http://www.imomath.com/othercomp/Aut/AutMO07.pdf

first problem, i can not do it...


[quote="silouan"]Do you have solution for this one  or you only created?[/quote]
I think arqady must have the solution.
Because he post it in "Inequalities Proposed & Own Problems " :)

I will a reply from him. We don't know ;)

[quote="silouan"]Do you have solution for this one  or you only created?[/quote]
Yes, I have nice proof!

I found a nice proof for $n\geq1,n\in\mathbb R.$ Sorry.

Of course, It's can be solved by C-S inequality. I found It's true for all $n \ge 0$.

[quote="hungkhtn"]Of course, It's can be solved by C-S inequality. [/quote]
C-S inequality. What is this? Thanks.

[quote="jasamper88"]I think it is Cauchy- Schwarz[/quote]
Thank you, jasamper88.
[quote="hungkhtn"]Of course, It's can be solved by C-S inequality.[/quote]
Yes!

[quote="zhaobin"]
nice,but may you need to remove some $\sum$ ;)[/quote]
Thank you, zhaobin.

The following more general statement holds:
$a>0,b>0,c>0, m>n>0,a^{m+n}+b^{m+n}+c^{m+n}=3.$ Prove that:
$\frac{a^m}{b^n}+\frac{b^m}{c^n}+\frac{c^m}{a^n}\geq3.$


inequalities
trigonometry
inequalities unsolved

[quote="NikolayKaz"]Prove that for
 $ a,b,c \in (0;1) \\
ab \plus{} bc \plus{} ac > abc$
I tried with substitution a=sin^2x but failed. Hope you can help me.[/quote]

$ ab\plus{}ac\plus{}bc > ab >abc$

I think have stronger Inequality
[code]begin{array}{l}
a,b,c \in (01){\rm{W}}ehave \\
 ab + bc + ca > 3abc(*){\rm{u}}\sin g.cauchy - sch{\rm{w}}art - {\rm{w}}ehave \\
 \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \ge \frac{9}{{a + b + c}} > 3 \Leftrightarrow ab + bc + ca > 3abc \\
 \end{array}[/code]

Sounds quite trivial;

$ ab\plus{}bc\plus{}ca > abc \plus{} abc \plus{} abc \equal{} 3abc$


Asymptote
limit
calculus
calculus computations

For each question below, I need the following:  NOTE: f = function.

*determine the number at which the f is discontinuous.
*determine if f has any removable discontinuities
NOTE: What do you they mean by [color=red]removable discontiuities?[/color]
*find vertical asymptotes
*determine the limit of f at a vertical asymptote

Question 1:

f(x) = (x^2-3x-10)/sqrt{x+2}

Question 2:

f(x) = (x^3+x^2-16x-16)/(x^2 -16)

Definition of "removable discontinuity":

$f$ has a removable discontinuity at $a$ when $L=\lim_{x\to a}f(x)$ exists and is finite, but either $f(a)$ is not defined or $f(a)\neq L$.
By defining $g(a)=L$ and $g(x)=f(x)$ for all other $x$, we remove the discontinuity; $g$ is continuous at $a$.

To solve these problems, factor the expressions.


[quote="Steve Adrian"]Date: Thu, 8 Sep 2005 11:36:14 EDT
From: SLADRIAN1@aol.com
To: (list of ARML coaches)
Subject: ARML and NCTM Team up on "Problem of the Month"

After considerable negotiations, ARML now has another interesting
challenge to offer. This is on a trial basis, for 6 months. If it works,
it will be continued. NCTM has agreed to post two ?^?^?ARML problems of
the month?^?^? at the web site
http://www.nctm.org/high/challenge/sept05chall.htm . These have been
written by ARML questions writers and will be graded by them as well.
Please direct any interested person to this web page and encourage them
to participate. And thank Paul Dreyer, Chris Jeuell and P.J. Karafiol for
volunteering to manage this site.

Please help us pass the word.
[/quote]

The link takes me to a page that says that that page does not exist.

Umm...I get a message that the connection was refused.

All right, as a service, I'll post the two September problems here:

[quote][b]September Challenge Problems[/b]

High School

Problem 1.

In trapezoid $TRAP,\,\overline{TR}\, || \,\overline{AP},\,\overline{TA} \perp \overline{PR},\, TR=6,\, TA=19,$ and $PR=180.$
Compute $AP.$

Problem 2.

Let $A$ and $B$ be opposite corners of a unit square, and

let $R=\{P: PA^2+PB^2\le3\}.$ Compute the area of $R.$

Hints will be provided on Sept. 19. Solution and solvers provided Oct. 7

Submit solutions to armlsolutions@gmail.com by [b]October 2, 2005.[/b]

Problems Developed by [url=http://www.arml.com]American Regions Math League[/url][/quote]

[b]DO NOT DISCUSS THESE PROBLEMS HERE. DO NOT POST SOLUTIONS HERE.[/b]


A sequence $ (x_n)$ is given by $ x_1\equal{}2$ and $ nx_n\equal{}2(2n\minus{}1)x_{n\minus{}1}$ for $ n>1$. Prove that $ x_n$ is an integer for every $ n \in \mathbb{N}$.

[quote="moldovan"]A sequence $ (x_n)$ is given by $ x_1 \equal{} 2$ and $ nx_n \equal{} 2(2n \minus{} 1)x_{n \minus{} 1}$ for $ n > 1$. Prove that $ x_n$ is an integer for every $ n \in \mathbb{N}$.[/quote]
We can easily prove that $ n!.x_n \equal{} 2^n.\prod_{k \equal{} 2}^n (2k \minus{} 1)$,so it suffices to show that $ n!|2^n.\prod_{k \equal{} 2}^n (2k \minus{} 1)$ which is true using induction and the fact that every natural number $ n \geq 2$

can be written in the form $ 2^p.(2q \plus{} 1)$ where $ (p,q) \in \mathbb N^2$,$ p \leq n \minus{} 1$ and $ q \leq n \minus{} 2$ which ends the proof.


Problem: Let Triangle $ABC$ be iscoceles with $AB=BC$, and let $D$ be the midpoint of $BC$. Let $E$ be the foot of the perpendicular from $D$ to $AB$ and $F$ the midpoint of $DE$. Prove that $AF$ is perpendicular to $CE$.

Very nice diagram, what program did you make that in?

Btw, either you did some wrong labelling (which I don't think you did) or your diagram is a little messed up and I don't see your reasoning behind EF=FD in the first line of the proof....

[quote="D.P.L"]Problem: Let Triangle $ABC$ be iscoceles with $AB=BC$, and let $D$ be the midpoint of $BC$. Let $E$ be the foot of the perpendicular from $D$ to $AB$ and $F$ the midpoint of $DE$ ... [/quote]I think $AB=AC\ !$

Sorry, it's missing a condition. $F$ is the midpoint of $DE$ :arrow: $DF=EF$ :wink:

BTW is this type of proof valid  :maybe:

I think you may have misinterpreted my intention for this thread. I was hoping someone could check if my proof is valid/ or correct

Sorry to say this, but to me, your proof doesn't seem like a proof at all...

[quote="D.P.L"]Let $\triangle ABN$ be iscoceles and $AF$ be it's altitude. We have $EF=FD$. Consider $AI$ to be the line in which point $E$ is reflected across to $D$. Thus basically $AI$ is the line of symmetry.  Since both $D$ and $H$ are collinear, we can translate $D$ up to $H$. Then we have $EG=GH$, :arrow:  $\measuredangle EAG\equiv\measuredangle GAH$, $GA=AG$. Implying $\triangle EAG\cong%Error. "tirnagle" is a bad command.
GAH$,  :arrow:  $\measuredangle AGE=90$.[/quote]

Why is AF an altitude of triangle ABN? (In fact, it isn't.) What do you mean by "the line in which point E is reflected across to D"? Etc... By the way, this problem belongs not into Intermediate Topics, but rather into Pre-Olympiad or Olympiad Section. It was also posted at http://www.mathlinks.ro/Forum/viewtopic.php?t=43097 .

  darij


Consider the roots of the quad. eqn $ z^{2}\plus{}az\plus{}9\equal{}0$. If $ z_{1}$ and $ z_{2}$ are the roots of this equation and $ 'a'$ is real, draw the locus traced out by the two roots in the complex plane as $ a$ takes on all real values

[quote="bos1234"]Consider the roots of the quad. eqn $ z^{2}\plus{}az\plus{}9 \equal{} 0$. If $ z_{1}$ and $ z_{2}$ are the roots of this equation and $ 'a'$ is real, draw the locus traced out by the two roots in the complex plane as $ a$ takes on all real values[/quote]

You just have to write that $ z\equal{}x\plus{}iy$ and you achieve :
$ x^{2}\minus{}y^{2}\plus{}ax\plus{}9\equal{}0$ and $ y(2x\plus{}a)\equal{}0$

So you have union of two locus :
$ 1).$ $ y\equal{}0$ and $ x^{2}\plus{}ax\plus{}9\equal{}0$ which gives $ y\equal{}0$ and $ x\in(\minus{}\infty,0)\cup(0,\plus{}\infty)$
$ 2.)$ $ x\equal{}\minus{}\frac{a}{2}$ and $ x^{2}\minus{}y^{2}\plus{}ax\plus{}9\equal{}0$, which gives the whole circle $ x^{2}\plus{}y^{2}\equal{}9$


How many integers, n, can be written as a+b+c where a, b, and c are distinct odd integers between 2 and 20?

I really ought to learn C stuff...

I'm guessing  the answer is [hide]84[/hide] by brute force, but I'm not sure.

[quote="Fierytycoon"][quote]I really ought to learn C stuff...[/quote]

This is a math problem, not a programming problem.[/quote]
The other C (the "chooses" one)

nope, this has nothing to do with combinatorics. it's a bit simpler...

Yeah, no combinatorics needed, just use logic.

Tare...your answer is a bit large. Perhaps you would like to explain your solution...

Well [hide]I figured since there's 9 numbers the solution would be 7*1+6*2+5*3+4*4+3*5+2*6+1*7 or 7(7+1)/2+6(6+1)/2+...+2(2+1)/2+1(1+1)/2 which should equal 84...maybe  [/hide]

[hide]

19 is the answer

The smallest is 15, the biggest is 51, and im willing to bet you can get everything in between

[/hide]

yes, mr. southpaw, you are correct, although [hide]i think you meant you could get any odd number in between[/hide]

Tare wrote:well isn't it the combination of [hide]a, b, and c within the set S={3,5,7,9,11,13,15,17,19}?[/hide]

tare...the reason this is incorrect is because [hide]you are counting some possibilities more than once. for example, you count both 3+7+11 and 5+7+9 while they are in fact the same thing[/hide]

[quote="Tare"]Ooooooh, I see. S[size=59]w[/size]o h[size=75]r[/size]o[size=59]i[/size]w d[size=75]t[/size]i[size=59]e[/size]d h[size=75]y[/size]e g[size=59]o[/size]e[size=75]u[/size]t t[size=59]r[/size]h[size=75]p[/size]e a[size=59]r[/size]n[size=75]o[/size]s[size=59]c[/size]w[size=75]e[/size]e[size=59]s[/size]r[size=75]s[/size] again?[size=59]![/size] *interlacing subliminal message* :)[/quote]

Tare...you're supposed to be a moderator.

*sigh* I know, I'm actually doing my job. I mean, I wanted to say "WRITE OUT THE PROCESS FOR THE LAST TIME!!!" but that would sound too harsh.

15 to 51.. 19 odd integers and you can generate all by shifting by two's when necessary [why?  :D ]


I never wrote a research paper before. And I just registered for this talent search thingy.
For one question,how does the solution supposed to look like(The Format) and how many pages or words (usually)?
do u still submit ur solution even if u know the answer is wrong?

Also, do they give you some certificate thingy so that u can show it on ur college application??

Thnaks for the Helpsss!

[quote="rrusczyk"]Look at the problems from past years on the website -- they have sample solutions.  Use those as models for your solutions.

Yes, submit your work even if you haven't finished the problem.  But don't submit scratch work.  Write up your findings neatly, even if you don't think you have completed the problem.

We provide medals for gold, silver, bronze, along with book prizes, and yes, you can list the USAMTS on your college application (but that shouldn't be the sole reason, or even one of the top 3 reasons, you are participating!)[/quote]

but how many medals are u guys giving out?

Read the website  :wink:

[url]http://usamts.org/Rules/U_RulesPrizes.php[/url]

Gold - 96-100
Silver - 75-95
Bronze - 60-74
Honorable Mention - 40-59

There's no "quota" of prizes you get.


Find integers $ a$ and $ b$ such that $ x^2 \minus{}x \minus{}1$ divides
$ a x^{17} \plus{} b x^{16} \plus{} 1$

Suppose that $ ax^{17}+bx^{16}+1=(x^2-x-1)Q(x)$ and let $ \varphi=(1+\sqrt 5)/2$. Verify that $ \varphi^2=\varphi+1$ and the identity $ \varphi^n=F_n\varphi+F_{n-1}$, where $ F_i$ is the $ i$-th Fibonacci number. Therefore $ a\varphi^{17}+b\varphi^{16}+1=0$ is equivalent to
\begin{align}a(F_{17}\varphi+F_{16})+b(F_{16}\varphi+F_{15})+1=0.\end{align}
Using the fact $ A+B\varphi=C+D\varphi\Leftrightarrow A=C, B=D$, where $ A,B,C,D$ are integers, which again follows from the fact that $ \varphi$ is irrational, $ (1)$ is equivalent to the following system
\[ aF_{17}+bF_{16}=0\\
aF_{16}+bF_{15}+1=0.\]
From the first equation, we get $ a=kF_{16},b=-kF_{17}$, for some integer $ k$. This, from the latter equation implies, $ k(F_{16}^2-F_{17}F_{15})=-1$. But from [url=http://en.wikipedia.org/wiki/Cassini%27s_identity]Cassini's identity[/url], $ F_{16}^2-F_{17}F_{15}=-1$. Hence $ k=1$ and we finally conclude that $ a=F_{16}=987,b=-F_{17}=-1597$.


geometry
MATHCOUNTS
Pythagorean Theorem
number theory
prime factorization

given a segment of length x, construct with standard geometry tools a segment of length $ x^2$.

given a segment of length x, construct with standard geometry tools a segment of length $ \sqrt x$.

 :D

i've seen those problems in a geometry textbook before, so it should be possible.  :)

It is possible if you have also have a segment of length 1.

[hide="/sqrt{x}"]Extend the segment with length 1.

Use compass to Also have a segment with length x next to it like:

_.____

Where the first one is the segment wiht length 1 and the second on eis segment with length 2.

Let the segment with length 1 be AB and the other segment be BC.

Construct the cirlce iwth diameter AC.

Then draw a perpendicular at B to AC, and let where it interesects the circle be D.

Then BD has length /sqrt{x}

 [geogebra]0949529ccf56e24a0e98123c59b778686f1139c2[/geogebra]

Why?

because ABD is similar to DBC, so DB^2 = 1*x.

.[/hide]

ANd it is not possible if you don't

[hide="hint"]Pythagorean theorem [/hide]

[size=75][color=indigo]LadyKn1ght: It's now hidden. When you're using hide tags, make sure to type "hide=..." not "hide:..." [/color][/size]

That's what I was thinking when I first saw it, Poincare. I just didn't know how to apply it to the actual problem.

Poincare, would u just stop with the hints and hiding and all that and just give the solution? LOL. Thx.  :)

Do the opposite to construct x^2

So...

You have x, double it, and perpendicular-bisect, mark 1 away from the segment on the bisector. Now construct a circle that intersects all 3 points.

define 1 and x. Blue segment is x^2.

[geogebra]33b15158afd712bfc05cb038c6fd219aef4f07bc[/geogebra]

[geogebra]1175f397368b0ed84010430fcee165a1158d3b46[/geogebra]
This can be generalized.
In the diagram, $ \angle{C}$ is the right angle of triangle $ \Delta{ABC}$, with perpendicular $ \overline{CD}$. Let $ AD\equal{}m$, $ BC\equal{}n$, so that $ CD\equal{}\sqrt{mn}$.
Doing blablabla we can see that we can find $ \sqrt{x}$, $ x^2$, etc. by randomly plugging in values for any 2 of $ AD,BC,CD$ and constructing the other side.

[geogebra]d125cd2067274080af14797565591f97051bcd48[/geogebra]

red = X/Y
green = Y/X
blue = X*Y

The power of "Power of Point"

[quote="toadoncart"].....

okay, well, i'm doing this old mathcounts nats problem right now. i know how to do it, but i forgot the trick for it.

what was that trick that helps u find the sum of the factors of some number x? (it has to do w/ prime factorization)
is there a trick for finding the sum of the "prime" factors of some number x?
compute $ 2^1 \plus{} 2^3 \plus{} 2^5 \plus{} 2^7 \plus{} 2^9$ without computing indivual exponent values.

Thx!  :)

(edited): thanks for the solution, james4l![/quote]

Trick for finding sum of factors:

Prime factorize it.

If it prime factorizes to $ a^bc^d$ (example), then the sum of the factors would be $ (a^0\plus{}a^1\plus{}...a^b)(c^0\plus{}c^1\plus{}...c^d)$.

I believe there is no trick for finding the sum of the prime factors of some number $ x$.

You don't really need a trick for adding prime factors - it's pretty straight forward adding once you prime factorize.

what about the $ 2^1 \plus{} 2^3 \plus{} 2^5 \plus{} 2^7 \plus{} 2^9$ question without expanding the exponents indivually?

what if u got like 72, and the question says "find the sum of the prime factors of 72.

so i prime factorize it and get $ 2^3 3^2$ Would i add 2+2+2+3+3 or just $ 2^3 \plus{} 3^2$?  :)

So we want to simplify $ 2 \plus{} 2^3 \plus{} 2^5 \plus{} 2^7 \plus{} 2^9$.

Note that to get from $ 2^a$ to $ 2^b$, we multiply by 2(if $ a$ and $ b$ differ by 1, and $ b$ is greater.) Make sure you see why!

So from $ 2$ to $ 2^2$, we multiply by 2. Then from $ 2^2$ to $ 2^3$, we multiply by 2 again. So to get from $ 2$ to $ 2^3$, we multiply by 4.

Now, to get from $ 2^3$ to $ 2^5$, we multiply by 2 twice(from $ 2^3$ to $ 2^4$, and then $ 2^4$ to $ 2^5$.) And we already found that from $ 2$ to $ 2^3$ we multiply by 4, so from $ 2$ to $ 2^5$, we multiply by $ 4 \times 4 \equal{} 16$. Do this for each term. Notice anything? To get each number, we multiplied by a power of four. First $ 4^0$, then $ 4^1$, and so on.

We can now write this expression in terms of $ 2$.

$ 2 \plus{} 2(4) \plus{} 2(16) \plus{} 2(64) \plus{} 2(256)$

Factor.

$ 2(1 \plus{} 4 \plus{} 16 \plus{} 64 \plus{} 256)$

We easily see this is $ \boxed{682}$.

oh yeah! that 2^2 rate reminds me of geometric series. so u could also do $ \frac {a_1 (r^n \minus{} 1)}{r \minus{} 1}$=$ \frac {2(4^5 \minus{} 1)}{4 \minus{} 1}$=$ \frac {2(1023)}{3}$=$ 2(341)$=$ 682$. lol. didn't think of that.  :)


calculus
integration
logarithms
trigonometry
calculus computations

Compute

$ I \equal{} \int_0^{\pi/2}\ln(\sin x) \cdot \ln(\cos x)\,dx$.


Find all functions $f : R \longrightarrow R$ satisfying
 the equation

 $f(x+y)+f(x)f(y) = f(xy)+2xy+1$,
 for all real numbers $x$ and $y$ .


n points A_1, A_2,...,A_n are chosen inside a unit square.
Find the probability of the situation when all points form a convex polygon.
The same for a unit circle.

Нет, не знаю:-)))

I'm sorry, it's in Russian.
I don't know any solution for this problem.
Moreover, I don't know from what side to start solving it...

In my opinion it is sufficient to know P(SABC \leq x) for all real x where A, B, C in unit square (circle).

[color=red][[ Da, mne do sih por interesno kak ona reshaetsja???!!! ]][/color]
Please edit this sentence and translate it into English. If you wish to discuss in another language, please do so in the [url=http://www.mathlinks.ro/viewforum.php?f=27]Babylon Tower[/url]

Admin Edit.

Similar problem were discussed in some old Kvant (199x).

That one seems pretty hard.  An easier one to start with might be this:  A stick of arbitrary length suddenly breaks into n pieces.  What is the probability the pieces form an n-gon (i.e. the probability that one piece is not longer than half of the stick)?

Alex

P.S.  I actually know the answer to this one!  And I think the one I know might relate to the above problem, I just figured I'd share this one before working out the other, because it seems rather tricky.


geometry
circumcircle
power of a point
radical axis
geometry unsolved

Let $ ABC$ be a triangle with circumcircle $ \omega$ . Point $ D$ lies on side $ BC$ such that $ \angle BAD \equal{} \angle CAD$ . Let $ I_A$ denote the excenter of triangle $ ABC$ opposite $ A$, and let $ \omega_A$ denote the circle with $ AI_A$ as its diameter. Cricles $ \omega$ and $ \omega_A$ meet at $ P$ other than $ A$. The circumcle of triangle $ APD$ meet line $ BC$ again at $ Q$ (other than $ D$). Prove that $ Q$ lies the excircle of triangle $ ABC$ opposite $ A$.

Let $ (I_a)$ be the excircle wrt vertices $ A$ of $ \triangle ABC$. Let $ A_0,B_0,C_0$ be the tangency point of $ (I_a)$ with $ BC,CA,AB$ respectively. Let $ (I)$ be the incircle of $ \triangle ABC$. $ AI_a\cap \omega\equiv K$, where $ K$ is defferent form $ A$. $ KA_0\cap \omega\equiv M$, where $ M$ is different from $ K$. First we note that $ K$ is the midpoint of the smaller arc $ BC$, it follows then $ KI \equal{} KB \equal{} KC \equal{} KI_a$, which implies $ K$ is the midpoint of $ II_a$. But $ (ADII_a) \equal{} \minus{} 1\Longrightarrow KI^2 \equal{} {KI_{a}}^2 \equal{} \overline {KD}\cdot \overline {KA}$. Now, for any point $ G\in [BC]$ such that $ KG\cap \omega\equiv J$, we always have $ \angle KJC \equal{} \angle KJB \equal{} \angle KCB\equiv \angle KCG$. Thus, it leads to the fact that $ KC$ is the tangent of $ K$ to $ (JGC)$, which implies that $ KC^2 \equal{} \overline {KG}\cdot \overline {KJ}$. Hence, for $ G\equiv A_0\Longrightarrow KC^2 \equal{} \overline {KA_0}\cdot \overline {KM}$. Therefore, $ \overline {KA_0}\cdot \overline {KM} \equal{} \overline {KD}\cdot \overline {KA}$, which implies that $ A,D,A_0,M$ are concyclic.

Now, since $ {KI_a}^2 \equal{} \overline {KA_0}\cdot \overline {KM}\Longrightarrow \angle DI_aA_0\equiv \angle KI_aA_0 \equal{} \angle A_0MI_a$. But $ \angle I_aDA_0 \equal{} \angle AMA_0$, $ \angle DI_aA_0 \plus{} \angle I_aDA_0 \equal{} 90^{\circ}\Longrightarrow \angle AMI_a \equal{} 90^{\circ}$. Hence $ M\in \omega_{a}$ . Thus, $ AM$ is the radical axis of $ \omega_{a},\omega,(ADA_0M)$, which implies $ M\equiv P$. The result is lead as follow.
Our proof is completed then.


Posted before $ a\equal{}\pm 1,x\equal{}0$.


summer program
Mathcamp
PROMYS
HCSSiM
Ross Mathematics Program
ARML
number theory

hi..

im wondering if anyone knows of any summer math programs, similar to Mathcamp, promys, etc.. that aren't as expensive as those.

thanks

As far as I know, all such camps are in that same general expense range ($2000, give or take several hundred dollars).  Hampshire might be a little cheaper, but no promises.  But to the best of my knowledge, all such programs have good financial aid.  You might look that part up.

Summer Math Programs for High School Students - http://www.ams.org/employment/mathcamps.html

yea, I know that the ROSS program gives out need based financial aid.

there's one in San Marcos, Texas called Honors Math Camp that is significantly cheaper because they receive some sort of external funding, I think from the company that makes Mathematica. Particpants get a free copy of Mathematica and spend some time at the camp learning to use it. Can't give you figures or a link off the top of my head, but it shouldn't be too hard to find their website. Many of the Texas ARML team members go, and I have a generally favorable impression although no personal experience. They are trying to increase their out-of-state participants and eventually be as well known as PROMYS et al, but the program is currently fairly new (4-5 years).

massachusetts (or maybe somewhere else) has one and it is only 500 dollars. however that is only for residents. it is about number theory and you get Math 3900 credit.

i planning to do that and calc 3 over the summer.

there are also some restricted to girls or minorities that are subsidized. Does that apply to you?

http://www.sciserv.org/stp/
This is a very good site if you want to find math or science programs. you can choose a program by what state you want, grade level, financial assistance, and other things. Some of the programs listed there are even free! Just follow the link at the bottom of a program information page to go to the official webpage because the ones there aren't always up to date. I hope this helps!

My friend was pretty poor, and he was living in a one bedroom apartment with his entire family.  He got all expense paid to go to HCSSiM... If that gives you any inspiration.

yea, one of my friends got full scholarship to the ROSS program because he did not have enough money.

I got a full scholarship to MMSS last year. I'm hoping  for one this year too.....


A pyramid $ OABC$ has equilateral triangle $ ABC$ as the base. If $ AB$=4, $ OA\equal{}OB\equal{}OC$, and the distance between line $ OA$ and line $ BC$ is $ \sqrt3$, find the length of $ OA$.

[quote="joshisunny16"]jelly, is the distance perpendicular or along AC or something else??[/quote]

Line $ OA$ and line $ BC$ are [b]skew[/b] lines. They don't have the same orientation and never intersect. If a segment is perpendicular to both $ OA$ and $ BC$, the length of this segment is defined to be the distance between $ OA$ and $ BC$.


For what value of n is 1/2+1/3+1/4+...+1/n an integer?

There is no such value.
[hide="Hint"]Write every number $ m\in \{2,3,...,n\}$ in the form $ m \equal{} 2^k\cdot l$ and prove that there is exactly one number with max k[/hide]

This is one of those problems that appears on the forums over and over and over again.  I think professors (or books, or whatever) are a little too fond of this fact.  See:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=37336
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=52930
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=212837
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=176128
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=162314
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=98535
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=40893

Goodness, I've posted in too many of those threads... well, let me tag this post to make it easier to reference in the future:  [i]the partial sums of the harmonic series are never an integer.[/i]


[quote="freemind"]In triangle $ ABC$ the bisector of $ \angle ACB$ intersects $ AB$ at $ D$. Consider an arbitrary circle $ O$ passing through $ C$ and $ D$, so that it is not tangent to $ BC$ or $ CA$. Let $ O\cap BC \equal{} \{M\}$ and $ O\cap CA \equal{} \{N\}$.
a) Prove that there is a circle $ S$ so that $ DM$ and $ DN$ are tangent to $ S$ in $ M$ and $ N$, respectively.
b) Circle $ S$ intersects lines $ BC$ and $ CA$ in $ P$ and $ Q$ respectively. Prove that the lengths of $ MP$ and $ NQ$ do not depend on the choice of circle $ O$.[/quote]
$ MP\equal{}NQ\equal{}2.DC.cos\frac{C}{2}$
Am i wrong?  :maybe:

Denote by $ X$ the intersection point of the perpendiculars in $ M$, $ N$ to the lines $ DM$, and $ DN$, respectively. Since the quadrilaterals $ DMXN$ and $ MCND$ are cyclic, we have that $ \angle{MXD} \equal{} \angle{MCD} \equal{} \angle{C}/2 \equal{} \angle{DCN} \equal{} \angle{DXN}$. The triangles $ DMX$ and $ DNX$ are now congruent, and therefore the segments $ XM$, and $ XD$ are equal. Thus, the circle with center $ X$ and radius $ XM \equal{} XD$ is tangent to the lines $ DM$, $ DN$ at points $ M$, $ N$, respectively. This proves a).

Next, notice that the triangles $ MXC$ and $ QXC$ are congruent (this is because $ XM \equal{} XQ$, $ \angle{XMC} \equal{} \angle{XNC} \equal{} \angle{XQC}$, and $ XQ$ is common), and thus $ MC \equal{} QC$. Similarly, the triangles $ XCP$ and $ XCN$ are congruent, and hereby $ CP \equal{} CN$. Since $ MP \equal{} NQ$, it is suffice to prove that $ MC \plus{} CN$ is constant. But this follows from Ptolemy's theorem, applied to the quadrilateral $ MCND$, since
\[ MC \plus{} CN \equal{} CD \cdot \frac {MN}{MD}
\]
(we know that $ MD \equal{} DN$ from the congruence of the triangles $ DMX$ and $ DNX$), and $ \frac {MN}{MD} \equal{} 2\cos{\frac {C}{2}}$. This proves b).


How many different integers can be written as the sum of two
distinct members of the set $ \{1, 3, 5, 7, 9, 11, 13\}$ ?

We can make only even numbers (odd + odd is even).  The largest we can make is $ 11\plus{}13\equal{}24$ and smallest we can make is $ 1\plus{}3\equal{}4$, so there are a total of $ \frac{24\minus{}4}{2}\plus{}1\equal{}\boxed{11}$ numbers we can make.


geometry
trapezoid
trigonometry
rhombus
circumcircle
perpendicular bisector
geometry proposed

Given a circle $\Gamma$ and the positive numbers $h$ and $m$,  construct with straight edge and compass a trapezoid inscribed in $\Gamma$, such that it has altitude  $h$ and the sum of its parallel sides is $m$.

Suppose that we have constructed the isosceles trapezoid $ABCD$ with $AB \parallel CD$

Let $\theta = \angle CAB$ and $k=\frac{m}{2}$

Let $C'$ be the projection of $C$ to the line $AB$. Then $AC' = \frac{AB}{2}+\frac{CD}{2}= k$

From the right triangle $AC'C$ we have $\tan\theta = \frac{CC'}{AC'}= \frac{h}{k}$

[size=125][color=blue]Construction[/color][/size]
Let $\Gamma = (O,R)$ be the given circle
We construct a right triangle with two perpendicular sides $h$ and $k$ and hypotenuse $w$.
Then the angle between $k$ and $w$ is $\theta$
We construct an arc $BC$ on the circle such that $\frac{BC}{2R}= \frac{h}{w}$. In other words it is $BC=2R\cdot\sin\theta$

We construct a triangle $BC'C$, right at $C'$, with $CC'=h \mbox{ }^{\star(1)}$

We extend $BC'$ to meet the circle at $A$
Through $C$ we bring a parallel to $AB$  which meets the circle at $D\mbox{ }^{\star(2)}$

[size=125][color=blue]Proof[/color][/size]
If the above constructed quadrilateral $ABCD$ is convex, then it is a trapezoid with altitude $h$, it is inscribed in $\Gamma$  and $\angle BAC$ satisfies the property $\tan (\angle BAC) = \frac{h}{k}$, so we have $AB+CD=2k$

[size=125][color=blue]Discussion[/color][/size]
$(1)$st condition $h<2R\cdot \sin\theta \iff h<2R\cdot \frac{h}{w}\iff \boxed{w<2R}$

$(2)$nd condition
The quadrilateral $ABCD$ must be convex. So, the points $A,D$ must be on the same side of $BC$
So, if $d$ is the perpendicular bisector of $AB$ then $C$ have to be in the same half-plane with $B$

So $AC>BC\iff$

$\angle{ABC}> \angle{BAC}\iff$

$\sin(\angle ABC) > \sin(\angle BAC) \iff$

$\frac{CC'}{BC}> \sin\theta \iff$

$\frac{h}{BC}> \sin\theta \iff$

$h>2R\cdot\sin^{2}\theta \iff$

$h>2R\frac{h^{2}}{w^{2}}\iff$

$\boxed{w^{2}>2Rh}$

[color=brown][b]Note[/b]
Another way to construct the point $A$, after finding the arc $BC$ is to use the fact that $BA=w$[/color]

Let  $ K,L,M,N$ be the midpoints of  $ AB,BC,CD,DA.$ It is clear that  $ ABCD$ is an isosceles trapezoid $ \Longrightarrow$ $ KLMN$ is a rhombus with known diagonals $ KM \equal{} h$ and $ LN \equal{} \frac {_1}{^2}(AB \plus{} CD) \equal{} \frac {_1}{^2}m.$ The construction of  a  rhombus congruent to $ KLMN$ is immediate and then we obtain the measure of the diagonal $ AC$ which is twice the length of its side. Triangle $ \triangle ACD$ with side length $ AC \equal{} 2KL,$ altitude $ h$ onto $ DC$ and circumcircle $ \Gamma$ is constructible, this is: Fix the chord $ AC$ in $ \Gamma$ and draw the circumference with diameter $ AC.$ Circumference centered at  $ A$ with radius $ h$ cuts the circumference with diameter  $ AC$ at the orthogonal projection $ H$ of $ A$ onto $ DC,$ then ray $ CH$ cuts $ \Gamma$ at $ D.$ The parallel line to $ CD$ passing through $ A$ cuts $ \Gamma$ again at  $ B,$ which completes the trapezoid.


geometry
trigonometry
incenter
perimeter
area of a triangle
Heron\u0027s formula
angle bisector

P is a point on the side BC of triangle ABC
r,u and v are the inradii of the triangles ABC,ACP and ABP respectively

Prove that 1/u +1/v -r/uv =2/h where h is the altitude from A to BC

Denote the sides, angles and the semiperimeter of the triangle $\triangle ABC$ as usual: $a = BC, b = CA, c = AB$, $\alpha = \measuredangle BAC, \beta = \measuredangle ABC, \gamma = \measuredangle BCA$, $s = \frac{a + b + c}{2}$. Combining Heron's formula and the usual formula for the area $\triangle$ of a triangle $\triangle ABC$,

$\triangle^2 = \frac 1 4 b^2 c^2 \sin^2{\alpha} = s(s - a)(s - b)(s - c)$

Using cosine theorem and substituting

$s(s - a) = \frac 1 4 (b^2 + 2bc + c^2 - a^2) = \frac 1 4 (2bc + 2bc \cos \alpha) = bc \cos^2{\frac \alpha 2}$

$\frac{(s - b)(s - c)}{bc} = \frac{\sin^2{\alpha}}{4 \cos^2{\frac \alpha 2}} = sin^2{\frac \alpha 2}$

Let $I, U, V$ be the incenters and $r, u, v$ the inradii of the triangles $\triangle ABC, \triangle ABP, \triangle APC$ and $h = AD$ their common altitude from the common vertex $A$ to the line $AB \equiv BP \equiv PC$. Obviously,

$r = (s - b) \tan{\frac \beta 2} = (s - c) \tan{\frac \gamma 2} = \sqrt{(s - b)(s - c) \tan{\frac \beta 2} \tan{\frac \gamma 2}}$

$r = AI \sin{\frac \alpha 2}$

Combining these equations,

$AI  =  \frac{\sqrt{(s - b)(s - c) \tan{\frac \beta 2} \tan{\frac \gamma 2}}}{\sin{\frac \alpha 2}} = \sqrt{bc \tan{\frac \beta 2} \tan{\frac \gamma 2}}$

The length of the altitude $h = AD$ is

$h = b \sin \gamma = c \sin \beta = \sqrt{bc \sin \beta \sin \gamma}$

$\frac{h}{AI} =  \sqrt{\frac{\sin \beta \sin \gamma}{\tan{\frac \beta 2} \tan{\frac \gamma 2}}} = 2 \cos{\frac \beta 2} \cos {\frac \gamma 2}$

WLOG, assume that $\beta > \gamma$. The angle $\measuredangle DAI$ between the altitude $AD$ and the angle bisector $AI$ is equal to

$\measuredangle DAI = 90^o - [ 180^o - ( \frac \alpha 2 + \beta ) ] = \frac{\beta - \gamma}{2}$

Hence, the altitude $h = AD$ is also equal to

$h = r + AI \cos{\frac{\beta - \gamma}{2}} = r + h\ \frac{\cos{\frac{\beta - \gamma}{2}}}{2 \cos{\frac \beta 2} \cos {\frac \gamma 2}} = r + \frac h 2 (1 + \tan{\frac \beta 2} \tan{\frac \gamma 2})$

$r = \frac h 2 (1 - \tan{\frac \beta 2} \tan{\frac \gamma 2})$

Denote $\theta = \measuredangle APB$. We can get similar formulas for the inradii $u, v$ of the triangles $\triangle ABP, \triangle APC$, because $h$ is the common altitude of these triangles and of the triangle $\triangle ABC$:

$u = \frac h 2 (1 - \tan{\frac \beta 2} \tan{\frac \theta 2})$

$v = \frac h 2 (1 - \tan{\frac{180^o - \theta}{2}} \tan{\frac \gamma 2} ) = \frac h 2 (1 - \frac{\tan{\frac \gamma 2}}{\tan{\frac \theta 2}})$

Multiplying the last two equations, we get

${uv = \frac{h^2}{4} (1 - \tan{\frac \beta 2} \tan{\frac \theta 2} -  \frac{\tan{\frac \gamma 2}}{\tan{\frac \theta 2}} + \tan{\frac \beta 2} \tan{\frac \gamma 2}}) =$

$= \frac{h^2}{4} (\frac{2u}{h} + \frac{2v}{h} - \frac{2r}{h}) = \frac h 2 (u + v - r)$

Dividing this equation by the product $\frac 1 2 uvh$ yields the desired result:

$\frac 2 h = \frac 1 u + \frac 1 v - \frac{r}{uv}$

[quote="pestich"]   Maybe someone  here can put more substance on both physics and projective geometry, for it is optics where they come so very close.

Lt. Pestich[/quote]

Lieutenant, you certainly can try to put more substance into any subject you wish, if you think it is lacking, but then you should do exactly that. This is quite different from putting vacuity into a subject, in fact, these two are mutually exclusive. We went through this, did not we?

[quote="pestich"] Maybe someone  here can put more substance on both physics and projective geometry, for it is optics where they come so very close.

Lt. Pestich[/quote]

I have to put a cog in these projective geometry attempts. Pity I did not quote the entire "solution" before it disappeared, it was exquisite.

Denote a = BC, b = CA, c = AB the sides and $p, s, \triangle$ the perimeter, semiperimeter and area of the triangle $\triangle ABC$. Furthermore, denote $a_1 = BP, a_2 = CP, d = AP$ and let $p_1, s_1, \triangle_1$ be the perimeter, semiperimeter and area of the triangle $\triangle ABP$ and likewise, $p_2, s_2, \triangle_2$ the perimeter, semiperimeter and area of the triangle $\triangle ACP$. The common A-altitude of the triangles $\triangle ABC, \triangle ABP, \triangle ACP$ is h and their inradii r, u, v, respectively. We have to show

$(?)\ \ \ \frac 2 h = \frac 1 u + \frac 1 v - \frac{r}{uv}$

The inradii just obscure the problem, they can be replaced by $\frac 1 u = \frac{s_1}{\triangle_1},$ $\frac{1}{v} = \frac{s_2}{\triangle_2},$ $r = \frac{\triangle}{s}$. Then we have to show

$(?)\ \ \ \frac 2 h = \frac{s_1}{\triangle_1} + \frac{s_2}{\triangle_2} - \frac{\triangle s_1 s_2}{\triangle_1 \triangle_2 s}$

$(?)\ \ \ \frac{2 \triangle_1 \triangle_2}{h^2} = \frac{s_1 \triangle_2}{h} + \frac{s_2 \triangle_1}{h} - \frac{\triangle s_1 s_2}{sh}$

Substituting $\frac{\triangle}{h} = \frac a 2 = \frac{a_1 + a_2}{2},$ $\frac{\triangle_1}{h} = \frac{a_1}{2},$ ${\frac{\triangle_2}{h} = \frac{a_2}{2}}$ leads to

$(?)\ \ \ a_1 a_2 = s_1 a_2 + s_2 a_1 - \frac{s_1 s_2 a}{s}$

This was only arithmetic and the rest is also only arithmetic. Substituting $a = a_1 + a_2$ and $s = s_1 + s_2 - d$, we get

$(?)\ \ \ a_1 a_2 = \frac{s_1 a_2 (s_2 + s_2) + s_2 a_1 (s_1 + s_2) - d(s_1 a_2 + s_2 a_1) - s_1 s_2 (a_1 + a_2)}{s}$

$(?)\ \ \ a_1 a_2 = \frac{s_1^2 a_2 + s_2^2 a_1 - d(s_1 a_2 + s_2 a_1)}{s}$

$(?)\ \ \ a_1 a_2 = \frac{s_1 (s_1 - d) a_2 + s_2 (s_2 - d) a_1}{s}$

Switching from semiperimeters to perimeters and substituting $p_1 (p_1 - 2d) = (c + a_1)^2 - d^2,$ $p_2(p_2 - 2d) = (b + a_2)^2 - d^2$,

$(?)\ \ \ 2 a_1 a_2 = \frac{[(c + a_1)^2 - d^2] a_2 + [(b + a_2)^2 - d^2] a_1}{p}$

$(?)\ \ \ 2 a_1 a_2 = \frac{2(c + b)a_1 a_2 + (c^2 + a_1^2 - d^2) a_2 + (b^2 + a_2^2 - d^2) a_1}{p}$

Denote $\theta = \angle APB$. Since $\angle APC = 180^\circ - \theta$, $\cos \widehat{APC} = -\cos \theta$. By the cosine theorem for the triangles $\triangle ABP, \triangle ACP$, we have

$c^2 - d^2 = a_1^2 - 2d a_1 \cos \theta,\ \ b^2 - d^2 = a_2^2 + 2d a_2 \cos \theta$

Substituting this to the last equation in question, we arrive to

$(?)\ \ \ 2 a_1 a_2 = \frac{2(c + b)a_1 a_2 + (2a_1^2 - 2d a_1 \cos \theta) a_2 + (2a_2^2 + 2d a_2 \cos \theta) a_1}{p}$

$(?)\ \ \ 2 a_1 a_2 = \frac{2(c + b)a_1 a_2 + 2(a_1 + a_2) a_1 a_2}{p}$

$(?)\ \ \ 2 a_1 a_2 = \frac{2p a_1 a_2}{p}$

which is an obvious equality. If someone claimed that the "forces" forcing the original relation are  [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=60833]strong[/url], then I would say yes.

Yetti

[i]Yetti[/i] and [i]M. Niland[/i], see the my proposed problem at
http://www.mathlinks.ro/Forum/viewtopic.php?t=50559

Because $2sr=ah_a$ results: $sr^2+ar_1r_2=sr(r_1+r_2)\Longleftrightarrow$

$h_a r+2r_1r_2=h_a(r_1+r_2)\Longleftrightarrow\frac{1}{r_1}+\frac{1}{r_2}=\frac{r}{r_1r_2}+\frac{2}{h_a}.$

Check this link: http://www.mathlinks.ro/resources.php?c=1&cid=155&year=2009&sid=8d659ee5c2b0426d6c54ca448a783a26
AFAIK, this is not a Zhautykov Olympiad problem.
I'll post the problem later, since I'm not holding the paper :D


function
induction
algebra unsolved
algebra

Let $ n \equal{} 2k \minus{} 1$ where $ k \geq 6$ is an integer. Let $ T$ be the set of all $ n\minus{}$tuples $ (x_1, x_2, \ldots, x_n)$ where $ x_i \in \{0,1\}$ $ \forall i \equal{} \{1,2, \ldots, n\}$ For $ x \equal{} (x_1, x_2, \ldots, x_n) \in T$ and  $ y \equal{} (y_1, y_2, \ldots, y_n) \in T$ let $ d(x,y)$ denote the number of integers $ j$ with $ 1 \leq j \leq n$ such that $ x_i \neq x_j$, in particular $ d(x,x) \equal{} 0.$  Suppose that there exists a subset $ S$ of $ T$ with $ 2^k$ elements that has the following property: Given any element $ x \in T,$  there is a unique element $ y \in S$ with $ d(x, y) \leq 3.$ Prove that $ n \equal{} 23.$

Of course, the typo "such that $x_i \neq x_j$" needs be corrected to "such that $x_j \neq y_j$". This is the Hamming distance between binary words.

Each element of $S$ is at distance at most $3$ from $N = 1 + \binom {n} {1} + \binom {n} {2} + \binom {n} {3}$ elements of $T$. Since $|T| = 2^n$, and given the conditions, we must have $|T| = N|S|$, which translates into $f(k) = 3\cdot 2{k-2} - k(2k^2 - 3k + 4) = 0$. The function $f(k)$ is strictly increasing from $k=12$ on (prove it by induction), yields $f(12)=0$, and is easily checked to have negative values for $6\leq k \leq 11$. Therefore $k=12$, so $n=23$, is the only solution.

In principle, we should also provide a model of such a set $S$; leave this open for the moment ...


Have you guys noticed that #26 and #29 on the HMMT Guts are wrong.

I frankly think the answers are 2009 and 139 respectively.

Did anyone else get this.

The set of 10 primes they give you in the solutions clearly shows that the answer to #29 is 141.

Can you tell us your reasoning?

[quote="azjps"]$ 1 \plus{} 4 \plus{} 1 \equal{} 6 \Longrightarrow 3|141$  :P[/quote]

Couldn't have said it better myself.

And for #29, if you look at the sigma notation, tehy get rid of the 2009 with no reason.  So it should be 2009*1=2009.

[quote="azjps"]$ 1 \plus{} 4 \plus{} 1 \equal{} 6 \Longrightarrow 3|141$  :P[/quote]

 :oops:  wow my math is becoming primitive

I really think they should let the students see the solutions to the guts round before the awards ceremony. There's no reason to have the ceremony 20 minutes after guts.

Did they really grade us according to these answers? That's too bad! :(

As the creator of the tests this year, I apologize for the errors on problems 26 and 29.

Luckily, these problems did not drastically change the scores of any team upon throwing them out.  Next year, we will be changing our scoring system so that we hang on to all of the submitted answer sheets during the Guts round, and so we can re-grade if a problem like this comes up again.

[quote="worthawholebean"][quote]I really think they should let the students see the solutions to the guts round before the awards ceremony. There's no reason to have the ceremony 20 minutes after guts.[/quote]Yes, there is - you don't want to keep people until all hours of the night, and you don't want downtime where people aren't doing anything (those who were at PUMaC know what I mean).[/quote]
LOLOLOL


If $x,y,z > 0$ then prove that:
$(x^2y^2+y^2z^2+z^2x^2) + xyz(x+y+z)+4(x^4+y^4+z^4)$
$\geq 12xyz\left( \frac{x^3}{x^2+yz}+\frac{y^3}{y^2+zx}+\frac{z^3}{z^2+xy} \right)$
(sorry it's too long)

$1=\frac{x^2 +yz}{x^2 +yz}\geq 2\frac{x\sqrt{yz}}{x^2 +yz}$
and
$4x^4 +y^2 z^2 +x^2 yz\geq 6x^3 \sqrt{yz}$
and the result follows! ;)


AB/_ _ _ _ is equal to BA. there is no remainder. what is A+B

This post probably doesn't make any sense (SORRY) I don't have LaTeX yet. I really need help with this, and I have the picture. You can look it up in George Lenchner's MOEMS book it's the one with the funky blue and white design on it.(Olympiad 24) I am not very deep into algebra so I don't know how to do this kind of stuff.

thanks

You don't need to "have LaTeX," just write the code and put it in dollar signs.

I don't understand the question.  Are you saying there's a four-digit number $wxyz$ and a two-digit number $AB$, and $\frac{AB}{wxyz} = BA$?  Because this is plainly impossible.  :(

its in george lenchner's olympiad book, the blue and white one, Olympiad 24, prob.3.

Why don't you just tell us what the problem means (I don't think many people have that book).  :?

In the division example at the [bottom] A and B represent different digits. What is the sum of A and B if he remainder is zero?

 [code]  ____[u]B A[/u]
AB)_ _ _ _
     [u]_ _ _[/u]
           _ _ 1
           [u]_ _ _[/u][/code]

THe three lines on the second row are supposed to line up wiht the three lines on the first row, the two lines on the third row are supposed to line up with the 2nd and 3rd lines on the second row, and the first two lines of the last row are supposed to line up with the third row's lines. Sorry if it doesn't make any sense

i maybe understood the question wrong but....
ab/___ = ba/1
so isnt ___ equal to 1? ... maybe theres something im missing...

to answer glorfs, its AB where A is the tens digit and B is the ones digit.
so, basically from what i understand,
AB/x = BA  where x = the four digit number
AB = xBA
x = AB/BA
think about this. How is a 2 digit number divided by another 2 digit number equal a four digit number? :?:

[quote="nonie"]here's the problem[/quote]

[hide="Click to reveal hidden text"]Let a and b be digits.

So $(10a+b)(10b+a)=xyz1$

$10a^2+101ab+10b^2=xyz1$

From looking at that, you realize regardless of what a and b are, $a^2$ and $b^2$'s one's digits will each be 0, and in order to satisfy the thing, $101ab$ must have a one's digit 1. Therefore $ab$ has a one's digit of 1.

Setting ab = 1 causes the total to be a 3 digit number.
Setting ab = 11 means either a or b is not a 2 digit number as 11 is prime. The same applies to 31 and 41.

Therefore $ab=21$

$21=7*3$ so $a+b=7+3=\boxed{10}$[/hide]

Answer: $37*73=2701$, so the answer is $10$

Solution:  The one is carried down.  This means that the last digit of the dividend

would be a $1$.  The only distinct last digits who have a product of last digit $1$

would be a $3$ and a $7$.  Therefore, $-3*-7=---1$.  Since the factors are inverse

numbers, it would be $73*37$, which would give an answer of $2701$.  Check with

long division; there will be a remainder of 0, and the first carrying-down will be a 1.

So, 3+7=10

 :gathering:  :rotfl:  :pilot:


Let ABC be a triangle, E and F the points where the incircle and A-excircle touch AB, and D the point on BC such that the triangles ABD and ACD have equal in-radii. The lines DB and DE intersect the circumcircle of triangle ADF again in that point x & y.prove that : XY||AB if & only  if AB=AC


Solve in the sets of integer the follow inequality: \[ a\plus{}b\plus{}c\equal{}abc\]

if you wanted to say "solve the equation" this is my solution for positive integers, you can use the same method for negative integers.

for $ a,b,c \neq 0$
suppose that, $ a\leq b \leq c$ then $ 1 \equal{} \frac {1}{ab} \plus{} \frac {1}{ac} \plus{} \frac {1}{bc}\leq\frac {3}{a^2}$
$ \Leftrightarrow a^2\leq3 \Leftrightarrow a \equal{} 1$
then $ 1 \plus{} b \plus{} c \equal{} bc \Leftrightarrow (c \minus{} 1)(b \minus{} 1) \equal{} 2$
$ c \equal{} 3$ and $ b \equal{} 2$
Finally, $ S \equal{} \big\{1,2,3\big\}$ and all their permutations(for positive integers).

What about when two of $ a,b,c$ have the same sign and the other has the other sign? For exemple when $ a,b>0$ and $ c<0$.

use the same methode.
suppose that $ a\leq b \leq c$ then $ a^2\leq3 \Leftrightarrow a\equal{}\pm 1$ or $ a\equal{}0$
if $ a\equal{}0$ then $ S\equal{}\big\{0,a,\minus{}a\big\}$ and all their permutations.
if $ a\equal{}1$ then $ 1\plus{}b\plus{}c\equal{}bc \Leftrightarrow (b\minus{}1)(c\minus{}1)\equal{}2$ then $ b\equal{}3,c\equal{}2$ or $ b\equal{}\minus{}1,c\equal{}0$
if $ a\equal{}\minus{}1$ then $ \minus{}1\plus{}b\plus{}c\equal{}\minus{}bc \Leftrightarrow (b\plus{}1)(c\plus{}1)\equal{}2$ then $ b\equal{}\minus{}3,c\equal{}\minus{}2$ or $ b\equal{}1,c\equal{}0$
Finally, all solutions in integers is, $ S\equal{}\Big\{(1,2,3);(a,0,\minus{}a);(\minus{}1,\minus{}2,\minus{}3)\Big\}$ and all their permutations.

this one is more strong,
solve in positive integers, $ \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\minus{}\frac{1}{abc}\equal{}1$

[quote="peine"]this one is more strong,
solve in positive integers, $ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \minus{} \frac {1}{abc} \equal{} 1$[/quote]
WLOG, $ a \le b \le c$
Obviously $ a, b, c \neq 1$, also $ a < 3$ or $ L.H.S. < 1$ or when $ a\equal{}1$ it's easy to check that it's impossible, so $ a \equal{} 2$
So $ \frac {1}{b} \plus{} \frac {1}{c} \minus{} \frac {1}{2bc} \equal{} \frac {1}{2}$, similarly $ b,c \neq 2$ and $ b < 4$, so $ b \equal{} 3$

$ a \equal{} 2, b \equal{} 3$ gives the only solution $ (2,3,5)$ and of course its permutations.


search AMC AIME Vieta algebra polynomial

Find $ax^5 + by^5$ if the real numbers $a$, $b$, $x$, and $y$ satisfy the equations
\begin{eqnarray*} ax + by &=& 3, \\ ax^2 + by^2 &=& 7, \\ ax^3 + by^3 &=& 16, \\ ax^4 + by^4 &=& 42. \end{eqnarray*}

I am just wondering if this (weird) logic could be applied

[hide]The fibbanocci (sp) sequence is the sum of two geometric sequences so it could make sense that the right side should follow a recursive pattern.

3*3+7=16
but
7*3+16=37

Since 7*7 is 1 greater than 16*3
When we change it to
3*10-7*2=16
7*10-16*2=38
the second part increases by one
applying the same step 4 times, we get
3*38-7*14=16
7*38-16*14=42
to get our answer
16*38-42*14=20

20[/hide]

If you search, I believe there is a possibility to find at least 2 threads concerning this specific problem.  I recall solving this problem twice before in this forum.

Masoud Zargar

Your idea is a good one, and during the actual AIME it would work...on a proof-oriented test, though, you'd have to actually prove it (which is much harder).

[quote="ehehheehee"][hide]The fibbanocci (sp) sequence is the sum of two geometric sequences so it could make sense that the right side should follow a recursive pattern.[/hide][/quote]

In fact, it [b]does[/b] make sense, and this is just the characteristic equation of a second-order homogeneous linear recurrence in disguise.  That's the solution I came up with.  :)

yeah, that totally makes sense.  So, it'll be easier (instead of try plugging in numbers) to say that:

$f(n+2) = af(n+1)+bf(n)$

and then we have:

$16 = 7a+3b$
$42 = 16+7b$

now just solve for $a, b$ and plug in to get $42a+16b$.

Note, of course, that towersfreak's solution leads you to exactly the same conclusion, and moreover it's completely rigorous.  (In fact, it provides a nice proof of the reason characteristic equations work in the first place.)

I just like how realizing that one very important fact gives an instant solution to an AIME #15.  :P

[quote="SplashD"]Can towersfreak's solution be applied to any series that's the sum of $n$ geometric series?[/quote]

That kind of sum would be the closed formula for an $n^{th}$-order linear recurrence (for example: a third-order recurrence might be $a_{0}= 1, a_{1}= 2, a_{2}= 3, a_{k+3}= a_{k+2}+a_{k+1}+a_{k}$).  Yes, it would still work in theory, I suppose.  You would have to use Vieta's formulas to figure out what to multiply by, and it would get confusing.

Again, fix your signature, the hide tag is hiding the quote button :P

More harder problem.

[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=1396658#1396658[/url]

[hide]Wow... this #15 is soooo easy (in fact it only took me about 15 mins to kill this one... perhaps I just got lucky in my trend of thought and started in the right direction right away i saw the problem :P)

We have $ ax^3 + bxy^2 = 7x$ and $ ax^2y + by^3 = 7y^^$
Summing gives $ 16 + xy(by + ax) = 16 + 3xy = 7(x + y)..............(1)$

Similarly we have $ ax^4 + bxy^3 = 16x$ and $ ax^3y + by^4 = 16y$
Summing gives $ 42 + xy(by^2 + ax^2) = 42 + 7xy = 16(x + y)..................(2)$

Similarly we have $ x^5 + y^5 + 16xy = 42(x + y)$

solving for $ xy$ and $ x + y$ in $ (1),(2)$ gives $ xy = - 38$ and $ x + y = - 14$
so $ x^5 + y^5 = - 42*14 + 16*38 = 608 - 588 = 20$

EDIT: Just wondering can this be generalized further........ and whats the solution w/ the characteristics eqn?
edit: never mind, got the solution w/ characteristics eqn..  :lol:[/hide]

edit again: hmm the recurrence method works REALLY well for more variables.... but somehow i couldn't see how towersfreak's (similar to mine) would work well in more variables...

$ (ax^n + by^n + cz^n)(x + y + z) = ax^{n + 1} + by^{n + 1} + cz^{n + 1} + \sum xy(by^{n - 1} + ax^{n - 1})$ where the last sum is cyclic through $ x,y,z$ and $ a,b,c$

but this relationship won't be that useful anymore... will it (or do we still have to work on the cyclic sum in the way similar to the two variables case)?

They could even be nonreal.

EDIT: Oops, never mind!

[quote="MSTang"]They could even be nonreal.[/quote]

It's given a,b,x and y are real numbers

[quote="Clotho"]I don't get how $ax^5+bx^5$ can suddenly decrease. I guess $x$ or $y$ would have to be negative?[/quote]

The answer is NO.

take x=0.5, y=0.5, you get 21, which is quite close to 20

Similar in spirit to towersfreak2006's solution, but differently presented: [hide="WOOT!"]
\[ \begin{eqnarray*} ax + by &=& 3, \\ ax^2 + by^2 &=& 7, \\ ax^3 + by^3 &=& 16, \\ ax^4 + by^4 &=& 42 \\ ax^5 + by^5 &=& ? \end {eqnarray*} \]Let $S_n=ax^n+by^n$. Then, the characteristic polynomial of the sequence is $ (t-x)(t-y) = t^2 - (x+y)t + xy $. Let $A=x+y$ and $B=xy$. Then, the characteristic polynomial can be re-written as $t^2-At+B$, so \[ S_n = A S_{n-1} - B S_{n-2}, \qquad \forall n \ge 3. \]For $n=3$, we obtain \[ 7A-3B=16. \]For $n=4$, we obtain \[ 16A-7B=42. \]Thus, solving for $A,B$, we have $A=-14$ and $B=-38$, so our desired result is \[ \begin {align*} ax^5 + by^5 &= S_5 \\&= AS_4 - BS_3 \\&= \boxed {20}. \end {align*} \]
[/hide]

PROBLEM
Find $ax^5+by^5$ given that:
$$ax+by=3$$$$ax^2+by^2=7$$$$ax^3+by^3=16$$$$ax^4+by^4=42$$
SOLUTION
We can multiply the middle two equations by $(x+y)$ to get:
$$(ax^2+by^2)(x+y)=7(x+y) \Longrightarrow$$$$ax^3+by^3+by^2x+ax^2y =7 (x+y) \Longrightarrow$$$$16 + xy(ax+by) = 7(x+y) \Longrightarrow$$$$16+3xy=7(x+y).$$
Similarly,$$(ax^3+by^3)(x+y)=16(x+y) \Longrightarrow$$$$ax^4+by^4+by^3x+ax^3y = 7(x+y) \Longrightarrow$$$$42 + xy(ax^2+by^2) = 16(x+y) \Longrightarrow$$$$42+7xy=16(x+y).$$
Using the two equations we just found, we know that $\frac{7}{16} = \frac{16+3xy}{42+7xy}$, so $294+49xy=256+48xy$. Hence, $xy=-38$. Plugging this into the equation $16+3xy=7(x+y)$, we have that $16-114=7(x+y)$ and $x+y=-14$. Hence, $ax^5+by^5=(ax^4+by^4)(x+y)-xy(ax^3+by^3)=$ $42(-14)-(-38)(16)=\boxed{20}$.

Woohoo, solved a 15! :)

We have $(ax^2+by^2)(x+y)=7(x+y) = (ax^3+by^3)+xy(ax+by) = 16+3xy$. Similarly, we have $16(x+y) = 42+7xy$. We want $ax^5+by^5 = (ax^4+by^4)(x+y)-xy(ax^3+by^3)=42(x+y)-16xy$. We can solve to get $x+y=-14, xy = -38$, for an answer of $\boxed{20}$.

So we have to find a Quadratic Equation with roots x & y. Let's assume the Equation looks something like this  t²-pt-q=0. We define, Tn=ax^n+by^n for n>2
Put n=3 we get,
T3=pt2+qt
   16 =7p+3q...........(i)
Put n=4
       42=16p+7q...........(ii)
Subtracting eqn 1 &2 we get, p=-14 & q=38
Put n=5=> T5=pt4+qt3
                T5=20 is correct ans
By Newton Sum Mah fav

We have that $(x+y)(ax^2+by^2)=7(x+y)=(ax^3+by^3)+xy(ax+by)=3xy+16.$ Also by similar computation $16(x+y)=42+7xy$, also $ax^5+by^5=(ax^4+by^4)(x+y)-xy(ax^3+by^3)=42(x+y)-16xy$, hence, $x+y=-14$, and $xy=-38$, so by some more computation we get $\boxed{20}.$


algebra
polynomial
function
algebra unsolved

prove that for all real number x,y,a,b satisfying x+y=a+b and x^4+y^4=a^4+b^4 then x^n+y^n=a^n+b^n for all n in N

$x^{n}+y^{n}$ can be written in terms of basic symetrical polynomials
$\sigma_{1}=x+y$ and $\sigma_{2}=xy$. So it's enogh to show that $\sigma_{2}=\sigma_{2}'$ where $\sigma_{2}'=ab$. We have
$x^{4}+y^{4}=\sigma_{1}^{4}-4\sigma_{2}\sigma_{1}^{2}+2\sigma_{2}^{2}$, so we must show that
$4\sigma_{2}\sigma_{1}^{2}+2\sigma_{2}^{2}=4\sigma_{2}'\sigma_{1}^{2}+2\sigma_{2}'^{2}$,
$(\sigma_{2}-\sigma_{2}')(2\sigma_{1}^{2}+\sigma_{2}+\sigma_{2}')=0$.
We just have to show that
$2\sigma_{1}^{2}+\sigma_{2}+\sigma_{2}'=0$ is impossible, so we can|t have
$2x^{2}+2y^{2}+5xy=-ab$. We have that
$-ab\geq-1/2(x^{2}+y^{2})$, so we need to show
$2x^{2}+2y^{2}+5xy<-1/2(x^{2}+y^{2})$,
$5x^{2}+5y^{2}+10xy>0$. The only case in which this is false is x+y=0 but then a+b=0 and x,y,a and b differ only in signs.

You can also use the fact that the function $y=x^{4}$ is concave all over $R$, and so for any $\delta\neq 0$,  $(x+\delta)^{4}-x^{4}=(y+\delta)^{4}-y^{4}\rightarrow x=y$, being $(x+\delta)^{4}-x^{4}$  strictly increasing.


[quote="aahmadi"]What is the diameter of a circle inscribed in an isosceles trapezoid with side lengths 20 m and 80 m?[/quote]Let us make the trapezoid as such:
http://newtonmusic4.googlepages.com/trap.bmp

Now, we can say that the area of the trapezoid is ${ \dfrac{base_1 + base_2}{2}height = \dfrac{a + c}{2}2r} = 100r$, since the height is twice the radius

We can also say that the area of the shaded triangle is $ \dfrac{rd}{2}$, where r=inscribed circle radius.  Therefore, we can say that the area of the trapezoid is also equal to:
$ \dfrac{ra}{2} + \dfrac{rb}{2} + \dfrac{rc}{2} + \dfrac{rd}{2} = r\dfrac{a + b + c + d}{2} = r(50 + b) = r(50 + \sqrt {(2r)^2 + (\dfrac{c - a}{2})^2}) = r(50 + \sqrt {4r^2 + 900})$
Now we se the two equal and say that:
$ r(50 + \sqrt {4r^2 + 900}) = 100r$
$ \sqrt {4r^2 + 900} = 50$
$ 4r^2 = 1600$
$ r = 20$
EDIT: oops, solving for diameter:
$ d=40$

[hide="More straightforward"]
[asy]size(300);
pair A=(0,0),B=(80,0),C=(50,40),D=(30,40);
pair F=(4/5*30,4/5*40), O=(40,20);
pair G=(30,0);
draw(A--B--C--D--cycle);
draw(Circle(O,20));
dot(A); dot(B); dot(C); dot(D); dot(O); dot(F);
draw(O--F); draw(D--G);
draw((C+D)/2--(A+B)/2);
label("\(d\)",(D+G)/2,E);
label("\(10\)",(D+F)/2,NNW);
label("\(10\)",(G+(40,0))/2,S);
label("\(40\)",(F+A)/2,NNW);
label("\(10\)",(D+(D+C)/2)/2,N);
label("\(30\)",(15,0),S);
draw(A--D--G--cycle,rgb(0,0,1)+linewidth(0.8));[/asy]
By the Two Tangent Theorem twice we find that the legs of the trapezoid have length $ 40 \plus{} 10 \equal{} 50$. By the Pythagorean Theorem on the right $ \textcolor{blue}{\triangle}$ (which in fact is a $ 3 \minus{} 4 \minus{} 5 \triangle$), we have $ d \equal{} \boxed{40}$.
[/hide]


AMC
AMC 12
trigonometry
geometric series

If $\displaystyle \sum_{n=0}^{\infty}\cos^{2n}\theta = 5$, what is the value of $\cos 2\theta$?

wow, i did not look at the problem very carefully, it was easier than i expected

[quote="Farenhajt"][quote="Altheman"][hide="hint "]you can make it a geometric series by doing cos +i sin, and taking the real part at the end[/hide][/quote]

Wrong - this already [i]is[/i] a geometric series, and not $\cos n\theta$ series.

[hide="Solution"]$S={1\over{1-\cos^2\theta}}=5\implies \cos^2\theta={4\over 5}\implies \cos 2\theta=2cos^2\theta-1={3\over 5}$[/hide][/quote]

I got the same answer, but I bashed it all out with cos inverses, etc, once I got to $cos^2\theta = \frac45$. How did you get $2\cos^2\theta-1=\cos{2\theta}$? Is this some identity I've never seen before? Can you show this? Thanks in advance :)

This is elementary trigonometry - the formulas for double angle and half an angle. Confer any trig textbook :)

Have you seen \[ \cos (2\theta) = \cos^2 \theta - \sin^2 \theta ? \] Compare that with \[ 1 = \cos^2 \theta + \sin^2 \theta, \] and you'll find \[ \cos (2\theta) = 2\cos^2 \theta - 1 = 1-2\sin^2\theta. \] :)

[quote="Farenhajt"]

[hide="Solution"]$S={1\over{1-\cos^{2}\theta}}=5\implies \cos^{2}\theta={4\over 5}\implies \cos 2\theta=2cos^{2}\theta-1={3\over 5}$[/hide][/quote]

How do you know that:

$S={1\over{1-\cos^{2}\theta}}=5$ ??

What happens to the $n$ in the original problem?

[quote="ehehheehee"]Because the summation is an infinite geometric series and since (cos(x))^0=1 we can apply the formula for the sum as
a/(1-r)
or
1/(1-cos^2(x))

which equals five[/quote]

Oh I see.  Thanks!  I didn't even think of an infinite series, even though I saw the summation...I don't know what I was thinking...

[hide]
This is an infinite series in the form:
1, cos^2, cos^4, ...
Therefore the sum is:
a/(1-r) = 1/(1-(costheta)^2) = 1/(sintheta)^2 = 5
(sintheta)^2 = 1/5
Using double angle formulas,
cos(2theta) = 1-2(sintheta)^2 = 1-2(1/5) = [b]3/5[/b]
[/hide]

you dont need to bump up new threads, the identical solution to yours exists right before your post.


HI everyone..... U all know less than a couple of months are left for NSEA... just wondering how much u people have done.....
I have done kinematics circular motion work energy collision and rotational from resnick halliday and some problems from HCV.... I wnted to know wat other topics are important.... In maths i have done straight lines some sets functions and trigo....

@skand how many questions were u able to do in last years senior paper....??

I observed that many questions are form electricity but they are kind of elementar... wanted to know that should i do electricity from resnick halliday ( it has 12 chapters for this :( )
What about other important topics....??
Kindly help...? :)

Read electricity from DJ Griffith, it is the 7th chapter I believe , then do problems from Resnick. That should do.

and what about heat !
i believe heat is a very important unit!

I have no idea about NSEA , I think the author means NSEP. And I think you know more than me about the syllabus n all but heat is imp in every physics exam. Heat and waves both imp.

@ conjurer
i meant nsea national standard exam in astronomy...

@ anurag I hadn't given it previous year(i was completely unaware of this exam in 10th)

Arnav was topper in it

I think heat and waves are important topic there also as there were ques from these in prev year paper(i checked on site of IAPT)

just discuss more abt 12th class chapters so dat i get even more discouraged............ :(  :mad:  :blush:

good

topper of this exam in 10th itself(scoring better than studs of 11th and 12th) is saying these words  :mad:

What will happen to me  :(

that was certainly not a joke......... :mad:  :mad:
well dont keep on exagerating nse......
now it has started making me depressd than motivating me......... :(

For NSEA senior you have to be through on 12th class topics like optics and waves. You can check last years paper. The problems from 12th are kinda of basic so just know the funda of them, you can work out the rest. Read book and magazines like Arihant's Olympiad series and the Spectrum magazine too. It will be helpful.


Try this out:

What this is: This is a fast paced 2-5 minute game you can play. Be sure you can switch between FTW Alpha and FTW Test version easily. In this "minigame", you set up/join two games: one in FTW Alpha and the other in the Test version. And yes, you will play both games at the same time. Note that FTW Test still has quite a few bugs, so if you happen to find one, post in thread or PM levans with the game number and how you think this happened.

Requirements: THIS GAME MUST BE UNRATED.
Number of players: 2+ (3-5 recommended)
Number of problems: 10 for Alpha & Test
Amount of time: 15 seconds for Test Version, 12 for Alpha
Scoring Method: Time or Response for either. Make sure that these are same in both games (no doing Time/Response)
Viewable: This is up to game creator.

Password needed? This is something I'm still thinking about, since there is really no need for it, except to tell others who aren't in that this is a half-tournament. The current password is [b]asfd[/b]. I did not typo that.

To game creator: Start the Alpha version first. Make both games start around the same time.

So organize this among the people online, seeing who wants to play and who doesn't. There will be no rating change whatsoever, so have fun!

Any name suggestion for this?

Reviving!!
Uh, you coud possibly name it: ALPHA MINIGAME


a rectangular garden with width of x and length of y is surrounded by a 2 inch wide walkway.  what is the area of the walkway in square feet?

[b]@ alextrainer:[/b] You need to tell us what units $ x$ and $ y$ are in. The solution provided by [b]Temperal[/b] assumes that $ x$ and $ y$ are in feet.

Actually, I realized my solution isn't correct. Since the walkway is on both sides, the fractions in the binomial multiplication should be $ \frac{1}{3}$. It's a trivial modification, though.


abstract algebra
group theory
combinatorics proposed
combinatorics

For non-empty subsets $ A,B\subseteq\mathbb{Z}$ let us define $ A \plus{} B$ and $ A \minus{} B$ by:
\[ A \plus{} B \equal{} \{a \plus{} b\ |\ a\in A,\ b\in B\}, \ \ A \minus{} B \equal{} \{a \minus{} b\ |\ a\in A, \ b\in B\}.
\]
In the sequel we work with non-empty finite subsets of $ \mathbb{Z}.$

Prove that we can [i]cover[/i] $ B$ by at most $ \displaystyle\frac {|A \plus{} B|}{|A|}$ [i]translates[/i] of $ A \minus{} A,$ i.e. there exists $ X\subseteq\mathbb{Z}$ with $ |X|\leq\displaystyle\frac {|A \plus{} B|}{|A|}$ such that:
\[ B\subseteq\displaystyle\bigcup_{x\in X} (x \plus{} (A \minus{} A)) \equal{} X \plus{} A \minus{} A.
\]
[i]Imre Rusza[/i]

Consider the sets of the form $ A \plus{} x, \, x \in B.$ Those sets are included in $ A \plus{} B,$ and have $ |A|$ elements.
Pick a maximal family of such sets where all the $ A \plus{} x, \, x \in X$ for some $ X \subset B$ are disjoint. (why (rather, "how") does it exist?)
By construction, there are at most $ \frac {|A \plus{} B|}{|A|}$ such sets and for any $ b \in B,$ the sets $ A \plus{} b$ and $ A \plus{} x$ (for some $ x$) intersect, which means $ B \subset X \plus{} A \minus{} A.$

Note that $ \mathbb{Z}$ is irrelevant here, and the result is true for any abelian group $ G.$


Maria has three identical apples and three identical oranges. How many ways are there for her to distribute the fruits among her four friends if Jacky won't eat oranges?

Balls and urns (search it if you don't know what it is) gives us $ \dbinom{3\plus{}4\minus{}1}{4\minus{}1}\dbinom{3\plus{}3\minus{}1}{3\minus{}1}\equal{}20\cdot10\equal{}\boxed{200}$.

[hide]First, just consider the apples. Imagine putting three dividers among the apples, so that Maria's first friend gets the apples to the left of the first divider, Maria's second friend gets the apples between the first and second dividers, Maria's third friend gets the apples between the second and third dividers, and Maria's last friend gets the apples after the third divider. With three dividers and three apples, there are $\binom{6}{3}=20$ ways to arrange the dividers. Now, consider the oranges. Since Maria won't give Jacky any oranges, Maria has to distribute the oranges among her other three friends. With three oranges and two dividers, there are $\binom{5}{2}=10$ ways for Maria to distribute the oranges. The total number of ways for Maria to distribute the fruit is $20\cdot 10=\boxed{200}$.[/hide]

wait why do we multiply $20$ and $10??$

cuz like aren't those independent cases? The number of ways to split the apples doesn't affect the number of ways to split the oranges . . .


AMC
AIME
number theory
greatest common divisor
conics
ellipse
geometry

Here are a couple from the older AIMES, (And yes I know that they are from the easier exams :lol: )

1984 AIME #10:

Mary told John her score on the American High School Mathematics Examination, which was over 80. From this, John was able to determine the number of problems Mary solved correctly. If Mary's score had been any lower, but still over 80, John could not  have determined this. What was Mary's score? (Recall that the AHSME consists of 30 multiple-choice problems and that one's score, s, is computed by the formula s=30+4c-w, where c is the number of correct and w is the number of wrong answers; students are not penalized for problems left unanswered.)

1985 AIME #13:

The numbers in the sequence 101,104,109,116,... are of the form a(n)=100+n^2, where n=1,2,3... For each n, let d(n) be the GCD of a(n) and a(n+1). Find the maximuum value of d(n) as n ranges through the positive integers.

Note: d(n) stands for d sub n, same for the others :)

1985 AIME #11:

An ellipse has foci at (9,20) and (49,55) in the xy-plane and is tangent to the x-axis. What is the length of its major axis?

1986 AIME #12:

Let the sum of a set of numbers be the sumj of its elements. Let S be a set of positive integers, none greater than 15. Suppose no two disjoint subsets of S have the same sum. What is the largest sum a set S with these properties can have?

Can anyone give me a different solution than those in Contest Problem Book 5?[/code]

The first problem was number 25 on Silverfalcon's AMC 2005 B. Check it out!

[quote="1/(ln x)"]
1985 AIME #13:

The numbers in the sequence 101,104,109,116,... are of the form a(n)=100+n^2, where n=1,2,3... For each n, let d(n) be the GCD of a(n) and a(n+1). Find the maximuum value of d(n) as n ranges through the positive integers.
[/quote]

[hide="Answer"]
401
[/hide]

[hide="Solution"]
Let x=max(gcd(n^2+100, n^2+2n+1)).  Clearly the gcd of two numbers must divide the difference between the two numbers.  Thus $x|2n+1$ and $x|n^2+100$.  (It follows that if these two conditions hold, then $x|n^2+2n+101$).  Converting these into congruences, we have:
$2n+1 \equiv 0 \mod x$  (1)
$n^2+100 \equiv 0 \mod x$ (2)
Subtracting 100 times the first from the second, we find
$n^2-200n=n(n-200) \equiv 0 \mod x$  (3)
$n$ and $2n+1$ are relatively prime, since any number dividing $n$ would also have to divide $n+1$, which is not possible.  Since $x|2n+1$, it follows that $n$ and $x$ are relatively prime.  Thus, from (3) it follows that $n-200 \equiv 0 \mod x$
If we subtract 401 from both sides of (1), we have
$2(n-200) \equiv -401 \mod x$
$\Rightarrow 401 \equiv 0 \mod x$
Thus the maximum possible value for $x$ is $\boxed{401}$.  But we still have to check that the maximum value is in fact reached.  Indeed, we find the maximum occurs when n=200.
[/hide]

Number 11 took me about 30 minutes.  :(

[hide]Since the sum of the focal radii is equal to the length of the major axis, let the two foci $(9, 20)$ be $F$ and $(49, 55)$ be $F_1$. And let the point where the ellipse is tangent to the x-axis be $A$. We know $FA + F'A = 2a$, where $a$ is the semi-major axis.

Now's where I wasted time. I tried analytical geometry, which I'm no good at, and failed, so this approach took a while.

We notice that if some other point outside the ellipse, say $B$, is chosen, that $FB + F'B > FA + F'A$. Now we need to figure out a length which is equivalent to $FA + F'A$ that can be determined. Reflect $F'$ to a point across the x axis called $F\"$. Let the intersection of $FF\"$ with the x-axis be that point $B$. So, $F'B = F\"B$. So, $FF\" = FB + F'B \ge  FA + F'A$. But! Notice by triangle inequality because $F'A = F\"A$ that $FA + F'A \ge FF\"$. The only condition that satisfies both inequalities is when $FF\" = 2a$. So after some math, you have $FF\" = 85 = 2a$. That's right I think.[/hide]

Uhhh can't you just say for #11, the point of tangency is the shortest path between the two focus points that touches the x axis?  So then take distance from 9, 20 to 49, -55 to get whatever it is...

[hide="1984 AIME 10"]We can subtract the 30 from the formula, since that does not change with the number of questions answered, correct, or wrong. So, we have $4c-w>50$. As $c+w\le 30$, the maximum possible number of questions answered to give a score of exactly $50$ is $c+w=30$, with $c=16$ and $w=14$. For every individual score, we can get the same score by adding one correct answer and four wrong answers. However, the lowest score for which we cannot do this is $(c,w)=(23,3)$, as that would exceed the number of problems on the test. To get Mary's score, we must remember to add back in the $30$ that we subtracted earlier, so we have $30+4\cdot 23-3=\boxed{119}$.[/hide]

[hide="1985 AIME 13"]$a_n=100+n^2$ and $a_{n+1}=100+(n+1)^2$
$GCF(a_n,a_{n+1})$ will be a factor of $a_{n+1}-a_n=2n+1$.

That's as far as I can get. :([/hide]

[hide="1985 AIME 11"]For the ellipse to be tangent to the x-axis, it must be tangent directly below the lower focus, or at the point $(9,0)$. Therefore, if we take the distances to this point from both foci, we get $20+\sqrt{40^2+55^2}=20+5\sqrt{185}$. :huh: I guess it isn't tangent directly below... I'm not sure why I thought it was, but how do you find the point of tangency otherwise?[/hide]

[quote="JesusFreak197"]For #12, what is a disjoint set?[/quote]

Two sets are disjoint if they have no elements in common.  With set theoretic symbols, two sets, $A$ and $B$, are disjoint iff $A \cap B=\emptyset$.

JesusFreak197, your answer for #10 is not correct.

Umm... then what is the correct answer? I don't see anything wrong with what I did.

[b]EDIT:[/b] Nevermind, I got it. Fixed.

This is a kind of cool way to do #11.

[hide]
It is a well known fact that if you are standing on the focus of a giant ellipse and you throw a ball that hits the edge of the ellipse and bounces off, it's path will intersect the other focus.  This means that the angle formed by a line through a given point on the ellipse and one focus and the line tangent to the ellipse at the given point equals the angle formed by a line through the given point and the other focus and the line tangent to the ellipse at the given point (angle of incidence equals angle of reflection).

Now back to the problem.  Call the point of tangency with the x-axis $A$.  Call  (9,20) and (49,55) $F_1$ and $F_2$, respectively.  The angle formed by the x-axis and $AF_1$ equals the angle formed by the x-axis and $AF_2$ (the acute angles in both cases).  Since the angles are equal, the tangents must also be equal.  Let $x$ be the distance between (9,0) and $A$.  Thus $\frac{20}{x}=\frac{55}{40-x} \Rightarrow x=\frac{32}{3}$.  Now to find the length of the major axis, all we have to do is find $F_1A+F_2A=$
$\sqrt{20^2+\left( \frac{32}{3} \right)^2}+\sqrt{55^2+\left( \frac{88}{3} \right)^2}$
$=\frac{68}{3}+\frac{187}{3}$
$=\boxed{85}$
[/hide]

another kind of cool way to do #11

[hide] Let F be the point (9,20) and F' the point  (49,55).
The projection of F on the x-axis is P, the one of F' is P'. Call the point which lies on the x-axis and the ellipse A. The reflection of F' across the x-axis is called G.
Now: angle FAP = angle F'AP' (because it's an ellipse)= angle GAP'
=> F, A and G lie on a straight line, and more: FG is the length of the major axis (FG=FA+AG=FA+AF'=major axis).
FG^2 = (20+55)^2 + (49-9)^2  (pythagoras)
<=> FG=85[/hide]

good evening or whatever

[quote="qweretyq"]sorry if this seems so basic, im bad with conics

does the sum of the distances from a point on the ellipse to the two foci equal the length of the major axis?
I'm thinking this based on your solutions...[/quote]

yeah, you know the sums of the lengths is equal for all points on the ellipse, so take either of the two on the major axis. if you look at the distances from each of the foci it's easy to see that they sum to the length of the major axis.

A visual way to do this is the following:

An ellipse can be formed by putting two tacks in a surface, tying a string around them, and drawing a line around the tacks to the full extent of the string. The tacks are the foci of the ellipse. When the pencil is positioned so that part of the string overlaps another part, you have the distance between the two foci plus two times the distance from the closer focus to the pencil. That is also the distance from the farther focus to the point opposite the pencil, so we can see that they sum to the length of the major axis.

I hope that made sense... :huh:

Is that not the definition of ellipse? (set of all points whose sum to two points are constant)

It's almost like a circle, except an ellipse has 2 points

An ellipse is the set of all points whose sums of the distances from two foci (points) are all equal.
i.e. for every point of the ellipse, their sums of the distances from the two foci are equal. This of course is true for a circle which is an ellipse with 2 overlapping foci.

Better? :P

1986 AIME #12:
I found the solution, but I don't now yet how to prove it precisely

[hide="my solution"]
I think the set {8, 11, 13, 14, 15} has the greatest possible sum (61).
In my opinion it's obvious to start with the greatest integer and checking each integer below whether it fits into the set or not.
[/hide]


We consider the sequences strictely increasing $(a_0,a_1,...)$ of naturals which have the following property :
For every natural $n$, there is exactly one representation of $n$ as $a_i+2a_j+4a_k$, where $i,j,k$ can be equal.

Prove that there is exactly a such sequence and find $a_{2002}$

Here is the solution I participated with (I translated it into English for the op02 project):

[color=blue][b]Bundeswettbewerb Mathematik 2002, Round 2, problem 2[/b]

Find all strictly increasing sequences $\left(a_0;\;a_1;\;a_2;\;\ldots\right)$ of non-negative integers with the property that every non-negative integer can be uniquely represented in the form $a_i + 2a_j + 4a_k$, where $i,j,k$ are (not necessarily distinct) nonnegative integers. Prove that there is exactly one such sequence. Determine $a_{2002}$.[/color]

[i]Solution.[/i] First we show that there can exist at most one such sequence $\left(a_n\right)_n$. In fact, it is clear that $a_0$ must be $0$ (else, since the sequence $\left(a_n\right)_n$ is increasing, it would not contain any $0$, so the number $0$ could not be represented in the form $a_i+2a_j+4a_k$). Also, if, for some integer $u\geq 0$, we know the first $u$ members $a_0$, $a_1$, ..., $a_{u-1}$ of our sequence, then the next member $a_u$ is uniquely determined as the smallest positive integer $T$ which cannot be represented in the form $a_i+2a_j+4a_k$, with $i$, $j$, $k$ being integers $<u$ (in fact, if the member $a_u$ were $<T$, then it would be representable in the form $a_i+2a_j+4a_k$, with $i$, $j$, $k$ being integers $<u$, and thus it would be representable in the form $a_i+2a_j+4a_k$ in two ways, what is a contradiction; if the member $a_u$ were $>T$, then $T$ would not be representable in the form $a_i+2a_j+4a_k$ at all, what is a contradiction as well; thus, we must have $a_u=T$). Thus, if there exists a sequence $\left(a_n\right)_n$ satisfying the condition of the problem, then its first member $a_0$ must be $0$ and it can be recursively constructed in a unique way. Hence, there can exist most one such sequence $\left(a_n\right)_n$.

Now, what remains in order to complete the solution is to construct a sequence $\left(a_n\right)_n$ satisfying the condition of the problem, and then to compute $a_{2002}$. Here is the construction of the sequence:

For any nonnegative integer $n$, let $n=2^k\cdot n_k+2^{k-1}\cdot n_{k-1}+...+2\cdot n_1+n_0$, with all $n_i$ being from the set $\left\{0;\;1\right\}$, be the representation of $n$ in the binary system. Then, $a_n=8^k\cdot n_k+8^{k-1}\cdot n_{k-1}+...+8\cdot n_1+n_0$. In other words, the number $a_n$ is the number whose representation in the digital system with base $8$ is identic with the representation of the number $n$ in the binary system. Or, in other words, the sequence $\left(a_n\right)_n$ is the sequence of all nonnegative integers whose representation in the digital system with base $8$ consists of only $0$'s and $1$'s (arranged in rising order).

Why does this sequence satisfy the condition of the problem? Well, every digit $d$ in the digital system with base $8$ is an integer with $0\leq d\leq 7$, and thus can be uniquely represented in the form $d_1+2d_2+4d_3$, with $d_1$, $d_2$, $d_3$ being integers from the set $\left\{0;\;1\right\}$. Thus, for any nonnegative integer $t$, by writing it out in the digital system with base $8$, representing each of its digits in this way, and combining the resulting $d_1$'s into one number $a_i$, the resulting $d_2$'s into another number $a_j$, and the resulting $d_3$'s into one number $a_k$, we have represented our integer $t$ in the form $t=a_i+2a_j+4a_k$. Also, such a representation is unique, since, when performing the addition $a_i+2a_j+4a_k$ in the decimal system with base $8$, carries cannot occur (since the numbers $a_i$, $a_j$, $a_k$ all have digits from the set $\left\{0;\;1\right\}$), so from the resulting number we can digit-wisely trace back the addition and obtain the original numbers $a_i$, $a_j$, $a_k$. Hence, every nonnegative integer $t$ can be uniquely represented in the form $t=a_i+2a_j+4a_k$. So our sequence $\left(a_n\right)_n$ indeed satisfies the condition of the problem.

Now it remains to find $a_{2002}$. This is straightforward. The representation of the number $a_{2002}$ in the digital system with base $8$ must be identic with the representation of $2002$ in the binary system. The representation of $2002$ in the binary system is $11111010010$; thus, the number $a_{2002}$ is $11111010010$ in the digital system with base $8$. Translating back into the decimal system, we get $a_{2002}=1227100168$.

  darij

[quote="darij grinberg"]For any nonnegative integer $n$, let $n=2^k\cdot n_k+2^{k-1}\cdot n_{k-1}+...+2\cdot n_1+n_0$, with all $n_i$ being from the set $\left\{0;\;1\right\}$, be the representation of $n$ in the binary system. Then, $a_n=8^k\cdot n_k+8^{k-1}\cdot n_{k-1}+...+8\cdot n_1+n_0$. [/quote]

How do you find this???

[quote="Labiliau"]How do you find this???[/quote]

If you are asking me how I proved this, just see the next paragraph. ;)
If you are asking me how I came up with this explicit construction: by looking up the sequence or a similar one in [url=http://www.research.att.com/~njas/sequences]Sloane's Encyclopedia[/url]. Not really fair, but the only way for somebody who knows nothing about combinatorics (what was surely true for me in 2002, and still doesn't quite miss the truth now).

  darij


analytic geometry
algebra proposed
algebra

I wonder how it will change its shape and special points if we have the graph (C1) of y=f(x) and then we plot the graph of y=f(2x) (C2)
Eg: A(3,2) on (C1) , then what is respective point of A on (C2) ?
Sorry if this question is silly.

Hi !

It's not silly but I think there is not his place....
Whatever, I'll answer you :for $x=3$ we have $2x=6$ so the "respective"
point of $A$ will be $B$ wich coordinates are $6$ and $f(6)$ and you can draw the graph of $f(2x)$ only if you know the hole graph of $f$.
Exemple 1) $f(x)=x^{2}$ so $f(2x)=4x^{2}$  :)
             2) $f(x)=x-1$ wich fixes with $A(3,2)$. Then $f(2x)=2x-1$.
I don't see a "Law" to change $f(x)$ into $f(2x)$... :maybe:


algebra
polynomial
inequalities theorems
inequalities

"Dumbassing" (or brute-force; see [url]http://www.artofproblemsolving.com/Resources/Papers/MildorfInequalities.pdf[/url]) is nice if you can actually [b]expand the polynomial[/b]. I'm wondering if there's a faster way to expand symmetric sums? For example:

\[\begin{eqnarray*}
\sum_{sym} a^2 (a+b)^2 (a+c)^2 &=& \sum_{sym} a^2 (a^2 + 2ab + b^2)(a^2 + 2ac + c^2)\\
&=& \sum_{sym} a^6 + 2a^5c + a^4c^2 + 2a^5b + 4a^4bc + 2a^3bc^2 + a^4b^2 + 2a^3b^2c + a^2b^2c^2\\
&=& \sum_{sym} a^6 + 4a^5b + 2a^4b^2 + 4a^4bc + 4a^3b^2c + a^2b^2c^2\\
&=& \text{extremely annoying}
\end{eqnarray}\]

Is there a way that takes you straight from $ \sum_{sym} a^2 (a+b)^2 (a+c)^2$ to $ \sum_{sym} a^6 + 4a^5b + 2a^4b^2 + 4a^4bc + 4a^3b^2c + a^2b^2c^2$?

It gets even worse with $ \sum_{sym} (a+b)^2 (b+c)^2 (c+a)^2$. How do you expand that?

[quote="Wikipedian1337"]"Dumbassing" (or brute-force; see [url]http://www.artofproblemsolving.com/Resources/Papers/MildorfInequalities.pdf[/url]) is nice if you can actually [b]expand the polynomial[/b]. I'm wondering if there's a faster way to expand symmetric sums? For example:

\[\begin{eqnarray*} \sum_{sym} a^2 (a + b)^2 (a + c)^2 & = & \sum_{sym} a^2 (a^2 + 2ab + b^2)(a^2 + 2ac + c^2) \\
& = & \sum_{sym} a^6 + 2a^5c + a^4c^2 + 2a^5b + 4a^4bc + 2a^3bc^2 + a^4b^2 + 2a^3b^2c + a^2b^2c^2 \\
& = & \sum_{sym} a^6 + 4a^5b + 2a^4b^2 + 4a^4bc + 4a^3b^2c + a^2b^2c^2 \\
& = & \text{extremely annoying} \end{eqnarray}\]

Is there a way that takes you straight from $ \sum_{sym} a^2 (a + b)^2 (a + c)^2$ to $ \sum_{sym} a^6 + 4a^5b + 2a^4b^2 + 4a^4bc + 4a^3b^2c + a^2b^2c^2$?

It gets even worse with $ \sum_{sym} (a + b)^2 (b + c)^2 (c + a)^2$. How do you expand that?[/quote]
I have discovered a nice tactic to multiply symmetric sums:
Let's take the example $ (a + b)^2 (b + c)^2 (c + a)^2$. Instead write it as:
$ (a + b)(a + b)(b + c)(b + c)(c + a)(c + a)$
Then you write up how many times $ a^4b^2, a^4bc, a^3b^3, a^3b^2c, a^2b^2c^2$ is in the product.
First $ a^4b^2: 1$ (We have to take all the a's, so it is obvious.)
$ a^4bc: 2$ (again we have to take all the a's)
$ a^3b^3: 2$ (We can't take any c's)
$ a^3b^2c: 6$ (Split up the cases from where you take the c)
$ a^2b^2c^2: 10$ (Again split up the cases or use that the total number should be $ 64$)

So $ (a + b)^2(b + c)^2(c + a)^2 = \sum_{sym} a^4b^2 + a^4bc + a^3b^3 + 6a^3b^2c + \frac {5}{3} a^2b^2c^2$

Edit:
Another nice thing is to know some expansions by heart. For instance I would always recognize that:
$ \sum_{sym} a^4b^2 - a^4bc + 2a^3b^2c - a^3b^3 - a^2b^2c^2 \ge 0$ since it is equal to $ (a - b)^2(b - c)^2(c - a)^2$.


calculus
derivative
calculus computations

Hi,

Differentiate

y= e^(k tan(sqrt(8 x)))

This is the problem that I am struggling with right now.

I came up with for my answer e^(k tan(sqrt(8x))) * k sec^2((sqrt(8x)) * 4x^-1/2

Im not sure what I am doing wrong, but the program webassign told me that I'm wrong and my instructor said that I am close but I messed up the last part of the chain rule. Im not sure what I did wrong so It would be great if someone could help me, especially since this is due within the hour.

Thank you

No need to be impatient: this is [b]not[/b] a math. help hotline. Besides, it seems like it is an assignment on which you are supposed to work by yourself ;), which, usually, would be a sufficient reason for not answering your post. Still, since the mistake you made is really stupid and it is quite apparent that you know what you are doing otherwise, I'll point it out for you.

The mistake is in the very end: $ (\sqrt {8x})' \equal{} \sqrt 8 (x^{1/2})' \equal{} \sqrt 8\frac 12 x^{ \minus{} 1/2} \equal{} \sqrt 2 x^{ \minus{} 1/2}$, not $ 4x^{ \minus{} 1/2}$. :).

P.S. And, please, don't double post! :spam: I deleted your identical post in Calculus Open Questions, which was a wrong subforum for your question anyway.

[quote="Kent Merryfield"]But you're differentiating with respect to $ x.$ There is no reason to assume anything other than that $ k$ is constant, hence there is no reason for the product rule.[/quote]crap true, sorry about that.  idk what i was thinking.

f(x)=c
f'(x)=0


Triangle ABC:
$AB=c, BC=a, AC=b$ and $a+b+c=3$
Prove that $3(a^2+b^2+c^2)+4abc \ge 13$

This inequality is even true for all non-negative real $a,b,c$  :) Since it can be proven by Schur and AM-GM without the assumption of $a,b,c$ being side of triangle .

[hide]
Yup. Homogenize and multiply through by $27$ to get

$27(a^2+b^2+c^2)+108abc \ge 13(a+b+c)^3$

$14(a^3+b^3+c^3)+30abc \ge 12(a^2b+a^2c+b^2c+b^2a+c^2a+c^2b)$

By Schur,

$10(a^3+b^3+c^3)+30abc \ge 10(a^2b+a^2c+b^2c+b^2a+c^2a+c^2b)$

so it remains to show

$4(a^3+b^3+c^3) \ge 2(a^2b+a^2c+b^2c+b^2a+c^2a+c^2b)$

which is obvious by Rearrangement or AM-GM.[/hide]


$ (A_1A_2A_3A_4...A_n)^{\minus{}1} \equal{} A_n^{\minus{}1} A_{n\minus{}1}^{\minus{}1}... A_2^{\minus{}1} A_1^{\minus{}1}$

$ (A_1A_2A_3A_4...A_n)^T \equal{} A_n^T A_{n\minus{}1}^T... A_2^T A_1^T$

where all the multiplications are defined, and all matrices are invertible and/or square where they need to be.

This is basically an exercise in associativity and definition.

$ M^{ - 1} = N$ if and only if $ MN = I$.

Is this satisfied?

\begin{eqnarray*} (A_1A_2\ldots A_n)(A_n^{ - 1}\ldots A_2^{ - 1}A_1^{ - 1}) & = & (A_1A_2\ldots )(A_nA_n^{ - 1})(\ldots A_2^{ - 1}A_1^{ - 1}) \\
& = & (A_1A_2\ldots )(I)(\ldots A_2^{ - 1}A_1^{ - 1}) \\
& = & A_1(A_2A_2^{ - 1})A_1^{ - 1} \\
& = & A_1 I A_1^{ - 1} \\
& = & A_1A_1^{ - 1} \\
& = & I \end{eqnarray*}

I probably should have written more steps, but you should get the idea.

I know the basic idea but I couldn't make the proof formal enough.

Both identities are a direct consequence of their $ n \equal{} 2$ cases by induction.  The first identity holds in any group; what's wrong with HilbertThm90's proof?  (It's undeniably rigorous when $ n \equal{} 2$.)

The second is somewhat subtle.  The $ n \equal{} 2$ case can be proven by rote using the definition of matrix multiplication, but I prefer to think about it in [url=http://en.wikipedia.org/wiki/Dual_space#Transpose_of_a_linear_map]somewhat more sophisticated terms[/url].


Putnam
Diophantine equation
number theory unsolved
number theory

Prove that exist $\infty$ integers $a$, such that $a, a+1$ $and$ $a+2$ can be expressed as the sum of two perfect squares.

Let's consider the Pell equation $x^{2}-2y^{2}=1$

It has $\infty$ solutions.

For each $y$ that satisfies that equation, $a=y^{2}+y^{2}$ can be expressed as a sum of two squares. $a+1=x^{2}+0^{2}$ can be expressed as a sum of two squares. $a+2=x^{2}+1^{2}$ can be expressed as a sum of two squares.

QED

This can be done without Pell and with a little algebra.

Note that there are infinitely many solutions in integers to the equation $z^{2}-1=x^{2}+y^{2}$. To see this, let $x=z-1, y=2k, z=2k^{2}+1$ for $k=1,2,3...$. Each such triple is a solution, since
$z^{2}-1=x^{2}+y^{2}$ gives $z^{2}-1=z^{2}-2z+1+4k^{2}$, so that
$2z=2(2k^{2}+1)$.
Now take $a=z^{2}-1=x^{2}+y^{2}$, so that $a+1=z^{2}+0^{2}$ and $a+2=z^{2}+1^{2}$. Since each $k$ produces a unique triple, we have infinitely many.


A five-digit positive integer is a ``mountain number'' if the first three digits are in ascending order and the last three digits are in descending order. For example, $ 35\ 761$ is a mountain number, but $ 32\ 323$ and $ 35\ 655$ are not. How many five-digit numbers greater than $ 70\ 000$ are mountain numbers?

the first three must be 789. if the fourth is 8, there are 8 choices for the fifth (0-8). if its 7, there are 7 choices. we have 1+2+3+...7, or 7(8)/2=$ 28$

Um, there are 8 choices for the fifth from [b]1 to 8, not 0 to 8[/b] and then it's $ 1\plus{}2\plus{}3\plus{}\ldots\plus{}\textbf{8}\equal{}\frac{8(9)}{2}\equal{}\boxed{36}$.


An assignment I do not understand:

Given the vector field $ \vec{F}\equal{}[y^2\minus{}xz,x^2\minus{}yz,z^2\minus{}xy]$

How do I find $ div\vec{F}$ and $ curl\vec{F}$?

Should I go for $ div\vec{F}$ first, than $ curl\vec{F}$? How? Could somebody please give me a push in the right direction.. :)

Look up the definition for div and curl.  Everything else is straightforward calculation.


inequalities
geometry
area of a triangle
triangle inequality
inequalities unsolved

Your question had been post in This Forum by Manlio. It generalizes the Hadwiger-Finsler inequality (see http://www.mathlinks.ro/Forum/viewtopic.php?t=31598 ).

[quote="keira_khtn"]If $ a$, $ b$, $ c$ are the sidelengths and $ S$ is the area of a triangle $ ABC$, then prove that
$ a^{2n}\plus{}b^{2n}\plus{}c^{2n}\geq 3\left(\frac{4}{\sqrt{3}}S\right)^{n}\plus{}\left(b\minus{}c\right)^{2n}\plus{}\left(c\minus{}a\right)^{2n}\plus{}\left(a\minus{}b\right)^{2n}$
for any positive integer $ n$.[/quote]

Let's revive some ancient topics. At least, this is a sure way to avoid helping cheaters with current olympiad/journal problems which seem to make out >50% of the newer posts...

Denote $ x\equal{}a^{2}\minus{}\left( b\minus{}c\right)^{2}$, $ y\equal{}b^{2}\minus{}\left( c\minus{}a\right)^{2}$, $ z\equal{}c^{2}\minus{}\left( a\minus{}b\right)^{2}$, $ X\equal{}\left( b\minus{}c\right)^{2}$, $ Y\equal{}\left( c\minus{}a\right)^{2}$, $ Z\equal{}\left( a\minus{}b\right)^{2}$.

Since $ a$, $ b$, $ c$ are the sides of a triangle, the triangle inequality yields $ c\plus{}a>b$ and $ a\plus{}b>c$, so that $ c\plus{}a\minus{}b>0$ and $ a\plus{}b\minus{}c>0$. Thus, $ a^{2}\minus{}\left( b\minus{}c\right)^{2}\equal{}\left( c\plus{}a\minus{}b\right)\left( a\plus{}b\minus{}c\right) >0$. In other words, $ x>0$. Similarly, $ y>0$ and $ z>0$. Also, trivially, $ X\geq0$, $ Y\geq0$ and $ Z\geq0$ (since $ X$, $ Y$, $ Z$ were defined as squares, and squares are $ \geq0$).

Obviously,

$ a^{2n}\equal{}\left( a^{2}\right)^{n}\equal{}\left(\left( a^{2}\minus{}\left( b\minus{}c\right)^{2}\right)\plus{}\left( b\minus{}c\right)^{2}\right)^{n}$
$ \equal{}\left( x\plus{}X\right)^{n}\equal{}\sum_{k\equal{}0}^{n}\binom{n}{k}x^{k}X^{n\minus{}k}$
$ \equal{}X^{n}\plus{}\sum_{k\equal{}1}^{n\minus{}1}\underbrace{\binom{n}{k}x^{k}X^{n\minus{}k}}_{\geq0\text{ since }x>0\text{ and }X\geq0}\plus{}x^{n}\geq X^{n}\plus{}x^{n}$.

Similarly, $ b^{2n}\geq Y^{n}\plus{}y^{n}$ and $ c^{2n}\geq Z^{n}\plus{}z^{n}$. Thus,

$ a^{2n}\plus{}b^{2n}\plus{}c^{2n}\geq\left( X^{n}\plus{}x^{n}\right)\plus{}\left( Y^{n}\plus{}y^{n}\right)\plus{}\left( Z^{n}\plus{}z^{n}\right)$
$ \equal{}\left( x^{n}\plus{}y^{n}\plus{}z^{n}\right)\plus{}\left( X^{n}\plus{}Y^{n}\plus{}Z^{n}\right)$.

Now, since $ n\geq1$ and the reals $ x$, $ y$, $ z$ are positive, the power mean inequality yields $ \frac{x^{n}\plus{}y^{n}\plus{}z^{n}}{3}\geq\left(\frac{x\plus{}y\plus{}z}{3}\right)^{n}$, so that

$ x^{n}\plus{}y^{n}\plus{}z^{n}\geq3\cdot\left(\frac{x\plus{}y\plus{}z}{3}\right)^{n}$.

But by [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=31598]the Hadwiger-Finsler inequality[/url], we have

$ a^{2}\plus{}b^{2}\plus{}c^{2}\geq4\sqrt{3}S\plus{}\left( b\minus{}c\right)^{2}\plus{}\left( c\minus{}a\right)^{2}\plus{}\left( a\minus{}b\right)^{2}$,

so that

$ \left( a^{2}\minus{}\left( b\minus{}c\right)^{2}\right)\plus{}\left( b^{2}\minus{}\left( c\minus{}a\right)^{2}\right)\plus{}\left( c^{2}\minus{}\left( a\minus{}b\right)^{2}\right)\geq4\sqrt{3}S$.

In other words, $ x\plus{}y\plus{}z\geq4\sqrt{3}S$. Hence,

$ x^{n}\plus{}y^{n}\plus{}z^{n}\geq3\cdot\left(\frac{x\plus{}y\plus{}z}{3}\right)^{n}\geq 3\cdot\left(\frac{4\sqrt{3}S}{3}\right)^{n}\equal{}3\cdot\left(\frac{4}{\sqrt{3}}S\right)^{n}$.

Thus,

$ a^{2n}\plus{}b^{2n}\plus{}c^{2n}\geq\left( x^{n}\plus{}y^{n}\plus{}z^{n}\right)\plus{}\left( X^{n}\plus{}Y^{n}\plus{}Z^{n}\right)$
$ \geq3\cdot\left(\frac{4}{\sqrt{3}}S\right)^{n}\plus{}\left( X^{n}\plus{}Y^{n}\plus{}Z^{n}\right)$
$ \equal{}3\cdot\left(\frac{4}{\sqrt{3}}S\right)^{n}\plus{}\left(\left(\left( b\minus{}c\right)^{2}\right)^{n}\plus{}\left(\left( c\minus{}a\right)^{2}\right)^{n}\plus{}\left(\left( a\minus{}b\right)^{2}\right)^{n}\right)$
$ \equal{}3\cdot\left(\frac{4}{\sqrt{3}}S\right)^{n}\plus{}\left(\left( b\minus{}c\right)^{2n}\plus{}\left( c\minus{}a\right)^{2n}\plus{}\left( a\minus{}b\right)^{2n}\right)$.

This solves the problem.

  Darij


Crazy movie trailer. Hahaha I wanna go see it. Tap dancing penguins wohooooooo.

I'll probably be the oldest person at the theatre seeing Happy Feet that isn't an adult bringing their kids.

.. beat casino royale in the US box office (for the premiere weekend).

GO ON THE HAPPYFEET WEBSITEEEE www2.warnerbros.com/happyfeet there is a new OOTP trailer!!!!

[quote="Ignite168"]That's becauase Casino Royale is a crappy movie and Happy Feet is the lesser crap and tons of soccer moms will bring their spoiled brats to see it.[/quote]

Wow ignite have you seen both?

Anyway, since I'm in the UK right now, bond seems to be one of the most popular movies series.

Your soccer mom comment is quite funny :-)


A sequence of integers $ a_1,a_2,a_3,...$ is such that $ a_1\equal{}2, a_2\equal{}3$, and

$ a_{n\plus{}1}\equal{}2a_{n\minus{}1}$ or $ 3a_n\minus{}2a_{n\minus{}1}$ for all $ n \ge 2$.

Prove that no number between $ 1600$ and $ 2000$ can be an element of the sequence.

posted before, but $ a_{n\plus{}1}\equal{}2a_n$ or $ a_{n\plus{}1}\equal{}3a_n\minus{}2a_{n\minus{}1}.$


linear algebra
matrix
linear algebra unsolved

It's pretty obvious that permutation matrix is a product of tranposition matrices, but how can we prove it?

The group of permutation matrices is isomorphic to $S_{n}$ in a canonical way: permutation \[\begin{pmatrix}123\dots n\\ i_{1}i_{2}i_{3}\dots i_{n}\end{pmatrix}\] corresponds to the matrix that has $1$ in the $j$th column and $i_{j}$th row. So the result follows from what we know about $S_{n}$: every permutation is a product of cycles, etc.