Given

$ x\plus{}y\plus{}z\equal{}1$
$ x^{2}\plus{}y^{2}\plus{}z^{2}\equal{}2$
$ x^{3}\plus{}y^{3}\plus{}z^{3}\equal{}3$,

show that

$ xy\plus{}yz\plus{}zx\equal{} \frac{\minus{}1}{2}$.

Show also that $ x^{2}y\plus{}x^{2}z\plus{}y^{2}x\plus{}y^{2}z\plus{}z^{2}x\plus{}z^{2}y \equal{}\minus{}1$

and deduce that

$ xyz\equal{}\frac{1}{6}$.

Hence find $ a, b$ and $ c$, independent of $ n$, such that

$ S_{n\plus{}1} \equal{} aS_{n} \plus{} bS_{n\minus{}1} cS_{n\minus{}2}$,

where $ S_{n} \equal{} x^{n}\plus{}y^{n}\plus{}z^{n}$.

[hide]
From the first two we can easily take that $ xy \plus{} yz \plus{} zx \equal{} \minus{} \frac {1}{2}$.(just square the first equation).

Then by euler's identity we take:

$ x^3 \plus{} y^3 \plus{} z^3 \minus{} 3xyz \equal{} (x \plus{} y \plus{} z)(x^2 \plus{} y^2 \plus{} z^2 \minus{} xy \minus{} yz \minus{} zx) \Longleftrightarrow 3 \minus{} 3xyz \equal{} 1 \cdot (2 \minus{} ( \minus{} \frac {1}{2})) \Longleftrightarrow xyz \equal{} \frac {1}{6}$.

Also $ \sum_{sym} x^2y \equal{} (x \plus{} y \plus{} z)(xy \plus{} yz \plus{} zx) \minus{} 3xyz \equal{} \minus{} \frac {1}{2} \minus{} \frac {1}{2} \equal{} \minus{} 1$.

For $ n\geq 3$ :

$ \boxed{x^{n} \plus{} y^{n} \plus{} z^{n} \equal{} (x \plus{} y \plus{} z)\left(x^{n \minus{} 1} \plus{} y^{n \minus{} 1} \plus{} z^{n \minus{} 1}\right) \minus{} (xy \plus{} yz \plus{} xz)\left(x^{n \minus{} 2} \plus{} y^{n \minus{} 2} \plus{} z^{n \minus{} 2}\right) \plus{} xyz\left(x^{n \minus{} 3} \plus{} y^{n \minus{} 3} \plus{} z^{n \minus{} 3}\right)}$[/hide]


geometry
area
combinatorics
combinatorial geometry
dissection
IMO Shortlist
IMO Longlist

A figure with area $1$ is cut out of paper. We divide this figure into $10$ parts and color them in $10$ different colors. Now, we turn around the piece of paper, divide the same figure on the other side of the paper in $10$ parts again (in some different way). Show that we can color these new parts in the same $10$ colors again (hereby, different parts should have different colors) such that the sum of the areas of all parts of the figure colored with the same color on both sides is $\geq \frac{1}{10}.$

Here's a solution. call the areas on the second side $A_1,A_2,...,A_{10}$, let the colors be $c_1,c_2,...,c_{10}$. now, consider the 10 colorings coloring $A_i$ with $c_{i+k}$, $k=0,...,9$, where indices are mod 10. summing all the areas up which are in common for the two sides of the paper, we immediately see that each point is colored exactly once with any color, thus we get exactly 1(every point on the back has a certain color and this color occurs exactly once and the area of the figure is 1). it immediately follows that for some coloring, the areas with same color are at least $\frac {1}{10}$, it even follows that there is some coloring where it is at most $\frac {1}{10}$.

Peter

:?

I don't get both:
[quote="Peter Scholze"]it immediately follows that for some coloring, the areas with same color are at least $\frac {1}{10}$[/quote]if you divide them 0.01-...-0.01-0.91, then there are at least 8 areas where the surface is only 0.02
[quote="Peter Scholze"]it even follows that there is some coloring where it is at most $\frac {1}{10}$.[/quote]if you make the division 0.1-0.1-...-0.1 then each color has 1/5 area regardless of how you color.

Or did you guys mean that you can also choose how to divide the paper up? :) Cuz then it's completely trivial :?


Find values for x that are solutions to the following inequality

[u]x/\3 - 10x +8[/u]
x/\2 - 3x -10

[u]<[/u]

[u]x/\2 - 4x +8[/u]
x/\2 - 3x - 10

[hide="partial solution"]they have the same denominator so we just look for values of x that satisfy $x^2-x^3-6x>0$.(It's the second numerator minus the first). I factored it and got x(-x+3)(x+2)>0...but i'm stuck there. :blush: [/hide]

[quote="surge"]$x^3-x^2-6x\le0$
$x(x-3)(x+2)\le0$
$(-\infty,-2] \cup[0,3]$[/quote]

$-2$ isn't included in the solution set.
$(-\infty,-2) \cup[0,3]$  :D

To prove that you have missing solutions, let $x=-1$ and $x=4$...
[EDIT] Counterexample: let $x=1$ and the inequality is not satisfied...

frt, there is one mistake in your final answer. See if you can find it :)

\boxed{(-\infty, -2)\cup (-2,0]\cup [3,5)} is that the answer?


what kind of music do you listen to? who are your favorite artists?

i like rock and pop, and some R&B and rap.
love Carlos Santana, Greenday, Michael Jackson, duran duran, 3 doors down, eminem, etc

i like hardcore thug rap music. representin' lil jon, luda, 50, 3-6. you know, i keep it crunk in herre.

[quote="stupidkid"]I think rap is just an african (or in the rare cases a white guy) guy talking with the accompaniment of a drum beat.[/quote]

Careful what you say. The very high majority of black people in the United States aren't African. Nationality is different from race.

I like most kinds of rock 'n' roll, especially the older stuff from the sixties. I like Simon and Garfunkel and Bob Dylan a lot. But I'll listen to almost anything, besides the really poppy stuff and rap. I even like listening to classical music  :?      :)

My favorite song? "The Sound of Silence" (Simon and Garfunkel). And may Ray Charles rest in peace. Great man, Ray Charles. And even better musician.

More and more, I come to appreciate good music that has stood the test of time and bands like U2, Metallica, (my mind is blancking) continue to sell albums (or singles if you are talking about online) even if they started 20 years ago.

I also like some newer bands like White Stripes, JET, Maroon 5, Coldplay, Red Hot Chili Peppers, The Offspring, and Rage against the Machine.

yay for rage against the machine.
(maybe somebody noticed my avatar)

I'll listen to just about anything (cept country).  Artists on my mp3 player/computer right now include Foreigner, Billy Joel, Hootie and the Blowfish, Phil Collins, U2, Pink Floyd, Metallica, Audioslave, 50 cent, Eminem, Ashanti, Jay Z, Nelly, 2pac, and so on.

John Adams
Short Ride on the Fast Machine

[quote="aznness"]GREEN DAY![/quote]

yaaaaaaaaaaaaaaaa

i listen to country music! it is so awesome. its not what some people think of like all that yodeling and stuff it is pretty up beat and fun! i listen to someother stuff too like green day. but country music is the best!!!!!!!!!!!

i listen to all kinds... just like joml88.

[quote="juicybooty911"]is Coldplay considered rock or pop?[/quote]
i would call it soft rock

[quote="juicybooty911"]is BoA a singer or movie star?[/quote]

Singer and model mainly.

She was invited to play Cho in Harry Potter but rejected. Thank goodness. Do not want her kissing Radcliffe. Yuck.

[quote="MithsApprentice"]
She was invited to play Cho in Harry Potter but rejected. Thank goodness. Do not want her kissing Radcliffe. Yuck.[/quote]

lol. u crack me up. she does look right for the part tho, better than the girl that got it, at least i think so.

[quote="MathFiend"][quote="MithsApprentice"]She was invited to play Cho in Harry Potter but rejected. Thank goodness. Do not want her kissing Radcliffe. Yuck.[/quote]lol. u crack me up. she does look right for the part tho, better than the girl that got it, at least i think so.[/quote]Probably I will find a lot of retractors here... but I think that Katie Leung is cuter than BoA (any way, both are very pretty asian girls), and I think both would do good for the part.

anybody here like 3 doors down?
i just brought their new cd seventeen days
awesome :D

mithsapprentice...my sister sez ur like obsessed with boA and i can see why *looks at avatar/sig* why do u like her so much music or looks??

anything except country.
jason mraz is great
usher is AWESOME
alicia keys, beyonce....
eminem, some ludacris.
train.

i never knew taht they asked BoA to play Cho Chang. interesting.

i mostly like rock, and i'm not too fond of rap and country. my favorite artists are Greenday and The Eagles.

[quote="noneoftheabove"]i hate country[/quote]
:D  :D  :D

I like 2Pac.I think he is the gratest rapper of all time,but I also listen to Rock,Pop and R-B.

Mostly rock - many types.. rock, punk rock, heavy rock, metal, nu-metal, heavy metal, alternative metal, and again, just rock.

I also like some rap, like 50 and eminem.. not my fav genre though.

Green day & 3 doors down are awesome, among others (tho i think green day doesnt have much musical talent)

Nightwish, Bloodflowerz, Within Temptation, Epica, Apocalyptica, Evanescence, Pantera, Nirvana, Iron Maiden, Rammstein, Darkseed, TT34, Amatory, etc
Also i like moldavian rock: Zdob si Zdub, Alternosfera, Gindul Mitei, ExNN

soundtracks, classical, rock, hard rock, classic rock, rap, hip - hop, all the hits nowadays, and a ton of others.....i'm really flexible

btw, anyone know where i can hear jamaican? its supposedly awesome!

Guster owns all.

Hellogoodbye also.  :D


Let $ x,y > \equal{} 0$ with $ x \plus{} y \equal{} 2$ Prove that $ x^2y^2(x^2 \plus{} y^2) < \equal{} 2$

$ x\plus{}y\equal{}2$   $ \iff$     $ (x\plus{}y)^2\equal{}4$

$ \iff$   $ x^2\plus{}y^2\equal{}4\minus{}2xy$

the inequality become :

$ x^2y^2(4\minus{}2xy)\,\leq\,2$
or :

$ 2x^3y^3\minus{}4x^2y^2\plus{}2\,\geq\,0$

wish is true  :)

[quote="mehdi cherif"]

$ 2x^3y^3 \minus{} 4x^2y^2 \plus{} 2\,\geq\,0$

wish is true  :)[/quote]

Why is this true?

yes . i think tha $ 2a^3\minus{}4a^2\plus{}2\,\geq\,0$

Yes very easy to prove first noticing that by AM-GM we have $ a\in (0,1)$
$ \sqrt{xy}\leq \frac{x\plus{}y}{2}\equal{}1$
Then easy to see that:
$ (a^{3}\minus{}2a^{2}\plus{}1\geq 0)\iff (a^{2}\minus{}2a\plus{}1\minus{}1\plus{}\frac{1}{a}\geq 0)\iff ((a\minus{}1)^{2}\plus{}(\frac{1}{a}\minus{}1)\geq 0)$ which is obvious :wink:

very good.
easy this ireland 2000  :!:  :!:


inequalities
real analysis
real analysis unsolved

Solve  $x^{356}\le 2x^{90}-7x+1$ using only Young's Inequality.
P.S: I need help so I do not actually have the solution.


analytic geometry
graphing lines
slope
inequalities

$ABCD$ is a square, and $E$ is a point on $CD$. Extend $AE$ and $BC$, and let $F$ be their intersection. Prove $AE+AF>2AC$.

[hide]Set up a coordinate plane where $ABCD$ is a unit square, with $A=(0,0),\ B=(1,0),\ C=(1,1)\ D=(0,1),\ E=(p,1),\ 0<p<1$. Line $AE$ has slope $\frac{1}{p}$, so extending to the line $x=1$ gives $F=\left(1,\frac{1}{p}\right)$. Then $2AC=2\sqrt{2}$, $AE=\sqrt{p^2+1}$, $AF=\sqrt{1+\left( \frac{1}{p}\right) ^2}$. Adding, $AE+AF=\sqrt{p^2+1}\left(1+\frac{1}{p}\right)$. Applying AM-GM Inequality (strict inequality since $\frac{1}{p}\neq 1$), $AE+AF>2\sqrt{\frac{p^2+1}{p}}=2\sqrt{p+\frac{1}{p}}$. Using AM-GM again, $AE+AF>2\sqrt{2}=2AC$.[/hide]

Wow, I didn't think of a proof using coordinate geometry. Nice job :)

[hide="Hint for non-analytical proof"]
The sum of the lengths of two sides of a triangle is always greater than the length of the third side ;) [/hide]


geometry
3D geometry
sphere
integration
analytic geometry
trigonometry
calculus

So I am working on this problem.
Find the surface area of that portion of the sphere $x^{2}+y^{2}+z^{2}=4$ that lies inside the cylinder $x^{2}+y^{2}=2y$.

So I used the formula for surface area.
$z=\sqrt{4-x^{2}-y^{2}}$  and $A=\int\int_{D}\sqrt{1+\frac{\partial z}{\partial x}^{2}+\frac{\partial z}{\partial y}^{2}}dA=\int\int_{D}\sqrt{1+\left(\frac{-x}{\sqrt{4-x^{2}-y^{2}}}\right)^{2}+\left(\frac{-y}{\sqrt{4-x^{2}-y^{2}}}\right)^{2}}dA$
=$\int\int_{D}\sqrt{1-\frac{x^{2}+y^{2}}{x^{2}+y^{2}-4}dA}$

This looks like the perfect problem for using polar coordinates! Therefore.
$\int\int_{D}\sqrt{1-\frac{x^{2}+y^{2}}{x^{2}+y^{2}-4}}dA=\int_{?}^{?}\int_{?}^{?}\sqrt{1-\frac{r^{2}}{r^{2}-4}}rdrd\theta$
But my problem is converting the cylinder boundary to polar. The cylinder is $x^{2}+(y-1)^{2}=1$ when you complete the square. I know a the boundaries for $x^{2}+y^{2}=1$ are simply: $0\leq r\leq1, 0\leq\theta\leq2\pi$ and that the boundary cylinder given is just that except the center point is move over to (0,1)...please help me move the cylinder.

Go back to $x^2+y^2=2y.$ Put in $x=r\cos\theta,\,y=r\sin\theta.$

$r^2=2r\sin\theta$ or $r=2\sin\theta.$

Sketch this circle; it lies entirely in the "upper" half plane $y>0$ (or $0<\theta<\pi.$)

The following should work as limits of integration:

$\int_0^{\pi}\int_0^{2\sin\theta}g(r,\theta)r\,dr\,d\theta.$

Also, as for the function you're integrating: you'd be a lot better off putting everything that's inside the square root over a common denominator. And since $r^2<4,$ you'll confuse yourself by writing $r^2-4.$ Leave that as $4-r^2.$

Don't forget that $dA=r\,dr\,d\theta.$

And that upper limit on the inner integral is $2\sin\theta.$ It can't possibly be $r\sin\theta$ - $r$ can't run over an interval whose boundaries depend on $r.$


calculus
integration
LaTeX
real analysis
real analysis unsolved

COmpute THE SUM OF the following two terms:

the triple integral of (sin x)/(x)
dx from y*pi to (2*pi)-z
dz from 0 to pi
dy from 0 to 1

and the triple integral of (sin x)/(x)
dx from y*pi to (2*pi)-z
dz from 0 to (2*pi) - (y*pi)
dy from 1 to 2

----------

I can't draw the region and I don't know how to make change of variables here, or Fubini's theorem??

Really, once was enough: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=257018

Also, learn some [[LaTeX]], either at that link or at http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690


"Another sort of material it might be worth collecting and making available is the [b]commentary on shortlist problems[/b] by individual Leaders and Observers (and sometimes comments by the PSC not included in the shortlist booklet) distributed to the Jury. (As the distribution of this material tends to be somewhat disorganised, it can be difficult to know how complete your set is for a given year.)
Another thing I would consider of interest to collect would be the [b]mark schemes[/b] themselves - though of course quite a bit ends up getting decided on precedent rather than being put in the written mark schemes.
Then there are the particular solutions that received [b]special prizes[/b], if known.  Once you go beyond just the statements of the problems, all of this forms part of the surrounding historical material." [url=http://www.srcf.ucam.org/~jsm28/]Joseph S. Myers[/url]

I am also interested in alternative solutions to IMO/ISL/ILL that participants/leaders/observers/coordinators figured out during or after the IMO. For example there is a very interesting solution that I first saw in the German IMO combinatorics training book "[i]Streifzüge durch die Kombinatorik (Excursions through Combinatorics)[/i]" by former long-term GDR IMO leader [i]Gustav Burosch[/i]. Thanks to [url=http://www.amt.edu.au/pjt.html]Peter Taylor[/url] who sent me the electronic copy from the original source: "[i]Maths Competitions, the Australian Scene, 1986. AMT Publishing, Canberra. pp 44-46. JM Campbell.[/i]" Apparently as [url=http://www.math.uni-rostock.de/~gronau/]Dieter Gronau,[/url] recipient of the [url=http://www.amt.edu.au/erdgronau.html]2008 Paul Erdös award,[/url] says there is also a forthcoming publication of Wegert-Reiher Discrete Appl. Math. that generalizes IMO problem 1986/3.

[b]Any contributions will be appreciated![/b]

Relaxation Procedures on Graphs by [b][url=http://www.mathe.tu-freiberg.de/~wegert/wegert.html]Elias Wegert[/url][/b] and [b]Christian Reiher[/b]:

[b]Abstract:[/b] [i]The procedures studied in this paper originate from a problem posed at the International Mathematical Olympiad in 1986. We present several approaches to the IMO problem and its generalizations. In this context we introduce a “signed mean value procedure” and study “relaxation procedures on graphs”. We prove that these processes are always finite, thus confirming a conjecture of Akiyama, Hosono and Urabe [1]. Moreover, we indicate relations to sorting and to an iterative method used in circle packing.[/i]


conics
parabola
calculus
derivative
analytic geometry
graphing lines
slope

The parabolas $ y\equal{}ax^2\plus{}bx\plus{}c$ and $ y\equal{}dx^2\plus{}ex\plus{}f$ have no common points, where $ a,b,c,d,e,f$ are real numbers with $ ad<0$. Prove that there is a line having no common points with either of the two parabolas.

I think the following works. The locuses of parabolas are $ P_1 \equal{} \left ( \minus{} \frac {b}{2a}, \minus{} \frac {b^2}{4a} \plus{} c \plus{} \frac {1}{4a}\right )$ and $ P_2 \equal{} \left ( \minus{} \frac {e}{2d}, \minus{} \frac {e^2}{4d} \plus{} f \plus{} \frac {1}{4d}\right )$. Let $ P_3$ be the intersection point of $ P_1P_2$ and $ y \equal{} ax^2 \plus{} bx \plus{} c$ and $ P_4$ be the intersection point of $ P_1P_2$ and $ y \equal{} dx^2 \plus{} ex \plus{} f$. Now the midpoint normal of $ P_3P_4$ is a line having no common points with either of the two parabolas.

As parabolas are continous, $ ax^2 \plus{} bx \plus{} c > dx^2 \plus{} ex \plus{} f$ for all $ x$. The derivatives are $ 2ax \plus{} b$ and $ 2dx \plus{} e$. There exists a point $ m \equal{} (e \minus{} b)/2(a \minus{} d)$ where derivatives gets the same value. There exits point $ (x,y)$ such that $ x \equal{} m$ and $ ax^2 \plus{} bx \plus{} c > y > dx^2 \plus{} ex \plus{} f$, let this be $ A$. Draw a line $ S$ which goes through $ A$ and has slope $ k$.

Now the derivative satisfies $ 2ax \plus{} b < k$ as $ x < m$ and $ 2ax \plus{} b > k$ as $ x > m$ or $ 2dx \plus{} e$ behaves vice versa. Thus parabola goes further from $ S$, which is the required line.


induction
modular arithmetic
number theory unsolved
number theory

Let $n$ be a positive integer. Find, with proof, the least positive integer $d_{n}$ which cannot be expressed in the form \[\sum_{i=1}^{n}(-1)^{a_{i}}2^{b_{i}},\]
where $a_{i}$ and $b_{i}$ are nonnegative integers for each $i.$

[quote="N.T.TUAN"]Let $n$ be a positive integer. Find, with proof, the least positive integer $d_{n}$ which cannot be
expressed in the form
\[\sum_{i=1}^{n}(-1)^{a_{i}}2^{b_{i}}, \]
where $a_{i}$ and $b_{i}$ are nonnegative integers for each $i.$[/quote]

Very nice problem. The answer is $d_{n}=2\frac{4^{n}-1}{3}+1$

Notations :
a) $\sum_{i=1}^{n}(-1)^{a_{i}}2^{b_{i}}=(a_{i},0,\cdot,a_{i-1},\cdot,0,\cdot,a_{1},0,0,\cdot)$, each $a_{i}$ being at the $b_{i}^{th}$ position from the right.
b) If k may be written with such exactly n nonzero positions, I'll write that k has a n_representation

[u]Lemma L1 [/u] : Obviously, any number equal to 3 modulus 4 ends with either $(1,1)$, either $(0,-1)$.
[u]Lemma L2 [/u]: If a number has a p_representation, it also has a q_representation for any $q>p$  :
Leftmost position is always $1$. $(1,$ anything with $p-1$ nonzero values$)=(1,$ $q-p$ values $-1,$ the same anything as LHS $)$
[u]Lemma L3 [/u]: If k has a p_representation ($p\geq 1$), $k+1$ has a $(p+1)$_reprentation. Easy with induction on k:
If 1 has a p_representation($p\geq 1$), 2 has a $(p+1)$_reprentation (since $2=(1,0)$ + Lemma L2)
If k=$2u$ has a p_representation ending with 0, $k+1$ has obviously a $p+1$_representation, (by adding a $1$ at righmost position)
If k=$2u+1$ has a p_representation ending with +1, $u$ has obviously a $(p-1)$_representation, so $u+1$ has a $p$_representation (induction), and $k+1=2u+2$ has a p_representation, and so a $(p+1)$_reprentation (Lemma L2)
[color=red][b]edited :[/b][/color] If k=$2u-1$ (we must consider that the number we are studying may end with 0, 1 or also -1)  has a p_representation ending with -1, $k+1=2u$ has a $(p-1)$_representation (suppress the rightmost -1), and so also a $(p+1)$_reprentation (Lemma L2)

Easiest proof now is with induction.
1) $n=1$
$1=(1)$
$2=(1,0)$
$3$ can't be written with just one nonzero value (see lemma L1 above).
And $d_{1}=3=2\frac{4^{1}-1}{3}+1$

2) assume the propery is verified up to rank $n\geq 1$. Let $m=d_{m}-1$

2.1) $1, 2$ and $3$ has a $(n+1)$_representation :
$1=(1)$ + Lemma L2
$2=(1,0)$ + Lemma L2
$3=(1,1)$ + Lemma L2 if necessary ($n>2$)

2.2) For any $q\in[4,4m+2]$ q has a $(n+1)$_representation :
If $q=4k$ with $k\leq m$, $k$ has a $n$_representation (induction), so a $(n+1)$_representation (Lemma L2), so has q (adding $(0,0)$ at the right.
If $q=4k+1$ with $k\leq m$, $k$ has a $n$_representation (induction), so q has a $(n+1)$_representation (adding $(0,1)$ at the right.
If $q=4k+2$ with $k\leq m$, $k$ has a $n$_representation (induction), so q has a $(n+1)$_representation (adding $(1,0)$ at the right.
If $q=4k+3=4(k+1)-1$ with $k+1\leq m$, $k+1$ has a $n$_representation (induction), so q has a $(n+1)$_representation (adding $(0,-1)$ at the right of the reprentation of $k+1$.

2.3) $4m+3$ don't have a $(n+1)$_representation
If $4m+3$ has a $(n+1)$_representation, it ends with $(1,1)$ or $(0,-1)$ (Lemma L1).
If it ends with $(0,-1)$, by deleting these 2 rightmost numbers, we have a n_representation of $m+1$, in contradiction with induction $m=d_{n}-1$
If it ends with $(1,1)$, by deleting these 2 rightmost numbers, we have a $(n-1)$_reprentation of $m$, so it exist a n_representation of $m+1$ (Lemma L3), in contradiction with induction $m=d_{n}-1$.

So $d_{n+1}=4m+3=4d_{n}-1=4(2\frac{4^{n}-1}{3}+1)-1=2\frac{4^{n+1}-1}{3}+1$

Q.E.D.

--
Patrick

[quote="Rust"]I don't undestand. For all natural k we have $k=\sum_{i=1}^{n}2^{b_{i}}$, were $b_{i}$ are distinct nonnegative integers.
$0=1-1$.[/quote]
Are you sure? Remember that $n$ is FIXED.
@pco: Thank you very much! I will check your solution later, i am busing.  :wink:

Yes. I don't read, that n fixed.
For these case obviosly minimal number $d_{n}=1010...1011=1+2*(1+...+4^{n-1})=1+2*\frac{4^{n}-1}{3}.$

Here is an inductive argument (which I hope works).

[b]Solution[/b]

Let a positive integer that can be expressed in the stated form be defined as $n$-good. We claim that the minimal number that is not $n$-good is $d_n = 2 \left( \frac{4^n - 1}{3} \right) +1$. We will prove this in two steps.

[b]Step 1:[/b] All $m \in \mathbb{N}$ such that $1 \le m \le 2 \left( \frac{4^n - 1}{3} \right)$ are $n$-good.

[i]Proof:[/i] Assume that the hypothesis holds for $n=k$. Therefore all $1 \le m \le 2 \left( \frac{4^k - 1}{3} \right)$ can be expressed in the described way. Since $1=2^{k}-2^{k-1}-2^{k-2} - \dots - 2^0$, $1$ is $k+1$-good. For any $m$ such that $1 \le m \le 2 \left( \frac{4^k - 1}{3} \right)$ consider the expressions $2^l \pm m$ where $l=0,1,\dots , 2k+1$. Since $2^{2k-1}<2 \left( \frac{4^k - 1}{3} \right)<2^{2k}$, by this method we achieve an expression with $k+1$ terms for each positive integer less than or equal to $2^{2k+1} + 2 \left( \frac{4^n - 1}{3} \right)= 2 \left( \frac{4^{k+1} - 1}{3} \right)$. Therefore all $m \in \mathbb{N}$ such that $1 \le m \le 2 \left( \frac{4^{k+1} - 1}{3} \right)$ are $k+1$-good. This completes the induction. $\blacksquare$

[b]Step 2:[/b] $2 \left( \frac{4^n - 1}{3} \right) +1$ and $\frac{4^{n+1} - 1}{3}$ are not $n$-good.

[i]Proof:[/i] Assume that both hypotheses hold for $n=k$. Note that any $n$-good number is $m$-good for all natural numbers $m \ge n$. This is because we may exchange a $\pm (2^l)$ in the expression with a $\pm (2^{l+1} - 2^{l})$ to increase the number of terms in the expression without changing the value. Therefore we may assume that there is only one $\pm 1$ since otherwise we can exchange any excess $\pm 1$ for $\pm 2$’s. Note that if a number is not $n$-good, then the minimum number of summands in the expression exceeds $n$. Now assume for contradiction that $2 \left( \frac{4^{k+1} - 1}{3} \right) +1$ is $k+1$-good. Then there must be a $\pm 1$ in the expression since it is an odd number. If it is a $+1$, then subtracting $1$ and dividing by $2$ yields that $\frac{4^{k+1} - 1}{3}$ requires $k$ summands minimum. This contradicts the fact that $\frac{4^{k+1} - 1}{3}$ is not $k$-good. Similarly if it is a $-1$, then adding $1$ and dividing by $2$ contradicts the fact that $2 \left( \frac{4^{k+1} - 1}{3} \right) +1$ is not $k$-good.  We arrive at the same contradictions for $\frac{4^{k+1} - 1}{3}$. This completes the induction. $\blacksquare$

Therefore the minimum value is $d_n = 2 \left( \frac{4^n - 1}{3} \right) +1$.

[hide="another inductive argument to show that all numbers less than the desired sum are n-good"]
[b]Lemma 1[/b] Any integer $n$ that can be expressed as $\sum_{i=1}^j{(-1)^{a_i}2^{b_i}}$ can also be expressed as $\sum_{i=1}^k{(-1)^{a_i}2^{b_i}}$ for all $k\ge j$.
[i]Proof[/i] We break into two cases: $k\equiv j\pmod{2}$, $\not\equiv j\pmod{2}$:
1. $k\equiv j\pmod{2}$:
Note that $n=\sum_{i=1}^j{(-1)^{a_i}2^{b_i}}=\sum_{i=1}^{\frac{k-j}{2}}{(-1)^{c_i}2^{d_i}}+\sum_{i=1}^{\frac{k-j}{2}}{(-1)^{c_i+1}2^{d_i}}+\sum_{i=1}^j{(-1)^{a_i}2^{b_i}}$ for any $\{c_i\}, \{d_i\}$.  Thus, we conclude.
2. $k\not\equiv j\pmod{2}$:
Note that $n=\sum_{i=1}^j{(-1)^{a_i}2^{b_i}}=\sum_{i=1}^{\frac{k-j-1}{2}}{(-1)^{c_i}2^{d_i}}+\sum_{i=1}^{\frac{k-j-1}{2}}{(-1)^{c_i+1}2^{d_i}}+((-1)^{a_j}2^{b_j+1}+(-1)^{a_i+1}2^{b_j})+\sum_{i=1}^{j-1}{(-1)^{a_i}2^{b_i}}$ for any $\{c_i\}, \{d_i\}$.  Thus, we conclude.

We claim that $d_n=2\left(\frac{4^n-1}{3}\right)+1$.
We show this by induction.  For base case, consider $n=1$.  Note that $2\left(\frac{4^1-1}{3}\right)+1=3$ is the smallest integer that is not a power of 2, and thus, $d_1=3$.  Thus, the conclusion holds for $n=1$.

Now, for the inductive hypothesis, assume that the result holds for $n=k$.  Then, we will show that it holds for $n=k+1$.
(We denote a number that can be expressed as $\sum_{i=1}^j{(-1)^{a_i}2^{b_i}}$ as $j$-good.  Thus, Lemma 1 can be phrased as "any $j$-good integer is also $k$-good, for all $k\ge j$.")
Now, because $2\left(\frac{4^k-1}{3}\right)+1=2^{2k-1}+2^{2k-3}+\cdot\cdot\cdot+2^1+2^0$ has $2k$ digits in base 2, any number with $\le2k$ digits is $k$-good and thus, by Lemma 1, $(k+1)$-good.  We now show that all numbers with $2k$ digits and $2k+1$ digits base 2 are $(k+1)$-good:
First, consider a number with $2k$ digits and $d$ 1's in base 2.  If $d\le k+1$, we can obviously express the number simply in its binary form, using Lemma 1.  If $d\ge k+2$, we can express the number as the difference of $2^{2k}-1$ and all the spots that have 0's.  Because there must be $2k-d\le k-2$ 0's, we can express the number in $\le2+(k-2)=k$ powers of 2, so it is $(k+1)$-good.
Now, consider a number with $2k+1$ digits and $d$ digits in base 2.  If $d\le k+1$, we can obviously express the number simply in its binary form, using Lemma 1.  If $d\ge k+2$, we can express the number as the difference of $2^{2k+1}-1$ and all the spots that have 0's.  Because there must be $2k+1-d\le k-1$ 0's, we can express the number in $\le2+(k-1)=k+1$ powers of 2, so it is $(k+1)$-good.
Finally, consider a number that with $2n+2$ digits that is less than $2^{2n+1}+2^{2n-1}+\cdot\cdot\cdot+2^1+2^0$.  Clearly, it must be of the form $2^{2n+1}+\cdot\cdot\cdot+2^{2n-x}+N$ where $x$ is odd and $N<2^{2n-x-2}$ (essentially, we take one power of 2 out of the sum above and replace the remaining spots with any smaller number).  Then, we use the exact same argument used above to show that $N$ can be expressed in at most $k-\frac{x-1}{2}$ powers of 2, after which we have that the number as a whole can be expressed in at most $\frac{x+1}{2}+k-\frac{x-1}{2}=k+1$ digits.
Thus, we have that any number less than $2\frac{4^{k+1}-1}{3}+1$ is $(k+1)$-good.

EDIT: I just realized my way of proving this is wrong...$\blacksquare$
[/hide]

How do you prove that $2\left(\frac{4^{n}-1}{3}\right)+1$ is not $n$-good?  I don't quite follow the explanation in the post above...

[quote="supercomputer"]How do you prove that $2\left(\frac{4^{n}-1}{3}\right)+1$ is not $n$-good?  I don't quite follow the explanation in the post above...[/quote]

Here is my own solution, hopefully it will help.

I will deal with everything in base $2$. The answer for $n$ is $1010...1011$ where there are $2n$ digits. The goal of the problem can be rethought of as taking a binary number and first adding powers of $2$ onto it (we'll call this part [i]building[/i]), then plucking off all the remaining $1$s so that you end up with $0$ (we will call this part the plucking part), (we will call this whole process [i]destroying[/i]).

[b]Lemma:[/b] You can re-destroy $0$.

[i]Proof:[/i] Let the number of moves required be $r$. If $2 | r$, then all you have to do is add $1$, then pluck it off. Easy. If $r$ is odd and $r > 1$, add a $1$, then add $1$ again to get $10$, then destroy it to get $0$ and now just alternate like before. If $r = 1$, then we have to retrace. The reason why we even include this lemma is if our destroy process hits $0$ before the $n$-th move. Thus, consider what the $n-1$-th move was. It had to be from the plucking stage, which means we were left with $2^a$ for some integer $a$. We have two moves left, so add $2^a$ and then subtract $2^{a+1}$.

$\Box$

[b]Part I:[/b] Anything less than $1010...1011$ ($2n$ digits) can be destroyed.

Assume that for all integers $k < n$, anything less than $1010...1011$ ($2k$ digits) can be destroyed in $k$ moves. The base case of $n = 1$ is quite obvious (there can be at most one $1$, so you just pluck that $1$ off. Now we employ the inductive hypothesis. If the number of $1$s is already at most $n$, then we can just skip the building part and pluck off all the $1$s. Otherwise, suppose we have $\ge n+1$ number of $1$s. If the number has $2n-1$ digits, then just change all the $0$s to $1$s in the building part, which takes at most $n-2$ moves, so we get $11111....1$, then add $1$ to turn it into $10000...0$ and we can pluck that $1$ off.

Otherwise, we have $10[\text{some number less than 10...1011 (2n-2 digits)]}$. That text portion, by the inductive hypothesis, can be destroyed in $n-1$ moves. Then just pluck off that last $1$ at the beginning.

[b]Part II:[/b] $1010...1011$ ($2n$ digits) can not be destroyed.

We will strengthen the induction a bit. I claim that, for all $k < n$, both $1010...1011$ ($2k$ digits) and $1010...101$ ($2k-1$ digits) cannot be be destroyed in $k$ moves and $k-1$ moves, respectively. Base case is easy to deal with. Suppose we have $1010...1011$ ($2n$ digits). At some point, we have to either pluck the last two $1$s or build a $1$ on top. If we pluck the last two $1$s, then we are left with $1010...1000$ ($2n$ digits), and now we need to destroy $1010...1$ ($2n-3$ digits) in $n-2$ moves, which is not possible. If we build a $1$, then we get $1010...101100$ ($2n$ digits), which of course, cannot be done due to the inductive hypothesis.

So we know that $1010...1011$ ($2n$ digits) cannot be destroyed. Now we need to show that $1010..101$ ($2n-1$ digits) cannot be destroyed in $n-1$ moves. The last step is either a pluck $1$ or build $1$ onto the last digit. You'll arrive at contradictions.

Call a number $x$ $n$-good if we can write $x$ as \[\sum_{i=1}^{n}(-1)^{a_{i}}2^{b_{i}}.\]
If a number $x$ is $n$-good then $2x$ is $n$-good because we can add $2$ to each $b_i$ and $x-1,x+1,x+2$ are $n+1$-good. By extension, $4x-1,4x,4x+1,4x+2$ are all $n+1$-good. Thus, if all $i\le x$ were $n$-good then all $i\le 4x+2$ are $n+1$-good.
\\$~$\\
Note that $1,2$ are $1$-good so all $x\le 10$ are $2$-good. Inductively we see that all $x\le 2(1+4^1+\dots+4^{n-1})$ are $n$-good. We claim that $1+2(1+4^1+\dots+4^{n-1})$ is not $n$-good and thus is our least positive integer $d_n.$

For each $i$ if $\pm 2^i$ appears more than once, we can merge two of them and create a shorter representation. Thus, we assume that each $\pm 2^i$ appears at most once. In particular, we can create a sequence $f(x)=(c_0,c_1,c_2,\dots)$ where $x=c_02^0+c_12^1+c_22^2+\dots$ and $c_i\in \{-1,0,1\}$ for all $i.$ When $x\equiv 3\pmod 4$ we must have $(c_0,c_1)=(0,-1),(1,1),$ or $(-1,1).$

Actually, we may replace $(-1,1)$ with $(0,-1)$, and the representation with $(-1,1)$ will be longer, so $(c_0,c_1)=(0,-1)$ or $(1,1).$ If it is $(0,-1)$ then we require $x+1 = 4(1 + 2(1+4^1+\dots+4^{n-2}))$ to be $n-1$-good, but this requires $\frac{x+1}{4}$ to be $n-1$-good, and so inductively this does not work.

If $(c_0,c_1)=(1,1)$ then we require $y=\frac{x-3}{4}=1+4^1+\dots+4^{n-2}$ to be $n-2$-good. This means $2y+1$ is $n-1$-good which is false, so inductively we reach a contradiction. Therefore, $1+2(1+4^1+\dots+4^{n-1})$ is not $n$-good.

Thus, our answer is \[d_n=1+2\left(\frac{4^n-1}{3}\right)=\frac{2^{2n+1}+1}{3}\]


A cereal company include two animal cards in each packet of cereal. There are 5 cards to collect.
1.a walrus card
2. a wolf card
3. a dolphin card
4. a frog card
5. a penguin card
The cards are sorted into packets randomly
a shop has 450 packets of cereal.
How many of those packets would be expected to have " at least one frog card".

To better understand, assume there are only two types of cards: frog and dog. Now, what does it mean the cards are [color=blue]randomly sorted into packets[/color]?
1. A first card is randomly put into a packet; clearly the probability to be a frog is $ 1/2$. If it is a dog (again with probability $ 1/2$), the probability of the second card put into the packet to be a frog is $ 1/2$. Calculating the total probability that the packet contains at least a frog card we get $ \frac {1} {2} \plus{} \frac {1} {2} \cdot \frac {1} {2} \equal{} \frac {3} {4}$.
2. The pair of cards is randomly put "in the same movement". Since there are three possible combinations:
two frogs/one frog and one dog/two dogs, the probability that the packet contains at least a frog card is $ \frac {2} {3}$.

Now, which is right?  :?:

i want the complete silution pls, ans is 162

In view of what I said before, using method 1. one gets the probability for one packet to be $ \frac {1} {5} \plus{} \frac {4} {5} \cdot \frac {1} {5} \equal{} \frac {9} {25}$, so there are $ \frac {9} {25} \cdot 450 \equal{} 162$ packets; while using method 2. one gets the probability for one packet to be $ \frac {5} {15} \equal{} \frac {1} {3}$, so there are $ \frac {1} {3} \cdot 450 \equal{} 150$ packets.
By your answer, they expected you to assume method 1.

If there were only one card in each, the probability would be $ \frac{1}{5}\cdot450\equal{}90$. So if there are two in each box, then multiply by two. However, you have to account for the case of getting two frog cards in one box. So $ \frac{2}{5}\cdot450\minus{}(\frac{1}{5})^{2}\cdot450\equal{}162$

just great explaination, which book do u prefer for probability? :blush:


quadratics
function
inequalities
inequalities theorems

I will post three theorems in inequalities. Please tell me if you like it and if there are usefull

[color=red]Theorem 1 . ACZEL ineq[/color]
If  $a_1^2-a_2^2-...-a_n^2>0$  and $b_1^2-b_2^2-...-b_n^2>0$ then

$(a_1^2-a_2^2-...-a_n^2)(b_1^2-b_2^2-...-b_n^2)\leq (a_1b_1-a_2b_2-....-a_nb_n)^2$

[color=blue]Theorem 2 . Bellman ineq[/color]
If  $a^n-b^n-c^n>0$  and $x^n-y^n-z^n>0$  then

 $(a^n-b^n-c^n)^{\frac{1}{n}}+(x^n-y^n-z^n)^{\frac{1}{n}}\leq [(a+x)^n-(b+y)^n-(c+z)^n]^{\frac{1}{n}}$

May I prove one of them? :D

[b]Theorem 1 . ACZEL ineq[/b]
If  $a_1,a_2\cdots ,a_n,b_1,b_2,\cdots ,b_n$ are real numbers such that $a_1^2-a_2^2-...-a_n^2>0$ [size=150][b]or[/b][/size] $b_1^2-b_2^2-...-b_n^2>0$ then
$(a_1^2-a_2^2-...-a_n^2)(b_1^2-b_2^2-...-b_n^2)\leq (a_1b_1-a_2b_2-....-a_nb_n)^2$.

[b]Proof[/b]
The following proof is similar to that of Cauchy-Schwarz's.
Let $A_n=a_1^2-a_2^2-...-a_n^2, B_n=a_1b_1-a_2b_2-....-a_nb_n, C_n=b_1^2-b_2^2-...-b_n^2$. Without loss of generality, let $A_n> 0$
Consider the quadratic function $f(x)=A_nx^2+2B_nx+C_n=(a_1x+b_1)^2-\sum_{i=2}^n(a_ix+b_i)^2$. Since it opens upwards and there exists $x_0=-\frac{b_1}{a_1}$ such that $f(x_0)\leq 0$, there exists $x=x_1$ such that $f(x_1)>0$.
Therefore the discriminant is non-negative.
$\Delta =4(B_n^2-A_nC_n)\geq 0$
or $B_n^2\geq A_nC_n$
It is easy to verify that equality holds iff $\frac{a_1}{b_1}=\frac{a_2}{b_2}=\cdots =\frac{a_n}{b_n}$.

[b]Remark[/b]
Silouan, why is there two Theorems 1 and where is your third theorem? :?

The second inequality can be derived from Minkovsky's Generalized Inequality (MGI):
$(\sum_{i=1}^n(a_{i}+b_{i})^p)^\frac{1}{p}\leq(\sum_{i=1}^n a_{i}^p)^\frac{1}{p}+(\sum_{i=1}^nb_{i}^p)^\frac{1}{p}$ , for $p,n\geq2$ and any $x_i,y_i\geq0 ;i=1,..,p$
by taking
$n\to 3,p\to n$ , $a_1=(a^n-b^n-c^n)^\frac{1}{n},a_2=b,a_3=c$ and $b_1=(x^n-y^n-z^n)^\frac{1}{n},b_2=y,b_3=z$
after playing around with the inequality you get Bellman.
I tried to remember the proof for MGI but so far I did not have any succes in remembering it or reproving it.
If somebody knows one I would be glad to see it.

Silouan these Inequalities are cute but ussualy wherever u could use one of them u can also use a more general inequality.

[quote="puuhikki"]I don't know what do you mean by the word "generalized" but I proved the Minkowsky inequality in post http://www.artofproblemsolving.com/Forum/viewtopic.php?t=58447 .[/quote]
Thanks


function
parameterization
calculus
analytic geometry
graphing lines
slope
algebra

1

A function is given: $ y\equal{}\minus{}x^{2}\plus{}x\plus{}2$  - find the tangent line $ y\equal{}ax\plus{}b$ to this function. This tangent line should pass through point $ P\equal{}(3,0)$.

2

For which values of parameter $ a$ line $ y\equal{}ax\plus{}b$ (which passes through point $ P\equal{}(3,0)$) crosses the function $ y\equal{}\minus{}x^{2}\plus{}x\plus{}2$ - but only for $ x>0$.

Thank you

Should Calculus-typed problem be in Intermediate Topics?

[hide="HINT"]
$ a \equal{} \frac{\text{d} y}{\text{d} x}$.
[/hide]

[hide="more hints"]
You know the slope (m=dy/dx), and you know a point (3,0).
so if you use the point slope formula y-0=m(x-3), you can write everything in terms of x and solve for x.
[/hide]

This is finding the tangent line of a polynomial, which does not have to be calculus. See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=179663]here[/url].

[quote="Invariant"]Please explain me how to do this problem - because I am not so advance in maths[/quote]
The solution implies by the hints in the link and the calculus solution are about as simple as you can get for this problem, so hopefully you're advanced enough to understand one.


[quote="Gib Z"]For which real values of b do the equations below have a common root?

$x^{3}+bx^{2}+2b-1=0$ and $x^{2}+(b-1)x+b=0$

I really have no idea, please help..[/quote]

You have $x^{3}+bx^{2}+2b-1=(x+1)(x^{2}+(b-1)x+b)-(2b-1)x+(b-1)$
So, if $x$ is root of $x^{3}+bx^{2}+2b-1=0$ and $x^{2}+(b-1)x+b=0$, it is also root of $(2b-1)x-(b-1)=0$ and so $b\neq\frac{1}{2}$ and $x=\frac{b-1}{2b-1}$
If we import this value in $x^{2}+(b-1)x+b=0$, we have $\frac{(b-1)^{2}}{(2b-1)^{2}}+\frac{(b-1)^{2}}{2b-1}+b=0$
So $(b-1)^{2}+(b-1)^{2}(2b-1)+b(2b-1)^{2}=0$
So $b(2(b-1)^{2}+(2b-1)^{2})=0$
And so the only possibility would be $b=0$ and we check easily that with $b=0$, we have $x=1$ commun root of the two polynomials.

quick and ingenious response, than you VERY much :D


What is the greatest three-digit number divisible by both 7 and 8?

We're just looking for numbers divisible by 56, so divide 1000 by 56, take the greatest integer function, and multiply that by 56.

This gives 17 x 56, which equals 952

for this problem we look for any # that is both divisible by both  7 and 8

this can be solved by finding the largest 3 digit # that can be divisible by 56

by finding the # 56 we just need to find the largest # below 1000

we can get the equation 1000>56x solving we get 17

so our equation is 56*17=952

so our answer is 952 $\boxed{952}$

PLEASE TELL ME HOW TO PUT A SIGNATURE AFTER EVERY POST I MAKE  THANKS!!

p.s you can see my signature when you click markoboy but you can't see my signature after everything i write please help me :(

Hello, At Markoboy I've Pm'ed you how to put a signature, Maybe it didn't completely go through. So, I'll tell you again :D Okay, goto Account at the upper right corner goto "Edit Signature"  type in your signature and click "Submit"

This should help. So, If it doesn't work your Account might actually have a bug in it. By your post It seems like you actually already sumbitted your signature and I don't know why it wouldn't show up in your post? You might have, to talk to a Mod, Admin about it.
Hope this helped?!?! :D


$A=\left\{\left(x^2+x-2\ ,\ 2x^2-5\right)|\ x\ge 1\right\}\Longrightarrow\left(\exists\right) f: [0,\infty)\rightarrow R$
so that $A=G_f$ (the graph of the function $f$) and in this case ascertain $Im(f)=\{f(x)\ |\ 0\le x\}\ .$

Farenhajt, see that I wrote what means $Im(f)$ (the image of the function $f$).

So isn't it just 2x^2-5, for x>=1, or [-1,infinity)?

[quote="JBL"]Im(f) should be the range (or image) of the function $f$.  I am having serious doubts that I understand what this question means, however.[/quote]

As far as I can tell, $f$ is given by parametric equations $x=t^2+t-2, y=2t^2-5, t\geqslant 1$ and what's asked for is the range of $f$ for $x\in[0,+\infty)$

[b]Definition.[/b] The correspondence $\{f,A,B\}$ is a function if for any $x\in A$ exists and is unique $y\in B$ so that $x\rightarrow y$, i.e. $f(x)=y$ [for any $x_1\in A$, $x_2\in A$ we have $x_1=x_2\Longrightarrow f(x_1)=f(x_2)$].
In my opinion, I think that you ([b]Bictor 717[/b]) made a mistake !

Sad to say, that wasn't a typo :blush: . I haven't done any arithmetic for months...
Let me try this again: 2(1)^2-5=2-5=-3
So final answer [-3, infinity).

[b]Bictor717[/b], $x: =1\Longrightarrow f(0)=-3$. It is O.K. What is the problem ?
You confound the domain with the image $Im(f)$ of the function $f$ !

[quote="JBL"]Okay, that was my initial response also, but doesn't that make the question incredibly simple (as Bictor pointed out, although with a typo)?[/quote]

Well, $x\geqslant 0$ gives $t^2+t-2\geqslant 0$ or $t\leqslant -2\lor t\geqslant 1$, but since $t\geqslant 1$ is given, we take just $t\geqslant 1$. Since $y=2t^2-5$ is increasing for $t\geqslant 0$, we conclude that the range of $y$ is $[2\cdot 1^2-5,+\infty)$ or $[-3,+\infty).$

Note: One shouldn't mix up the two meanings of $x$ in the statement of the problem (which admittedly can create confusion easily). The correct solutions springs up immediately if you put $t$ instead of $x$ in the parametric equations.

At first, must prove that:
$1^{\circ}\blacktriangleright x_1\ge 1$, $x_2\ge 1$, $x^2_1+x_1-2=x^2_2+x_2-2$ $\Longrightarrow 2x^2_1-5=2x^2_2-5$;
$2^{\circ}\blacktriangleright$ for any $x\ge 1$ we have $x^2+x-2\ge 0$;
$3^{\circ}\blacktriangleright$ for any $x\ge 0$ exists $t\ge 1$ so that $t^2+t-2=x$.
The all are truly.[hide="Remark."]The searched function is $f: [0,\infty)\rightarrow R$, where $\boxed {\ f(x)=2x-\sqrt{4x+9}\ }$. Indeed, for any $t\ge 1$, $f(t^2+t-2)=$ $2(t^2+t-2)-\sqrt{4(t^2+t-2)+9}=$ $2t^2+2t-4-\sqrt{4t^2+4t+1}=$ $2t^2+2t-4-(2t+1)=2t^2-5$, i.e. $f(t^2+t-2)=2t^2-5$ and $\boxed {\ Im(f)=[-3,\infty)\ }$. Indeed, for any $x\ge 0$ we have $f(x)\ge -3$ $\Longleftrightarrow$ $\sqrt{4x+9}\le 2x+3$ $\Longleftrightarrow$ $4x+9\le 4x^2+12x+9$ $\Longleftrightarrow$ $4x(x+2)\ge 0$, what is truly.[/hide]


Currently, I am an Aerospace Engineering major but this past week I have decided to switch to Pure Mathematics. I have been thinking about it a good long while and decided that Engineering was not for me. However, I would still consider working in the Aerospace industry if I could, just not as an engineer. I know that engineering firms hire not just engineers but also people such as physicists, chemists, mathematicians, and computer scientists. I know that APPLIED mathematics can land you a job in an engineering firm but what about pure mathematics? Do engineering firms hire pure mathematicians? If not, what else could I do with a pure mathematics degree besides teaching and doing research (not that I wouldn't want to do either of those, I am just curious)? Thanks in advance.

I assume that if you take some applied math courses with your pure math degree you'll have most/all of the job paths available to applied math majors open to you.

Although of course I'm completely unsure.

This post serves mostly to increase my post count.

Plenty of organizations will hire people with "pure" mathematics degrees if they are willing to do applied work.  You can demonstrate your willingness and abilities by taking on summer internships  doing mathematical applications.

There are some "pure" mathematics students who know with certainty that they will spend their lives in research and teaching.  The majority are probably like ROLEX4Life and have some doubts about whether they'll be able to succeed in a math graduate program and find interesting work as a college professor or mathematics researcher.

What if one goes down the "pure" math path and then fails to obtain a Ph.D?  What if one gets the Ph.D and then finds the work unsatisfactory or unavailable?  How easy will it then be to make a transition to mathematical applications or other work?

Most "pure" math students at the baccalaurate level end up making a fairly successful transition - especially if they have an interest in the application and are willing to invest some time learning programming and relevant modeling techniques.

Thanks for the feedback. I would most certainly enjoy doing internships learning mathematical applications. Out of the three math major sub options I had to choose from (Pure, Applied, and Statistics), I chose Pure because the classes seemed more interesting to me. I hope in the future to pursue a Ph.D not necessarily just because I would have to, but because I would want to out of pure interest for becoming an expert in the field. Also, being a mathematics professor in college is something that I could see myself doing in the distant future. I am not sure exactly what I want to do, but i DO know that what I DON'T want to become is an engineer  :P


function
real analysis
real analysis unsolved

Let $ \textbf{g}(k) \equal{} \sum_{n \equal{} 1}^{\infty}{\frac {\left\{(n \minus{} 1)k\right\}!}{(nk)!}}.$

Try to find $ \textbf{g}(5).$

$ g(5) \equal{} \sum_{n\equal{}1}^{\infty}\frac{1}{(5n)(5n\minus{}1)(5n\minus{}2)(5n\minus{}3)(5n\minus{}4)}$,

and one possible way for computing this is by using a generating function in the form of a power series.


[hide]The man's velocity in the upstream direction is 3000 meters/hour and his velocity toward the other bank is 4000 meters/hour. Therefore the magnitude of his velocity is 5000 meters/hour. He would finish in 6 minutes if he didn't swim against the current.[/hide]

[quote="Lazarus"]A 500 meter wide river is flowing downstream at 3 kph. A man swims across the river and he reaches the other side 7.5 minutes later, directly across from the point where he started. Had he allowed the current to carry him instead of swimming partly against the current, he would have ended up further downstream; furthermore, he would have finished quicker. How many minutes would he have taken under these circumstances? Assume that he swims at a constant rate.[/quote]
[hide="solution"]
Imagining the river as not moving and everything else as moving in relation to the river, in the first swim he went diagonally by going $ 375$ meters upstream and $ 500$ meters across the river. Using the pythagorean theorem, he went $ 625$ meters diagonally. If he were to go directly to the other side, it would've been $ 500$ meters. He would spend $ \frac {4}{5}$ of the time it took him ($ \frac {500}{625} \equal{} \frac {4}{5}$).
$ (7.5)\left(\frac {4}{5}\right) \equal{} 6$ minutes.
It would take $ 6$ minutes for him to finish if he went with the current.
[/hide]


Prove that :

$ \sqrt{ab}\ \le\ \frac{\sqrt[n]{(a\plus{}nb)(2a\plus{}(n\minus{}1)b)\dots(na\plus{}b)}}{n\plus{}1}\ \le\ \frac{a\plus{}b}{2},$ $ \forall\ a,b>0,\ n\in\mathbb{N},\ n\ge 2$ .

[quote="Mateescu Constantin"]Prove that :

$ \sqrt {ab}\ \le\ \frac {\sqrt [n]{(a \plus{} nb)(2a \plus{} (n \minus{} 1)b)\dots(na \plus{} b)}}{n \plus{} 1}\ \le\ \frac {a \plus{} b}{2},$ $ \forall\ a,b > 0,\ n\in\mathbb{N},\ n\ge 2$ .[/quote]

For the right part. Let $ X\equal{}(a\plus{}nb)\cdots(na\plus{}b))$.
\[ (n\plus{}1)\sqrt[n]{X}\le\frac{n(n\plus{}1)a\plus{}n(n\plus{}1)b}{2}\]
Now
\[ \frac{\sqrt[n]{X}}{n}\le\frac{a\plus{}b}2\]
But
\[ \frac{\sqrt[n]{X}}{n\plus{}1} \le \frac{\sqrt[n]{X}}{n}\]
And we have done.

[quote="mathson"]For the right part. Let $ X = (a + nb)\cdots(na + b))$.
\[ (n + 1)\sqrt [n]{X}\le\frac {n(n + 1)a + n(n + 1)b}{2}\]
[/quote]
This is wrong - e.g. $ a=b$. Using AM-GM you get: $ n\sqrt[n]{X} \le \frac{n(n+1)a+n(n+1)b}{2}$, when we use $ 1+2+...+n = \frac{n(n+1)}{2}$. Which is equivalent to $ \frac{\sqrt[n]{x}}{n+1} \le \frac{a+b}{2}$.
$ (n+1-k)a+kb \ge (n+1){a^{(n+1-k)/n}b^{k/n}}$ by AM-GM. So $ (na+b)...(a+nb) \ge (n+1)^na^{(1+2+..+n)/(n+1)}b^{(1+2+..+n)/(n+1)} = ( (n+1)\sqrt{ab} )^n$, again using $ 1+2+..+n = \frac{n(n+1)}{2}$. And this is equivalent to:
$ \sqrt{ab} \le \frac{\sqrt[n]{(na+b)...(a+nb)}}{n+1}$

[quote="Mateescu Constantin"]Prove that :

$ \sqrt {ab}\ \le\ \frac {\sqrt [n]{(a \plus{} nb)(2a \plus{} (n \minus{} 1)b)\dots(na \plus{} b)}}{n \plus{} 1}\ \le\ \frac {a \plus{} b}{2},$ $ \forall\ a,b > 0,\ n\in\mathbb{N},\ n\ge 2$ .[/quote]

\[ \frac{{\sqrt[n]{{\left( {a \plus{} nb} \right)\left( {2a \plus{} \left( {n \minus{} 1} \right)b} \right) \cdots \left( {\left( {n \minus{} 1} \right)a \plus{} 2b} \right)\left( {na \plus{} b} \right)}}}}{{n \plus{} 1}}\]
\[ \equal{} \sqrt[n]{{\left( {\frac{1}{{n \plus{} 1}}a \plus{} \frac{n}{{n \plus{} 1}}b} \right)\left( {\frac{2}{{n \plus{} 1}}a \plus{} \frac{{n \minus{} 1}}{{n \plus{} 1}}b} \right) \cdots \left( {\frac{{n \minus{} 1}}{{n \plus{} 1}}a \plus{} \frac{2}{{n \plus{} 1}}b} \right)\left( {\frac{n}{{n \plus{} 1}}a \plus{} \frac{1}{{n \plus{} 1}}b} \right)}}\]
\[ \ge \sqrt[n]{{\left( {a^{\frac{1}{{n \plus{} 1}}} b^{\frac{n}{{n \plus{} 1}}} } \right)\left( {a^{\frac{2}{{n \plus{} 1}}} b^{\frac{{n \minus{} 1}}{{n \plus{} 1}}} } \right) \cdots \left( {a^{\frac{{n \minus{} 1}}{{n \plus{} 1}}} b^{\frac{2}{{n \plus{} 1}}} } \right)\left( {a^{\frac{n}{{n \plus{} 1}}} b^{\frac{1}{{n \plus{} 1}}} } \right)}}\]
\[ \equal{} \sqrt[n]{{a^{\frac{1}{{n \plus{} 1}} \plus{} \frac{2}{{n \plus{} 1}} \plus{}  \cdots  \plus{} \frac{n}{{n \plus{} 1}}} b^{\frac{1}{{n \plus{} 1}} \plus{} \frac{2}{{n \plus{} 1}} \plus{}  \cdots  \plus{} \frac{n}{{n \plus{} 1}}} }}\]
\[ \equal{} \sqrt {ab}\]

\[ \frac{{\sqrt[n]{{\left( {a \plus{} nb} \right)\left( {2a \plus{} \left( {n \minus{} 1} \right)b} \right) \cdots \left( {\left( {n \minus{} 1} \right)a \plus{} 2b} \right)\left( {na \plus{} b} \right)}}}}{{n \plus{} 1}}\]
\[ \le \frac{{\left( {a \plus{} nb} \right) \plus{} \left( {2a \plus{} \left( {n \minus{} 1} \right)b} \right) \plus{}  \cdots  \plus{} \left( {\left( {n \minus{} 1} \right)a \plus{} 2b} \right) \plus{} \left( {na \plus{} b} \right)}}{{n\left( {n \plus{} 1} \right)}}\]
\[ \equal{} \frac{{\left( {1 \plus{} 2 \plus{}  \cdots  \plus{} n} \right)\left( {a \plus{} b} \right)}}{{n\left( {n \plus{} 1} \right)}}\]
\[ \equal{} \frac{{a \plus{} b}}{2}\]


x^2+25x^2 over (x+5)^2 = 11
can anyone detail how to solve this?

but now u left out that x^2 that is at the beggining...

ok.. i got to tha now i have x^4 + 10x^3 + 39x^2 - 110x - 275 = 0

and i couldnt do anything with it


[quote="mdk"]Five points are situated inside an equilateral triangle whose side has length one unit. Show that two of them may be chosen which are less than one half unit apart. What if the equilateral triangle is replaced by a square whose side has length of one unit?[/quote]

[hide="Triangle"]Draw a triangle $ ABC$ pick $ D$ on $ AB$ to be midpoint, pick $ E$ to be midpoint of $ BC$, pick $ F$ to be midpoint of $ AC$. Connect $ DE$ and $ BF$. There are 4 sections. Now given $ 5$ points two end up in the same region. So the maximum distance between those two in this region is $ 1$.

NOTE: I didnt actually do the computations on this triangle, but I am sure it works. :)
[/hide]

[hide="Square"]Divide the square into 4 squares. By pigeonhole 2 end up in the square. So the max distance between them is $ \sqrt{2}/2$.
[/hide]


A triangle has sides $ a$, $ b$ and $ c$ and the medians of this triangle are used as sides of a new triangle. If $ r_m$ is the inradius of this new triangle, prove that

$ \frac{3abc}{4(a^2\plus{}b^2\plus{}c^2)}\geq r_m$

[quote="Ligouras"]A triangle has sides $ a$, $ b$ and $ c$ and the medians of this triangle are used as sides of a new triangle. If $ r_m$ is the inradius of this new triangle, prove that

$ \frac {3abc}{4(a^2 \plus{} b^2 \plus{} c^2)}\geq r_m$[/quote]

For any  triangle with sides $ a$, $ b$ , $ c$ and  medians  $ m_a$ ,  $ m_b$ , $ m_c$ we have triangle with sides $ m_a$, $ m_b$ , $ m_c$ and  medians  $ \frac{3}{4}a$, $ \frac{3}{4}b$ , $ \frac{3}{4}c$.

Area of this treangle : $ S^*\equal{} \frac{3}{4}S$

$ r_m \equal{} \frac{3S}{2(m_a\plus{}m_b\plus{}m_c)}\equal{} \frac{3abc}{8R(m_a\plus{}m_b\plus{}m_c)} \le \frac {3abc}{4(a^2 \plus{} b^2 \plus{} c^2)}$

$ (m_a\plus{}m_b\plus{}m_c)R \ge \frac{1}{2}(a^2\plus{}b^2\plus{}c^2)$

( $ m_aR \ge \frac{b^2\plus{}c^2}{4}$ , $ m_bR \ge \frac{c^2\plus{}a^2}{4}$ , $ m_cR \ge \frac{a^2\plus{}b^2}{4}$)


Let a,b>0 and 0<x<1. Prove: (a+b)^x<a^x+b^x

Let $ p\equal{}\frac{a}{a\plus{}b}$        , $ p<1$  then  $ p^x\plus{}(1\minus{}p)^x>p^1\plus{}(1\minus{}p)^1\equal{}1$


The surface area of a sphere with radius $ r$ is $ 4\pi r^2$. Including the area of its circular
base, what is the total surface area of a hemisphere with radius 6 cm? Express your
answer in terms of $ \pi$.

[asy]fill((0,10)..(-10,0)--(10,0)..cycle,gray(0.8));
draw((0,10)..(-10,0)--(10,0)..cycle,linewidth(0.7));
draw((0,-0.9)..(-10,0)--(10,0)..cycle,linewidth(0.7));
draw((0,1.3)..(-10,0)--(10,0)..cycle,linewidth(0.7));
fill((0,-0.9)..(-10,0)--(10,0)..cycle,gray(0.8));
draw((0,-7)--(0,-1),Arrow);
draw((10,10)--(5,5),Arrow);
label("half of sphere",(10,10),N);
label("circular base",(0,-7),S);[/asy]

SA of a hemisphere would be 3πr^2 (right?) 2πr^2 for the curved surface + πr^2 for the circle on the bottom

plug in the #'s

3π6^2

would be 108π

this marks the end of my 52nd post. (applause is neccessary)

[i] this is not guaranteed to be right by all means [/i]


search
number theory proposed
number theory

Prove that the product of for whole natural consecutive and nonnull cannot one be  perfect square

Do you mean "Prove that the product of 4 consecutive natural numbers cannot be a perfect square" or "Prove that the product of 2 consecutive natural numbers cannot be a perfect square"?

Actually, both ways, it doesn't really matter. I believe this and its generalisation has been discussed many times on this forum already. There is a general theorem that states that the product of any 2 or more consecutive natural numbers cannot be a perfect power, by Erdos I think... (You can do a search on the forum)

If it's the 2nd you are asking, it's really very trivial. If it's the first qn, then we just note that ${(a^2 + 3a)}^2 < a(a+1)(a+2)(a+3) < {(a^2+3a+1)}^2 \ \forall a \in \mathbb{N}$.


In a barn with cows and chickens, the number of legs was
16 more than twice the number of heads. How many cows
were in the barn?

If it were all chickens then there would be exactly twice the number of heads. But each cow adds 2 more legs. Thus there are 16/2=8 cows.


I've noticed that many colleges require 3 SAT II's for admission, so I was planning on taking Math 2c, Physics, and Chemistry before. I've already taken Math 2c, and I'm pretty sure I got 800. But I realized I don't like chemistry at all, so I need a third subject test. [b]Can I take Math1c, or does it look bad to take the both maths?[/b] I am pretty sure I can easily get 800 on Math1c, and 800 on Physics, but only a 700ish on Chemistry right now.
So basically, I will have taken Math2c, Math1c, and Physics under this scenario.

Thanks.

You could take one of the languages, or whatever else there is an SAT II test on that you like more than chemistry. Take something you like--that will make the studying more bearable.

Math 1c and Math 2c is unacceptable to most colleges.  Moreover, it looks bad - what logical reason is there to take 1c after 2c?

Take whatever subject you have the best chance of scoring well in for your third subject - my experiences, based on what little prep I did, was that any studying for the SATs was unbearable.

I thought that the sat II writing test had to be taken.....

Doesn't exist anymore - the new SAT has a writing portion, remember?

many colleges required 3 subject tests before the new sat was adopted so that it would include writing also. I don't think it's any use taking math 1c. if you don't think you're prepared for another subject test just don't take any, otherwise just try a foreign language or something. the subject tests aren't that hard if you have a minumum of preparation(although I messed things up on my French one and got only a 740 :(  :P )

[quote="1234567890"]I am pretty sure I can easily get 800 on Math1c[/quote]
Don't underestimate it, I have seen many smart people make mistakes.

Anyway, from experience..
1. 700s is totally fine - I have 760 physics
2. having both IC and IIC doesn't hurt you (I don't know how much it would help, though)- I have both, and I took them in one day, and that didn't prevent me from getting into MIT.

Geehoon, is that all you took? Physics, Math2c, Math1c

640 writing, but I took it when the perfect score for the SAT was 1600.


For all positive integers n prove that

(n+1)^(2n)  \geq 2(n^(4/3) + n^(5/3))^n


Prove that for all positive integers $n$ and all positive reals $a,b,c$ the following inequality holds: \[\frac{a^{n+1}}{a^{n}+a^{n-1}b+\ldots+b^{n}}+\frac{b^{n+1}}{b^{n}+b^{n-1}c+\ldots+c^{n}}+\frac{c^{n+1}}{c^{n}+c^{n-1}a+\ldots+a^{n}}\\ \ge \frac{a+b+c}{n+1}\]

It is so classical.We have $\sum_{cyc}\frac{a^{n+1}}{a^{n}+a^{n-1}b+..+b^{n}}=\sum_{cyc}\frac{b^{n+1}}{a^{n}+a^{n-1}b+..+b^{n}}$>because if you substract  them you have $a^{n+1}-b^{n+1}=(a-b)(a^{n}+a^{n-1}b+..+b^{n})$.Now it remains to prove that $(n+1)(a^{n+1}+b^{n+1})\geq (a+b)(a^{n}+a^{n-1}b+..+b^{n})$ which is easy :lol:

Indeed, this problem is easy if one sees that the inequality is symmetric. However, you don't need to make it symmetric, just sum up \[\frac{x^{n+1}}{x^{n}+x^{n-1}y+\ldots+xy^{n-1}+y^{n}}\geq \frac{(n+2)x-ny}{2(n+1)}\label{4}\] which is equivalent to your inequality :D


(Prove ):y=sqtr(x+27)+sqrt(13-x)+sqrt(x) max  and  min

[quote="xifan"]y=sqtr(x+27)+sqrt(13-x)+sqrt(x) max  and  min[/quote]
$ Y_{max}\equal{}11;$

$ Y_{min}\equal{}3\sqrt{3}\plus{}\sqrt{13}.$


Here is some advice from someone who made mistakes and has learned from them. I am now in graduate school for math.
First I am making an assumption, that you are looking to major in math.
If you are going to be in the field of mathematics and you do not want to teach high school you will have to go to graduate school and get a Ph.D.
So knowing this you can make some wise choices at this stage in your academic career.
Don't bother so much with big expensive schools, just because they have that prestigious name. Look for schools that have very strong math departments and good undergraduate programs. If you do well in the your undergrad and stand out there, then when you look at grad school this is where you can worry about prestige. If you get into a good grad program and do significant research there, then no one cares where you went for your undergrad. There are a number of very good state schools that have very good math programs. Look for a school that has a very well balanced program, i.e. pure and applied math. One of the most important things that they must have is research oppertunities at the undergraduate level. This will help you find what type of math you want to consentrate on and it helps on the grad school applications. Also, this will give you the oppertunity to see if being a mathematician is really what you want. You may really like the math classes you take but if you don't like doing research then this may not be the field you want to go into.
I hope that this helps in your search for schools.

thanks for the advice. so where did you do your undergrad?

any tips for being an efficient researcher?


calculus
integration
trigonometry
special factorizations
calculus computations

Evaluate the integral. I think we need to use completing the square, but I'm not sure

[img]http://img35.imageshack.us/img35/5277/17a81c796cbe343c2f8d07c.png[/img]

Yes, complete the square. Your answer, which should only be valid on the interval $ (0,13),$ should include an arcsine.

i try to complete the square but for some reason it didn't work

What "didn't work" about it? Show what you tried, so we can see your mistake.

Let $ (zx)^2 \equal{} 13x \minus{} x^2$. Then $ \frac{\minus{}2}{1 \plus{}z^2}dz \equal{} \frac{1}{\sqrt{13x \minus{} x^2}}dx$. On integrating, we get $ \minus{}2 \tan^{\minus{}1} \left(\frac{\sqrt{13x \minus{}x^2}}{x}\right) \plus{} C$ as the answer to the original integral.

Unnecessarily complicated. That's not really what we're waiting for Aafour do to. (And the answer we'll expect Aafour to arrive at will be expressed using arcsin rather than arctan.)


This just looked a little odd to me, but now every AoPSWiki page is in the form of
http://www.artofproblemsolving.com/Wiki/index.php/Whatever_The_Page_Is
I don't think the "index.php" part should be there, or should it?
 :huh:

It's always been like that forever and ever.

Wikipedia configured their server to get rid of the index.php, but that's not the default MediaWiki setting.

OK.  I didn't know if it was a new path because I never noticed it until today.
:D

[quote="#H34N1"]Just enable mod rewrite and map it so[/quote]There are already a few rules and I won't put more load onto the server just for someone's eye candy. ...


calculus
integration
trigonometry
real analysis
real analysis unsolved


[quote="GameBot"]Bryan has some 3 cent stamps and some 4 cent stamps. What is the least
number of stamps he can combine so the value of the stamps is 33 cents?[/quote]

To minimize the number of stamps, we want as many 4 cent stamps as possible. If we have 8 4-cent stamps, that's only 32 cents. If we have 7 4-cent stamps, we still need 5 more cents, which we can't make out of 3-cent stamps. If we have 6 4-cent stamps, then we need 9 more cents, which we CAN make out of 3-cent stamps. Thus our answer is $ \boxed{9}$ stamps - 6 4-cent ones and 3 3-cent ones.


[quote=victornumber]I was wondering if anyone has any recommendations for websites or books that helps to stimulate ideas and possible topics in chemistry so I can think of an original chemistry research question to pursue for the high school IB programme 4000 word extended essay in chemistry. Thanks in advance and suggestions are greatly appreciated.

Victor[/quote]

academia.edu, researcher.com are plentiful of ideas you can start working on.


Γεια σας,
έχω ενα πρόβλημα γεωμετρίας και το οποίο θεωρώ αρκετα δύσκολο:
"Ένα ισοσκελές τρίγωνο έχει δύο γωνίες 72 μοιρών και δύ πλευρέ μήκους 1 η καθεμία.Ποιο είναι το μήκος της τρίτης πλευράς;"
-Στο πρόβλημα να μη χρησιμοποιηθούν τριγωνομετρικές συναρτήσεις.
Όταν δείτε το πρόβλημα 8α σας φανεί εύκολο αλλα δεν είναι.
Όποιος δύναται να βοηθήσει ας μου στείλει.
Ευχαριστώ.

Πλευρά κανονικού δεκαγώνου εγγεγραμμένου σε κύκλο ακτίνας 1

Εχουμε το ισοσκελες τριγωνο ΑΒΓ με ΑΒ = ΑΓ = 1, ΓΒ = x και γωνιες Β = Γ = 72 μοιρες, τοτε η τριτη γωνια Α ειναι Α = 180 - 144 = 36 μοιρες = Γ/2. Θεωρουμε τη διχωτομο της γωνιας Γ που τεμνει την ΑΒ στο Μ, τωρα το ΓΜΑ ειναι ισοσκελες αφου ΜΓΑ = ΓΑΜ = 36 μοιρες, αρα ΜΑ = ΜΓ = y, τωρα ακομα εχουμε ΒΜ = 1 - y και το τριγωνο ΒΜΓ ειναι και αυτο ισοσκελες με ΜΒΓ = ΒΜΓ = 72 μοιρες (η ΒΜΓ ειναι εξωτερικη του τριγωνου ΓΜΑ), αρα ΒΓ = ΜΓ <=> χ = y (1), ακομα τα τριγωνα ΑΒΓ και ΓΒΜ ειναι ομοια αφου εχουν τις γωνιες τους ισες, αρα $ \frac {x}{1 \minus{} y} \equal{} \frac {1}{x} < \equal{} > x^2 \equal{} 1 \minus{} y$. και επηδη απο την (1) εχουμε x = y, τελικα εχουμε $ x^2 \equal{} 1 \minus{} x < \equal{} > x^2 \plus{} x \minus{} 1 \equal{} 0 < \equal{} > x \equal{} \frac {\sqrt {5} \minus{} 1}{2}$ (H αλλη λυση απορριπτεται διοτι ειναι αρνητικη και δεν μπορει να παριστανει μηκος πλευρας τριγωνου), αρα τελικα ΒΓ = $ x \equal{} \frac {\sqrt {5} \minus{} 1}{2}$

Ελπιζω να μην υπαρχει πουθενα λαθος...

Η λύση που πρότεινες Nicknafplio είνα πολύ καλή.
Το πρλημα που αντιμετώπιζα ήταν στο ότι δεν έφαιρνα τη διχοτόμο της Γ.
Η αλήθεια είναι ότι είχα καταλάβει ότι είναι εγγεγραμμένο σε κύκλο δεκάγωνο αλλά εκεί δεν μπορούσα να συνεχίσω.
Έτσι λοιπόν θα με ενδιέφαιρε να μου πει ο r_βορις πως 8α συνέχιζε το πρόβλημα αν ακολουθήσουμε τη δική του μέθοδο.
Ευχαριστώ.

Έτσι περίπου είναι γραμμένο στο βιβλίο της Β΄ Λυκείου (κανονικά πολύγωνα)


I didn't do the AMC 12, just the 10. I made AIME, but like only by 9 points due to 2 really stupid mistakes and 2 questions I omitted for absolutely no reason cuz I was like asleep on test day. :(

[quote="Aryth"]am I the only high schooler on this forum anymore? D:[/quote]

Nope :p. I'm also a high schooler.


If you don't any calculus except simple derivatives and integration, what should you be studying next for the calculus part of this competition?

Also, what topics or problem solving strategies should we practice that we wouldn't find in an AP Calc class?

Sequences and series from my limited experience show up a lot.

[quote="aufha"]Sequences and series in calculus problems? What would be an example?[/quote]Power series, generalized binomial theorem, convergence, etc.

A good place to start is to look at the problems and solutions to some past HMMTs. Of course, you'd want to try the problems yourself, first. More recent contests are more representative, because the problem writers are current students.

It helps to be comfortable with series and the generalized Binomial Theorem. But, just because of the way the problems are often written, you may find that your solutions are completely different from the official solutions; that really depends from person to person. In my case, (perhaps more than in other peoples), I find that many of my solutions are quite different from the official ones.

Especially with problems that aren't given in terms of a series, but are intended to be solved with series, theres usually many other ways of doing them as well.

In short, you'll be fine


Stiind ca $\pi < \frac{22}{7}$ (o istorica aproximatie prin adaos a numarului $\pi$) sa se stabileasca semnul sumei $\sin 5+\cos 6$, unde unghiurile sunt exprimate in radiani, adica $\{5;6\}\subset \left( \frac{3\pi}{2},2\pi \right)$.

$\pi < \frac{{22}}{7} \Rightarrow 7\pi < 22 \Rightarrow 14\pi < 44 \Rightarrow \frac{{7\pi }}{4} - 5 < 6 - \frac{{7\pi }}{4} \Rightarrow$

$\Rightarrow \left| {\sin 5} \right| < \left| {\cos 6} \right| \Rightarrow - \left| {\sin 5} \right| + \left| {\cos 6} \right| > 0\mathop \Rightarrow \limits_{\mathop {\scriptstyle \sin 5 < 0 \hfill \atop \scriptstyle \cos 6 > 0 \hfill}\limits^ \uparrow } \sin 5 + \cos 6 > 0$

[u]Metru[/u], your solution is nicely ! The my solution is directly.
[u]
The first method.[/u]
$\sin 5+\cos 6=\sin 5+\sin \left(\frac{\pi}{2}-6\right)=2\sin \left(\frac{\pi}{4}-\frac 12\right)\cos \left(\frac{11}{2}-\frac{\pi}{4}\right).$
$\frac{\pi}{4}-\frac 12\in \left(0,\frac{\pi}{2}\right),\ \frac{11}{2}-\frac{\pi}{4}\in \left(\frac{3\pi}{2},2\pi\right)\Longrightarrow \sin 5+\cos 6\ge 0$.

[u]The second method.[/u]

$0<2\pi -6<2\pi -\frac{11}{2}<\frac{\pi}{4}<2\pi -5<\frac{\pi}{2}$ a.s.o.


quadratics
function
limit
search
algebra solved
algebra

Let xN be the positive root of g(x)=g2x 2 + g1x+g0=N for large integer N. It is fact xN+1-xN \downto 0.

Let MN be a value f(xN)=f2x2 +f1x+f0.

We have
(1N) g2(xN+12 - xN2) + g1(xN+1-xN)=1
(2N) f2(xN+12 - xN2) + f1(xN+1-xN)=MN+1-MN.
Since xN+1+xN-> \infty we obtain
\lim (2N)/(1N) = f1/g1 but (2N)/(1N) is an integer for all N ==> limit f1/g1 is integer too.
Denote m=f1/g1.
(2N)-m(1N)=(f1-m.g1)(xN+1-xN) is an integer  for all N and tends to zero ==> it is zero and f1-m.g1=0.
And finally
MfN=f(xN)=m.g(xN)+f0 - m.g0 ==>   k = f0 - m.g0 is an integer

[quote="iandrei"]Here is another nice Bulgarian problem. It's not difficult but it's very nice:)

If f,g are two quadratic functions such that if for every x>0 , if g(x) is an integer then so is f(x). Prove that there exist the integers m and n such that f(x) = m*g(x) + n.[/quote]
Let $ f(x) \equal{} ax^2 \plus{} bx \plus{} c;g(x) \equal{} px^2 \plus{} qx \plus{} r$ assume without loss of generality $ p > 0$ and $ q \equal{} 0$(by the change of variable $ x \to x \minus{} \frac {q}{2p}$)
Let $ k$ be a integer such that $ k > s$ and $ t \equal{} \sqrt {\frac {k \minus{} s}{p}} > \frac {q}{2p}$
SInce $ g(t) \equal{} k \in Z$,so is $ f(t) \equal{} \frac {a(k \minus{} s)}{p} \plus{} bt \plus{} c$,as is:
$ f(\sqrt {\frac {k \plus{} 1 \minus{} s}{p}}) \minus{} f(\sqrt {\frac {k \minus{} s}{p}}) \equal{} \frac {b}{\sqrt {p}}.\frac {1}{\sqrt {k \plus{} 1 \minus{} s} \minus{} \sqrt {k \minus{} s}} \plus{} \frac {a}{p}$
This tends to $ \frac {a}{p}$ as $ k$ increases,so $ \frac {a}{p}$ must be an integer,moreover $ b$ must equal $ 0$,or else the above expression will equal $ \frac {a}{p}$ plus a small quanlity for large $ k$,which cannot be an integer .Now put $ m \equal{} \frac {a}{p}$;$ n \equal{} c \minus{} ms$ then $ f(x) \equal{} mg(x) \plus{} n$

For trivial reasons, we may set $ g(x), f(x)>0$ for all sufficiently large $ x$-s. This is: $ f_2, g_2>0$. Now, again by trivial, we may consider $ g_2 > f_2$. If that's not the case, we divide $ f$ by $ g$ and use the remainder $ r$ instead of $ f$.
Let $ \Delta F\equal{}F(y)\minus{}F(x)$ for some function $ F$ and sufficiently large numbers $ y>x$. Then if $ F$ is quadratic with positive leading coefficient, then $ \Delta F>0$ for sufficiently large $ x$.
Now, using the assumptions for $ f,g$, the function $ h\equal{}g\minus{}f$ is quadratic with positive leading coefficient. Then: $ \Delta g\minus{}\Delta f \equal{} \Delta h > 0$. Setting $ g(x), g(y)$ to be two sufficiently large consecutive integers, one gets:
$ \Delta f < 1$, and thus $ f(x)$ and $ f(y)$ can't be integers simultaneously. Then it must be:
$ \Delta f\equal{}0$, or:
$ f \equiv c (\text{mod} g)$, which is exactly the asked.

See also http://www.mathlinks.ro/viewtopic.php?search_id=9481000&t=18474 with general statement.


Already posted before: http://www.mathlinks.ro/Forum/viewtopic.php?p=292205#p292205

[quote="enescu"]An interesting lecture [url=http://www.imo-register.org.uk/2005-report.html]here[/url]![/quote]Geoff Smith (the author of the report) is going to be an instructor for [url=http://www.artofproblemsolving.com/Classes/AoPS_C_WOOT.php]WOOT[/url].

[quote="DPatrick"]Geoff Smith (the author of the report) is going to be an instructor for [url=http://www.artofproblemsolving.com/Classes/AoPS_C_WOOT.php]WOOT[/url].[/quote]
No IMO gold medalist in your WOOT yet, right?

Brian Lawrence (perfect scorer) was at MOP and thus is automatically in WOOT.

I think he means a gold medalist as a tutor in WOOT...


inequalities
geometry
circumcircle
trigonometry
function
trig identities
Law of Sines

Given are $\triangle ABC$ prove that
$\frac{a(b+c)}{m_{a}}+\frac{b(c+a)}{m_{b}}+\frac{c(a+b)}{m_{c}}\le 12R$
Please help me! Thank you! :)

You always have $2Rm_{a}\geq b^{2}+c^{2}$. Call $A'$ the intersection of the median with the circumcircle. One can prove that $AA' = \frac{b^{2}+c^{2}}{m_{a}}$. Thus the inequality follows.

Treeforne and fitim can you show me clearly, how to apply your inequalities for my problem??? ,I don't understand so much! Thank you! :)

I have questions: In every triangle is true ineq:

$\frac{1}{m_{a}}+\frac{1}{m_{c}}\geq \frac{2}{h_{b}}?$
($m_{a},m_{b}$ are medians and  $h_{b}$ is altitudes)

Fitim I use cabri II I see it is true but I can't prove it,can we infer my problem from your inequality?
Hope everybody help me!  :)

Fitim really I drawed wrong the inequality $\frac{1}{m_{a}}+\frac{1}{m_{c}}\ge \frac{2}{h_{b}}$ is false thus we can't prove my problem by this inequality,
Can treegoner say me clearly how to use your inequality to prove my problem thanks! :)

Treegoner, I think you use $4Rm_{a}\geq b^{2}+c^{2}$ rather than $2Rm_{a}\geq b^{2}+c^{2}$, and I must disappoint you that your proof is wrong because $\sum\frac{a\left(b+c\right)}{b^{2}+c^{2}}\leq 3$ is a wrong inequality, even for the sidelengths of a triangle...

Fitim, your inequality $\frac{1}{m_{a}}+\frac{1}{m_{c}}\geq \frac{2}{h_{b}}$ is wrong, and the opposite inequality $\frac{1}{m_{a}}+\frac{1}{m_{c}}\leq \frac{2}{h_{b}}$ is wrong as well (we have $\frac{1}{m_{a}}+\frac{1}{m_{c}}\geq \frac{2}{h_{b}}$ for a = 2, b = 2, c = 3, while $\frac{1}{m_{a}}+\frac{1}{m_{c}}\leq \frac{2}{h_{b}}$ for a = 7, b = 8, c = 5).

Anyway, I have a proof of the original inequality now.

[color=blue][b]Problem.[/b] For any triangle ABC with circumradius R and medians $m_{a}$, $m_{b}$, $m_{c}$, prove the inequality

$\frac{a\left(b+c\right)}{m_{a}}+\frac{b\left(c+a\right)}{m_{b}}+\frac{c\left(a+b\right)}{m_{c}}\leq 12R$.[/color]

[i]Solution.[/i] By the extended law of sines, a = 2R sin A, b = 2R sin B, c = 2R sin C, so that the inequality in question becomes

$\frac{2R\sin A\left(b+c\right)}{m_{a}}+\frac{2R\sin B\left(c+a\right)}{m_{b}}+\frac{2R\sin C\left(a+b\right)}{m_{c}}\leq 12R$.

Upon division by 2R, this simplifies to

$\frac{\sin A\left(b+c\right)}{m_{a}}+\frac{\sin B\left(c+a\right)}{m_{b}}+\frac{\sin C\left(a+b\right)}{m_{c}}\leq 6$.

Now, according to [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=235025#235025]http://www.mathlinks.ro/Forum/viewtopic.php?t=37178 post #3[/url], we have $m_{a}\geq\frac{b+c}{2}\cos\frac{A}{2}$, and similarly $m_{b}\geq\frac{c+a}{2}\cos\frac{B}{2}$ and $m_{c}\geq\frac{a+b}{2}\cos\frac{C}{2}$. Hence,

$\frac{\sin A\left(b+c\right)}{m_{a}}+\frac{\sin B\left(c+a\right)}{m_{b}}+\frac{\sin C\left(a+b\right)}{m_{c}}$
$\leq\frac{\sin A\left(b+c\right)}{\frac{b+c}{2}\cos\frac{A}{2}}+\frac{\sin B\left(c+a\right)}{\frac{c+a}{2}\cos\frac{B}{2}}+\frac{\sin C\left(a+b\right)}{\frac{a+b}{2}\cos\frac{C}{2}}$
$=\frac{2\sin A}{\cos\frac{A}{2}}+\frac{2\sin B}{\cos\frac{B}{2}}+\frac{2\sin C}{\cos\frac{C}{2}}=\frac{2\cdot 2\sin\frac{A}{2}\cos\frac{A}{2}}{\cos\frac{A}{2}}+\frac{2\cdot 2\sin\frac{B}{2}\cos\frac{B}{2}}{\cos\frac{B}{2}}+\frac{2\cdot 2\sin\frac{C}{2}\cos\frac{C}{2}}{\cos\frac{C}{2}}$
$=4\left(\sin\frac{A}{2}+\sin\frac{B}{2}+\sin\frac{C}{2}\right)$.

Now, since the function $f\left(x\right)=\sin x$ is concave in the interval [0°; 90°] (this interval contains the angles $\frac{A}{2}$, $\frac{B}{2}$, $\frac{C}{2}$), the Jensen inequality gives us

$\sin\frac{A}{2}+\sin\frac{B}{2}+\sin\frac{C}{2}\leq 3\sin\left(\frac13\cdot\left(\frac{A}{2}+\frac{B}{2}+\frac{C}{2}\right)\right)$
$=3\sin\frac{A+B+C}{6}=3\sin\frac{180^{\circ}}{6}=3\sin 30^{\circ}=3\cdot\frac12=\frac32$,

so that

$\frac{\sin A\left(b+c\right)}{m_{a}}+\frac{\sin B\left(c+a\right)}{m_{b}}+\frac{\sin C\left(a+b\right)}{m_{c}}$
$\leq 4\left(\sin\frac{A}{2}+\sin\frac{B}{2}+\sin\frac{C}{2}\right)\leq 4\cdot\frac32=6$,

and we are done.

  Darij

[quote="darij grinberg"]
$m_{a}\geq\frac{b+c}{2}\cos\frac{A}{2}$[/quote]
$m_{a}\geq\frac{b+c}{2}\cos\frac{A}{2}\Leftrightarrow(a+b-c)(a+c-b)(b-c)^{2}\geq0,$
but your proof of this fact is very simple.

To Darij

You are completely right. My solution is obviously wrong. Too careless!!! :oops:

Your solution is quite nice using a lemma that I think I have seen before but I don't remember. Thank you anyway.

[hide="To gemath"]

Sorry for this late reply. I didn't check this section and YM.
[/hide]

Thanks Darij for your good solution!  :)
[hide="To treegoner"]No problem treegoner, thank you for your interest  :) [/hide]

[quote="darij grinberg"][b]Problem.[/b] For any triangle $ABC$ with circumradius R and medians $m_{a}$, $m_{b}$, $m_{c}$, prove the inequality
\[\sum\frac{a\left(b+c\right)}{m_{a}}\leq8R\sum\sin\frac{A}{2}\leq 12R. \]
[/quote]
The following inequality is also ture:
\[\sum\frac{a\left(b+c\right)}{m_{a}}\leq\frac{8R}{\sqrt{3}}\sum\cos\frac{A}{2}\leq 12R. \]


[quote="Z-coli"]Larry can swim from Harvard to MIT (with the current of the Charles River) in $ 40$
minutes, or back (against the current) in $ 45$ minutes. How long does it take him to
row from Harvard to MIT, if he $ rows$ the return trip in $ 15$ minutes? (Assume that the
speed of the current and Larry’s swimming and rowing speeds relative to the current
are all constant.) Express your answer in the format $ mm: ss$.

Source: HMMT $ 2006$[/quote]
$ v_c \equal{} \text{speed of current}$
$ d \equal{} \text{length of river}$
$ v_s \equal{} \text{speed of Larry swimming}$
$ v_r \equal{} \text{speed of Larry rowing}$
$ t_d \equal{} \text{time to row downstream}$

Note that when Larry is traveling downstream, the current's speed is added to his "natural" speed; likewise, the current's speed is subtracted when he is going upstream.

$ d\equal{}rt \rightarrow d \equal{} (v_s \minus{} v_c)45 \equal{} (v_s \plus{} v_c)40$
$ 45v_s \minus{} 45v_c \equal{} 40v_s \plus{} 40v_c$
$ 5v_s \equal{} 85v_c \rightarrow v_s \equal{} 17v_c$
$ d \equal{} (17v_c \minus{} v_c)45 \equal{} 720v_c$

$ d \equal{} (v_r \minus{} v_c)15 \equal{} 720v_c$
$ v_r \minus{} v_c \equal{} 48v_c$
$ v_r \equal{} 49v_c$
$ d \equal{} 720v_c \equal{} (v_r \plus{} v_c)t_d \equal{} (49v_c\plus{} v_c)t_d \equal{} (50v_c)t_d$
$ t_d \equal{} \frac{720v_c}{50v_c} \equal{} 14.4$
$ 14.4 \text{ min} \equal{} \boxed{14: 24}$


logarithms
inequalities
superior algebra
superior algebra unsolved

let make all power of 2 together :124816...
show that evrery integer can be seen in this representation some where.
I hope that some one can helpe me to solve it , i'm not sure that's in algebra !

The decimal representation of $2^{n}$ begins with $2007$ if and only if $2007\cdot 10^{k}\le 2^{n}<2008\cdot 10^{k}$ for some integer $k\ge 0$. This can be restated as $\log_{10}2007+k\le n\log_{10}2<\log_{10}2008+k$. Since the number $\log_{10}2$ is irrational, the inequality holds for infinitely many values of $n$ (with $k$ dependent on $n$).

Thank you a lot , but there was some thing not clair , i know you use the fact that ZZ+µIN is dense in IR when µ is irrationnal, but we need to show that the "entire part" ( if x=p+µq in ZZ+µIN so i speek about p) is a négative integer to satisfay the inequality and to déduice that the number begin with ....i think that we shold take the inequality strictly , in that case the "entire part" ( p in p+qlog(2)) will be necessery négative and the conclusion dosen't change, i hope not doing stupidy now  :oops: ?


Let a, b, c are positive real numbers and a+b+c=1.
Prove that  $ (1\plus{}a)\sqrt{\frac{1\minus{}a}{a}}\plus{}(1\plus{}b)\sqrt{\frac{1\minus{}b}{b}}\plus{}(1\plus{}c)\sqrt{\frac{1\minus{}c}{c}}\geq\frac{3\sqrt{3}}{4}\frac{(1\plus{}a)(1\plus{}b)(1\plus{}c)}{\sqrt{(1\minus{}a)(1\minus{}b)(1\minus{}c)}}$

[quote="Luba1123"]Let a, b, c are positive real numbers and a+b+c=1.
Prove that  $ (1 \plus{} a)\sqrt {\frac {1 \minus{} a}{a}} \plus{} (1 \plus{} b)\sqrt {\frac {1 \minus{} b}{b}} \plus{} (1 \plus{} c)\sqrt {\frac {1 \minus{} c}{c}}\geq\frac {3\sqrt {3}}{4}\frac {(1 \plus{} a)(1 \plus{} b)(1 \plus{} c)}{\sqrt {(1 \minus{} a)(1 \minus{} b)(1 \minus{} c)}}$[/quote]
Dear Luba1123,you can use Holder inequality :) .
I will post my solution then some days :)

[quote="Luba1123"]Let a, b, c are positive real numbers and a+b+c=1.
Prove that  $ (1 \plus{} a)\sqrt {\frac {1 \minus{} a}{a}} \plus{} (1 \plus{} b)\sqrt {\frac {1 \minus{} b}{b}} \plus{} (1 \plus{} c)\sqrt {\frac {1 \minus{} c}{c}}\geq\frac {3\sqrt {3}}{4}\frac {(1 \plus{} a)(1 \plus{} b)(1 \plus{} c)}{\sqrt {(1 \minus{} a)(1 \minus{} b)(1 \minus{} c)}}$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=187375


anybody that can say a brief explanation of mod?  :lol:

just go on the search function at the top of your screen..there have been so many posts on this topic, or just go on the AoPS wiki.

Try this:
[url=http://en.wikipedia.org/wiki/Modular_arithmetic]Modular Arithmetic[/url]


group theory
abstract algebra
superior algebra
superior algebra unsolved

Can anyone help me solve this problem?

Suppose H, K are normal subgroups of G with K=<H.  Define a map phi:G/K-->G/H by phi(Ka)=Ha.  Then prove the following:

(a)  The map phi is well-defined (ie. if Ka= Kb, then phi(Ka)=phi(Kb)
(b) The map phi is a homomorphism
(c)  The map phi is onto
(d)  Ker(phi)=H/K

Hence conclude by the fundamental homomorphism theorem that

[u](G/K)[/u] ~ [u]G[/u]
(h/K)                 H

here is nothing to help, because every step is totally trivial. so please try it yourself and tell us your advances.

Is there a particular part that you're having trouble with?

For part (a),

If $ Ka \equal{} Kb$, then $ ab^{\minus{}1} \in K$. Since $ K \subset H$, $ ab^{\minus{}1} \in H$, so $ Ha \equal{} Hb$.

i havent done a lot of them.. so i need help with the steps, and explaining why each step is being done, so that next time ill understand it.. the logic of proofs just,goes over my head


Determine the shape of triangle $ABC$ such that $\sin C=\frac{\sin A+\sin B}{\cos A+\cos B}.$

[hide="Solution"]$\sin C=\sin(\pi-(A+B))=\sin(A+B)=\sin A\cos B+\cos A\sin B$, which gives

$\sin A+\sin B=(\cos A+\cos B)(\sin A\cos B+\cos A\sin B)$

$\sin A+\sin B=\sin A\cos A\cos B+\sin B\cos^{2}A+\sin A\cos^{2}B+\cos A\cos B\sin B$

$\sin A-\sin A\cos^{2}B+\sin B-\sin B\cos^{2}A=\sin A\cos A\cos B+\cos A\cos B\sin B$

$\sin A\sin^{2}B+\sin B\sin^{2}A=\cos A\cos B(\sin A+\sin B)$

$(\sin A+\sin B)(\cos A\cos B-\sin A\sin B)=0$

$(\sin A+\sin B)\cos(A+B)=0$

Since $\sin A+\sin B\neq 0$ for the angles in a triangle ($B\neq-A, B\neq A+\pi$), it follows $A+B={\pi\over 2}=C$, hence the triangle is right.[/hide]

Your answer is correct. :)
You can also use the theorem of sine, cosine rule.

hmmm

[hide="intuition"]
If A and B add up to 90 degrees, then sin A + sin B = cos A + cos B (sin A = cos B, sin B = cos A), making the RHS equal to 1, and if angles A + B are equal to 90, then angle C must be 90 degrees as well. The sine of 90 degrees is 1, which is equal to the RHS.

(It's a contest problem - I assumed it had a simple solution and it worked out from there.)
[/hide]


One friend ask me?
What day of the week was July 1,1949?

He said it was a math contest question held in his school,
does it has fomula to do this question?

It’s pretty well known that the cycle of days of the week repeat every 28 years. Hence 1 July 1949 fell on the same day of the week as 1 July in the year 1949 + 2×28 = 2005. I suspect the contest was held on or very close to 1 July 2005, so savvy contests would get the right answer straightaway. :roll:

FYI, it was a Friday. :P

Look like the 28 Years Cycle has some conditions?

By looking at up calendar we know that July24 1837 was monday

1837+28=1865, July 24 Monday
1865+28=1893, July 24 monday
1893+28=1921, july 24 was sunday!

So Here is another question,

We know that today 2008,Jan 21, is Monday, so what day of the week was
july04, 1867?

[quote="Jane_Bennet"]It’s pretty well known that the cycle of days of the week repeat every 28 years.[/quote]

Considering the cycle of leap years repeats every $ 400$ years, this seems doubtful.  [i]In general[/i], I don't think this is true.  Fortunately, $ 2000$ was a leap year as is expected by the naive $ 4$-year cycle, so I think your answer is still valid.

John Conway is an expert on this. watch his video, "calendar conundrums".

Actually, there is no need for any formula.
U just have to use ur logic.

I'm sorry - but the doomsday theory (not so much formula) allows you to figure out random dates within a matter of seconds.

Doomsday Theory? How does it allow you to figure out random dates?

Sorry to bring this back, but I totally forgot about answering this question

anyways, year x and x + 0 (mod 11) are on the same day

hopefully that helps you out

Today (2008.08.24) is 237th day of a year (31*4+30*2+29+24=237) , so 2008.01.01 was before 236 days
236=33*7+5; Today Is Sunday, before 5 days was Tuesday, So 2008.01.01 is Tuesday. 364=52 weeks, so we can hold that Common year is +1 week day, Leap-years - +2 week days.

Counting from 1949 to 2008 there were 14 Leap-years (1952...2004) and 59-14=45 Common years.
14*2+45=73=7*10+3. Tuesday-3 day = [b]Saturday[/b]

[quote]59-16=43[/quote]

Why -16? If you subtract 14 (it is the number of leap years you've calculated, isn't it?) instead, you'll obtain the correct result.

[quote="Bugi"]You should have computed when was [b]July 1.[/b] not [b]January 1.[/b].
July 1. 2008. was on Tuesday. Now I take that year starts on July 1.
From 1949. there were 15 Leap years, and 43 common years so $ 15\times2 \plus{} 43 \equal{} 73 \equal{} 7\times10 \plus{} 3$
Tuesday-3 days=Saturday.

But again January 1. and July 1. were on the same day tthis year...[/quote]

Actually July 1st is off the point  :lol:

Hsiljak found my mistake  :)

dont you have to use modular stuff to solve problems like this?

EDIT: oops sorry.

Also, it cycles every 2800 years, but that doesn't help. There is some formula to help do these things.

actually ducky revived it...from august.

[quote="mewto55555"]I know he did, but I was telling POincare.[/quote]

ducky = girl...

But anyways, I have seen a lot of reviving going on.  Please just make a new thread instead.


Hi,there!
 There are some interesting Number Theory problems but I haven't got any solution. Please help me!

 Problem 1:
 The sequence $P_n$ is defided by
 $\displaystyle P_n=1+\frac{1}{2}+\cdots+\frac{1}{n}$
 Where $P_n$ is a fraction and it is expressed in lowest terms.
 The question : Find all $n$ such that the numerator of $P_n$ is divisible by 3

 Problem 2
 Let $a,b$ be two possitive integer such that $a^n-1$ is divisible by $b^n-1$ for all $n \in N$.
 Prove that $a$ is a power of b

 Let go !!!

For 2), you may have a look here :
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=A.M.M.&t=4556

Pierre.

Oh very fast  :D
 Thank you very much !

 MathLinks is the best! :P

Yes but how about the first one !

 $n=2,7,21 ?$

  :(  :(  :(

I think that third value is $22$. It's been discussed nefore. It was used in the Bulgarian olympiad last year, I think. Try looking for it.


Do you recommend taking notes when reading books such as AOPS 2, ACOPS and PSS? Recently, I have found that without notes sometimes I forget to consult a certain concept when working through possible strategies for problems. But also, notes tend to be time consuming and eventually become too large in volume to review entirely before a contest.

What I have decided to do is read a chapter and record the major theorems so that when I am approaching a problem which concerns a certain field (for example take circles) I am able to ensure that I exhaust all possible approaches which I know before giving up. Also, if I get any problems wrong or am close to not seeing a solution, then I write down a quick reminder of how to solve that problem or all problems which deal with the part of the problem where I became stuck.

Should I continue with this method, alter it or abandon it entirely? Also, do you take any notes when reading math textbooks such as the ones listed above? Any tips or feedback will be greatly appreciated.

When I do AoPS texts, I always make my own solution manual (kind of). I write my own solution for each problem I do. I try to do all the problems and stick to this method unless I am just trying to pick up a few concepts right before (3-5 days before) a competition.

I dont know about taking notes though. If you really understand a concept, you shouldn't have to look back at your notes to be able to solve it. You could just look back at the book and that should trigger your memory.

Hope this helps  :lol:

Thanks!   :)

I find that when I study for anything (school tests, etc.) that making not necessarily notes (not explanations) but bullet points which name and help me organize multiple concepts in a list-like format is useful. Do you not do something like that?

Also, almost never write out full solutions. Sometimes I try to do problems in my head. Does this put me at a disadvantage, and should I change to writing out full solutions?

well, math comps are way different from school tests. For one thing, they actually stimulate if not challenge the mind.

That said, whenever I see a theorem or w/e on aops ii, or an easy usamo problem (which is a challenge for me) i look at the text's proof or solution, and sit there thinking of another way to prove it. I write it down, so that pretty much i could glance at it  and be like "hey, thats how i came to that theorem"...
I do recommend jotting the bombs on aops. They're useful tips that could really save u points (mostly cautioning u not to assume sucha nd such). But, understand why these misconceptions can cause mistakes, and why these misconceptions occur in the first place.

Taking notes on math books for the sake of reviewing them later seems unnecessary.  In a history textbook, the key ideas are buried under overwhelming amounts of facts, and writing down the key ideas is important because you might not be able to find them again otherwise.  In a math book, especially the ones you mentioned, the important theorems and ideas are neatly laid out for you, and if you want to review them, you can just open up the book one more time.  Having a list of possible approaches to each type of problem would only be a distraction, and using such a list would require a significant amount of luck to run into a solution.  A good, challenging problem won't be solvable by pulling a technique off a predetermined list.  You should instead focus on why certain techniques are effective, and which aspects of a problem make it conducive to a certain approach.  Your solutions will then be motivated by the aspects of the problem, not stumbled upon randomly.

All of that said, taking notes can be useful.  I would at least have a piece of paper out at all times when you're working through any of these books.  If a book shows a derivation or proof, write out the key steps on a piece of paper as you read through it.  You could take more detailed notes too.  I take very detailed notes for most of my classes, but not with the intention of ever looking at them again.  I do it just so that I can remember the material better.  My dad tells me he used to do the same thing in college, but he would throw out his notes while walking out of the class.

About writing out complete proofs, you should do it enough so that you're confident you could write out a complete proof if you needed to.  It's also good to practice writing up proofs quickly.  But you don't need to do it for every problem you solve, unless you really want the practice.


Suppose the medians AA' and BB' of triangle ABC intersect at right angles.  If BC = 3 and AC = 4, what is the length of side AB?

[quote="Mathalete"]Pythagorean Theoruem to get 3^2+4^2=sqrt25
sqrt25=5[/quote]

No no, it's almost that simple but that's too much. It's the medians that are orthogonal not the sides of the triangle

Daniel

[hide]Let $x=AB$. From the theorem of the medians we have:

$x^{2}+AC^{2}= 2A'A^{2}+\frac{BC^{2}}{2}$

$x^{2}+BC^{2}= 2B'B^{2}+\frac{AC^{2}}{2}$

adding the two equations we get

$2x^{2}+\frac{AC^{2}+BC^{2}}{2}= 2(A'A^{2}+B'B^{2}) \Rightarrow$

$4x^{2}+(AC^{2}+BC^{2}) = 4(A'A^{2}+B'B^{2}) \ \ (1)$

Let $G$ be the centroid of the triangle. Then $\boxed{AG^{2}+BG^{2}=x^{2}}$

$A'A = \frac{3}{2}\cdot AG \Rightarrow \boxed{A'A^{2}= \frac{9}{4}\cdot AG^{2}}$

$B'B = \frac{3}{2}\cdot BG \Rightarrow \boxed{B'B^{2}= \frac{9}{4}\cdot BG^{2}}$

adding the last two (boxed) equations we get

$A'A^{2}+B'B^{2}= \frac{9}{4}\cdot (AG^{2}+BG^{2}) = \frac{9}{4}\cdot x^{2}$

$(1) \Rightarrow$

$4x^{2}+(AC^{2}+BC^{2}) = 9 x^{2}\Rightarrow$

$x^{2}= \frac{AC^{2}+BC^{2}}{5}= \frac{25}{5}=5 \Rightarrow x=\sqrt{5}$[/hide]


The Malfatti circles of a triangle are the three circles inside the triangle,mutually tangent to each other,and each tangent to two sides of the triangle. Express the inradius of the triangle w.r.t. the inradii of the Malfatti circles. I know the formula and only one complicated proof. But I want to know how to derive the formula as easily,precisely as possible. May I ask you a favor ?  :)

No one seems to be able to solve this, so please post your proof (or the strategy at least).


A sequence is defined recursively by
$a_1=1$ $a_{n+1}=1+\frac{1}{1+a_n}$
Show that $a_n$ is convergent by using the theorem of bound, and monotonic sequence.

   I don't know how to do this one since for odd terms, and even terms are totally different. Can we say that this sequence are made by 2 small sequences. One sequence is converge (monotonic increase), the other is also converge (monotonic decrease). Therefore the orginal sequence converge?

If the limits of the two subsequences are the same, that should work.

[quote="Gibbenergy"]

   I don't know how to do this one since for odd terms, and even terms are totally different. Can we say that this sequence are made by 2 small sequences. One sequence is converge (monotonic increase), the other is also converge (monotonic decrease). Therefore the orginal sequence converge?[/quote]

Say that $\lim_{n \to \infty }a_{2n}= \ell$ and $\lim_{n \to \infty }a_{2n-1}= m$

Now from $a_{n+1}=1+\frac{1}{1+a_n}$ we have that ..... :)

[quote="kunny"]Solve the given recurrence, we have $a_n=\sqrt{2}\cdot \frac{1+r^n}{1-r^n}\ (n\geq 1),$ where $r=\frac{1-\sqrt{2}}{1+\sqrt{2}}.$ Since $|r|<1,$ yielding $\lim_{n\to\infty} a_n=\sqrt{2}.$

This is related to the continual fraction of $\sqrt{2}.$[/quote]

  To find that limit $\sqrt{2}$ we can solve the equation
$L=1+\frac{1}{1+L}$ so it is very easy.
   Your formular $a_n=\sqrt{2}\cdot \frac{1+r^n}{1-r^n}\ (n\geq 1),$ where $r=\frac{1-\sqrt{2}}{1+\sqrt{2}}.$ is nice. May you let me know how you find it?
   In this problem, I don't think your solution is nice since it is a homework problem and can be done in the easier way.
   About Ramanujan's hint, I am perfectly understand that we have to prove two subsequence have the same limit. But... I admit that I don't know how to prove.

     [quote="Ramanujan"]Say that $\lim_{n \to \infty }a_{2n}= \ell$ and  $\lim_{n \to \infty }a_{2n-1}= m$
Now from  we have that $a_{n+1}=1+\frac{1}{1+a_n}$..... [/quote]
But how?
$l=1+\frac{1}{1+m}$ ???????

Also you have it with $m$ and $l$ in the opposite positions.


[quote="panos_lo"]Suppose that $ P_n$ is the n-th prime. Solve: $ 32\plus{}31(P_1)^5\plus{}...\plus{}31(P_n)^5\equal{}(32)^{n\plus{}1}$ [/quote]

Clearly $ n\equal{}1$ is a solution.

And for $ n \ge 2$, we can prove $ (P_n)^5<32^{n}$ by Bertrand's Postulate with induction.

So for $ n \ge 2$, $ 32\plus{}31(P_1)^5\plus{}...\plus{}31(P_n)^5<32\plus{}31(32)^1\plus{}...\plus{}31(32)^{n}\equal{}32^{n\plus{}1}$

$ \Rightarrow n\equal{}1$ is the only solution.


Let $ ABC$ be a nonisosceles triangle and $ D\in BC$ , $ AD\perp BC$ . Denote the incircles

$ C(I_1,r_1)$ , $ C(I_2,r_2)$ of $ \triangle ABD$ , $ \triangle ADC$ respectively. Prove that $ AD^2\equal{}DB\cdot DC$ $ \Longleftrightarrow$ $ \frac {r_1}{r_2}\equal{}\frac cb$ .

[quote="Virgil Nicula"]Let $ ABC$ be a nonisosceles triangle and $ D\in BC$ , $ AD\perp BC$ . Denote the incircles

$ C(I_1,r_1)$ , $ C(I_2,r_2)$ of $ \triangle ABD$ , $ \triangle ADC$ respectively. Prove that $ AD^2 \equal{} DB\cdot DC$ $ \Longleftrightarrow$ $ \frac {r_1}{r_2} \equal{} \frac cb$ .[/quote]

The triangles $ \triangle ADB$ , $ \triangle ADC$ have the same right angle, so $ AD^2\equal{}DB\cdot DC\iff\frac{AD}{DC}\equal{}\frac{DB}{AD}\iff \triangle ADB\sim \triangle ADC\iff \frac {r_1}{r_2} \equal{} \frac{c}{b}$.


On a table $ 10$ x $ 10$.We fill numbers from $ 1$ to $ 100$ such that : from $ 1$ to $ 10$ lie on the first row from left to right;from $ 2$ to$ 20$ lie on the second row from left to right,....Anna cut the table by $ 1$x$ 2$ rectangles.After that she add $ 50$ products of two numbers in a that rectangle.Determine thae way to Anna can cut the table such that above sum get the minimun value.

Russia MO: http://www.mathlinks.ro/viewtopic.php?p=832718#p832718

notice that in each domino with numbers a,b. We can express the product as $ ab \equal{} \frac {a^2 \plus{} b^2 \minus{} (a \minus{} b)^2}{2}$. since the sum of all such a and b is constant, we are trying to maximize the sum of all $ (a \minus{} b)^2$, which is easy since each $ (a \minus{} b)^2$ is at most 100(in fact it is either 100 or 1). :lol:


analytic geometry
geometry
vector
complex numbers
combinatorics unsolved
combinatorics

We have three numbers $a,b,c$ written on a blackboard. We execute an operation on these numbers.
We choose one of them and substitute it for the number equal to the sum of this number and difference of the two others multiplied by any rational number. Prove, that starting from the $0,1,\sqrt{2}$ we cannot obtain
$0,2,\sqrt{2}$ and any permutaiotn of it.

$a,b,c$ if we choose $a$, then we have $a+(b-c)\frac{p}{q} , b,c$

If $a,b,c$ are in the form $p + q \sqrt 2$, with $p,q \in \mathbf Q$, then applying the given transformation, we remain with numbers of the same form.

Moreover, if $p + q \sqrt 2 = p' + q' \sqrt 2$, then necessarily $p=p',q=q'$ (otherwise $\sqrt 2$ would be rational); so the representation as a rational linear combination of $1$ and $\sqrt 2$ is unique.

In this way, we can represent our numbers in a coordinate [rational] plane, by associating $p + q \sqrt 2$ with the point $(p,q)$.

With this geometric representation, our three values become the vertices of a triangle, and the move becomes: [i]choose an existing vertex and move it by an arbitrary [rational] quantity along the line parallel to the opposite side.[/i]

Clearly, the area of the triangle is not changed by the move (two triangles with same bases and altitudes), so it's invariant. But starting with $0, 1, \sqrt2$ you find a triangle with area $1/2$, and $0,2,\sqrt 2$ has area $1$. []

If you made up this solution on your own you are great :) , it's original solution. I knew it from the book and I wondered whether exists another approach to this. But still thanks

Yes, and no...

The solution is indeed original. But to tell all the truth, the technique is quite standard, especially if you have seen at least once gaussian integers (and their fractions: complex numbers with rational Re/Im parts).

You simply apply the same idea of plane representation. And the rule a+something(b-c) should flash the idea "Point A translated by vector CB" (think about affine geometry). The rest is putting the ideas together and verifying the definitions make sense.

Anyway, thx.

M.