symmetry
algebra
polynomial
linear algebra
matrix
inequalities

How many ordered four-tuples of integers $(a,b,c,d)$ with $0 < a < b < c < d < 500$ satisfy $a + d = b + c$ and $bc - ad = 93$?

Interesting question, so we need to find some law for these four-tuples, obviously, there will be many such four-tuples. They will occur in specific order:

(1, 2, 94, 95) (1 < 2 < 94 < 95 and (1 + 95) = (2 + 94) and 2*94 = 1*95)
(1, 4, 32, 35)
(2, 3, 95, 96)
(2, 5, 33, 36)
(3, 4, 96, 97)
(3, 6, 34, 37)
....
but at the end, symmetry will broke at:
...
(406, 409, 437, 440)
(407, 410, 438, 441)
(408, 411, 439, 442)

and the end value, will be
(465, 468, 496, 499)

I have count 870.

$a+d=b+c$ so $ac+cd=93+ad+c^2$ . $\implies (c-a)(d-c)=93$

Hence $(c-a,d-a)=(1,93),(3,31),(31,3),(93,1)$

Solve them in tems of $c$ to get
$(a,b,c,d)=(c-93,c-92,c,c+1),(c-31,c-28,c,c+3),(c-1,c+92,c,c+93),(c-3,c+28,c,c+31)$

Easily check that the last two solution doesnt follow $a<b<c<d$ . So we only need to consider the first two solutions .

First solution gives us $c-93\geq 1$ and $c+1\leq 499$ $\implies 94\leq c\leq 498$
second one gives us $32\leq c\leq 496$

So the total orderr pair of solution is $405+465=\boxed{870}$

[hide="hint"]
put $b=a+x$, $c=b+y$, $d=c+z$ or something like that (i can't remember right now) and factor
[/hide]

[quote="Arvind_sn"]Hmm... this is strange. If I set

$b+c = a+d$,
$bc = ad+93$

and then I let $X^{2}-(a+d)X+(ad+93)$ be the polynomial whose roots are $b, c,$ then we have that $(a+d)^{2}-4(ad+93) = (a-d)^{2}-4 \cdot 93 = k^{2}\Leftrightarrow (a-d+k)(a-d-k) = 2^{2}\cdot 3 \cdot 31.$

Since $a-d+k$ and $a-d-k$ must have the same parity, we have the only solutions to be:

$a-d-k = 2, a-d+k =186$ and $a-d-k = 6, a-d+k = 62.$

When we solve for the first part, we get that $a-d = 94$ and $k = 92.$ But a contradiction exists, since we get $a = d+94,$ and $a < d$... what did I do wrong?[/quote]

$a-d-k$ and $a-d+k$ should be negative, I think. So $(a-d-k,a-d+k)=(-186,-2)$ and $(-62,-6)$ are possible given $k$ is positive.

Can we use the fact that 93 is equal to the determinant of the following matrix?
|b      a|
|d      c|

Just a thought, after seeing those pairs together I thought determinants..

$b,a$
$d,c$

R2-R1->R2

$b,a$
$d-b,c-a$

then we are like OH SNAP $d-b=c-a \iff a+d=b+c$

so we can factor that out

$(d-b)*$
$b,a$
$1,1$

Then that last thing can just be evaluated...so we get $(d-b)(b-a)$.

[I don't remember how to determinants in latex...]

@4everwise: congratulations, very elegant solution! There's just a little typo:
[quote="4everwise"] $bc-ad=(1+m)(1+m+n)-a(a+2m+n)$ [/quote]
You should substitute those $1$'s with $a$'s!...  :wink:

I thought all the solutions above sort of complicated the problem:

[hide]
Let $a = b + c - d$ so that $bc - (b+c-d)d = 93$
Simplify and factor to get $(c-d)(b-d) = 93$
From the given ($0 < a < b < c < d < 500$) we see that there are two cases:
case $d-c = 1$ and $d-b = 93$

So solve for everything in terms of d, (note a = b + c - d) $a = d-93 + d - 1 -  d = d - 94, b = d - 93, c = d - 1$
From the inequality (a term) we see 0 < d - 94, so d>94 and d < 500, so for this case there are $499-95 + 1 = 405$ occurrences

case $d-c = 3$ and $d-b = 31$

do the same thing to get 0 < d - 34 < d - 31 < d - 3 < d < 500, we see d > 34 so there are $499 - 35 + 1 = 465$ examples

To finish, we just add both cases to get $405 + 465 = \ans{870}$

[/hide]

[hide="Alternate Solution"]
Since $a+d=b+c$, let $b=a+k$ and $c=d-k$ for some nonnegative integer $k$.  Then \[(a+k)(d-k)-ad=93\implies k(d-a-k)=1\times 93=3\times 31.\]If $k=1$, then $a-d=94$, so any quadruple $(a,b,c,d)=(a,a+1,a+93,a+94)$ works.  We must have $a+94<500\implies a<406$, so $a$ can range from $1$ to $405$ for a total of $405$ solutions.
If $k=3$, then $a-d=34$, so any quadruple $(a,b,c,d)=(a,a+3,a+31,a+34)$ works.  We must have $a+34<500\implies a<466$, so $a$ can range from $1$ to $465$ for a total of $465$ solutions.
If $k=31$ or $k=91$, then the condition that $a<b<c<d$ does not hold.  Therefore there are a total of $405+465=\boxed{870}$ solutions.[/hide]

I found taking advantage of the fact that $(a,d)$ and $(b,c)$ had the same averages quite useful.  If we let the average be $x,$ and $a=x-n,d=x+n,b=x-m,c=x+m,$ then we arrive at the equation $(n-m)(n+m) = 93.$  In taking advantage of the average of the variables, we essentially reduced it down to just 2 variables, which was easier to work with in my opinion.

[hide=Solution]We have $a+d=b+c$, so upon substituting we get $(c-a)(d-c)=93$. We can hence have $(c-a, d-c)=(1,93), (3,31), (31,3), (93,1)$. The last two do not satisfy $a<b<c<d$, so disregard them. Upon solving for $c$, the first two are $(a,b,c,d)=(c-93, c-92, c, c+1), (c-31, c-28, c, c+3)$ respectively. The first gives $405$ ways and the second gives $465$, for a total of $\boxed{870}$.[/hide]


More questions from Munkres.  Would appreciate if someone can provide link to the answers from the book.

Let $ \bold{x}$ and $ \bold{y}\in \mathbb{R}^\omega$.  Equip $ \mathbb{R}^\omega$ with the box topology.  Show that if $ \bold{x}\minus{}\bold{y}$ is not eventually zero (ie there does not exist an integer $ N$ such that $ x_n\equal{}0$ for all $ n>N$), then there is a homeomorphism $ h$ of $ \mathbb{R}^\omega$ with itself such that $ h(\bold{x})$ is bounded and $ h(\bold{y})$ is unbounded.

Munkres deliberately never published solutions from his book, because it's used to teach classes. Don't
count on anyone posting a list of solutions here, either.


rotation
combinatorics solved
combinatorics

Consider p and n positive integers and a coloring of the plane in p colors. Prove that one can find two congruent regular n-gons having the vertices of the same color.
plz help me with this darn good problem
best regards

this is romania 1995 selection test problem.
so ne of the romanian guys having the solution??
cheers

I belive that you have the question wrong (again :D) galois - just to be sure I've looked up the problem and I found that we need to find only two (well actually an infinity of polygons but that's easy once you found a pair) CONVEX polygons  (not regular :) ).

On that problem the solution would be considering a circle and np+1 points on the circle which divide it into equal arcs. Obviously there are n points collored in the same color making a mono-color sub-polygon. There are a finite number ( np+1 choose n ) of such polygons, and if we rotate the polygon into the circle, we can find an infinite number of positions in which the sub-polygons have the same color and same shape.

A connected problem is the following (again in Romania '95 :D)

Let n,p be positive integers, n\geq 6, 3\leq p\leq n-p. The vertices of a REGULAR (this time :D) n-gon are colored with two colors: p vertices with white and n-p black vertices. Prove that there exist two congruent polygons, having each at least [p/2]+1 vertices, one of them with white vertices and the other with black vertices.

Have fun :D

hey
i'm really sorry abt the wrong problems.actually i had done proving the convex bit,but was wondering how cud one prove that there are regualr polygons.the site from which i got romania 1993,1994 and 1995 "explicitly"
states that there "might" be some mistakes.
so i guess,all of us have to wait till valentin puts up the romanian olympiads of all the years
cheers mate

yeah - sorry for keeping you waiting so much but I am not finished working with the site yet - a new surprise is awaiting. :D

Ok!
The rule of contraries
We will rotate p vertices n times. Each time [p/2] or less white points will be coincide with black points. Thus there are *less* than [p/2]*n coincidence. On the other hand each of p white points will meet each of n-p black points  ==> there are exactly p(n-p) coincidence. Contradiction as [p/2]*n \leq p(n-p)


search
number theory proposed
number theory

find all prime numbers $ p,q$ such that : $ pq|(5^p\minus{}2^q)(5^q\minus{}2^p)$

http://www.mathlinks.ro/viewtopic.php?search_id=108505887&t=18694 :)
http://www.mathlinks.ro/viewtopic.php?search_id=108505887&t=50838  :)


vector
linear algebra
matrix
linear algebra unsolved

Let A be a 5 x 3 matrix of rank 3 and let {X1, X2, X3} be a basis of R^3.

a) Show that N(A) = {0}
b) Show that if Y1 = AX1, Y2 = AX2, and Y3 = AX3 then Y1,Y2,Y3 are linearly independent
c) Do the vectors Y1,Y2,Y3 from part b) form a basis for R^5? Explain.

For part a) I know that I can show it by using definitions. However, I don't understand where to go with b) and c)??

well for (b)you have the linear map $f_A$ related to the matrix $A$ which is from $R^3$ to $R^5$ the fact that $rk(f_A)=3$ means that there are  $v_1,v_2\& v_3$ linearly independent in $R^5$   and $u_1,u_2\&u_3$ in $R^3$ such that for $0<i\leq 3$ we have $f_A(u_i)=v_i$ thus $f_A$ is an isomorphism from $R^3$ to $<v_1,v_2,v_3>$ then the answer to the question b is a simple conclusion .
for (c) $3<5$ so how can it be a basis ;)?
if this is the first time u do algebra u av to know that it's just with lil practice u'll be able to master it ,good luck :)


let P be a convex polygon with n vertecs.
for each point like X,draw line L wich goes through X,and cut the convex polygon in two pices(AandB)in wich:
the area of A/the area of B=1/n

:rotfl: i SUGGEST U ,THAT YOU GO TO PRE_OLYMPIAD SECTION,WITH YOUR EASY problems. :!:

Hilbert, BEHAVE YOURSELF. I've told you the same thing before.


[color=darkred][u][b]Problema 2[/b] (Prof. Nicolae Pavelescu).[/u]  In patrulaterul $ABCD$ avem $AD\parallel BC$ si notam $O\in AC\cap BD$. Pentru un punct $E\in (BD)$, $E\not\equiv O$ notam simetricul $F$ al punctului $C$ fata de punctul $E$. Paralela dusa prin punctul $F$ la dreapta $AD$ taie dreapta $AB$ in punctul $G$ iar paralela dusa prin punctul $F$ la dreapta $AB$ taie dreapta $AD$ in punctul $H$. Sa se arate ca punctele $E$, $G$, $H$ sunt coliniare daca si numai daca patrulaterul $ABCD$ este un paralelogram.

[u][b]Problema 4[/b] (Prof. Virgil Nicula).[/u] O dreapta $d$ care trece prin centrul de greutate $G$ al triunghiului $ABC$ taie laturile acestuia in punctele : $D\in BC$, $C\in (BD)$; $E\in (CA)$; $F\in (AB)$, $AF=AG$. Notam mijlocul $N$ al laturii $[AC]$ si centrul $O$ al cercului circumscris triunghiului dat. Sa se arate ca $ED=EF$ $\Longleftrightarrow$ $2b^{2}=a^{2}+c^{2}$ $\Longleftrightarrow$ $OH\perp BN$.[/color]

[quote="Virgil Nicula"][color=darkred][u][b]Problema 2[/b] (Prof. Nicolae Pavelescu).[/u]  In patrulaterul $ABCD$ avem $AD\parallel BC$ si notam $O\in AC\cap BD$. Pentru un punct $E\in (BD)$, $E\not\equiv O$ notam simetricul $F$ al punctului $C$ fata de punctul $E$. Paralela dusa prin punctul $F$ la dreapta $AD$ taie dreapta $AB$ in punctul $H$ iar paralela dusa prin punctul $F$ la dreapta $AB$ taie dreapta $AD$ in punctul $G$. Sa se arate ca punctele $E$, $G$, $H$ sunt coliniare daca si numai daca patrulaterul $ABCD$ este un paralelogram.[/color][/quote]
[color=darkblue][b][u]Demonstratie 1 (metoda vectoriala).[/u][/b] Alegem originea sistemului de vectori in varful $A$, adica pentru orice punct $X$ din planul patrulaterului dat vom nota $\overrightarrow{AX}=X$. Asadar : exista $c\in \mathrm R$ astfel incat $C=B+cD$ deoarece $BC\parallel AD$; exista $m\in (0,1)$ astfel incat $E=(1-m)B+mD$ deoarece $E\in{BD}$; exista $\{g,h\}\subset \mathrm R$ astfel incat $G=gD$, $H=-hB$; $F=G+H=-hB+gD$ deoarece patrulaterul $AGFH$ este paralelogram; $2E=C+F$ deoarece punctul $E$ este mijlocul segmentului $[CF]$, adica $2E=2(1-m)B+2mD=$ $B+cD-hB+gD$ $\Longrightarrow$ $\{\begin{array}{c}h=2m-1\\\ g=2m-c\end{array}\|$ $\Longrightarrow$ $\{\begin{array}{c}G=(2m-c)D\\\ H=(1-2m)B\\\ F=(1-2m)B+(2m-c)D\end{array}$. Prin urmare, $E\in HG$ $\Longleftrightarrow$ $|\begin{array}{ccc}0 & 2m-c & 1\\\ 1-2m & 0 & 1\\\ 1-m & m & 1\end{array}|=0$ $\Longleftrightarrow$ $m(1-c)=0$ $\Longleftrightarrow$ $c=1$ $\Longleftrightarrow$ $C=B+D$ $\Longleftrightarrow$ patrulaterul $ABCD$ este paralelogram.

[b][u]Demonstratie 2 (metoda sintetica).[/u][/b] $\{\begin{array}{c}S\in AF\cap HG\\\ L\in \overline{BED}\cap HF\end{array}\|$ $\Longrightarrow$

$\{\begin{array}{c}SA=SF\\\ EB=EL\ ,\ \boxed{BC=FL}\end{array}\|$. Therefore, $E\in HG$ $\Longleftrightarrow$ $S\in HE$

(semidreapta $[HE$ este mediana in triunghiul $\triangle HBL$ si $SA=SF$) $\Longleftrightarrow$

$AF\parallel \overline{BEDL}$ $\Longleftrightarrow$ patrulaterul $AFLD$ este paralelogram $\Longleftrightarrow$

$\boxed{AD=FL}$ $\Longleftrightarrow$ $BC=AD$ $\Longleftrightarrow$ patrulaterul $ABCD$ este paralelogram.[/color]

pentru
Concursul Mathematica - Modus Vivendi, Rm. Valcea, Romania,
este si pantru clase mai mari??, adica nivel mai avansat??

Ce intelegi prin nivel mai avansat? Daca vrei sa vazi subiecte si clasamente intra pe http://mircea.unet.ro/ .

propun crearea unui topic ( Stickie ) in care sa se anunte fiecare concurs interjudetean ....de ex eu as fi dorit sa particip la concurs(daca si aflam de el )

si eu as fi vrut sa merg [b]'la scoala cu ceas' [/b]daca aflam la timp... de mmv am aflat prin ianuarie si am putut participa.
Apropo are cineva subiectele de la a opta/baraj juniori de la scoala cu ceas?

[quote="icx"]Profesori insotitori? Eu nu am cu cine sa ma duc :( Felicitari Andrei.[/quote]

si eu am mers cu maica-mea, trebuia cu profesorul de mate (dirigintele meu), dar nu a mai putut...

[quote="Virgil Nicula"]
[u][b]Problema 4[/b] (Prof. Virgil Nicula).[/u] O dreapta $d$ care trece prin centrul de greutate $G$ al triunghiului $ABC$ taie laturile acestuia in punctele : $D\in BC$, $C\in (BD)$; $E\in (CA)$; $F\in (AB)$, $AF=AG$. Notam mijlocul $N$ al laturii $[AC]$ si centrul $O$ al cercului circumscris triunghiului dat. Sa se arate ca $ED=EF$ $\Longleftrightarrow$ $2b^{2}=a^{2}+c^{2}$ $\Longleftrightarrow$ $OH\perp BN$.[/quote]

Propun celor interesati urmatoarea problema oarecum intarita, inspirata din aceasta [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=135517[/url]

Daca tot a venit vorba de La Scoala cu Ceas, poate posta cineva subiectele?

Oricum, sustin ideea unui Sticky cu informatii despre concursurile astea.

[quote="Virgil Nicula"][color=darkred][u][b]Problema 4[/b] (Prof. Virgil Nicula).[/u] O dreapta $d$ care trece prin centrul de greutate $G$ al triunghiului $ABC$ taie laturile acestuia in punctele : $D\in BC$, $C\in (BD)$; $E\in (CA)$; $F\in (AB)$, $AF=AG$. Notam mijlocul $N$ al laturii $[AC]$ si centrul $O$ al cercului circumscris triunghiului dat. Sa se arate ca $ED=EF$ $\Longleftrightarrow$ $2b^{2}=a^{2}+c^{2}$ $\Longleftrightarrow$ $OH\perp BN$.[/color][/quote]
[color=darkblue][b][u]Demonstratie.[/u][/b] Notam mijlocul $M$ al laturii $[BC]$. Din relatia $AF=AG=\frac{2m_{a}}{3}$ obtinem $FB=\frac{3c-2m_{a}}{3}$ si din teorema lui Menelaus aplicata transversalei $\overline{DGF}$ in triunghiul $ABM$, adica $\frac{DB}{DM}\cdot\frac{GM}{GA}\cdot\frac{FA}{FB}=1$ se obtine $CD=\frac{a(4m_{a}-3c)}{6(c-m_{a})}$. Aplicam teorema Menelaus transversalei $\overline{AEC}$ in triunghiul $BFD$: $\frac{AF}{AB}\cdot\frac{CB}{CD}\cdot\frac{ED}{EF}=1$. Prin urmare,
$1.\blacktriangleright\boxed{EF=ED}$ $\Longleftrightarrow$ $\frac{AF}{AB}\cdot\frac{CB}{CD}=1$ $\Longleftrightarrow$ $\frac{2m_{a}}{3c}\cdot\frac{6(c-m_{a})}{4m_{a}-3c}=1$ $\Longleftrightarrow$ $4m_{a}(c-m_{a})=c(4m_{a}-3c)$ $\Longleftrightarrow$ $4m^{2}_{a}=3c^{2}$ $\Longleftrightarrow$ $2(b^{2}+c^{2})-a^{2}=3c^{2}$ $\Longleftrightarrow$ $\boxed{a^{2}+c^{2}=2b^{2}}$.
$2.\blacktriangleright \boxed{OH\perp BN}$ $\Longleftrightarrow$ $OG\perp BN$ $\Longleftrightarrow$ $OB^{2}-ON^{2}=GB^{2}-GN^{2}$ $\Longleftrightarrow$ $OA^{2}-ON^{2}=4\cdot GN^{2}-GN^{2}$ $\Longleftrightarrow$ $AN^{2}=3\cdot GN^{2}$ $\Longleftrightarrow$ $\left(\frac{b}{2}\right)^{2}=3\cdot\left(\frac{m_{b}}{3}\right)^{2}$ $\Longleftrightarrow$ $3b^{2}=2\left(a^{2}+c^{2}\right)-b^{2}$ $\Longleftrightarrow$ $\boxed{a^{2}+c^{2}=2b^{2}}$.[/color]


How many line segments have both their endpoints located at the vertices of a given cube?

$\text{(A)}\ 12 \qquad \text{(B)}\ 15 \qquad \text{(C)}\ 24 \qquad \text{(D)}\ 28\qquad \text{(E)}\ 56$

[hide]
This should be the sum of all the edges, the diagonals of each face and then the inner diagonals of the cube.
There are 12 edges, 4 inner diagonals and then $6(2) = 12$ face diagonals.
$12 + 12 + 4 = 28$
(D)
[/hide]

True. But there is another way (I think quicker).

How is line defined? In another word, how many points do you need to form a line?

[hide="I think this is the other way..."]
A cube has 8 vertices. A line segment has 2 endpoints. So the number of line segments with vertices as endpoints would be $7+6+5+4+3+2+1=28$.

[/hide]

[quote="frt"][hide]
You're choosing a pair of vertices from 8 vertices, so it's simply $\binom{8}{2}=28$ ;) [/hide][/quote]

Oh, right. I knew there was a simpler way to put it. Thanks.  :P


At a party there are only single women and married men with their wives.
The probability that a randomly selected woman is single is $ \frac{2}{5}$.
What fraction of the people in the room are married men?

Or:

This means 3/5 women are married. Thus there are 2 parts single, 3 parts married W, 3 parts married M so 3/8

Suppose 5 women. Then we have 2 single, 3 married women, 3 married men hence $\frac{3}{8}.$


Vieta
conics
parabola
analytic geometry
symmetry
special factorizations

Suppose $ k\in \mathbb{R}$ and $ \alpha ,\beta$ are two real roots of the equation $ x^2\minus{}(k\minus{}2)x\plus{}(k^2\plus{}3k\plus{}5)\equal{}0$,
then find the maximum value of $ \alpha^2 \plus{}\beta^2$.

In order for both roots to be real, we must have $ (2 \minus{} k)^2 \minus{} 4(k^2 \plus{} 3k \plus{} 5) \equal{} \minus{} 3k^2 \minus{} 16k \minus{} 16\ge0$

$ \implies(k \plus{} 4)(3k \plus{} 4)\le0\implies \minus{} 4\le k\le \minus{} \frac43$.

Also, by Vieta's, $ \alpha^2 \plus{} \beta^2 \equal{} (2 \minus{} k)^2 \minus{} 2(k^2 \plus{} 3k \plus{} 5) \equal{} \minus{} (k^2 \plus{} 10k \plus{} 6)$, whose maximum occurs when $ k \equal{} \minus{} 5$.

Since this is a downward opening parabola with peak at $ k \equal{} \minus{} 5$, the smaller $ |k \plus{} 5|$, the larger $ \alpha^2 \plus{} \beta^2$.

Thus, the maximum occurs when $ k \equal{} \minus{} 4\implies\alpha^2 \plus{} \beta^2 \equal{} \minus{} (k^2 \plus{} 10k \plus{} 6) \equal{} \boxed{18}$.

[quote="math154"]In order for both roots to be real, we must have $ (2 \minus{} k)^2 \minus{} 4(k^2 \plus{} 3k \plus{} 5) \equal{} \minus{} 3k^2 \minus{} 16k \minus{} 16\ge0$

$ \implies(k \plus{} 4)(3k \plus{} 4)\le0\implies \minus{} 4\le k\le \minus{} \frac43$.

Also, by Vieta's, $ \alpha^2 \plus{} \beta^2 \equal{} (2 \minus{} k)^2 \minus{} 2(k^2 \plus{} 3k \plus{} 5) \equal{} \minus{} (k^2 \plus{} 10k \plus{} 6)$, whose maximum occurs when $ k \equal{} \minus{} 5$.

Since this is a downward opening parabola with peak at $ k \equal{} \minus{} 5$, the smaller $ |x \plus{} 5|$, the larger $ \alpha^2 \plus{} \beta^2$.

Thus, the maximum occurs when $ k \equal{} \minus{} 4\implies\alpha^2 \plus{} \beta^2 \equal{} \minus{} (k^2 \plus{} 10k \plus{} 6) \equal{} \boxed{18}$.[/quote]

Can you explain clearly?

Also, by Vieta's, $ \alpha^2 \plus{} \beta^2 \equal{} (2 \minus{} k)^2 \minus{} 2(k^2 \plus{} 3k \plus{} 5) \equal{} \minus{} (k^2 \plus{} 10k \plus{} 6)$, whose maximum occurs when $ \boxed{k \equal{} \minus{} 5}$.

Since this is a downward opening parabola with peak at $ k \equal{} \minus{} 5$, the smaller $ \boxed{|x \plus{} 5|}$, the larger $ \alpha^2 \plus{} \beta^2$.

Thus, the maximum occurs when $ \boxed{k \equal{} \minus{} 4}\implies\alpha^2 \plus{} \beta^2 \equal{} \minus{} (k^2 \plus{} 10k \plus{} 6) \equal{} 18$.

[hide="Box 1"]
It is well known that the extreme point of the parabola $ y\equal{}ax^2\plus{}bx\plus{}c$ occurs when $ x\equal{}\minus{}\frac b{2a}$.

This can be shown by completing the square: $ y\equal{}ax^2\plus{}bx\plus{}c\equal{}a\left(x\plus{}\frac b{2a}\right)^2\minus{}\frac{b^2}{4a}\plus{}c\ge\minus{}\frac{b^2}{4a}\plus{}c$ (by the Trivial Inequality).
[/hide]
[hide="Box 2"]
Sorry, I typoed here.

But if you draw the parabola on a $ k\minus{}y$ plane, you will see that the closer the $ k$ coordinate of a point is to the axis of symmetry $ k\equal{}\minus{}\frac b{2a}\equal{}\minus{}5$, the closer the $ y$ value is to the extreme point; in this case, the smaller $ |k\minus{}(\minus{}5)|\equal{}|k\plus{}5|$ is, the larger $ y$ is and the closer it is to the maximum.

This fact can easily be proven algebraically if we compare the values produces by $ k\equal{}\minus{}5\minus{}a$ and $ k\equal{}\minus{}5\minus{}b$ if $ a>b$.
[/hide]
[hide="Box 3"]
Since $ \minus{}4\le k\le\minus{}\frac43$ in order for $ \alpha,\beta\in\mathbb R$, $ \frac{11}3\ge|k\plus{}5|\ge\minus{}1$, and since we want to minimize $ |k\plus{}5|$, we see that this occurs when $ k\equal{}\minus{}4$.
[/hide]

[quote="math154"][hide="Box 1"]
It is well known that the extreme point of the parabola $ y \equal{} ax^2 \plus{} bx \plus{} c$ occurs when $ x \equal{} \minus{} \frac b{2a}$.

This can be shown by completing the square: $ y \equal{} ax^2 \plus{} bx \plus{} c \equal{} a\left(x \plus{} \frac b{2a}\right)^2 \minus{} \frac {b^2}{4a} \plus{} c\ge \minus{} \frac {b^2}{4a} \plus{} c$ (by the Trivial Inequality).
[/hide]
[hide="Box 2"]
Sorry, I typoed here.

But if you draw the parabola on a $ k \minus{} y$ plane, you will see that the closer the $ k$ coordinate of a point is to the axis of symmetry $ k \equal{} \minus{} \frac b{2a} \equal{} \minus{} 5$, the closer the $ y$ value is to the extreme point; in this case, the smaller $ |k \minus{} ( \minus{} 5)| \equal{} |k \plus{} 5|$ is, the larger $ y$ is and the closer it is to the maximum.

This fact can easily be proven algebraically if we compare the values produces by $ k \equal{} \minus{} 5 \minus{} a$ and $ k \equal{} \minus{} 5 \minus{} b$ if $ a > b$.
[/hide]
[hide="Box 3"]
Since $ \minus{} 4\le k\le \minus{} \frac43$ in order for $ \alpha,\beta\in\mathbb R$, $ \frac {11}3\ge|k \plus{} 5|\ge \minus{} 1$, and since we want to minimize $ |k \plus{} 5|$, we see that this occurs when $ k \equal{} \minus{} 4$.
[/hide][/quote]

Fully unterstood.Thanks for your nice demo.  :lol:


Would it be easy to implement? Maybe something like
[list]
[*]Eenie (15)
[*]Meenie (23)
[*]Minie (4)[/list]

..Joe

You do realize that with a title like this the chances of me doing what you are asking me are slim to zero and instead you might have your blog taken away from you?  :o

ehh.. Sorry, I didn't realize it'd be offensive. :oops:

EDIT: I just looked up the word; I never knew it had those other meanings (ergh *reminds self*). Sorry again.


inequalities
calculus
integration
function
calculus computations

Let $ f$ and $ g$ two continous functions in $ [a,b]$ and $ p,q$ two INTEGERS satisfaying $ \frac {1}{p} \plus{} \frac {1}{q} \equal{} 1$ then prove that:

$ \sqrt [p]{\int_{a}^{b}f^p (x) dx}\sqrt [q]{\int_{a}^{b}g^q (x) dx} \ge |\int_{a}^{b}f(x)g(x) dx|$

I could only prove the case p=q=2

If $ p \equal{} q \equal{} 1$ then how is $ \frac{1}{p} \plus{} \frac{1}{q} \equal{} 1$?

in fact it's p=q=2 (cauchy shwarz ineq)

thank you for your correction

The classic argument:

1. The inequality is homogeneous. Multiply each of $ f$ and $ g$ by positive scalars, and the inequality remains essentially the same. That means that if we can prove it under the assumption $ \|f\|_p\equal{}1$ and $ \|g\|_q\equal{}1,$ then we can do so in general.

2. Prove and make use of this inequality: for $ a,b>0,\ ab\le\frac1p\,a^p\plus{}\frac1q\,b^q.$

I'll let you fill in the details from there.

Two minor issues with your statement:

1. $ f,g$ continuous is essentially irrelevant. All you really need is that $ f$ and $ g$ are integrable under whatever theory of integration you are using. Certainly it is true for $ f\in L^p,g\in L^q.$

2, But since you didn't require $ f$ and $ g$ to be positive and since $ p$ and $ q$ are allowed to be arbitrary real numbers in $ (1,\infty),$ then the left hand side of what you wrote should be:

$ \left(\int_I|f(x)|^p\,dx\right)^{\frac1p}\left(\int_I|g(x)|^q\,dx\right)^{\frac1q}$

In other words, the absolute values are vital; without them, it wouldn't even make sense.

[quote="Kent Merryfield"]The classic argument:

1. The inequality is homogeneous. Multiply each of $ f$ and $ g$ [u]by positive scalars[/u], and the inequality remains essentially the same. That means that if we can prove it under the assumption $ \boxed{\|f\|_p \equal{} 1}$ and $ \|g\|_q \equal{} 1,$ then we can do so in general.

2. Prove and make use of this inequality: for $ a,b > 0,\ ab\le\frac1p\,a^p \plus{} \frac1q\,b^q.$

I'll let you fill in the details from there.[/quote]

In fact I still don't know this notation.
I think your proof is the one given here:
http://en.wikipedia.org/wiki/Hölder's_inequality

And I only know integration of a continous function

Is there an elementery solution?

You can prove it for sums and use Riemann sums

I think you mean:

$ (\sum f^p(a \plus{} k \frac {b \minus{} a}{n}))^{\frac {1}{p}}(\sum g^q(a \plus{} k \frac {b \minus{} a}{n}))^{\frac {1}{q}} \ge |\sum f.g(a \plus{} k \frac {b \minus{} a}{n})|$

and then we take the limit $ n \to \plus{} \infty$

You need to divide by $ 1/n$ on both sides (note that $ 1/p \plus{} 1/q \equal{} 1$ so you can move $ 1/n$ in to both sums on the LHS). You also need to add modulus of $ f$ and $ g$ on the LHS, and for the proof also on the RHS. So basically you need to show this inequality for a sum of a finite number of arbitrary positive numbers, and then it's proved.

Note though that this way is not really any more elementary than what Merryfield suggested. You can use Young's inequality as he suggested and it falls right out. You can prove the finite sum case exactly like that as well (in fact, this inequality is true for integrals w.r.t to any measure, and with an appropriate measure finite and infite sums can be described as integrals w.r.t. to that measure).


MATHCOUNTS AMC AIME USA(J)MO USAMO geometry trigonometry

So, does anyone have any AMC test-taking tactics to share? The test is 75 minutes long, with 25 problems, so there are 3 minutes to do each problem. Of course, some of the earlier problems probably take less time to solve, so are most of you, by, say problem 15, quite ahead of the gun (though 45 minutes would have passed if you strictly allot 3 minutes per problem)? Also, how do you guys deal with the problem of careless mistakes? Do you folk just recheck each problem carefully right after you do it, or do you allot, say, 5 minutes at the end of the test to recheck everything?

Fierytycoon

Some of my old strategies are [url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_Mistakes.php]here[/url].  I'm curious to hear others'.

thanks for the tips. ill try to execute them and stuff like that. but in some competitions like mathcounts you cant write anysphink on your paper beforehand. so should i still go along with the boxes and numbers thing. cause that might take a minute or two.  in those one or two minutes i could have solved like 2 problems at least. but if i were organized i could probably stop a few mistakes. and in nationals 1 question can make all the difference. so which strategery should i use in mathcounts. but in amc ill do the organizing thingie for suree. thanks. adios. nuff said

Hmm, where can I find the exact rulings on what can be written down beforehand on scratch paper for the AMC tests? I mean, if I can draw boxes, could I write down actual equations as well?

Fierytycoon

[quote="Fierytycoon"]Hmm, where can I find the exact rulings on what can be written down beforehand on scratch paper for the AMC tests? I mean, if I can draw boxes, could I write down actual equations as well? [/quote]

No, I am quite certain that you are not allowed to write down equations  beforehand.  Because otherwise, one could argue that any sort of math notes are allowed, thereby making the test "open-book" when the AMC10/12 competitions are clearly meant to be "closed-book," like virtually all standardized tests.  These tests are made to test your math knowledge and skill; allowing any sort of notes would completely undermine that purpose!

Plus, it says explicitly "no aids are permitted other than scratch paper, graph paper, ruler, compass, protractor, erasers and calculators that are accepted for use on the SAT."  Boxes on a paper make it graph paper, therefore as you said, boxes are okay.

Oh, and does anyone know about the rules on calculator programs?  Because it's possible to put a bunch of notes in a program on a calculator such as the TI-83+.  While this clearly contradict's the intention of the exam writers, it never actually says "no programs allowed" or "your calculator must be cleared of all programs" or "all programs must be approved by the exam manager and... <insert long boring list of things not allowed>"  My belief is that, in perspective, it doesn't matter if students CAN use notes on the AMC10/12 exams that much because if they only move on (to AIME) by using those notes, they are likely to get weeded out during the AIME (because calculators aren't allowed, so they wouldn't have notes...if you follow my logic) and won't make USAMO.

I have often wondered about that too. I don't think it's really allowed, but knowing a few extra random formulae probably won't help you anyway, so there's really no point.

Technically, I still have not seen no calculator programs allowed at such competitions as the AMC and MATHCOUNTS.  However, if programs are allowed than you could make the arguement that notes should be allowed too.  Also, protractors and graph paper are also NOT among the things that are not allowed, so take some with you.  Graph paper could come in quite handy.

I haven't even heard of MATHCOUNTS, AMC, etc. even [i]mentioning[/i] Calculator programs.  Perhaps they prefer to ignore their existence in order to simplify things.  Besides, there are certain times (i.e. Sprint Round, AIME, USAMO) when you CAN'T use calculators, so you won't be able to cheat all the way by using calculators.

For that matter, since they aren't allowed, it isn't even [i]cheating[/i].  But we won't get into moral/ethical arguments over whether it's right to "cheat" using calculator programs - we've already debated about whether it's "cheating" to take the SAT more than once.

In the 2002 National Mathcounts competition, there was a debate about calculator programs.  Many of the coaches and administrators, I think, had a meeting to discuss whether calculator programs should be banned. I forgot whether you had to or not, but I know that one of the mathletes on my team did delete all of the programs on his graphing calculator.

-interesting_move

[color=cyan]I am 99% sure that programs are allowed on the contests.  The basic reason for this is administrative -- it's simply too difficult to prevent people from using them, once they are allowed to use a calculator.  However, in my personal opinion, it doesn't really matter much -- particularly in the AMC, where there is a major time pressure, there almost always will be a nicer way to do the problem than using a calculator program.  For example, a program for Heron's formula will take you longer to impliment on many triangles (such as a 13, 14, 15) than remembering the Pythagorean triples (or, in that particular case, that the area is 84, which turns up a lot).  In general, no one making it to the highest levels of competition will do it based on the power of their calculator.  I believe this is also the view that Mathcounts administrators take.  Personally, I think they should just get rid of calculators all together, or at least graphing calculators, but I think the TI corporation gives out too much money to really make that a feasable option.[/color]

[quote="Tare"]I wondered the same question because some people make and/or get tons of math programs for AMC on their graphing calculators and I personally think it's cheating.
[quote="JBL"]I think they should just get rid of calculators all together, or at least graphing calculators[/quote]
I agree. Even though I have a graphing calculator, I don't really think it's necessary to use it on tests. In most cases any calculator with +-*/, maybe with  :^2: or :sqrt: is enough.
I also think it's not fair for people who doesn't have graphing calculator/ can't afford one because graphing calculator have trig function, rational function, exponential function, logarithmic function, and other things that normal calculators doesn't have, thus giving people with graphing calculators an advantage.

With that said, I question the fact that graphing calculators are allowed on SATs because graphing calculators aren't just calculators.

P.S. I asked the AMC10 coordinators why graphing calculators are allowed, and they just said "It's allowed on SAT so we allow them too" (note that it is not exact words of the coordinator)[/quote]

The AMC is simply trying to be consistent with other tests.  Many of the standardized tests use the same rules governing calculator use, and it is simply easier to go along with those rules than make up new rules prohibiting, say, certain kinds of calculators.  A list would have to be distributed, people would inevitably use the wrong kind of calculator - some would be disadvantaged and some would use an illegal one - without the test administrator or, certainly, the AMC, being aware of it.

As I've said in another thread regarding potential "cheating" on the AMC, when you look at that particular test in retrospective, it doesn't matter that much.  If you cheat your way to get to the AIME (either by using programs or some 'other' form), then you have earned a free chance to take the AIME.  Period.  You can't use calculators there, and it's much harder to cheat on it than the AMC.  You get nowhere.  The End.

No one is going to make USAMO solely based on their calculator.  Eliminating calculators in AIME reduces this chance even more.  And besides, many math courses use a calculator a lot; allowing a calculator makes the contest more friendly to many students, thereby reaching more students than it would if it were harder and didn't allow one.

Besides, calculators take time to use, so if you can do something in your head faster, good for you.  You have an advantage.

Another point: is using programs on your calculator cheating?  No.  The AMC explicitly states that you may use a graphing calculator such as the Ti-83+, and they aren't stupid; they know that you can (and that most good math students will) make programs to solve quadratic equations, programs to find areas and volumes, programs to find logarithms, etc.  Truth is, they accept this.  So in their opinion, it's not cheating.

I know someone will say this, so I will first: not a lot of cheating goes on in the AMC. Although it's a small venture by a couple of UNL professors who have no hope of identifying every source of cheating, this little community has integrity.

So in summary, if you don't want to use a calculator or you don't want to use programs, that's your decision, not mine.  But you are certainly allowed to; countless other students do.

And to address the "some students don't have graphing calculators issue", keep in mind that most teachers have a bunch of graphing calculators that they loan out to students.  Schools make accommodations for needy students.  And if cheaper calculators were the only ones allowed?  Then students who used graphing calculators would have to go out and buy another one just for the AMC.  Instead, the AMC accommodates all sorts of calculators, so whichever one you have, you may use.

[color=cyan]I agree that programs are allowed.  And I agree that banning specific types of calculators can be an administrative pain.  I disagree that "most good math students will make programs to solve quadratic equations, programs to find areas and volumes, programs to find logarithms, etc."  I guarantee you that the people who do really well on these contests never used such programs.  I certainly would be able to solve a quadratic equation on paper faster than my TI-89 could.

Finally, one comment on the issue of calculator cost.  There are really two issues at play.  One is that many students simply could not afford the $139 it takes to buy a new graphing calculator.  Furthermore, while many schools do make some calculators available to their students, the numbers are generally small, and there is a definite advantage in taking a test using your own calculator.  So, this is bad.
The second issue is really worse, though -- most of the students for whom the cost of a calculator is a problem will never get remotely near having an opportunity to take the AMC or any other math contest.  I am sure that a quick survey of the students who took the AMC would place the majority all in the (relatively) higher ranges of family income.  I think if anything, this is a far more serious issue, although not one that pertains to whether or not calculators should be used on exams.[/color]

i agree with most of the people that no calculator should be used to solve equations, systems of equations, etc, but i dont think it matters on just graphing. someone who uses his calculator to graph a function and uses that graph to figure something out clearly knows what hes doing; the calculator just saves the time of drawing a graph. personally, having taken a few math classes where our teacher told us not to use a calculator at all, i think the time save factor is a fair use of graphing calculator, even on a contest.

There are, I believe, some other reasons that calculators are allowed on the AMC exams. I read in the preface or introduction or something to the Contest Book #5 (1983-1988 AHSME and AIME) that they would not help students get problems wrong by making the wrong answers very tricky. (For example, suppose you solve a problem right, but it asks you for the diameter of the circle, but you give the radius. That answer probably won't be there.) Calculators help people avoid solving a problem correctly and then adding some numbers wrong to get the wrong answer (something I have done way too many times myself). The point of the competition isn't to see who makes the fewest arithmetical errors. So that may be one of the reasons they allow calculators.

[quote="JBL"][color=cyan]One is that many students simply could not afford the $139 it takes to buy a new graphing calculator.[/quote]
I don't like to nitpick, but the price is not that extravagant.  The Ti-83+ cost $83.00 at Office Depot the other day.  Used ones can go for $20 to $60.  I think most people who are that serious about math can spare 20 bucks to get a good calculator.  That's how much you'd spend on a math book.

And while it's true that not all good math students make programs, it's true of many that I know.  For them, it's somewhat like a mini-programming-language.  There are plenty of people who do C/C++ programming for fun, and these people do this on their calculators.

And while good math students can solve a quadratic equation quickly, can they find the real roots of a fourth-degree one, for instance?  Graphing calculators can certainly aid in this.

[color=cyan](a) $139 is the price Texas Instruments asks on their website for a new TI83+ Silver Edition.  (b) Where could everyone get access to a used calculator?  There aren't many I've seen for sale, at any price.  (c) You overestimate the ability of many people to spend money.  A significant portion of the American population cannot afford $150 for something like a calculator.  That's really just a fact.  Some people can't afford math books.  Many, many people can't afford less than $3,000 dollars the City University of New York charges for tuition at its colleges.  There are a lot of people with no money.  Most people like that never get a chance to find out whether they might want to be serious about math.

When are you ever asked to find real roots of a 4th degree equation that you can't factor?  Competition writers specifically aviod writing questions like that.[/color]

[quote="mathfanatic"][  I think most people who are that serious about math can spare 20 bucks to get a good calculator.  That's how much you'd spend on a math book.[/quote]

I don't agree. If you are really serious about math, you will have learned the methods and tactic to tackle problems, such as polynomials, WITHOUT a calculator. In most cases, especially on competitions, graphing calculators are NEVER NECESSARY. tHEY CAN just be a great aid, like having a dice program that rolls dice to find the probability.......

[color=cyan]Haha.  For those of you who don't get that last comment, we held a competition at the math camp where I TA and darkquantum is a student, and his team got a question right because in the last 5 minutes, one team member pulled out a random number program and started doing a probability problem and came up with the right answer that the whole team hadn't been able to get by "real" means in almost half an hour.[/color]

This is just my opinion and I know many people disagree, but I prefer that calculators not be used on such exams.  I don't think an exam like the AMC leaves a serious time deficit and there are not many people (if any) who have ever been capable of taking the USAMO (and doing well) who have missed the cut because of a single arithmetic error.

Calculators are very advanced these days and I do feel that a good calculator can be an advantage to some students that outweighs the factor of cutting down on arithmetic errors.  It's certainly hard to get data on such a theory, but it's not just an issue of more advanced calculators -- it's also an issue of some students understanding calculator machines to such a greater degree that the advantage is no longer in just understanding the mathematics.

Just my two cents.  But if you learn problem solving well, the calculator is not the difference.

[quote="JBL"][color=cyan](a) $139 is the price Texas Instruments asks on their website for a new TI83+ Silver Edition. [/color][/quote]
True.  But most people would be looking at the cheaper model.  > 100 bucks is a lot for a calculator.  But you can get calculators on eBay, for instance, for much cheaper.
It's definitely true that there are kids who just learn how to use a calculator and end up depending on it.  You definitely learn much more by learning the method or art of solving a problem than, say, by plugging and chugging.  But calculators are nice to check your work, to save you from forgetting that 91 isn't prime, and for grinding out long things when you can't think of a faster way.  And as long as calculators are there, they're there, and I intend to use mine in conjunction with using my brain.
And JBL, I know that people are generally unwilling to spend $150 on a calculator.  My parents were.  I were.  I thought it was completely extravagant (of course, that was back when I was "normal", before I became corrupted.   :twisted:  anyway...) so we got a cheaper one.  Then, when I went to mathcounts nationals I got a free Ti-83+.  I like my calculator, and I find it useful from time to time.  But I couldn't solve all the problems JUST using a calculator.

I realize that this is a rather old topic, and the most recent entries are about something totally different, but I have some tactics I find to work as far as eliminating stupid mistakes (I still make far too many stupid mistakes, but I'm getting better about it). I think the original post asked about this.

Rusczyk has a lot of great tips, so certainly take a look at that. The thing I've found to be most useful, though, is to get into the right mindset. You have to be totally focused on what you are doing and have an absolute conviction that you are not going to falter.

One thing that I do which I have found to be very effective, even if you don't think it would be, is to practice working on very easy problems, being careful to get every single one of them. For example, I wrote a calculator program that quizzes me on multiplication of two-digit numbers, and I will just set it to quiz me on the numbers 12 through 19 multiplied by each other (in a random order) and time myself, seeing how quickly I could get all 64 questions right. While basic multiplication is not that important a skill, exercises like this actually help problem solving ability, from what I can tell. I think the reason for this is that they teach you to get into a mindset in which you are trying to be fast, but also meticulous and deliberate, and never to falter in your logic.

The ability to get into that sort of mindset has helped me on everything from trivial local contests to the USAMO, despite the fact that the USAMO requires a totally different type of problem solving. I think that is why many of the greatest Olympiad competitors- Reid Barton, Po-Ru Loh, Tiankai Liu, etc., also did very well at Mathcounts. While Mathcounts may rely on very simple and very fast problem solving, the mental wiring necessary to quickly and accurately mow down a Mathcounts sprint round is very similar to that which is necessary to have the mental toughness to carefully and relentlessly investigate and solve Olympiad-level problems, as well as everything in between.

In any case, this is all just my own speculation, and I'm sure many people strongly disagree with some of the hypotheses I have presented above. But they are based on my personal experience, and also what I have seen in other people.

Oh, and on the topic more recently discussed here, I don't think calculators should be allowed on most contests. My primary reason is simply that they disqualify so many potentially good problems. There are a lot of very good problems that involve a very clever technique and have an elegant solution, but can be done right away with a calculator, so such problems can't be used on contests that allow calculators.

And of course, sometimes there are problems trivialized by a calculator that somehow still make it onto the contest, which is most curious. The most obvious example I can think of was a contest I went to once in which a question was to take the GCD of two very large numbers, and graphing calculators were allowed. Stupid test writers...


geometry
circumcircle
parallelogram
geometry theorems

I'm having some trouble proving the following theorem: Let ABC be a triangle, H its orthocenter, M the midpoint of BC, and A' the point diametrically opposite to A in the circumcircle. Prove that H, M, and A' are collinear. How do you do this?

Let H' be symmetric to wrt BC. It's easy to prove that H' lies on the circumcircle of ABC => H'A'||M'B (M' is the intersection point of A'H and BC so A'M'=HM'). We also see that BH||A'C (both perpendicular to AC). Hence A'CHB is parallelogram => M' is intersection point of diagonals => CM'=BM' QED


The sequence $ (x_n)$, $ n\in\mathbb{N}^*$ is defined by $ |x_1|<1$, and for all $ n \ge 1$, \[ x_{n\plus{}1} \equal{}\frac{\minus{}x_n \plus{}\sqrt{3\minus{}3x_n^2}}{2}\] (a) Find the necessary and sufficient condition for $ x_1$ so that each $ x_n > 0$.
(b) Is this sequence periodic? And why?

a) $ x_1>0$. $ x_2>0$ gives $ \minus{}x_1\plus{}\sqrt{3\minus{}3x_1^2}>0$ so $ x_1\in(0,\sqrt3/2)$. We can easily prove that $ x_n\in(0,\sqrt3/2)$ for all $ n$.

b) Easy to check that $ x_{n\plus{}1}\equal{}x_{n\minus{}1}$ for $ n\ge2$.

[hide]
Define a new sequence $ \theta_n$ such that $ x_n \equal{} \sin(\theta_n)$ and $ \theta_n \in [\minus{}90^\circ, 90^\circ]$ for all $ n \in \mathbb{N}$.

Then the given recursion becomes:
$ sin(\theta_{n\plus{}1}) \equal{} \frac{\sqrt{3}}{2}\cos(\theta_n) \minus{} \frac{1}{2}\sin(\theta_n)$

$ sin(\theta_{n\plus{}1}) \equal{} \sin(60^\circ)\cos(\theta_n) \minus{} \cos(60^\circ)\sin(\theta_n)$

$ sin(\theta_{n\plus{}1}) \equal{} \sin(60^\circ \minus{} \theta_n)$

$ \theta_{n\plus{}1} \equal{} 60^\circ \minus{} \theta_n$

Therefore, $ \theta_{n\plus{}2} \equal{} 60^\circ \minus{} \theta_{n\plus{}1} \equal{} 60^\circ \minus{} (60^\circ \minus{} \theta_n) \equal{} \theta_n$.

So, $ \theta_n$ is periodic, and thus, $ x_n \equal{} \sin(\theta_n)$ is also periodic.

Note that for all $ k \in \mathbb{N}$ we have: $ \theta_{2k\minus{}1} \equal{} \theta_1$ and $ \theta_{2k} \equal{} 60^\circ \minus{} \theta_1$.

So, if $ x_n \equal{} \sin(\theta_n) >0$ for all $ n \in \mathbb{N}$, then $ \theta_n > 0$ for all $ n \in \mathbb{N}$.

Therefore, $ \theta_1 \in (0^\circ,60^\circ)$, and thus, $ x_1 \in \left(0, \frac{\sqrt{3}}{2} \right)$ is nessesary and sufficient for $ x_n > 0$ for all $ n \in \mathbb{N}$.
[/hide]


Can we change our password on the USAMTS without changing our password on AoPS?

Yes - the two sites are no longer related in any way.


Find all solutions for $ x$ and $ y$ natural numbers for the following equation:

$ 2^{2x} \minus{}3^{2y}\equal{}55$

Please explain it step-by-step... thanks!!

[hide="solution"]You can factor the equation to give $ (2^x\minus{}3^y)(2^x\plus{}3^y)\equal{}55$.  Now the sum of two numbers whose product is 55 gives $ 2\cdot 2^x$ or a power of two.  The only factorizations of 55 are $ 55\cdot1$ and $ 11\cdot5$.  Adding these two shows that only the latter works.

Thus $ 11\plus{}5\equal{}2\cdot 2^x$ or $ 16\equal{}2^{x\plus{}1}$.  We have that $ x\plus{}1\equal{}4\Rightarrow x\equal{}3$.  From here, we can manipulate the given equation to give:

$ 2^{2x}\minus{}3^{2y}\equal{}55\Rightarrow 2^{2\cdot3}\minus{}55\equal{}3^{2y}\Rightarrow 64\minus{}55\equal{}9^y\Rightarrow 9\equal{}9^y\Rightarrow y\equal{}1$.

Thus the only solution is $ (3, 1)$.[/hide]


function
calculus
derivative
trigonometry
real analysis
real analysis solved

Find all real functions with continuous derivative such that $f(0)=-\frac{1}{2} $  and $f(2x)=2f^2(x)-1$ for all real number x.

Yup, you are fast Moubi!I will not ask for a proof, because I know you have it. Let the others enjoy the pleasure of proving this.

After we derive it we get $f'(x)=f'\left(\frac x{2^n}\right)\cdot (2^n f\left(\frac x2\right)f\left(\frac x{2^2}\right)\ldots f\left(\frac x{2^n}\right))$.

It's easy to see from $f'(2x)=2f(x)f'(x)$ that $f'(0)=0$. Since $f\in C^1$, it means that all we need to do in order to obtain $f'(x)=0,\ \forall x$ is show that $2^n f\left(\frac x2\right)f\left(\frac x{2^2}\right)\ldots f\left(\frac x{2^n}\right)$ is bounded (when $x$ is fixed).

Let $f\left(\frac x{2^n}\right)=\cos(u\left(\frac x{2^n}\right))$ (we can do this for large enough $n$, since $f$ will then take values which are very close to $-\frac 12$). $u(x)=\arccos (f(x))$ for $x$ close enough to $0$. We get $f'(x)=f'\left(\frac x{2^n}\right)\cdot \frac{\sin(2^n u\left(\frac x{2^n}\right))}{\sin(u\left(\frac x{2^n}\right))}$. We have thus found $a_n(x)=\frac{\sin(2^n u\left(\frac x{2^n}\right))}{\sin(u\left(\frac x{2^n}\right))}$. This is bounded because the denominator goes to $\sin(\arccos(-\frac 12))\ne 0$ as $n\to\infty$.

I hope it's correct.


I am going to be attending many mathematical activities next year and need to get better at number theory, which is my main weakness. Any tips or good resources you guys have in mind?

I know a good book :[b]Introduction To The Theory Of The Number [/b] by Ivan Niven,Montgomery ,... .published WIE.

Number Theory is a vast topic. What exactly do you need help with?

We will be publishing an Introduction to Number Theory book here at AoPS sometime in the next few months.  The difficult of the topics will carry students from a pre-algebra understanding of integers all the way up through perhaps 80% or more of AMC level number theory problems and even a few easier AIME level number theory problems.


All the problems are difficult. I can only solve IMO 2001/3 and USAMO 1999/3. But I take more than 1 month to solve these two problems. I cannot solve all the other problems. Can I vote all of them are difficult?

This poll is very interesting!
wonderful idea! :D

imo shortlist 1999 combinatorics7 is a very difficult problem

i think that the problem which appeared 6th in the IMO 2002 was not so difficult(i mean that there are many more harder problems in the IMOs like 1988\6 2003\6 etc.)


Functional Analysis
real analysis
real analysis unsolved

Here it comes in some extended form:
Consider an Euclidean space with:
A. finite dimension $m$.
B. infinite dimension.
How many balls with radius $1$, can you fit into another ball of radius $n$, where $n \in N$?

At first glance, the special case of $n=2$ may be the most interesting.

For $n\ge3$ in an infinite dimensional space - and here I am interpreting "Euclidean space" to mean inner product (Hilbert or pre-Hilbert) space - the answer is, of course, infinitely many balls. Just let $\{e_k\}$ be an infinite orthonormal set, and take the balls of radius 1 centered at $2e_k.$ They're disjoint.

But the $n=2$ case needs a second look.

For $n=2$, the answer is 2, independent of dimension.
Let $a$ and $b$ be the centers of two balls in our set. We have $||a||\le1$ and $||b||\le1$. We know that $||a-b||^2=||a||^2+||b^2||-2\langle a,b\rangle\le (||a||+||b||)^2\le4.$ We must have $||a-b||\ge2$ for the balls to be disloint, so we must have equality in all parts of the chain. The first equality holds when $a=-cb$ for some positive real number $c$, and the second holds when $||a||=||b||=1.$ Combining these, we must have $||a||=1$ and $b=-a.$ This clearly allows only two balls.

If the radius, $n$, is 3 and the dimension, $m$, is 2, then there is a configuration that fits 7 balls into the larger ball. Computing the optimal packing in other cases appears to be a very complex problem.

Indeed it is a very complex and still open problem.

For radius $n=3$ and the dimension $m=3$, the answer is $15$.
What about the radius $n=3$ and the dimension $m=4$ ?!?  :?


For which positive integer values of $n$ can the set $\{ 1, 2, 3, \ldots, 4n \}$ be split into $n$ disjoint $4$-element subsets $\{ a,b,c,d \}$ such that in each of these sets $a = \dfrac{b +c +d} {3}$.

All even $n$.  First note that the sum of the elements in each subset $\{a,b,c,d\}$ is a multiple of 4, since (mod 4) $a+b+c+d\equiv a+3a \equiv 0$.  So the overall sum $1+2+\ldots +4n=\dfrac{4n(4n+1)}{2}$ is also a multiple of 4, and so $n$ is a multiple of 2.

If $n$ is even, the numbers $1, 2, \ldots, 4n$ can be divided into $\dfrac{n}{2}$ subsets of the form $\{8k+1, 8k+2, \ldots, 8k+8\}$.  As
\[ 8k+4=\dfrac{(8k+1)+(8k+3)+(8k+8)}{3}\text \]
and
\[ 8k+5=\dfrac{(8k+2)+(8k+6)+(8k+7)}{3}, \]
the result follows.


geometry
inequalities
absolute value
combinatorics proposed
combinatorics

Colour each point of the plane according to the following rule: the colour of the point $ (x; y)$ is white if $ [x] \plus{} [y]$ is even, and is black otherwise. For arbitrary positive integers $ a$ and $ b$, consider the triangle with vertices $ (0; 0), (0; a)$ and $ (b; 0)$; and denote by $ f(a,b)$ the absolute value of the difference between the areas of the white resp. black regions. Prove inequality $ \frac {n}{10} < f(u_{3n},f_{3n \plus{} 1})<10n$, where $ u_n$ denotes the n-th term of the Fibonacci sequence, that is, $ u_1 \equal{} 1, u_2 \equal{} 1, u_{n \plus{} 2} \equal{} u_n \plus{} u_{n \plus{} 1}$.

I posted this problem before. You can see here
http://www.mathlinks.ro/viewtopic.php?p=916125#916125
Does anyone has a solution to this problem?


I am an international student that currently study in a California high school. I made USAMO, but I can't make MOSP as i don't have U.S passport (though with my score(9) I can't make it). So can I be USAMO winners/Honorable mention? I notice that 25 students in USAMO winners/Honorable mention are all in MOSP. I plan to work hard this summer but I am quite confused by this.

I'm an international student studying in the US, and I went to blue MOP last year. I guess you can go to MOP if you are either in the US or a US citizen.

Many thanks. Do you have permanent residency?
So only IMO needs citizenship, MOP doesn't?  I doubt it needs permanent residency that I don't have.

Unless it's an exceptional case, I am somewhat sure that being able to attend the IMO and being able to go top MOP go hand in hand. Also not all of the winners/HMs are allowed in to MOP.

This is a question better asked to the AMC directly, because you probably aren't going to get a sure answer from people here (unless AMCDirector replies). (EDIT: Well, look at that.  :roll: )

Too many criteria are involved here (residency status, age, grade in school, USAMO score) to answer your question in this forum.  Please contact the AMC office by any of the means on our printed materials or website and we can give a definitive answer to your particular case.

Steve Dunbar
MAA Director, American Mathematics Competitions

[quote="wuyonglin2004"]Does anyone knows the email of AMC office? Thanks![/quote]

I believe they can be reached here:
amcinfo@maa.org


Bac xem cai file loi giai day, trong do co du ca :)

Chẳng muốn nhìn mấy bài toán trong đề thi ĐH VN nhưng mà khổ lỗi là năm nào cũng phải ...

Bất đẳng thức xuất hiện trong kỳ thi này, thực tình tôi thấy, chưa ưng ý. Vẫn cứ loanh quanh biến cái nọ thành cái kia, không đi trực tiếp vào vấn đề. Bệnh này thì có trong nhiều nhiều sách. Cả mấy trò giải phương trình, hệ này kia nữa. Cứ xuất phát từ cái đơn giản, sau đó biến đổi lung tung lên, rồi bắt con người ta đi giải. Cách làm này ít nhiều ảnh hưởng bởi sách của Nga, và được sinh ra từ các lò luyện thi, học theo dạng hết sức mẹo mực. Nhưng xin thưa là người Nga bây giờ đi xa khỏi cái này nhiều rồi. Đợi đến lúc các bác nhà ta "chán" trò này thì chắc lâu lắm.

Nhìn đi nhìn lại thì những cái cơ bản vẫn cứ lung linh nhất.

[color=red]Edited by N.T.TUAN[/color]

Nói chung là bài toán thì như thế còn lời giải loằng ngoằng và nguỵ tạo :D Ông sửa lại font đi, sao để chữ to tướng thế?  :maybe:


search
real analysis
real analysis unsolved

let Fn be Taylor's  polynom of e^x  where deg (Fn)=n

how many  roots does Fn has?

For $n>1$, it has no rational root. Actually, it is even irreducible. I've heard of a proof by samn (maybe a search on the forum would help you).

If you're interested in real roots then
for odd n $F_{n}(x)$ has exactly 1 root,
for even n $F_{n}(x)$ has no roots.
It may be easily checked by induction.


[quote="rookworm"]Sorry if this is the wrong forum, but I have been racking my brain to remember an old china math olympiad problem. It is roughly like:

A teacher gives some facts about a natural number that he is thinking of to two students (The facts are what I forget!). The students have perfect reasoning ability. then the following discussion occurs:

S1: I don't know the number
S2: I don't know it either.
S1: I still don't know the number.
S2: Now I know it!
S1: So do I.

The question is, "what is the number?". My question is, "What is the question?"![/quote]

Go to [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=37312&start=41[/url]. It might be a similar problem.

Yes, thanks for that, but it's not the exact one I was thinking of.

Further details, I believe the number was an integer between 1 and 100, and the teacher tells them d(n) (?) (although [i]we[/i] aren't told!). I'm pretty sure it was a chinese MO problem, and was published in an old issue of CRUX MATHEMATICORUM, from the 80's or 90's (my library has them all!). That's all I can remember right now. I'm not sure if the dialougue is exactly right.


Find $ r$ if $ 3(r\minus{}5)\equal{}24$.

divide 3 both sides so
$ r\minus{}5\equal{}8$ and add 5 so $ r\equal{}13$


inequalities
geometry
circumcircle
geometric transformation
homothety
parallelogram
complex numbers

Note OA_1/AA_1=[OBC]/[ABC](1)
By homothety of pole O and power 1/2 we transform ABC into a triangle with vertices on sides of O_1O_2O_3 (the centers).
By using identity R=abc/4S and (1) the consition trasforms to sum(OA*AB*AC)>=AB*BC*CA which is a well-known beatiful problem

[quote="iura"]By using identity R=abc/4S and (1) the consition trasforms to sum(OA*AB*AC)>=AB*BC*CA which is a well-known beatiful problem[/quote] well known you say, but be aware that andreis, mzero, ral and pachitariu marius, all team members of the Balkan MO, or IMO, this year, did not solve this "well-known" problem during the contest.

the simplest solution can be derived using complex numbers: (we have already used the fact that $O$ is inside the triangle $ABC$ while using the areas to transform the inequality), so let us suppose that $O$ is the origin of the complex plane, and $A,B,C$ have afixes $a,b,c$ respectively.
In fact we want to prove that (I belive that you did not replace the variables carefully :?)
\[ \sum OA \cdot OB \cdot AB \geq AB \cdot BC \cdot CA \quad \Leftrightarrow \]
\[ \sum |bc(b-c)| \geq \left| \sum bc ( b-c ) \right| = | -(a-b)(b-c)(c-a)| \] which is obviously true by the triangle inequality.

I didn't say it's easy. But it's well-known.
It's China(1996 I think) and should be known by everyone due to its particular beaty(IMHO). Other solution involves construction of two parallelograms and use of Ptolemy's inequality. It's certainly nicer than the one with complex numbers (though that is more reachable)

$%Error. "sumS" is a bad command.
_{OBC}*R_1\ge S_{ABC}*R$,then $\sum AB*AO*OB\ge AB*BC*CA$
it's a result from 1998 CMO

[quote=Valentin Vornicu]Three circles $\mathcal{K}_1$, $\mathcal{K}_2$, $\mathcal{K}_3$ of radii $R_1,R_2,R_3$ respectively, pass through the point $O$ and intersect two by two in $A,B,C$. The point $O$ lies inside the triangle $ABC$.

Let $A_1,B_1,C_1$ be the intersection points of the lines $AO,BO,CO$ with the sides $BC,CA,AB$ of the triangle $ABC$. Let  $ \alpha = \frac {OA_1}{AA_1} $, $ \beta= \frac {OB_1}{BB_1} $ and $ \gamma = \frac {OC_1}{CC_1} $ and let $R$ be the circumradius of the triangle $ABC$. Prove that
\[ \alpha R_1 + \beta R_2 + \gamma R_3 \geq R. \][/quote]


For positive real numbers $ a,b,c$, where $ abc\equal{}1$, prove that

$ \frac{1}{1\plus{}a^2}\plus{}\frac{1}{1\plus{}b^2}\plus{}\frac{1}{1\plus{}c^2}\ge\frac{3}{2}.$

It's hard to prove because it's not true.  Take $ a \equal{} b \equal{} n$, $ c \equal{} n^{ \minus{} 2}$, the sum is $ \frac {2}{1 \plus{} n^2} \plus{} \frac {1}{1 \plus{} n^{ \minus{} 4}}$, which approaches $ 1$ as $ n\to \infty$.

Thanks. It's not hard, it's impossible. Thanks again for opening my eyes.

Dear [b]Shanku[/b],
Your inequality is true for the additional condition $ a \plus{} b \plus{} c \leq 3$. Of course, this is a different problem!
You can try to solve my problem (which is posted at http://www.mathlinks.ro/viewtopic.php?p=1732120#1732120).
I'm sure that your idea of proving for your inequality is not the same as mine.
The new problem is a particular case of my inequality (for $ n \equal{} 3,k \equal{} 1$).


how do you prove that three altitudes ha, hb, hc : 1/ha + 1/hb + 1/hc = 1/r ?


A triangle ABC is drawn. A line segment BD is drawn so angle ABD = angle DCB. Prove that AB squared = AD*AC
[url=http://www.imageshack.us][img]http://img264.echo.cx/img264/8960/scan00026ka.jpg[/img][/url]

Ues the ratio of areas can be a simple solution.

Imagine that you have a circle that contains points B,C, and D but not A with $\overline{AB}$ as a tangent and $\overline{AC}$ as a secant.

First, we know that $\angle ABD = \angle BCD$. Also, notice that $\angle A$ is in both $\triangle ABD$ and $\triangle ABC$. So, we have $\angle A = \angle A$ by reflexive property and by AA similarlity, we have $\triangle ABD$ ~  $\triangle ABC$.

Now, using proportion, we have $\frac{AB}{AC} = \frac{AD}{AB}$ and cross-multiply, we get $AB^2 = AC \cdot AD$.

QED.

I hope this helps!

Ok, great! I had gotten as far as proving similarity, but could not find the ratio...

That is actually a theorem that comes handy on many geometry problems.

See if you or anybody can find the name of that theorem. :P

trying to guess this name is like trying to guess out of 5000 90-year-olds which one will die first.

maybe I don't know this theorem.


If $ a \plus{} b \equal{} 7$ and $ 2a \minus{} b \equal{} 17,$ then $ a \minus{} b$ equals

$ \textbf{(A)}\ 1 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 7 \qquad \textbf{(E)}\ 9$

[hide]$ (a,b) \equal{} (8, \minus{} 1)\ \Rightarrow\ a\minus{}b\equal{}9\ \Rightarrow\ \mathrm{E}$[/hide]

Above answer is correct. But that's not the solution.

There are two ways to solve this. I'll give hints to how you guys can solve this.

[hide="Hint 1"]
We can use simple algebra. What can we do with those two equations? Add? Subtract? Multiply? Divide? [/hide]

Or, for more many equations, we can solve them using matrices.

[hide="Hint 2"]
$ \left( \begin{array}{cc} 1 & 1 \\
3 & \minus{} 1 \end{array} \right) \left(\begin{array} {c} a \\
b \end{array} \right) \equal{} \left(\begin{array}{c} 7 \\
17 \end{array} \right)$

By adding 1st and 2nd row, we get:

$ \left( \begin{array}{cc} 1 & 1 \\
3 & 0 \end{array} \right) \left(\begin{array} {c} a \\
b \end{array} \right) \equal{} \left(\begin{array}{c} 7 \\
24 \end{array} \right)$

From here, we see that $ 3a \equal{} 24$ so $ a \equal{} 8$. From here, we see that $ b \equal{} \minus{} 1$ as well.
[/hide]

The simple algebra

\begin{align*} a + b & = 7 \\
2a - b & = 17 \end{align*}
$ \implies a = 8,b = - 1\implies a - b = \boxed{  9}$

[hide]Multiplying the second equation by $ 3$, we have
\[ 4a \minus{} 2b \equal{} 34
\]
Subtracting the first equation from this, we have
\[ (4a \minus{} 2b) \minus{} (a \plus{} b) \equal{} 3a \minus{} 3b \equal{} 34 \equal{} 7 \equal{} 27
\]
Then, dividing both sides by $ 3$, we have
\[ a \minus{} b \equal{} 9 \Rightarrow \boxed{E}
\]
[/hide]@hunter34: Your final answer is incorrect.

[quote="i_like_pie"][quote="Silverfalcon"]Above answer is correct. But that's not the solution.[/quote]
:?: I omitted the first part where I solved the system of equations; I don't see what other method you can use.

[quote="Silverfalcon"]...
From here, we see that $ 3a \equal{} 24$ so $ a \equal{} 8$. From here, we see that $ a \equal{} \minus{} 1$ as well.[/quote]
Small error there...[/quote]

I meant $ b \equal{} \minus{}1$. Thanks.

No, you basically provided answer but you did not post how you got $ (a,b)$ so what you posted was not solution.


Hi,

[hide]I guess the answer is the golden ratio and its conjugate.

It is clear from the equation that x = 1+ 1/(1+ 1/(1+ 1/(1+ 1/(.......... continue infinitely,

which means x = 1 + 1/x implies x 2  - x - 1 = 0; the solution to which are

(1 + :sqrt: 5)/2 and (1 - :sqrt: 5)/2, the former being the golden ratio and the latter is its conjugate.

[/hide]

**********************************************************************************

Sambit

eum sambitnayak, there's nowhere "..." so I think he's asking for a finite number :|

We got the recursion $x_n = 1 + \frac{1}{x_{n+1}}$ and we must solve $x_1=x_{13}$

...though I got no clue how to do that :?

Or did you mean the infinite one, diogene? :)

Note that if $X=1+\frac{1}{X}$, then we keep substituting we get $X=1+\frac{1}{X}$, so $X=1+\frac{1}{X}$ is true. So solve for that X we have
$x^2-x-1=0, $
$x=\frac{1 \pm \sqrt{5}}{2}$

yes... but how do you prove it's the only solution? ;)


Gauss
Euler
algorithm
calculus
geometry
abstract algebra
conics

So, of all the famous mathematician's there have been throughout history, who is your favorite? Some suggestions might be: Pythagoras, Euclid, Abel, Galois, Riemann, Dirac, Hilbert, Newton, Ramanujan, Hardy, Gauss, Euler, Fourier, and many more that I can't think of off the top of my head.

Personally, I'm a fan of Ramanujan. I have a book about him, and it fascinates me how he was able to reinvent so much of known modern mathematics without any help, until later of course. It's a shame he died young.

What do you guys think?

John Conway is the coolest mathematician I've met. Out of the historical mathematicians, I don't have an opinion since I don't know much about mathematical history.

myself :smile:
actually, i know almost nothing about mathematical history, so i prolly can't say

[i]Sir[/i] Isaac Newton. Maybe it's just the personal ironies of Newton. He seems to be everywhere. And even more fascinating after reading [i]The Da Vinci Code[/i].
Einstein counts as a mathematician right? He's the coolest. Maybe his eccentricities. I know he's the most famousest (yes, I did that on purpose) of all mathematicians/scientists, but really, he was a cool man. No wonder [i]Mental Floss[/i] picked him as their coverboy. *laughs*

Mine would be Riemann, followed closely by Gauss and Euler.

Galois - hes nothing like any other mathematician. When he applied to the Academie, he finished his interview by throwing his eraser at the intervier's head and leaving. I feel a lot like doing that these days. :)

Maybe also the descendant of Abel :wink:

My favourite is either Euler or Paul Erdos.  Both of them could notice subtle things no one else would even have thought of (similar to Ramanujan).

Gauss.  I share his birthday ;-)  Erdos is cool too.

Newton - mathematician, alchemist, physicist, astronomy  :lol: .

Is it just me, or did i post the same question a while back but it just got ignored? My favorite mathmatitian is probably the dude who did lots of geometry and found the five basic axioms or somthing like that

[quote="Meatman"]Iurie Boreico, I think.[/quote]

I think he is no mathematician (yet)...

Ramanujan, Euler, or Newton

Who's the guy who wrote The Elements? If he isn't above, him too.

I think you mean Euclid (genius).

[quote=Meatman] Iurie Boreico, I think.[/quote]

Anyone know what he's up to after high school?

I still say $ \text{Ramanujan}$. I mean he grew up in poor conditions yet he still manages to re-discover a great deal of mathematics for himself. When I read an account of his life, I not only marvel at his incredible mathematical skills but only his perseverance. As a kid, he didn't even have paper (I think, or is my memory failing me)!

My favorites:

Ramanujan,
Rieman,
Galois,
Cauchy,
Poincaré,
Alkhawarizmi,
Gauss,

and david blaine  :P

I would say that my favorites are Gauss and Newton. Newton was the king of calculus, and Gauss the prince of math, so that sounds cool to me, yeah.

My favorite mathematicians are the proposers of IMO problems.

I think euclid is the best one because was one of the first ones... can you imagine? CONSTRUCT MATH BY YOURSELF EVERYDAY??? Euler is also very good, very polifacethic....


Determine an equation for the locus of pts equidistant from the pts on a graph of y = 2(x-3)^2 + 2.

This question doesn't make any sense at all.  "Equidistant from the points on a graph" appears to mean that you are looking for points which are the same distance from every point on a parabola.


group theory
abstract algebra
superior algebra
superior algebra solved

Prove that if N is a normal subgroup of G and a is an element of G, then the order of Na divides the order of a (where the order of Na denotes the order of Na as an element of G/N).

Please help!   :)

If r is the order of a, then $(Na)^r = N$, so the result follows.


function
inequalities
calculus
calculus computations

Let $ f: [0,1] \to \mathbb{R}$ be a continuous function with $ f(0)\equal{}0,$ and for $ 0<x<1,$ $ f$ is differentiable and $ 0 \le f'(x) \le 2f(x).$

Prove that $ f$ is identically zero.

Multiply by $ e^{-2x}:$

$ 0\le e^{-2x}f'(x)\le 2e^{-2x}f(x)$

Rearrange the second inequality to

$ e^{-2x}f'(x)-2e^{-2x}f(x)\le0$

$ (e^{-2x}f(x))'\le0.$

Hence $ g(x)=e^{-2x}f(x)$ has the following properties:

$ g(x)\ge0$ on $ [0,1]$

$ g(x)$ is nonincreasing on $ [0,1]$

$ g(0)=0.$

The only function $ g(x)$ satisfying these conditions is the identically zero function. It follows that $ f(x)$ is identically zero.

Although this is a differential inequality rather than a differential equation, the standard multiplying factor method for solving first order linear differential equations works just as well. That's where the idea to multiply by $ e^{-2x}$ comes from.

This inequality is very closely related to the uniqueness theorem for ODEs. See [url=http://en.wikipedia.org/wiki/Gronwall%27s_inequality]here[/url] for more.


find th sum af all 3 digit numbers whose digits are 1,2,3,4

the digits can be repeated.

[quote="10000th User"]chess64 use brute force method?

$4^2$ numbers start with digit 1 then 2 then 3 then 4

same with other digits...

Answer is $(1+2+3+4)4^2\times 111=17760$[/quote]

good solution

[quote="10000th User"]$4^2$ numbers start with digit 1 then 2 then 3 then 4

same with other digits...

Answer is $(1+2+3+4)4^2\times 111=17760$[/quote]

I don't understand how you got that last expression :?

[quote="10000th User"][quote="Pakman2012"][quote="gunes"]there is another.
tha smallest 111
the highhest 444
and there are 64 numbers such that.
((111+444)/2)*64=17760[/quote]

Thats how I did it, this way generally works for all palindromish style questions... although this didn't directly deal with them.[/quote]
But this method does not work if I ask: Find the sum of all 3-digit numbers whose digits are 2,3,5,8

To chess64: Think of it this way:    1xx, 2xx, 3xx, 4xx. How many start with 1? $4^2$ Do the same for 2,3,4

Now, think next of this way: x1x, x2x, x3x, x4x. How many have 2nd digit 1? $4^2$ Do the same for 2,3,4

Do same with this way xx1

Each number is 3 digit so 100+10+1=111. There are $4^2$ numbers of each digit in each place. So $1\times 4^2\times 111+2\times 4^2\times 111+3\times 4^2\times 111+4\times 4^2\times 111=(1+2+3+4)4^2\times 111=17760$[/quote]

You're right... I dont know what I was thinking.  But to solve the problem you gave what I would do is first: Find the number of arrangements $4*4*4=64$

There will be 16 200's, 16 20's, and 16 2's
and so on for each number. So we do $(2+3+5+8)*16*(100+10+1)=18*16*111=31968$

btw, very nice explanation.


For solving instructions, please refer to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=13586]Puzzle 1[/url].

---

[b]The Art of Puzzle Solving[/b]

Sometimes, when solving a puzzle, it's essential that you get feedback. Some discussion is nice too, as long as you don't change the subject and get bored. I bet I can get help on solving this puzzle if I just go mingle (and discuss various topics, math-related or not) with the great folks at AOPS...

14336, 6, 7
13319, 2, 9
281, 9, 11
10189, 49, 15
12933, 2, 22
12027, 14, 190
13831, 21, 10
5546, 4, 14
14346, 1, 7
3665, 19, 5
6611, 11, 153
8138, 2, 4

[quote="TripleM"]Am I the only ones doing these puzzles? Numbers 3 and 4 are a lot easier than 1 and 2.. how about some more people have a try..[/quote]
This is the first one I haven't known something about before it was posted.


Let $x, y, z>0$ such that $xyz=1$. Prove that the following inequality holds true:

                   \[ 9+(x+2z)(y+2x)(z+2y)\geq 6(x+y+z+xy+yz+zx) \]

Put $x=a/b, y = b/c, z = c/a$ to remove the constraint. The inequality is then the same as

\[ \sum_{sym} 2a^3b^3 + a^4bc - 6a^3b^2c + 3a^2b^2c^2 \ge 0  \]

Now we have two inequalities:

\[ \sum_{sym} a^3b^3 - 2a^3b^2c + a^2b^2c^2 \ge 0  \]

\[ \sum_{sym} a^4bc - 2a^3b^2c + a^2b^2c^2 \ge 0  \]

Both are Schur's inequality in disguise. Use the substitution $u = bc, v = ca, w = ab$ to get the first from $u^3+v^3+w^3+3uvz-u^2(v+w)-v^2(w+u)-w^2(u+v) \ge 0$ and the second follows from $abc(a^3+b^3+c^3+3abc-a^2(b+c)-b^2(c+a)-c^2(a+b)) \ge 0$.

Do you have a different idea?

Let $x, y, z>0$ such that $xyz=1$. Prove that the following inequality holds true:$$21+(x+2z)(y+2x)(z+2y)\geq 8(x+y+z+xy+yz+zx) $$$$\frac{129}{5}+(x+2z)(y+2x)(z+2y)\geq \frac{44}{5}(x+y+z+xy+yz+zx) $$


function
real analysis
topology
geometry
geometric transformation
invariant
real analysis theorems

Rudin has an entire section in Ch 3 called Approximation by Continuous Functions. He states that if $ X$ is locally compact Hausdorff, then for $ 1\leq p<\infty$, $ C_c(X)$ is dense in $ L^p(\mu)$.

He seems to make a big deal out of this and shows a lot of relationships that come from it, but out of the 26 problems in this chapter, essentially none of them deal with this.

So this is sort of backward of how things are done here, but does anyone have any good problems that require you to approximate by continuous functions to help me get a handle on the technique?

Thanks.

Many of the uses I know require a little more structure - for instance, that $ X$ is also a (locally compact) group and the measure we have constructed is translation-invariant (that is, a Haar measure). In that case, $ \|f(\cdot\minus{}t)\minus{}f(\cdot)\|\plus{}p\to0$ as $ t\to0$ in the group. That in turn can be used to obtain certain convolution results, such as approximate identity results.


[quote="parinaz.sa"]let $x,y,z>0 , x+y+z=xyz$ prove this:
$xy+yz+zx \ge 3+\sqrt{x^{2}+1}+\sqrt{y^{2}+1}+\sqrt{z^{2}+1}.$[/quote]
It's obviously wrong. Try $x=y=z\rightarrow0^{+}.$ :wink:

But then we cannot have $x+y+z = xyz$.  And if $x, y \to 0^{+}$ then $z = \frac{x+y}{xy-1}$ is negative.

You are right, t0rajir0u. Thank you.
[quote="parinaz.sa"]let $x,y,z>0 , x+y+z=xyz$ prove this:
$xy+yz+zx \ge 3+\sqrt{x^{2}+1}+\sqrt{y^{2}+1}+\sqrt{z^{2}+1}.$[/quote]
Let $a=\frac{3}{yz},$ $b=\frac{3}{xz}$ and $c=\frac{3}{xy}.$
 Then $x=\sqrt{\frac{3a}{bc}},$ $y=\sqrt{\frac{3b}{ac}}$ and $z=\sqrt{\frac{3c}{ab}},$  $a+b+c=3$ and
$xy+yz+zx \ge 3+\sqrt{x^{2}+1}+\sqrt{y^{2}+1}+\sqrt{z^{2}+1}\Leftrightarrow$
$\Leftrightarrow3\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\geq3+\sum_{cyc}\sqrt{\frac{3a}{bc}+1}\Leftrightarrow$
$\Leftrightarrow(a+b+c)(ab+ac+bc)\geq3abc+\sum_{cyc}\sqrt{abc(3a^{2}+abc)}\Leftrightarrow$
$\Leftrightarrow\sum_{sym}a^{2}b\geq\sum_{cyc}\sqrt{abc(a^{3}+a^{2}b+a^{2}c+abc)}\Leftrightarrow\sum_{sym}a^{2}b\geq\sum_{cyc}a\sqrt{bc(a+b)(a+c)}.$
But $\sqrt{bc(a+b)(a+c)}\leq\frac{c(a+b)+b(a+c)}{2}.$
Thus, remain to prove that $\sum_{sym}a^{2}b\geq\sum_{cyc}\frac{a^{2}b+a^{2}c+2abc}{2},$ which triviality.

umm...

if we do like $x=\tan A$, etc, then $\tan A=-\tan B+C$ ... $A+B+C=k\pi$... $\frac{\pi}{2}>A,B,C> 0$, $k=1$...so we have an acute triangle...

does this help?

$\tan A\tan B+\tan B\tan C+\tan C\tan A\ge 3+\sec A+\sec B+\sec C$

after some algebra/identities...

which is like

$(\sum \tan A)(\sum \tan^{-1}A)\ge\sum (1+\sec A)$

...i ono...cauchy? the RHS has an okay factoring thing going on with the triangle side substitution...

Let $x,y,z>0 , x+y+z=xyz$ and if
$k(xy+yz+zx) \ge 9k-6+\sqrt{x^{2}+1}+\sqrt{y^{2}+1}+\sqrt{z^{2}+1}$ is true then $k\geq\frac{7}{16}.$ :wink:
If $k=\frac{1}{2}$ it's still easy.

[quote="parinaz.sa"]Let $x,\,y,\,z>0,\, x+y+z=xyz.$ Prove that:
\[xy+yz+zx \ge 3+\sqrt{x^{2}+1}+\sqrt{y^{2}+1}+\sqrt{z^{2}+1}\]
[/quote]
We have:
$xyz=x+y+z\ge 2\sqrt{xy}+z\Longleftrightarrow z\left(\sqrt{xy}\right)^{2}-2\sqrt{xy}-z\ge 0\\ \Longleftrightarrow \sqrt{xy}\ge\frac{1+\sqrt{1+z^{2}}}{z}\Longleftrightarrow z\sqrt{xy}\ge 1+\sqrt{1+z^{2}}$
Hence: $xy+yz+zx\ge x\sqrt{yz}+y\sqrt{zx}+z\sqrt{xy}\ge 3+\sqrt{x^{2}+1}+\sqrt{y^{2}+1}+\sqrt{z^{2}+1}$


For any integer $ a$ show that $ gcd(2a \plus{} 1, 9a \plus{} 4) \equal{} 1$.

you could either
see [url=http://www.artofproblemsolving.com/Wiki/index.php/Euclidean_algorithm]Euclid's algorithm[/url]
or try to follow [hide="this"]
suppose an integer b greater than 1 divides both of those numbers
then $ 2a\plus{}1\equal{}kb$ and $ 9a\plus{}4\equal{}lb$ for integers k and l
and then $ 9(2a\plus{}1)\minus{}2(9a\plus{}4)\equal{}(9k\minus{}2l)b$
expanding/collecting like terms gives
$ 1\equal{}(9k\minus{}2l)b$
since b, k, and l are integers, (9k-2l) and b will both be integers
since b is positive, both b and 9k-2l will equal 1, which contradicts the first line[/hide]

Euclid's Algorithm:
[hide]$ \gcd(a, b) \equal{} \gcd(a,b\minus{}an)$ where $ n$ is an integer.
$ \gcd(2a\plus{}1,9a\plus{}4)\equal{}\gcd(2a\plus{}1, 2a) \equal{} \gcd(1,2a) \equal{} 1$.
[/hide]

let $ d|2n\plus{}1$ and $ d|9a\plus{}4$.
then: $ d|(9a\plus{}4)\minus{}4(2n\plus{}1) \Rightarrow \boxed{d|a}$.
So: $ d|(2n\plus{}1)\minus{}2(a) \Rightarrow \boxed{d|1}$.
$ \max{d}\equal{}\gcd(2n\plus{}1,9a\plus{}4) \equal{} 1$


Alcumus MATHCOUNTS probability AMC AIME Support

What does Alcumus help you with? Does it prepare you for Mathcounts, help you with schoolwork, or just broaden your horizon of math?

Well, if you try to learn how to do each problem, then I guess it would greatly benefit you in some math competitions like MATHCOUNTS or AMC.  It is more specialized in Counting and Probability, by the way.

Alcumus is a huge help in trying to understand probability and combinatorics, the reasoning behind them, and how they work. For me, probability was always a weak point, but after doing about 500 problems of Alcumus, I found that probability was now what helped me make it into AIME, and score fairly okay on it. It's amazing; it can be a really great learning tool. Hilariously, it's free. Then again, some of the best stuff in the world is free.

Alcumus is a tool to help you sharpen your C&P skills, I guess......

Well, that is what I think.


geometry
geometric transformation
reflection
circumcircle
cyclic quadrilateral
perpendicular bisector
geometry unsolved

We have a quadrangle $ABCD$
Point $M$ is on diagonal $AC$ and $AM=MC$
$\angle{BMC}=\angle{CMD}=\angle{BAD}$.
Prove, that we can inscribe quadrangle $ABCD$ in a circle.

Take $D'$ on the ray $(MD$ s.t. $ABCD'$ is cyclic. We still have $\angle BMC=\angle CMD=\angle CMD'$, and if we show that this common value is also equal to $\angle BAD'$ we're done, becasue then the lines $AD,AD'$ would coincide, and we would have $D'=AD\cap MD=D$.

The problem is now this: given a cyclic quadrilateral $ABCD'$ with $M$ - the pidpoint of $AC$ s.t. $\angle BMC=\angle CMD'$, the common value of these two angles is $\angle BAD'$.

Reflect $D'$ in the perpendicular bisector of $AC$ onto $D''$, which also lies on the circle. Furthermore, $B,M,D''$ are collinear. Since the angles $\angle BAD'',\angle BCD''$ are supplements and the triangles $BAD'',BCD''$ have equal areas, we get $AB\cdot AD''=CB\cdot CD''$, meaning that $AB\cdot CD'=CB\cdot AD'$, i.e. $ABCD'$ is a harmonic quadrilateral. This means that the tangents in $A,C$ to its circumcircle meet on $BD'$, or, in other words, $BD'$ is the $B$-symmedian of $ABC$, so it's the isogonal conjugate of the median $BM$. We thus get $\angle ABD'=\angle MBC$, and since $\angle AD'B=\angle MCB$, the triangles $ABD',MBC$ are similar, so $\angle BAD'=\angle BMC$, as desired.

Let $O$ be the circumcenter of $\triangle ABC$ and $D'$ the point where the circumcirle of $\triangle ABC$ intersects the ray $MD$. To prove that $ABCD$ is cyclic it suffices to prove that $D = D'$ or that $\angle BAD' = \angle BAD = \angle BMC$.
Let $D''$ be the second point of intersection of the line parallel to $AC$ passing through $D'$ with the circumcircle of $\triangle ABC$. It is clear that $B$, $M$ and $D''$ are aligned because $\angle AMD''=\angle CMD'=\angle CMB$. Now $\angle MBO=\angle MD''O$ because $\triangle OBD''$ is isosceles, and $\angle MD''O = \angle MD'O$ because $D'$ and $D''$ are symmetric with respect to the line $OM$, so $MBD'O$ is cyclic. Then we have $\angle BAD'=\frac 1 2\angle BOD'=\frac 1 2\angle BMD'=\angle BMC$.

QED


A connected grph G has 500 vertices,each with degree 1,2 or 3.We call a black-and-white coloring of these vertices good if more than half of the vertices are white but no two white vertices are connected.Prove that it is posible to choose several vertices of G so that in any good coloring more than half of the chosen vertices are black.


search
linear algebra
linear algebra unsolved

Let $ trA\equal{}0$. Show that there exist matrices $ X$ and $ Y$ such that
              $ X$ is $ Hermitian$, $ trY\equal{}0$ and $ A\equal{}XY\minus{}YX$.

Nice Ibrogimov, but (almost) already posted by harazi http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1525734767&t=101774


limit
real analysis
real analysis unsolved

Is it true that the sequence $\{x_{n},\,n\ge1\}$ of real numbers is convergent if and only if $\ds\lim_{n\to\infty}\mathop{\overline{\rm lim}}_{m\to\infty}\mid x_{n}-x_{m}\mid =0?$
(O. Nesterenko)

Well, I'm still wondering where is the trap ...

If $(x_{n})$ is convergent, then in particular it is Cauchy and the direct implication is obvious.

Reciprocally, let $\epsilon>0$. There is some $N \in \mathbb{N}, \overline{\lim}_{m \to+\infty}\ |x_{N}-x_{m}| \leq \epsilon$. Now there is $M \in \mathbb{N}, \forall m,m' \geq M, \ |x_{N}-x_{m}| \leq 2\epsilon$. Eventually, $\forall m,m' \geq M, |x_{m}-x_{m'}| \leq |x_{N}-x_{m}|+|x_{N}-x_{m'}| \leq 4 \epsilon$. This proves that $(x_{n})$ is Cauchy in $\mathbb{R}$, whence it is convergent.

There was no trap.

Trough this problem you can see if the students  understood well
the notion of convergence, Cauchy criterion, and limite superior.
Very nice problem of O.Neterenko.


[quote="ciprian"]Fie k pozitiv si a,b,c, cu c<a+b pozitive distincte ( nu includ pe 0 la pozitive :wink: ). Sa se demonstreze ca egalitatea:
$a^{k}+b^{k}-c^{k}=(a+b-c)+c^{k}$
are loc daca si numai daca k=1.[/quote]

$a^{k}+b^{k}-c^{k}=(a+b-c)^{k}+c^{k}$,adica puterea lui $(a+b-c)$ este $k$ sau $1$?


hey guys
need help with these two inequalities
1)find the minima of
 x_1 + (x_2)^2/2 +(x_3)^3/3 +.......... + (x_n)^n/n
subject to the condition
 \sum 1/(x_i) = n

can be done using lagrange multipliers but that doesn't make a good olympiad solution,does it

2)let P(x) = a_0 + a_1*(x) +a_2*(x^2)+....... a_n*(x^n)
with complex coefficients supposse roots of P(x) are b_1,b_2....b_n
with /b_1/ > 1 , /b_2/ > 1...  /b_j/ >1 and /b_(j+1)/<=1 , /b_(j+2)/<=1 till /b_n/ <=1.
prove that product of
/b_i/ {i ranging from 1 to j)  \leq   \sqrt ( \sum /a_i/^2) {i from 0 to n}

please help (ps:/ a/ denotes modulus of a)
i hope i have made the question clear
best regards

The first one is very simple, the second one is very hard.
   For the first one use the fact that (x_k)^k>=1+k(x_k-1), so
 sum (x_k)^k/k>=sum 1/x_k+sum x_k-n. But it's easy to see that sum x_i>=1, so the minimum is n.
   For the second one, I tried to find a solution long time ago, but I wasn't able.

I don't think you are right Harazi.

The first can be solved with the power mean inequality :
Let H = 1 +1/2 ... + 1/n and w_i = 1/(i*H) for i = 1,...,n.
clearly w_i > 0 and um w_i = 1.

Let S be the given expression. Then :
S/H = sum w_i*(x_i) i   \geq  prod (x_i) i*w_i  = (prod x_i) 1/H \geq (n/sum 1/x_i) n/H = 1.
Then S  \geq  H, and equality occurs iff x_i = 1 for each i.

Pierre.

With what am I not right? Isn't my solution good?

No it is not good Harazi, since the minimum is 1 + 1/2 + 1/3 + ... + 1/n.

Now I realised. Typical for me:
  In fact from (x_k)^k>=1+k(x_k-1)) we have sum (x_k)^k/k>=sum 1/k+sum x_k-n>=sum 1/x_k. Sorry!


Let triangle $ABC$, $M$ is midpoint of $AB$. Let $D$ is a point in the opposite ray of ray $CA$ such that : $CB$=$CD$.
Let bisector of $\widehat{ACB}$ meet $MD$ at $K$.Prove that: $\widehat{KBC}$=$\widehat{BAC}$

Denote $a = BC, b = CA, c = AB.$ The case $a = b$ is trivial, because then $M \equiv K$ are identical and $\angle B = \angle A.$ Assume $a < b.$ Let $F \in AB$ be the foot of the bisector of the $\angle C,$ $O$ the circumcenter of the $\triangle ABC.$ Let a normal to $CO$ from $B$ meet the bisector $CF$ at $K'$ and the circumcircle $(O)$ again at $L.$ $CL = CB,$ hence $\angle K'BC \equiv \angle LBC = \angle BAC \equiv \angle A.$ Also $\angle ACF = \angle BCF,$ hence the triangles $\triangle AFC \sim \triangle BK'C$ are similar with the similarity coefficient $\frac{CK'}{CF}= \frac{CA}{CB}= \frac{b}{a},$ so that $\frac{K'F}{K'C}= \frac{CF-CK'}{CK'}= \frac{b-a}{a}.$ Then

$MF = AF-AM = \frac{bc}{a+b}-\frac{c}{2}= \frac{c}{2}\cdot \frac{b-a}{a+b}$

$\frac{DC}{DA}\cdot \frac{MA}{MF}\cdot \frac{K'F}{K'C}= \frac{a}{a+b}\cdot \frac{a+b}{b-a}\cdot \frac{b-a}{a}= 1$

so that the points $M, K', D$ are collinear, which means $K' \equiv K$ are identical. The case $b > a$ is similar.

See also http://www.mathlinks.ro/Forum/viewtopic.php?t=19500 .

  darij


OK so like, now we have "organized" tourney mania, in that izzy can turn on and off the spigot to the fountain of tournament generation (yeah metaphors!!!).

So like, what do these tournaments accomplish?

More to the point, what does any tournament accomplish? It determines who is the best.

But there is a problem. When there is like a bazillion (actually, 5!=120) tournaments in a relatively short time period, we don't know who is the best. Moreover, sometimes the best doesn't get to join, because tournaments are too restricted in size and good players don't make it within 5 hours (see: tourney 120).

So I am proposing a NEW SYSTEM.

In NASCAR, the system used for ranking people is giving them points based on their finishes in give races. This is because there are so many races. Why not apply a similar system here?

Basically, open up the tournaments to anyone and everyone who wants. That is, give a set singup-time-period (3 days?) and take everyone who signs up then.

Then, give points based on place. For example (I doubt a tourney will go over 32 people=5 rounds)

Winner= 50 pts
2nd =30
SF= 20
QF=10
2nd round=5
1st round=0

Obviously then, if there only needs to be 4 rounds, give everyone who signs up five points. Then, we can give top seeds byes. The original seeding can be done based on rating, then later seeding can be determined by total points.

Whaddya think?

We already have a working system.  The 16 player maximum was enforced so we don't have tournaments that would take an excessive amount of time, so a player who loses does not have to wait very long until a new tournament starts.  Signups usually start at a time of the day when most people are awake (ok, not Tourney 5 which started before 4:30 AM here in CA) so it's the player's responsibility to sign up if they want to.  Chances are (there are obviously exceptions to this) that if a player doesn't check the FTWOS forum that often, they won't play their games.

Also, players who go to math camps or vacations would have a disadvantage, as they would not be able to play in some tournaments.

That's just my three cents :P

[quote="AIME15"]We already have a working system.[/quote]

That's what they said last year.

[quote="AIME15"]The 16 player maximum was enforced so we don't have tournaments that would take an excessive amount of time, so a player who loses does not have to wait very long until a new tournament starts. [/quote]

If more than 16 are active, what be teh point? And more than one tourney can run at once.

[quote="AIME15"]
Signups usually start at a time of the day when most people are awake (ok, not Tourney 5 which started before 4:30 AM here in CA) so it's the player's responsibility to sign up if they want to.  Chances are (there are obviously exceptions to this) that if a player doesn't check the FTWOS forum that often, they won't play their games.[/quote]

Do we have ANY SHRED of evidence for this sweeping generalization?

[quote="AIME15"]
Also, players who go to math camps or vacations would have a disadvantage, as they would not be able to play in some tournaments.

That's just my three cents :P[/quote]

Simple, set up a system where only the top X scores of yours count to the championship. Or, average all of yours (but only if you have >X games).

It's a common phrase, why pollute it by adding one?

The purpose of a tourney is not to necessarily see who's best, but just like an organized bunch of games. For example, AIME15 won tourney 3, and I came in second place, but clearly I am the best. :P Just kidding.

But seriously, I don't see why it has to get so complicated? Things are working fine right now.

[quote="AIME15"][quote="Mewto55555"][quote="AIME15"]We already have a working system.[/quote]

That's what they said last year.[/quote]

How do you know your system will work?  The current system is working.  Since the "true" start of summer, tournaments are going on at a good pace, and are finishing.

[quote="Mewto55555"][quote="AIME15"]The 16 player maximum was enforced so we don't have tournaments that would take an excessive amount of time, so a player who loses does not have to wait very long until a new tournament starts. [/quote]

If more than 16 are active, what be teh point? And more than one tourney can run at once.[/quote]

There are already multiple tournaments running simultaneously.  Also, tournaments with over 16 players tend to take longer, and those eliminated earlier in the tournament won't have a tournament to participate in (especially if all the running tournaments are large).  Furthermore, larger tournaments are harder to manage.

[quote="Mewto55555"][quote="AIME15"]
Signups usually start at a time of the day when most people are awake (ok, not Tourney 5 which started before 4:30 AM here in CA) so it's the player's responsibility to sign up if they want to.  Chances are (there are obviously exceptions to this) that if a player doesn't check the FTWOS forum that often, they won't play their games.[/quote]

Do we have ANY SHRED of evidence for this sweeping generalization?[/quote]

Not checking the FTWOS forum implies not checking the tournament thread, and thus not getting the game assignment.

[quote="Mewto55555"][quote="AIME15"]
Also, players who go to math camps or vacations would have a disadvantage, as they would not be able to play in some tournaments.

That's just my three cents :P[/quote]

Simple, set up a system where only the top X scores of yours count to the championship. Or, average all of yours (but only if you have >X games).

It's a common phrase, why pollute it by adding one?[/quote]

X would have to be large to prevent huge ties, and if X is large, then players who go on vacation or to camps have a disadvantage.

6 is a perfect number, so I'll add that to 2.  That's just my eight cents.[/quote]

I know I've just joined AoPS recently, but here's my take on the situation:  Both Mewto5555 and AIME 15 have valid opinions, and large and small tournaments both have their merits and drawbacks.  However, I think that whoever founds the tournament has the right.  Also, since kids don't really spend the ENTIRE summer at a math camp, the "X" system works just fine.  I don't want to spark up any controversy, but AIME15 is mostly right.  However, many might find a large tournament more fun(more rounds to go through, more exciting, etc.)  Furthermore, it's not AIME15's decision.  If enough people want to change the system, that's how it's going to be.  Of course, what's really needed here is a compromise.  Just putting in my 1 cent.

Ahhhh pyramid quoting!

Anyway, it'd be nice if you had a little bit of logic behind your opinion :wink:

Hm I have a small question regarding the logic of this:

The only changes required would be:

1. A new thread to keep track of points.

2. Expanded tournament size.

1 is like a non-issue. As for 2, we could just set, and stick to, the time limits. There needs to be some new method of deciding who wins when both people are inactive, or neither of them, but they don't get their game done, but we can cross that bridge when we come to it.

As for the claims that it would be unfair to certain individuals who aren't there, I would like to throw your logic back at you.

Either said individuals

a. Do care about tourneys results

b. Don't

If b, then it clearly won't matter to them if the tourney is somehow unfair to them. As long s they have fun and play games.

If a, then they should embrace the new system, because it actually puts a premium on doing well as opposed to merely shrugging off the results of tournaments.

So exactly what's wrong with shrugging off the results or tournaments, may I ask?

What about those who care? You are just dismissing their point of view with one sweep of the proverbial hand.

[quote="vallon22"]why not have a results page
like, we could have a thread (or single post) listing pplz who have finished in the top three and what they finished in
that collection of info could potentially tell us who is the best

in addition, [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=267086]this[/url] has attempted to find a best player
i thought it was an ok idea, but unfortunately, it sort of died away[/quote]

That incidentally, happened to be my inspiration. Then a thought hit me: Why not use the existing tournament structure?

remind me again mewto why you or anyone needs to know they are the best? and this is just my cynical opinion but are you trying to start another CTC using this new tournament system as a basis?

[quote="BOGTRO"]I think this is a valid idea. One possibility is to have this forum and a seperate sub-forum for mewto's idea.

However, there is no need to fix something that isn't broken. Yet. I like the way it is now, but this will not last for more then a few weeks at most (in all likelyhood).

Mewto, why can't you start this yourself?[/quote]

Why can't I start it myself? That's easy.Within 5 hours, I would have >10 posts on the thread flaming me saying I was trying to subvert the system, that I'm too far back on the queue, and that I am a mulri.

Oh and of course, mathblitz's perennial allegations that Iam attempting to steal the yoke of power and remake ctc. note: izzy remade ctc and put her in charge.

Also, I am on vacation, and I can start this up when I get back.

[quote="Mewto55555"] note: izzy remade ctc and put her in charge.
[/quote]

Please, do explain.

mewto you still havent answered why you or anyone needs to know your the best.

[quote="BOGTRO"]I think this is a valid idea. One possibility is to have this forum and a seperate sub-forum for mewto's idea.

[/quote]

only admins can make sub-forums and i doubt levans would want to spend time making something that, in his eyes, is quite pointless


Buenos dias,

solo para informar que los examenes selectivos para la IMO
en el PERU
se realizaran como sigue:

[b]1er examen: sabado 05 de amyo (clasificatorio, solo pasan los 10 primeros)

2do examen: lunes 07 de mayo (se consideraran los 6 mejores promedios)[/b]

Para mayor informacion..
ir al Blog de nuestro compañero Jorge Tipe:

http://selectivos-peru.blogspot.com/

Saludos,

¿Cada uno califica para tomar ambos exámenes? ¿O solamente el 10 superiores en el examen primer?

Por los leido en el blog,

el primer exam es selectivo...
solo pasan las 10 mejores notas...

eso quiere decir q solo 10 jovenes daran el 2do exam..

y para elegir la seleccion (6), se considerara el promedio obtenido en los
dos exams.

saludos,

Mayor info, consultarlo a JTIPE (jorge tipe) ;)

saludos,