-c$ for all positive integers $c$, which gives our final set of solutions $(8x^4-x, 2x)$. $\blacksquare$

[b][color=#f00]Remarks: [/color][/b]Nice discriminant bounding problem. This problem was quite straight forward and wasn't as contrived as I thought it would be.

When in doubt, do some bounds.

Let $a^2=k(2ab^2)-b^3k+k$ then $a^2-(2b^2k)a+b^3k-k,$ so by the quadratic formula $a=b^2k\pm \sqrt{b^4k^2-b^3k+k}$

For $a$ to be an integer, $b^4k^2-b^3k+k$ must be a perfect square. If $b=1$ then clearly all even $a$ work, so let $b>1.$

Let $r=b^2k-\frac{b}{2}+c$ then $r^2=b^4k^2-b^3k+\frac{b^2}{4}+2b^2kc-bc+c^2.$ If $b^4k^2-b^3k+k=r^2$ then $\frac{b^2}{4}+2b^2kc+c^2=bc+k.$

If $c$ is positive, then $b^2kc\ge bc$ and $b^2kc\ge k$ so $\frac{b^2}{k}+c^2\le 0$ implying $c=0$ a contradiction.

If $c$ is negative, then let $c=-c_0$ note that $\frac{b^2}{4}-bc+c^2=k(1-2b^2c)$ so $1+2b^2c_0\le \frac{b^2}{4}+bc_0+c_0^2.$ Note that for $c_0\ge 1$ we have $2c_0< \frac{1}{4}$ while $1\le bc_0+c_0^2$ so we have another contradiction. Thus, $c=0.$ This implies that $k=\frac{b^2}{4}.$

We can plug in for $a.$ Thus, we have the following solutions:

$(2k,1),(k,2k),(8k^4-k,2k)$ for positive integer $k$ which all work.

#2 in $\LaTeX$.

First note that we must have $$1\leq 2ab^2-b^3+1=b^2(2a-b)+1$$
And so, $$1\leq b \leq 2a.$$
Now, we'll prove that the pairs that solve this problem are $$(a,b)=(n,1),(n,2n),(8n^4-n,2n),$$
where $n$ is positive.

Indeed, if $b=1,$ we get the first pair, while if $b=2a$ we get the second
pair.

Now assume that $1<b<2a$ and set $r=\frac{a^2}{ 2ab^2-b^3+1}.$
Then the quadratic equation

$$x^2-2xrb^2+r(b^3-1)=0$$

has two solution that are positive integers, since one of them is $a$ and their product is $r(b^3-1)>0.$ Evaluating the discriminant $D$ we have
$D=4(r^2b^4-rb^3+r).$ Thus $r^2b^4-rb^3+r=m^2$ for some natural
number and the above equation has solution $rb^2+m$ and $rb^2-m.$ Now note that

$$m^2 = r^2b^4-rb^3+r > r^2b^4-2rb^3+b^2=(rb^2-b)^2,$$

with $rb^2-b>0.$ Hence, $b>rb^2-m.$
However, since $a^2> b^2(2a-b) \geq b^2$, we have $a>b.$ Therefore,
$a=rb^2+m > b$ and the other (also positive) root, say $c,$ is $c=rb^2-m < b.$
Furthermore, if $2c>b$, then we have

$$r=c^2/(b^2(2c-b)+1 \leq c^2/b^2 <1,$$

a contradiction. Thus $b=2c$ and $r=c^2$. Since $ac=r(b^3-1)$, we get
$a=c(b^3-1)=8c^4-c$. It is easy to verify that if $c\geq 1$, then the integers
$b=2c$ and $a=c(b^3-1)$ solve the problem, since

$$\frac{a^2}{2ab^2-b^3+1}= \frac{c^2(b^3-1)^2}{(b^3-1)(2cb^2-1)}=c^2.$$

The proof is complete.

Let $(a,b)$ satisfy $$\frac{a^2}{2ab^2-b^3+1}=m, ~~~ m\in \mathbb{N}.$$
It can be simplified to $$(2(a-b^2m))^2=(2b^2m-b)^2+4m-b^2. ~~~~~(1)$$

------

[u][b]Case 1: $4m-b^2>0.$[/u][/b]

By (1) we have \begin{align*}4m-b^2=(2(a-b^2m))^2-(2b^2m-b)^2 \\ \geq (2b^2m-b+1)^2-(2b^2m-b)^2=2(2b^2m-b)+1 \end{align*}

This is true because we can set $$M=2(a-b^2m), N=2b^2m-b, ~~~ M,N \in \mathbb{N}.$$
Due to $M^2-N^2>0$ and $N>0,$ \begin{align*} |M| \geq N+1 \\ \implies M^2-N^2 \geq (N+1)^2 -N^2=2N+1 \\ \implies 2(2b^2m-b)+1+b^2-4m \leq 0 \\ \implies 4m(b^2-1)+(b-1)^2 \leq 0. \end{align*} So it must be that $b=1,$ giving $\boxed{(a,b)=(2m,1)}$ as a satisfying pair.

--------

[u][b]Case 2: $4m-b^2<0.$[/u][/b]

By (1) we have, \begin{align*} 4m-b^2&= (2(a-b^2m))^2-(2b^2m-b)^2 \\ \leq (2b^2m-b-1)^2 -(2b^2m-b)^2=-2(2b^2m-b)+1 \\ \implies (4m-1)b^2-2b+(4m-1)\leq 0~~~~~ (2) \\ \implies D=4(1-(4m-1)^2)< 0, \end{align*} where $D$ is the discriminant of the quadratic polynomial of $b$.

Therefore because $4m-1>0,$ (2) is not true. And thus there is no pair $(a,b)$ satisfying the property in this case.

------
[u][b]Case 3: $4m-b^2=0.$[/b][/u]

So, $$(2(a-b^2m))^2=(2b^2m-b)^2 \implies 2a-2b^2m=\pm (2b^2k-b).$$

[u]Subcase 1:[/u]

$2a-2b^2m=-2b^2m+b$ implies that $b=2a.$ So we get the pair $\boxed{(a,b)=(k,2k)}$ where $k$ is any natural number, as a satisfying pair.

[u]Subcase 2:[/u]

$$2a-2b^2m=2b^2m-b \implies 4b^2m-b=2a.~~~~~(3)$$ However $4m-b^2=0$ implies $b$ is even. So there exists a $c$ such that $b=2c$ and $m=c^2.$ By (3) we get $a=8c^4-c.$

So another pair is $\boxed{(a,b)=(8c^4-c, 2c)}$ and we are done satisfying all cases.

Solved with [b]pi271828[/b] a while back.

The answer is $\boxed{(2n,1), (n,2n), (8n^4-n,2n)}$ for any positive integer $n$. These work. We now prove they are the only solution.

Let $\frac{a^2}{2ab^2-b^3+1}=k$.

Clearly if $b=1$, then $k=\frac{a}{2}$, so $a$ must be even. This falls in the solution set ($(2n,1)$).

Then we have \[a^2=2ab^2k - kb^3 + k = a^2 - 2kb^2 \cdot a + (kb^3 - k) = 0\]

Claim: Either $b=2a$ or $a>b$.
Proof: Suppose $a\le b$.

Note that $a^2\ge 2ab^2 - b^3 + 1 > 0$.

So $a^2 > b^2 (2a-b)>-1$.

This implies $2a-b\ge 0$.

If $2a-b>0$, then we must have $a^2>b^2$, contradiction. So $b=2a$. $\square$

Now, consider the quadratic \[X^2-2kb^2 X + (kb^3 - k) = 0\]

Let the two roots of this equation be $a_0$ and $a_1$, with $a_0\ge a_1$.
We have \begin{align*}
a_0+a_1 = 2kb^2 \\
a_0\cdot a_1 = kb^3 - k = k(b^3-1) \\
\end{align*}

If $a_0= a_1$, then $a_0=a_1=kb^2$, so \[(kb^2)^2 = k(b^3-1)\implies kb^4 = b^3 - 1,\]not true.

So we have $a_0>a_1$.

This implies $a_0>kb^2$.

If $b=2a_0$, then $b\ne 2a_1$, which implies $a_1>b$, however $b=2a_0>2a_1$. So $b\ne 2a_0$.

Case 1: $b=2a_1$.
Then we have $a_0 = 2kb^2 - \frac{b}{2}$, so $\left(2kb^2 - \frac{b}{2}\right)\cdot \frac{b}{2} = kb^3 - k$, which implies $kb^3 - \frac{b^2}{4} = kb^3 - k$., so $k=\frac{b^2}{4}$.

So $a_0 = \frac{b^4-b}{2}$.

Setting $b=2a_1$ gives the solutions $(a_0, b) = (8a_1^4 - a_1, 2a_1)$ and $(a_1,b) = (a_1, 2a_1)$, both of which fall into the solution set described above ($(8n^4-n,2n)$ and $(n,2n)$).

Case 2: $b\ne 2a_1$.
In this case, $a_0>a_1>b$.

Since $a_0>kb^2$, we have $a_0\cdot a_1 > kb^2 \cdot b = kb^3$, which implies $a_0\cdot a_1\ne k(b^3-1)$, contradiction.

This is a pretty nice Vieta jumping problem.

The answers are $(2n, 1), (n, 2n), (8n^2-4, 2n)$ for all positive integers $k$. They obviously work. Assume $b>1$ in what follows.

Fix $k$ to be the value of the expression, and assume it is an integer. Then, the condition is equivalent to $$a^2-2ab^2k+b^3k - k = 0.$$ In other words, if $(a, b)$ is a solution, then so is $\left(\frac{b^3k-k}a, b\right).$

[b][color=#f00]Claim.[/b][/color] We will have $b^3k-k < a^2$ unless $2a=b$ or $a \leq b$.

[i]Proof.[/i] Note that $b^3k-k < b^3k <\frac{a^2b}{2a-b}$, so the condition simplifies to $a > b$. $\blacksquare$

On the other hand, if $a\leq b$ in the original equation, then we have $a^2 > b^2(2a-b) > -1$, which implies that $b=2a$ as otherwise $a^2>b^2$. As a result, this means that the minimal solution $(a_0, b_0)$ to this equation for fixed $k$ takes the form $(n, 2n)$ for some positive integer $n$.

Plugging back into the recursive sequence of solutions, we notice that the solutions alternate between $(n, 2n)$ and $(n, 8n^4-n)$. Thus, these yield the entire curve of solutions.

Nothing new of a concept here; imo these problems should be less frequent because they're similar to a computational problem where you literally know the approach, you just need to follow through, whereas in problems like sequences or a combo game you need to be creative.

    Note that $b=2a$ works, and $(a,b)=(2m,1)$ both work. We claim that the only other solution is $(a,b)=(8m^2-4,2m)$; henceforth assume the first two are not true. Then $$\frac{a^2}{(2a-b)b^2+1}:=k,b^3k-k < b^3k <\frac{a^2b}{2a-b}<a^2\iff a>b;a^2-2ab^2k+b^3k-k=0\implies(a,b)\Leftrightarrow\left(\frac{b^3k-k}a,b\right)$$ yields a smaller solution as long as $a>b$. If $$a\le b,b\ne 2a,a^2\ge2ab^2-b^3=1>0,\text{ then }a^2>b^2(2a-b)\ge0\stackrel{\implies2a-b\ge0}{a\not>b\implies}2a-b=0,$$ which we've already accounted for. Plugging back in, we find that the solution set alternates between $(m,2m)\leftrightarrow(m,8m^4-m)$, which is all the solutions by our Vieta jumping, as desired.

What is a quadratic

The answer is $(n,2n)$, $(8n^4-n,2n)$, $(2n,1)$ only for any $n \in \mathbb{Z}^+$, which clearly work. Obviously if $b=1$ then $a$ must be even, so suppose $b>1$.

Rewrite the denominator as $b^2(2a-b)+1$. For the quantity to be positive, we need $2a-b \geq 0$. If $2a=b$ then we get the first curve of solutions. Otherwise, I claim that $a>b^2$. Indeed, if $b \geq a$ then the denominator is at least $b^2+1$ which is too big, and if $\sqrt{a} \leq b \leq a$ then the denominator is at least $(\sqrt{a})^2a+1$ which is also too big.

Now, if $2ab^2-b^3+1 \mid a^2$ then $2ab^2-b^3+1 \mid a^2(2b^2)^2$, and by the Euclidean algorithm it follows that $2ab^2-b^3+1 \mid (b^3-1)^2$. Obviously, $2ab^2-b^3+1$ is $1 \pmod{b^2}$, and since $(b^3-1)^2 \equiv 1 \pmod{b^2}$ as well it follows that $\tfrac{(b^3-1)^2}{2ab^2-b^3+1}$ is a $1 \pmod{b^2}$ positive integer as well. On the other hand, since $a>b^2$ we have
$$\frac{(b^3-1)^2}{2ab^2-b^3+1}<\frac{(b^3-1)^2}{2b^4-b^3+1}\leq b^2 \impliedby b^6-2b^3+1\leq 2b^6-b^5+b^2 \iff b^6+2b^3+b^2 \geq b^5+1,$$
which is obvious since $b$ is positive integer. Hence we need
$$\tfrac{(b^3-1)^2}{2ab^2-b^3+1}=1 \implies b^3(b^3-1)=2ab^2 \implies a=\frac{1}{2}b(b^3-1).$$
Plugging this back into the original, for this curve of solutions we would need
$$\frac{\frac{1}{4}b^2(b^3-1)^2}{b^3(b^3-1)-b^3+1}=\frac{b^2(b^3-1)^2}{4(b^3-1)^2}=\frac{b^2}{4} \in \mathbb{Z},$$
which clearly implies $b=2n$ for some $n \in \mathbb{Z}$ and $a=8n^4-n$. $\blacksquare$


Solve for p:

$\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$

I couldn't figure it out. I tried isolating P, but I end up getting more Ps around the equation. Seems like a no-win situation.

Are the variables all different? Assuming they are, I believe there are multiple solutions. Well it is possible to isolate them.
I will do it in a basic way to make it easy to understand.

Multiply $F$ to both sides.
It becomes, $1=\frac{F}{p}+\frac{F}{q}$
now just multiply p to both sides, then it becomes
${p=\frac{Fq}{q-F}}$

I think this is all you need, since there will be multiple solutions depending on the variable.

That's what I did, but P is still on the other side of the equation.

Here are the possible solutions:

a) p = f - q
b) p = q - f
c) p = fq/q-f
d) p = q+f/fq
e) none of these

[quote="Genecks"]How do I get the reciprocal? That's what I've been trying to figure out.[/quote]

\begin{eqnarray*}p &=& \frac{1}{ \frac{1}{F}-\frac{1}{q}}\\ &=& \left( \frac{1}{ \frac{1}{F}-\frac{1}{q}}\right) \cdot \left( \frac{Fq}{Fq}\right) \\ &=& \frac{Fq}{ \frac{Fq}{F}-\frac{Fq}{q}}\\ &=& \frac{Fq}{q-F}\\ \end{eqnarray*}

[quote="8parks11"]thanks for giving me the choices.
basically make it so its

$\frac{1}{p}=\frac{1}{F}-\frac{1}{q}$
now if you combine them and find the reciprocal,
you get $(c)$[/quote]
Taking the recipocal is indeed quite a bit easier, but not necessary:
[hide] $\frac{1}{f}=\frac{1}{p}+\frac{1}{q}\Rightarrow \frac{1}{f}=\frac{p+q}{pq}\Rightarrow \frac{f(p+q)}{pq}=1\Rightarrow pq=fp+fq\Rightarrow pq-fp=fq\Rightarrow p(q-f)=fq\Rightarrow p=\frac{fq}{q-f}$ [/hide]


modular arithmetic
function
number theory

What is the difference in meaning of notation between

[tex]  1)    11=10 + 1 (mod5)[/tex]

[tex]  2)    11= 10 (mod5) + 1 (mod5)[/tex]

*treat "=" as a congruent sign (I don't know the code for it yet)

I think (1) and (2) denote exactly same fact.

Right?

Well, I suppose they denote the same thing. But in #2, I wouldn't put (mod 5) twice in the same sentence. It's not exactly wrong, but it's redundant and rather pointless. In both of your examples, since 11 = 10 + 1 anyway, the equations don't really say anything for which you need this notation. There's a lot more point to saying that $11\equiv 1 \pmod 5$ or $11\equiv -4 \pmod 5$. (By the way, that's "\equiv" to get the symbol you want.) Or to saying that $11^{2003^{813}}\equiv 1\pmod 5$ since it's really just 1 raised to that ridiculous power.

Don't get too hung up in the notation - the whole point of modular arithmetic is that it frees you to say things very quickly and very efficently. A good saying is, "Commit to the clock." Once you've decided that you're working mod 5, you know that 11, 1, and -4 are all the same and that you stopped thinking that they are different from each other in any way. You're free to use the most convenient name, whatever it is.

[quote="Ultimatemathgeek"]What is the difference in meaning of notation between

[tex]  1)    11\equiv 10 + 1 \pmod 5 [/tex]

[tex]  2)    11\equiv  10 \pmod 5 + 1 \pmod 5[/tex]

I think (1) and (2) denote exactly same fact.

Right?[/quote]

(I put the equivalence signs you wanted into your equations.)

For modular congruence the (mod 5) modifies the equivalence sign,  :equiv:, telling which kind of equivalence it represents. Since the equation has only one equivalence sign, it needs only one modular marker, which traditionally goes at the end of the equation. As Kent Merryfield said, "I wouldn't put (mod 5) twice in the same sentence. It's not exactly wrong, but it's redundant and rather pointless." If I were inventing the notation today, I would have put the modular marker right next to the equivalence sign it modifies:
[tex] 31 \equiv_{\pmod 5}  6, [/tex]
but the modular notation was invented in olden days when subscript fonts were very difficult to use.

However, in more recent decades, some mathematicians have used "mod" to denote a binary operation on integers: [i]x[/i] mod [i]y[/i] means the remainder when [i]x[/i] is divided by [i]y[/i]. For example, 19 mod 4 = 3. That version of mod does not use parentheses around it. Since it is an integer function, equations with it are typically integer equations, which use equals signs rather than equivalence signs; though it would be possible to use the integer mod function inside a modular congruence equation.

Writing
[tex]   11\equiv  10 \pmod 5 + 1 \pmod 5[/tex]
runs the risk that the reader might think you wrote
[tex]   11 =  10 \bmod 5 + 1 \bmod 5,[/tex]
which is wrong since 10 mod 5 = 0 and 1 mod 5 = 1 but
[tex]   11 \neq  0 + 1.[/tex]

Erin Schram

What Erin Schram said can be put in terms of grammar. In the sentence $11\equiv1\pmod5$, "(mod 5)" is an adverb, not an adjective. If it were an adjective, modifying the noun "1", then the noun-adjective phrase "1 (mod 5)" might have independent meaning - but it doesn't. As an adverb, "(mod 5)" modifies the verb "$\equiv$", and that's why you don't need to say it twice.

Erin also mentions a second meaning for "mod" as a function. If I'm going to use that meaning, I'll go ahead and express it in the notation used in some spreadsheets and some programming languages. (It is a built-in function in Excel). In that notation, mod(11,5) = 1, and "mod(11,0)" would get a "division by zero" error message.

Incidentally, there's no reason to confine the use of mod to integers. There's nothing at all wrong with saying $-4.763\equiv.237\pmod1$ or $\displaystyle \frac{7\pi}3\equiv\frac{\pi}3\pmod{2\pi}$. There are certainly times in which you want to think that two angles that are equivalent mod $2\pi$ "might as well be the same angle."

Thanks.

Yes, it was a good thing you pointed out the difference between mod and integer mod function.

The reason I posted the original question about the notation is that in AoPS Volumn 1, the author uses mod quite a few times in a single line of equation.

In Chapter 5, there are a whole bunch of simple identities such as this:

[tex](a+b)(mod n) \equiv (a (modn))+(b (mod n))[/tex]

I think in that case, (mod n) is put for each term to emphasize something?

[quote="joml88"][quote="Ultimatemathgeek"]In Chapter 5, there are a whole bunch of simple identities such as this:

[tex](a+b)(mod n) \equiv (a (modn))+(b (mod n))[/tex]

I think in that case, (mod n) is put for each term to emphasize something?[/quote]

They are defining the basic laws of mods.  These may seem obvious but they must be defined first.  A lot of things that seem obvious are sometimes not even true!  The point of showing this is so that if you had something like (10+1) (mod 10).  You wouldn't have to add 10 and 1 first.  You could just note that 10 = 0 mod 10.[/quote]

So,
Was I right about the fact that the book uses (mod 10) after each term to emphasize that each term can be treated seperately? (My question is worded wierdly :) )


I am new to Asymptote and currently learning how to use from the AoPSWiki page on Asymptote. I followed the instructions in "Interactive Mode": I created the hot key and created 'test.asy'. I hit the hot key and get the following error message:

error:could not load module 'C:\Documents'
error:could not load module 'and'
error:could not load module 'Settings\.."
...

It's trying to find a file under `C:\Documents and Settings\...'. How do I solve this problem? FYI, I am using Windows. Thank you.

Use double quoted path, not single quoted.

I do have double quoted paths as instructed in the "Interactive Mode" on AopsWiki page. I just do not know if asy.exe gets double quoted paths all the time. I think asy.exe is getting paths without double quotes and parsing the paths by blanks and end of lines. All this is internal. The problem still persists.

Can you post every step you do and your sources also?


Go on [url=http://www.google.com/trends]Google Trends[/url] and search for "George W. Bush".  Look at which city the most searches come from :D

For some reason this makes me even more suspicious of Sean  :ninja:

Edit: some other fun ones:

- http://www.google.com/trends?q=terrorist&ctab=0&geo=all&date=all
- I knew
Canada was plotting something: http://www.google.com/trends?q=world+domination&ctab=0&geo=all&date=all

Yes, but let me point something else out to you, my MIT-attending liberal friend:
http://www.google.com/trends?q=John+Kerry&ctab=0&geo=all&date=all

Haha British people, enough said:
http://www.google.com/trends?q=porn&ctab=0&geo=all&date=all
http://www.google.com/trends?q=dentists&ctab=0&geo=all&date=all

I saw the first British one :D

Of course, we can't leave France out of this:

http://www.google.com/trends?q=retard&ctab=0&geo=all&date=all

Speaking of the French... [url]http://www.google.com/trends?q=aargh&ctab=0&geo=all&date=all[/url]


Prove that:$\sum\frac{(a^{2}-a)}{b}\geq 0$
where $a;b;c$ are positive real numbers satisfying $abc=1$


function
limit
complex numbers
complex analysis
complex analysis theorems

Let $z \in \mathbb{C}$ such that $z = a+bi$ (where $a,b \in \mathbb{R}$), then

$\Re(z) \equiv a \qquad \Im(z) \equiv b$

and
$\mod(z) \equiv (\Re(z)^{2}+\Im(z)^{2})^{\frac{1}{2}}$

$\arg(z) \equiv \arctan\Big(\frac{\Im(z)}{\Re(z)}\Big)$ for $-\pi < \arg(z) < \pi$

Can these be considered [b]formal[/b] definitions of $\Re(), \Im(),\mod(),$ and $\arg()$?

The only issue with defining real and imaginary parts and modulus "formally", as you say, might lie in the definition of what a complex number is. I think the easiest way to formalize it is to say that a complex number is a function $z$ from $\{1,2\}$ to $\mathbb{R}^{2}$. Then again, to formalize what a function is we need to formalize what an ordered pair is (you could of course also let a complex number be an ordered pair), and then you need to formalize what a set is and then you need to take a course in circle-jerking, 4 credits.

The conclusion is that worrying about whether the first three definitions are formal or not is quite pointless. It's intuitively obvious that they are good working definitions, no?

The definition of argument however is wrong. Not for some silly technical detail, but because it doesn't work. First of all, the real part might be zero. Secondly, arctan's range does not contain $(-\pi,\pi)$. Furthermore, even if you construct a function that will for any z give an argument in, say, $-\pi < \arg z \le \pi$, there is really no good motivation for why this would be the best interval to pick arguments in. This way $\arg$ will be a discontinuous function.

Rather, I'd like to say that for a $z \neq 0$ an argument of $z$ is a number $\theta$ such that $z = |z|e^{i\theta}$. You prove in complex analysis that there is no way to define an argument function that is continuous in an entire (punctured) neighborhood of 0. Instead you have to pick a branch of it for each application so that the branch is continuous around the values you are considering. The branch with $-\pi < \arg z < \pi$ is sure important, but it's not always the best. Having to deal with different branches of functions is a major theme in complex analysis.

[quote="Kalle"]The definition of argument however is wrong. Not for some silly technical detail, but because it doesn't work. First of all, the real part might be zero. Secondly, arctan's range does not contain $(-\pi,\pi)$. Furthermore, even if you construct a function that will for any z give an argument in, say, $-\pi < \arg z \le \pi$, there is really no good motivation for why this would be the best interval to pick arguments in. This way $\arg$ will be a discontinuous function.[/quote]
Even though I understand that there can't be a formal definition of $\arg(z)$. Would the following definition work better?

$\arg(z) \equiv \lim_{w \to z}\arctan\Big(\frac{\Im(w)}{\Re(w)}\Big)$

No, that would not work better. That definition would give you a result in $\left[-\frac{\pi}2,\frac{\pi}2\right],$ which still leaves out the second and third quadrants.

You can make a perfectly serviceable piecewise definition. For instance:

$\text{arg}(x+iy)=\begin{cases}\arctan\left(\frac{y}{x}\right),& x>0\\ \arctan\left(\frac{y}{x}\right)+\pi,& x<0\\ \frac{\pi}2,&x=0,y>0\\-\frac{\pi}2,&x=0,y<0\end{cases}$

BUT: That's only one possible definition. I've made this particular definition discontinuous on the nonpositive imaginary axis. You might want to move the discontinuity to some other location, which would change some of the details. And, as Kalle said,
[quote="Kalle"]...there is no way to define an argument function that is continuous in an entire (punctured) neighborhood of 0.[/quote]


Consider a isosceles triangle $ ABC$ $ (AB \equal{} AC)$ and let $ D$ the midpoint of $ BC$. If $ DE\bot\ AC$ prove that $ AF\bot\ BE$ where $ F$ is the midpoint of $ DE$.

[quote="galois01"]Consider an [color=red][b]isosceles [/b][/color] triangle $ ABC$ $ (AB \equal{} AC)$ and let $ D$ the midpoint of $ BC$. If $ DE\bot\ AC$ prove that $ AF\bot\ BE$ where $ F$ is the midpoint of $ DE$.[/quote]

 It is enough to take isosceles triangle .

[b]NOTE[/b]
Really nice problem. It was given at a competition in ... USSR. I think I have seen it in the book of Gegorov ect.
 The problem is already discussed here , but there are always chances for different solutions. Enjoy it !  :)

[u] Μπάμπης[/u]

[quote="stergiu"]
 It is enough to take isosceles triangle .[/quote]

You are right mr. Stergiou I wanted to write isosceles but I made a typo :blush: I edit it :)

[quote="galois01"]If $ DE\bot\ AC$[/quote]
How do you define the point E? If the only condition is that E must belong to the perpendicular to AC that passes by D, this theorem seems to be false...

[quote="GioMott"]
How do you define the point E? If the only condition is that E must belong to the perpendicular to AC that passes by D, this theorem seems to be false...[/quote]

Why this? $ DE$ is the perpendicular to $ AC$. The problem is correct see it again.

Let $ M$ be the midpoint of $ AD$, connect $ MF$. Then $ MF \perp DE$, $ DF=EF$. Since $ \angle{MDE}=\angle{ABC}$, so $ \triangle{MDE} \sim \triangle{ABC}$. So there is:

$ \frac{AD}{BC} = \frac{MF}{DE} \Longrightarrow \frac{AM}{MF}= \frac{\frac12 AD}{MF}= \frac{\frac12 BC}{DE}=\frac{BD}{BE}$

Because $ \angle{BDE}=\angle{AMF}$, so $ \triangle{BDE} \sim \triangle{AMF}$. So $ \angle{EBD}=\angle{MAF}$, which gives that $ A, B, D, F$ are concyclic. By $ AD\perp BC$, we have $ AF \perp BE$.

The problem is valid for any triangle:

[color=#400080]Given the triangle $\triangle ABC$, $AD$ its altitude, $DE$ the altitude of the triangle $\triangle ACD$ and $F\in\overline {DE}$ with the property that $\frac{DF}{EF}=\frac{DC}{BD}$; $AF\bot BE$ then.[/color]

[i]Proof:[/i]
[color=#408000]$\frac{BD}{CD}=\frac{AE}{CE}=\frac{EF}{DF}$ and, since $\triangle ADE\sim\triangle BCA$, having the homologue sides perpendicular, hence $DE$ and $AC$ are homologue, while $E$ and $F$ divide them at the same ratio, being homologue as well, consequently $BE$ and $DF$ are homologue and consequently perpendicular, done.[/color]

Best regards,
sunken rock


calculus
integration
function
calculus computations

This may seem annoying to you experts, but I am a self learner in mathematics.  I am interested in studying problem solving in advanced calculus as opposed to (well actually in addition to) the rigorous theory provided by most textbooks.  I want to know all the ways one can evaluate a definite integral. I know some such as the principle method of finding the primitive function, or replacing a part of the function with an infinite sum or another definite integral and rearranging, or parameterization.  Can some one give me a general run through of all the methods with an example of how to use each.

very define about it:
1/Reimamn
2/ Darboo
....

contour integration can be used to solve lots of definite integrals, but I'm not sure exactly how it works
also, sometimes there are obvious identities or simplifications you can make
other than that, every 'special' definite integral seems to require a different method


[quote="mayhemkerim"]Let  x,y,z  be positive  integer numbers satisfy    x - y + z =1   and  any of them not  equal others.Show   that  there are infinitive x,y,z   positive  integer  numbers  that   multiply  two  of  them  can divisible other.[/quote]

I don't understand your question. $ (k,k,1)$? :wink:

[quote]I don't understand your question. (k,k,1)?[/quote]

It is given that
[quote]any of them not equal others[/quote]

Let x,y,z be "different" positive integer numbers satisfy x - y + z =1 .Show that there are infinitive x,y,z positive integer numbers that multiply two of them can divisible other.
Sorry,that is the true sentence.

"are infinitive x,y,z positive integer numbers that multiply two of them can divisible other."

I didn't understand this. Do you mean $ xy|z$ or $ xz|y$ or $ yz|x$?

Btw, this should be in the number theory section.

Yes,you understand correctly.I wiil carry out that question  number theory section.

[quote="feliz"]
I didn't understand this. Do you mean $ xy|z$ or $ xz|y$ or $ yz|x$?
[/quote]
I think all three of these have to be true or else we could take (2k,2k+1,2)


The digits 2, 3, 5 and 7 are arranged randomly to form a
four-digit number. What is the probability that the number is
odd? Express your answer as a common fraction.

there are $ 4!$ or $ 24$ ways to arrange the numbers
6 ways result in even integers (3572, 3752, 5372, 5732, 7532, 7352)
therefore, there are 18 ways of getting an odd integer

so the probability is $ \frac{18}{24}\equal{}\frac{3} {4}$

our answer is $ \frac{3}{4}$

In order for a number to be odd, the units digit has to be odd. There are 4 ways to choose the units digit, and 6 ways to choose the rest of the number. Out of the 4 units digits, 3 are odd. Since there are 6 ways to pick the rest of the number for each units digit, the 6's cancel, leaving 3/4.

[quote="rightysportsfan1"]Or, you can just look at the problem and see that out of 4 numbers, 3 are odd.[/quote]

Is this a valid way to do this? Cuz that's cool. Never thought about it like that.

Counting by symmetry, sir. Good day sir.


Suppose $P(x)$ is a polynomial such that $P(1) = 1$ and
$\frac {P(2x)} {P(x + 1)} =8-\frac {56} {x + 7}$
for all real $x$ for which both sides are defined. Find $P(-1)$.

(Edited by pbornsztein.)

If I'm not wrong I found $P(-1)= - \frac 5 {21}$.

Rewrite the given relation as $(x+7)P(2x) = 8xP(x+1)$. (*)
Note that this relation between two polynomials holds for an infinite number of values, so it holds for all $x$.

Clearly $0$ and $-6$ are roots of $P$, so that $-2$ is a root too.
Now assume that $a \notin \{-6,-2,0,2 \}$ is a root of $P$. It is easy to verify that $2a-2$ is also a root of $P$ with $2a-2 \notin \{-6,-2,0,2 \}$. In that case, we can construct inductively an infinite number of distinct roots of $P$, so that $P(x)=0$ for all $x$, which contradicts that $P(1)=1$.
Thus the roots of $P$ are $-6,-2,0$ and, perhaps, $2$.
It follows that
$P(x) = \lambda (x-2)^{\alpha} \cdot (x+2)^{\beta} \cdot x ^{\gamma} \cdot (x+6)^{\delta}$ for some integers $\alpha \geq 0$ and $\beta, \gamma, \delta > 0$, and real $\lambda \neq 0.$
Back to (*) and identifying the exponents, we deduce that :
$\alpha + \beta + \gamma + \delta = 3$ and $\delta = 1 , \beta = \delta , \beta = \gamma, \gamma =1$.
Thus $\alpha = 0, \beta = \gamma = \delta = 1$, and $P(x) = \lambda x(x+2)(x+6)$.
From $P(1)=1$, it follows that $\lambda = \frac 1 {21}$, and then $P(-1) = -\frac 5 {21}.$

Pierre.


Hi, I was trying to put something like this:
[url=http://www.artofproblemsolving.com/Forum/weblog.php?w=1737] $ \text{\LaTeX} M\sigma\eta\kappa\varepsilon\gamma$ [/url]
In my sig (I would turn on BBCode when I actually put it in), but this does not react the way I expect it to. Instead it drags me to the site with all that LaTeX magic on it. How do I get the BBCode to win?

The $ \text{\LaTeX}$ in your post renders as an image. It is automatically linked to the code (where it says what you typed). I don't think you can make the image a link to somewhere else.

You can do something like this:
[url=http://www.artofproblemsolving.com/Forum/]This is $ \text{\LaTeX}$[/url]
and the code used is this:
[code][url=http://www.artofproblemsolving.com/Forum/]This is $\text{\LaTeX}$[/url][/code]