superior algebra
superior algebra unsolved

Suppose $R$ is a commutative ring with $1$ which there exist $a\in R$ which $a^3-a-1=0$. Prove that if $J$ is an ideal of $R$ which $R/J$ has at most $4$ items then $R=J$.

I have no idea if this is ok....but....
So, consider this ring, $R/J$. It contains $0$. Also, it has $1$, because otherwise it would be the null ring, hence $R=J$. Now, the relation we have also means that $a$ can't be a $0$ in this ring, since if it were, it would follow again that $1=0$ in the ring, hence $R=J$. So we have 3 elements up to now: $0,1,a$. One thing: $a \neq_{R/J} 1$, because again, plugging in, $1=_{R/J} 0$. So we have 3 distinct elements $0,1,a$. Now, $1+a$ and $a^2$ mut be in the ring. $1+a$ is obviously different from the 3 elements we had (same contradictions). So we have that $a^2$ must be one of those elements. But, if $a^2=0$ then $a^3=0$ and thus $a=-1$ which brings to the contradiction $1=0$. If $a^2=1$, similiarly $a^3-a=0$ and $1=0$, and if $a^2=a$ then $a^3=a^2=a$ hence $a^3-a=0$ hence $1=0$. So we got our contradiction....
Is it ok?

Your idea is correct, but you have to work with cosets or rather calculate modulo I:

Assume $R \neq I$. This means $0 \not\equiv 1$, and hence $0 \not\equiv -1$. The elements $0,1,a,a+1$ are distinct modulo $I$, which can be seen by the following contradictions:

$0 \equiv 1$
$0 \equiv a \Rightarrow 0 \equiv a^3 - a - 1 \equiv -1$
$0 \equiv a+1 \Rightarrow a \equiv -1 \Rightarrow 0 \equiv a^3 - a - 1 \equiv -1$
$1 \equiv a \Rightarrow 0 \equiv a^3 - a - 1 \equiv -1$
$1 \equiv a+1 \Rightarrow a \equiv 0$
$a \equiv a + 1 \Rightarrow 0 \equiv 1$

Because of the assumption $|R/I| \leq 4$, we have one of the following cases:

$a^2 \equiv 0 \Rightarrow a^3 \equiv 0 \Rightarrow a \equiv a^3 - 1 \equiv -1 \Rightarrow 0 \equiv a^2 \equiv 1$
$a^2 \equiv 1 \Rightarrow 0 \equiv a^2 a - a - 1 \equiv -1$
$a^2 \equiv a \Rightarrow 0 \equiv a a^2 - a - 1 \equiv a^2 - a - 1 \equiv-1$
$a^2 \equiv a + 1 \Rightarrow 0 \equiv a(a^2-1) -1 \equiv a^2-1 \equiv a$

in each case a contradiction.

Well, that's what I essentially did, I mean I said we work in the ring $R/J$ .....I just didn't write up all the implications....


Dyo mathimatikoi,o Giannis kai o Dimitris,synantiountai meta apo polla xronia.O Giannis thimotan oti o Dimitris exei treis gious alla den mporei na thimithei tis ilikies stis opoies mporei na vriskontai tora.Opote o Dimitris ton voithaei kai tou leei:

                                TO GINOMENO TON ILIKION TOUS EINAI $ 36$

                                 TO ATHROISMA TON ILIKION TOUS ISOUTAI ME TON ARITHMO TON PARATHIRON TOU APENANTI SPITIOU

           KAI                  O MEGALYTEROS MOU GIOS EINAI KOKKINOMALIS.

 Einai arketa ta dedomena etsi oste na vrei tis ilikies ton paidion o Giannis?

Όμορφη άσκηση

Αφού το γινόμενο των ηλικιών είναι 36, εύκολα βρίσκουμε ότι (με την προϋπόθεση πως οι ηλικίες είναι ακέραιοι αριθμοί) αυτές θα είναι:
1 1 36
1 2 18
1 3 12
1 4 9
1 6 6
2 2 9
2 3 6
3 3 4 ...(sorry αν μου ξέφυγε κάτι  :oops: )
Οπότε, τα αθροίσματα των ηλικιών των παιδιών του μπορεί να είναι 38, 21, 16, 14,13, 13,11, 10 αντίστοιχα. Αν το απέναντι σπίτι έχει 38, 21,16, 14,11 ή 10 παράθυρα, τότε μπορούμε πολύ εύκολα να βρούμε τις ηλικίες των παιδιών. Αν το άθροισμα αυτό είναι 13, τότε τα παιδιά θα έχουν ηλικίες 1,6,6 ή 2,2,9. Αλλά στην τριάδα 1,6,6, δεν υπάρχει ο μεγαλύτερος γιος, για τον οποίο ξέρουμε πως έχει κόκκινα μαλλιά. Έτσι, οι ηλικίες των παιδιών του Δημήτρη θα είναι 2,2,9.
Σε όλες τις παραπάνω περιπτώσεις οι ηλικίες μπορούν να βρεθούν.

op op....poios einai edo.....oraios pano....!

[quote="silouan"]Καλωσορίζω το νέο μέλος στο φόρουμ ! Μικρέ it's time for problem solving !  :wink:[/quote]

It's showtime!


Find the 357th odd number. Then find the sum of all the first 357 odd numbers. Then find the sum of the sum of the first 357 odd numbers and the sum of the first 468 even numbers. Then find the product of the first 357 odd numbers. Then find the odd factors of 357. Enjoy!!! :D

WOW.
[hide]$357^{\mathrm{th}}$ $\textrm{odd number}$ $=713$

$\textrm{sum}= 357^{2}$

$\textrm{sum of sum}= 357^{2}+2(234)(235)$

$\textrm{product}= \textrm{I have no idea}$[/hide]

the product is obvious

1 x 3 x 5... x 357 :lol:

also, the odd factors of 357 are 1,3,7,17,21,51,119,357

Now:

Find the sum of the cubes up to 468.
Find the sum of the squares up to 468.

Simplify $\sqrt{468}$

If I type questions at the speed of 468 words/minute, and you type the answer at 234 words/minute:

a.)Find the average of the speeds of typing, if possible.

b.)If we both type for two hours, then what will be the lapse of time between I eat dinner and the time when you land on Mars, if right after we finish typing we are able to do the following activity?

What is the 468th cube?

Find the volume of a cube with a diagonal of length 468.

Without using a calculator, find the square of 468, and add the square of 357 to it.

Have fun!!!! :D


geometry
rectangle
trapezoid
perpendicular bisector

A rectangle intersects a circle as shown: $AB=4$, $BC=5$, and $DE=3$. Then $EF$ equals:

[asy]size(200);
defaultpen(linewidth(0.7)+fontsize(10));
pair D=origin, E=(3,0), F=(10,0), G=(12,0), H=(12,1), A=(0,1), B=(4,1), C=(9,1), O=circumcenter(B,C,F);
draw(D--G--H--A--cycle);
draw(Circle(O, abs(O-C)));
label("$A$", A, NW);
label("$B$", B, NW);
label("$C$", C, NE);
label("$D$", D, SW);
label("$E$", E, SE);
label("$F$", F, SW);

label("4", (2,0.85), N);
label("3", D--E, S);
label("5", (6.5,0.85), N);
[/asy]
$\mathbf{(A)}\; 6\qquad \mathbf{(B)}\; 7\qquad \mathbf{(C)}\; \frac{20}3\qquad \mathbf{(D)}\; 8\qquad \mathbf{(E)}\; 9$

There is no rectangle in the questiopn...hope the question isn't giving away any hint :)

Well, if you draw it out, that's obviously impossible, so I don't think it's giving anything away... but you probably shouldn't be giving hints, anyway, unless people don't know how to do it. Give them a little more time. ;)

Well, the only reason I put title "Rectangle and circle" was because the original problem had rectangle in it. But if I wanted to use original wording, I needed to draw a picture so I just changed the wording little bit so you don't need to have picture.

It doesn't matter on solving it though.

[hide]

First of all, the two chords $BC$ and $EF$ are parallel. By drawing a line between $B$ and $E$ and also between $C$ and $F$ it will form an isosceles trapezoid.

Now draw a perpendicular between $AB$ and $DE$ from point $E$ to where it intersects $AB$. Call this intersection point $G$. This will form a right triangle $BEG$. It will also form a rectangle outside the circle $AGED$.

$AG = DE = 3$ so $BG = 1$

Now draw a perpendicular between $BC$ and $EF$ and call the point where the perpendular intersects $BC$ extended, point $H$. This will form a right triangle on the opposite side of the inscribed isosceles trapezoid, which will be called triangle $CHF$.

We now have two equal triangles, $BEG$ and $CHF$, since $BE = CF$ and $GE = HF$.

Therefore, if $BG = 1$, then $CH = 1$.

$BG + BC + CH = EF$ so $EF = 1 + 5 + 1$ or $7$

[/hide]

[hide] Form centre of the circle draw the two perpendicular bisector of secant BC and EF forming a rectangle, Call the point that bisects GC K and the point that bisect EF L. AK = 4+5/2 = 6.5, since its a rectangle, EL = 6.5-3 =3.5.
Since EL bisects EF,  EF = 7
 [/hide]


Wait...
What [i]is[/i] the purpose of this topic???


1)How many 6 digit numbers can be made from 2 and 3 alone such that at least 1 digit is different.

2)In how many ways can 12 identical items be distributed to 7 people,each getting atleast 1.

2. Is this right?

Let each one get an object, there are 5 objects remaining and 7 people, each object can be given to any of the 7 people. Therefore, there are 7^5 (edited later) ways of doing it .
So,  7^5(edited later) is the answer.

2.  Since this is a combination with repitition we can use:

$x_1+x_2+x_3+x_4+x_5+x_6+x_7=5$

Since each person has to have at least 1, we can subtract 7 from 12 leaving us with 5 items to be distributed.

$\frac{11!}{5!(6!)}$

$462$

I'm not sure what you mean in question one, but I'm going to assume, that you just mean:  How many ways are there to permute the digits 2 and 3 among 6 containers, with each container having only one digit.

1.  $2^6$ since there are two choices for each slot.

[quote="shadysaysurspammed"]1)How many 6 digit numbers can be made from 2 and 3 alone such that at least 1 digit is different.[/quote]There can be 1, 2, 3, 4, or 5 twos, and the opposite number of threes. The permutations for each one yield $2\frac{6!}{5!} + 2\frac{6!}{4! \times 2!} + \frac{6!}{3! \times 3!} = 12+30+20=\boxed{62}$. Or, this can be done by inclusion-exclusion, where we include all combinations minus the combinations that include no 2's or no 3's, which is $2^6-2=\boxed{62}$

Please can anyone explain the mistake in my argument?

First, I think you meant 7^5, not 5^7.  For example, if there were just one item, there would be 7, not 5, different ways of giving it to the 7 different people.

Second, the question states the items are identical, and your argument assumes that they are distinct.

Thanks for  pointing out the flaw, :blush:

$1$ every 6 digit number of 2's and 3's contains atleast one different digit except for 222222 and 333333. So just count the total number of 6 digit numbers using only 2's and 3's, and then subtract 2.
$2^6-2=62$

$2$ Give exactly one item to each of the 7 people to satisfy the condition. Then, one at a time give the remaining 5 items to anyone in $7^5$ ways

I tried looking at $\#2$ with generating functions. Anybody get anything? And what's wrong with Petoria's answer to #2? I understand what he's getting at, but there must be some subtlety we're forgetting.

I think Petoria's answer to #2 is right.  Has anyone written something to suggest otherwise...?

[hide="Problem 1"]
There are $2^6$ possible numbers without restrictions, but because 222222 and 333333 don't satisfy the condition, there are $2^6-2=\boxed{62}$ possible numbers.
[/hide]
[hide="Problem 2"]
First, give each person the 1 item he is entitled to. There are $12-7=5$ remaining items left to be distributed. 5 items can be divided among 7 in $ 11 \choose 6=\boxed{462}$ ways (it is a correspondence).
[/hide]

Kokosai,
In my proof where have I assumed its distinct? Anyway I think drunner's proof is right for 2., the answer is 7^5.

Hmm, maybe you're not assuming they're distinct afterall.  But $7^5$ makes it matter when a person receives an item.  For example, let's say individual [i]A[/i] receives an item first and [i]B[/i] receives an item second; Petoria's approach assumes that this is the same as [i]B[/i] receiving an item first and [i]A[/i] receiving an item second, but your approach takes them as two different events.  For example, work out the problem for 2 people and 2 remaining items: are there 3 or 4 ways to distribute the items?  It depends on how you read the question, I guess.  If it matters when they receive it, you're right, but if it only matters how many items each individual has at the end, Petoria's right.


What if there was a triangle that had a degree of 179.999, and you were standing right at the vertex of that triangle, and then you just stood still and dabbed a bit of paint on all three sides of the triangle. There isn't a path shorter than that, because you don't have to move at all.... :rotfl:

Seriously, I think that the shortest path would be from the vertex of the obtuse angle to the altitude from the obtuse angle to the other side, or something like that.

Well, you have to trace out a nondegenerate triangle inside the obtuse triangle that has the smallest perimeter.

Sorry about that; I was really sleep-deprived when I wrote this problem. And it's still probably wrong...

[quote="mathnerd314"]I think I remember reading a proof in a book that there is no shortest path in an obtuse triangle that touches all three sides. That is, if you were imprisoned in an obtuse triangle and had to walk around the room, putting a dab of paint on all three walls, you would always be able to find a shorter path. Unfortunately, I can't find the book (which really stinks because it was a library book :roll:)

Prove or disprove.[/quote]
You have something always decreasing but bounded below. So wouldn't there be a shortest path?

[quote="cincodemayo5590"]Wouldn't the shortest path just be the altitude from the vertex with the obtuse angle?

or am i misunderstanding something...[/quote]
That sounds good, can we prove it?

[quote="cincodemayo5590"]Wouldn't the shortest path just be the altitude from the vertex with the obtuse angle?[/quote]
I suppose the book doesn't accept the degenerate case where touching a vertex means touching two sides. If this would be the minimum and the book doesn't accept it, then of course you can never get the shortest path, only approximate it more and more.
[quote="PenguinIntegral"]You have something always decreasing but bounded below. So wouldn't there be a shortest path?[/quote]
If a function is decreasing but bounded below, then it doesn't necessarily have a reachable minimum, but only a "greatest lower bound".


Here is a question that would be helpful for Mathcounts particularly if you need to work on operations.

The operation @ is defined by $a@b=\frac{3a-2b}{2ab}$ for $a,b \neq 0$.  If $x@y=\frac{1}{4}$ and $y@x=-1$, then $x@x$ is equal to

(A)$-1$
(B)$\frac{-1}{2}$
(C)$\frac{-1}{4}$
(D)$\frac{1}{2}$
(E)$1$

[quote="mathgeniuse^ln(x)"]Could you please type a solution?[/quote]

[hide]I simplied each equation and put it in terms of y

Then I set them equal to each other.
$\frac{2x}{2x+3}=\frac{6x}{x+4}$

I solve to get x=0,-1 but $x\le0$, so $x=-1$

Then I sub it into x@x to get what I got.[/hide]

Hey, mathgeniuse^ln(x), why do you have multiple choice for a MATHCOUNTS question?

[quote="SplashD"]Hey, mathgeniuse^ln(x), why do you have multiple choice for a MATHCOUNTS question?[/quote] It's probably a problem from a competition with multiple choice but it's at the MathCounts level.

[quote="mathgeniuse^ln(x)"][quote="SplashD"]
Hey, mathgeniuse^ln(x), why do you have multiple choice for a MATHCOUNTS question?[/quote]

Um, the source says that it is MMPC which is not MathCounts, right?[/quote]

What's MMPC?..and just so you know, I don't notice or ead those things.

[quote="mathgeniuse^ln(x)"]MMPC is Michigan Mathematics Prize Competition.  It is only allowed to be taken by people from Michigan.[/quote]

What do you win and did you win?

[quote="mathgeniuse^ln(x)"]It is a competition for high schoolers, so hopefully I didn't win. :lol: I got 12th Place out of 14000+ who took it. :lol:

There are 2 parts Part I and Part II.

If you get 50th-11th Place you get 500 dollars.
10th-6th Place you get 700 dollars.
5th-4th Place you get 900 dollars.
3rd Place gets 1200 dollars
2nd Place gets 1900 dollars
1st Place gets 2600 dollars.

I believe this is correct and I think I answered your question.[/quote]

I think I should move to Michigan, jk. :P
Wow, do you have to pay to compete?

It is only 3 dollars for part I, and if you qualify for part II, you don't have to pay any money.  What makes you say that you can win money easily jk.

[quote="bpms"]Is there any place to see old mmpc's. Might as well see how I would do :).[/quote]

Yeah, I was wondering that too.

On there website.  They don't have last years for some reason.  If you want to know your hardest year, it would be 2005, but it is not posted, and the easiest would be one of the earlier years.

But, to be honest, the questions are really simple if you just know what to do.  I average a 37/40 on the Part I and a 30/50 on Part II.  That is pretty good.


Given that all roots of $ x^6\minus{}6x^5\plus{}ax^4\plus{}bx^3\plus{}cx^2\plus{}dx\plus{}1\equal{}0$ are positive , find a,b,c,d.

[quote="Ognjen09"]Given that all roots of $ x^6 \minus{} 6x^5 \plus{} ax^4 \plus{} bx^3 \plus{} cx^2 \plus{} dx \plus{} 1 \equal{} 0$ are positive , find a,b,c,d.[/quote]

Let the positive roots be $ a_1,a_2,...,a_6$, then

product of root $ \equal{}1 \Rightarrow \sqrt[6]{a_1a_2...a_6}\equal{}1$

sum of root $ \equal{}6 \Rightarrow 1\equal{}\frac{a_1\plus{}a_2\plus{}...\plus{}a_6}{6} \ge \sqrt[6]{a_1a_2...a_6}\equal{}1$

Equality holds $ \Leftrightarrow a_1\equal{}a_2\equal{}...\equal{}a_6\equal{}1$

So $ a\equal{}15, b\equal{}\minus{}20, c\equal{}15, d\equal{}\minus{}6$


Let a,b,c be positive real numbers such that abc=1
Prove that:
${ \frac{1}{\sqrt{2a+2ab+1}}+\frac{1}{\sqrt{2b+2bc+1}}+\frac{1}{\sqrt{2c+2ca+1}}\geq \ 1}$

[quote="small"]Let a,b,c be positive real numbers such that abc=1
Prove that:
${ \frac {1}{\sqrt {2a + 2ab + 1}} + \frac {1}{\sqrt {2b + 2bc + 1}} + \frac {1}{\sqrt {2c + 2ca + 1}}\geq \ 1}$[/quote]
$ \sum_{cyc}\frac {1}{\sqrt {2a + 2ab + 1}}\geq\sum_{cyc}\frac {1}{a + ab + 1}=1.$

Dear Mr Namdung,
We have $ (x\plus{}1)^2\minus{}(2x\plus{}1)\equal{}x^2>0$, which shows that
$ \frac{1}{\sqrt{2x\plus{}1}} \ge \frac{1}{x\plus{}1}.$ :)

[quote="small"]Let a,b,c be positive real numbers such that abc=1
Prove that:
${ \frac {1}{\sqrt {2a + 2ab + 1}} + \frac {1}{\sqrt {2b + 2bc + 1}} + \frac {1}{\sqrt {2c + 2ca + 1}}\geq \ 1}$[/quote]
Another solution is to use the substitution $ a=\frac{x}{y},b=\frac{z}{x},c=\frac{y}{z}$, then our inequality becomes
$ \sqrt{\frac{x}{x+2y+2z}}+\sqrt{\frac{y}{y+2z+2x}}+\sqrt{\frac{z}{z+2x+2y}} \ge 1$,
which is trivial as
$ \sqrt{\frac{x}{x+2y+2z}} =\frac{x}{\sqrt{x(x+2y+2z)}} \ge \frac{2x}{x+(x+2y+2z)} =\frac{x}{x+y+z},$
:)

[quote="can_hang2007"]Dear Mr Namdung,
We have $ (x \plus{} 1)^2 \minus{} (2x \plus{} 1) \equal{} x^2 > 0$, which shows that
$ \frac {1}{\sqrt {2x \plus{} 1}} \ge \frac {1}{x \plus{} 1}.$ :)[/quote]

Oh, I see. Thank you can_hang2007.

So, I guess that we dont have equality?


geometry
circumcircle
inradius
3D geometry
prism
Pythagorean Theorem

[quote="t0rajir0u"]A bicentric quadrilateral is a quadrilateral that can be both circumscribed around and inscribed in a circle.  Show that the area of a bicentric quadrilateral with sides $a, b, c, d$ is given by

$K = \sqrt{abcd}$[/quote]
[hide]
It is well known that the sum of opposite sides in a quadrilateral that you can inscribe in a circle is equal.
So we have a+c=b+d.
Using Bhramagupta's, we get $\frac{\sqrt{(a+b+c-d)(a+b-c+d)(a-b+c+d)(-a+b+c+d)}}{4}$
$\frac{\sqrt{2b \cdot 2a \cdot 2c \cdot 2d}}{4}$
$\sqrt{abcd}$[/hide]

[hide]For a quadrilateral that has an inscribed circle, sum of opposite sides are equal (from Power of a Point).
So $a+c=b+d$.
Since the quadrilateral is cyclic, the area given by Bragmahgupta is $K=\sqrt{(a+b+c-d)(a+b-c+d)(a-b+c+d)(-a+b+c+d)}/4=\sqrt{2b\cdot2a\cdot2d\cdot2c}/4=\sqrt{abcd}$.[/hide]

Two questions:
1) Can anyone do it without resorting to Brahmagupta?
2) Suppose I take $d = 0$, so our quadrilateral is actually a triangle.  Well, triangles have both circumscribed and inscribed circles, but this formula gives us that the area is 0.  Why?

[hide="b"] The inscribed circle is nowhere near the spot where the side with length $d$ "used to be."  In other words, as $d$ becomes very small the condition that the inscribed circle needs to touch the side with length $d$ forces the quadrilateral's area to approach zero. [/hide]

a quadrilateral has an incircle iff a+c=b+d...but d=0 gives us a+c=b...i.e. a degenerate triangle

so the formula is correct since a degenerate triangle has no area...

just obviously the incircle of a triangle is nowhere near to tangent to the side that is approaching 0 (the incircle would have to pass through a vertex... AI>r is relatively easy to show [hint: pythagorean theorem])

the circumcircles are the same though

Thats quite a good question, and the answer is no.

The reason is that if you have two circles, $ r$ and $ R$ with $ r<R$, and the distance between the centers of these circles is $ \sqrt{(R)(R\minus{}2r)}$ then for any point on the large circle, you have a triangle that has an incircle as the smaller circle and the larger circle as it's circumcircle.

I think it is a remarkable fact (the proof isn't that bad).

[quote="gabriella"]Is it possible to express the area of a bicentric quadrilateral only by its circumradius and its inradius?[/quote]
See Poncelet's Porism at [url]http://members.ozemail.com.au/~llan/bimovie.html[/url] for a visual demonstration of different bicentric quads all with respect to the same two circles.


trigonometry
geometry
circumcircle
geometric transformation
reflection

Triangle $ ABC$ is such that $ AB > BC$ and $ \hat{BAC} \equal{} 45^{\circ},\hat{ABC} \equal{} 30^{\circ}$.Point $ M$ is located inside of the triangles that $ \hat{MAB} \equal{} \hat{MBA} \equal{} 15^{\circ}$.Find $ \hat{BMC}$

[hide="Solution"]
Let the circle with center $ P$ and radius $ AP \equal{} PC$ meet $ AB$ at $ Q$.  Let $ AP$ meet $ QC$ at $ R$ respectively.  Now, $ \angle PAC \equal{} 15 \equal{} \angle PCA$, so $ \angle PAB \equal{} 30$.  Thus, $ \angle PQA \equal{} \angle PAQ \equal{} 30$.  Then, $ \angle APQ \equal{} 120$ and $ \angle APC \equal{} 150$, so $ \angle QPC \equal{} 90$.  Thus, $ \angle PCQ \equal{} 45 \equal{} \angle PQC$.  Since $ \angle PCB \equal{} 30 \minus{} 15 \equal{} 15$, we have that $ \angle BCR \equal{} 30 \equal{} \angle BAR$, so $ ACRB$ is cyclic.  This gives that $ \angle PRB \equal{} \angle ACB \equal{} 30 \equal{} \angle BQP$, so $ PBQR$ is cyclic.  Then, $ \angle PBR \equal{} \angle PQC \equal{} 45$.  Then, $ \angle PBC \equal{} 45\minus{}15\equal{}30$, so $ \angle BPC \equal{} 135$. [/hide]

[color=blue][i]This nice solution needs a triangle $ ABC$ with $ \angle A \equal{} 45 , \angle C \equal{} 30 , \angle PAC \equal{} \angle PCA \equal{} 15$. This is a little different from the initial wording, but it is only a little typo.So, delight in  the solution :[/i][/color]

[quote="The QuattoMaster 6000"][hide="Solution"]
Let the circle with center $ P$ and radius $ AP \equal{} PC$ meet $ AB$ at $ Q$.  Let $ AP$ meet $ QC$ at $ R$ respectively.  Now, $ \angle PAC \equal{} 15 \equal{} \angle PCA$, so $ \angle PAB \equal{} 30$.  Thus, $ \angle PQA \equal{} \angle PAQ \equal{} 30$.  Then, $ \angle APQ \equal{} 120$ and $ \angle APC \equal{} 150$, so $ \angle QPC \equal{} 90$.  Thus, $ \angle PCQ \equal{} 45 \equal{} \angle PQC$.  Since $ \angle PCB \equal{} 30 \minus{} 15 \equal{} 15$, we have that $ \angle BCR \equal{} 30 \equal{} \angle BAR$, so $ ACRB$ is cyclic.  This gives that $ \angle PRB \equal{} \angle ACB \equal{} 30 \equal{} \angle BQP$, so $ PBQR$ is cyclic.  Then, $ \angle PBR \equal{} \angle PQC \equal{} 45$.  Then, $ \angle PBC \equal{} 45 \minus{} 15 \equal{} 30$, so $ \angle BPC \equal{} 135$. [/hide][/quote]

[img]http://img186.imageshack.us/img186/5717/immaginelr9.jpg[/img]

$ \widehat{ABM} \equal{} \widehat{BAM} \equal{} \widehat{MBC} \equal{} 15$
$ \widehat{CAM} \equal{} 30$

Call $ \widehat{BCM} \equal{} x$ and $ \widehat{ACM} \equal{} y$

For Ceva we have

$ \sin15 \cdot \sin15 \cdot \sin y \equal{} \sin15 \cdot \sin30 \cdot \sin x$

$ \sin15 \cdot \sin y \equal{} \sin30 \cdot \sin x$

and $ x \plus{} y \equal{} 180 \minus{} 45 \minus{} 30 \equal{} 105$

So it's easy to see that

$ 2 \sin15 \cdot \sin y \equal{} \sin x$
$ 2 \sin15 \cdot \sin 75 \equal{} 2 \sin15 \cdot \cos 15 \equal{} \sin 30$

So $ x \equal{} 30 \Rightarrow \widehat{BMC} \equal{} 180 \minus{} 30 \minus{} 15 \equal{} 135$

Take $P$ the circumcenter of $\triangle ABC$, see that $\triangle APC$ is equilateral, while $M$ is the reflection of $P$ in $AB$, i.e. $AM=AC, \angle MAC=30^\circ$. It implies $\angle AMC=75^\circ$; with $\angle AMB=150^\circ$ we get $\angle BMC=360^\circ-\angle AMB-\angle AMC=135^\circ$.

Best regards,
sunken rock


I was told by my chemistry teacher that the dissolving of NaCl in water is not considered a chemical reaction.

I am very confused because isn't a chemical reaction any change that involves the formation of new substances? When dissolving the NaCl into H2O, Na+ and Cl- are formed. Last time I checked, Na+ and Cl- are not the same thing as NaCl.

Unless the bonding of ions due to electroweak force is considered a physical attraction rather than a chemical change. If this is true I think I understand why dissolving would not be considered a chemical reaction as it would just be rearranging ions physically.

But that would mean that the formation of ionic bonds is not a chemical reaction, being just the rearrangement of ions, thus being contradictory.

I am confused. Someone please explain.

Ionic bonds are when you have ions held together by electrostatic forces.

So, NaCl is really just Na+ and Cl-.  The Na+ and Cl- ions are just held together by the electrostatic forces, which *is* a physical attraction.

Say you want to make NaCl by putting sodium chunks in a container of chlorine gas (cool reaction).  This would be a chemical reaction because the sodium and chlorine originally aren't ionized, but they become ionized after the reaction.

That was kinda awkward.  I hope it helps.

[quote="breez"]I was told by my chemistry teacher that the dissolving of NaCl in water is not considered a chemical reaction.

I am very confused because isn't a chemical reaction any change that involves the formation of new substances? When dissolving the NaCl into H2O, Na+ and Cl- are formed. Last time I checked, Na+ and Cl- are not the same thing as NaCl.
[/quote]

Well, your chemistry teacher and my chemistry textbook disagree.  The formation of aqueous $NaCl$ has an associated enthalpy of reaction like any other reaction - it changes the potential energy of the system, so it should rightfully be considered a reaction.

(Technically, the products aren't just ions - rather, they're ions surrounded by the respective polar ends of $H_2 O$, decreasing the potential energy of the system, which is why the reaction is exothermic.)

To support my viewpoint, http://en.wikipedia.org/wiki/Chemical_change states that the breaking or formation of chemical bonds is a valid chemical reaction, and when $NaCl$ is dissolved in $H_2 O$ the ionic bond is broken and new ion-dipole 'bonds' are formed.

I don't think that your chemistry teacher is right either. The dissolving of salt is a simple decomposition reaction (at least it looks a lot like one), and, like you said, new substances are formed (and like t0rajir0u said, heat is released).

Classically, a chemical reaction would be considered one in which the ions were formed, not the seperation of the ions.

I think your teacher is correct.

How would you answer the same question to the process of pouring a solution of NaCl into a solution of KOH?

The ions are changing, the enthalpy is increasing, there is a heat of mixing.  However, most would not consider it a chemical reaction.  If this is than almost every combination would be considered a chemcial reaction, depriving the noun of its value in differentiating some things from others.

In the final analysis, it is a definition, not a reasoning question.


[quote="Ligouras"]Let $ a,b,c$ positive reals such that $ abc \equal{} 1$. Prove that

$ \frac {ab \plus{} c}{a \plus{} b} \plus{} \frac {bc \plus{} a}{b \plus{} c} \plus{} \frac {ca \plus{} b}{c \plus{} a}\geq2 \plus{} \frac {3}{ab \plus{} bc \plus{} ca} \plus{} \frac {ab \plus{} bc \plus{} ca \minus{} 1}{2(a \plus{} b \plus{} c)}$[/quote]
$ a\equal{}b\equal{}c\equal{}1$ gives $ LHS \equal{} 3$ and $ RHS \equal{} 3\plus{}\frac{1}{3}$ if I'm not mistaken

[quote="Mathias_DK"][quote="Ligouras"]Let $ a,b,c$ positive reals such that $ abc \equal{} 1$. Prove that

$ \frac {ab \plus{} c}{a \plus{} b} \plus{} \frac {bc \plus{} a}{b \plus{} c} \plus{} \frac {ca \plus{} b}{c \plus{} a}\geq2 \plus{} \frac {3}{ab \plus{} bc \plus{} ca} \plus{} \frac {ab \plus{} bc \plus{} ca \minus{} 1}{2(a \plus{} b \plus{} c)}$[/quote]
$ a \equal{} b \equal{} c \equal{} 1$ gives $ LHS \equal{} 3$ and $ RHS \equal{} 3 \plus{} \frac {1}{3}$ if I'm not mistaken[/quote]

Sorry my Friend:

$ \frac {ab \plus{} c}{a \plus{} b} \plus{} \frac {bc \plus{} a}{b \plus{} c} \plus{} \frac {ca \plus{} b}{c \plus{} a}\geq2 \plus{} \frac {3}{ab \plus{} bc \plus{} ca} \plus{} \frac {ab \plus{} bc \plus{} ca \minus{} 3}{2(a \plus{} b \plus{} c)}$

[quote="Ligouras"]Let $ a,b,c$ positive reals such that $ abc \equal{} 1$. Prove that

$ \frac {ab \plus{} c}{a \plus{} b} \plus{} \frac {bc \plus{} a}{b \plus{} c} \plus{} \frac {ca \plus{} b}{c \plus{} a}\geq2 \plus{} \frac {3}{ab \plus{} bc \plus{} ca} \plus{} \frac {ab \plus{} bc \plus{} ca \minus{} 3}{2(a \plus{} b \plus{} c)}$[/quote]

only  to prove

$ abc \equal{} 1$

$ \frac {(\sum{(ab \plus{} c))^2}}{\sum{(ab \plus{} c)(a \plus{} b)}} \geq2 \plus{} \frac {3}{ab \plus{} bc \plus{} ca} \plus{} \frac {ab \plus{} bc \plus{} ca \minus{} 3}{2(a \plus{} b \plus{} c)}$


The Wonderlich family has 3 children. At least one of them is female. If having a female or male is equally likely, what is the probability that at least one of the 3 kids is a male?

Answer is [hide]6/7[/hide] but I really don't see how you get that..

[hide]There are 8 combonations BBB,BBG,BGG,GGG,GBB,BGB,GBG,GGB and the 1st one is impossible because it is known that there is one male so 6 out of 7 of them is the probability that there is one boy[/hide]

Yeah you have
[hide]$7/8$ but you subtract 1 from top and bottom leaving $6/7$[/hide]


calculus
integration
trigonometry
logarithms
function
calculus computations

More generally we can say the following:

$\int_{\alpha}^{\frac{\pi}{2}-\alpha}(2\cos^{2}x)\phi(\sin 2x)dx = \int_{2\alpha}^{\frac{\pi}{2}}\phi(\sin x)dx$ for a "nice" function $\phi$. Set $\alpha = 0$ and $\phi(x) = \ln x$ to obtain the given problem.


integration
limit
calculus
algebra unsolved
algebra

how do I calculate $F_n=\frac1{1^2}+\frac1{2^2}+...+\frac1{n^2}$
I tryed in many ways but I didn't succeded...give it a try.thank you!!!

afterwords I need to find the integer part of $F_{2005}$

Is it from a contest or you just wanted to calculate it yourself ??
what I know is that we can calculate $\lim_{x \rightarrow \infty} F_{n}$ we find that it is equal to $\frac{\pi^2}{6}$

I am preparing for an exam.
... and how do I get to have that limit ?

You don't need that limit to calculate an integer part. Just show that $1<F_n<2$ ;)

Yes by showing $1<F_{n}<2$ we can calculate the integer part.
We can get the limit by using the method of integral to calcuate the limit of sequences

The integralmethod is not useful. here is how you get the sum:
Consider the series:
x(1-xpi/1)(1+xpi/1)(1-xpi/2)(1+xpi/2).... =sin(pi^2 x).
Equating coefficients of x^3 we get: sum (pi/n)^2 =pi^4/6
so we get the result.
Of course, this is not my solution. I read it years ago and its by L.Euler.
-a

It would look niver if you wouldn't add an $\pi$ everewhere where an $x$ occurs  ;)

But that only deduces it to proofing $sin(\pi x) = \pi x \prod_{i=1}^\infty \left( 1-\frac{x^2}{i^2} \right)$, which is not that easy when trying it on the strongly correct way, and to show that you can compare those coefficients.


Dear all;

Can anyone offer help in the proving of the below inequality?

Let $ x,y \epsilon \mathbb{R}$ such that

$ \frac{x^2\plus{}y^2}{2} \geq (\frac{x\plus{}y}{2})^2$

There are multiple ways to do so:

- write the inequality as sum of squares $ \ge0$
- use the Inequality of Cauchy-Schwarz, which works for all reals
- use trivial inequalities such as $ |a|\plus{}|b|\ge|a\plus{}b|$ to get the variables positiv, and then use other known inequalities
- (Jensen Inequality)

With these Hints you should also be able to proof:

$ x_i\in\mathbb{R}$ for $ 1\le i\le n$

$ \frac{\sum{x^2_i}}{n}\ge \left(\frac{\sum{x_i}}{n}\right)^2$

Dear Rafikafi;

Can you show me how to prove if I were to write the inequality as sum of squares  $ x \geq 0$

Thanks.

Since you want it to be "Sum of some squares" $ \ge 0$ take everything to the left side and to make things simpler multiply with 4.
Just try i think you can hardly fail.

[hide="Try before looking ;-)"]
$ (x \minus{} y)^2\ge 0$[/hide]

Are you saying I should first

$ \frac{x^{2}\plus{}y^{2}}{2}\minus{}(\frac{x\plus{}y}{2})^{2} \geq 0$

then if multiply by 4, I will obtain:

$ 2x^2\plus{}2y^2\minus{}(x\plus{}y)^2 \geq 0$

Further expansion will result:

$ x^2\plus{}3y^2\minus{}2xy \geq 0$

If I am right above, how should I proceed?

[quote="Symbian"]
Further expansion will result:

$ x^2 \plus{} 3y^2 \minus{} 2xy \geq 0$

If I am right above, how should I proceed?[/quote]

check your subtraction  :lol:

Oopps.... Thanks for pointing out my error my friend..... Got messed up....

The final equation should reads:

$ x^2 \plus{} y^2 \minus{} 2xy\geq0$

now the square part is missing...

maybe i give an other example:

$ a^2\plus{}\frac{1}{a^2}\ge 2 \Leftrightarrow a^2\minus{}2\plus{}\frac{1}{a^2}\equal{}\left(a\minus{}\frac{1}{a}\right)^2\ge0$

I see.

If the square are maintain the equation will be

$ 2[x^2\plus{}y^2]\minus{}[x\plus{}y]^2 \geq0$

but $ x^2\plus{}y^2 \neq (x\plus{}y)^2$

Can you offer directions in solving this problem?

It's hard to give a direction without giving the solution. maybe this helps:

$ x^{2}\minus{}2xy\plus{}y^{2}\geq0$

Isn't this equation the same I post ealier? Just a shift in location?

My equation in previous post: $ x^{2}\plus{}y^{2}\minus{}2xy\geq0$

Your equation: $ x^{2}\minus{}2xy\plus{}y^{2}\geq0$

$ x^2 \plus{} y^2 \minus{} 2xy \equal{} (x \minus{} y)^2\geq0$  :lol:

I know, I thought you would see it that way.

The only thing you can do if you don't see it is looking at the solution I gave before. If you do so then you should solve this one:

$ \frac{x^2\plus{}y^2\plus{}z^2}{3}\ge \left(\frac{x\plus{}y\plus{}z}{3}\right)^2$

$ \frac {1}{2} \cdot ((x \minus{} y)^2 \plus{} (y \minus{} z)^2 \plus{} (z \minus{} x)^2) \equal{} x^2 \plus{} y^2 \plus{} z^2 \minus{} xy \minus{} yz \minus{} zx$

[quote="Symbian"]Ah I think I get what you mean now Rafikafi. When Luba1123 point it out, I can't believe I am such a simpleton for you have already shown me.  :blush:

As for your example; can it be solve since there is now 3 variable x, y, z? Previously if there are 2 variable,  I can easily apply quadratic formula.

Thank you Rafikafi and Luba1123 for deepening my knowledge.[/quote]

A lot of Inequalities with as much variables you want can be written in sum of squares. For example

[quote="geniusbliss"]$ \frac {1}{2}\cdot ((x \minus{} y)^{2} \plus{} (y \minus{} z)^{2} \plus{} (z \minus{} x)^{2}) \equal{} x^{2} \plus{} y^{2} \plus{} z^{2} \minus{} xy \minus{} yz \minus{} zx$[/quote]

does prove $ \frac {x^{2} \plus{} y^{2} \plus{} z^{2}}{3}\ge\left(\frac {x \plus{} y \plus{} z}{3}\right)^{2}$

the trick is to look at your negative terms and try to get them in a square. For example the term $ \minus{} 2xy$ could be in $ (x \minus{} y)^2$ or $ \left(ax \minus{} \frac {y}{a}\right)^2$ or $ (x \minus{} y \plus{} z)^2$ and also equality cases give hints for squares.
then all you have to do is start somewhere and hopefully come to an end.

Of course there are more complicated ones where you have to add and substract terms such that you can make squares:

$ 2a^4 \plus{} b^4 \plus{} c^4 \minus{} 4a^2bc \equal{} a^4 \minus{} 2a^2b^2 \plus{} b^4 \plus{} a^4 \minus{} 2a^2c^2 \plus{} c^2 \plus{} 2a^2b^2 \minus{} 4a^2bc \plus{} 2a^2c^2$

$ \equal{} (a^2 \minus{} b^2)^2 \plus{} (a^2 \minus{} c^2)^2 \plus{} 2a^2(b \minus{} c)^2\ge0$

and still there are examples even more complicated, but all you need is practice and you'll get step for step better.


Putnam
function
induction
calculus
integration
college contests

For positive integers $ m$ and $ n$, let $ f\left(m,n\right)$ denote the number of $ n$-tuples $ \left(x_1,x_2,\dots,x_n\right)$ of integers such that $ \left|x_1\right| \plus{} \left|x_2\right| \plus{} \cdots \plus{} \left|x_n\right|\le m$. Show that $ f\left(m,n\right) \equal{} f\left(n,m\right)$.

Trivial.
clearly the number of solutions to |x1| + ... + |xn| = k is:

sum (nCj) 2^(j) (k-1 C j-1)  where k>=j>=1.

Hence f(m,n) = [sum sum (nCj) 2^j (k-1 C j-1)] +1 where k>=j>=0 and m>=k>0
Now m-1Cj-1 + m-1Cj-1 +... + j-1C j-1 = mCj
Hence
f(m,n) = [sum nCj 2^j  mCj] +1 =f(n,m)

You might want to add a little more explanation of why your formula is true- what's the combinatorial interpretation that proves it? I doubt this "solution" would get much credit.

The proof I know for this problem works by showing that $f$ and its flip satisfy the same recursion, with obvious equality in the base cases where one argument is zero.

That's interesting.  I proved it by creating a bijection from the n-tuples to the m-tuples.  Then since the two functions count the number of such tuples, they must be equal.

Since my argument is the one jmerry mentions (satisfying the same recursion both ways), I'd really like to see Sly Si's bijection in detail.

Well of course im putnam paper i explained more, but here it goes:

the number of solutions to y1+...+yj= k where yi's >=1 is k-1Cj-1
the number of solutions to |x1|+...+|xn-j| = k where all xi's are nonzero is
clearly 2^n-j k-1Cn-j-1
So the number of solutions to |x1|+...||xn| = k with j zero's is the above formula * nCj
the rest is clear.
-Ali

Here's a fun solution: All you need to do is find a bijection between the sets counted by f(m,n) and the sets counted by f(n,m).

If the n-tuple in question is (0, 0, ..., 0), then we send it to (0, ..., 0). Otherwise, let the n-tuple be S_1,S_2,...,S_k,0,...,0, where S_i is a group of z_i - 1 zeroes that ends in the nonzero number 2. Add as many zeroes at the end as necessary to bring the number of elements to n.

For example, (0, 2, 4, 0, 0, 1, 0) has S_1 = 0, 2 (so z_1 = 2 and x_1 = 2), S_2 = 4 (so z_2 = 1 and x_2 = 4), and S_3 = 0, 0, 1 (so z_3 = 3, x_3 = 1).

You then set up a m-tuple so that the z's are swapped with their respective x's. The reason this works is that x_1 + ... + x_k <= m, and z_1 + ... + z_k <= n (the zeroes at the end allow for inequality). We then want T_1, T_2, ..., T_k, 0, ..., 0, where T_i is a string of x_i - 1 zeroes followed by the nonzero element z_i, with as many 0's at the end as necessary.

Examples:
(0, 2, 4, 0, 0, 1, 0) --> (0, 2, 0, 0, 0, 1, 3) from f(7, 7) to f(7, 7)
(0, 2, 4, 0, 0, 1, 0) --> (0, 2, 0, 0, 0, 1, 3, 0, 0, 0) from f(10, 7) to f(7, 10).
(1, 3, 0, 5, 1, 2) --> (1, 0, 0, 1, 0, 0, 0, 0, 2, 1, 0, 1) from f(12,6) to f(6, 12).

This is clearly a bijection (which happens to be its own inverse, since what we're doing is finding a way to swap x's and z's), therefore, the cardinality of the sets of n-tuples are equal.

You can show that

\[ f(m,n) = f(m-1,n) + f(m, n-1) + f(m-1,n-1). \]

We do this by correspondence.

Suppose that $|x_1| + \dots + |x_n| \le m$. Then if $x_n = 0$, then we can get a correspondence between these solutions and solutions to $|x_1| + \dots + |x_{n-1}| \le m$.

Otherwise, $x_n \ne 0$. Consider the shrinking map

\[ \rho(x) = x+1 \textrm{ if } x < 0, \quad x-1 \textrm{ if } x > 0. \]

Then we get a solution $x_1, \dots, x_{n-1}, x'_n$ for $|x_1| + \dots + |x'_n| \le m-1$ from $x'_n = \rho(x_n)$.

Note that this mapping is invertible iff $x'_n \ne 0$, and is a two-to-one correspondence if $x'_n = 0$. There are $f(m-1,n)$ solutions in general, and $f(m-1,n-1)$ of these must be counted in again, so we have shown

\[ f(m,n) = f(m,n-1) + f(m-1,n) + f(m-1,n-1). \]

Then it is a trivial matter to show $f(m,1) = f(1,m) = 2m+1$, and then by induction using the above relation, $f(m,n) = f(n,m)$ for all $n,m > 0$.

I solved this using generating functions, which took me about 10 minutes:

For $f_1(m) =$ number of x with $|x| = m$, \[ \sum_{m=0}^\infty f_1(m) x^m = 1+2x+2x^2+\cdots = 1+\frac{2x}{1-x} = \frac{1+x}{1-x}.  \]
For $f_n(m) =$ number of $(x_1,x_2,\ldots,x_n)$ with $\sum_i |x_i| = m$, then \[ \sum_m f_n(m) x^m = \left[ \sum_m f_1(m) x^m \right]^n = \frac{(1+x)^n}{(1-x)^n}.  \]
Thus \[ \sum_m f(m,n) x^m = (1+x+x^2+\cdots) \sum f_m(n) x^m = \frac{(1+x)^n}{(1-x)^{n+1}}.  \]
Therefore, \[ \sum_{m,n} f(m,n) x^m y^n = \sum_n y^n \frac{(1+x)^n}{(1-x)^{n+1}} = \frac{1}{1-x} \sum_n \left[\frac{(1+x) y}{1-x}\right]^n = \frac{1}{1-x} \cdot \frac{1}{1 - (1+x) y/(1-x)} = \frac{1}{1-x-y-xy}.  \]
Since this expression is symmetric with respect to interchanging x and y, $f(m,n) = f(n,m)$ follows.

(Note that this calculation reproves Nukular's recurrence.)

here is the easiest way:
It's well known that the equation :a_1+..+a_k=m has exactly k-1Cm-1 positive intergral k-tiple.
So $f(n,m)=\sum_{k=0\to n}\sum_{i=n-k\to m}2^{n-k}*C^{n-k-1}_{i-1}*C^{n-k}_{n}=\sum_{j=0\to n}2^{j}*C^{j}_{n}*C^{j}_{m}$
here k denotes the number of 0-s!
And now it's easy!

Well here is my weird solution by bijection.

Let $g(a,b)$ denote no of sol to $s_1+s_2....s_a=b$ where $s_i$ are nonnegative. We will show that $g(a,b)=g(b,a)$ by a bijection between $a$ tuple summing to $b$ with a $b$ tuple summing to a such that both tuples have exactly the same no of positive integers. Hence we will also have a bijection between $f(a,b)$ to $f(b,a) $and we would be done with problem.

Let $S_1$ be the set corresponding to $g(a,b)$ and $S_2$ the other. Here is how we establish bijection between $S_1$ and $S_2$

Take a $a$ tuple from $S_1$. We divide it into several groups by scanning from left. If a number $a_i$ is not succeded by $0$s then $(a_i)$ is a group. While if some number $a_j$ is succeded by $k$ zeroes then group is $(a_j\underbrace{ 000}_{\text{k zeroes}})$. Finally leading zeroes, if present from very start, are grouped with last positive number (cyclically). For example $0098070$ is grouped as $(9)(80)(7000)$.
Now we define $T$ as an operation on the tuple according to foll rules

$1)T((a00)(d)(e0))=T(a00)T(d)T(e0)$ that is, $T$ acts groupwise.

$2)T(k000(\text{l zeroes}))=l+1 000(\text{k-1 zeroes})$

There are two striking things that appear...

.first is that an $a$ tuple is transformed to a $b$ tuple with sum atleast $a$( because in 2nd operation sum of sequence on LHS=no of terms in sequence of RHS, and vice versa too)....

Second thing is that $T(T(x)=x$ so operation $T$ indeed establishes bijection and we are done.

For example
$T(230100)=T(2)T(30)T(100)=(10)(20)(3)=10203$ and
$T(10203)=T(10)(20)(3)=(2)(30)(100)=230100$

We make the convention that $f(m,0)=f(0,m)=1$. In counting the number of solutions of $|x_1|+\ldots+|x_n|\le m$, note that if $|x_n|=k$ then there are $f(m-k,n-1)$ options for $(|x_1|,\ldots, |x_{n-1}|)$ and $2$ options for $x_n$ if $k>0$ and only one if $k=0$. Thus,
$$ f(m,n)=f(m,n-1)+2(f(m-1,n-1)+\ldots +f(1,n-1)+f(0,n-1))$$

Keeping $m$ fixed and summing up all these relations where the second entry of the LHS varies from $1$ to $n$ gives
$$f(m,n)=f(m,1)+2\displaystyle\sum\limits_{\substack{0\le k\le m-1 \\ 1\le l \le n-1 }} f(k,l)=1+2\displaystyle\sum\limits_{\substack{0\le k\le m-1 \\ 0\le l \le n-1 }} f(k,l)$$
because $f(m,1)=f(1,m)=2m+1$. The claim follows now by strong induction on $m+n$.

Let $\mathcal N$ denote the set of all $n$-tuple of integers $(x_1,\dots x_n)$ such that $$|x_1|+\dots+|x_n|\le m$$And $\mathcal M$ denote the set of all $m$-tuples $(x_1,\dots,x_m)$ such that $$|x_1|+\dots |x_m|\le n$$
We want to prove, $|\mathcal M|=|\mathcal N|$.Denote by $X$ the set of all $(m+n)$ length $0-1$ sequences with $n$ ones and $m$ zeroes.The number of such sequences is $\frac{(m+n)!}{m!n!}$.Also let us assume that $N,M$ denote the set of respectively $n$-tuple and $m-$ tuple of non-negative integers whose sum of components are respectively less than equal to $m,n$ .($M,N$ and $\mathcal M ,\mathcal N$ are different.)

[b][color=#f00]Claim:[/color][/b] There is a bijection $\sigma_1$ from $X$ to $N$ and there is a bijection $\sigma_2$ from $X$ to $M$.

[i]Proof.[/i] For any sequence $a\in X$,let  $\sigma_1(a)$ denote the $n$-tuple whose $i$th $(i=1,2,\dots,n)$ entry is the number of 0's between $i-1$-th and $i$-th 1 in $a$.(First entry of $\sigma_1(a)$ is the number of 0's before the first 1 in $a$).
Similarly $\sigma_2(a)$ denote the $m$-tuple whose $i$th $(i=1,2,\dots,m)$ entry is the number of 1's between $i-1$-th and $i$-th 0 in $a$.(First entry of $\sigma_2(a)$ is the number of 1's before the first 0 in $a$).These are bijetions which are easy to verify.$\blacksquare$

For a tuple $v$ ,let $f(v)$ denote the number of nonzero entries of $v$.As there are 2 choice for plus or minus sign in each nonzero entry of a tuple of $\mathcal M$,or $\mathcal N$ so, $$|\mathcal N|=\sum_{v\in N} 2^{f(v)},\quad |\mathcal M|=\sum_{v\in M} 2^{f(v)}\qquad(*)$$So we ony need to prove that the 2 above sums are equal.

Consider a $0-1$ sequence $a$ from $X$ whose starting and ending digit are same.Then it is easy to check that $\sigma_1(a)$ and $\sigma_2(a)$ have same number of nonzero enties.($\sigma_i$ are defined as the claim.)Indeed,if $a$ starts and end with $0$ (for example) then both $\sigma_i(a)$ has the number of nonzero digit equal to number of continuous block of 1 in $a$.

Now consider $a\in X$  such that it starts and ends with different digits.$a'$ denote the sequence of $X$ writing $a$ in reverse order.(e.g. if $a=00011001001$ then $a'=10010011000$).We have $a=(a')'$.Then we claim that,

[b][color=#f00]Claim:[/color][/b] $f(\sigma_1(a))=f(\sigma_1(a'))$
[i]Proof.[/i]If $a$ starts with 0 and ends with 1 then $f(\sigma_1(a))$ is equal to the number of blocks of 1 in $a$ and $f(\sigma_2(a))$ is equal to the number of blocks of 0 in $a'$.By the definition of $a,a'$ they are equal.
If $a$ starts with 1 and ends with 0 then  $f(\sigma_1(a))$ is equal to the number of blocks of 1 in a minus 1. And $f(\sigma_2(a))$ denote the number of blocks of 0 in $a'$ minus.Again they are same.$\blacksquare$

So the multisets $\{f(v)|v\in N\}$ and $\{f(v)|v\in M\}$ are equal.So the 2 sums in (*) are infact equal.$\blacksquare$


The zeros of the quadratic equation x^2 + ax + b = 0 are a and b (b can't equal 0).  Find a and b.

[quote="Art of Owna"]use vietas theorm..[/quote]

which is...?

[hide]

$-a=a+b$ therefore $b=-2a$
$b=ab$ therefore $a=1$ which means $b=-2$

thus giving us the ordered pair for $(a,b)=(1,-2)$.[/hide]


function
integration
algebra
polynomial
calculus
calculus open

Find the general solution of the differential equation \[\frac{d^2y}{dx^2}+x^2y=0\qquad x>0\]

There's a perfectly good power series expression: $a(1-\frac{x^4}{3\cdot4}+\frac{x^8}{3\cdot4\cdot7\cdot8}-\dots)+ b(x-\frac{x^5}{4\cdot5}+\frac{x^9}{4\cdot5\cdot8\cdot9}-\dots)$, but I don't know whether these functions have names.

Where doez it come from? I don't know it also. Does a solution exist in closed form?

This is a standard method; assume a power series solution, then use the differential equation to generate a recursion for the coefficients.

I'm pretty sure there's no closed form solution in elementary functions, but this probably has a name- something in a very broad family. It's not exactly a Bessel function, but it's somewhat related.

$y\"+x^2y=0$ has solution of the form $y=\sqrt{x}Z_{\frac{1}{4}}(\frac{1}{2}x^2)$

where $Z_{v}(x)$ is Bessel functions solutions of

$Z_{v}\"+\frac{1}{x}Z_{v}'+(1-\frac{v^2}{x^2})Z_v=0$

It may seem obvious, but why not solve this equation as you would any other?
$\displaystyle \frac{d^2y}{dx^2}+x^2y=0$
$\int{\int{\frac{1}{y}}}d^2y=\int{\int{-x^2}} dx^2$
$yln(y)-ay=-\frac{x^4}{12}+bx+c$
where a, b, and c are constants.
This is equivalent to:
$\displaystyle (\frac{y}{a})^y=\frac{bc^x}{^{12}\sqrt{e{}^x{}^4}}$
It becomes apparent you will not be able to solve only in terms of y.

That method is nonsense, diophantus- only first order equations can be solved as separable.

Why is it nonsense jmerry? It seems plausible to deal with a second order function as you would a first, though I may be wrong as I have just begun differential equations in school.

Nevermind.  I have realized the error in this method.  As I said, I'm new to DE and am surprised my teacher hasn't corrected me yet.  I've been using similar methods to take shortcuts on my homework for several weeks.  Thanks for putting me straight.

I suppose we'll just have to brute force this one. Given $\frac{d^{2}y}{dx^{2}} + x^{2}y = 0$. This is a second order homogeneous diff eq with nonconstant coefficients. we'll use the power series method.

let $y=\sum_{n=1}^{\infty} a_{n}x^{n}$

$y^{'}=\sum_{n=1}^{\infty} n a_{n}x^{n-1}$

$y^{''}=\sum_{n=2}^{\infty} n(n-1) a_{n}x^{n-2}$

we plug these right into our equation, and we have

$\sum_{n=2}^{\infty} n(n-1) a_{n}x^{n-2} + \sum_{n=0}^{\infty} a_{n}x^{n+2} = 0$

we can mess around with the second sum a bit

$\sum_{n=2}^{\infty} n(n-1) a_{n}x^{n-2} + \sum_{n=4}^{\infty} a_{n-4}x^{n-2} = 0$

$2(1)a_{2}x^{0} + 3(2)a_{3}x^{1} + \sum_{n=4}^{\infty} \left[n(n-1) a_{n} + a_{n-4}\right] x^{n-2} = 0$

so our solution is given by $a_{2} = 0$, $a_{3} = 0$, and the recurrence relation $n(n-1)a_{n} = -a_{n-4}$

hmm... did i do this one right? it's late... i think i screwed up somewhere. anyways. for those of you interested in physics also, this diff eq is reminiscent of the time-independent schrodinger equation given by

$-\frac{\hbar^{2}}{2m} \frac{d^{2}\psi}{dx^{2}} + \frac{1}{2}m \omega^{2}x^{2} \psi = E \psi$

which is just a nonhomogeneous version of this diff eq at the top of this post, with some constants thrown in, and this can be solved using a similar power series method, and another cleverer algebraic method involving the use of operators.

anyways feel free to correct me... i really think im wrong, but it's late, and im out.

I've already posted a power series, back in post #2.

That's interesting about the Schrödinger equation, but that's an eigenvalue problem.
That would be the one-dimensional quantum harmonic oscillator, and the physically maningful solutions are polynomials times Gaussians, none of which have zero energy. Also, it's the wrong sign of $x^2$, and that means that we're essentially on the imaginary axis of a nonphysical solution.
Alternately, we could have an inverted harmonic potential, but there's just no physical insight for that, and we're not at a nice energy value anyway.

In short, the solutions to this problem are not elementary functions.

yea thats right.see it this way diophantusf"(x)=-x^2f(x).how then are u going to integrate f"(x)/f(x)?


Prove that at most $ \frac{n(n^2\minus{}3n\plus{}8)}{3}$ regions of space can be produced by $ n$ intersecting spheres (not necessarily of the same size).

(When there is only one sphere, 2 regions of space are said to have been produced.)

This problem seems very difficult. I think I 'm close to a solution by induction but I heven't finished it yet. :|

I look forward to your solution.
I am reading a book that says this result can be guessed at by finite differences and then proved by deductive methods, but it doesn't provide a proof.  :(


geometry AMC AIME USA(J)MO USAMO HMMT Duke

I currently have covered precalculus taking calc next year (grade 11)
I've obtained a 4 in AIME.

I now want to order the AoPS books but I'm wondering whether its worth it to order volume 1 when you already have sufficient basic knowledge (or at least I hope I do).

I noticed that Volume 1 was recommended for 'Getting Started' thus I'm asking this question.  Solving extra problems won't hurt?  I'm not concerned about the money.  What do  you think?;;;

Skimming thru the content the only area which I think I need a review on is the "prove it" section.  How in-depth does the first volume go into?  i see "introduction to trig"...  is it for starters?

Volume 1 is good, because it will help you get to around 6-8 AIME and very good on the AMCs.

There is a famous remark in one of Polya's books--I think it is [i]Mathematics and Plausible Reasoning[/i]--in which he says that people who skip the easy parts because they have already learned them will miss more than people who skip the hard parts because they aren't ready for them yet. In other words, it is always a good idea to review the basics from a fresh perspective. School mathematics courses don't begin to teach everything that is in AoPS Volume One. If money is not an issue, your time will be well rewarded by working through much of Volume One, and reading all the expository parts of that book.

ah thank you.  Maybe my unstable base has led to my trouble in solving 22+ AMC's.

I'll definately buy the two books!

Given what you've described, I'd recommend Vol 1 mainly for the geometry chapters (and the number theory if you don't yet know modular arithmetic).  Spend a lot of time working the geometry problems in the book, and you'll get pretty good at AMC/AIME geometry.

volume 1 is probably a bit basic, but if money isn't a conern it won't hurt to have an extra problem resource. volume 2 would probably be okay.

[quote="bubala"]If this year I got a 136 on the AMC 12 and a 10 on the AIME, is AoPS volume 1, 2, both, or neither right for me?[/quote]

thats USAMO qualifying scores, so you're probably better off going with AOPS Volume 2. However, Volume 1 most certainly won't hurt.

The great thing about Volume 1 is tha tit covers the basics (and then some) but gives you advanced problems. I highly recommend it.

Yes!  I've placed my order two days ago and I'm looking very forward to receiving it.

Geometry; thats my weakest area  :P

[quote="peter"]Volume 1 is good, because it will help you get to around 6-8 AIME...[/quote]

whatever. last year i got a 3 on the AIME. during the summer, i went through volume 1. and i have somewhat started volume 2. this year i managed to get a:
[hide]4 on the AIME. what an improvement.[/hide]

everyone doesn't perform to their full potential all the time.  mathfanatic for example (you dont mind being used as an example do you?   ;) ) had made to USAMO when he was only in 8th grade.  Evidently, after another year of preparation he should've made to another round of USAMO or at least achieve an reasonable AIME score.  But during grade 9?  He made a 2.  This year he made a 12.

I won't say much more.

This thread makes me scared. furious studied the whole summer and only increased his score by 1! I'm not as optimistic as I was before about next year's AMC/AIME, especially after hearing about mathfanatic.

These people are just the extremes, there will always be people who don't improve as much as they should, or ven deproove.  And there are always people who increase from 1 to 11, or 1 to 13 (joml88, and dpopov I think).

[quote="cooljoe"]h_s_potter, do you know what those guys did to improve so drastically? (I got a 2 on the AIME I) :?[/quote]

Simply put, work their a## off. There's really no substitute for hard work. I know that one of those people that you mentioned did a lot of work over the summer; doing contest after contest really makes you get so accustomed to the problems and introduces you to so many new topics, that eventually, when you take one of the AMCs or AIMEs, it would seem like you've done every type of question before, if not the exact question with just numbers changed around! It's really all about the effort you put into it. I really doubt that someone he never worked AIME-type math problems in his life just went ahead and took the AIME, and miraculously pulled off a 10 or above ;)

But here's contests that I would suggest looking at (in order)

1. Past AMCs/AHSMEs which you can buy contest problem books published by MAA, or look here: http://www.math.ksu.edu/main/events/hscomp/samples/amc12/sample.htm
And ofcourse the contest resources page on AoPS!

2. Past UGA math contests and UNC math contests that are here:
http://www.math.uga.edu/mathmeet.html/
http://www.math.uncc.edu/~hbreiter/problems/UNCC.html

3. North Carolina State Math Contests found here:
http://192.154.43.167/goebel/statecon/tests/tests.htm

4. HMMT (General Tests) found here:
http://web.mit.edu/hmmt/www/datafiles/

5. Duke Math Meet Contests found here:
http://www.math.duke.edu/dumu/contests/problems/

6. HMMT subject tests found here
http://web.mit.edu/hmmt/www/datafiles/problems/2006.shtml

7. AIME problems found on the contest resources for AoPS

8. HMMT Guts Round problems also found here:
http://web.mit.edu/hmmt/www/datafiles/problems/2006.shtml

These sound like A LOT of problems to do, and they are, but as you do more of these problems (IN ORDER), they start to become a lot easier and you become better and better. If you finish all of those problems and understand everything, I seriously don't see what can hold you back from USAMO qualification ;)  :D

[quote="cooljoe"]Thanks for the links, iversonfan. However, do you think that doing problems or actually reading the AoPS books would be better for preparing?[/quote]

Read the books first, then do the problems :)
The book shows you good approaches how to do each type of problem whereas the contest solutions kinda already expect you to know those approaches. So if you look at some of the solutions to those problems beforehand, you'll be like WHOA! HOW DID THE KNOW HOW TO DO THAT?


Can anyone prove the brianchon`s theorem in easy way

hello, i hope, this will help you
http://www.cut-the-knot.org/Curriculum/Geometry/Chasles.shtml
Sonnhard.


limit
function
complex numbers
calculus
calculus open

$f(x)=\sum p_nx^n= 1+2x+3x^2+5x^3+7x^4+11x^5+...+$
nth term has a coefficient equal to the nth prime

define $q(x)=\frac{1}{p(x)}=\sum q_kx^k$

Then N.Backhouse conjectured that

$\lim |\frac{q_{n+1}}{q_n}|=1.4560749485826896....$

Does any one have seen recent results about this constant ?

It has been calculated to 1,300 decimal places by now (see all the digits [url=http://www.worldwideschool.org/library/books/sci/math/MiscellaneousMathematicalConstants/chap9.html]here[/url])  On the other hand, I do not see what is so exciting about knowing that a certain  analytic in the unit disk function with positive coefficients has a negative zero closer to the origin than any pair of its complex conjugate zeroes (this is all there is to the Backhouse constant, isn't it?) even if this function is somehow related to primes. After all, if such a fact is true, it can be easily demonstrated by a brute force computation using the argument principle...


conics
parabola
calculus
calculus computations

i was asked to compute of disprove the existence of the  lim of $ \frac{x^4}{y+x^2}$ as (x, y) approaches (0,0) where limit is taken along all possible paths that never intersect the parabola y=-x^2 except at common end point (0,0)

does it mean i cant take $ y=x^4-x^2$ t and compare to $ y=x$ to prove its non existence, i am so confused about this question, thanks in advance
lims appear to be very subjectiive ot me as we have to choose specific cases to prove non-existence but how do we know if a limit does not exist, we cannot test every single case to find a case that does not work!!??

To prove that a limit does indeed exist, you could either compute the limit along every possible path (which is pretty much impossible) or you can use the $ \varepsilon$-$ \delta$ definition.  To prove there is no limit, you just need one counterexample (or you can use $ \varepsilon$-$ \delta$).

[quote="JRav"]To prove there is no limit, you just need one counterexample (or you can use $ \varepsilon$-$ \delta$).[/quote]
Really, two counter-examples: choose two paths along which the expression has different values.  Anyhow, what's wrong with $ y \equal{} x^4 \minus{} x^2$ and $ y \equal{} x$?  (Well, you might need to make an adjustment "far away" from 0, but that doesn't matter.  The $ y \neq \minus{} x^2$ is just to say that you're taking the limit through points that actually belong to the domain.

if you take y=o and x approach 0 then the limit tends to 0 and if you take y=x^4-x^2 the llimit tends to 1


What should be the basis on which we should decide which form is more stable?

usually
keto form is less stable only if it contains an active methylene group :P

In general, the keto form is more stable because a carbon-oxygen double bond is stronger than a carbon-carbon double bond. In fact, the enol is more stable only when its double bond can be stabylized by conjugation with another double bond, like active methylene compounds.


On a circle there are $n$ points such that every point has a distinct number. Determine the number of ways of  choosing $k$ points such that for any point there are at least 3 points between this point and the nearest point. (clockwise) ($n,k\geq 2$)

I think it is the number of ordered pairs (a,b,c) so that (a,b+3,c+6)  contains no equal numbers. Now something like inclusion-exclusion would do the job, because let A12 number of ways , so that a = b + 3, and so on.

if $M$ is the number of way of choosing $b_1,b_2,....,b_k\ge 1$ such that $b_1+b_2+...b_k=n-3k$
then the solution is $4M=4{{n-3k-1}\choose {k-1}}$.

in fact, if $A(1),A(2),....,A(n)$ are those point and $A(a_1),A(a_2),....,A(a_k)$ are the $k$ points satisfying the condition
we have just to the take $b_i=$[the number of point between $A(a_{i+1})$ and $A(a_i)$]$ - 2$. in this case $b_1+b_2+...b_k=n-3k$
so for each choosen $b_i$ we have a solution containing the point $A(a_1)$
then we do a translation to have the other $3$ solutions $A(a_1+h),A(a_2+h),....,A(a_k+h), h=1,2,3$
that's why we have to mutiply $M$ by $4$


As a moderator, I'm interested in seeing who are taking what this year. Since I can't seem to lock the topic without locking the poll, I would very appreciate [b]much[/b] if you just vote and not post. That's a way to reduce any spammers who only say, "Yeah I'm taking @@@@@."

Thanks.

Happy Clicking!  :)

Um, you can delete this post when you see it, but I have to point out that there are no options for 10A/10B or 12A/12B and that those are common choices as well.

Thank you.

Another kind user also pointed out this to me. I fixed it.

i don't mean to go against your wishes, but there are no options for people that are taking AMC 12A/AMC 10B

*edit=my stupidity

[quote="mihail911"]i don't mean to go against your wishes, but there are no options for people that are taking AMC 12A/AMC 10B[/quote]

Yes there is. I have it up there. You just didn't see it.  :idea: It's with AMC 12B/AMC 10A.

Whats the difference between AMC 10A and AMC 10B?  How can you take both 10 and 12?

[quote="unimpossible"]im taking 10A and 12B.[/quote]

[quote="Silverfalcon"]I would very appreciate [b]much[/b] if you just vote and not post. That's a way to reduce any spammers who only say, "Yeah I'm taking @@@@@."[/quote]

Please read the first post before posting yourself.

[quote="Brut3Forc3"]I know I'm not supposed to post, but my choice (12A and 10B) isn't up there...[/quote]

If you had read the topic you would have found this  :)   !  But I am making it easier by putting it all into my post.
[quote="mihail911"]i don't mean to go against your wishes, but there are no options for people that are taking AMC 12A/AMC 10B [/quote]
[quote="Silverfalcon"]Yes there is. I have it up there. You just didn't see it.  It's with AMC 12B/AMC 10A.[/quote]

[quote="PI-Dimension"]I'm just curious, what would happen if you took 10A and 12A or 10B and 12B (like one in the morning and one in the afternoon or something)[/quote]

I don't think that's allowed. :wink: They share problems.

[quote="PI-Dimension"]I'm just curious, what would happen if you took 10A and 12A or 10B and 12B (like one in the morning and one in the afternoon or something)[/quote]
That's explicitly against AMC rules. From the Teacher's Manual: "A student may only take one exam on a given day ..." And it would be an obvious violation of the rules, and would imply that a contest manager wasn't following the rules. The AMC does not want to see contest managers fail to follow the rules. I don't think you want to find out what the AMC would do in such a case; it might not be good for the entire school.


geometry
trigonometry
angle bisector
Pythagorean Theorem
trig identities
Law of Cosines
power of a point

I took a contest and the questions are supposed to be based on triangle similarity and proportionality theorems like the angle bisector theorem.  And well, I couldn't get the answer to this question during the contest, but I brought it home and I [i]think[/i] I solved it.  The thing is, I didn't use similarity or anything.. and it looks like an angle bisector problem... so I'm thinking there is probably an easier way to do it :

In triangle PQR, PQ has length 6, QR has length 15, and angle PQR is 60 degrees.  S is a point on PR such that QS bisects angle PQR.  Find the length of PS.

Try it first, but here's how I did it (and I don't even know if this is right...
[hide]First I found the triangle's area = (45 :sqrt: 3)/2.  Then I dropped two altitudes, from P to QS and from S to QR, so we've got 2 30-60-90 triangles.  since PQ = 6 then the altitutude to QS = 3.  Let QS = x so the altitude from S to QR = x/2.  Then I just added the areas (3*x)/2 + ((x/2)*15)/2 and set that equal to the total area, (45 :sqrt: 3)/2.  Solving for x I got x = (30 :sqrt: 3)/7

(ok, so I don't have to keep saying this, let point T be the point on QS where the altitude from P lands).  From the 30-60-90 triangle, we know QT = 3 :sqrt: 3 so TS = (30 :sqrt: 3)/7 - 3 :sqrt: 3 = (9 :sqrt: 3)/7.  Then I used the pythagorean theorem and got  [b]PS = (6 :sqrt: 19)/7[/b]  Any simpler ways?? This was a no calculator contest, too, so all those high radicals were fun  :P [/hide]

And I'm hungry, so I'll post the second question after someone replies to this one.

[hide]

law of cos: $(PR)^2=15^2+6^2-2*6*15*\cos(60)$, so $x$=$3\sqrt{19}$.  Then use angle bisector theorem.  $\frac{6}{PT}$=$\frac{15}{3\sqrt{19}-PT}$, solve for $PT$.[/hide][/code]

[quote="djshowdown2"]woah thats.. pretty cool  :) I've never used the law of cosines before, but I've known it from those crazy animations on the left bar  :P

Ok thanks, the second one is this:

AB is a diameter of a circle O.  DA and BC are tangents circle O.  DB and AC intersect at point E, which is a point on circle O.  If AD has length 10 and BC has length 8, what is the length of BE?

Now this one I didn't get an answer to.  I've had some scattered thoughts, but I doubt they'll mean much.
[hide]Um, well, at first I thought it was a two-poles problem, so I found  that the distance from point E to AB is 40/9.   Then I had lots of similar triangles.. but I didn't get anywhere.  And I found that the key point in this problem (or so I think) is that E is on the circle... but that still didn't mean much.

But now I just had a thought.  hmm.  oh, I might have figured it out.   hold on  :lol:

EDIT: haha, yeah, I figured it out while I was typing this  :P  but A, maybe there's an easier way, and B, maybe I"m wrong.  Here's what I got for my answer.
[hide](8 :sqrt: 5)/3[/hide][/hide][/quote]

I'm getting the ans. as 40/9..is that correct??

[quote="djshowdown2"]AB is a diameter of a circle O.  DA and BC are tangents circle O.  DB and AC intersect at point E, which is a point on circle O.  If AD has length 10 and BC has length 8, what is the length of BE?
[/quote]

It is "power of a point to the circle" problem. If you have a circle (O) and and a point P (inside or outside of the circle) and if you draw 2 secants intersecting the circle at points K, L (the first secant) and M, N (the second secant), then the product (called power of the point P to the circle)

$PK.PL = PM.PN$

does not depend on the secant, but only on the circle and the point. To see this, just realize that the quadrilateral KLMN is cyclic, $\angle KLM = 180^o - \angle NMK$ and consequently, the triangles $\triangle PKM \sim \triangle PNL$ are similar and

$\frac{PK}{PM} = \frac{PN}{PL}$

One of the secants can be a tangent and power of the point becomes square of the tangent length from the point. In your problem, power of the point D to the circle with diameter AB is

$DA^2 = DE.DB = (DB - BE).DB$

On the other hand the lines AD, BC are parallel, the triangles $\triangle DAE \sim \triangle BCE$ are similar and

$\frac{BE}{CB} = \frac{DE}{DA} = \frac{DB - BE}{DA}$

$DB - BE = BE\ \frac{DA}{CB}$

$DB = BE \left ( \frac{DA}{CB} + 1 \right )$

Substituting this to the "power of the point D to the given circle" equation,

$DA^2 = BE^2\  \frac{DA}{CB} \left ( \frac{DA}{CB} + 1 \right )$

$BE = \frac{DA.CB}{\sqrt{DA^2 + DA.CB}} = \frac{80}{\sqrt{100 + 80}} = \frac{40}{3 \sqrt{5}} = \frac {8 \sqrt{5}}{3}$

Note that there is a right angle at E because it's inscribed in a semicircle. Also, A and B are right angles because they are formed from a centerline and tangents. You have a bunch of right triangles that are similar, so make a bunch of proportions.
I got: [hide="click here"]$\frac {8\sqrt 5}{3}$ as well.[/hide]


AMC USA(J)MO USAMO inequalities algebra polynomial geometry

This is a somewhat broad topic but I'll need both opinions and facts that anyone can give.

For my contest's last round, I plan to use other national olympiad problems to mimic USAMO because for one, I cannot make Olympiad problems because I haven't seen that many of it. I plan to see a lot of them as I can and select the good ones to use for this round but to do that, I'll need some information.

Since this topic could go anywhere, I'll ask several questions in hide and you can just put # in your answer to address any that you know, whether from your practice or from participation.

[hide="1. Inequality"]
There are many inequality problems on USAMO. Now, generally what type of inequality do they focus? Algebraic? Geometric? All of them? Also, do proofs involve direct application of Cauchy and Jensen and so on or combinations of them? What number does inequality problem usually take? In Day 1 or Day 2? [/hide]

[hide="2. Algebra"]
Do most USAMO questions solely have proving some polynomial, algebraic expression, and so on or do they ask for actual answer with proof? If so on either of them, which is more common? And is it like 1 problem out of 6 or 2 problems out of 6? [/hide]

[hide="3. Geometry"]
Can you name some topics that appear in the area of geometry? I know there are some for locus and cyclic polygons but that's about it... Any lattice ones? [/hide]

[hide="4. Number Theory"]
What are usual types of problems regarding NT? [/hide]

[hide="5. Combinatorics; Counting; Probability"]
Do these appear? I have not seen too many counting problems in USAMO level though... :? [/hide]

[hide="6. Game"]
I know sometimes olympiad questions have game-style questions (about asking strategy to win certain game or stuff). How are they? Do they appear popular and worth appearing in the mimic competition or will they just be useless? [/hide]

[hide="7. Etc"]
Feel free to mention anything else or areas of problems that need to be addressed and yet not mentioned.. Help me out  :)  :) [/hide]

THANK YOU!  :lol:

Let's not forget graph theory  :)  Although, I'm pretty terrible at graph theory questions.

For inequalities, rarely do you just have to directly apply Cauchy or Jensen. There's usually some tricky manipulation you have to do before that. But there are some exceptions (like IMO 2001/2, which is straight up Jensen, then AM-GM). The problem with many inequalities is a simple word: dumbass. It's tough to come up with a non-dumbassable inequality in a set number of variables (particularly, 3). This is why there haven't been many three variable symmetric inequalities on the USAMO and IMO in recent years (from what it looks like, a large number of people at the IMO 2005 multiplied out on #3, which was extremely easy to do).

So if you're going to put an algebraic inequality on, it should probably either be:
-asymmetric
-have lots of radicals in it so it's really difficult to dumbass
-have n variables,

or any combination of the three. As for geometric inequalities, anything really is fair game. The thing with geometric inequalities is that they can be very geometric in nature (like USAMO 2007.6), where the hard part is finding the geometric relations, then the inequality part is just like AM-GM or whatever. Or, the easy part may be calculating the lengths or angles in question, then the hard part being the algebra. It's kinda up in the air I guess. I'm not too experienced with (non-trivial...) geometric inequalities.

As for the rest...most of these questions can be easily answered by like, looking at olympiad problems. As for combinatorics, there are definitely a large number of combinatorial problems, and certainly there are problems in which you have to count things. I mean, take USAMO 2008.4 (that's probably not the best example). However, there aren't generally AMC-type problems that are just all like "Count the number of ways to do this blah blah."

So yeah, just look at problems.

On the other hand, I don't think it would be a good idea to just take national olympiad problems, even if they're from other countries, at least a few people have probably already seen them. If you don't feel comfortable with coming up with six olympiad problems, you should ask for submissions from more experienced people (Altheman comes to mind as being particularly good at coming up with problems).

For combinatorics, you'll often see things involving sets, and then either drawing some kind of correspondence (like a bijection) or a pigeonhole-type argument.  There's no simple way to just classify these problems...

For regular geometry, probably the two shapes which appear most often are triangles and circles.  Often there will be some angle chasing or length chasing which needs to be done, but some big topics which appear are spiral similarity/homothety and cyclic quadrilaterals.  I find geometry problems are, with sketchpad available, not too difficult to generate...

For algebra...I suppose it's largely working with polynomials, but then consider this year's USAMO #5...

NT...Well you'll have stuff involving divisibility, and diophantine equations aren't all that rare (ex. TST #4)

You might also want to find some good "hybrid" problems: ie, those which involve more than one general area of olympiad math.  I'm talking about, say, something which involves both divisibility arguments (NT) as well as polynomials and/or counting, or, say, combinatorial geometry.

If I happen across some good problems, I can send them over :)

Thanks for tips about inequality. I'll definitely have to look into Mildorf's Inequality pack (and I remember there's one by Kiran Kid...I can't spell the last name :P).

So far, really great! Keep them coming  :D

P.S. About using other national math olympiad problems, yes, I understand the risk of having someone seen them before but well, I don't know... Asking for good problems is a difficult thing (for one thing, everyone that I know well are in college now and will be busy).

Yeah, what Carl said for inequalities is very, very true. Or just make it nonhomogeneous without a condition  :P But for those, it's usually clear that you should use polynomials to homogenize or something.

Functional equations are good, IF you can find a "good" one. And "good" ones are PRETTY HARD to find. I find that functional equations that don't let you plug in 0 are harder...

Geometry - well for geometry, there are a lot of "advanced" techniques and stuff. On the IMO, you never "need" inversion, but it often makes problems dealing with lots of circles, or collinearity/concurrency, easier. (Pole-polar can do the same for the latter too.) So with geometry, if it is not a harder problem, try to make sure there's no easy way to kill it with advanced theorems...
Also, what Krishanu said is very good. But problems with quads, or n-gons, are maybe less common, but they are still interesting; the real USAMO committee could choose such a problem to throw people off, maybe.

Combo - counting in two ways is pretty common; estimating what you count is rather annoying, but appears quite a lot. Unfortunately I can't say much about combo because I suck at it :(
Graph theory is fun too!

NT - yeah diophantine and divisibility are very common; how about integer polynomials? Those are pretty cool. Or number-theoretic functions (phi, etc)

Algebra - aside from ineqs/functional equations, all you have left (basically) are ... polynomials and weird systems of equations.   They are sometimes exciting, but algebra in general is just kind of dry (even though I like it.)

Hmm. I've heard of most terms I've read so far, and I think I'm starting to get the general idea. It'll take more time for me to really know what problems are used though.

Now with "advanced" geometry theorems, what are they? Like... Nine-point circle? I don't know too many geometry theorems that are considered advanced. I have Geometry Revisited but I haven't really touched that much.  :D

I don't know too much about graph theory although I think I've seen some problems involving both graph theory and PHP? (like about towns and some number of roads and stuff)


probability
MATHCOUNTS
geometry
perimeter
rotation
geometric transformation
trigonometry

meh. Felt like starting one.

Read this topic to get a feel for what a marathon is:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=70276

OK.

[size=150]PROBLEM 1[/size]
Four people walk into an empty building, which has 7 exits. Three people walk three different doors and lock each door they go through. The fourth person goes to a random door and tries to open it. If it is unlocked, he opens it. If it is locked, he doesn't go through it. If he is only allowed to pick one door, then what is the probability that he exits the building?

[hide]
This is probably making it too simple, but is it $ \frac{4}{7}$?
There are 4 unlocked doors and 7 doors total.
[/hide]

If that's right:
[size=150]Problem 2[/size]
Simplify the following complex fraction:

$ \frac{1}{1\plus{}\frac{1}{1\plus{}\frac{1\plus{}\frac{1}{2}}{2}}}$

[hide]$ \frac{7}{11}$  It it is easy to find once we set a common denominator.  Since I don't have a question of my own, I'll let someone else post a full solution and then their own problem.[/hide]

It's easy. Start with the original problem, and work starting from the inside out.

1+1/2=3/2

(3/2)/2=3/4

1+3/4=7/4

1/(7/4)=4/7

1+4/7=11/7

1/(11/7)=7/11

If that was correct, then what is (a-x)(b-x)(c-x)...(z-x)?

P.S. Showing solutions like the above would be a lot easier if someone could tell me what to do if I want it to be like how mathcountsfreak has it.

[quote="mathcountsfreak"]
[size=150]Problem 2[/size]
Simplify the following complex fraction:

$ \frac {1}{1 \plus{} \frac {1}{1 \plus{} \frac {1 \plus{} \frac {1}{2}}{2}}}$[/quote]

Meh, I my abnormally retarded brain sees this as [hide]brute force[/hide].

[hide="asdf"]
Start from the bottom, work your way up. We have 1 over 7/4. This is equivalent to 4/7. 1/11/7 is our answer. Now this simplifies to

$ \boxed{7/11}$

Bleh, that obviously wasn't very clear.[/hide]

[size=150]Problem 3[/size]

Solve the following cubic equation:

$ 5x^{3} \plus{} x^{2} \minus{} 6x \minus{} 15 \equal{} x^{3} \minus{}7x^{2} \plus{} 3x \plus{} 3$

[hide="ANSWER"]
Subtract x^3-7x^2+3x+3 from both sides, to get
0=4x^3+8x^2-9x-18.
Then, just factor, to get
(2x+3)(2x-3)(x+2)=0
Therefore, x=3/2, -3/2, -2
[/hide]
Correct me if I'm wrong

No problem :(

Correct

[/size=18]Problem 4[/size]

Bob and Joey are both part of the 7 person math club. The math club has one circular table that holds 7 people. Bob and Joey are enemy arch-rivals and refuse to be placed together. How many possible seatings are there?

Guys. Let's keep this at Classroom Math level.

I don't want to do problem 4, so I'll post the next problem, since that problem is not mathcounts level.

[size=150]Problem 5[/size]
A square has a perimeter of 4, and a right triangle has legs of length 10 and 24. Figure $ \mathbb{R}$ is made up of those two figures. What is the area of figure $ \mathbb{R}$?

You just add the areas so it should be 121.

Correct me if im wrong.

[quote="Nerd_of_the_Ages"]
[size=150]Problem 3[/size]

Solve the following cubic equation:

$ 5x^{3} \plus{} x^{2} \minus{} 6x \minus{} 15 \equal{} x^{3} \minus{} 7x^{2} \plus{} 3x \plus{} 3$[/quote]

um, dude, factoring cubics is not classroom math level...

and do the figures that make up $ \mathbb{R}$ interest?

[quote="Nerd_of_the_Ages"]

[/size=18]Problem 4[/size]

Bob and Joey are both part of the 7 person math club. The math club has one circular table that holds 7 people. Bob and Joey are enemy arch-rivals and refuse to be placed together. How many possible seatings are there?[/quote]

I'm still a little insecure about my problem solving abilities, but i think it's 3,360.

Hey guys,

Reading through this thread, most of the problems wouldn't be considered classroom math.  Might wanna move this to MATHCOUNTS, although some of the problems have even been above that level.

I'm sorry, I'll try to post easier problems.
hardc0rerock, I got the same answer.

Is the difficulty levels of these problems okay?

[size=150]Problem 6[/size]
[size=84]An Easier Counting Problem[/size]

The math club is made up of 7 people. They have a circular table. How many possible different seatings are there?

[size=150]Problem 7[/size]
[size=84]Order of Operations practice[/size]

Find
$ 6/3\plus{}2*(4\minus{}2)\plus{}1$

[hide="To Problem 7"] Parenthesis first...4-2=2. Then, division. 6/3= 2. Multiplication, 2*2=4..So it makes:

2+4+1=7[/hide]

[quote="Nerd_of_the_Ages"]I'm sorry, I'll try to post easier problems.
hardc0rerock, I got the same answer.

Is the difficulty levels of these problems okay?

[size=150]Problem 6[/size]
[size=84]An Easier Counting Problem[/size]

The math club is made up of 7 people. They have a circular table. How many possible different seatings are there?

[size=150]Problem 7[/size]
[size=84]Order of Operations practice[/size]

Find
$ 6/3 \plus{} 2*(4 \minus{} 2) \plus{} 1$[/quote]

Problem 7 is fine.  You really wouldn't see Problem 6 in a regular classroom math class - It's more of a MATHCOUNTS type of problem.

Erm... Me not being that smart, can you post an easier problem? Or I'll do it.

Arite placed a 5 at the unit's place of a 3 digit number. THis increased the number by 3074. What was the original number?

[hide]
If you stick a 5 on the end of a number, you multiply the number by 10 and add 5. Call the original number $ x$. We have $ (10x\plus{}5)\minus{}x\equal{}3074$. Solving, we have $ 9x\equal{}3069$, so $ x\equal{}341$.[/hide]

[size=150]Problem 16[/size][size=75]I think[/size]
Flavor Flav can mow a large lawn in 5 hours. Chuck Norris can mow that same lawn in 20 minutes. Flavor Flav starts mowing the lawn. He mows for 2 hours, and then Chuck Norris arrives and starts mowing the lawn with him. If Flavor Flav started mowing at 1:00 pm, what time, to the nearest minute, will they be done mowing the lawn?

So which one am i doing!!!Fine ill do both!!!

[hide="Zhang Peijins problem"]
Ok so its kinda alot to read at once!!!So lets do each person independentally!!!ME starts walking at lets say 1:00 Pm So ill exit the moving side walk in $ \frac{100}{12}$ seconds which is $ \frac{25}{3}$ seconds or $ 8\frac{1}{3}$ seconds.  Ajain will exit the the moving stairs in $ 2$+$ \frac{100}{18}$ or $ \frac{68}{9}$ which is $ 7\frac{5}{9}$ seconds and and the creep is going to exit the moving starway(running beside) 11 seconds after I enter the stair way so Ajain will be $ 41\frac{1}{3}$ feet ahead of the weird guy so divided by 8 we get that he will catch ajain in $ 5\frac{1}{6}$ seconds so then we add $ 7\frac{5}{9}$ and we get $ 12\frac{13}{18}$(i did the addition in my head--might be wrong)  Umm im probably wrong![/hide]

[hide="other problem"]
ok so our number is just $ 100a \plus{} 10b \plus{} c \plus{} 3074\equal{}1000a\plus{}100b \plus{} 10c \plus{} 5$
And we want to solve for $ 100a \plus{} 10b \plus{} c$ so we have $ 900a\plus{}90b\plus{}9c\equal{}3069$...but he look at this...if we divide everything by $ 9$ we get $ 100a \plus{} 10b \plus{} c \equal{} 341$ and thats our answer becuase we wanted to solve for $ 100a \plus{} 10b \plus{} c$ so our answer is $ \boxed{341}$!!!!!![/hide]

Darn like a minute too late...ill post my solution anyway!!!

and this just came up so ill do it too!!!

[hide="mathcounts freaks problem"]
Ok--just a suggestion try to make the problems with nice numbers!!!!!!!AGHH
ok so we know that by the time chuck norris(why not some body else) comes the lawn will be 2/5 done and so working alone chuch norris can do 3 lawns per hour and you cna do 1/5 lawns per hour so we have that togather we can do $ 3\frac{1}{5}$ lawns per hour so then we divide $ \frac{3}{5}$ by 3.2 and we get that it will take 11.25 more minutes to mow the rest of the lawn so we have 2 hours and 11 minutes(rounded you will be working) so your answe is $ \boxed{3: 11}$!!!
[/hide]

My problem!!!!!!

Mount cylender is called that becuase it is a cylender(excuse my spelling) with a radius of 5000m and a hight of 12000m. If you plan to reach the very tip of Mount cylender by nightfall which is 10hours away, what is the minimum speed that you can travel at(in m/s) to acheive this goal pleas round to the nearest thousanth of a meter per second!!!

ok so to reach the very tip(which in my opinion is top center of the cylinder) you add the radius to the height. To find the average speed in m/h, you divide by 10. then divide by 60. then divide by 60 again.
It shoukd come out to:
12000+2500=14500/10=1450/60=24.16(6 repeating), 24.16/60 is bout .403 meters per second. If the tip isnt top center, then this is wrong. Otherwise, it should be right.

About that problem i put up earlier... I got that from a holt geometry worksheet that i chose not to do in favor of the harder worksheet. Sorry if it was hard, but it seemed like classroom math because it is geometry.

[quote="hardc0rerock"]Area of a regular polygon= 1/2ap[/quote]

What does a and p stand for?  Is this true for ALL regular polygons??

[quote="hardc0rerock"][size=200]Problem 17?[/size]

A regular pentagon with side length 4 is in a circle with radius 3.4. The parts of the circle not covered by the pentagon are shaded, What is the area of the shaded region?

Hint:
Area of a regular polygon= 1/2ap[/quote]

The apothem is the radius of the inscribed circle, not the circumscribed one. :wink:

The apothem can be calculated by $ \frac{s}{2 tan \frac{\pi}{n}}$

Where
s=side length
and n= # of sides
(tan is used on a value in radians)
So $ a\equal{}\frac{4}{2 tan \frac{\pi}{5}}$

$ a\equal{}\frac{2}{tan \frac{\pi}{5}}$

So if $ A_p$=area of pentagon
Then $ A_p\equal{}(\frac{1}{2})(\frac{2}{tan \frac{\pi}{5}})(20)$

$ A_p\equal{}\frac{20}{tan\frac{\pi}{5}}$

If $ A_c$=area of circle,
Then $ A_c\equal{}3.4^2\pi$
$ A_c\equal{}11.56\pi$

So the area of the shaded region=$ A_c\minus{}A_p$

$ \equal{}\frac{20}{tan\frac{\pi}{5}}\minus{}11.56\pi$

Maybe this solution is wrong, seems a bit complex; did u round the apothem? Anyways, that's what I got.

considering this is classroom math, we don't use trig.  Anyways, let's assume that he meant that the circle was inscribed in the pentagon.  (btw, if we used trig, the situation would be impossible)

Idk... I didn't have time to write my own formula so I used Google  :P .

In fact, I was about to suggest my sister about locking this topic. It seems that making the problems within a relatively easier skill range in fact, takes a lot of skill.

Express the sum of the following sequence as a common fraction.
\[ \frac{1}{2}\plus{}\frac{1}{6}\plus{}\frac{1}{12}\plus{}\frac{1}{20}\plus{}...\plus{}\frac{1}{9900}\]

[quote="levans"]Please keep problems and solutions to Classroom Basics level (no greater than AMC8).[/quote]

That problem isn't Classroom math level, Peijin.

The answer is 99/100 btw. You find a pattern that when it is added, it is n/n+1 when n(n+1)= the denominator of the last number.


If the smallest value of y satisfying the equation y=3x^2+6x+k is 4, find the value of k

the vertex is c - b^2/4a for the y value...and the parabola opens up.


function
real analysis
real analysis unsolved

Use
$ x\geq \ln(x\plus{}1)$
1. Consider the function :
$ f(x)\equal{}\ln(x\plus{}1)\minus{}\frac{x}{x\plus{}1}$


geometry
ratio
trigonometry
circumcircle
geometry solved

Denote by $L$ midpoint of segment $CD$. Then by area-ratios we have $\frac{[QDL]}{[QLC]}=1=\frac{QD \cdot \sin{\beta}}{QC \cdot \sin{\alpha}}$ where $\angle DQL= \beta$ and $\angle LQC= \alpha$. From this relation we conclude that $\frac{\sin{\beta}}{\sin{\alpha}}=\frac{QC}{QD}$. Now let $\angle BQP=x$ and $\angle PQA =y$. Also denote by $N$ intersection point betwen lines $AB$ and $PQ$, and by $M$ intersection point betwen circle $S_{1}$ and segment $PN$. It's very easy to see that $\angle BAM= \angle PBM=x$ and $\angle MBA= \angle MAP=y$. Then by sine-Ceva theorem in triangle $\triangle ABP$ we have $\frac{\sin{ \angle BPM}}{\sin{ \angle APM}}=(\frac{\sin{x}}{\sin{y}})^{2}$ (this is $*$). Denote $AP=BP=t$. By sine theorem in triangles $\triangle AQP$, $\triangle QBP$ we have $\sin{\angle APN}=\frac{QA \cdot \sin{y}}{t}$ and $\sin{\angle BPN}=\frac{QB \cdot \sin{x}}{t}$. Seting this in ($*$) we have $\frac{QB}{QA}=\frac{\sin{x}}{\sin{y}}$. However by the power of the point theorem we have that $\frac{QB}{QA}=\frac{QC}{QD}$ therefore  $\frac{\sin{\beta}}{\sin{\alpha}}=\frac{\sin{x}}{\sin{y}}$ and  $\alpha+\beta=x+y$. From this it folows that $x=\beta$ and $y = \alpha$ and points $Q,P$ and $L$ are colinear.

q.e.d.

[quote="superbhn"]Two circles $S_{1}$ and $S_{2}$ intersect at two points $A$ and $B$. Let $Q$ be a point on $S_{1}$. The lines $QA$ and $QB$ meet the circle $S_{2}$ at $C$ and $D$, respectively (apart from $A$ and $B$), while the tangents to $S_{1}$ at $A$ and $B$ meet at $P$. Assume that $Q$ lies outside $S_{2}$, and that $C$ and $D$ lie outside $S_{1}$. Prove that the line $QP$ goes through the midpoint of the segment $CD$.[/quote]

With some triangle geometry, everything is obvious:

The circle $S_{1}$ is the circumcircle of triangle QAB, and the tangents to this circumcircle $S_{1}$ at the vertices A and B intersect each other at the point P. Hence, according to Theorem 1 of [1], the line QP is the Q-symmedian of triangle QAB.

According to [2], Theorem 17, implication $\mathcal{E}_{7}\implies\mathcal{E}_{9}$, we have:

[color=blue][b]Lemma.[/b] Given a triangle ABC and a line g which doesn't pass through its vertex A. Let this line g intersect the lines CA and AB at P and Q, respectively. If there exists a circle which meets the line CA at the points C and P and meets the line AB at the points B and Q, then the midpoint of the segment PQ lies on the A-symmedian of triangle ABC.[/color]

In our problem, consider the triangle QAB and the line CD which intersects the lines AQ and QB at C and D. There exists a circle (namely, the circle $S_{2}$) which meets the line AQ at the points A and C, and meets the line QB at the points B and D. Hence, after the Lemma, the midpoint of the segment CD must lie on the Q-symmedian of triangle QAB. Since we know that this Q-symmedian is the line QP, it follows that the midpoint of the segment CD lies on the line QP. Problem solved.

[b]References[/b]

[1] [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=564645#564645]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=99571 post #2[/url].

[2] Darij Grinberg, [i]Isogonal conjugation with respect to a triangle[/i], can be found at http://www.cip.ifi.lmu.de/~grinberg/geometry2.html and http://www.artofproblemsolving.com/Forum/viewtopic.php?t=18472 .

  darij


algebra
polynomial
modular arithmetic
Euler
function
quadratics
quadratic formula

[hide="#1"]$ x^2\plus{}x\plus{}1\equiv 0\pmod{11}$[/hide]
[hide="#2"]$ x^3\plus{}x\plus{}1\equiv 0\pmod{13}$[/hide]
[hide="#3"]$ x^4\plus{}x^3\plus{}2\equiv 0\pmod{7}$[/hide]
I had rather unorthodox solutions. Maybe I'm missing something?

[hide="Potent First Step"]$ x \equal{} 1$ isn't a solution for the general case unless its in modulo 3, so multiply by $ x \minus{} 1$ to get $ x^3 \equiv 1 , \; x \neq 1 \pmod n$. While this does not directly give the answer, it leads to an important filter, and also makes computation a bit lighter in general.[/hide]

[hide="Filter"]Observe that we must have $ 3| \lambda (n) \implies 3| \varphi(n)$ as a necessary condition, where the two are the Charmichael and Euler-Totient functions, respectively. This eliminates many cases.

For example, in given #1, we have $ \varphi (11) \equal{} 10 \not \equiv 0 \pmod{3}$. Thus, there are no solutions.[/hide]

[hide="Other Two Answers"]$ n\equal{}13 \implies x \equiv 3$ and $ n\equal{}7 \implies x \equiv 2$ by trial and error... I'll post a more relevant solution later if I feel any better[/hide]

I don't see why you have to do anything fancy when you can just test all values 0,1,...,n-1

Until you get the right number of roots.  Even better, once you find one root you can divide out (if you know how to divide polynomials over a finite field, which isn't too bad).

This is for cubic polynomials and higher.  The quadratic formula works for quadratic polynomials $ \bmod p$.  (Can you see why?)


If I'm going to be driving to UTD, to what address should I be going (for mapquest purposes)?
Also, during what time range should I be ariving there?

Titu should have sent you an email about the building location and a campus map, so you can mapquest UTD (probably) and quickly find it

In the mail earlier he sent (included with packing list), it said be there from 3-5 so that you can set up your room and everything